A TextBook of Fluid Mechanics and Hydraulic Machines - Dr. R. K. Bansal.pdf

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ATextbook of

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Dr. :R.K. Bans t

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Author : Dr. R.K. Ban~ .. l Compiled by : Smt. Nirmol Ba1l6Al Q All rights reserved with Author and the Publishers. No part of this publication may be reproduced, stored in a retrieualsystem, or transmitted in any form or by any means, electronic , mechanical, photocopying, recording or otherwise without the prior written permission ofthe publisher.

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CONTENTS Chapter

Pages

Chapter L PrOllerties of Fluids

1-34

1.1. Introduction 1.2. Properties of Fluids 1.2. 1. Density or Mass Density 1.2.2. Spedfic Weight or Weight Density

1.2.3. 1.2.4.

Spocific Volume Spedfic Gravity Solved Problems 1.1- 1.2

2 3

Kinematic Viscosity Newton's Law of Viscosity 1.3.3. 1.3.4. Variation of Viscosity with Temperature 1.3.5. Types of Fluids Solved Problems 1.3-1.19 Thermodynamic Properties 1.4. 1. Dimension of R 1.4.2. Isothermal Process 1.4.3. Adiabatic Process Universal Gas Constant 1.4 .4. Solved Problems 1.20-1.22 Compressibility and Bulk Modulus Solved Problems 1.23-1.24 Surface Tension and Capillarity 1.6. 1. Surface Tension on Liquid Droplet 1.6.2. Surface Tension on a H"llow Bubble 1.6.3. Surface Tension on a Liquid Jet Solved Problems 1.25- 1.27 Capillarity 1.6.4. Solved Problems 1.2&--1.32 Vapour Pre~sure and Cavitation Highlights Exercise

1.3 .2.

1.5. 1.6.

1.7.

1 1

2 2

l.3. Viscosity Units of Viscosity 1.3. 1.

104.

1

Chapter 2. Pressure and Its Measurement

3 5 5 6 6 6 17 18 18 18 19 19 21 22

23 23

"2424 25 26 29 30 30

3....8

2 . 1. Fluid P ressure at a Point 2.2. Pascal"s Law 2.3. Pressure Vllristion in a Fluid at Rest Solved Problems 2.1-2.7

35 35 36 37

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IL 2.4. Absolute. Gauge, Atmospheric and Vacuum Pressures Solved Problem 2.8 2.5. Measurement of Pressure 2.5.1. Manomet
3.t. Introduction 3.2. Total Pressure and Cl!n tre of Pressure 3.3. Ve rtical Plane Surface Sub-merged in Liquid Solved Problems 3.1-3.12 3.4. Horizontal Plane Surface Sub· merged in Liquid Solved Problem 3.13 3.5. Inclined P lane Surface Sub·merged in Liquid Solved Problems 3.14(a}-3.21 3.6. Curved Surface Sub-merged in Liquid Solved P roblems 3.22-3.31 3.7. Total Pressure and Centre of Pressure on Lock Gates Solved Problems 3.32-3.33 3.8. Pressure Distribution in II Liquid Subjected to Constant HorizontaVVertical Acceleration 3.S.1. Liquid Containers Subject to Constant Horizontal Acceleration Solved Problems 3.34-3.36 Liquid Containers Subjected to Constant 3.8.2. Vertical Acceleration Solved Problems 3.37-3.38 Highlights Exercise

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41 42 42 42 43 43 43 43 44 48 50 50 50 51 53 53 56 56 56

58

59 60 64 65 69-130

69 69

69 72

85 86

66 66 97

99 107 109 112 112 115

120 122

124 125

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Ch a pte r 4 . BUOYlincy and Floatlltion

131 - 162

4. 1. Introduction 4.2. Buoyancy 4. 3. Centre of Buoyancy Solved Problems 4.1-4.6 4 .4 . Meta-centre 4.5. Meta-cent ric Height 4.6. Analytical Method for Meta-Centre Height Solved Problems 4.7-4,11 4.7. Conditions of Equilibrium of a Floating and Sub-merged Bodies 4.7. 1. Stability of a Sub-merged Body 4.7.2. Stability of a Floating Body Solved Problems 4.12-4.18 4.8. Experimental Method of Determination of Me ta-centric Height Solved Problems 4.19-4.20 4.9. Oscillation (Rolling) of a Floating Body Solved Problems 4.21-4.22 Highlights Exercise Chapte r 5 . Kinematic .. of Flow and Ideal Flow

A. KINEMATI CS OF FLOW 5. 1. Introduction 5.2. Methods of Describing Fluid Motion 5.3. Types of Fluid Flow 5.3.1. Steady and Unsteady Flows 5.3.2. Uniform and Non·uniform Flows 5.3.3. Laminar and Turbulent Flows 5.3.4. Com pressible and Incompressi ble Flows 5.3.5. Rotational and lrrotational Flows 5.3.6. One, two and Three-Dimensional Flows 5.4. Rate of Flow or Discharge (Q ) 5.5. Continuity Equation Solved Problems 5.1-5.5 5.6. Continuity Equation in Three-Dimensions 5.6.1. Continuity Equation in Cylindrical Polar Co-ordinates Solved Problems 5.5A 5.7. Velocity and Acceleration 5.7. 1. Local Acceleration and Convective Acceleration Solved Problems 5.6-5.9 5.S. Velocity Potential Function and Stream Function 5.8. 1. Velocity Potential Function 5.8.2. Stream Function 5.8.3. Equipotential Line 5.8.4. Line of Constant Stream Function

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131 131 131 131 136 136 137 138 143

143 143 144

1"

155 156 158 159 160

163- 258 163 163

163 163 164

1" 164

165 165 165 165 166 170 171

173 174 175 175 181 181

182

183 183

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Fl ow Net Relation between Stream Function and Velocity Potential Function Solved Problems 5.10---5.17 5.9. Types of Motion Linear Translation 5.9.1. 5.9.2. Linear Deformation 5.9.3. Angular Deformation or Shear Deformation 5.9.4. Rotation 5.9.5. Vorticity Solved Problems 5. 1S----5. 19 5. 10. Vortex Flow 5.10.1. Forced Vortex Flow 5.10.2. Fre
5.8.5. 5.8.6.

184 18.

184 191 191 191

192 192 192 192 193 193 194

195 196 197 202 202 209

210

(B) lOU!. . ' LOW (POTENTIAL FLOW)

5. 1 1. Introduction 5. 12. Important Cases of Potential Flow 5 .13. U n iform F low 5.13.1. Uniform Flow Parallel tox·Axis 5.13.2. Uniform Potential Flow P afaliel toy-Axis 5. 14. Sou rce Flow 5.15. Sink Flow Solved Problems 5.33- 5.35 5. 16. Free-Vortex Flow 5. 17. Supcr·!mposed Flow 5. 17.1. Source and Sink Pair Solved Problems 5.36-5.37 5.17.2. Doublet Solved Problem 5.38 5.17.3. A Plane Source in a Uniform Flow (Flow Past a Half-Body) Solved Problems 5.39-5.41 5.17.4. A Source and Sink Pair in a Uniform F low (Flow Past a Rankin e Oval Body) Solved Problem 5.42 5.17.5. A Doublet in a Uniform Flow (Flow Past a Circular Cylinder) Solved Problems 5.43- 5.44

Highlights f.xercise

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210 211 211 211

213 21.

216 216 219 221 221 225

228 231 233 237 241

2.. 246 250 252 254

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Chapter 6. Dynam ics

or Fluid

259-316

Flow

6.1. Int rodu ction 6.2. Equ a tion s of Motion 6.3. Euler's Eq uation of Motion 6.4. Bernoulli's Equation from Euler's Equation 6.;'\. As s um ptions Solved Problems 6. 1- 6.6 6.6. Bernoulli's Equation fo r Real Fluid Solved Problems 6.7- 6.9 6.7. P ractica l Applications of Bernoulli's Equation 6.7.1. Vc nlurimctcr

Soh,('od Problems 6.10---6.21 Orifice Meter or Orifice Plate Solved Problem s 6.22 - 6.23 6.7.3. Pilot-tube Solved Problems 6,24---6,28 6.S. The Momentum Equation Solved Problems 6.29- 6.35 6.9. Moment of Momen tum Equation Solved Problems 6.36-6.37 6 .10. Free Liquid J ets Solved Problems 6.38-641 Highlights 6.7,2.

E"en;;ise

7.5.

7.6.

7.7. 7.8.

283 285

266 288

289 298 298 301 303 307 309 317-3;'14

Chapter 7. Orifices and Mouthpi eces 7.1. 7.2. 7.3. 704.

259 259 260 261 261 261 265 266 268 268 270 261

Int roduction Classifications ofO rifiees Flow Through an Orifice Hydraulic Co-efficients 7.4. 1. Co·efficient of Velocity (Col 7.4.2. Co·efficient of Contraction (Cel 7.4.3. Co-efficient of Di~charge (C d) Solved Problems 7.1 - 7 .2 Experimental Dete rmination of Hydraulic Co·efficients 7.5.1. Determination of Co-efficient or Discharge ( C d) ' 7.5.2. Determination of Co-efficient or Velocity (C.) 7.5.3. Determination of Co-efficient of Contraction (C , ) Solved Problems 7.3-7.10 Flow Through Large Orifices 7.6.1 . Discharge Through Large Hectangula r Orifice Solved Problems 7. 11- 7.13 Discharge Through Fully Sub·merged Orifice Solved Problems 7.1 4-7. 15 Discharge Through Partially Sub-merged Orifice Solved Problem 7. ]6

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317 317 317 31' 318 319 319 319 320 320 321 321 32 1 327

328 328 330 331 331 332

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7.9 . Time of Emptying a Tllnk Through an Orifice at its Bottom Solved Problems 7.17- 7.18 7. 10 . Tim e of Emptying a Hemispherica l Tank &JIved Probl<)ffis 7.19---7.21 7.11. Time of Emptying a Circular Horizontal Tank Solved Problems 7.22- 7.23 7.12. Classification of Mouthpieces 7.13. Flow Through an External Cylindrical Mouthpiece Solved Problems 7.24-7.25 7.14. Flow Th rough a Convergent-Divergent Mouthpiece Solved Problems 7.26-7.28 7.15. Flow Through Internsl or Re-entrant on

332 333

335 336 338 339 341 341

342 344 345 347 349

&>rda's l\fouthpiere

Solved Problem 7.29 Highlights Exercise

350 352

Ch a p ler 8. Notches an d We ir s

355-386

8.1. l ntroduction 8.2. Cl a ssification of Notches and Weirs 8.3 . Discharge Over a Rectangular Notch or Weir Solved Problems 8.1-8.3 8.4. Discharge Over II T riangula r Notch or Weir Solved problems 8.4-8.6 8.5. Advantages of Triangular Notch or Weir over Rectangular Notch or Weir 8.6. Discharge Over a Trapezoidal Notch or Weir Solved Problem 8.7 8.7. Discharge Over a Stepped Notch Solved Problem 8.8 8.8. Effed on Discharge Over a Notch or Weir Due to Error in the Measurement of Head 8.8.1. For Rectangular Weir Or Notch 8.8.2. For Triangular Weir or Notch Solved Problems 8.9-8.11 8.9. (a) Time Required to Empty a Rese rvoir or a Tank with II Reetan!,"Ular Weir or Notch (b) Time Required to Empty a Reservoir or a Tank with a Triangular Weir or Notch Solved Problems 8.12-8.1 4 8.10. Velocity of Approach Solved Problems 8.15-8.19 8.1 1. Empirical Formulae for Discharge Over Rectangular Weir Solved Problems 8.20---8.22 8.12. Cipolletti Weir or Notch Solved Problems 8.23-8.24 8.13. Discharge Over a Broad-C rested Weir

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355 355

356 356 358 359 361 361

362 362 363

364 364 364 365

366 367 368

370 370 374 374 376 377 378

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8. 14. Discharge Over a Narrow·Crested Weir 8.15. Discharge Over an Ogee Weir 8. 16 . Discha rge Over Sub-merged or Drowned Wei r Solved Problems 8.25-827 Highlights Ex ercise

Chapter 9. Viscous Flow

379 379 379 380 381 383 387-432

9.1. Introduction 9.2. Flow of Viscous Fluid Through Circular Pipe Solved Problems 9.1- 9.6 9.3. Flow of Viscous Fluid between Two Parallel Plates Solved Problems 9.7-9.12 9.4 . Kinetic Energy Corn)dion and Mom!)ntum Correction Factors Solved Problem 9.13 9.5. Power Absorbed in Viscous Flow 9.5.1. Viscous Resis tance of Journal Bearings Solved Problems 9.14-9.18 Viscous Resistance of Foot-step Bearing 9.5.2. Solved Problems 9.19--9.20 Viscous Resistance of Collar Bearing 9.5.3. Solved Problems 9.2 1- 9.22 9.6. Loss of Head Due to Friction in Vi scous Flow Solved Problems 9.23---9.24 9.7. Movement of Piston in Dash·pot Solved Problem 9.25 9.8. M(lthods of Dt)t..,rmination of Co-t)fficit)nt of Viscosity 9.8 , l. Capillary Tube Method Falling Sph..,re Resistance Method 9.8.2. Rotating Cylinder Method 9.8.3. Oriflce TyPtJ Vi scometer 9.8.4. Solved Problems 9.26---9.32 Highlights Exercise

Chapter 10. Turbulent Flow

387 387

391 397

400 404 404 407 407 408 411 412

412 413 414 415 417 418 419 419

420 421 422

423 427 429 433-464

10.1. Introduction 10.2. Reynold s Experiment 10.3. Frictional Loss in Pipe Flow 10.3.l. Express ion for Loss of Head Due to Friction in Pipes 10.3.2. Expression for Co-effici..,nt of Friction in Terms of Shear Stress 10.4. Sh..,ar Stress in Turbulent Flow 10.4 .1. Reynold s Expression for Tu rbulent Shear Stress 10.4 .2. Prandtl Mixing Length Theory for Turhulen t Shear Stress

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433 433 434

434 436 437

437 438

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10.5 . Velocity Distribution in Turbulent Flow in Pipes 10.5.1. Hydrodynamically Smooth and Rough Boundaries 10.5.2. Velocity Diijtribution for Turbulent Flow in Smooth P ipes 10.5.3. Velocity Dis tribution for Turbulent Flow in Rough Pipes Solved Problems 10.1- 10.4 10.5.4. Velocity Distribution for Turbulent Flow in Te rms of Average Velocity Solved Problems 10.5-10.6 10.5.5. Velocity Dis(.ributi()n for Turbul ent Flow in Smooth Pipes by Power Law 10.6 . Res istance of Smooth and Rough Pipes Solved Problems 10.7- 10.13 Highligh ts Exerc ise

Chapter 11. Flow Through Pipell

438 440 441 442 442

446 448 450 450

453 461 462 465-558

11.1. Introduction 11 .2. Loss of Energy in Pipes 11 .3. Loss of Energy (or head ) Due to Friction Solved Problems 11.1- 11.7 11 .4. Minor Energy (H ead) Losses 11.4.1 . Loss of Head Due to Sudden Enlargement 11.4.2. Loss of Head Due to Sudden Contraction Solved Problems 11.8--11.14 11.4.3. Loss of Head at the Entrance of a Pipe 11.4.4. Loss of Head at the Exit of Pipe 11.4.5. Loss of Head Due to an Obstruction in a Pipe 11.4.6. Loss of Head Du e to Bend in P ipe 11.4.7. Loss of Head in Various Pipe Fittings Solved Problems 11. 15-11.2 1 11 .5. Hydraulic Gradient and Total Energy Line 11.5.1. Hydraulic Gradi en t Lin e 11.5.2. Total Energy Line Solv<..>d Problems 11.22-11.26 11 .6. Flow Through Syphon Solved Problems 11.27- 11.29 11 .7. Flow Through Pipes in Series or Flow Through Compound P ipes Solved Problems 11.30- 11.30A 11 .8. Equivalent Pipe Solved Problem 11.31 11 .9 . Flow Through Parallel Pipes Solved Problems 11.32- 11.41 11 . 10. Flow Through Branched Pipes Solved Problems 11.42-11.44

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465 465 465

467 471

471 473 474 482 482 482

483 483

483 491

491 491

491 498 498

502 503 507

508 508 509

524

525

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11 . 11 . Power Transmission Through Pipes 11.11. 1. Condition for Maximum Transmission of Power 11.11.2. Muximum Efficiency of Transmission of Power Solved Problems 11.45-11.47 11.12. Flow Through Nozzles 11.12.1. Power Transmitted Through Nozzle 11.12.2. Condition for Maximum Power Transmitted Through Nozzle 11.12.3. Diameter of Nozzle for Maximum Transmission of Pow..,r Through Npzzle Solved Problems 11.48---11.51 1l.13. Water Hammer in Pipes 11.13.1. Gradual Closure of Valve 11.13.2. Sudden Closure or Valve and Pipe is Rigid 11.13.3. Sudden Closure of Valve and P ipe is Elastic 11.13.4. Time Taken by Pressure Wave to Travel from the Valve to the Tank and from Tank to the Valve Solved Problems 11.52-11.55 11.14. Pipe Network 11.14.1. Hardy Cross Method Solved Problem 11 .56 Highlights Exercise

Chapte r 12. Dime n s ional and Mo d e l Ana lysis 12.1. Introduction 12.2. Secondary or Derived Quantities Solved Problem 12.1 12.3. Dimensio nal Homogeneity 12. 4 . Methods of Dimensional Analys is 12.4.1. Rayleigh's Method Solved Problems 12.2-12.7 12.4.2. Buckingham's 1!-Theorem 12.4.3. Method of Selecting Repeating Variables 12.4.4. Procedure for oolving Problems by Buckingham's 1!-Theorem Solved Problems 12.8- 12.14 12.5. Model Analysis 12 .6. Similitude-Types of Similarities 12.7. Types of Fort.'Cs Acting in Moving ~'luid 12.8. Dimensionless Numbers 12.8.1. Reynold's Number (R .) 12.8.2. Froude's Number (F, ) 12.8.3. Euler's Number (E ") 12.8.4. Weber's Number (IV, ) 12.8.5. JI.!ach's Number ( M )

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530 531

531 531 535 537 537 538

539 541 542 542

543 545 545 547 548

549 552

554 559-610

559 559 560 561 561 561 562 565 566 566

568

578 579 580 581 581 582 582

582 582

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12.9. Model Laws or Similarity Laws 12.9.1. Reynold's Model Law Solved Problems 12.15-12.18 12.9.2. "roude Model Law Solved Problems 12.19-12.27 12.9.3. Eu ler's Model Law 12.9.4. Weber Model Law 12.9.5. Mach Model Law Solved Problem 12.28 12.10. Model Testing of Partially Sub-me rged Bodies Solved Problems 12.29-12.32 12.11 . Classification of Model s 12.11.1 . Undistorl.ed Model s 12.11.2. Distorted Mod els 12.11.3. Scale Ratios for Distorted Models Solved Problem 12.33 Highlights Exercise

Chapter 13. Boundary Layer Flow 13.1. Introduction 13.2. Definitions 13.2.1. La minar Boundary Layer 13.2.2. Turbulent Boundary Layer 13.2.3. Laminar Sub· layer 13.2.4 . Boundary Laye r Thicknes~ (0) 13.2.5. Displacement Thickness (0· ) 13.2.6. Mome ntu m T hickness (9) 13.2.7. Energy Thickness (lin) Solved P roblems 13. 1- 13.2 lS.3. Drag Force on a Flat Plate Due to Boundary Layer 13.3.1. Local Co·efficient of Drag [C v -1 13.3.2. Average Co·efficient of Drag [C v 1 13.3.3. Boundary Cond itions for the Veloci ty Profiles Solved Problems 13.3- 13.12 13.4 . Turbulent Boundary Layer on II Flat Plate Solved P roblem 13.13 lS.5. Analysis of Turbulent Boundary Layer lS.6. Total Drag on a Flat Plate Due to Laminar and Turbulent Boundary Layer Solved Problems 13. 14-13.17 lS.7. Separation of Boundary Layer 13.7.1. Effect of Pressu re Gradient on Boundary Layer Separation 13.7.2. Location of Sepa ration Point Solved Problem 13.18

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583

583 58. 587 590 595 596 596 597

598 600 60.

6" 605 605

606 606

607 611-656 611 612

612 613 613 613 613 615 615 616 619 622 622 622 622

638 638 641 641 642 646

648 649 650

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13.7.3.

Methods of Preventing the Separation of Boundary Laye r Highlights Exercise

Chapter 14. Forces on Sub-merged Bodies 14.1. Introduction 14.2. Force Exerted by II Flowing Fluid on II Stationary Body 14.2.1. Drag 14.2.2. Lift 14.3. Expression for Drag and Li ft 14.3.1. Dimensional Analysis of Drag and Lift Solved Problems 14. 1- 14.15 14. 3.2. Pressure Drag and Friction Drag 14.3.3. Stream-lined Body 14.3.4. Bluff Body 14.4. Drag on II Sphere

Solved Problem 14.16 14.5 . Terminal Velocity of II Body Solved Problems 14. 17- 14.20 14.6. Drag on II Cylinder 14.7. Development of Lift on II Circular Cylinder 14.7.1. Flow of Ideal Fluid Over Stationary Cyl inder 14.7.2. Flow Pattern Around the Cylinder when II Constant Circulation r is Imparted to the Cylinder 14.7.3. Expression for Lift Force Acting on Rotating Cylinder 14.7.4. Drag Fonce Acting on a Rotating Cylinde r 14.7.5. ElI"pression fo r Lift Co·efficient for Rotating Cylinder 14.7.6. Location of Stagnation Points for a Rotating Cyli nde r in a Uniform Flow_field 14.7.7. Magnus E ffect Solved Problems 14.21- 14.23 14.8. Development of Lift on an Airfoil 14.8.1. Steady-state of a Flying Object Solved Problems 14.24-14.25 Highlights Exencise Chapter 15. Compressible Flow 15.1. Introduction 15.2 . Thermodynamic Relation s 15.2.1. Equation of Sta te 15.2.2. ElI"pansion and Compression of Perfect Gas

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651 651 653 657-692

657 657 658 658 658

659 660 670 671 671 671

672 673 673 677 677 678

678

680 682

682 683 683 683 686 687 687 689 690 693- 736

693 693

893 69.

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IL (u, 15.3. Basic Equations of Compressible Flow 15.3.1. Continuity Equation 15.3.2. Bernoulli's Equation Solved Problems 15.1- 15.3 15.3.3. Momentum Equations 15.4. Velocity of Sound or Pressure Wave in II Fluid 15.4.1. Expression for Velocity of Sound Wave in II Fluid 15.4.2. Veloeity of Sound in Terms of Bulk Modulus 15.4.3. Velocity of Sound for Isothe rmal Process 15.4.4. Vdoci ty of Sound fo r Adiabatic Process 15.5. Mach Number Solved Problems 15.4-15.7 15.6 . Propagation of Pressure Waves (or Disturbances) in II Compressible Fluid 15.6.1. Mach Angle 15.6.2. Zone of Action 15.6.3. Zone or Silence Solved Problems 15.8- 15.10 15.7. Stagnation Properties 15.7.1. Expression for Stagnation Pressure (p.) 15.7.2. Expression for Stagnation Density (p, ) 15.7.3. Expression for Stagnati.m Temperature ( T ,) Solved Problems 15.11- 15.12 15.8. Area Velocity Relation ship for Compressible Flow 15.9. Flow of Compressible Fluid Through Orifices and Noz~les Fitted to a Large Tank 15.9.1.

695 695 695

697 702

702 702 704 705 705

705 706

708

709 710 710 710 711 711 715

715 716 718

719

Value of II or !!l. for Maximum Value p,

721

15.9.3.

of Mass Rate of Flow Value of V 2 for Maximum Rate of Flow of Fluid Maximum Rate of Flow of Fl uid Through No~zle

722

15.9.4.

Variation of Mass Rate of Flow of Compressible

15.9.2.

Fluid with Pressure ratio

(~ )

Velocity at Outlet of Nozzle for Maximum Rate of Flow is Equal w Sonic Velocity Solved Problems 15. 13- 15. 15 15. 10. Ma ss Rate of Flow of Compressible Fluid Through Venturimeter Solved Problem 15. 16 IS.H . Piwt·Static Tube in a Compressible Flow Solved P roblem 15.17 Highlights Exercise

721

723

15.9.5.

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723 724 727 728

730 731 731 734

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IL (xxil Chapter 16. Flow in Open

ChRnnel~

16.1. Introduction 16.2. Classification Qf flow in Channels 16.2.1. Steady Flow and Unsteady Flow 16.2.2. Unifonn Flow and Non-uniform Flow \6.2.3. Laminar Flow and Turbulent Flow \6.2.4. Sub-critical, Critica l and Super-Critical ~'low

16.3. Disc harge Through Open Channel by Chezy's Formula Solved Problems 16.1- 16.7 \6.4. Empirical Formulae for the Value ofChczy's Constan t Solved Problems 16.8-16. 12 16.5. Most Economi cal Section of Channels 16.5.1. Most EronomiClii Roctangular Channel Solved Problems 16.13- 16.15 16.5.2. Most Economical Trapezoidal Channel Solved Problems 16.16- 16.22 16.5.3. B(lst Side Slope for Most Economical Trapezoida l Se<;tion Solved Problems 16.23- \6.24 16.5.4. Flow Through Circular Channel Solved Problems 16.25-16.29 16.5.5. Most Economical Circular Section Solved Problems 16.30- 16.32 16.6. Non-Uniform Flow through Open Channels 16.7. Specific Energy and Specific Energy Curve 16. 7.1. Cri t ical Depth (he ) 16.7.2. Critical Velocity (Ve ) 16.7.3. Minimum Specific E nergy in Terms of Critical Depth Solved Problems 16.33-16.35 16.7.4. Critical Flow 16.7.5. Streaming ,"'low or Sub-c ri tical Flow or Tranquil Flow 16.7.6. Super-Critical Flow or Shooting Flow or Torrentia l Flow 16.7.7. Alternate Depths 16.7.8. Condition for Maximum Discharge for a Given Value of Specific Energy Solved Problems 16.36-16.37 16.8. Hydraulic Jump or Standing Wave 16.8.1. Exprf'ssion for Depth of Hydraulic Jump 16.8.2. Expression for Loss of Energy Due to Hydraulic Jump 16.8.3. Exprf'ssion fo r Depth of Hydraulic Jump in Te rms or Upstream Froude Number

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737-802

737 737 737 737 738

738 739 740 744 745 749

7" 750 752 754 762 763

766 766 771 775 777 777 779 779

780 780 781 782 782 782 782 782 783 784 786 787

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IL (xxii)

16.8.4.

Length of Hydraulic Jump Solved Problems 16.38- 16.42 16.9. Gradually Varied Flow (G.V.F.) 16.9. L Equation of Gradually Varied Flow Solved Problems 16.43- 16.44 16.9.2. Back Water Curve and Affux 16.9.3. Expression for the Length of Back Waler Curve Solved Problem 16.45 Highlights Exerc ise

Chapter 17. Impact of Jets a nd Jet Pro)Ju\s ion 17.1. Introduction 17.2. Force Exerted by the J et on II Stationary Vertical Plate 17.2.1. Force Exerted by II Jet on Stationary Inclined Flat Plate 17.2.2. Force Exerted by II Jet on Stationary Curved Plate Solved Problems 17.1- 17.6 17.3. Force Exerted by II Jet on II Hinged Plate Solved Problems 17.7- 17.10 (a) 17.4. Force Exerted by II Jet on Moving Plates 17.4.1. Force on Flat Vertical Plate Moving in the Direction of J et 17.4.2. Force on the Inclined Plate Moving in the Direction of the Jet Solved Problems 17.ll- 17.13 17.4.3. Force on the Curved Plate when the Plate is Moving in the Direction of Jet Solved Problems 17. 14-17.17 17.4.4. Force Exerted by a Jet of Water on an Unsymmetrical Moving Curved Plate when Jet Strikes Tangentially at one of the Tips Solved Problems 17.18- 17.23 17.4.5. Force Exerted by a Jet of Water on a Series ..,fVanes 17.4.6. Force Exerted on a Series of Rad ial Curved Vanes Solv(.od Problems 17.24-17.26 17.5 . Jet Propulsion 17.5.1. Jet Propulsion ofa Tank with an Orifice Solved Problems 17.27- 17.28 17.5.2. Jet Propulsion of Ships Solved P roblems 17.29-17.33 Highlights Exercise

I I

787 787 790

790 792 793 794 795 796 799

B03-&;2

803 803 804

805

807 809 810 814 815 815 816

818 819

823 826 833

834 837

840 841 843

843 844 849 850

Ii

~ I

IL (r:ciii) C hapter 18 . Hy d r a u li c Mlic hin es--Turbines 18.1. 18 .2. 18.3. 18.4. 18.5. 18.6 .

18.7.

18.8.

18.9 .

l S. 10.

18. 0 .

18. 12.

IS. 13.

I I

Introduction Turbines General Layout of a Hydroe lectric Power Plant Definitions of Heads and Efficiencies of a Turbine Classification of Hydraulic T urbines Pelton Wheel (or Turbine) 18.6.1. Velocity Triangles and Work Don", for Pelton Wheel 18.6.2. Points to be Remembered for Pel ton Wheel Solved Problems 18. 1- 18.10 16.6.3. O<)sigrl ofPeilon Wh eel Solved Problems 18.1 1- 18. 13 Radial Flow Reaction Turbines 18.7.1. Ma in Parts of a Radial Flow Reaction Turbine IS.7.2. Inward Radial Flow Turbine IS.7.3. Degree of Reaction IS.7.4. Definitions Solved Problems 18. 14- 18.20 18.7.5. Outward Radia l Flow Reaction Turbine Solved Problems 18.21- \8.22 Francis Tu rbine 18.8.1. Important Relations for Francis Turbines Solved Problems 18.23- 18.26 Axial Fl()w Readion Tu rbine 18.9.1. Some Important Point for Propeller (Kaplan Turbine) Solved Problems IS.27- 1S.33 Draft-Tube IS.1O.1. Types of Draft Tubes 18. 10.2. Draft-Tube Theory IS.10.3. Efficiency ()f Draft-Tube &lIved Problems IS.33 (0 l-- I S.35 Specific Speed IS. 11. 1, Derivati()n of the Specific Speed IS.II.2. Significance of Specific Speed Solved Problems IS.36- 1S.41 Unit Quantities 18.12.1. Uni t Speed 18.12.2. Unit Discharge IS.12.3. Unit Powe r 18.12.4. Use of Unit Quantities (N . , Q., Po) Sol\·(,..1 Problems IS.41 (0 )- 18.45 Characteristic Curves of Hydraulic Turb ines 18.13. 1. Main Characwristic Curves or Constant Head Curves 18.13.2. Ope rating Characteris tic Curves or Constant Speed Curves

853-944

853 853 853 853 856

857 859 861 862 873

87. 877 877

878 880

884 88' 892 893 895 896 896 903

905 905 915 915 916 916 917 920 920 921 921 927 927 927 928 928 929 933

933 93.

Ii

~ I

IL (xxiv)

18.13.3. Constant Efficiency Curves or Musche! Curves or Iso-Efficiency Curves 18.14. Governing ofl'urbines Highlights Exercise

Chapler 19. Centrifugal Pumps 19.1. Introduction 19.2. Main Parts ofa Centrifugal P ump 19.3. Work Done by the Centrifugal Pump (or by Impnler) on Waler 19.4. Definitions of Heads and Efficiencies of 1\ Centrifugal Pump Solved Problems \9. 1- 19.12 19.5. Minimum Speed for Starting a Centrifugal Pump Solved Problems 19. 13- 19. 15 19.6. Multistage Centrifugal Pumps 19.6.1. Multistage Centrifugal Pumps for High Heads 19.6.2. r.Iultistage Centrifugal Pumps for High Discharge Solved Problems 19.16---19.17 19.7. Specific Speed of a Centrifugal Pump (N, ) 19.7.1. fo:xprossion for Specific Speed for a Pump 19.8. Model Testing of Centrifugal Pumps Solved Problems 19. 18- 19.22 19.9. Priming of a Centrifugal Pump 19.10. Characteri~tic Curve~ of Centrifugal Pumps 19.10.1. Main Characteristic Curves 19.10.2. Operating Characteristic Curves 19.10.3. Constant Efficiency Curves 19. 11 . Cavitation 19.11 . 1. Precaution Against Cavitati(1n \9.11.2. Effects of Cavitation 19.11.3. Hydraulic Machines Subjected to Cavitation 19.11.4. Cavitation in Turbines 19.11.5. Cavitation in Centrifugal Pumps Solv<.od Problem 19.23 19.12. Maximum Suction Lift (o r Suction Height) 19.13. Net Positive Suction Head (NPS H) 19.14. Cavitation in Centrifugal Pump Solved Problem 19.24 Highlights Exerc ise

Chapter 20. Recil,rocating Pumps 20.1. Introduction 20.2. Main Parts of a Reciprocating Pump 20.3. Working of a Reciprocating Pump

I I

935

936 937 939 945-992

945 945

947 948 951

965 966

966 966 969 969 971 971

972 973 976 978 978 979 979

960 960 961 981 981

961 982 983 985 965

966 987 989 993-1040 993 993 994

Ii

~ I

IL 20.3.1. 20.3.2. 20.3.3.

20.4. 20.5.

20.6.

20.7.

20 .8.

20.9 .

20.10.

Discharge Through a Reciprocating Pump Work Done by Redprocating Pump Discharge, Work Done and Power Required to Drive a Doubl e-acting Pum p Slip of Reciprocating Pump 20.4.1. Negative Slip of the Reciprocating Pump Classification of Reciprocating Pumps Solved Problems 20.1- 20.2 Variation of Velocity and Acceleration in the Suction and De livery Pipes Due to Acceleration of the Piswn Effect of Variation of Vel ocity on " riction in the Suction lind Delivery Pipes Solved Problem 20.3 Indicator Diagram 20.8.1. ldcal l ndica.lor Diagram 20.8.2. EtTect of Acceleration in Suction and Delivery Pipes on Indicator Diagram Solved Problems 20.4-20.9 20.8.3. Effect of I'riction in Suction and Delivery Pipes on Indi cator Diagram 20.8.4. ElTect of Acceleration and Friction in Suction and Delivery P ipes on Indicator Diagram Solved Problems 20. 10-20.12 20.8.5. Maximum Speed ofa Reciproca ting Pump Solved Problem 20.13 Air Vessels Solved Problems 20. 14-20. 18 Comparison between Centrifugal Pumps and Reciprocating Pumps Highlight.'< Exercise

Chapter 21. F l uid System 21.1. Introduction 21.2. The Hydraulic Press 21.2.1. Mechanical Advantage 21.2.2. Leverage of the Hydraulic Press 21.2.3. Actual Heavy Hydraulic Press Solv(.od Problems 21.1 - 21.5 21.3 . The Hydraulic Accumulator 21.3.1. Capacity of Hydraulic Accumulator Solved Problems 21.6-2 1. 11 21.3.2. Differential Hydraulic Accumulator 21.4. The Hydraulic intensifier Solved Problems 21. 12-21. 13 2 1.5. The Hydrauli c Ram Solved Problems 2 1.14-21.15

I I

99<

995 995 996 997

997 997 998 1001 1001 1003 1003 1004 1004 1012

1013 lOIS 1019 1020 1021 1030 1037 1037 1038 104 1- 1070 1041 1041 1042 1042 1042 1043 1045 1046 1047 1051 1051 1053 1053 1055

Ii

~ I

IL (xxui)

21 .6. The Hydraulic Lift 21.6.1. Direct Acting Hydraulic Lift 21.6.2. Suspended Hydraulic Lift Solved Problems 21.16- 21.17 21.7. The Hydraulic Crane Solved Proble ms 21.18-21.20 21.8. The Fluid or Hydraulic Coupling 21.9. The Hyd raulic Torque Converter 21.10. The Air Lift Pump 21.11. The Gear-Wheel Pump Highlights Exercise

Objective Type Questions Appendix

Subject Indel<

I I

1056 1057 1057 1058

1061) 1061) 1063

10'" 1065

IOEm 1067 1068 1071- 1094 1095-1000 1097_1102

Ii

to 1. 1

INTRODUCTION

Fluid mechanics is that branch of sc ience which deals with the behav iour o f the fluids (liquids or gases) at rest as well as in motion. Thus this branch of scienl'C deals WiTh the stat ic. kinematics and dy namic aspecTs of fluids. The study of fluids at rest is called fluid statics. The study of fluids in Illotion. where pressure forces are not considered. is c all ed tluid kinematics and if the pressure forees are also considered for the flu ids in motion. that branch of science is callt>d fluid dynamics.

to 1.2

PROPERTIES OF FLUIDS

1.2. 1 Density or Mass Density. DensiTy or mass densi ty of a fluid is defincd as the ratio of the mass of a fluid TO its volume. Th us mass per unit vo lum e of a Iluid is called density. It is denoted by the symbol p (rho). The unit of mass density in SI urlit is kg per cubic metre. i.p .• kg/m 3. The density of liquids ma y be cOrlsidercd as COrlSlant while that of gases changes with th e variation o f pressure and temperature. Mathematically. Illass de nsity is wr ill en as

p= The value of density of water is I gm/crn l or I()()(} kg/m 3. 1.2.2 Specific Weight or Weight Dens ity. Specific weig ht or weight density of a fluid is the ratio between the weight of a fluid to its volume. Thus weight per unit volume of a fluid is called weight density and it is denoted by the symbo l w. Thus mathematically.

IV = Weight o f fluid '" ,(,Mc"",co,fcnc'o;cd~);X'i-AC'C"'C""'O"C;OC"7'd"c,c,cQ"g~,c'c'c",-Y Volumc of fluid Vo lumc of fluid

Mass of fluid ) =p Volu me o f flu id

=p>
..•

...( 1.1 )

IV= pg

1

I I

Ii

~ I

IL 12 Fluid MC(;hanics The va lue of specific weight or weig h! density (w) for water is 9.81 x WOO Newlon/II]"' in Sl units.

1.2.3 Specific Volume. Specific volu ille of a fluid is defi ned as the volume of a fluid occupied by a unit mass or volum e per unit mass of a fluid is callcd specific vo lu11Ie. Mathemat ically. il i~

expressed as Specific volume

=

Vo lume of fl uid

~M,!","",,,o"r~n~""d;r; '" p

M ass of fl uid

Volume of flu id Thus specific vol um e is the rec iprocal of mass density. It is c.'pressed as m' fk g. It is commonly applied 10 gases.

1.2.4 Specific C;uvity. Specific gravity is defined as the rali o of th e weight densit y (or de nsity) of a fluid to the weight de nsit y (o r den sity) uf a standard fluid. For liquids. the standard fluid is Taken water and for gases, th e standard fluid is taken air. Specific gravity is nlso called relntivc dens it y. h is dimension less quantit y and is de noted by tlie symbol S. Mathem aticall y, S(for liquids) = -;,w",'cig~hc'cd;'e"c'7h~YC(CdC'""e·'ChlYC)Oo"r"tioq~""id" We igh t de nsi ty (de ns ity) of water

:;;:'"igehc'CdC'C·"C'Ch"YC(~dO'C"C'Ch"YC)OOcf~gc~ o S(for gases) = -;_W We ight de nsity (de nsi ty) of air Thu s wrig ht density of n liqui d = S x =Sx The densi ty of a liquid =Sx =Sx

Wei ght dt' nsit y of water lOOOx9.81 N/m J Density of water 1000 kg/ml .

.. .(\.]A )

If tlie specific gra vity of a fluid is known. then the density of the fluid will be equal to specific grav ity of fluid multiplied by tli e density of water. For exa mple. tlie specific gravity o f mercury is 13.6. hence density of mercury = 13.6 x 1000 = 1]600 kg/ml.

Problem 1.1

Co/m/ill" tl/<' sfH'ciflc weigili. delisiTy om/ spnific gral'ily of 0111' lifr" of (/ liquid

... IIid, II'pigll,' 7 N.

Solution. Given: I

Volume = I litre = - - m llJOO Weight = 7 N (i) Spedfk weight (w)

c-_ (1~) Illl

w7000

I

;

or I li tre

= 7000 N/llr'. AilS.

1

,

= - = - - kg /m = 71.\.5 k g/Ill ' . Ans . g

(iii) Specific grav ity

.

IIlTe = 1000 m

7CN = Weig ht = T.c. Volum e

(ii) Density (p)

,

9.81

'" Densi ty of liqu id '" 7 135 Density of water 1000

I .:

De ns ity of water = 1000 kglm11

'" 0 .7 1.\5. AilS.

I I

Ii

~ I

IL Properties of Fluids Problem 1.2

31

Ca/culale tile dellsity. specific weigh! and weight of one lilre of Pe/rot of sped/it

gra..!ty '" 0.7 Solution. Gi ven:

D"'I~'ily

I~ 10

rn 3 =0.00 1 rn 3

5 = 0.7

Sp. gravil y (il

x lOOOem 3 =

Volume: [Iitre =

(p)

Using cq uJlion (I. IA), = S x 1000 kg/ml '" 0.7 x 1000 == 700 kg/hi ), Am.

Dellsif), (p)

(ii) Specific ...eight (w)

Using equation (1.1),

w

= p x g = 700 x 9.81 Nlrnl = 6867 Ntm J , Ans.

(iii) IVeigl1l ( IV)

We know [hal sp"cific weig h!

,,,,,;,,:hC ' = .-w Volume

IV \I' ... = - - or6867 = - 0.00 1 O.OO[

IV ", 6867 x 0.00 1 = 6.867 N. An s. ..

1,3

VISCOSITY

Vis<:osi ly is d efi ned as the property of a fluid whi<:h offers rcsiswnce tu the movement of o ne layer of fluid over imolhcr adj ace nt layer of ihe fluid. When twO laye rs of a fluid. a dislanl'C 'dy' aparL move one over the other at different velocities, say u a nd u + du as show n in Fig. 1.1. Ihe viSC(}~ity togclher wi th relative velocity cause~ a shear stre~s acting between Ihe nuid laye rs. Th e lOp layer eauses a shear Slress on the adjace nt lower layer while the lowe r layer causes a shear stress on th~ adjacent top layer. This shear stress is propo rtional to the rate of c han ge of velocit y with respec t to y. It is denoted by symbol VELOC ITY PROFILE t (Tau).

"

Mathe maticall y.

,~

d, dy

-

t~)J

Fig. 1.1 Velocity variation near a mlid boundary.

"

... (1.2)

-

dy where )J (calk'd mu) is the (.unstant of proportionality and is known as the (.u-cfficient of dynamic viS«)sity d, oronl y viM-usity. - represents the rate of she
].I

1

= -,( d,

.. .( 1.3 )

If)"

Th us viscosity is a lso defin ed as th e she ar stress required to produce unit rate of shear strain. 1.3.1 Unit$ of Vi$co$ity. Th e uni ts of viscosity is obtained by putt ing the dimcnsions o f the quantities in cq uation (1.3)

I I

Ii

~ I

IL 14

Fluid MC(;hanics

r orce! Area

Shear st ress

". ~~~~~.~~~~Change o f velocity 1

(Lengeh )

Change o f distance

Time

x Leng th

'" r"'Ofcc/(l.c ngth)' = Force x T ime 1 TI me In MKS system. force is represented by k.gf and length by InClre (m). in eGS system. force is represented by dyne and length by em and in SI system force is represented by NewlOn (N) and length

by metre (m). M KS unit of viscosity

= kgf -scc

m1 = dyne-sec

eGS unit o f viscosity

cm

l

In the aho"c expression Nlm l is also known as Pascal which is represe nted by Pa. Hence N/rn2 = Va

= Pasca l SI unit of viscosity = Ns/m 2 = Pa s. Newton sec

SI unit of viscosity

Ns

The uni t of viscusity in eGS is ~ISQ called Poise which is eq u al 10

-sec "'dyne ' 'em! C;:=-

The nume rical conversion oflhc unit ofviscosi l y from MKS unit to eGS unit is given beluw: 9.8 1 N-sec

BUl one Newton = one kg (mass) x one

=

(

I':

I kgf=9.8 1 New ton)

-m, ) (acceleration) ~'

(1 000 g m) x ( 100 c m) scc

l

= 1000 x 100

gm-cm sec

l

dyne = gm x -c m-, )

= 1000 x 100 dy ne

='

one kgf -sec =9.81 x 100000 dyne-sec nI l

e l11

2

= 9.81 x 100000

====--, dyne-sec

100 x IOOxem 2

• 98 . 1 dyne-sec ,d"y,"",~,:c'=-c = , = 98. 1 poisc c m' em ' Th us for solving numerical prob lems. if viscosity is given in poise. it must be divided by 9&.1 ils equivalent numerical value in MKS.

1.:

1

8m

ml

one Ns

~

I I

=+,"

one kg f -scc • 9.8 1 Ns Ill "

POiSC) 10

get

= 98.1 poise

9&.1 • - - poise = 10 poisc 9.81

One poise =

1 N, 10 ml

Ii

~ I

IL 51

Properties of Fluids

.

Alt rrnllil' Mdhod . One po lS<: '"

dyncxs em

gill

But dyn e

l

'"

( l gm X I Cm ) s

,

X- , el11

[ em X -

,-

" I

I gm '" ;;::;; '"

O ne poise

l

lOiXI

k

g I ' -- m 100

=

I

IOpoisc.

"'

1000

1"'otr . (i) In $[ un;l< second i. represented by's" and nOl by 'sec', (iil If \'iscosity i. gi,'cn in poi,." it mus, be divided by 10 to gel its equivalent numerical ..alue in SJ unil", Somclimcs a unit of v;"''OSily as centipoise is used where

. , I I ccntlpo,,,, '" The visco,ily of water at

20~C

, 1 P POIse or ICP. 100 100 is D.OJ poise or 1.0 centipoise.

rtf' '" Centipoise. P '" Poise 1

1.3 .2 I(in e m~ti c Viscos ity . h is deFined as the ratio between Ihe dynamic viscosity and density of fluid. 11 is de uoted by lhe Greek symbol (v) c a lled 'no . Thus, ma thematica ll y. \I '"

Vi scos ily De nsi ty

jJ

...( 1.4)

'" -

p

The units of kinematic viscosity is oblai llcd as UnitsufjJ

,. URilS of P

Force x Time

--'=~~ = (Le ngth) ' x Mass J

Force x T ime M ass

Length

(Length )

Mass x

=

=

Length , x T ime

(Time)"

1

--7-':';"'--;--

M,,, ( Length

r

Force"" M ass x Ace. .

= M ass )(

)

Leng th l

Time

(Le ngth )!

S="Tim e

In MKS and SI. the unit of kinematic viscosity is metre'!scc or m'/sec while in CGS unilS it is written as (;111'/s. In CGS units. killcm~lic viscosity is also known as stoke. Thus, olle sloke

=clI1 '/5=

Cen tistoke means

=

1

100

c~or

4 O1'/S= 1O- 01'/s

sloke.

1.3 . 3 Newton ' s Law of Visco s ity. It SImes that1h" she ar sIres.-; (l) on a fluid clemcnt layer is dircClly proportional to the ral c of shear strain. The 1:01151:1111 of proportionality is called the cocfficielll of viscosi ty. Mathematically, ;1 is c~prcssed as givell by equation (1.2) or as till t = !-I - . (/ )'

I I

Ii

~ I

IL 16 Fluid MC(;hanics Fluids which obey the above relation arc known as NewtonL'l1i fluids and the fluids which do not

obey tlie above relation arc callcd No n-Newto nia n fluids.

1. 3.4

Variation of Viscosity with Temperature .

Temperature affects the viscosity. The

viscosity of liquids dcneasc s with Ihe increase of tcmpcrmurc while Ihe viscosity of gases inncascs with the increase of tempermure. This is duc to reason that Ihe viscous forces in a fluid arc duc to cohesive forces and molecular momentum lransfcr. In liquids. the cOhesive forces predominates the Illolccular 1110111cn lurn lraJlskr, duc 10 closely packed Illolcculcs and with Ih e increase in temperature. Ihe cohesive forces decreases wilh the resuh o f decreasing viscosity. Bul in case of gas.cs the cohesive forces arc small and molecular momentum transfer predominaks. With the increase in tempera ture, molecular fllom<"nturn transfer increases and h<"nce vis.cosily increases. The relation between viscosity and temperatu re for liquid.~ and g ases are: (I) Fo r liquid s, (.I

whe re

(.I

= (.10 (

,l

... ( l.olA)

I I +O:I+llr

= Viscosity of liquid at

rc, in poiSt:

1-10 = Viscosity of liquid at O°C . in poise 0:..

P " Constants for the

liquid

ror water. (.10 = 1.79 x 10 J pois.c. (I = 0.03368 and P= 0.000221. Equation (1.4A) shows that wilh the increas.c of tcmperature. the viscosity dcr rcases. (ii) For a gas , J.I" (.10 + o:r _ PI 2 9 where for air J.lo " 0.000017. 0:= 0.000000056. P'" 0 .1189 X 10- .

... ( I.4B )

Equation (I.4B) shows that with the increase o f temperature. the viscosity increases. 1.3 . S Types of Fluids. The fluids may be classified into the following five types: I. Ideal nuid. 1. Real fluid. 3. Newtonian fluid. 4. Non -New tonian fluid. and 5. Ideal plastic fluid. I. Id"al Fluid. A fluid, whic h is ;ncorll pr<"ssiblc and is having no I'is.cosily. is known as an ideal fluid. Id~al fluid is only an imaginary nuid as all the nuids. which exis!. ha ve some viscosity. 2. R"nl Fluid. A fluid. whic h possesses viscos il y. is known as real nuid. All lhe nuids. in aClual practice, are real fluids. Ne wtonlun )<' Iuld . A real fluid. in which lhe shear 3. stress is direc tly proportional to the rate of shear strain (or velocity gradient). is know n as a Newtonian nuid. 4. Non -Newtonian Fluid . A real fluid. in whic h the shear stress is not proportional tu the r:lte of shear strain (or velocity gradient), known as a Non-Newtonian nuid.

1

IDEAL FLUtD -

VELOCITY GRADIENT (~~)

Fig . 1.2

Typt$ of flm·d$ .

5. Ideal Plastic Fluid . A fluid. in which she ar st ress is more than the yield value and shear stress is propor1ional to the rate of shear strain (or velocity gmdient), is known as ideal plastic fluid.

Problem 1.3

,

If IIII' re/ociry dislribuliOlr O)'e ~ (I pialI' is girol by u = .:. y J

relocily ill mefre per seCOIrd aI

y " 0 alld y = 0./5

I I

111.

(I

-l

in "'hieil U is lire

disfmrce )" melre ab()re lilt! plme. de/amilre Ille

.~lIe(lr

J·lres.!

{If

Take dYllamic I'iscosify offtllid (IS 8.63 poises.

Ii

~ I

IL 2

Solution. Giwn :

till

,

II= -Y-Y

dy

3

(~:.)~ y_O 01(;:;),.0'"

("") -

,/ Y

-

"' ",,_0."

("")

Properties of Fluids

,

71

'" ..:. - 2y 3

2

2

- - 2(0)= - =0.667 3 3

2

= - -2x.15=.667-.30=0.367 3

d Y ,_ 0' 5

Value of).l = 8.63 roisc = 8.63 51 units = 0.863 N s/m 2

10

Now shear stress is given by equat ion (1.2) as t =).l

(/11

,/ )'

(I) Shear stress Jt y = 0 is given by

(~) '" 0.863 x 0.667 '" 0.5756 N/m~. Am. (Iy y. Q

t o = I-l

(ii) Shear strcs.~ at y = 0.15 III is given by

(l:)y

. O_l~

'"

(;;1y') '" 0.863

Il

x 0.367 '" 03 167 N/m l, AilS.

y_ O l ~

Problem 1.4 A plm/' 0.025 """ (/i.
Distance bel ween pl ates. Velocity of upper

= .025 x 10-] m 1/ = 60 em/s = 0.6 m/s

pl:u~.

N F= 2.0 2 , m

('orce on upper plate.

This is lhe value of shear Slr~ss i.t', •• lei the fiuid vi.'>Cosily between the plales Using lhe equalion (1.2). we whe re

f

dy = .025 nlln

ha,,~ t

==

~

T "'i"'''''''''''' FIXED PLATE

Fig. 1.3

is~.

-d" .
Ii" = Change of velocily = II - 0 = II '" 0.60 IIl/s

dy '" Change of distaJicc '" .025 x IO- J III t '" Force per unil area'" 2.0

N

IIll

-;:;~0~.6~0 '" c

2.0 x .025 X lO-l == 8.33 x 10 .s Ns 111 2 0.60 .025 x 10 5 '" 8.33 x 10- X 10 poise '" tl..~ .~ x 1O -~ IlOi~(' . Ans. Problem 1.5 A jlal pimp of
' '('11/

I I

Ii

~ I

IL Is Fluid MC(;hanics Solution. Given: Area uf the plate. Speed of plate relative 10 anOlhcr plale. "" '" 0.4 tn/s Distance ~Iwccn [he plales.liy = 0.15 mill = 0.15 x 10- J III Viscosity

)..1= I

Using eq uation (1.2) we have '!: " )..l

(r):. (ii)

Shear force,

Powcr~

F

=

I

N,

poise: - ,. 10 Ill '

d"

I

dy

10

! X area

.4=' x -:~~O~ ) = _66.66 2N .l5xlO III

= 266.66 x 1.5 = 400 N. AilS.

required to move the pialI' at thc speed 0.4 mlsec

== Fx II =400x0.4 = 160 W . AilS. Problem 1.6 De/ermilll' Ilu' il//(,lls;l), of shnlr of {III oil/r{/\'illg I'iscosily '" I poiSt'. 'I'llI' oil is used for illbricwillg Ih(' c/I'lIrlW('I' bl'l"'''1'11 {/ shaft of(liallieler /0 rill {l1II/ its jOllnwl bl'arillg. '/'1,1' c/nlr
Solullon. Giv"l1 : Dia. of shaft.

10

D= IOcm=O.1

Ill ' III

Distance be tween shaft and journal bearing. dy = 1.5 nun = J.S x to

1 III

Speed of shaft. N = 150 r.p.m. Tangell1ial speed of shaft is given by II

=

It DN

== It x 0. 1 x 150 == 0.785 mls

60

60

Using equation ( 1.2). where (III '" change of ve locity betwee n s haft and bearing'" I

0.785

10

l.5 xlO

==-x

II -

0 '"

II

.. ! J =.2 ..BNfm. AIIs.

Problem 1 .7 ,a/rllia/(' II", (Iy"m"ic I·iscmily of w, oil ....hid, is uSNI for lubricatioll bNW"CII a 1,Iatp of sic,' O.S III x O.S III amI (III i"diflNI philiP wilh allgl/' of ifldi""titm .woflJ sho",,, ill Fig. lA. The w..ighl of the S'lllfl'.. plat .. is 300 N (Illd il slilln (IOwlIl{,r illclill~d plmlf Wilh" IIl1iform I'elocil), ofO.J /Ills. 'fI", thicJ.:III'ss of oil fillll is J.5 111111.

J'IU"rt'

Solution. Given: Area of plate, Angle of plane. Weight of plme. Velocity of p late.

A = 0.8 x 0.8 = 0.64 111 2

e = 30" IV= 300 N II == 0.3 111/s Fig. 1.4

*

I I

Power=l'x"Nmls=FxulV ( '.· NmJs=Wal1 j

Ii

~ I

IL Properties of Fluids

91

I" lly" 1.5 111 III " 1.5 x 10 3 III leI the ViSl:os ity of nuid between plate and inclined plane i s~. Component of weight IV, along the plane" IV cos 60~ '" 300 oos 60° '" 150 N Thus the she ar force. F • on the bunum su rface of the plate'" 150 N Thick.ness of oil film.

F <= - - " 150 Nfm~

and shear stress.

Area

0.64

Now us in g eq uation ( 1.2). we have

rill

T =).I -

dy where tlu = change of ve locily = 11 - 0 = " = 0.3 mfs dy'" [ = 1.5 X IO-J 1ll

150

0.3

O.64 =).I I.5xIO

J 3

150 15 10"= "''''';';'::-;;c;-'" 1.17 N 0.64 x 0.3 x

Problem 1.8 wi,II oil

x

2

s/m '" 1.17 x 10 " 11.7 Jlois('. Ans.

Two IlOri~ollllll l'/mf'S (Iff' plll("'f(1 1.25 em (l1'MI. I/W 51'11("(' bnwulI IIII'm bf'illg jillnf / 4 l>oi5"5. CII/ru/ml' 110(' S/WlI' 5/ rl'55 ill oil if UPP'" phil" ; .5 mm'~d wilh fI \-<'10";1)'

('I \'i5,.OS;'Y

0[2.5 ",Is. Solullon. Giv<,)11 : Dista nce between plates.

Viscosity.

dy = 1.25 em = 0.0 125 III Il" 14 poise ==

.!.i

N sJIll 2

10

Ve loc ity of up pe r plate.

1/ "

2.5 Ill/seC.

Shear stress is given by equ ati on (1.2) as, t " Il where til!" Chan ge of ve locit y be twee n p lmes"

I! -

(I I!

dy 0" "" 2.5 Ill/seC.

dy=0.0 125 m. 14

T"

10 x

25 z .0125 ,, 2MO N/m . Ans.

Problem 1.9 '11", sP""" bn",rr/l IWo s'I"''''' JIm IJamU,,1 plain is fillt'd \\,;111 (Iii. Eacll shit' of II,,, plait' ;s 6V UII. 1'11" Illidllt'ss of IIIr oil film is 12.5 111m. n,,, IIPP"f pllllr, ",!ticll I/W ..<'S <1/ 2.5 mt'lfr pr f ,'re r('
:. Area. Th ic kn ess of oil film. Ve loc ity of upper plate.

~ I

,,60 elll '" 0.60 1ll 1 A " 0.6 x 0.6 " 0.36 m d y " 12.5 mm '" [2.5 x 10- 3 1ll 11" 2.5 mIse"

I~

~ I

IL 110

Fluid Me
Ctmllgc of veloc it y between plm es.
Force F 98.1 N ,. - - = - = , Area A 0.36 m-

She ar st ress.

(i) LeI )l '" Dynamic viscosity o f oil

II"

Usin g eq uatio n (1.2).

t =).l -,,, )l '"

0'

98.1 -0.3-6

25

"')l)(

98.1 )( 12.5)( 10-

1 '"

CI2~'~X";;'O'" 1.3635

N~

0.36 2.5 rn ' '" 1.3(35)( 10", I.t6.\5 puist'o Ans.

Cln~lS

'"

to poise )

( ii) Sp. gr. of oil. S '" 0.95 leI v'" kine m 3lic viscosity o f o il

Usi ng equation ( 1.IA).

x 1000 '" 0.95 x 1000 '" 950 kg/Ill}

p'" S

Mass de ns it y of oi l.

1.3635 ( Using the relatio rl . v

=

1:. we ge l v =

N~l III -

P

=.00 1435 m 2/sec = .()() 1435 x 104 C111 2/S

950 (": cm 2/s '" sIDke)

= 14 ..'5 s lok~s. Ans.

Problem 1.10 "PO;/ll in oil

j"

j

Filll/,11l' killfl/Uuir visrosily of 1111 oil/w\'illg (/l'1!5ily 98/ kgllll , TIll' S/W(If 0.2452 NI",l ' Illd ",,'/oril), grad;f'JIl lU I/U" p"ill/ is 0.2 pn .w("(md.

5Iri'.' S (1/

Solutio n. Gi ven: p" 98 1 kg/lll ' 1: = 0.245 2 N/m!

Ma ss densit y. Shear stress. Velocity grad ien t. Usi ng Ihe equation ( 1.2),

~

"y

=0.2 s (I "

t =j.l -

,/ y

orO.2452"j.l x O.2

j.l " 0.2452 " 1.226 Nslm! 0.200 Kincmatic viS\:osil y v is given by v=

1:." p

1.226 =. 125x 10- 2 m !/scc 98 1

" 0. 125 x 10- 2 X 104 I:m 2 /s" 0. 125 x 10 2 cm'/s = 12.5 en/Is" 12.5 stokl'. Ans.

Problem 1. 11 ,·iu(}.lity 0.0.'15

(: cm1/s" stoke)

Dnnmill" Iii" S{J"c ijic gr(l\'ily of" JI"hllllll'ing .. i.l("(uil), 0.05 /1Oi.I" mill kin"",atic stok~ •.

Solution. Givcn :

W

V'Is.cOSI' 1y. j.l =. 005' POISC = 0.05 N sJm 2

~ I

I~

~ I

IL Prope rties of Fluids

11 1

v:: 0.035 sto kes :: 0.035 cm1ls :: 0.035 X 10- 4 m 1/s

Kinematic viscosity.

I.l

_4

Usi ng thc relation v:: - .wcgclO.035xIO P

0.05

I

10

p

= -- x -

0.05 X ~~'--C4" = 14 28.5 kg/m3 10 0.0]5 x [0

Problem 1.12

Density of liquid

14285 :: - - :: 1.4285 ::: 1.4.\. AilS. Density of wate r 1000

::

Sp. gL of liljuid

/Jell'rmilll' IIIl' I'iscosily of (/ hlfllid havillg til/i'lI/lUil' \-;Seosil)' 6 s/okn 1111<1 specific

gm.-it)' 1.9.

Solution. Given: Kinematic viscosi ty Sp. gL of liquid LeI the v iscosity of liq uid

v:: 6 stokes = 6 cl11'/s:: 6 X 1O- 4 11//S :: 1.9

.p D<:nsity of lh ~ liquid Density of water

Now sp. gr. of a liquid

1.9 :: c"' ='c"="c'Yso~rcn,q='::: ;d 1000

kg

= 1000 x 1.9 = 1900 - ,

Ik nsi ly of liquid

m

v::

.. Usi ng th e relation

1:., we get p

4

6x 10- = - " -

1900

J.I=6x 10- 4 X 1900= 1.14 Ns/m 1 :: 1.14 x 10:: 11.40 Il oise. Ans. Problem 1.13

Tilt' l'e/oeiIY distriblliiou for jlow OI'e r " JIM p/lIlr is gil'l'lI by " = -1- )'

II ;5 IIII' \'e/oeil)' ;111111'1'1' IN" secQlul aJ)'

a (/;5/(IIICI' y 1111'1'" abore IIII' 11/(1/1'. DrlulII;nr IIII' silell' SI'I'SS

(1/

= 0. 15 Ill . Takl' (/),II(11l1ic \'is("o,lily ofjluid ( 1.1 8.6 poiSl'.

Solution. G iven:

11=

(I u

-

If)'

At .1'=0. 15.

Viscosity.

I I

-l ill which

(Iu

-

If .I'

1-1

~

1' _ 1,2

4 '

.

3

= - - 2)' 4 3

= - - 2xO. 15=0.75 - 0.30=0.45 4 = 85 poisc = 8.5 N ~ [0 m-

N~ l ( .: 10 poisc = 1 m

Ii

~ I

IL 112

Fluid Me
r = '" -

Using eq uatio n ( 1.2).

85

-

~

10

If )'

N

N

x 0.45 - : = 0.,\1'1 25 - : . AilS. m

In

Problem 1.14 11/1' (/pwlllic viscosity of WI oil. lISt'll for II/brim/ioll bnwrl'u (I SIUlfi 01111 5/1'1'1'(' is 61"';u, TI", .,'mp i., of di"'"N(,T 0.4 III "",I roW/P.' 1lI 190 T.p.m. Cuk ulmr 1111' I}{)""P T /0.'/ ill IIii' brllrill8 fo r ti slnl'(' /""8,11 oJ90m",. TIIr lilickllrH of/lie oil film is 1.5 "''''. Solution. Gi ve n : 1.5 mm Viscos ity j.I=6poi se 6 Ns ~

Dia. o f sha ft. Speed o f shaft.

f)

10 m

= 0.4

l

m-

y o_~ m

,C': ';::J lKAFT

III

N= 190r.p. m

Sleeve k ng1h. Thic kn ess of oil fi lm,

L = 90 mm = 90 x 10- )

Tangc llti al vel oc ity of shaft .

/I

Using the re lation

t=l-I dy

where

d

Ns

= 0.6 - ,

SLEEVE III

I'" 1.5 111m '" 15 x 10- .1 m

Fig. 1.5

'(tDN IT x 0.4 x 190 = ~= 60 = 3.98 IIl!S tlu

'/11 = Change of veloc ity = II - 0 = /I '" 3.98 m!s Ify = Change of d istance = I = 1.5)( 10- 3 1ll

--:-3~.9~8;,. '" 1592N/m' .=lOx -;l.5x lO J This is shear

st res.~ On

shaft

Shc ar for(;c o n thc s haft. F '" Shcar stress x Area '" 1592x1tDxL", 1592x1tx.4x90x lO -3 = 180.0SN D 0 0.4 T= Forcc x - '" 18 .05 x '" 36.0 1 Nm

Torquc on th e sha ft ,

2

2

_ 21tNT _2 1tX I90 X36.oJ - 7164SW

-

60

-

60

-

.

.

. "'n~ .

Problem 1. 15 If lilt' r('/ocily profil~ of {/ jlllid Ol't'f {/ (,((l/r is p
OJ _

211'N 2 11' NT T)( - - Wan _ - - - Wall

60

I I

u • , 2(l em/Sf!(:

11 '" 85 poise = 8.5 N ~ '" 0.85. lO m '

Viscosity .

• Power in S.1. unit_ T·

y

Fig. 1.6

'"

Ii

~ I

IL Prope rties of Fluids The ve locity profile is g ive n parabolic and eq uation o f ve locity profile is II =
131 ... (1)

,/u

(c) illy=20cm.-=O.

"y Substitutin g boundary conditi on (ll) in equation (,) . we gd c = 0. Boundary cond iti on (b) o n sub stituti o n in (i) gives 2 J 20 = a(201 + b(20) = 400" + 20b Boundary co ndition (c) 0 11 substi tution in equ ati on (il gi ves

,II. =l,,)' +b

-

dy

... (ii)

...(iii)

-

0= 2 XII X 20

+ b = 4{M + b

Solving L'ijua ti ons (ii) and (iii) for a and b From eq uation (iii), b = - 40<1 Substitutin g this va lu e in equati on (ii) , we get

120 = 400<1 + 20 x (- 40a) = 400(/ - 800<1 : - 40011

120

3

tI= - - = - - = - 0.3 - 400 10 b= - 40)« - 0.3)", [2.0

Subslilulin g the va lu es of a. band (' in equat io n (i),

0.3/ + 12y.

II" -

Vrlodty Gradient -(/ " = - 0.3 )( 2y + 12=-0.6y + 12 Ify

"' Y

(""] (""]

= O. Veloc ity grad ient. (/ y

at y '" 10c m.

at y : 20 c m.

= - 0.6 )( 0

= - 0.6)( 10+1 2= - 6+ 12: 6/s. AD S.

-

(f y

( ~".] )

+ 12 = Ills. Ans.

1*°

1 - 10

= - 0.6 )( 20 +1 2: - 12+12= 0. Ans. y _ 20

Shear Str('sses Shear stress is g iven by.

I I

t : fl

till

(/ )'

Ii

~ I

IL 114

Fluid Me
(i) Shear stress at y

(ii) Shear s[rc ....~ at y

- [d"d ) t=J.l[:;"]

= O.

t - lJ

= 10.

y

y

(iii) Shear stress at y '" 20.

t=)J

_.o5x l '0_._l,~Nfrn. -0" 0' '

~.o

=O.85x6.0=5.1 N/m

2

•

1 - 10

=O.85xO= O. Ans. [d"] Ii)' ,-. -

Problem 1 .16 A New/ollillll Jill;" is fill ..,} ill IIII' c/CII'/'/1 {( shaft IIml" collCl'IIlrir sIN'I'e. Till' s/,,{'\,I' IIIll1illS {/ sP('('(/ of 50 ollis, W/1l'1I II lorn" of 40 N is IIPIJ/inf 10 the S/I'I'I'I' parallel 10 IIII' slmfl. /){'/('flllillf IIIl' SpUI/ if {/ jora of 200 N is (If'l'lin/. Solu t ion. Gi ven: Speed of sleeve, fit = 50 em/s when force. FI =: 40 N.

Let speed of sleeve is "2 when force. F2 = 200 N. Using relation

t =

I.l

till dy

rorce F where t = Shear stress'" - - • Are a A
/I -

0="

-F =1.1 -" A

Y A jJ u

I.

F=--."

A. j.l JIlO yare constanT}

Y

F f~ ...l= ....:.

",

Substituting val ues. we gel 40 ==

200

50

50x 200 40

'" 50 x 5 == 250

cml.~.

Ans.

Problem 1.17 A 15 nil diol/w/f r "erlirol c)'lil/der rotales COI/C<'III rical/)' illsitit' ((lIatller cylilllirr oJ !lilli/WIer 15.10 nil. 80111 cylindt'rs ort' 25 rill lIigli. Til<' SP((C<' bnw<'<'n III<' c)'lilUlers is jillnl willi 0 liquid whost' I'ismsil), is unknowll. If " torqut' of J2.0 Nm is r..quired 10 rOUlt .. II, .. illll'" cylillll'" lit 100 r,p.III .• dr/erll/illt' Iht' viscosity of lilt' fluid.

Solution. Given: Diameler of cyli nder Dia. o f oUler cylinder Lenglh of cy li nders, Torque.

I I

== 15cm==O.15m == 15.10 cm == 0. 151 111

L == 25 cm == 0.25 r == 12.0 N1II

111

Ii

~ I

IL Prope rties of Fluids

151

N", 100 T.p.m.

Speed, Let th e viscusit y

""

. . . nDN IlxQ.15x100 Tan genllal ve loc ity of cyl inder." '" - - '" '" 0.7854 mfs

60

A = nD x L = It x 0.15 x 0.25 = .1178 rn 2

Surface area of cy linder.

till

Now usin g relation

where ,/" =

1/ -

60

l: :p. -

dy

0 = " = .7854 rnls

15,,',,--,,0,,.' "' eo: III = .0005 dy = -"0.,, 2

m

I-l x .1854 .0005 Sheaf force.

F = Shear stress x Area =

Torque.

T= Fx -

).1x .7854 x . 11 78 .0005

lJ

2

12.0",

I-l x .7854

X. 11 78 x

.0005 12.0 x .0005 x 2

.7854x .11 78 )(.l 5

.15

2

= 0.864 N shn!

'" 0.864 x 10 = 1i.64]loisc. An s.

Problem 1 . 18 Two Illrgt' phll/P sIIr/tln's (I'f' 2.4 rill apart. Thl' Spllcr b(,/I<'I'I'II ,h" slIr/act's is filln! willi giY("f'rilll'. II'lul/ fora 13 rrq"irnllo drag /I 1't"Y Ihill plllll' of sur/act' (If('ll 0.5 squarr /lleI,." b"I»''''''' /\1'd largl' plm,,' sII ,fa,.. s til II sprn/ of O. 6 m/s, if:

I",.

(i ) lill' 1i1ill [,/alr is ill lh" middll' oflhl' 11>'0 p!t11l1' surfllCl's. mul fii) IIII' Illill 1'1(1/(' is at rl disliIllCl' of 0.81'111 frO/II 011(' of Ille plllll<' sIOrfll(,('3 ? Take IIII' dYliamic "ism sil), ofgl),urill(, '" 8.10 x uri N .!1m?

Solution. Given: Distan ce Iwtw", ... n two large ~urfac es '" 2.4 em Ar~a of th in plate. A = 0.5 [11 ! Velocity of thin plate. Viscosity of g lyce rine.

/I

= 0.6

mf.~

!-! = 8.10 X 10- 1 N shn l

C:tw I. Wh en the thin plate is in the middle of the tw o plane surfaces 1Re fer to Fig. 1.7 (n)1 FI '" Shear force on the upper s id e o f the th in plate ""2 = S h~ar forc~ On th~ low~ r s id~ o rth e th in pl~te "" = Total force required to drag Ih~ plate Then The , hear

I I

S1r~ss

1.2 em 2.4 em

.,

1.2 em

Fig . 1.7 (a)

F",F1+F1 (t l ) on th e upper side o f the thin plate is give n by equation,

Ii

~ I

IL 116

Fluid Me
tl '"

~C~:~),

where 1111 '" Relative ve locit y between thin p late and upper large plane surface

'" 0.6 mlsec d y '" Distance between thin plale and upper large plane surface '" [.2 em '" 0.012 III (p late is a thin one and he nce thickness of plate is neglected)

11=8.IOx 10 x(.012 0·')=40.5Nllll1 I

Now shear force.

FI '" Shear stress x Are a = t l

xll=40.5xO.5= 20.25 N

Similarly sllear st ress (t 2) on the lower si de of the 1hin plate is given by 12 =1.1

(~) dy !

= 8.IOx 10 I x

(~) 0.012

=40.5 N/ml

Sliear force. Fl '" t 2 )( A '" 40.5 x 0.5 '" 20.25 N Total force. F = FI + F1 = 20.25 + 20.25 = 40.5 N. Ans . CaSt' II. When the thin plalc is at 11 distanceofQ.8cm from onco r the plane surfaces [Refer to Fig. 1.7 (b)]. Lei Ihe lltil1 plate is al a distance 0.8 em from Ih e low er pia""

Ulem

surface. Th<"n

Thc

I

disl~nec

sh~ar

of thc plalc from lh e upper

2.4 em

pl~nc sun~c<"

I

:2.4 -0.8: 1.6"m=.016m (Negltt ting thkkllcss of the plate) forc~ 011 th~ upper side of lhe thin plate, FI = Shear stress x Area: I I X II

[(Iy"")

"" The shear force

011

xll:S.IOxIO I

.,

0.8 em

•

Fig. 1.7 (b)

- -) xO.5:15. ISN ,x (0.' 0.016

the lowe r s ide of the thill plme.

:IUOx 10

I X (

0.6

0.81100

) xO.5= 30.36 N

Tot al fo rce required = PI + P z : 15. 18 + 30.36 '" 45,54 N. AilS. Problem 1 .19 II "p nil'al gal' 2.2 rm ",id~ of infinile "X/rill ""lIIaills a fluid of \·is.'osiIY 2.0 N JIm! aud sp('djir gr
I I

" 2.2 em. viscosity, Jl" 2.0 N s1m 2

" 0.9

Ii

~ I

IL Prope rties of Fluids

,

Weight density of fluid

= 0.9 x 1000 '" 900 kgflm J

'"

I

900 x 9.81 N/m J

(.: I kgf=9.81 N) Volume of plale

= 1.2

III

x 1.2 I1IxO.2 em

'" 1.2 x 1.2 x .(X)2 Ill) = .00288 Ill)

Thickness o f plate Veloc it y of p late

171

"

= 0.2 em = 0.15 mf.'\Cc = 40 N.

'"

"" ,. 0.2cm

Weight of pla te When plale is in the m iddle of the gap. the distance o f tlie plate

from vertical surface of the gap

Fig. 1.8

= ( Width of gap- T; iCkn ess of Plate) =

(2.2 - 0.2)

2

=lcm= _Olm.

Now the shear force On the left side of tlie nw tallic plale.

FI '" Shear stress x Area

=" (:!:!.], (f )'

."J (O .01

xArca=2.0x - - xl.2xL.2N (.; Area = 1.2x 1.2m 2)

= 43.2 N. Similarly, Ihe shear force on Ihe Tight side of the metallic plale,

(0."]

F, = Shear stress x Area = 2.0 x - - x 1.2 x 1.2 = 43.2 N

•

.01

Total sheaf force = FI + F z = 43.2 + 43.2 = 86.4 N. In this case lh~ weight o f plate (wh ich is acting verti cally dOwnward) and upward thrust is also to be taken into account. The upward thrust = Weight o f nuid displaced = (Weigh t density of nuid) x Volume of nuid displaced =9.81 x900x .00288 N ( .: Volume of nuid displaced = Volume of plat~ = .(0288) The net

forc~

= 25.43 N. acting in the downward di rection duc to weight of the plate and upward thrust

= Weight ofplatc - Upw ard thrust = 40 - 25.43 = 14.57 N Tot al force required to lift the plate up = Total s hear force + 14.57 = 86.4 + 14.57'" 100.97 N. Ans.

... 1.4

THERMODYNAMIC PROPERTIES

Fluids consist of liquids or gas.:s. Bul gas.:s are compressible fluids and hence thermody namic prope r1i es play an important role. With the cha ng e of pressure and I~mpcrature. the gases undergo

I I

Ii

~ I

IL 118

Fluid Me
large var iati on in de nsit y. T he rel ati onship be twee n press ure (absolut e), s pecific vo lum e and temperatu re (absolute) of a g as is give n by the equatio n of state as p,;/= RT or'£' = RT

.. .( 1.5)

P

wh ere p = Ab so lu te pressure o f a gas in Nlm ' '
I

p R = Gas L'Olls tam T= Abso lute temperatu re in oK p = Densi ty of a g as.

1.4. 1 Dimension of R. The gas constant. R. depe nds upon th e pan icul a r gas. The d imension of R is obtained fro m c qu~ti o n ( 1.5) as

R= ~

pT

( 0 In MKS units

R=

kgf/ m 2 ii5C'--

kgf -m

kgoK

(~} K

(ii) In 5 1 un its, p is ex pressed in Ncw ton/m ' or N/m!,

R = N/ m' kg ~X K

= For air.

Nm louie -- = -kg -K kg-K

lJ oule = Nm I

J kg- K

R in MKS = 29.] kg f m kg c K

R in SI = 29.3 x 9.81

~ = 287 kg¢K

J

kg- K

1.4.2 botherm~1 Procen. If Ihe c hange in de nsity occu rs a1 co nstanl temp erature. the n the process is called iMlthe nnal and relatio nship be twee n pre.'i.~ ure (p ) and densit y ( p) is give n by J.!... == Constant p

...( 1.6)

1.4 . 3 Adiabatic Process. If the c hange in d~ n s ity occ urs with no heat exc han ge to and fro m the gas. the process is c all ed adiabatic. And if no h<":at is ge nerated within th e gas due to fri ct ion. the relatio nship be twttn press ure and densi ty is give n by

...!!.... == Constant p'

... (1 .7)

where k == Rati o o f specific hea t of a g as at co nstant pressure and constan t volum e. == 1.4 for air.

I I

Ii

~ I

IL Prope rties of Fluids 1.4.4

191

Universal Cias Constant Le,

1/1"

Mass of a gas in kg

V '" Volume of gas uf mass

1/1

J! == Absolute pressure T= Absolute temperatu re

Then. we have /IV == mRT .. .( 1.8) where R == Gas eonM:mL Equation (1.8) can be made universal. i.I' .. applicable 10 all gases if it is expressed in moll'·bas is. If

== Number of moles in volul11e of a ga s

'T/

== Vo lume o f th e gas

M ==

"c'c"o'coof,;,,'ho'C""c'co"'oOol'«"'olc'c'

M ass of a hydrogc ll atom 1/1" Mas;; of a gas in kg Then. we have II x M == III. Substituting the val ue o f 1/1 in equation ( 1.8), we get

...( \.'))

1''V=nxMxRT

The produl'! M x R is called universal gas <:OIlStanl and is equal to 848

-=-

_.k" ',"' ' '" cc in MKS units

kg -mo le o K

and 8314 l/kg -rnolc K in 51 units. One kilogram mole is defined as the product of One kilogrlllll mass oflhe gas and its molecular weight.

Problem 1.20

A gns wl'iglis /6 NI",J lU 25°C

Wid fII WI

IIbsolu/f prl'SslIrt' of 0.25 NIl/III':. Dnfr-

111;111' Ille glls ("OIIS/(/1/I (l1II! dflls;ry of IIII' gas.

Solution. Given: '" '" 16 N/m~ I " 250C T ", 273 + I"" 273 + 25 '" 2RROK

Weight den s ity. Temperature.

P" 0.25 Nlmm~ (a bs.) '" 0.25 x 106 N/m 1 '" 25 x 104 Nlm 1 (i) Using relation", "" pg. density is obtained as

16

w

P = - '" - - '" g 9.8 1 (ii) Using equation ( 1.5).

J.(i .~

,

kg/m· . "ns.

.!!.. = NT P p N = -p-T =

Problem 1.21

A ("y/ind", of 0.6

25xlO"

CI.6~3~XC2~8"'8

" S.l~.SS

Nm -. kg"

AilS.

mJ ill "01111111' ("01l/a;1IS air a/50°C allil 0 J Nlmm! lIbsollllr

"rI'HI"I'. '1111' lIir is ("omprnsed 10 0.3 IIr'. Find (i) prl'.,wrl' in~· idl' till' ("yli"dl'r ".'.IlImi"8 i.IlII/lI'rlt!(1I pro'·I'.'.' (11"/ ( ii) prnsllrp ",,,lll'Inpnfllllrf' ('.,3,,,,,in8 ",li"ba/i' ·I,,"'·n.<. T"k" k = 104.

Solution. Give n Initial

I I

vo lum~.

Ii

~ I

IL 120

Fluid Me
Temperatu re

I,:SO°C 1', '" 213 + 50 '" 323°K

Pressure

P, == 0.3 N/rnm 2 '" 0.3 x 106 Nlm 2 '" 30 x 104 N/m 2 'If 2 == 0.3 m}

Final voluille

k= 1.4 (i) Isolhe r lllJj[ Ilroccss :

~ '" Constant or p'V '" Constant.

Using equation (1.6),

p

1','<1, "'1'/
0.3

'\f 2

(ii) Adiabatic

Ilroccs_~

: -I'. '" Constant or I' V• '" Constant p

Using equation ( 1.7),

I' I '\f~ "" 1'1'\f~.

f'2:1":~ =30XI04x(~~r' :30x104x21.4 :0.791 x lOb N/m l ", 0.791 N/ mm !. Ans.

For temperature. using equ.uiull (1.5), we gel plt;j '" 81' and also f' Vi '" Constant NT

NT

P '" -;;;- and -;;- x

RTv.

I '"

'
Constant

r '\fl ~ 1 '" Constant T 'It i _ I _ T '\tk- I I

I

(.

R is also tonstanll

1 T2=T,(;:r-

2

2

I: '" 426.2 - 273 '" IS.l.rc. AilS.

Problem 1 .22 (',,!cu/wi' tile pr('ssllfI' (,Xf'rI~d by 5 I.:g ofllilrogell gas 11/ (l1l'IIIPl'''III''l' of 100e if Ihe \'0111111(, i5 0.4 III "'. Molnll/(lr "·e iglll of uilrogl'u is 28. A ssuml'. irinll 8"5 III\\'5 (lrl' "I'P/ iC(lblt'. Solu t ion. Given: Mass of nitrogen Temperature_

'" Skg 1= lQoe

T=273+ 10=2S3°K Volume of ni trogen. Molecular weighl

'V = 0.4 m l

= 28

Using equat ion (1.9). we have p'V '"

I I

tJ

X M x RT

Ii

~ I

IL Prope rties of Fluids

where M x R" Universal gas cOnstan'" 83 14

21 1

Nm

,----''-'''-c"C"

kg-mole 0 K and one kg-mole = (kg -ma·,s) x Molecul ar weight" (kg -mass) x 28

, 8314 Nm R for nnrogcn = - - '" 296.9 - - 28 kg " K The gas laws for nitrogen isp'V = mRT, whe re R = Characteristic gas conSIant

or

pxO.4 =5x296.9x283

5 x 296.9 x 283 (1=

... I .S

0.4

" 1050283.7 N/m 2

"

LOS N/mml. Ans.

COMPRESSIBILITY AND BULK MODULUS

Compressibility is Ihe reciprocal of the bulk modulus uf dasticily. K which is defined as Ih", ratio of compressive stress

v

•

-- d'V' J-o PISTON

10 \'olurne\ric strain.

Consideracylindcr fi lled wi th a piston as shown in Fig. 1.9. LeI 'V" Volume of a g as cnclo.\.Cd in the cy l inder

,

(J '" Pressure of gas when vo lume is V

LeI the pressure is increased to f! + d/!. the volu me o f gas decreases frolll V to V - lil:l. Then increase in pressure = dp kgflm 2 CkcreaSt! in volume = dV

CYLINDER

Fig. 1.9

dV

Vulumetric strain

- ve sign mea ns th e volume decreases with increase of pressu re. Bulk modulus

Increase uf pn:ssure K ~ ~~~';"""C" Vol umetri c !)Irain _ dp _ - dp V - - dV-dV

... ( I . I 0)

V

I

Compressibility

K

... ( 1.1 1)

Relationship between Bulk Modu lus (K) and Pressure (p) for a Gas The relationship betwee n bulk modulus of elastici ty (K) and pressu re for a gas for two different processes of compression arc as: (,) For Iso thermal Proces!l. Equation (1.6) gives the relationship between pressure (1') and density (M of a gas as

J!.. p

I I

= Constant

Ii

~ I

IL 122

Fluid Me
Diffcrc",i~ti n g

this equation. we gel (I' and '1/ both arc pdV + 'ltdp '" 0

or

vari~blcs)

l'd'v '" - \:tlfp

Substituting this value in equation ( LlO). we get ...( 1.12)

K= p (ii) FUT Adillha til.: I'rcKt'SS. Usi ng equat ion (1.7) for adiabatic process

-li'" Constant or f' p Differentiating, we gCl l'd(Vl) + 'cil(d!,)

or

pxkxV·- I,/'
or

ph/V + ViiI' '" 0

'I;Il '" Constant

=0 ICanceliing \;It-! to both s id es ]

1'/;(1 '1", - ';ltlp

or

'It dl'

I'k=-

dV

Hence from equation (1.10), we have K= pk where K

= Bulk

...( 1.13)

modulus and k = Ratio of spc<:ific heats.

Problem 1 .23 Dnl' nnilll' IIIf' bllik moliulus of elasticity of a Iiquitl. if IIII' (",,!SUrf of lilt' liquid is inf'rl'ased from 70 Nkm 2 /0 130 Nkn/. nit' "O/lIIlIe of ,II" liqllid durNlus by O. /5 I'er a/ll. Solu llon. Given:

Initial pressure

= 70 Nfcm!

Pina l press ure

" 130 Nlc m 2

,/1' '" In crcasc in prcssure " 130 - 70 '" 60 Nlcm 2 Decreasc in vo lum e "O.IS'l>

,/ "i

O. IS = + -V 100 Bulk modulus. K is given by cqu31io n ( 1.1 0) as

K:~,,60N/cm2 d "i V

.IS 100

,,60x l00 .15

"' 4xIOJN/cm2. ATl.~.

Problem 1.24 IV/Ill I is I/U' bll//I; 1I/0(/lIlu5 of "/lI5/icitv of II /iqllid ",/lid, is ('olllprnsed ill (I (' }'/il/d,'r frolll (/ \'olwIU' of 0.0125 111 3 (1/ 80 Nklll! prnSllrf 10 (/ \'011,1111' of 0.0124 1/1" (1/ 150 Nklll! I'rnSllrl' ? Solution. Given:

Initial \·o lum e. Final vol ume

,,0.0 124 m)

Decrcasc in vo lum e. ,/"i" .0 125 _ .0124 '" .0001 10 3

I I

Ii

~ I

IL Prope rties of Fluids d'::l ';J

Initial pressure

=

23 1

nOOl

.0125

= 80 Nlcrn l

Final pressu re = 150 Nlcrn' Increase in pressure. dp = (150 - 80) = 70 N/cm 2 Bulk modulus is given by equation (1.10) as

lif! 70 2 K= ---;t;j= .0001 = 70 x 125 Nlcm

.0 125 = 8.75 x 10- N/cm '' . Ans . ';J

... 1.6

,

SURFACE TENSION AND CAPILLARITY

Surface tension is defmed as the te nsile force acting on the su rface of a liquid in contact with a gas or on the surfal:c between twO immis.:iblc liquids such that the contact surfa"c behaves like a membrane unde r tension. The I1wgnitudc of this force per uni t icngth of Ih.: free surface will have the same val ue as the surface energy per unil area. It is denoted by Greek Idlef 0 (calkd sigma). In M KS units. it is cxprcs~d as kgflm while in SI uniLs as N/m. The phenomenon of surface tension is explained by Fig. 1.10. CO rl sider three mol~cules A, ll. C of a liquid in a mass of liquid. The molec u le A is allmcled in all directions ~'{jually by the surrounding molecules of Ihe liquid. Th us the resultant force actirlg on the molecule A is zero. But the molecule IJ. which is situat~d ncar the free surface. is acted upon by upward and downward forces which arc unbalanced. Thus a net result.1r1t fo rce on molec u le II is acting in the downward direction. The molecule C. situated on the free su rfa.:e of liquid. does experience a resultalll downward force. Fig. 1.10 S"r/au unsion. All the molecules on the free surface experience a downward force. Thus the free surfa.:e of the liquid acts like a very thin film under tension of the surface of the liquid act as though it is an elastic membrane under tension.

-- - -- - ---------~~~~~~~~~~~~~~~-

1.6. 1 Surfilce Tension on Liquid Droplet. Consider 11 small spherical droplet of a liquid of radi us .,. On the enlire surface oflhe droplel, the tensi le force due 10 surface tension will be acting. Let 17= Surface tension of the liquid p " Pressure intensity inside the drople t (in excess o f the ou tside pressure intensi ty) (/" Dia. of droplet. Let the droplet is cut into two halves. The forces acting on one half (~y left half) will be (J) (ensile force due to surface tension acting around the ci rcumference of th~ cut portion as shown in Fig. 1.11 (b) and this is equal to '" cr x Circumference =crxTld

I I

Ii

~ I

IL 124

Fluid Me
(ii) pressure force on Ihe area

~ll ! = l'

x

~ ti!

as shown ir>

4 4 Fig. Lli (e). Thc~ two forces will be equal and opposite under equilibrium co nditions. I.P ..

(9) DROPLET

(b) SURFACE TENSION

f'X ~ ,r=aXT(,! 4

(J

x

40

Ttll

...(1.14)

"

(e) PRESSURE FORCES

Fig. 1.11

Foret's on dropln.

1.6 .2 Surface Te nsion on
"x.::. rr=2x(crxrrd) 4

2 and

Scr

.::. d!

d

p= - - - = -

... ( 1. 15)

4 1.6.3 Surface Tension on
= p x :Irea of semi jcl ==pxLxd Force due 10 surface tension == a x 2L. Equating the forces. we have Force due

10

pressure

pxLxd==ax2L a x2 L f' == Lx"

, •

... (1 . 16)

"tf

and lenglh 'L' as

•

•

• • • • • •• • • • • • •• ••

•• •• • ••• •• •• •• •

•

I" Fig. 1.12 Forcts 011 liquid jn Pro blem 1 .25 1111' sHrfll("(' IntsiOIl of ""iliff ill ("011/(1("1 will! lIir (1/ 20°C is O . 0725 NIl/!. Till' prl'Ssllr<' ilrshlr II drop/("/ of ""fila is /Q h" 0.02 Nkm 1 grNiI"r (//(III IIII' ouuid" pre Hllr". Ca/ru/(I/(" IIII' I/imll("/(" r of IIII' l lw/,In of "·Il/ff. Solut ion. Given: Su rface tensioll.

(J

I"

= 0.0725 Nlm

Pressure intensity. I' in exc ess of outside pressure is ,

• N

p = 0.02 Nlcm '" 0.02 x 10 - , m-

1/ = dia. of the droplet

I I

Ii

~ I

IL Prope rties of Fluids

2s 1

Using equation (1.1 4), we gctp" 4cr or 0.02 X 104 ", 4 )(0.0725 If d 4 x 0.0725 d", " ,00145 III " ,00145 x 1000" 1.45 nUll. Ans. 0.02 X(lOt Problem 1.26

Fim/lhr sllrfu("/' II'IIS;OIl ill a soap bllubl" of 401111/1 dilllll<'la wll"l1 lilt' imide

pnssurl" is 2.5 NIIII" abm'(' (!III/osp/wrie prnsllre.

Solution. Gil'en :

(/=40111111=40 x IO- l m Dia. of bubble. Pressure in excess of outside./,,, 2.5 Nlm 2 ['or

3

soap bubble. using equation (1. 15), we gel

80 d

p=

8xo

2.5= ~~" 40xlO 1

25)( 40 x 10- ) 8 Nlrn '" 0 .0125 N/m . AJL~.

Problem 1.27

TlIi'

I'rnSllrf

VUlilid" IIii' dropli'/ O[wlII'" of diamfiPr 0.04 "'m is 10.32 Nlrl// ((1/. - i.! girt'1! (IS

mospltt'rjr prns"'.-). C(II,u/(l/p /Ilf p'l'SSlI re witilill II", tlrop/l'I if

0.0725 Nlm "j ....lIIPr. Solution. Gil'i.:n : Dia. of droplet.

Pressure outside the droplet

0.04 mill '" .04 x 10- 1 111 ,,10.32 Nlcm l " 10.32)( 104 Nlm l.

(f",

Surface tension. (J " 0.0725 Nlm The pressure inside the droplet. in excess of outside pressure is given by equation (I 14) 4 x 0.0725 , : J ,,7250Nlm-= 7250 N = 0.725 Nkml ,I .04 x 10 10' cm ~ Pressure inside the droplet = p + I'ressure outside the droplet 4(1

P" -

= 0.725 + 10.32 = 11.045 N/em !. A il S. Capillarity. Capillarity is defined as a phenomenon of rise or fall of a liquid surface in a small!U~ relative to the adjacent gcnerallevcl of liquid when the tube is held vertically in the liquid. The rise of liquid surface is known as capi lla ry rise while the fall o f the liquid surface is known as capillary depression. [t is ex presscd in Te rms of (;m or mm of liquid. Its " .. 9 1" val ue depends upon the specific weight of the liquid. diameter of Ihe tube and surface tension o f til.: liquid. E)(IJress ion fur Cljllillary Risc. Consider a glass lube of small diameter 'd' opened at ooth ends and is inserted in a liquid . .~y water. The liquid wi ll rise in the tube aoove the level of the liquid. L<:t II" heig ht of the liquid in the lUbe. Under a state ofeq uilihrium. : LIQUID the weigh t of liquid of height II is balanced by the force at the surface of The liquid in the tube. But the force at (he surface of the liquid in Ihe F ig. 1.13 Capjflary rilc. tube is due to surface Tensioll . 1.6 .4

,

Let

(J

= Surface tension of liquid

9" Ang le of contact betwee n liquid and glass tu be. The we ight of liquid of heighT II in the tube" (Area of tube x III x p x g

I I

Ii

~ I

IL 126

Fluid Mechanics

=~,PxlJxpxg

.(1.1 7)

4

where p '" Density of liquid Vertil:al \:ump<)llcm of the

surf~cc

tensile fuTl'C

'" (0 x Circumference) x cos 0

=crx/tdxcosO

•..( 1. 1H)

For equi librium. equating (1.17) and (1.18), we get It

,Pxllxpxg=axlf.I/xcosO

4 crXltl{XCOSi}

II '" ";-"'''--'-'=-O Il

,

4 d - xpxg

4 crL'OsO

.. (1. 19)

pxgxd

The value of 0 betwee n water and clean g lass tube is approximately equa l to zero and hence cos equal to unity. Tlicn ri se of water is given by II =

e is

40

---''''--c

... ( 1.20) pxgX(' EXll rcss ion for f:a llili llry Fall . Iflhe g lass lube is dipped in mercury. the level of mercury in Ihe lube wi ll be lower than the general level of the olllsidc liquid as shown in Fig. 1.14. U:i II '" Height of depression in tube. Theil in equilibrium. twO forces arc a<:ling un the mercury inside the tuhc. First one is due to surf~ce lension ~cting in Ihe downward direction ,md is eqn~1 to (I x 1((/ x cos a. SCl"Ond force is due to hydrostatic fo rce acting upward and is equal 10 inlensity of pressure al a deplh '11' x Area '" P X

1(

, 1( , IF '" pg )( h )( - d" 1 .. P '" pgll) 4 4

-

Equal ing the IWO. we gel

•

O"X1(dxcos9=pghx II

Value of 9 for mercury and

= -'4cO':"eOc'~' pgd

gl 3 s~

lube is 128".

... ( 1.21)

"'

,"

¥ ,"\' \

MERCURY

Fig. 1.1 4

Problem 1 .26 C{I/m/III,. Ihl' copillllry riS/' ill (I glass /UbI' of 2.5111111 dimlll'lrr ",hell illllll,.rsed \'f'rlicoll y itl (II) wllil'rlmd (b) mff("w, '. Take surfllce lellsiollS 0'" 0.0725 Nlm for II"O/,.r IIl1d 0 =- 0.52 Nlm for ml'rmry ill ("011/(/("1 lI"illl lIir. Ti,e 51'''(" iji(" g "wiry for mnmry is gil"<'l1 1/5 13.6 mill IIl1g/e of ("011/1/("1 = 130". Solution. Given: Dia. of tu be. ,/ '" 2.5 mill '" 2.5)( 10- 3 111 Surface lension. 0" for water = 0.0725 N/Ill

fo r mercury Sp. gr. of me rcury

0"

~ I

= 0.52 N/m

=- 13.6

I~

~ I

IL Prope rties of Fluids

271

'" 13.6 x 1000 kglln ·1.

Density

«(I) Capillary ris(' ror water (9 '" O ~)

Using equa1ion (1.20), we gd II =

40pXgxd

'"

x 0.0725 '"'=--'''''''''''''-___ =, lOOOx9.8 1x 25xlO l 4

'" .0 118 m = 1.18 em. Ans. (b) For mereur}'

Angle of conlact between ulcrcury and glass lu be, 9 ", 130

Q

4a cos6 4 x0.52xcosI30" = ,,-;-~",,"''''Ci-C':-';'''';-;;0 pxgxd 13.6xlOOOx9.Rlx2.Sx lO = -.004 III = - 0.4 em. Ail S. The negalive sign ind icmcs the capill:lry dcpn:ssion.

Usmg equation ( I 2 1), we gel Ii '"

Problem 1.29 ill (i)

ill " glaH IlIbl' of 4 1111/1 ilialllf'/a, 1>'11('11 (ii) III"'CIO,)'. Till' /('IIIIN'wlI>rl' ofll,rlilfllill is 20 0 IIIU/Illt' I'll/III'S a/llw slIrf{/("(' (rllsioll Ofll"(l/fr I/m/ IIII'rCllry til 20"(" ill ('Oll/ac/ w;lh lIir (Iff 0.073575 NIIli (llIti 0.51 Nlm rnpl'Cliw'/y. Tltf augl" of rOIl{(j(,/ for 1\'(1/'" is :I'rO 11/1(/1/1(1/ for mnTury i~ 130°. Tllia' drllsity of ....me r m 20°C liS rqUIII to 998 kgll,,-'.

illlllll'r....d

CalC/,/(I/(' III,. Cflpill
e

lI'aln, (///(/

Solution. Given: Dia. of tube.

(1=4m m ",4x 10- 3 11\

The cap illary effect (i.e .• capillary rise or depressio n) is given by equation ( 1.20) as

where

11= 4(J oosa pxgxd (J == surfucc tension in N/m a"" anglc o f c()nt:lct. and p "" density

(i) ClI l)illary efTect for wattr

(J = 0.073575 Nlm. e '" 0 " p '" 998 ~g/m '\ 31 20°C

I,,,, 4 x 0.073575 x cos 0° ==7.5 1 x 10 998 x 9.8 1 x 4 x 10- 3

J

m ",7.51 mm. Ans.

(ii) atl)illary effect for mt'rcury (J '"

0.51 Nlm. e", 130° and

p '" sp. gr. x 1000 '" 13.6 x 1000 '" 13600 kg/m 2

The

n~gative

sign

indica t~s

4x05lxcos130o It '" C-;Oc;:'=;:';;;~C'=C"'" 13600x9.8lx4xIO the capi llary depression.

'" - 2.46 x 10-

3

m '" - 2.46 mm. AilS.

Problem 1 .30 T/u' rapilhlry riu in II", glass llIj,r i .• ,wi /() ".H·rN/ 0.2 mm "fwlUrr. DNrrm;'IP milli",,,,,, si~". gil'''n that sIIrface 1""Si,," for wain ill umtart with air'" 0.0725 Nlm.

il~

Solution. (j iven : Capi lI ary rise. Surface tension.

I I

I,,,, 0.2 mm '" 0.2 x 103 III (J '" 0.0725 Nlm

Ii

~ I

IL 128

Fluid Mechanics

" ,/

Let dia. of tube

The angle e for water " 0" Dt:nsily (1') for water " 1000 kg/Ill) Using equatiun (1.20). we get

40 ="=

I," p xg x ii

or 0.2 x 10-) "

:;;~4iXCOc·,0;.72C5 , ':-, lOOO x9.81xd

4"X,::°C·O~7~2C5c::",J ,,0.1 48111 == 14.S em. Ans. (/= -;:",c . IOOOx 9.8 1x.2 xl O

Th u ~

minimum diameter o f the lube shou ld be 14.8 em .

Problem 1.31

Filld U!/f II",

minimu",

S;:;I'

of glllH

IIlI' l:" l'illary riu iu 1/'" IIIlu' is W he rPslril'tt'd 10 wilh nir a~ O.U7J575 NIl/!.

Solution. Given: Capi Ilary risco

Surface tension.

if

II = 2.0 lllm = 2.0 x 10-) 111 a = 0 .073575 Nfm

Lei dia. of tube The angle

/UbI' I/UII "'/II hi' U$nllO IIINuur" II'mn Il'wl

2 111m, COl!sitll'y .lllr/ucr 11'11S;"" "iw{l/p r in ,'O/lUwl

=d

e for wa ter

au

==

The densi ty for wakr, p = 1000 kg/m l Using cqualioll ( 1.20), we get

40 or 2.0 x 10- 1 = ,,= .,-,,"'--;px g x d 4 x 0.073575

if= -;-~""':",'="-''c~

lOOOx 9.8 l x2xlO J

Thus minimum diameter of the

l u~

shou ld

~

4 x 0.073575

lOOO x 9.8 1 x d

= 0.015 III = 1.5 em. AilS.

1.5 ern.

Problem 1.32 All oil of l'is ("Osil), 5 l",i3l' is l.s<'(1 for IIIU,i",lioll Ul'lln'l'lI II slw/l wltl £1,,<'1'1'. Tilt' diwlI"I'" of III<' $1111/1 i:l" 0.5 m lIml il 'Ol/lles m 2()() '.p.m. ClI/cH/mp IIII' po,,"'" losl ill oil for (/ sll'el'e Ip1IgIII of I()() """. Tile IIIirk1l1'S.1 of oil jil'" is 1.0 "''''. Solution. Given: Vi~osi ty.

~

= 5 poise

Dia. of shan. Speed o f shan. S I ~e\'c lengt h.

/. =

I{)() rnm: I{)() X IO- J III : 0.1 III

Thick ness of oil film .

I : 1.0mrn: I X IO- J m

Tangc mi a l veloc ity of shan.

I I : 11:

Using the re lation.

I I

2.

= 0.5 N s/rnl 10 D=O.5 rn N = 200 r,p.m. =

t:

DN : 60

II x 05 x200

60

= 5.235

Ill/S

""

~ -

Ii,

Ii

~ I

IL Properties of Fluids

wltcrc .

29 1

ClilIngc of velocity = 11 - 0 = Il = 5.235 tnls dy'" Cllange of distance = I = 1 x 10-3 III

dll '"

05 5.235 '" """""i~ = x x

1 10

This is lhe

sh~ar Slre.
J

26175 N/m"

on the shaFt

Shear forct;! o n [he shaft. F = Shear stress x Area = 2617.5 x nO x I. = 2617.5 x It x 0.5 x 0.1 = 410.95 N

Area = nO xL)

D 0.5 = 4 10.95 x - = 102.74 Nm 2 ,

T= FOl"<:c x -

Torque on th e sitar!.

Power* lost'" T x ro Walts = T x

2rr N

60

w

, 2n x 200 = 10_.74 x 60 = 2 150 W == 2. IS kW. .. 1.7

(":

AilS •

VAPOUR PRESSURE AND CAVITATION

A cbange from Ih e liquid slale to lhe gaseous Slale is know n as vaporiza tio n. The vaporization (whkl! depends upon lhe prev ail ing pressure and temperature C{)udition) occurs because of continuous

escapinll of lhe molecules lh rough llie free liquid surfac". Consider a liquid (say waler) whic h is confined in a c losed vesse l. Lellhe temperature of liquid is 20°C and pre.'\S urc i~ atmospheric. This liquid will "aporise at JOO°c. When vaporization lakes place. Ihe molecules escapes from Ihe free surface of Ihe liquid. These vapour molecules ge t accumulated in Ihe space between th e free liquid s urface and top of the vessel. These accumula ted vapou rs exen a pressure on Ihe liquid su rface. This pressure is know n 3S vapo u r press u ~ of the liqu id or This is Ihe pressure al wh ich Ihe li quid is convened into vapours. Agai n \'onside r the same liqu id al 20°C at almosphe ric pressure in the c losed vessel. If the pressur" above th e liquid surface is reduced by so me means. the boi ling Temperature will also reduce. If Ihe pressure is reduced to such an exten t that it bccotnes equal to or less than the vapour pressure. Ihe boil in g of the liqu id wi ll stan. Though th e temperature of the liquid is 20°e. Thus a liquid may boil even at onlinary lcnlperature. if the pressure above the liquid surface is reduced ,' \0 as to h.: equal or less than the vapou r pressur~ of the liquid m that tempe rature. Now consid~r a flowing liquid in a syste m. If The pressure at any point in this flowing liqu id becomes equal to or less Ihan the vapour pressu re, th~ vapori za tion of Ih e liquid starts. The bubbles of these vapouT5 arc carr ied by the flowing liquid into th e region of high pressure where they collapse. giving risc to high impact pressure. The pressure developed by the collapsing bubbles is so high that the material from The adjoining boundaries geTS eroded and c aviTi es are formed on them. This phenomenon is kno wn as ( " ,·lIalioll. Hence the cav itat ion is the phenomenon of fornwtion of vapo ur bubbles of a flowing liquid in a region where Ihe pressure of the liquid falls be low the vapour press ure and sudde n co llapsing of Ihese vapour bubbles in ~ region o f higher pressure. When Ih e v;lpour bubbles collnpsc, n very high pressure is created. The metallic surfaces. above which the liquid is flowing. is subjeCTed to these high pressures. which c~usc pilling action on the surfnce. Thus cavi ti es arc fonned on th~ metallic surface ~nd hence the name is cav itati on. • Power in case of S.l. Unit

I I

~

,

2rtNr

2rr.fff

60

60.000

.

1 X (0 or - - Watts or - - - kW. The angular wloclty

(0 ~

2nN -60 .

Ii

~ I

IL 130

Fluid Mechanics

HIGHLIGHTS I . The weight density Or sp<.>eific weight of a fluid is equal [0 weight per unit volume. It is also cqualto,

w"'pxg. 2. Specific volume is the reciprocal of mass density. ,I, The shear stress is proponiOllal 10 the velocity gradient

""

"'

4 . Kinematic viscosity V is giwn by V ~ ~ .

P 5. Poise and sto~es are the unil .• of viscosity and kinematic viscosity respccti,-cly. 6. To convert the unil of viscosity from poise to MKS units. poise should be divided by 98.1 and 10 convert poise i!l1o SI unils. the poise should be dividcU by 10.51 unit of "iscosity i< Nslm l or Pa s, where Nlm l

.. Pa '" Pascal. 7. For a perfect gas. the eq ualion of state is

!!... = RT P

. kg f-Ill where R .. gas conSI:m! :md for aIr .. 29.3 - - - ., 287 Jlkg oK.

kgo K

8. For isolhcmlal process, 9.

Bul~

J!.. '" Constant whereas for adiabatic process,

modulus of elasticity is given as K

~

- lip

(":r

10. Compressibility is the reciprocal of bulk modulus of elasticity or" II . Surface tension is

.. constanl.

p

c~presscd

~.

in N/m or dyne/em. The relation between surface tension (0 ) and difference

of prcssurc (p) betwcen the inside and outside of a liquid drop is givcn as p _

80

For a S(lap hubble.

p" ,/

For a liquid jet.

P~d '

40

"

20

02. Capillary risc or fall of a liquid is given by /, m

40eos9 "'''''!-'

""

The value of a for water is taken equal to zero and for mercury equal to 128°.

EXERCISE (A) THEORETICAL PROBLEMS I. Define the following fluid propenie, :

Density. weight density. specific volume and specific gravity of a fluid. 2. Differentiate between: (i) Liquids and gases . (ii ) Real fluids and ideal fluids. (iii) Specific weight and specific volume of a fluid . •1. What is the difference between dynamic viscosity and kinematic viscosity ? State their units of mcasuremcnlS.

I I

Ii

~ I

IL Properties of Fluids

311

4 . Explain the len". : (i) Dynamic vi>cosily. and (ii) Kinema!ic viscosity. Give their dimensions. 5. State the Newton's law of viscosity and Kjve examples of its application. 6 . Enunciale Newton's law of ,·;""osi1y. Explain the importance of viscosity in n uid motion. W hat ;s the effect of temperature on viscosity of water and that of air? 7. Define Newl
,

-

while for a perfect gas undergoing isenlropi<" compression is - .

"V

p

Ill . Define surface tension . l>rovc thallhc reblionship between surface tension and pressure inside a droplet of

40

liquid in exee .. of outside pressure is given by p '" -

"

.

II . EXplain the phenomenon of capillarity. Obtain an expression for capillary rise of a liquid. 12. (1/) Distinguish betwcen ideal fluids and real fluids. Explain thc importance of compressibility in fluid flow. (b) Define the t'nTIS , density. specific volume, specific gravity. va.uum p""sure, mmp""sible and incompressible fluids. (R.G.P. Vishw(lI"i(/Y(l/a)"(I. IJlwp(l/ S 200Z) 13 . Deflne and explain Newton 's law of viscosity. 14. Conver! I kgls-m dynamic viscosity in poise. 15. Why does the viscosity of a gas increases with the increase in temperature while that of a liquid decreases with increase in temperature ? 16. (a ) ~Iow doe, viscosity of a fluid "ary with temperature " (b) Cite examples where surface tension effects playa prominent role. (J.N.TV.. Ifyilaahad S ZOOZ) 17 . (i) Develop the expression for the relation between gauge pressure P inside a droplet of liquid and the surface tension. (ii) Explai n the following: Newtonian and Non-Newtonian fluids . vapour pressure. and mmpressibility. (R.G.P. V.. Bhopal S ZOOI)

(B) NUMERICAL PROBLEMS 1. One litre of crude oil weigh' 9.6 N. Calcu late ilS specific weight. den,ity and .
'2 Y_111.

3. 4.

5..

6.

I I

"elocity in metre per second :n a dislan'"" y metre abwpamting them is havin,!! visrosity 1.5 poise. [An s. 300 N. 89.8 W I An oil film of thickness 1.5 nun is used for lubrication between a square plate of size 0.9 m x 0.9111 :md an incli ned plane having an angle of indination 20~. The weight of thc square is 3'n.4 Nand it .• lides down the plane with a unifonn velocity of 0.2 mi•. Find the dynamic viscosity of the oil . IAn .•. 12 .42 poisel

Ii

~ I

IL 132

Fluid Mechanics

7. In a stream of glycerine in 1I101ion. at a certain point the "ciocily gradient is 0.25 metre pcr sec pcr melre. The mas~ density of fluid is 1268.4 kg pcr cubic melre UTl d kinematic viscosity is 6.30 x 10 • square metre pcr St-><:ond, Calculate lhe shear StreSS al1he poin!. [An •• 0.2 NIm' 1 8 . Find the kinematic viscosity of an oil having density 980 kg/m' "hen al a certain point in Ihc oil. Ihe shear

stress is 0.25 Nfm' and velocity gradie1ll is 0.3/5.

[

Ans. O,IXlO849 rn' or 8.49

SlOkCS]

"'

'I. Determine the spttitic graYi1}, of a fluid h,wing viscosily 0,07 poise and kinematic viscosity 0.042 stokes. IAns. 1.667[ 10. DClcmlinc 1he viscosily of a liquid having kincmalic ";seosily 6 slokes and specific gravity 2.0. [Ans. 11.99 poise] II . If the velocity distribution of a fluid over a plate is gi,'en by u ~ (3/4) Y -;. where u is the velocity in melre per scwnd al a dislance of )' melres above the plale. delennine Ihe shear slress at y • 0,15 metre. Take [A"".. 3.825 X 10--1 kgJlm' j dynall1ic "is:cosity of the fluid as 8.5 X 10 3 kg_sec/m Z, 12 . An oil of viscosity 5 poise is used for lubrication belween a shaft and slecve. The diametcr of shaft is 0.5 m and it rotates at 200 r.p.m. Calculale the power losl in Ihe oil for a sk'"CYe lenglh of 100 nnn, The thickness of the oil fihn is 1.0 mm. jAn s.. 2, 15 kWI

,

U . The ,'elocity disuibulion over a plate is gi"en by u ~ ~ y - ; in Which u is the "elocily in mlsee at a distance of y m above the plaTe. [kterminc the shear stress at y., 0, 0.1 and 0.2 m. Take ).l = 6 poise. IAn s.. 0,4 .. 0.028 and 0, 159 N/ml j 14. In 'luestion 13.. find Ihe dislance in melres above Ihe plate .. al which Ihe shear stress is zero. IAns .. 0.333 ml 15 .. The ,'elocily profile of a viscous fluid m'er a plate is parabolic wilh venex 20 cm from Ihe plate. where Ihe velocity is 120 cm/s. Calculale the "clocity gradic111 and shear stress at distances of O. 5 and 15 cm from the plate. given the viscosity of the fluid = 6 poise. IArrs. I2Is. 7.18 N/m': 9/s. 5.385 N/m ' : 3/., 1,7<)5 Nfm' j 16. The weight of a gas is given as 17.658 N/m l at 30°C and at an absolule pre>sure of 29.43 Niem', \)elcr. J. 8k S 539.55N ml kgo K [ A ns' --;;;-r-'

mine Ihe gas constant and also Ihe density of the gas,

17 . A cylinder of 0.9 Ill} in volume conlains air at O°C and 39,24 NI~rn l absolule pressure. The air is cornpres.""d 10 0045 ml . Find (i) Ihe pressure inside the cylinder a~suming isothemwl process. (ii) pre,sure and lempem ture assumins mJiabalic proce.~ure exe" ed by 4 kg lna>< of nitrogen gas at a lemperature of IYC if Ihe volume is 0.35 ml . Molccular weight of nitrogen is 28. [Ans. 97.8 Nlem' j 19 . The pressure of a liquid is increa«ed from 60 Nlem' 10 100 Nlcm' and volume decreases hy 0.2 per cenL Detennine the bulk modulus ofelasticily. IAns. 2 X 10" Nleml l 20. DCle"nine Ihe bulk modulus of elaslicity of a fluid which is compressed in a cylinder from a volume of 0.009 ml at 70 Nlcm l pressure to a volume of 0.0085 m l at 270 Nlcm l pre .. ure . [An s. 3.6 X 10' N/cm l l 2 1. The surface lension of watcr in contact with air at 20"C is gh'cn as 0.0716 N/m. The pressure inside a droplet of waler is 10 be 0.0147 N/em l grealer than the outside pressure. calculate Ihe diameter of the
I I

-!::,) , em"

Calcu lale the pressure "'ithin the droplct of [An s. 11.77 Nlcm' l

Ii

~ I

IL 33 1

Prope rties of Fluids

24 . Calculate the capillary risc in:l glass tube of 3 .0 nlln diameter when hnll1cr>ed vcnically in (,,) waler. and (1)) mercury. Take surface len.ions for mercury and water as 0.0725 N/m and 0.52 Nirn respectively in contact with air. Spcdfic gravity for mercury is given as 13.6. IAn •. 0.966 em. 0.3275 eml

25. The ~apillary risc in the glass lube used for mc"suring waler level is not (0 exceed 0.5 min. Dctennin" ils minimum size. given thaI surface lension for water in conlae! wilh air,. 0,07112 Nlm. [Ans. 5.8 em] 26 . (SI LTnits). One lilre of crude oil weighs 9 .6 N. Calculmc its specifIc weight. density and specific gravity. [Ans. %00 Nlm'; 979.6 kg/ln ' : 0.971\61 27. (SI "nits). A piston 796 mill diameter and 200 nun IonS works in a cylinder of 800 mill diameter. If the annular space i. filled with a lubricating oil of viscosity 5 cp (c.·nti -poi,.,), cakulate the speed of de>cem of the piston in vertical po~jtion. The weight of the piston and 3 .• iaII03d are 9.1\1 N. [,\ns. 7.84 m/s[ 2!1, (SI lTnit~). Find the capillary risc of water in a wtlc 0.03 em diameter. The surface tension of water is 0.0735 Nlm. [A tls. 9.99 eml 2', Calculate the specific weight. density and specific gravity of two litres of a liquid which weight 15 N. [AilS, 7500 N/m l . 764 .5 kg/mI. 0.764 1 30. A I SO mm diameter vertical cylinder rotates concentrically inside another cylinder of diameter lSI mm. Both the cylinders are of 250 mm height. The space between the cyl inders is filled with ~ liquid of viscos· ity 10 poise. Detennine the torque required to rotate thc inner cylinder at 100 T.p.m. IAns. 13.S7 Nml .~ 1 . A Shaft of diameter 120 nlln is rotating inside a journal bearing of diameter 122 mm at a speed of 360 r.p.m. The ']XIce between the Shaft and the bearins is f,lIed with a lubricating oil of viscusity 6 poise. Find the [Ans. 115.73 WI power absorbed in oil if the length of bearing is 100 mm. .\2. A shaft of diameter 100 mm is rotating inside a journal bearing of diameter 102 mm at a space of J60 r,p ,m. The space between the shafl and bearing is filled with a lubricating oil of viscosity 5 poisc. The [Ans. 111.58 W[ length of the t1c:.ring is 200 mm. Find the power absortlcd in the lubricating oil. .\,\ . Assumi ng that the bu lk modulus of elasticity of water is 2.07 x 106 kN/m l at standard atmospheric conditions. detcnnine the increase of pressure necessary to produce I % reduction in volume at the same tcmperature. [l1int . K

~

• , 2.07 x 10 kNlm

Increase in pressure (dl')"

-dV

;~

0 -

1

100

.0.01.

Kx ( - ~V) ~ 2.07 X10" x 0.01 = 2.07 x 10' kNhn 1.]

.\.4. A square plate of size 1 m x 1m and weighing 350 N slides down an inelined plall e with a unifonn velocily of 1.5 mls. The inclined plane is laid on a slope of 5 venicalto 12 horizonlnl and has an oil film of 1 mm thickness. Caiculutc the d)l11amic viscosity of oil . [J.tV. T.V., lIy,lembm/. S 2002[

' 5 =Be [ Ihnl. A= I x I ~ 1m l . W_3SON.u", 1.5m/s. tan 0 .. 12 AB Component o f weight along the plane = W x sin (I 3C _ _'

where sin (I .,

AC

[, ..

13

I' = W sin (I .. 350 x Now

r.c:o-:, .. JI2~+52 . 13

1

A

"

12

W

~ '"' 134.615

8 E

350N

Fig. 1.15

13

,,"dy whcre du _ u_O = u = 1.5 mlsandd.v = 1 mm= I x 10 T=).I~. :. 1-1

I I

AC"'JAB1+BC~

c

' m

F dy 134.615 IxlO ' Ns = -x - = x '.";"i--= 0.0897 -:::T" 0.S97 poisei A

du

1.5

ttl

Ii

~ I

IL

I I

Ii

.. 2 . 1

fLUID PRESSURE AT A POINT

Consider a sma ll area liA in large mass of fluid. If tile fluid is stationary. tllen tile force exerted by the surrounding fluid on tile area IiA will always be P'Crpcndicular to the su rface (LA. Let ,IF is tile force
acting on the area dA in the normal direction. Tllen the ratio of -

dA

pressure or simply pressure and this ratio is reprcsented by p. Hence point in a fl uid at rest is p =

is known as the intensity of

math~lllatically th~

pressure at a

dA

If the forl"C (1-1 is uniformly distributed ove r the area (A). then pressure at any point is given by F

p

Force Area

= -A =""'''-

Force or pressure force. F = I' x A. The unils of pressure ar~: (i) kgffm" and kgffclll 2 in MK5 units, (i,) N~wtonfrl1 lOr Nfm l and Nfnl1n 2 in 51 units. Nfm 2 is known as Pascal and is represented by Pa. Other common ly used units of pressure arc : kilo pascal = 1000 Nfm ~ ' 100 kPa '" 10 Nflll-.

,

.. 2.2

PASCAL'S LAW

II states that the pressure or intensity of pressure at a point in a .'\Iatic fluid is equal in all directions. This is proved as : Tile flu id clement is of very small dimensions i.P .. l/X. tly andl/$. Conside r an arbitrary fluid clement of wedge shaP'C in a fluid mass al rest as show n in Fig. 2.1. Let the width of tile cle ment P'Crpendicular to the plane ofpaP'Cr is unity and p~.

P. · dy .,

>0,

•

, p) .L\)(.\

Fi g. 2. 1 ForcN on a flll id dement.

35

I I

Ii

~ I

IL 136

Fluid Ml>chanics

f\. and fI: arc the pressufCs or inlcilsity o f pressure acting on Ihe face AB. AC and Be respectively. Let LABe",

a. 1'lIcll

th e fo rces acting Oil the clc ln elll arc:

L Pressure forces nonnallo the surfaces. and 2. Weight of c lemen t in Ihe vcnical direction. The forces 011 the faces arc: Force on the face All

'" fI, x Area offacc AlJ

""p" xdyX I Simi larly force on Ihe face AC '" p,. x d.f X I Force on the face UC '" fI , x ds x 1 Weighl of clement '" (M ass of element) x g '" (Volume x p) xg '"

(AB; AC x I) x px g.

where p '" density of fluid. Resol ving Ihe forces in .r-direction. we lI ave p, xdyx I - p(dsx I )si n (90 Q - 9) "0 p" xdyx l - f!: d~' X 1 cosO '" O. 0' BUl from Fig. 2. 1. ds cos e '" All "" lIy p, xdyx l - p,- xdyx I "0

...(2.1 )

P."'Pz

or

Similarly. resolving the forces in y-dircction. we get p,Xdxx l-p, XtlJ'X I

cos(900 _e) _~dc·'~'cdCY'xl 2

XpXg=O

tI_(dy

P-, x dx - P, tis s in e - - , - x p x g = O. But ds sin e = d.l :md also the elemem is very small and hence weight is negligible. P/u--p , xdx=O or P. =P, From equ:uiOlls (2 I) alld (2.2). we have

..(2 _2)

... (2.3) p" = p" = P, The above equation s hows that the pressure aI any po int in x.)' and Z directions is equal. Since the choice of fluid element was co mple tely arbitrary. wh ich means the pressure aI any poim is the same in all direc tions. ~

2 .3

PRESSURE VARIATION IN A FLU ID AT REST

The pressure at :1I1y poin! in a fluid at rest is obtained by the Hydrostntic Law which slates thnt the rate of increase of pressure in a ,"e n ically dow ll ward direction must be ~-qu al to the SJXocifie weight of the fluid at Ih at point. This is prt)l'ed as: Consider a small fl uid cleme n! as shown in Fig. 2.2 Let M= Cross-sectional area of element ~Z'" Height of fluid c lement P '" rres~ure on face AS Z", Distance of fluid elemellt from free surface.

The forces acting on tile fluid clement are:

I I

ri::::-

{ REE SURFACE OF FLU ID

------------- -- ---- J ---- ---

:::::::::::::~ ' ~:(::::_ B .,--_-_-_-_-_-_-_-------------_"_-_-_-_-_-_ ·tJZ _-_-_-_ -_-_-_ -_-_-_-_-_- A -

------ --------:-:-:-:-:-:' ? -ilp-£t:-::-:-:-:-:-: ---------- ----------.{ p+

-_-_-_-.-_" -:k Fig. 2.2

tJZ)M.--------___-_-.-_-_-

Forces on a fluid e!emeT/l.

Ii

~ I

IL Pressure ilnd its Mea surement

37 1

L Pressure forc e on AB '" P x All. and acting perpendicu lar to face Ali in the downward dircClioli. 2. Pressure force on CD =(1' +

~~ llZ) x dA. actin g perpendicular 10 face CD. w rti cally upward

direction.

3 . Weight of fluid c lement = Density x g )( Vo lume = p)( g x (All. x AZ). 4. Pressure forces on surfaces IJC and AD arc equa l and opposite. For equilibriu m o f nuid c le men!. we have pM -

(1'+ ~~IlZ) M

I'M - I'M -

0'

~

+ pxgx (M

x~

=0

1l7..&1. +pxgxM xZ =0

-~~~+PXgX.MllZ=O ap = p x II [cancelling MAZ on both :~IlZ6A = p Xg xMIlZ or Jz

"I'

aZ

... (2 ...1 )

(':pXg=w)

=PXg=w

sid~sl

where w = Weigh! density of fluid.

Elj uati on (2.4) Slat es Ihal rate of im;rcasc of pressure in a vert ic al density of the fluid at that poin t. This is Hydrostatil: La w. By integrating tile above equation (2.4) for liquids. we ge t

dir~'(; tjon

is equa l to we ight

{dp={pgdZ

= pgZ

...(2.5) whe re II is the pressure iIOO\'e mmosphcrie pressure imd Z is the height uf the point frum free surfaces. p

From equation (2.5). we have

Z'" _ 1_'-

px,

..•(2 .6)

Here Z is called ]Irl'ssurl' hl'ad . Problem 2.1 A hydrmdie prt'ss lUIS /I W/II of 30 elll dilllnt'lrr mul a pillnger of 4.5 e/ll dillllll"lrr. Find Ihe !>'t'ighl fijlt'd by lilt' h)'llftllllie prt'JJ ",ht'll Iht' foret' applinl (1/ lilt' plllllger is 500 N. Solulion. Give n : Dia. of ram. /) '" 30 em = 0.3m Dia. of plunger. d", 4.5 em = 0.045 m

Force on plunger. Find weight lifted

"'",5OON

"W

Area of ram. Are a of plunger.

I I

Ii

~ I

IL 138

Fluid Ml>chanics

Pressure il1! cnsity due to plun ge r '"

Force o n pl un ger A rea of plun ger

F 500 2 = - = - - - N/m .0 0 159

(l

PLU NGER

Due to Pasc al's law . the inte ns it y o f pressu re will be eq ua ll y trans mi u ed in a ll di rcl:1 io ns. He nce the pressure inte ns it y at the ram 500 ::: 3 14465.4 Nfrn 1 .00 159

: -:-_W""";'~'hC'_ '" ~ '"

But pressure intensit y a! ram

Area of ram

w

.07068

A

• Fig. 2.3

_'_V_ N/m 1 .07068

= 3 14465.4

We ight = 3 [4465.4 x.07068 = 22222 N '" 22.222 kt'l. Ans . Problem 2.2 A Il)'liralllir pre SJ bas (/ ralll of 20 011 ,Iimll" fPr (11111 (I pi IIl1ga of 3 Til! (liw/lFler. It is usn i for liftillg (/ wi'iglll of 30 liN. Pilld lilt' fOITe rt' qllirn/ mille p/1I1I8e r.

Solution. Give n : /) '" 20 em = 0.2

Di a. of ram.

Area of ram. Dia. of plunger Area of plun ge r. Weig lll lifted.

A '"

III

~ /)2 '" ~ ( . 2) 2 '" 0.03 14 111 2

4 4 ll=3cm =O.03 m

"'" ~(.03) 2 = 7.068 X 10- 4

rn !

4 W =30 kN = 30x IOOON =30000 N.

See Fig . 2.3. Fo rce F Pressure int ensi ty develo ped due 10 p lun ge r ", - - == - . Area " By Pasca l" s La w. th is pressure is tralls mi ltcd equa ll y in all d irect io ns Hence press ure tran sm itted at th e ram == Fo rce actin g o n ram

!...

" inte nsity X Area o f ram = Pressure F

= - x A ", But force al:tin g o n ram

"

F x .03 14 N 7.l168 x 10-4

= Weig ht li fK"d '" 30000 N

30000 :

Fx .03 14 7.068 x 10-4

F: 30000 x 7.068 X 10-4

.03 14

'" 675.2 N. Ans .

Problem 2 .3 Ca/n'/(/((' f/If prnsllr(, (/II(' to a CO/UIIIII of0.3 of(n ) w(/(a. (b) (III oil ofSI'. (c) Ilwrcur )' of sp. gr. 13.6. Til/.;(' i/('Il$ify ofw(lfa. p '" /000 kg/IIIJ. Solution. Give n : Height o f liq uid co lumn. Z=O.3 m.

I I

,~ r.

0.8. (II ld

Ii

~ I

IL Pressure ilnd its Measurement

39 1

The press ure al an y point in a liquid is given by equation (2.5) as p'" pgZ «(I) For

wal~r.

p

= 1000 kg/Ill)

I' = pgZ '" 1000 x 9.&1 x 0.3 '" 294) Nlm 2 -_ ['(j'T 294] Ntem !-_ O"94\NI ! A liS. . ._ . {fll. (b) For oil of sp. gr. 0.8.

From eq uation ( 1.IA). we know lhat Ihe densit y of a fluid is eq ual to specific grav ity of Iluid multiplied hy density of water.

Density of oil. Now pressu re,

Po = Sp. gr. of oi l x Dens it y of wakr = 0.8 x P '" 0.8 x HXXl '" 800 I;g/m3

(Po= [k nsi ly of oil)

1'=PoxgxZ = 800 x 9.8 1 x 0.3 = 2354.4

~

= 23s:.4

10

Ill '

= 0.2.\54

~. em'

~. AilS.

,m

(c) For mercury. sp. gr.

'" 13 .6 Fro m equa tion (l. IA) we know that tile den s ity of

J

fluid is eq ua l to specific g ravi ty of fluid

multiplied by densit y o f water

Densi1y of mercury.

p. = Spcl:ific grav it y of mercury x Density of water = 13.6 x 1000 = 13600 kg/m l P=P.xgxZ

'" 13600 x 9.81 x 0. ] " 40025 ~

40025 - - , - " 4.002

10

N

--!'

em

N

~ , 111 -

Ans.

Problem 2.4 Till'" prnSlln illlnlsily (1/ (I poim ill (lfluid is gil"nl 3.924 Nlnl/. Find Iltf ("orrnpolll/illg Iti'ighl offluid ...hi'llilti' fluid i~ ; ( II) W(I(Fr. lind (b) oil OfS!" gr. 0.9. Solullon. Given: N , N Pressure imensity. 1''' ].924 - , "].924 x 10 ~,. cm -

m-

The cU lTcsponding heigh1. Z. of 1hc fluid is givcn by eq ua1i u ll (2 .6) as

z== - " -

I'X ,

(u ) For watcr.

p " 1000 kghn·l

z== ~"

I'X,

(b) For uil. s p. gr.

:. Density of oil

lOOOx 9 .81

- 4 III uf wa ter. An s.

" 0.9 l Po" 0.9 x 1000 " 900 kg/m·

Z

I I

3.924 x 104

__ _ ,_, _ __ 3 o~ 92~4~X~'~O~' -:; == 4.44

Po xg

900 x9.81

11\

uf uil. Ans.

Ii

~ I

IL 140

Fluid Ml>chanics

Problem 2.5

All oil of sp. gr. 0.9 is

("OIlWillt'd

ill

(I

"<'asd. AI


poim 11Il' heig/II of oil is 40

tIl.

Find

Ihe corrt'Sl'olUlillg ileighl ofwO/n at lil" porlll. Solution. Given: Sp. gr. of oil,

So = 0.9

Hciglll of oil.

Zo=40m Po = Sp. gL of oil x Dt:nsily of walcr = 0.9 )( 1000 = 900 kglm 3 N p = Po x g x 7-0 = 900 x 9.81 x 40 - ,

Density of oi l. lntcnsi ly of press ure.

'"

Corresponding liclght of water'" -;::c~_cIC' _ __ De nsityo f water xg

~

9OO)(9.8 I x40 "",:C""',:," = 0.9 x 40 = .\6 10 of w ~tc r. Ans . 1000)(9.8 1

Problem 2.5 All 01'1'11 {(Ink romulus W(l/er IIpto (I df/NII of 2 III (md (lboI'<' il (III oil of SI'. gr. 0.9 for d"plil of J m. Filut 1/,1' 1'("('8811'1' ill/fluiry ( jim /11i' inlf rfowl' "llIw ""0 liquills. llIui (ii) (If /Iw hOI/om

(I

of 'h" "IIlk. Solution. Given: Hciglll of water. Height of oil. Sp. gr. of oi l. Density of water. Density of oi l.

Pressure intensity at

~ny

ZI",2m Z2'" 1m So'" 0.9 PI '" lOOO kg/m ' Pl'" Sp. gr. of oil x Dens ity of water '" 0.9 x 1000== 900 kg/m' poim is given by Fig. 2.4

p =pxgxz' (i) At interface. i.l' .. at A

P"'P2 x g X1 .O ",900x9.81 x 1.0

N 8829 • '" 8829 - , ~ - ,- = 0.8829 Nfcm -. Ans. III lO (ii) At Ihe bottom. i." .• at B

l' '" P2 X gZ2 + PI X 8 X ZI = 900 x 9.81 )( 1.0 + [000 x 9.81 x 2.0 1

28449 ,

•

= 8829 + 19620 = 28449 N/1I1 = ~ Nkm' = 2.8449 Nfcm -. Ans. Problem 2.7

TI", dilllllt'/prs of (I sm(ll/ pi.
Solution. Given: Dia. of small pislOn. Area of small

I I

pi~IOn,

(1=3cm 1'1',2

II'" -

4

(,

11

,

= - x (3)' = 7.068 e lll 4

1

Ii

~ I

IL 41 1

Pressure ilnd its Mea surement Dia. of large piston.

IJ=

A", P )( (1 0)2 '" 7&.54 cm 1 4 Pc: 80 N

Are a o f larger piston. Force on small piston.

,

= W. When thl' Ilistons arc 31 thl' same It'HI

Ld the load

(a)

10 em

Prcs~urc

lin~d

~W~~~;i'G' P ISTON ~.-.-.-.-.-.-.

in[cllsity on small piston

F

80

II

7.068

------. .-----__ . _-. -----_ -. -. -. -. -. -.-

This is Iransmincd equ all y on thc large piston. Pr".'\Sure intensity on Ih" large piston

Force on the large piston

s~,

PISTON

f i g. 2.5

= -7.068 '" Pressure )( Area

80 = - - )( 78.54 N '" 8tlX.'J6 N. AilS. 7.068 (b) Whl'n the small Ilision is 40 em abo ve th(' tar~(' I)iston

,

Pressure intens ity on Ihc small piston F

80

(I

7J168

I

Pressurc ill1cnsity at sectio n A-A F '" - + Pr~ssurc i ntensity due [0 height o f 40 ern of liquid.

"

But pressure intensity due to 40 cnl of liquid =pxgxh= IOClOx 9.8 1 xOANlm 2 1000 x9.8 1 x.40 Nlt:m! = 0.1924 Nlcm!

10'

Pressure

int~nsity

w

------" --_._--------------------------.-----------------------

at sect ion A -A

Fig. 2.6

80 = - - + 0.3924 7068 = 1l.J2 + 0.3924 == 11.71 Nlcm 1 Pressure intensity transmitted to the large pislOn = 11.71 Nkm 2 Force on the large pislOn = Pressure x Area of the large piston =11.71 x A == 11.71 x 78.54 = 9 19.7 N . .. 2 .4

ABSOLUTE, GAUGE , ATMOSPHERIC AND VACUUM PRESSURES

The pressure on a fluid is measured in twu different syste ms. In one system. it is measured above the absolute zero or complete vacuum and it is called the absulute pressure and in other sys tem. pressure is measured above the atmospheric pressure :md it is called gauge pressure. Thus: I. AI.soluh' 1H"t'sS UH' is defined as the pressure which is measured with refe rence to absolute va;:uu1t1 pressure. 2 . C,lIU/tC l.rCSSUrl' is defined as th e pressure whid is measured with the help uf a pressure measuring instrument. in which the atmospheric pressure is wken as datum. The atmospheric pressure on the s<:a le is marked as zero.

I I

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IL 142

Fluid Ml>chanics

.\ . Vacuum IIrl'ss un' is defin ed as the pres- III sure be low the atmos phe ric pressure. The re lati onship bctwcc ll lhc abso lut e ptcS1;urc. g au~c pressu re and vac uum pressure arc shown in Fig. 2.7. Mathe mati call y: (i) Ahso lute pressure == Atmosph eric pressure + Gauge press orI'

•

~

1

, - I ./ f'"'

GAUGE PRESSURE / ~TMOSPHERIC

/"

PRESSURE

~r VACUUM PRESSURE ASSOlUTE PRESSURE .....

II., = P.. ," + p~",V' "' (ii) Vac uum pressure

,

ABSOLUTE ZERO PRESSURE

Fig. 2.7 Relationship bl'llllUn pr/'S!IIrt'S. == Atmosph eric press ure - Absol ute pressure. Not e. (;) The atmospheric pressure at sea level al 15°C is 101.3 kNlml or 10.13 Nlcm 1 in 51 unit. In case of

M KS units. it is equal to 1,033 kgf/em 1, (ii) The atmospheric pressure head is 760 mm of mcr<;ury or 10.33 m of water.

Problem 2.8

IV/WI (Iff liff gllUgf prrsslIrf lind {/bsollllf prrs5" ff (If II poi", 3 III bflow Ilw frt'f 511 rf ll('f (If II liquhllUl\'ill/l II dnl5i1y of J.5J x /lY k,~/1Il3 if IIw UllIl05p/wrir prrssurf i5 fqllil'lIlml 10 7501111/1 ofmnrllry? 17lf Silfrijir g fll l'i1y flfllwrrlln' i5 13,6 (II1t1 dfllsily of WUlff = lOOO kg/m 3.

Solution. Give n : Dt: ptll of liq ui d. Dt:nsi ty of liquid.

2 , ,, J m

At mosph eric pressure head .

20

where

mO

P," 1.53 X 10' kg/m' 750 mm of Hg 750 " WOO " 0.75 In o f Hg "

Atm osph e ric pressure. P..on" Po x g x Zo Po" Densi ty of Hg " Sp. gr. o f me rc ury x Density o f water " 13.6 x 1000 kg/m' Zo" Pressure Il ead in te nn s of merc ury. /''''m'' (13.6 x 1(00) x 9.81 x 0.75 Nfm2

(.,'

Zo" 0.75)

" 100062 Nfml Pressure at a poin t. which is at a de pth of 3 m fro m th e free surface of the liquid is give n by. p=plxgxZ, ,,(1.53 x 1(00) x 9.8 1 x 3" 45028 Nlm l

G au ge pressure. Now absol ute pressure

... 2 .S

11= 45028 N'm ~. AilS. " Gauge press ure + Atm osplle ric pressure = 45028 + 100062 = 145090 N/m 2 . Ans .

MEASUREMENT Of PRESSURE

The pressure of a fluid is measured by [he fo ll ow in g dev ices: I. Man o me[e rs 2. Mec han ic al Gau ges. 2.S. 1 Milnom~ten . Mano meters are defi ned as [he dev ices used for measuri ng th e pressure at a point in a fluid by ba lancin g th e co lu mn of fl uid by tile sa me o r anothe r co lumn of the fl uid . T il ey are d assified as : ( b) Di ffe renti al Manomders. (tl) Simpl e M:lllometers.

I I

Ii

~ I

IL Pressure ilnd its Measurement

431

2. S.2 Mechanical Gauges. Mechan ic al ga uges arc defmed as th e dev ices used for me as urin g Ih e press ure by balanc in g llie nuid co lumn by Ih e spring or dead we ig ht. The commonly used mcc hani (;al pressure g aug es arc: (a) Di:lphrag m pressure gauge. (b) Bourdol1 1Ubc pressure gauge, (r) Dead-we ight press ure gauge , and (d) Bellows pressu re gauge.

to 2.6

SIMPLE MANOMETERS

A s im ple manometer cu nsiscs of a gl ass tube having one of its e nds connected 10 a poinT where pressure is to be measured and mh er end re main s open to atm usphe re. Co mm on types of simple ma nometers arc: I . PiczomclCr: 2. U· lubc Man ome ter. :mll 3. Single Column Man ometer.

T

2.6. 1 Piezometer. It is Ihe s implest form of mano meter used fOf measurin g gauge pressures. One end of this mano meter is cu nn ected to the point where pressure is to he measured and ot her end is open to the atmusph erc as shown in Fig. 2.8. The ri se of liquid g ivcs th e pressure head at that point. If at a point A. th e height of liquid say wate r is II in piezometer tu he. the n pressure at A N =p x 8 xlI - , . m-

Fig. 2.8

PinomfuT.

2. 6 .2 U-tube Manometer. It cons ists o f g lass tube bent in U-shape. one e nd uf which is cunnected to a point at whi ch press ure is to be measured and other e nd re main s upe n to th e atmosphere as shown in Fig. 2.9. T he tuhe gene rall y contains mercury u r any oth er liquid whose specific gravity is grea te r than th e specific gravity uf the liquid whu se press ure is to he me asured.

T

1 (al For ga uge pressure

(b) For vacuum p<8SSUf8

U·tube Manometer. (
I I

m~asured.

whose va lue is,).

Ii

~ I

IL 144

Fluid Ml>chanics

As th e pressure is th e same for the horizontal surface. Hence pressure above th e horizontal datum line A -A in the Ieli co luilln and in the right co lumn of U-Iube manometer shou ld be same.

Pressure above A-A in tile left co lumn " P + PI X g X hi Pressure above A-A in the right column " Pl x g x II, p + Pig/II'" p,-g l,! Hence equating th e two prc!iSUrCS

P" (pUS/I, - P I X g x III)'

.. .(2 .1 ) (b) ~-or V"cuum P ....SS UI'f'. For rn ~ asuri n g vacuum pressure. the k vc l of th e heavy liquid in the mallomC lc r wi ll be as shown ill Fi g. 2.9 (b). T hen Pressure above A-A in the left co lumn Pressure h"ad in Ihe ri ght column arovc A-A

"P!8"2 + Plgh l + f! " 0

P181!2 + Pig/II + II" 0 ...(2.8) fI" - (P;oKiJ 1 + Plg" I)' Tile right limb of a ,~imple U-/Ilbe mOI/Olllet'" CQllfaillitlg mercury is ope,1 to Ille almO.fpllere while Ihe left limb iI connected 10 a pipe in which a fluid of sp. gr. 0.9 i.f flowing. Tile cenlre of IIII' pipe is J1 cm below IIII' lere! of mercury in IIII' riglll limb. Find IIII' pre.l.wre of fluid in III<' pipe if IIII' difference of mercury lel'e! in Ihe two limbs is 10 cm. Solution. G ivcn : Sp. gr. of fluid, 5, = 0.9 Densi!y of fluid. p , =5, x 1000=0.9x 1000 = 900 kglm 3

Problem 2.9

Sr. gr. of me rcury, Densi ty of mercury.

5~

r

= 13.6

" j

p~ = l3.6 x 1000 kgfmJ

Difference of mercury level, II~ = 20 c m '" 0.2 m Heigh! o f fluid from A-A. h, = 20 - 12 = Scm = O.OS m Let P = Pressure of fluid in pipe

,

Equating Ihe pressure above A-A. we gel

"'

P + p ,Sh, = P ~gl' 2 P + 900 x 9.81 x 0.08 = 13.6 x 1000 x 9.81 x .2

Fig. 2. 10

P= 13.6x IOOOx 9.81 x.2-900x9.S1 xO.OS

= 26683 - 706 = 25977 Nfm2 '" 2.597 NIemI. ADS, Problem 2.10 A simple lj·llIbe mOl/omeler cOilwiliing mercury is connected 10 a pipe in wllich 0 fluid of sp. gr. 0.8 lIlIIflnn·ing l'lICUUIII pressure is flowing. The OIher em f of lhe mllltollieler is open 10 almosphere. Find IIII' melwm pl"l'Hllre ill pipe, if the differellce of mercury 1{"I·el ill IIII' two limbs i.f 40 cm w,d the heighl of fluid in IIII' left frolll Ihe celllre of pipe iJ 15 cm below.

SolutIon. Givcn : Sp. gr. of fluid. Sp. gr. of mercury. Density of fluid. Cknsity of mercury,

S, '" 0.8

Sl= 13.6 PI '" SOO

P2 '" 13.6 x 1000

Difference of mercury kvel. hl = 40C1O = 0.4 nl. Height of liquid in Idl limb. I" = 15 cm = 0. 15 m. Lei Ih e pressure in pipe = 1'. Equat ing pressure above datum line A ·A . we ge l

I I

1

,,= T

, "-'- , Fig . 2.11

Ii

~ I

IL Pressure ilnd its Measurement

45 1

I P28i'2 + Plg h ,]

Ii '" -

= - 11 3.6 x 1000 x 9 .31 x 0.4 + 800 x 9.81 x 0 .1 5 ] = - [53366.4 + 11 77.2 1 = - 54543.6 N/m2 '" _ 5.454 Nie m I, Ails. Problem 2 .11 A U-Tube m(mum eler is used /0 measure Ihe preS~'''re oj ....aler in II pipe /ine, .... hiel! is in <'... cess of IlIIrIOJ'pileric press" re. The righl lim" of tile mUllome/a COll la;ns mercury and iJ' open /0 almasp/wu:. The con/ac/ be/ween woler (lIId mercury is ill Ihe left limb. Delermine Ihe pressure of ,nUer in the main lille. if 111<' difference in leI'e! of mercury ill IiiI' limbs of V-Illbe is JO em Ulll/lhe free surface of mercury is in Jerel willi Ihe eenlfe of rhe pipe. If Ihe preSSIIrI' of warer in pipe line is redllced In 98 10 Nlml, cu/cu/(lie the 1Iell" difference i ll Ihe le,-e! of mercury. Sketch Ihe arrangements ill bOlh ea.ies. Solution. G ive n : Diffe rence o f merc ury'" 10 em '" 0 .1 m The ~rran ge men t is show n in Fig. 2. 11 (a ) 1st I'art u:t p). " (pressure o f wa te r in pipe line (i.e .. at po int A) The poi nts B and C li e o n th e sam e ho ri zonta l line. Hence pressure ,11 B s hould be equ,lil o pressure at C. Bu t pressure at B '" PreSl;ure at A + Pressure d ue 10 10 c m (or 0. 1 m) RIGHT LlMB _ of wate r

;~ER

"'p~ + pxgxll

where

p '" 1000 kgl m

3

and II '" 0. 1 m

({T -

'" 1'). + 1000 x 9.8 1 x 0.1 2 "/I). + 98 1 N/m

...( i)

Pressure at C", Pressure at D + I'reSl;ure due 10 10 em o f merc ury "'O+PoxgX/lo where Po for mercury '" 13.6 x 1000 kg/Ill} and 110 '" 10 em '" 0. 1 111 Press ure at C", 0 + ( 13 .6 x 1000) x 9 .&1 x 0. 1 = 1334 1.6 N ...(ii) But pressure at B is equa l to pressure at C. Hence equat ing the equati ons (il and (ii). we ge t 1'.. + 9 8 1", 13341.6 fI).::

-

-LEFT lIMB_

,

O

1

'T

C

MERCURY _

-

Fig . 2.11 (a)

13341.6 - 9 81 N

= 12360.6 -2 . An s.

'"

lin d Pa rt Gi ve n. 1'.. = 9 8 10 N/m 1 Find ncw di ffe re nce of merc ury level. The arrangemcnt is shown in Fig. 2. 11 (b) . In this case the pressure at A i5981 0 Nlm '"w hie h is Jess lhanlh e 12360.6 N/ml. He nce merc ury in Je fllimb will ri se. The ri se o f merc ury i n Jeft lim b will be equ a l 10 l he fall o f mercury in righl lim b as lhe tOlal vo lu me of merc ury re mai ns sam e. Lel .1 '" Ri se of merc ury in left linlb in (;m Then fall of mcr(;ury in rig ht limb ", x cm Th e poi nts S , C and D show lhe initial cond itions wlwre as poi nts 8 *. C* and D* show lhe fin al conditi ons.

I I

Ii

~ I

IL 146 or

Fluid Ml>chanics

The pressure at 8* '" Press ure at C· Pressure at A + Pressu re duc to (10 - xl e m of WJ\cr '" J>rcssure at J)* + Pressore due to

(10 - 2x) ern of mercury

1910 + 1000 x 9.8 1 x

i'OO ("""'10-")

..--1+"' ~

to em

(10-2.,)

",0 + (13.6 x I OOO) x 9.8 1 x

or or

--------

-- -- - -

100

J,j ,-r

( 10 _ 2.)

11,. __ ,,- _L

~B

- -1:0,- -

Di vid illg by 9.81. we get 1000 + 100 - lOx = 1360 - 272x

272x - l0-,= 1360-1100 262x'" 260

or

2 6() ,= -'" 0.992 em 262

New diffcfCIlCC of mercury '" 10 - 2 l em =10 - 2 x 0.992

Fig. 2.11 (b)

'" IWI6 em. Ans.

U-t"""

Problem 2.12 Fig. 2.12 .\110;"$ "colliC(IIl'l'Sse/ /I(I\'illg iu 011//1'1 til A /0 wllirh fI mlUlOllli'll'r i.1 rOlllll'("/I'l/. H", rNII/ill8 oflhl' /II"HomNer g;,'''" ill IIIl' figllre S/Wlt'" wlwllll", ""$.W'I is ('/"1'1)'. Filld 1/'" rl'{I{lillg "1'/'" ""IIl"III('/l'r Wlt,," Ih" w ssd i. (""",[,Indy filled ,..ith .m/('T.

Solu tion. Vcssd is elllll' )'. Given: Difference of mercury level Le t ill "" He igh t of water abo ve X·X

Sr . gL of mercury, Sr . gr. of watu.

Sl = 13,6 SI

= 1.0

Density of nU'rcury. P2 '" 13.6 x lOoo Dcnsity o f wmc r. PI '" 1000 Eq uatin g the pressure above datum line X-X. we have

P2 xgxil 2 ""Plxgxil l 13 .6x IOOOx9.8 1 x 0.2"" lOoox9.8 1 xiii

or V~ssel

,

,M I

1 I

"

1

III "" 2.72 111 of water. is full uf wa te r. When ves.w l is full of water. the

Fi g . 2. 12 pressure in the right limb will increase and mercury leve l in the rig ht limb will gu down. Let the distance throu gh which mercury goes down in the right limb be . ), cm as shown in Fig. 2. 13. The mercury wi ll rise in the left hy a d ist an ce of)' Clll. Now the datum line is Z-z. Equat ing the pressure aoove th e datum line Z·z. Pressure in left lim b"" Pressure in righ t limb 13.6 x WOO x 9.81 x (0.2 + 2)"1100) '" lOOOx9.8 1 x(3+II I +yI I 00)

I I

Ii

~ I

IL Pressure ilnd its Measurement

13.6 x (0.2 + 2y1lOO) '" (3 + 2.72 + yllOO) 2.72 + 27.2yf1OO '" 3 + 2.72 + )'/ 100 (27.2)' - y)l100 '" 3.0 26.21' = 3 x 100 = 300 300 y ~ - - = 11.45clIl 26.2 The d ifference of mc n; ury le\'e l in twO li mbs = (20 + 2y) e m of men:ury

47 1

hi = 2.72 e rn )

-- -1 3m

1 1,

= 20 + 2 x 11.45 = 20+22.90

.J

1- -

- T

1x (2{l + 2y)cm i - -'-

T

1",

F T'

"',

= 42.90 em of mercury

Re ading of manometer'" 42.90 em. AilS.

Problem 2.13 A pressure gmlge cOll sis /s of ","0 c)'lilUlrim/ bulbs IJ /1111/ C r(/cl, of /0 sq. em crossseclio/l11/ oretl, which ""- cOIIIIl'Cled b y /I U-lUhl' Willi va/jell/limbs ('(wl, of 0.25 sq. em cross-sUliO/wl /lrn l. A rnl/iqllid of sl'l'cifk gral'il), 0.9 is filled ill/a C 1/1111 c/('(l f Will'" is jilin/ imo B, IIII' S II'ftirF of Jf'/Ulrtlliu/l bring ill Ihl' limb '1I/Ilci1n{ 10 C. Filii/II/(' ili SplaCf'1II1'I1I of 1/11' sur/art' of sepllrm ;011 will'll IIII' pressure 011 lilt' sur[net' ill C i s gr,.fllu 1IIfIllilru i illlJ by wr WlIOIIIIII'qUlliio / olr Irpm/ o[II'/II,.r. Solution. Give n : Ar~ a o f eac h bulb Band C, Ar~ a

of eac h v.:rtical li mb ,

1\= IOc rn 1 (I

= 0.25 ern 1 Its d~n si l y = 900 kg/m3

= 0.9

Sp. gr. of red liquid

X·X = In ilial se parali on leve l

Lo<

11(." = Heighl of red liquid above X-X

11[1 = He ighl of watc r abo l'c X-X Press ure above X-X in

lh~

Idllimb = 1000 x 9.81

X "B

Press ure above X-X in the ri g ht limb = 900 x 9.81 x Ire Eq uatin g the lWO prcssure . we get IOOO x 9 .81 xll ll ",900x9.81 Xll c

1111 = 0.9

.. ·(0

II C

lJ40

Wh en the pressure head o\'er lh e surface in C is inc reased by I cm of wale r. let the separat ion leve l falls by an amo unt equa l 10 Z. Th en Y- Y b<:co m ~s lh e fi nal separat ion le l'''l. Now fall in surfac" IeI'd of C mul tip lied by crosssec li o nal area o f bulb C mUSl be equ a l 10 the fall in sc parallon level mullipli ed by cross-scclion al arc a o f limb. FI NAL. SEPARATION LEVEL Fa ll in surface lel'.:1 of C = Fa ll in separalion kl"eJ x a

A

I I

lJ40

",

- , 1:1--- -

INITIAL SEPARATED LEVEL

Fig . 2.14

Ii

~ I

IL 148

Fluid Ml>chanics

z X0.25

Z

10

40

A

Also fall ill surfncc level of C '" Rise in surface level of H

Z

40 The pressure o f I Cln (or 0.0 I In) of water = pgh = 1000 x 9.81 x 0.0 I'" 98.1 N/m2 Consider final separation lev e l Y-Y Pressure above Y- ¥ in Ihe Jeft limb '" [000 x 9.81

( + "e + Z) + !) Z

Pressure ahove y. Y in the rig ht limb = 900 x 9 .81 ( Z

40

Ire -

+ 98.1

Equating Ihe two pressure, we get

]000 x 9.81 (Z + 11 8 +

! )'" (2 +

he -

~ ) 900 x 9.S! + 98.1

Dividing by9.81, we get 1000 (Z+h8 +

! ) '"

Dividing by 1000. we gel Z + lIy + Btu from cqu3lion (i),

~

900 ( Z+hc -

! )+ ! )+

'" 0.9 ( 2 + he -

10 0.01

"N '" 0.9 he

Z+O. 9h c + Z '" 39Z x O.9+0.9h c + O.QI

40

40

4 !Z

39

--"-x.92+.01 40 40

"'

Z(:~ - 394~·9) = .01 _

Z

=

Of

40xO.OI

z(41~t·J) = .01 = O.067H m = 6.78 ern. Ans.

5.9 2. 6 . J Single Column Manometer. Single colulnn manometer is a 1I1odificd form of a U- IUOC manometer in wllich a r,,~ r voir. having a large cross·scctional area (about ]00 limes) as compared to tile area of the tube is connl....:kd 10 one of tile limbs (say left limb) of the manometer a~ shown in Fig. 2. 15. Due to Jarge cross-sectional are a o f the reservoir. for any va riation in pressure. tile change in tile liquid level in the reservoir will be very s mall whicll may be neglected and hence tile pressure is given by thc height of liquid in the other limb. The other limb ma y be vcnical or inclined. Thus thcre are two types of single column manometer as: I. Ven icaJ Sin,gle Column Manometer. 2. Inclined Single Column Manom eter.

I. Vertical Single Column Manometer rig. 2. IS s hows tile vertical single column manometer. Lei X-X bc the datum line in the reservoir ,md in tile right limb of the manomete r. when it is not connected to the pipe. When the manom eter is

I I

Ii

~ I

IL Pressure ilnd its Measurement

49 1

conn~ctcd to the pipe. due to high pressure at A. tile heavy liquid in the reservoir will be pushed downward and will risc in the right limb.

Let 1'>11 '" Fall of heavy liquid in reservoir

r

I,! '" Rise of heavy liquid in right limb h,

= Heighl of centre of pipe above

X-X

T

/I" = Pressure at A. w hich is \0 bll measured

",

A = Cross-section!11 area of the reservoir (I '"

1

Cross-sc(;t;onal area of Ihe right limb

T

S, = Sp. gr. of liquid in pipe $1

_,r!;~'::'~_:'1R~Yo~:_=~"'Qj y

y

'"

= Sp. gr. of heavy liquid in reservoir and right limb

Fig. 2.15 V"lical si'lgie coll/mn

PI = Density of liquid in pipe

malwmeur. Pl = Density of liqu id in reservoir Fall of heavy liquid in rc!,crvoir will c ause a rise of heavy liquid kvcl in lh", right limb. A

X ~,

= (/

llh =

X

(1)(

hl

112

... ( i)

A

Now consider the datum line Y-Y as shown in ri g. 2.[5. Then pressu re in the right limb above Y-Y.

Pz x g x (llh + /'2) Pressure in lhe [eft limb above Y-Y = p, x g x (1'111 + Il,) + IJ" =

Equaling these pressures. we ha ve Pl xg x(l'1h + " 1) = p, Xg x {l'1h +/i,)+p"

1'" = P!8 (t'1I! + "0

"'

-

p,g(tlh + ",)

= I'1h lPlg - p,g] + "ZPlg - ",p,g

But from eq uation (i),

I'1h = a x "1 A

.. .(2 .9)

As the area A is very large as compared 10

fl.

hence ratio!!... becomes very s mall and can be A

neglecled. Thcnp,,="lP!8 - ",P,g ... ( 2. 10) From equation (2. 10), it is c lear lhal as ", is known and hence by knowing /'1 or rise of heavy liq uid in the right limb. lhe pressure at A call be ca lcu laled . 2. Inclined Single Column Manometer

Fig. 2.16 shows the incli l1ed sin gle (o lumn mal101l\e\Cr. This manometer is more sensitive. Due to in(lination th r dis\ance moved by the heavy liquid in the rig ht limb will be more.

I I

Fig. 2.16

Inclined single coll/mn manometer.

Ii

~ I

IL Iso

Fluid Ml>chanics

Let

L", Length o f licavy liquid moved in right limb from X -X

e = Inclination ofrighllirnh Wilh horizontal 11 2 '" Vertical rise of heavy liquid in ri gh t limb from X-X ", L x sin

e

From equ ation (2.10), Ihe pressure at A is II... = "~P2g - "IPlg· Substituti ng the val ue of 11 2, we gCI

e

... ( 2 . 1 I ) /I,., = sin x P28 - h lPlg· Problem 2.14 A single coluIIIn manometer is cOIlller/ed /0 (I pipe comaillins a liquid of sp. gr. 0.9 as slloll'n ill Fig. 2./7. Find lite pressure ill the pipe if the orea of Ihe reserl'Oir is 100 limes Ihe area of lite rube for IIIe mmwme/er reading ShOIl"1I in Fig. 2.17. The specific grol'ity of mercury is 13.6. Solution. Gi ve n:

Sp. gr . of liquid in pipe.

St = 0.9

I"

PI = 900 kglm )

Density Sp. gr. of heavy liqu id.

S 2 = 13 .6

Density.

p, '" I1.6 x 1000

f

1

Area of reservoi r '" ~ '" 100 Area of right limb Hcigtlt of liquid. Ri se of mercury in right limb.

a

hi::: 20 ern '" 0.2 rn f".g . 2 .17

h Z =40un=0.4m

Let Using equation (2.9). we get

PA = Pressure in pipe

"

PA = -A IhlpzR - PIS! + hop,S • - - "IPIS 1 : X 0.4[13.6 x 1000 )( 9.81 - 900 x 9.81 J + 0.4 x [3.6 x 1000 )( 9.8 1 - 0.2 x 900 x 9.81 100

== 0.4 [1 3341 6 _ 8829 ) + 53366.4 _ 1765.8 100 = 533.664 + 53366,4 - 1765.8 Nlmz = 52 [34 Nlmz = 5.21 Niem I. .. 2 . 7

A ilS •

DIFFERENTIAL MANOMETERS

Diff.:rcmial manomcters arc th e d.:viees used for mcasuring th e differencc of pressures betwee n two poims in a pipe or in two differelll pipes. A differential manometer l'Ons iSls of a U· lUbe. coma in· in g a heavy liqu id. whose two ends are con nected to lhe poinls. whose difference of pressure is 10 be measured . Most commonly types of differential manometers arc: !. U·tube differential lIlano lll eter and 2. In ve n ed U·tu bc diffe rent ia l manometer. 2 . 7 . 1 U - tube Differential Manometer. Fig. 2.18 shuws the differential manumeters of U·tubc type.

I I

Ii

~ I

IL Pressure ilnd its Measurement

T , 1

, t , -- J" ,

t

T

(a)Two rHpes at differenllevels

Fig. 2.18

511

'1-- ' (b) A aJ'ld B ale at the sa me Ieve1

U-wbe difftr..nlial manomtll"l'$.

In Fig. 2.18 (ll). the tWO points A and 8 are at different level and also contai ns liquid s of different sr. g r. These point~ are C{lonecled 10 the U -luhe differential manometer. Le t th e pressure al A "fld B arc I'A and PB" Le\

,,=

Difference of mercury level in the U -tune.

Y'" Distance o f th e cCl1m: o f B, from the mercury level in the right limh.

x'" Distall(;c of the l-emrc o f A. from thc mercury level in the right limb. PI

= Density of liquid at A.

P 2 = Density of liquid al B. p & = Density of he avy liquid or m<:rcury.

Taking dmum lin e m X-X. Pressure abo ve X-X in Ihe left limb '" Plg(1J + x) + 1',\ wherc p,\ '" pressure at A. Pressure above X-X in the rigllt limb '" P~ x g X II + P~ X g X Y + 1'8 wllere 1'8 '" Pressure 3t 8. Equ ating tile two pressure. we Ilave

Plg(1! + x) + I'll '" PJ X g x I! + P28Y + P/I

P.t - pB '" P, x g X II + P!G)' - Plg(1I + x) '" It x g(P J - PI) + P2g), - Plgx Difference of pressure at A and 8 '" I! x g(PJ - PI) + Pig)' - Plg.t·

... (2. 12)

In Fig. 2. 18 (b), the two points A and lJ are at the same le ve l and contains tile s.ame liquid of den sity PI' Tlien Pressure aboveX·Xin rigllt limb '" p.xg xl, + PI xg xX +PB Pressure above X -X in le ft limb

'" PI x g x (II + xl + p ...

Equati ng the two press ure

P, x g X II + Plgx + 1'/1 '" PI X g x (II + xl + JIll I'" - JIB'" p. X g x I! + p ,gx - p,g(1I + x) ... ( 2 . 13) '" g x lI(p~ - PI)' Problem 2.15 A pip .. COIII{lillS (11/ oil of sp. gr. 0.9. A diff"'''l!Iia/ 1I/1II1
I I

Ii

~ I

IL 152

Fluid Ml>chanics

Solution. Given: Sp. gr. of oil.

Density. PI "" 0.9 x I()(K)" 900 kg/rn

3

I,: lSClll=O.ISm

Difference in rncn:ury level.

Sp. gr. of rncffury. Ss = 13.6 Density, p~ = 13.6 x [000 kglm J The diffcl\)ncc of pressure is given by equation (2. 13) f',t - PB" g X h{P, - PI)

'" 9.8! x 0. 15 (13600 - 9(0) '" 186K!1 N/m!. Ans .

Problem 2.16 A difJrTl'1IIiai m"""",,,'l''' i .• (',,""N"/I'll (l/ flIP Iwo Iminl s A ",,,18 "f I ... " l'ipl's {I$ SilO,.." ill Fig. 2./9. Til .. p'iN' A . 'omaills a Ii'll/itf of 31" gr. = 1.5 ... /,i/" [}ip" lJ collwillS 1I liquid "I w- gr. = 0.9, Til" prnmrn a/ II mnlH (If" I kgflrm ! IIlIff /,80 kgJ7m/ rl'.!prrlil"r/y. Fi"d I/U' dijJrrf'IICf ill mereu,)' /","/'I ill Ih" (Iiffrrnllitd """!(Jllli'lrr.

Sp . gr." ' .5

PA =1 kg/1mi'

SolutIon. Given: Sp.gr.ofliquidatA, SI = I.S

..

PI"" 1500

Sp. gr. of liquid at /J. 52 "" 0.9 Pl"" 900 Pressure at A. PA = I kgflcm 1 = 1 x 10· I.:gffm 2 = 10"x9 ,81 Nfm!('; 1 I.: gf=9.81 N) Pressure al 8,

1'8'" 1.8 ~gflcm 2

2.0m

,

~

= 1.8 x 104 kgffrn l lknsity of mercury

[ ~

=1.8xlD"x9.81 Nfm2 (": Ikgf = 9.81N) x = 13.6 x 1000 kgfm J

Sp, gr.=0.9

Pe = 1.8 kgflan'

r

Taking X-X as datum line. Fig. 2.19 Pressure above X-X in the left limb = 13.6 x lO00x9.8 1 xh+ 1500x9.8 1 x(2+3)+/J,\ '" 13.(; x 1000 x 9.8 1 x II + 7500 x 9.81 + 9.8 1 x 104 Pressure above X-X in the right limb '" 900 x 9.81 x (11 + 2) + Pfl =900x9.8 1 x(h+ 2)+ 1.8 x 10"x9.81 Equ at ing the two pres.<;ure. we get 13.6 x 1000 x 9.81// + 7500 x 9.81 + 9.81 x 10· '" 900 x 9.81 x (II + 2) + 1.8 x 10" x 9.81 Dividing by 1000 x 9.81. we gel 13.6/1 + 7.5 + 10 = (h + 2.0) x.9 + 18

"'

13,6/1+ 17.5=0.911+ 1.8 + 18=0.9h+ 19.8 {l3.6 - 0.9)11 '" 19.8 - 17.5 or 12.711 '" 2.3

23 ,,= -=0.181 12.7

III

= 18.1 em .

AIlS .

Problem 2 .17 A dijJfrl'mi(l/lIwlwllwli'r i~ ("",,,w("li'd (If IIU' 1"'0 Imims A (/Jut H (IS sho"'" ill Fig. 2.20. AlII "ir prnsurl' ;,' 9.8/ NiemI ((11M). find Ihl' "b,
"'

I I

II '" 9.8 1 N/cm l

1'8=9.81 x lO"N/m l

Ii

~ I

IL Pressure ilnd its Measurement IXnsity of oil '" 0.9 x 1000 '" 900 kg/m 3 Density of me rcury '" 13.6 x 1000 kg/fill Let the pressure at A is P,1 T:lking datu m line ~t X·X Pressure above X-X in the right limb = 1000 x 9.81 x 0.6 + I'll '" 5886 + 98100 = 103986 Pressure above x-x in Ihe Icfllimb '" 13.6x I OOOx9.81 xO.1 +900 x9.81 xQ.2+PA = 1334 1.6 + 1765.8 +

53 1

Oil OF Sp. gr. ~O . 9

,

I'"

Equating Ihe two pressure head s 103986 = 1334 1.6 + 1765.8

+ PA p,\ '" 103986 - 15107.4 '" 88876.8

Fig. 2.20

2

88876.8N N , = 8.887 -~,. lOQOOc m " e ll) " A = 8.887 Nf(m l , Ans.

p", '" 88876.8 N/rn =

Absolute pressure at 2. 7.2 Inverted U-tube Differential Manometer. [[ consists of an inverted U-Wht:, cont ainillg a light liquid. The two ends of lh~ tube arc connected to the points whose diffacncc of pressure is to be measured. h is used for me asuring difference of low pressures. Fig. 2.21 shows an invc n ed U-tube different ial manometer connected to thc two point!; A and 8. Lct the pressurc at A ,s more than the pressure at 8. Let II, = Height of liq uid in left limb below the datum line X·X II! = Heigh t of liquid in right limb II == Differe nce of light liquid p, == Density of liquid al A P 2 = Densi ty of liquid al /J P, == Densi ty of light liquid p" '" Pressure at A PB = Pressure at /J. Taking X-X as dat um li ne. Then pressure in the left limh below X-X

,

, ~

,

==p,,- p,xgxiI,. Fig. 2.2 1 Pressure in the right limb be low X-X "'PB - P2 xg X /1 2 - P.xg X II Equating the two pressure 1'" - p, x 8 X II, == I'B- p! X 8 X /12 - P. X g X II . .. (2 . 14 ) I',,-I'B = 1', xg X " I - P! xg x1l 2 - p. xg x II. Problem 2.18 m(l1lOm~l~r Iwvillg '111 fJii fJfsp. gr. 0.8 is CO"'''''·I(>(/. Th~ p'f'.""'~ /"'(1(/ ill 1/'" pip~ A i .• 2 /11 of»"fIIf". jim/I/lf' pressure ill Ilu' pip" II fo r Ih" ",UllfJmNl'r re(l(lillg.1 a.1 sllo»"11 ill Fig. 2.22. Solulion. Given:

"'

Pressule head at

A == p" =2 III ofwal~r

pg

1',j==jJxgx2= IOOOx9.81 x2= 19620Nlm~

Fig. 2.22 shows the arrangement. Taking X-X as datum linc. I>rcs;;urc below X-X in the left limb '" p" - PI X g X /' 1

I I

Ii

~I

IL 154

Fluid Ml>chanics

6''''>o, OIL of Sp. gr. 0.8

'" 19620 - 1000 x 9.81 x 0.3 '" 16677 N/m!. Pressure below

x-x in

the

ri~hl

-,

limb

=PB - 1000x9.81 xO.I - 800x9.81 xO. 12 =PR - 98 1 - 941.76=PR - 1922.76 Equati ng Ih e t wO pressure. we gel 16677 =PR-1922.76

or

I'll '" 16677 + 1922.76 ", 18599.76 Nfrnl

or

I'll '" I.KS\l9 NIcOl!, AilS.

WATER

Fig . 2.22

Problem 2.19 III Fig. 2.23. (III im'fr/n/ (flffarmio/liumOIlJl'lfr i~ romlfclnJ IQ ''''0 pi"t'S A /IIullJ ",/lid, COIIW)' "'tUa. TIll' jlI,id ill IHmWlllflFr is oil ofsp_ gr. 0.8. Ff" 1/,(, IIWIWl/lnU rl'(IIIill85 showll ill ,II .. fig"'''. jiml IIU' l'rf $S/lrf difJrrl'lu'f' hp/wi'('11 It (111(/ B.

Solution. Given:

,-

= 0.8

Sr. gr. of oit

OIL of

Difference of oil in Ihe two li mbs

Sp. g'. 0.8

I:

-x

H

=(30+ 20) - 30 = 20cI11 Taking datum lin e 31 X-X

11

Pressure in th e left limb tJ..,]ow X-X

=PIl - IOOOx9.81 xO

Fig. 2.2.3

'" p" - 2943 Pressure in the ri ght limb be low X-X

"'p,, - IOOO x9.8 1 xO.J - 800x9.8 1 xO.2 Equating the twO pressure 1',,- - 2943

"'I',, - 2943 - 1569.6"'p" - 45[2.6 '" 1'" - 4512.6

p" - p,, '" 4512.6 - 2943 '" 1569.6 N/m ~. A"-~ .

Problem 2 .20

Fimf 011/ Ilw dijJl'rfllli(ll rnulillg 'iI' of 1111 im'erletl U-Iube 1I1lI1I01II<'11'r collwillillg oil of sl",cijic grm-ity 0.7 <'5 A IIIIIILJ lire lorflled III Ihe 5WI/<' lerel lIIul a55W"" Ihe 1'11'551111'5 (/I A (IIu/ B 10 be eqlla/.

Solution. Given: Fig. 2.24 shows th e arrangement. Taking X-X as datum [Inc.

p,,- '" Pressure at A 1'" '" Pressure at B

Density of liquid in pipe A

'" Sp. gr. x 1000 '" 1.2 x 1000

'" 1200 kglm Density of liquid if] pipe B Density of oi l

2

'" I x [000 '" 1000 kg/m) '" 0.7 x 1000", 700 kgfm3

,

&"""' j'.".~.'

-r-

, gr·t .O

"

~-

w=

~-

Fig. 2.24

II

Ii

~ I

IL Pressure ilnd its Measurement

55 1

Now pressure below X-X in the left limb

"p" - 1200 x 9.81 x 0.3 - 700 x 9.81 x" rrc~urc

below

x-x in

the right lilnb =P8 - 1000 )(9.81 x (h + 0.3)

Equati ng the two pres,ure. we get 1'.-I-1200x9.81 xO.3-700)(9.81 Xh=Pn-1QO(}x9.81 (11+0.3) p", = 1'8 (given)

But

1200 x 9.81 )( 0.3 - 700 x 9.81)( II = - 1000 x 9.81 (II + 0.3) Dividing by 1000)(9.8 1 - 1.2 xO.3 - 0.711 = - (II + 0.3)

0.3 x 1.2 + 0.711 = " + 0.3 or 0.36 - 0.3 = Ir - 0.7/1 = 0.311

"'

0.06 II = 0.)6 - 0.30 "--m 0.30 0.30 I

1l1=~X

5

100 = 20 em. Ans.

Problem 2.21 All illw'rll'd U-II/be 1/1(1/10111('/'" i~ colIIlN"/ed /0 /11'0 lror;:ollllli pipn A /lml IJ Ihrollgh ...flich Wain is flowitlg. Thl' W'rrical (/iSlalln' bnwl'fll Ihl' (/.\"t'S of tllest' pipn is 30 ('III. II'hl'lI WI oil ofspeciftr grm'il)' 0.8 is 1151'(/ (U (I glluge f/llitl, I/U' "l'rlicollwiglw ofwlI/a ro/mlllls ill I/W two limbs of Ow ill"erln! m(IHOIllflU

( ""WI! /IINI.,urn/

from Ihl' rl'Sl'f("I;"(' {'('mr('

lillI'S

of Ilw ,,'PI'S)

lin'

fmll/(Ito hI' SlI/IIf (IIu/ I'(I,wl to 35 Oil. Dnrrmillf till' l/ijJfr('ll{,(, of prnSllrf hl'n"ffll till' pipl'S,

SolutJon. Given : . gr. =O.B

S..,.,cific gravity of measuring liq uid "" 0.8 The arrallgelll<'nl is shown in Fig. 2.24 (ll). leI

PA = pressure at A PI!" pressure at R.

The poilUs C and /) lie on th e same horizomal line. Hence pressure at C should be equal to pressure at D. But pressure at C

""fl,, - pgh

And pressure at /)

'" 1'" - 1000 x 9.81 x (0.35) '" PB - p,gh, - P28 11 2 "" jiB -

WATER

1000 x 9.81 x (0.35) - 800 x 9.81 x 0.3

BUI pressu re at C", pres;;urc at D

B Fig . 2.2~

JiA - 1000 x 9.&1 x .35

fa)

"'p/i - lOOOx9.81 xO.35 - 8oox9.81 x 0.3 or

"'

I I

8OOx9.&1 X 0.3 "'PB - PA

JiE - PA'" 800 x 9.81 x 0.3 == 2J54.4

N on

-I.

Ans.

Ii

~ I

IL 156

Fluid Ml>chanics

.. 2.8

PRESSURE AT A POINT IN COMPRESSIBLE FLUID

For compress ible fiu ids . densit y (p) changes wilh tlie c han ge of pressure and te mperature . Such pro blems arc e nco un te red in aeronautic s. ocean og raph y and meteoro logy where we arc con ce rn ed with atm os ph eric * air where dens it y. press ure and temperature changes with e le vati o n. Thu s for fluid s wilh varia bl e densi ty. equatio n (2.4) canno t be integ rated. unless th e relatio nship betwee n p and p is kn ow n. Fo r g ~ scs th e equatio n of state is

1!.. '" NT

... (2. 15)

P -

p

p - NT dp

-

No w equat io n (2 .4) is

ilZ

=w=pg = -

P

NT

Xg

"I' = ...1...,/Z

... (2.1 6)

P RT In equat ion (2.4 ). Z is meas ure d ve rticall y dow n ward . But if Z is me asured ve rti call y up , th en d[1

-

"l

"" - pg and hence e quation (2. 16) beco mes

= .=L tlZ ... (2. 17) I' RT CotSI' I. [f temperature T is constant which is true for isot hC'rmalllroctip

2 .S. 1 Isothermal Process. tOSS. equ ati o n (2 .1 7) ca n be ill1cgral~ d as p

I

P.

til' '"

_jZL(fz= _L 1" RT

f' I'

log Po

~

-g

-

JZd:

RT Z.

IZ - 2;. 1

NT whcrepo is th e pressure where he ight is Zo. Ir the d atum lin e is Hlke n at th", pressure at d:llulil linc.

log L Po

=

Zo. th en Zo = 0 and 1'0 bo:co mcs

::.!. z NT

L = ('- ..11FT p, or pressure at a hciglit Z is give n by I' = PrI'- S7/KT' • • •( 2.18) 2 . B. 2 Adiabat ic Process. If tc m pc ralllrc T is not constant bu t the process follows adiabatic law th en the relati on be twee n pressure and density is g i ve n by I'

-t

= Conslanl = C

...(il

P • The standard atmospheric pressure. temperawre and dcnsily referred 10 STP at {he sea-level are : Pressure .. 101.325 k.N/m l; Temperature", 15°C and ();:,nsily = 1.225 I;:glm J.

I I

Ii

~ I

IL 57 1

Pressure ilnd its Measurement where k is ratio of specific constant.

".

. .. ( i i) Then equation (2.4) for Z measured vert ically up becomes. til' III tip

(~l Inlcgrating. we gel ) "

em I'

Ilk

=_pg= _(,,)II. g C

_ =-gdZur e

III

til'

~I~

_

=-8'/Z

pc

III: ill' = j Z_glfZ

'"

"

".

[C is 3 constant. can be laken inside I

But from eq uation (;), Substituting this va lue of

above. we gel

] ' [ -""' , ,. [ p~l'• = - g[Z - ZoJor [-'~ !C1P p

'"

-~

p

or

e llt

lZ - Z-o J = - g-

p 1 - - +1

l

!

/.:-1 P

P •

".

= - g[2 - 201

~

' [£,_PO] =-gIZ-ZoI

k-l

P

Po

If datum line is taken :Il 4" when: prcs.<;ure. temperature and density arc Po. To :md Po. then

_'~ [.f.._PO] = _ gZ k-IPP o

or

~I

or

4

= O.

l!.. _ ES!..= _ gZ(k-l) PPo

k

l!..=PO_gZ(k-l)",ft [ l _k-l gZfu] P Po k po k po

I~

~ I

IL Iss

Fluid Ml>chanics

i!.. xPn

0.

P Po But from equ at ion (il,

=[1+/::- 1gZfu] Po

l' '" Po

?p~

Substituting th e va lu e o f

£.£.

... (iii)

k

[&.)k '" ElLporP Po " ( "'-)'" P I'

or

in equation (iii). we gel

P

P

] (P )'" [ ] [ )-'" [ (Ll'-i"[L)';' =[, _'-:c',z&.] /::- 1 - Po " 1- - , - gZ 1'0

Po

I'a x

-

p

Po

0.

, -

P

/:: - 1

P

k

Po

,, 1- - - gZ.....2..

l'o

Po

l'o

.l!...."

k po

,

[1_/::-1 gZPo]N

Po k Po Press ure at a he ight Z from ground leve l is g iven by p '"

In equat io n ( 2.19), P o '" pressure

al

,

Po[1 - I fu]G

gZ k Po gro und leve l. where 1 0 '" 0 k-

... (2 . 19)

Po '" densiTy of air at ground leve l Equation of Slale is

Po : R1 0 0r££.: - ' Po

Po

RIo

Substitutin g the va lu es o f Po in eq uatio n (2. 19), we ge l Po

2

,

8- ]'-' 1'=1'0 1- [ '-k-' -RTo Temper
... (2.20)

2. B. 3 For the adiab at ic process. the lemperature al any height ill air is \' alc ulm cd as : Equ ati o n of stale at gro und leve l and al:t heig ht Z from grou nd levd is wri llc n as

Po -- Ri'0

-

Po

3n dJ!... -

Ri'

P

Dividing thcse cqualio ll s. we get

" O)+J!...=RTo = l n ( Po PRTT

".

I I

!..IQ

or

-Po x -P =-To

Po =&. x J!...=..E.... x Po po

P Po

I'

T .. .(i)

P

Ii

~ I

IL Pressure ilnd its Measurement But

L from ~qualion (2.20) is given by P.

591

•

L== [l_lr.- l /1Z ]i-=i k

Po

Also for adiabatic process

4 = p~

or

P Po

RTo

(!2]t

== ['0 J!

P

P; . [I;:]'.

0'

[:J'

. [1-' - 1,2](: ,].( ;) . [1-' - 1,z] k

Substituting the values of

L

RTo

£..2.. in cqu
:md

Po

1'0

· ·,

R ro

(I). we get

P

~

k

, =[1_ /.: - 1gZ ]t-i X[1_ k-1 /lZ ]-t:1 k

RIo

k

RTo

' [l _~gZ ]i=I-i-=i = [J_k-J k

Kin k

k

/lZ ]

K/0

1 2]

T= "' "o [ I- '-;-!To ...(2.21) Lapse-Rate ( L) . [\ is defined as the rate at which the te mperature

2. 8 .4 T~mperiiture changes with elevation. To obtain an expression for Ihe tcmpc ralllrc lapsc-ralc, Ihe IC lll pcrJllIrC given by cqumioll (2.21) is diffcrcmimcd with respect to Z as

:~: ,;~ [To (l_k;J :~J] ==

where To_ K, II and R arc constant

X70= -:(";1) -1I (k-l)

;~~ = _ k;IXR~o

The t
... (2. 22) (JT In cquation (2.22). if (i) k '" I whIch mcans isothermal process. '" 0, which means temperature dZ is constant with height. (ii) If k > 1. the lapse· rate is negative which mca[lS temperature decreases with the increase in heigh!. In atmosphere. the value of k varies with height and hence the valu~ ofte11lpcrJlure Ics unifonnly at the ratc of 0.0065"Chll. from 11000 1\1 10 32CXJO Ill, the telllp"rature rellla ins ronst3nt at - 56.5°C and hence in this range lapse·rate is :tero. Temperature rises again after 320Cl0 10 in air.

I I

-k

Ii

~ I

IL 160

Fluid Ml>chanics

Problem 2 .22 (SI Units) if rhe a/1II()Jpilere pres,~ure (1/ .~ea /el'e/ is 10. 143 Nkm ! , derami/Ie Ihe pressure m a ''''igM of 2500 III aS$lwling Ille pressure mriarirm follows (i) Hydros/a1ie lim', and J (ii) isolhermal law. The density of oir is gil'en as 1.208 1r.g1m • Solution. Given : Press ure at se a-level. Po = 10.1 4 3 N/c1111 = 10. 14 3 x I O~ N/rn l 2 =2500 111 1 !knsi ty of air. Po = 1.208 kg/nr (I) Press u re by hyd rostatic h, w. For hyd ro static law. p is assumed constallt and hence p is g ive n . dp by equalloll = - pg dZ Height.

In teg ratin g . we get

fPdp '" f - pg dz= - pg f .Zd Z Po

0'

For da\tJlll lin c at sea- le ve l.

70

P - Po= - pg IZ - Zol Zo = 0 P - Po =- pgZ or P=Po - pgZ = 10.143 x Itr - I.208x9.81 x 2500 [ ": P=Po= 1.208]

N m

7 ]804 10

'" 10 1430 - 29626 = 71 804 ---y or __,_ N/c m 2

= 7. 18 Ni em I. Am . ( ii) I'rcss ur e by Iso therm a l Law. Pressure at an y h ~ i g ht Z by isoth cnnal Inw is give n by equ ation (2. 18) as

= 10. 143 x 1 0~ e

-~

r.

-~

= 10. 143 X 10 4 e I'l> = 10 . 143 x I 0 4 e (_ lloOO .

) ,1(IB x 9,8) JllO.)43. )If'

= IOl4 30xe .291 = 10 1430x - ' - = 7574 3 Nlm2 1.339 1

75743

=~

Nlc l11

1

,

= 7.574

Z

N/cm • An s.

Problem 2 .23 The barOlllelric p,,·s,m re ar .fea lel'e/ ;,5 7601111/1 of lIIercury ..... hile Ihm on a 1II00WIlIin J lOp i.f 735 111111. If lite d<'fu;ry of air is a.f.fljllled COl/SWill m 1.2 kg/III , ..... har ;s Ihe eiel'arion of Ihe IIIOll1l1ain lOp? Solution. Gi ve n : Pressure· at s.c a ,

Po = 760 111111 of Hg = 760 x 13.6 x 1000 x 9.8 1 N/m2 = 10 1396 N/ml

'000 • Here pressure head (Z) is given as 760 mm of Hg Hence (pfpg) .. 760 mm of Hg. The density (p) for mercury '" 1J.6 x 1000 ~glmJ. Hence prcssure (p ) will Dc cqual to p x g X Z i.f .. 13.6 x 1000 x 9.81 x : : NI1111.

~ I

I~

~ I

IL 61 1

Pressure ilnd its Mea surement Pressure al mount ain.

p '"

735

11lI1i

of Hg

735 , = - - x 13.6 x Hloo x 9.81 = 98060 N/m-

lOoo

Iknsity of ai r .

,,=

p = 1.2 kg/ml

Let Height of the mountain from sea-level. We know Ihm as the elevation above the sea-level increases. the atmospheric pressure dc.::rcascs. He re the density of air is given conSlallt. hence the pressure 31 any heighl .". above the sea-Icvel is given by the equation. 1'=Po- pXgxh It = Po - II", 101396 -98060 = 21B ..B m . ADS. pXg 1.2x9.8 1

Problem 2.24 Ca/r"/ll/(' Ih" prnsllrt' at {/ 11I'iglll of 7500 II! nbol'f 51'(1 Inri if IIIf' 111lI1osplwricpress",.. is JO.14.i Nkm~ lind IfillpUaillre is 15°C (1/ IIlI' se a ·/{'\·e/. (lsslImillg (;) tli, is illcompn'Mib/(', (ji) pnSSllrl" WIr;m;OIl follows iso/hal/wi /m>l. mllf (iii) f"fS5Jf 1"1" \'(,,;111;011 /01/0,,"5 (II/;lIb(l/;(' low, T nkf' Ihe tlnlsit)' of
10 '" 15°C To" 273 + IS" 288°K P = Po = 1.285 kg/ml

(I) Prn.wn ,../1f/! (lir is i,u'OIll/'U'Hibll':

"p

,'Z = - pg

f Pdp = - f ZpgdZ '0 4 1'''l'o-pgZ

or I' - ['0 = - pg[Z - lol

[.:

lo"datumline"O)

= 10.143)( 104 - 1.285)( 9.81 )( 7500 = 101430 - 94543 = 6887 N/rn l " O.6~~~ , Ans, em "

(ii) f'nsslITl' l"(lrimioH follows iSOlIiI'rIll(lI/ow:

Using equation (2.18), we have

/, = PrI'

Jl/II.T

= f'o e ,lPn'po

= 101430 ,,-

{-:

~: = R T

:.

~: =

;A

Jl.(>rjPo = 101430,,-7500~ 12'!5~9811101~.\(I

= 101430 ('- .9.120 = 101 4 30 x .39376 = .W9.W N/m~ or .t99 .~ N/em! . Ans. (iii) Prl'.f$Urp mrim;,m fl,lI",,"s ",/i,,/wli(" I"",,: [k = 1.41

'I Using equation (2. 19), we have

I I

, where Po = 1.285 kg/m

l

Ii

~ I

IL 162

Fluid Ml>chanics

p = I 0 1430

[

1-

,. (1.4 - 1.0) x9.81)( -,( 7"5"00,,X .:,',.,2:::8," 5) J' .

_1.0

1.4

101430

'" 10 1430 II - .2662 1' ''''· == 10 1430 X (.7337) 3.5 , N '" .\4.'10 Nfm - or .'.431 - -, . AIl~ .

,m'

Problem 2.25 Ca/ndllll' lilt' prnsllre (//1(/ dn/sit)' Of air (1/ II Iwight of 4000 III from 5I'1I-/PI'l'lwl!frt' prf ss,,,/' (I/UI {('III/N' ralllrf' of /111' (lir (/rl' 10. 143 Nkm- (jill! J5°(.' rnpaliw/),. TIll' Il'lIIpff(llllrl' 1111'S(' raIl' is g;1'I'1I as OJXJ65°CIIII. Take
Tc mpc rmu rc at sea-leve l.

10= 15°C To'" 273 + IS == 288° K tiT L= = - O.0065° Kfm dZ Po::: 1.285 kg/ml

Te mperat ure lapse- rate.

Using

cq u ~t ioll

(2.22). we

=_1.(' -')

L= tiT

h ~l VC

dZ

R

k

_ OJI065= _ 9.81 ( k - ' ).WhcrcR= ~ = R

PaTo

k

10 1430 =274.09 L185x 288

x(k-')

- 0.(1065: -9.81 274.09 k k - ] = 0.0065 x 274.09 =0. 181 5 k

9,8]

kl]- . 18 15]: I - ' - '" 1.222 .1815 .8 184 This means that the va lue o r powe r ind ex k'" 1.222. (il I'r~ss ur~ at 4000 m he igh t i ~ givcn by cq uation (2. 19) as k=

k- '

pJ,"

P"'l'o I - - - ,Z - o

[

k

,w herck= 1.222 and Po'" 1.285

1'0 un

P

'" 10 1430 [ , _ ( 1.222 - 1.0 ) x 9.&1 x 4000 x 1.285 ]1.222 - [.0 1.222 101430 '" 1 0 1 430 [I - O.0915.~ ", I0 1430x.595 , N '" 60350 Nfm '" (j.O.~5 - • . Ans.

cm -

I I

Ii

~ I

IL Pressure ilnd its Measurement

63 1

(ii) I)" nsil y. Us in g eq uation of s tate. we get

,

.e = RT f' == Pressure at 4000 111 heigh t P == Density al 4000 III height T", Te mperatu re at 4000 III lic igh t Now T is calc ulated from temperatu re la pse-ralc as

where


lat4000m

= 10 +

-

P

Densi ty is given by

60350

3

3

P == RT = 274 .09 x 262 kglll1 '" O.K4 k glm . AilS.

Problem 2.26 All II1'rUpillllt' is flyillg III WI {,I/;Im'" of 5000 III. C,lIn,/ale IIlI' prf'SSIIrf' tlrol/llI/liI" Ql'rop/mw, g;\'1'1I liIl' IlIpSI'-rll/(' ill IIIl' atlllosphrrl' liS OlXJ65° Kim. N('gll'c/ "'tria/ioll of g wilh III/iIUdl'. TakF puss/lfe 11m! umprr(l/ilre (1/ grol/lld lewl (IS 10.143 Nlclll~ (///(I 15°C (llid dellsity of air (IS 1.285 kg/ow'. Solution. Given: Height.

Z=5000m

,rr

Lapse-rate.

1. == = - .0065 KJm dZ 1'0 = 10.143 x IO~ N/ml 15°C To = 273 + 15 = 288" K Po = 1.285 kglm)

Pressure at ground leve l.

Q

'0=

Density.

Te mperature at 5000 m hright " To + af x Height" 288 - .0065 x 5000

elZ ,,288 - 32.5" 255.5°K

First find the va lu e or power indcx k as From equat ion (2.22). wc have

__ .If = _ _ 1.. (k- l ) 1_ _ (rz H. k

_.0065 =_ 9!I ( k ~ l )

or where

H.

",...EE.....", Po To

10 1430 ,,274.(]9 1.285 x 288

_ .0065=_~(k-l)

The pressure is giv~n

274.09 k", 1.222 by equation (2.19) as

k

k-I

, ]1.',)

"=1'0[ 1--- gZ~ k

I I

1'0

Ii

~I

IL 164

Fluid Ml>chanics l,ll2

== 10 [430

[I_( 1.222 -

1.0 ) x 9.8 1 x 5000 x 1.285] '-:12-

1.222

'-0

10 1430 l.lll

'" 101430 [ 1- .222 x9.81 X 5000 X 1.28S j ,222 1.222 10 1430 S '" 101430 [I - Q.112S81 ,S
HIGHLIGHTS I . The press ure at any point in a fluid is defined as the force per unit area , 2. The Pascal's law ,tates that intensity ()f prc.. ure for a fluid .11 rest is equal in all direction •. ~ . l>ressure variation at a point in a fluid at reSt is given by the hydrostatic law which siate, thai the rate of

incre:''''' of pres,ure in the vertically downward direction is equal 10 the specific weight of the fluid.

tip _w*pxg.

dl 4. The pressure at any point in a incompressible fluid (liquid) is equal to the produ(,! of density of fluid at Ihm point. :lcccleralion duc to grnvity and verlical heigh! from free surface of nuid. p .. pXgxZ. S. Absolute pressure is the pressure in which absolute vacuum pressure is taken as datum while gauge pl'C'ssure is the pressure in which the atmospheric pressure is taken as datum. p_ = p _ + 1'_1< 6. Manomeler is a device used for measuring pressure at a point in a nuid. 7. Manomelers arc classified as (n) Simple manometers and (b) Differentia l manometers. 8. Simple manometers all' used for measuring PIl'SSUIl' al a point while differentialmanomelers all' used for measuring the difference of pressures belween the IWO poims in a pipe. or two different pipes. 9. A ~ingle column manomeler (or micromeler ) is used for measuring small pressures, where accuracy is required. 10. The pressure al" poim in SIalic compressible fluid is oblained by combining Iwo equ~lions. i.e .. equal ion of Slale for a gas and equ~lion given by hydroslalic law . I L The pressure at a height Z in a stalic compressible fluid (gas) under going isolhennal compression

p " Po c

where

I'~ "

rZlRT

Absolute pressure at sea-Iewl or al ground le\'el

Z .. Height from sea or ground le"el

R .. Gas conSlam T ~ Absolute lemperalure. 12. The pressure and lempcmlurc al a heighl Z in a SIalic compressible fluid (gas) undergoing adiabalic compression (plpi = consl. )

,

,

[ k-I . P"j" [k-I gZ ]I,'i

I'=Po l - - , - gZ 1'0

II

= /'oI- - '-RT

o

Ii

~ I

IL Pressure ilnd its Measurement T ~ To

and temperature,

65 1

'Z-l [1- -Hk- -RTo

"here Po. To are pressure and tcmpcmture at sea-level k" 1.4 for air. 13. The ratc at which the tcmrcr~ture changes with elevation is known as Tempermure Lapse-Ratc. It is given by

-'R ('-') ,

L .- -if

(i) k

E

I. temperature is Zero.

(ii) k > 1. lcmpemturc decreases with the increase of height.

EXERCISE (A) THEORETICAL PROBLEMS I . Define pressure. Ohtain an expression for the pressure intensity at a point in a fluid.

2. Slate and pro"c Ihe Pascal's law. J. What do you understand by Hydrostalic Law ? 4. D ifferentiate between: (r) Absolute and gauge pressure. (;/) Simple manometer and differential nmnom c!cr. and (iii) Piezometer and pressure ga uges. 5. What do you mean by vacuum pressure ? 6. Whm is a manometer ? How arc they dassified ? 7. What do you mean by single colum n manomClers ? Ilow are they u>ed for the mea,urement of pressure ?

8. What is the differencc between U·tube differential manometers and inverted U-lUbe differential manometers? Where arc they used ? 9. Distinguish between manometcrs and mechanical gauges. What are tlie different types of mc.:hanical pressure gaugcs ? 10. 1)erive an expres.
I'U ~PO [I_k-1 *

gZrlland T",TO[I_k-1 gZl

*

RTo whcre 1'0 and To are the pressure and temperature at sea-level. 12 . What do you understand by the tcnn. ·Temperature L1pse -Rate ? Obtain an expression for the temperature Lapse-Rat" . 13. What is hydrostatic pressure distribution ? Give one example where pressure distribution is non -hydrostatic. 14 . Explain hriefly the working principle of Bourdon Pressure Gauge with a neat sketch. [(fo

(J.N.1:U. , Hyderabad. S

lool)

(B) NUMERICAL PROBLEMS J. A hydraulic press has a ram of 30 em diameter and a plunger of 5 em di:unet,·r.l'ind the weightlificd by the h)·draulic press when the force applied althe plunger is 400 N. [Ans. 14.4 kNI 2. /I hydraulic press has a r~m of 20 em diameter and a plunger of 4 em di~lI1eter. It is used for lifting ~

weight of 20 kN. find the foree required al the plunger.

I I

[Ans. SOONI

Ii

~ I

IL 166 .\ .

Fluid Ml>chanics Cakul~tc th~

pressure due to a col umn of 0.4

In

of (u) wuter. (b) an oil of >1>. gr. 0.9, and (coJ mercul)' of!>p. gr.

"

13.6. Ta~e density 0( .....:l1cr. p _ 1000- 3 ,

m

jAns. (a) 0.3924 Nlcm', (b) 0.353 NiemI, (el 5.33 Nlemll

,

-" The pressure imensily at a point in a fluid i.< g;,-en 4,<) Nlcm', Find the corresponding height of fluid when il is ,(n) water. and (I,) an oil o f sp. gr. OX I AI~~. (n):; m o f waler. (b) 6 .25 In of oill

S. An oil of sp. gr. 0.8 is contained in a vess.cl. At a point the height of oil is 20 m. Find the correspondi,,)! height of ....:!tcr at that point. jAilS. 16 ml 6. An open I,mk contains waler UplO a depth of 1.5 m and above it an oil of sp. gr. 0.8 for a depth o f 2 m. FioJ the pressure intensity: (il at the interface of the two li'luids. and (ii) al Ihc bonum of the (a"k. IAn_,. (il 1.57 Niem I. (ii) 3.04 Nleml j 7. The diameter> of a ...mal l pislon and a largc pi'ton of a hydraulic jacl are 2 em and 10 cm respectively. A fo("(;e of60 N is applied On the small piston . Find the lo.1d lifted by the large piston. when: (tI) the pistons me at Ihc salHe level. and (I,) slllal] pislon is 20 Cllt above Ihe large pislon. The density of Ihe liquid in the jack is given as

II . ':I .

10.

II .

12.

1.1 .

".

1000~.

jAns. (,,) 1.500 N. (v) 1520.5 NI m Oclcnn;ne the gauge and absol ute pre"ure at a point which i'i 2.0 III below lite free -,urface of walcr. Take almo. . pheric pressure as 10.1043 Nlem'. IAns. 1.962 Nlcm l (g.1uge). IU)66 N"-1nl (abs .) I A simple manometer is used 10 measure Ihe pressure of oil (51'. gr. '" 0 .8) flowing in a pipe line. lis righl limb is open to the atmo'iphere and lefll imb;s connecled to Ihe pipe. The centre of the pipe is 9 cm be low the le'·cl of mercury (sp. gr. 13.6) in the right limb. Jfthe differcnce of mercury level in Ihe two limbs is 15 ~m. dclcnnine the absolute pressurc of the oil in the pipe in N/em". IAns. 12.058 Nlem'l A simple manomet.:. (U·wbe) ~ontaining mercury is connected to a pipe in which an oil of sp. gr. 0.8 is flowing. The pressure in the pipe is vacuum . The other end of the manometer is open to the atmosphere. Find the vaCUUIII. preS'iure in pipe. if the difference of IIIcrcur)' le,·el in the Iwo limbs is 20 cm and he ight I,\IIS. _ 27.86 Nlcm1 j of oi l in tbe left limb from the cenlre of the pipe is 15 em below. A single column vertical manometer (i.e .. micrometer) is conn«ted 10 a pipe containing oil of sp. gr. 0.9. The area of Ihe reservoir is 80 times the area of Ihe manometer lube. The reservoir C()Iltains lIIercury of 51'. gr. 13 .6. The level of mercury in the reservoir is at a height of 30 em below the centre of the pipe and difference of mercury levels in the reservoir and right limh ;s 50 CIII. Find the pressure in the pipe. IAlls. 6.474 Nlcm1 1 A pipe contains an oil of sp. gr. O.g. A differenlial manometer connected al tbe two points A and B of tbe pipe shows a differcnce in mercury level '15 20 cm . Find the difference of pressure at lhe lWO points. IAlls. 25113.6 Nlml ] A V·lube di fferential manometcr connects 11>'0 pressure pipes A and E. Pipe A contains cmbon tetrachloride baving 3 specific gmvity 1.594 under a pressure of 11 .772 Nlcm 1 and pipe R contains oil of '1'. gr. 0.8 undcr a pressure of I ].772 Nkm". The pipe A lies 2.5 m abo"e pipe 8 . Find the difference of pressure measured by merc"ry as fluid filling U·tubc. IAlls. 3 1.36 cm of mercury1 A diffcrcmi al manometer is conn«tcd at the two point.' A and R a., shown in I' ig. 2.25. Al fJ air pre"ure is 7.848 Nkml (abs .j. fi nd the absol ute pressure at A. IAns. 6.91 Nlcm1 j

OIL Sp.

gr. ~O.8

, -'" ,';-1 "'=

,~

i ,,~

1 -M ERCURY

Sp. g'.En.6

Fi g . 2.25

I I

OIL Sp .

T "''' = I

""

T

. '

~i~=

1.

~ ~Ei

~

WATER

Fig . 2.26

Ii

~ I

IL Pressure ilnd its Measurement

67 1

15. An in"crled differential manomcl~r containing an oil of sp. gr. 0 .9 is connected to find the diffuence of pressures at twO poinB of a pipe comaining waler. If the manometer rcading is 40 em, find the difference of prc«urcs. IAn_s, 392.4 N/m'l 16. In above ''is' 2.26 shows an in\'ened differential rna""mc!e, connected 10 lwo pipes A and B containing water. The fluid in manometer is nil of sp. gr. O.H. For the manometer readings shown in the figure. find the di fference of pressure head octween A and B. [,\US. 0 .2 6 m of waterl 17. If the atmospheric pressure at sea · lc"cI is 10.143 Nlcm l , dClcnninc the pressure at a height of 2000 m assl!ming that (he pressun: variation fol lows: (i) Hydrostatic law, and (U) lsothennallaw. The density of

air is given as 1.208 k8/mJ.

[,\ns. (i) 7.77 Niem I. (ii) 8.03 Nlem Jj

III, Calculate the pressure at a height of 8CXXl m abo,·c sea-level if the atmospheric pressure is 101 .3 kN/ml

19.

20.

2 1.

U.

2.1.

24 .

ZS .

Z6_

~ I

and tempemlUre is 15°C at the sea-level assuming (i) air is incompressible. (ii) pressure variation follows ~diabatic law. and (iii ) pressure vari"tion follows i>othermailaw. Take the dcn.ity of air at the sea -Ic,'el a. equal to 1.285 kg/m). Neglect "ariation of K with altitude. jAilS. (i) 607.5 Nlml. (ii) 31.5 kNlm l (iii) 37.45 kNlml ] Calculate the pressure and density of air at a hcight of 3000 m above sea-leve l where pressure and temperature of the air are 10.143 Niem I and IS oC res""clively. The lemperalUre lapse-rate is given as O.()(I6S" KIm. Take dCl1Sily of air at sea-level equal to 1285 kg/111 ). jAl1 S_ 6.896 Nfcml . 0.937 kg/1n J I An aeroplane is flying at an altitude of 4000 m. Calculate the pressure around the aeroplane. given the lapse-rate in the atmosphere as 0.0065°KJm. Neglect varimioll of 1/ with altitude. Take pressure and temperature at ground Ie"el as 10.143 Nlcm l and 15°C resp«tivcly. The density of air at ground level is gi"en as 1.285 kglnl '. [,\ ns. 6.038 NiemI] The atmospheric pressure at the sea"level is 101.3 kN/ml and the temperature is 15°C. Calcul"te the pressure 8CXXlm abo.'c sea-Iewi. aSSum ing Ii) air is incompressible. (ii) isothcnnal .·"riation of pressure and density. and (iii ) adiabatic variation of pressure and density. A,sume den,it y of air at sea-level a, 1.2!\5 kg/1n J . Ncgl""t varialion of '8' with ~Ititude. ]Ans. (i) 501.3 Nlm ' , (ii ) 37.45 kN/m '.liii) 31.5 kN111l1] An oil of sp. gr, is 0.8 under a pressure of 137.2 kN/ml (i) What is the pressure head cxpressed in melre of water ? (ii) What is the pressure head cxpressed in metre of oil? [AtlS. (i) 14 m. (ii) 17.5 m] The atmospheric pressure at the sea-level is 101.3 kN/m l and tC11l""ralUrC is I soc. Calculate the pressure 8000 m above sea-level. assumi ng (i) isolhennal variatio!1 of pressure and dcnsity. and (ii) adiabatic variation of pressure and density. Assume density of air at sea-level as 1.285 kg/m J . Neglect variation of 'g' with altitude . Derive the fonnula that you may use. IAII". U) 37.45 kNI11l'. (ii) 31.5 kN111l'1 What are the gauge pres.,ure and absolute pre.",ure at a point 4 m below the frec surface of a liquid of specific gravity 1.53_ if atmospheric pre.<>ure is equivalent 10 750 mm of mereury. [,\Ils. 60037 Nfm ' and 160099 N/m ' ] Find the gauge pressure and absolute pressure in Nltn ' at a point 4 In below the free surface of a liquid of sp. gr. 1.2. if the almospheric prcssure is equivalent to 7~ mm of tnCrcury [Ans. 47088 NI11l '; 147150 NI11l ' j II tank contains a liquid of specific gravity 0.8 Find the absolute pressure and gauge prcssure at a point. which is 2 m below the free surface of lhe liquid. The atmospheric pressure head is equivalent 10 7(1) mm oi" mercury. [Ans. 11 7092 Nf11l' ; 15696 Nfml j

I~

~ I

IL

I I

Ii

... 3. 1

INTRODUCTION

This chapter de~ls with the fluids (i.e .. liqu ids ~nd gases)
all a)"

wil l also be Zero. Then the forces actin g

on the flu id particles wi ll be : I. due to pressure of fluid normal to the surface. 2. due to gravity (or

... 1.2

sclf~weigh t

of fluid particles) .

TOTAL PRESSURE AND CENTRE OF PRESSURE

Total pressure is defined as Ihe force exerted by a sta ti c fluid on a surface either plalle or curved when the fluid comes in contact with the surfaces. This force always acts nonnal to th~ surface. Ce nt re o r pressu re is defined as the point of application o f the total pressure on the surf~ce. There are four ,",~ses of su bmerged surfaces on which the total pressure force and ,",entre of pressure is to be dctemlined. The submerged surfa,",es m,ty be : I. Venical plane surface.

2. Horizontal plane surface. 3. Inclined plane surface, and 4. CUfved surface .

... 1.1

VERTICAL PLANE SURFACE SUBMERGED IN LIQUID

Consider a plane venical surface of arbitrary shape immersed in a liquid as show ll in Fig. 3.1. Let

A", Total area of the surface

II '" DiMaiKe of C.G. of the area from free surface of liquid G '" Centre of gravity of plalle surface

P '" Centre of pressure h· '" Diswnce of centre of pressure from free surface of liquid.

69

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IL 170

Fluid Mechanics

(a) Tot a l P ress u ~ (1') . The total pressure 011 tlie surface may be determined by dividing the emire surface into II numbe r of small parallel strips. The force 011 small strip is then calculated and the tOlal pressure force on the whole area is calculated by illtcgrating Ihe force on smnll Slrip. Consider a strip of thickness til! and width /J al II depth of II from free surface of liquid as shown in Fig. 3.1 Pressure intensity On Ihe strip. I' == pgll

(See e quation 2.S) Area of the slrip.

dA",bxdl!

Total pressure force on Sirip.

dF=pxArca =pghxbxdh

FREE SURFACE OF lIaUID

"'T""'" ,-,',',',','"'"''1''1' ' , 1/ '\ " "L J, ! ,. , b

_

A

G.

Fig. 3.1

Total pressure force on lite whole surface. F= f dF= jpghXbXdh==pg j bXhXdh

""'

J bX/,Xdh '" J hXdA

'" Moment of surface area about the free surface of liquid

= Arc~ of sorfa<:e x Distan<:e of e.G. from free surfa<:e =Axll

F", pgAh ... (3 . 1) For wmcr th e value of p = 1000 kglm 3 alld g '" 9.81 mfsl. The force will be ill Newtoll. (b) C CIl I rt' of P ress u re (h t ). Centre of pressure is cal<:ulntcd by using the "Principk of Moments", which states thatlhe mOlllent of the resultant force about an axis is equal 10 the sum of mom en IS of the components about the same axis. The resultan t force F is acting at P. at a distance h· from free surface of the liquid as shown in Fig. 3.1. Hence moment of the force'" about free surface o f the liquid'" F X /1* ...(3.2) Moment of force dF. acting on a strip about free surface of liquid I ': dF=pghxbxdh! '" dF x II "'pghxbxdllx/i Sum of mom e nts of all such forces about free surface of liquid '" f pgllXbXdllXII=pg f bxhxlldh

",pg f bh~dh = pg f h ~dA

B",

(.:

bdh",dA)

fll ldA '" f billdh '" Moment of Inertia orthe s urface about free surface of liquid

'" '0

Sum of moments about fn"C surface

= pglo

I I

...(3 .3)

Ii

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IL Hydrostatic Forces on Surfaces

711

Equatin g (3. 2) and 0 .3), we get

F x /!*=pg ' o

B",

F = pgAl1 pgAl1 X h*

= pg fa

1,·:= pg'?.. pgAh

0'

=~

...( 3 .4 )

All

By Ihe th eorem o f parallel ax is. We have I O= IG + Ax h 2

where I e "M omc rll of ln cnia of ar(! 3 about an axis pass ing through the C G . of Ihe are a and parallel \0 th e fre e s urface of liquid. Substitutin g 10 in equat ion (3.4), we get

,_

•'o ,',-,: +~A~I<_' ~ + -II ... ( 3 .5 ) AI< AI< In equat ion (3.5), h is the di stan ce o f c.G. o f the area o f Ih e ve nica l surface from fre e surface of Ih e liqu id. He nce frolll equati on (3.5), it is c lear that : (i) Cent re of pressure (i.e .• 11*) lies be low the centre of gra vit y of th e ve rti cal surface. (il) The di stance of ce ntre of pressure from free s urface of liquid is independe nt of th e de ns it y of Ihe liquid . h* =

Tab le 3 .1

The m o ments o r in.orlia a nd olile r

Plime ,"utfOi'e

g~'tIm etric

CG. frum lilt

propcrliC5 or so m e im port an t p la ne surfaC<.'!l

Arto

,=, I

Moment of inerTia abowl ott axis passing through CG. (mil

Moment of inenia aboUi hal'/' (101

parallel to hme (ld

Rectangle

I

T --- -, I

r i 1 ,. G

d

x ~ -

2

bd

>d' --

>d' -J

,

bh' -36

-

1--' ----0<

"

Triangle

,/1\i . ~

x~

•

-

3

-bh

bh'

" COllld ..

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IL 172

Fluid Mechanics

Plull/'

c.G. from lhe

sur/au

1M",

I

Area

Momem of i"alia

/IIu"",,,,1 of

I'lbouIIITI axis lUlSS;IIK

i,jfrtiu aboul

rhrough C G. lInil

base (lu)

puralleilO b" se

(/01

3 Circle

)/ ,

Gj 1 ~

j"

.w'

d

--

.1'=-

2

4

.w' -64

-

4. Trapezium

fzL~IG~ I

"

x_(2l1+b)~ IHb

3

(u +b )

---d

2

1 1 (u +4ab +b )Xh1

36(o+b)

-

If

Pro blem 3 .1 A ,ec/(lIIg/liar plane .mrface is 2 III wide ami 3 In deep. II lie.! in I'errical plane in W
upper edge is horizollIal and (a) coillcides ",ill, waler surface. (b) 2.5 m below II,e f ree water surface. Solution. Given : Width of plane surface. b=2 01 d=) 111 [)cplll of plane surface. (a) Uppu edg e coincides with water s urfa ce (Fig. 3.2). Tot:,1 pressure is ~ i\'c n by equati on (3.1 ) f" = pgAI.

wh ere

P == 1000 kg/m

A ", 3 x 2 = 6

J

,

FREE WATE R SURFACE

11 == 9. 81 I11 /S1

,h

111 •

"' "2I (3) '"

15 m.

F = lOOOx9.81 x 6 x 1.5 '" 88290 N. A ns .

Depth of l'C rme of pressure is give n by eq uati on (3. 5) as I" AI<

-

11· =~ + ll

where

'G= 1'.1 .0.1. about e.G. o f the area of surface

Fig. 3.2

J

J

bd 2x3 4 " - " --=4.5m 12

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12

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IL Hydrostatic Forces on Surfaces

73 1

4.5 h· == - - - + 1.5 == 0.5 + 1.5 == 2.0 m. " ns. 6 x 1.5 (b) Uppe r l..t g e is 2.5 III bel ow water s urface ( Fig. 3.3). T01aJ prcssur~ (F) is given by (3. 1)

F == pgAh wliere

ii == ==

WATER SURFACE

Distance of CG. from free surface of wate r 2.5 +

'23 =4.0 III F = IOOOx9.8 1 x6x 4.0

= 235440 N. Ans. I

-

Centre of pressure is given by 11·==-4r+1l Ah whe re IG == 4.5. A == 6.0,

Ii

--- -.=- -..

11 °'-0'-'-- 1

.-,

i

2.5m

L:J=0G

3.0m

~O'--:-i J

== 4.0

f--2 m------l

--;;4~.5';-;, +4.0 11·= -;6.0x 4.0

Fig. 3.3

== 0.1875 + 4.0 == 4.[875 == 4.1H7S m. Ao s. Problem 3 .2 Determine the 10101 pressure on (l circu/(lr p/ale of di(lmelu /.5 m which is pluced reTfieally ill wuler in sitch II way lillI/ Ihe cenlre of 111/: plale is J In hdow I/ll! free ~'urface of ,.."Ier. Find the pUJ'ilion of en,lre of pressure (I/SO. FREE SURFACE

rU ---umuum ------ -- - j ---------

Solution. Given: Dia. of plate, II == 1.5 m

Area.

.-

It = 3.0 III Total pressure is give n by eq uati on (3. J).

F = pgAh

= 1000 x 9.81 x 1.767 x 3.0 N

3.0 m

GOL\r - - o ,

+

<-

= 52002.11 1 N. Ans .

Position of centre of pressure (11·) is g iven by equation (3.5). I"

-

h·=~ + h

Fig. 3.4

Ah

4

whe re IG

rtd It X 1.5 =-= 64 64

1----1.5 m --..j

4

= 0.2485 Ul~ 0.2485 1. 767 x ].0

+ 3.0 = 0.0468 + 3.0

'" 3.04611 m . ADS.

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IL 174

Fluid Mechanics

Problem 3.3 A reC/lwgl/lar .fluice gme is silUllleil QII lite I'u/ieal \\'01/ of a lock. Til<' I'errical side oJfhe sluice is 'd ' metres ill lellglh ami deplh ofcel1lroid of tile area is 'p' III below Iile W
Pro"e Ilwl Ille depth of pre:>5j,re is eq"ai /0 Solution. Gi ve n : [kplli of vc n kal gale

lei (h e w idth of gate

1

FREE SURFACE

' . p + -d ( /2p

11

t--- b

i Ll= ~

P

h'

r ij

= lI m ",b m

Are a. A=b x d m 2 [krill of e.G. fro m free surface

Fig . 3.5

h =p lll .

Le i 11* is the depth o f ce ntre of pressure from free surface. whic h is give n by equati on n.5) as bd l I,· '" ~ + h . where I e '" - '2 AI> I

_

[bd'/

11·= -

12

/J x,/ xl'

1+ P= - - +' d' 12p

d'

p + - - . Ail s.

12,

Problem 3 .4 A circular opening , J //I diameter. in (/ ref/rcal side of a lan k is closed b.y a disc of J //I diameler y,·/tich cem fo/ule about (I 1I0ri2O"I(,' diameter. Calculale : (i) Ihe fOTn~ UI! Ihe disc. alld (ii) Ihe IOrque required 10 maiMail! Ihe disc il! equilibrillm ill Ihe "erlie,,1 POSilioll whell the head of"''''er abol'e Ihe IwriWIII(II diameter is -I m. Solution. Given :

Dia. of openin g. Are a.

d=3m

"

A = - x3 = 7.0685 m-' 4

Depth. of e.G.. 1I = 4m (i) ['orce on (he di sc is given by eq uation (J .I) as f' =

pgAh = 1000 x 9.8 1 x 7.(l685 x 4.0

= 277368 N = 277.368 kN. Ans. (il) To find th e torqu e requ ired to ma inta in th e di se in equilibrium , first c alculate th e poi nt o f appli cati on o f force ac ting on th e di sc. i.e" ce ntre of press ure of th e force F. Th e depth o f cent re of pressure (//. ) is give n by equatio n 0 .5) as

I"

-

.!!.- d" c<64", · __

h·= ~ + ll = -;;

All d'

: d l )(4.0

+ 4.0

J~

: ~'--o,,+ 4.0= ~'--o" + 4.0 = 0.14+ 4.0 = 4. 14 m 16x4,0 16 )(4.0

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IL Hydrostatic Forces on Surfaces

B U

75 1

----------'1- TT

------------------------------ - --0_::-- 'm

".-

10 ,

,-

3m -' -

1..

,

fig. 3.6

The for(;c F is ad;"g at a distance of 4. 14 m from free surface. Momcll1 uf this fOKe about horizontal diam eter X-X

'" F x (h ·-ii) " 277368 (4. 14 - 4.0) '" 3HIBI

Nm. A IlS.

Hence a torque of )883 1 Nm mu st be applied o n the di sc in Ihe c loc kwise di rect ion. Problem 3 .5 A pipe line II'lIic/1 is
Dia. of pipe.

"=4 rn

OJ

19.6

Fig . 3.7

Area,

A=~x 42 =4 1ln1 2 4

S = 0.87 Density o f oiL Po = 0.87 x 1000 = 870 kg/m "' Weight dcnsily of oil. "'o"Po xg,,870x9.8 1 Nfm 3 Pressure at Ih~ ct;! ntre of pipe. P " 19.6 Nfcm 2 " 19.6 x 104 Nfm2

Sr . gr. of oil.

Pressure head al Ih e

cenlr~

p

19.6 x I0·

11'0

870 x 9.8 1

=-:

'" 22.988 m

The heig ht of equiva lent free oi l surface from the ce ntre of pipe" 22.988 Ill. The depth o f e.G. of Ih e gate va lve from free oi l surface II = 22.988 (i) Now Ihe force exef(",d by Ihe oil on Ih e gate is given by

nl.

F = pgAh where p = densily of oil = 870 kg/m 3 F = 870 x 9.8 1 X 4n x 22.988 = 2465500 N " 2.465 r.I N. Ails. (ii) Posi li on of !;enlre of pressure (".) is given by (3.5) as

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IL 176

Fluid Mechanics

'" 0.043 + 22.988 '" 23.031 m. Ans . Or celltre o f pressure is be low the ce lllTC o f Ih e pipe by a di st anc e of OJl43 1ll. AilS. Problem 3 .6 Determine tile lowl pressure atld celllre of pre.uure all WI i,I'osedes Irilmgular plale afbase" In and allilude -I In wllell il is immu.ied I'erlically ill WI oil of 5p. gr. 0.9. Tire base of Ille plme coincides ",ill, Ihe free .m rface of oil. FREE OIL t-- 4 m-------i

Solution. Given: Base o f plale. Height of pl~IC, Area.

Sp. gr. of oi L

,

b=4m 11=4111

~

A: bxh '" 4)( 4 =8.0 m2

2

2

s= 0.9

Fig. 3.8

l DenS it y of oil. P '" 900 kg/m , The dis tance o f e.G. from free surface of oil.

I I II = -xh=~)(

3

3

4

'" 1.33m.

Total pressure (F) is give n by F = pgAT. '" 900 x 9.8 1 )( 8.0 x 1.33 N '" 9597.6. N. AilS. Ccmrc of pressure (11* ) from free surface o f oi l is given by

II+= 'G..+h AI<

where fc = M.0.1. of triangular section about its e.G.

bI/ 4 x4 l = - - = - - - =7. 11 m4 36

II. =

7.1 1

36

8.0 X 1.33

+ 1.33 = 0.6667 + 1.33 = 1.99 m . AilS.

Problem 3 .7 A "ulical sluice gale is used IQ CQ l'er an opl'liilig in a dam. The opnlillg is 1 m wide and 1.1 m /Iigll. 011 111l' IIp_lIream ofl/l e gale. I/Ie liquid of sp. gr. 1.45. 1il'.1 "1110 {/ Iwig/II of 1.5 m abO"e I/Ie lOp of I/Ie gate. "'/Ierea.lon I/Ie dO"'II_lIream .fide Ihe waler is ami/able IIplO {/ heiglll 101lch· illg Ihe lap of Ihe gate. Filld Ihe resulumr force acring on Ille gate alld pQ.lirioll of celi lre of pressure. Find also Ihe force aCling horizolltally at rhe fOp of Ihe gale .... hich is capable af opening ir. Assume Ilwl rile gille is hinged all/U' bOl/om. Solullon. Given: Width of gale. b = 2 III [kplll of gate. d=1.2 m Area.

Sr . gr. of liquid

I I

A =b xd= 2 x 1.2= 2.4m 1

= 1.45

Ii

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IL Hydrostatic Forces on Surfaces PI = 1.45 x 1000:: 1450 ~ gJllI '

De ns it y of liqui d .

The force P, is g ive n by where

77 1

FI

= Poree exerted by the fl uid of sp. gr. 1.45 un ga te

F~

= POfce cxen cd by wa ter on the gate.

F(:: Plg)( A x

h,

PI'" 1.45 x 1000 '" 1450 kgl m2 h , '" D epth of e.G. of gate f rom free surface o f liqui d

:: 1.5 +

21. 2 :: 2. 1 m. FREE SURFACE OF LIQUID

UQUIOOF

Sp . gr.=1.45

FREE SURFACE

F OF WATER

UPSTREAM

DOWN STREAM HINGE

Fig. 3.9

F,= 1450 x9.8 1 )(2.4 x2. 1 =7 169 1 N Similarly. l where p! '" 1.000 kglm

F2 :: p~ . A;;l

/,!:: Depth oren . of gmc fro m free surface uf watcr ",.!.xl.2=O.6 m 2 F 1 :: 1000 x 9.81 x 2.4 x 0.6:: 141 26 N ( i) Rl."sultanl fo r ce un th e gule '" F, - F2 = 7 169 1 - 14 126 = 57565 N. A ns. (ii) Pos ition of ce ntre of pl't'ssure of ...,s uUan! fo rce. Th e fo rce PI will be acting al a de plh of 11, * from free surface of liqu id. g iven by the rel ati o n

hl* =

I,

-

~ +h , Alii

where

hl*=

.288 +2. 1 =0.057 1 +2.1=2.157 1m 2. 4 x 2.1

Di sta nce of FI from hinge = ( 15 + 1.2) - h 1* '" 2.7 - 2. 1571 = 0.5429 m The force F2 will be actin g at a dep th of 11 2- f ro m free surface of wate r and is given by

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IL 178 where

Fluid Mechanics

4

-

IG = 0.288 rn , /'1

= 0.6

JIl.

II!- '"

A = 2.4

2 10 •

0,288 + 0.6 = 0.2 + 0.6 = 0.8 2.4 xO.6

III

Distance of F1 from hinge '" 1.2 - 0.8 = 0.4 III The r~sul ta nt force 57565 N will be acting at a distance given by

= ,7,1069clcXc.o54~2,9~-ci-l4cl,206cXcO",.4 57565 =

3392 1- 5650.4

m above hi nge

57565

= 0.578 m abo ....~ the hin ge. An s. (iii) Force at th e 1011 of gale ",hich is capa ble of o pe nin g the ga le. Let F is tile force required on Ihe lOp of the gmc to open ;1 as shown in Fig. 3.9. Taking the moments of F. 1', and F1 about Ihe hin ge. we get Fx [. 2 + 1'2)(0.4 =F, )(.5429

F=

or

=

f; x .5429 12

,.~

-

x 0.4

7169 1 x .5429 - 141 26 x 0.4

38921 - 5650.4

1.2

1.2

= 27725.5 N. Ail S. Problem 3.8

A caisson [or closing tile elllrance 10 Ii dry dock iJ' oflrapezoidalform 16 III wide a/ Ille

rop wId 10 III wide a/ lire bOl/om and 6 III deep. Filltlille lotal pre~'SlIre allli celltre of pres.~I1'" Otl Ille caissoli if IIII' water 011 IIII' o"lsidf' is j"~'1 /ne/ ...ill> IIII' lop and tlock is emply.

Solution. Given: Width al top = 16 m Width al bottom =lO m Depth. tI=6 m Are a of tr apezoidal ALJCD.

WATER SURFACE

,..._ _ _ t 6m A

,

=:\ ' 1

'0

A = ,I~ B C,-;;+~A~D") x If 2 Fig. 3.10

"

180 + 36 78 = 2.769 m from water surface.

(i) Total P....,ss ure (F). Totill pressure. F is given by

I I

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IL 79 1

Hydrostatic Forces on Surfaces ,.. '" pgA h

'"

1000 x 9.8 1 x 78 x 2.769 N

'" 2118783 N '" 2. 11 87113 MN. A ns . (ii) Cen trt' of ]'ressurt' (ht). CcmTe of pressure is II~ '"

I

~iven

by

(3.5) as

cq u ~lion

-

c:.. + II

AI,

where fG = M.O .I. oftra[X:zoidal ABeD about its c.G.

'c;, '"

M.O. I. of rectangle FBeE about its c.G.

IG, '" M.O.1. of t wO tos ABF iHld ECD

bd J

'r;, '" [2=

Then

IOx6 12

~bout

its CG.

~

J

= 180

In

1<;, is the M. O.1. of the recltmgle about the axis passing through G,. M.O. I. o f the rectangk atJ.oul the axis passing th rough the e.G. oflh.e trape zoida l 1(;, + Area of I1'c[ang le x .f,l where .r , is distance between the CG. of rectang le and e.G. of trapezoidal '" (3.0 - 2.769) '" 0.231 III M .0.1. of FIJCE passing through CG . of trapezoi dal '" 180 + 10)(6)«0.231 )2 ", 180+3.20= 183.20m4

Now

IG, = M.O.1. of MBD in Fig. 3. 11 about

=

(16 - 10) X6 3 3b

=36 m

G~

=

bd' J6

~

The distance between the e.G. of triangle and e.G. of trapeZOidal '" (2.769 - 2.0)" 0.769 M.O. I. of the two ~~ about an axi s passing through C.G. of trape zoi dal == la, + Area of triangles x (.769)2 6 x6

= 36.0 + - - x (.769)

I

10~64

Ie

== 36.0 + = 46.64 = M.O.1. of trap.:widal about its e.G. = M.O.1. of rectangle about the CG. of trapezoidal

=

F. E

A == 78.

Ii

I ,

-

-4. + II Ah

0

;---T----;~ 2

G.,.----- ____L I

j

6m

+ M.O.1. of triang les about the e.G. of the trapezoidal 4 183.20 + 46.64 = 229.84 m II" ==

where

A

_

"C Fig. 3.11

== 2.769 h* "

78x2769

+ 2.769" 1.064 + 2.769" 3.833 m . Ans .

Alter nate Met hod The diSlance of th e C.G. of the trapezoidal channel from surface AD is given by (re fer to Table 3.1 on page 7 1)

I I

Ii

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IL Iso

Fluid Mechanics x=

=

(2(1 + /1) (a + b)

II x-

3

(2xlO + !6)

(10 +1 6)

6 )( 3

a= IO,/>= 16andlr=6)

(.:

'" 36 x2=2.769In 26

This is also equal ro th e di stance of III" CG. o f the trapezoidal from rn:" surface of water.

Ii '" Total pressu re.

2.769 m

F = pgAiI

(":

A = 7S)

'" 1{lOO x 9.81 x 78 x 2.769 N '" 211H783 N. AilS. CCIHrc o f Pressu re.

Now Ie; fro m Tab le 3 .1 is g ive n by.

(a 1 + 4ab + b 1) } (10 1 +4 )(IO XI 6 + 16 1) l IG", 36(a + b) xii '" 36 (10+ 16) x6 '"

(100 + 640 + 256) 36x26

x216=229.846m ~

229.846 + 2.769 78 x 2769 '" 3.1B3 m. AilS. A lrapezoidol channel 2 III wide at Ille borrom (lnt! I m deep lla.! side slopes I : I. 1/. '"

Problem 3 .9 De/ermine: (i) Ihe lOW/ pressure. and (ii) the cenlre of pressure 011 the renieal gme closing rhe chalille! whell il is fil II o/water. Solution. Gi ve n : Width al bonom !kplh. Side s lopes .. Tupwidlh. Arcil uf rectangle FBEe,

=2 m

t - --

d= lm '" I : I

.

m

-

A WATER SUR FACE

-<

,

AD==2 + 1+1==4m AI'" 2 x l '" 2 m!

(4 - 2) 2

,

Area of TWO Trian g les ABF and ECD. A, '" - - - x I '" I 111-

.

Area uf Trapezo idal ABeD , A == A, + A1 == 2 + I == J m 1 DepTh uf e.G. of rl'CT angl e FBEe from waTer surface , ..

/1 1

I I

Fig. 3.12

I

"' "2 ", 0.5 m

Ii

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IL Hydrostatic Forces on Surfaces

811

IXplh o f CG. of two triangles AUf' and ECD from water surface. I I II I " "3 x I '" III

"3

Depth of CG. of trapezoidal ABeD from free s urface of wmcr -

-

A,xh , +A , x/1l II == •

(AL +A!)

2 x 0.5+ I x 033333 " .44444 (2 + I)

(i) Totu! Press ure (Fl. TOlal pressure F is given hy

F == pgA h

= IOOOx9.81 x3.0 x 0.44444 = 13079.9N. Ans. (ii) Cenl ..., of Pressure ( h*). M.O .I. of rectangle FB eE aboul its CG., bd J 2x] J I fo = -- = -- = - Ill~ I 12 12 6

M .0 .1. of FBeE abou t an axis passi ng through the e.G. of trapezoidal

'c,· 'c, ==

+ AI x [DiSlancc he tween CG. of rec tangl e and CG. of lra p.:zoidalf

I

,

'" - + 2 x 10.5 - .4444 1- " .1666 + .006 182" 0. 1727

6 1'.1.0.1. orl he twO triang les ABF am.! ECD about th eir C.G .•

M.O.I. of the

IWO

triangle s aboul1hc CG. o f trapezoidal. I e: == I e, + A2 x [Distan ce bet wee n CG. of triang les and e.G. of lrapczo idall l

'"

1~+ IX [h-h2r "'~+IX [.4444 -*r

'"

l~

+ (.1 111 )1",0.0555+ (. 111l)1

'" .0555 + 0.01234 '" 0.06789 m 4 M .a.]. of lite lrape zoi dal aboul il~ e.G.

' o '"

10,

. + 10 , • '" . 1727 + .06789 '" 0.24059 m

4

Tlten centrc of pressu re (h*) on lltc vcrtica llrapczoidal.

11* '" "'- + Ii Air

'"

0.24059 + .4444 '" 0.18()4{, + .4444 '" 0.6248 3x.4444

'" 0.625 m. AilS.

I I

Ii

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IL 182

Fluid Mechanics

Altt'rnate M eth od

The diswncc oflhe C.G. oflhe trapezoidal channel frum su rfa ce AD is given by (refer to Tab le 3.1 on page 7 1).

x=

(2a + b)

«(I +b)

II (2)<2+ 4) I x -= x-

3

(2+4)

( .;

3

a=2,b=4andh= I )

== 0.444 m

h =X=

0.444 III

F=pgAh= 1000><9.81 >< 3.0 x.444

Total pressure.

(":

A

= 3.0)

= 13079 N. Ans .

'"

h*=-b+I, AI,

Cemre of pressu re. where 1(1 from Tab le 3

is given by Ie;'"

(a l + 4I1b + b'-) 36(II+b)

0.2 407 3.0 x.444

3

xII =

(Zl + 4xZX4+4 l) 36(2+ 4 )

3 52 4 x l = - - - =01407111 36x6

+ .444 '" 0.625 m. Am .

Problem 3.10 A sqlwre oper7llre in the rerrical side of a ((III/.: Iws Dill' diagonal I'errical and is camp/cleiy col'ered by a p/alle plate Ilil/ged a/ollg olle of Ihe upper sides ollhe aperlllre. The diagonals of Ihe oper/ure (Iff! 2 III /ollg and II,e Ilmk COll10illS Ii liquid of specific gral'ily 1./5. Th e cerllre of aperture is /.5 /II heloK' r/,,,, f ree surface. Co!c,,/rlle the thrusl exerled Oil Ille p/ale by 1111: liql/it! (lnt! posililm of ib aI/Ire of pressu re. Solution. G ive n : Diagonals of aperture. AC", lJD", 2 m Area of square aperture. A = Area of ~ClJ + Area o f 6ACD ACxlJO

2

ACxOD

+

2

Sp. gr. of liqu id = 1.15 Density of liqu id. p: 1.15x 1000= 1150kgfm 3 Depth of centre of aperture from free surface.

Ii '" - .,

' . '. ' . '_

1.5 Ill.

,

.. -" ' . ' , '

SQUARE .6.PERTURE

1.5m

Sp . !/f.,1.15

Fig. l.U

I I

Ii

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IL Hydrostatic Forces on Surfaces

83 1

(i) The thrust on th e plate is give n by

F", pgAh = 1150x9.8 1 x 2 x 1.5 = 33844.5. Am. (ii) Centre o f pressure (/1·) is given by

Ie I,· '" ---4" + II AI<

where

'G= M.O.I. of ABeD about diagonal AC = M .O.l . of triangle ABC about AC + M.O.! . of triangle ACD about AC ACxOB l

=

="",,'-+ 12 2 X 13

="""'ACxOOJ

2 x] J

(.: M.a .l.of a tria ngle aooulits base '" b/;J)

12 I

1

12

12

6

6

~

I

= --+--=- + - = -

3

h· '" - '-

III

2x l.5

+ 1.5 '"

3x2xl.5

+ 1.5", 1.6 11 m. An s.

Problem 3.11 A /allk cO/lwills waler lipta a heighl 0[0.5 m obare fh e base. All immiscible liquid of sp. gr. 0,8 is filled an ,lie lop a/ willer "plo I m height. C(l/cuhlle ; (i) lollli presJ'ure Oil Olll' ~'ide ofille Illllk, (ii) the pOSilioli of C,,"lre of prcn'ure for 011t' 'iide vf Ihe /,mll., which is 2 In wide. Solution. Gi ven: [krIll of wate r '" 0.5 m ", 1m De pll! of liqu id Sp. gr. of liquid '" 0.8 [kosity of liqu id, P I '" 0.8 x 1000 '" 800 kg/m l 1 [kosity of water. P2 '" 1000 kg/m ",2 m Width of tank (i) Tota l press ure on on e side is ca lculaK'd by drawing pressure diagram. wh ic h is shown in Fig. 3.1 4. Intensity o f pressure 00 tOp. P.. '" 0 Intensity of pressu re on D (or DE). PD'" Pl g.h l ", 800x9.81 x 1,0=7848 N/n/

,

, Fig. 3.14

Intensity of pressure Now force

00

hasc (or

no.po '" PlgiJ l + P2E

x 0.5

'" 7848 + 1000 x 9 .8 1 x 0.5", 7848 + 4905 '" 12753 N/m 2

F t '" Area of MDE x Width of tank

I I '" - x AD x DE x 2.0 = - x 1 x 7848 x 2.0 '" 7848 N

2

I I

2

Ii

~ I

IL 184

Fluid Mechanics F 2 = Area of rectangle DI1fT x Width of lank

Force

=O.5x 7848 x 2 = 7848 N F J = Area of !:.EFe x Width of lank

'" "2I )( EF x Fe)( 2.0 '" "2I x 0.5 x 4905 x 2.0 '" 24525 N F = F, + F, + F}

:. Total pressure.

'" 7848 + 7848 + 2452.5 '" 18148.5 N. AilS. (ii) Cen tre of Pressu re (h. ). Ta kin g the Illoments of all force abou t A, we get

Fx

,,* '" F, x ~3 AD + F,- (AD + ~2 80)+ F,[AO+ ~ 8D1 " 3

18148.5 X /1* = 7848 x

~

x 1 + 7848 (1.0+

O~5 ) + 24525 ( LO + ~X.5)

= 5232 + 9810 + 3270 = 18312 •

183[ 2 == 1.009 m from top. Ail S. 18148.5 Problem 3.12 A cubicol /(Ilik l/as sides of 1.5 III, II coll/oins '\"(Iter for Ihe lower 0.6 In depllt. The upper remaining pori is filled wilh ai/ of specific gT(I\';I), 0.9. Cu/cr/Ii'/e Jor aile rerticol side of I/Ie lallk: II =

(u) tolill pT(!SSrtre, Ulld

(h) /l05;lioli 0/ c<,,, lre of preSJ'" reo

Solutio n. Given:

Cubi<.;al wnk of sides 1.5 I:kpth of wa ter I:kpth of liquid

In

means the dimensions of the tank :Ire 1.5 m x 1.5 m x 1.5 tn.

Sp. gr. of liquid I:knsity of liqu id .

=O.6m = 1.5-0.6 =0.9 m = 0.9 PI = 0.9 x 1000 = 900 kglm J

I:knsity of water.

Il l '" 1000 kg/m>

(a) 1'01,11 pressure on one veni cal side is ca lcu lated by drawing pressu re diagram. which is shown

in Fig. 3. 15. A

OIL OF SP. GR . ~O . 9

,

--

.

- -

- -..

--- - ---- - ~- .

I:

I'

,

,_

7946.1

IF 5686

1383i .t

,

I

:,

Fig. 3.15

I I

Ii

~ I

IL Hydrostatic Forces on Surfaces

85 1

hucnsity of press ure a1 A. p~ = 0 Intensity of pressure ~t D,PD = P ig X II '" 900 x 9.81 x 0.9 '" 7946.1 N/m 2 Intensit y of pressure al 8, Po = P igill + P2gill '" 900 x 9.8 1 x 0.9 + 1000 x 9.81 x 0.6 = 7946. 1 + 5886 = 13332. 1 Nlm l Hence in pressure diagram: DE: 7946.1 Nlml . BC = 13332. 1 Nlml, FC= 5886 Nlm l The press ure diag ram is split into Iriangle AD£. rectangle BDEF and trian g le EFe. The total pres· sure force consists of (he following components: (i) Force FI = Area of lrian glc ADE x Width o f tank

= (t XADx DE) x 1.5 ==

(t x 0.9)( 7946.1) x 1.5 N = 5363.6

(,:

Width ", 1.5 m)

N

,

This for<:e will be acting at the CG. of tile triangle ADE. i.e .. al a di>!ancc of ~ xO.9 = 0.6 m below A (ii) Force

Fl = Area of rectangl e BDEF x Wilhh of tank = (BD x DE »( 1.5 = (0.6 x 7946. I) x 1.5 ", 7151.5

Th is force will be acting at the CG. of th e rectangle RDEF i.e .. at a di s tance of 0.9 + 0.6 '" 1.2 m

2

oclow A . ( iii) Force

F3

",

Area of triangle EFC x Width of tank

'" (t x EFx

FC)

x 1.5 '" (t xO.6x58&6) x 1.5 '"

2648. 7

This force will be acting at the CG. of th~ triangle EFC, i.e .. at a di stance of 0.9 +

N

"32 x 0.6 '" 1.30 m

below A. Total pressure force on one ve ni caJ face of the tank. F", F, + f'2 + FJ '" 5363.6 + 7 151.5 + 2648.7 '" 15 163.H N. Ans. (b) I'os ition of ce ntre of pressure Let the lOtal force F is acting at a depth of 11* from th e free surface of liquid. i.e .. from A. Taking the mom ent s of all forces :Ioout A. we gd Fxh* = FI xO.6 + F!x 1.2 + F3X 1.3 0'

11* '"

FI x(t6+ F: x 1.2+ fj x 1.3 F

5363.6)( 0.6 + 7 151.5 x 1.2 + 264& 7 x 1.3 15163.8 '" 1.005 III from A. An s .

=

.. 3.4

HORIZONTAL PLANE SURFACE SUBMERGED IN LIQUID

Consider a plane hori7.ontal surface immersed in a stalic nuid. As every point of the surface is at the same depth from the free s urface o f the liquid. the pressure intensity will he equal on the e ntire s urface and equal 10. {J '" pgll, where II is depth of surface.

I I

Ii

~ I

IL 186

Fluid Mechanics FREE SURFACE

A", Total area of surface

'-'F'-'-'-'-'-'-'-'-'-,-,-,-,-'F

Then total force. F. on the surface

h

1\'

"'pxArc3= pgx/! xA '" pgAh where II '" Deplh orCG. from fr~c surface of liquid == /, also Depth of centre of pressure from free surface", II. Problem 3.13 fig. 3.17 5//011'5 a /(lIIkfull a/water. Find: (i) Towl pressure on rhe bOI7OI11 of tank.

I,' '"

Fig. 3.16

(il) Weight oj lrlljer ill the I(I/Ik. (iii) Hydros/alie parado;r be/wee" 1/11' THlllls of (i) (md (i;). lVidlil of /(mk is 2 III.

Solution. Given:

OAm

[Rplll of Wala on bonom of

tan~

----

h,,, 3 + 0.6 '" 3.6m Width of lank

'" 2 m

Length of tank at bottom

'" 4 rn

Area 31thc bonom. A : 4)(2" 8m 1 (i) Total pressure F, on the boltOlli is F:pgAI!" (000)(9.81 x8x3.6

I·

= 282528 N. An s. (ii) Weight of water in tank", pg x Volume of tank

•I

'm I 1

'm Fig . 3. 17

'" 1000 )(9.81 x P x 0.4 )( 2 + 4 x.6 x2] = 1000 x 9.&1 12.4 + 4.&1 '" 70632 N. Ans . (iii) Fro m the results of (i) and (ii). it is observed tlial the total we ight of water in tlie tan~ is much less than the toml pressure at the bottom of the tank. Tliis is known as Hydrostatic paradox .

... l .S

INCLINED PLANE SURFACE SUBMERGED IN LIQUID

Consider a plane surface of arbitrary shape immersed in a liquid in sucli a way that the plane of the su rface mak es an angle 9 witli the free surface of the liq uid as sliown in Fig. 3.1 &. FREE LIQU ID SURFACE

Let

Fig. 3.18 A '" Total area of inclined surface

Indined immersed

~urface.

" '" Dcptli ofCG. of inc lin ed area from free surface ". '" Dist ance of centre o f pressure from free surface of liquid 9 '" Angle made by the plane of the surface with free liquid surface.

I I

Ii

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IL Hydrostatic Forces on Surfaces

87 1

Let th e plane of the surface. if prod uced meet the frce liquid surface a1 O. Then 0-0 is the axi s perpendicular to the pl3n c of the s urface .

y = distance of the e.G. of Ihe inclined surface from 0-0 y. = distance of th e centre of pressure from 0-0. Consider a s mall strip of area dA at a depth 'II' from fret: surface and at a distance y from tlie axi s 0 -0 as shown in Fig. 3.18. Pr~ssure ill1cnsil y 011 th e strip_ p = pgh Press ure force. dF, on the strip. dF = p x Area o f slrip = pglt x dA

Total pressure force o n the who le area, F",

But from Fig. 3.18,

J J tlF =

pghdA

~ = ~=~ :sine

)'

y

y.

h=ysin 9

F = j pgx yxSin9xdA : pgsin9 f Yi/A

jFM=AY

B" whae)'

= Distance orCG. fro m axi s 0 -0 F = pgsin9 yxA

= pgA l1

(.;

h= y sin 9) . .0 .6)

Cenl .... of Press ure (h·) Pressure force o n lhe strip.dF = pglldA = pgy sin dA Moment of the force. IIF. ahout axis 0·0 = IIF x Y = pgy si n e dA x y = pg sin Sum of mom e nts of all such forces about 0-0

e

J

= pg ~i n

B"

J/ dA =

e yl dA = pg sin eJi

y*

Now

= 'P~'c'c;'~'cBc ,. f

F

".

e]

e lilA

dA

M .O.1. o f th e surfac.: about 0-0

Sum of moments of all forces about 0-0 = pg sin e 10 Mom.:nt of th.: total force. F. aoou t 0 -0 is a lso given by = Fx y* where y. = Distance of centre of pressure from 0-0. Equati ng the two va lues give n by equatio ns (3.7) and (3.8) Fx y"= pgs in9,o

III = y sin

= 10 ..•(3.7) ... (3.8)

...(3.9)

y. = - - . F= pgAI!sin 9

and ' 0 by th e theorem of parallel axis = IG + A~, l .

I I

Ii

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IL 188

Fluid Mechanics

Substituting llicsc va lu es in eq uatio n (3.9), we get

~ -_ PKSine[1<) + A.>C'j sin

e

pgAII

si n ~ e -, I,· = --_[Ie + Ay - ] AI,

I<

B",

==si n9 y

I,· '"

or

I, y= - si n

""'o[

~

AI<

e e1

IG + Ax - .-h',Sm

• Ic si n ' S II = + 11

"'

... (3 . 10)

AI<

If

e '" 90 °, equatio n (3.10) hecomes same as eq uation (3.5) which. is app licab le to vert ic all y plane

submerged surfaces.

In equation (3.10), Ie '" M.O.1. of inc lined surfaces about an ax is passing th rough G ami parallel to 0 -0 . Problem 3.14 (a) A fec/ung,,{ar pit", .. J"ur!"ce 1 m wide and 3 m d"ep!ies in ,mler iii J'uc/! tI w"Y tiull il$ plmle makes ' m (lIIgle of 3~ " Will, tile free SlIr/llce of waler. Determine Ihe total preSSlue and 1'0,\';1;011 of [enlre of pre.ISllfe when lile Ilflper edge is /.5 III below IIIl' free water surface. Solu t ion. G iven:

FREE WATER SURFACE

Width of plJne su rface. b = 2 m Depth. 11= 3m Angle, = 30" Di~tance of upper edge from free wa la surface = 1.5 m (,) Total pre._~ ure fo rce is gil'en by equa li o n (3.6) as

e

o

".

F= pgAIi where p = 1000 kglm 3 z A=bxd=3x2=6m

Ii

= Depth of c.G. fmm free wate r surface

'" 1.5+ 1.5 sin 30°

Fig. 3.19

'" 1.5+ 1.5xt",2.25m F '" 1000 x 9.81 x 6 x 2.25 '" 132435 N. Ans. (ii) Centl't' or press ure (h·)

Using eq umion (3. 10). we ha ve bil l 2X]3 where I e = - - " - -.- = 4.5 m4 12 12

" C',,30,-' +.5= 22 /r*= -,40.5C'C'O'C

1 4.5x 4

6x2.25 6 x 2.25 = 0.OS33 + 2.25 = 2.3333 m. An s.

I I

+ 2.25

Ii

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IL Hydrostatic Forces on Surfaces

89 1

Problem 3.14 (b) A rec/angI4/ar pia/ie surface J III wide alld 4 III deep lies ill water ill sudl a way (lwl ils plalle makes atl ollgle of 30" willi Ille free SlIrface a/water. Determine rhe lotal pressure force and position of cenlre of pressure. when the upper edge is 2 III below II,e free surface. Solution. Gi ve n :

Free sufface of water

b", 3 m . d == 4

Ill.

e" 30°

---:".-.

Dist an ce of uppe r edge from rrce surface of wa ter == 2 III (i) Tolal press u re fon:e is give n by eq uati o n (3.6) as F= pgAh

where

and

V_ """""I

J

p = 1000 kglm , A :bxd = 3x 4 =

h '"

to plate [2 m 2

Depth of e.G. of platc fro m rrce wate r s urface

=2 + 3 1:::2 + Il Csi n 0

=2 + 2 ~in

30o= 2 +2x ~ = 3 m 2

Fig. 3. 19 (a)

F", 1000 x 9.8 1 x 12 x 3 '" 353167 N" 353. 167 kN . An s.

Using equat ion (3. 10). we have h· =

where

'e= ,

bd ' 3 x41 ~ - - = - - - = 16 m 12 12

""""7'""_',,-9 + II A"

• 16xsin 2300 16x 4 II = "''C;=-''''~ + 3: - 36 + 3: 3.111 rn. An s. 12 x 3 Problem 3.15 (a) A circ14/af plate 3.0 m diameter is immersed in Il'IIter ill such a Imy ,Iwt its gre(llesl and lea~'1 deplh be/ow Ihe free ~'urface are 4 m alJd 1.5 m fHpecli)'ely. lJelt'rm;IIe '''e loMI prHSIlre 011 0111' fat'e of Ihe pla/e (lilt! pOS;/;Oll of the celilre of pressu re. Solution. Giv" n : FREE WATER SURFACE Dia. of plale. Area. Distance

E


i

d2 :

0 1,5 m

~ (3.0)l: 7.0685 m l

,

DC= 1.5 m.BE= 4m

Dist an ce ofeG. from free s urface

: h :CD +GCsin9 : 1.5+ 1.5sin9 Om

AB

s in 9 =

BE - AE

-8-C= ""-iB~C""- =

4.0 - DC

3.0

=

4.0 - 1.5

3.0

'" 2.5 '" 0.8333

3.0

;; '" 1.5 + 1.5 x .8333 '" 1.5 + 1.249", 2.749 m

I I

Fig. 3.20

Ii

~ I

IL 190

Fluid Mechanics

(i) 1'01,,1 pNss ure (F)

F", pgA/'

'" 1000 x 9.81 x 7.0685 x 2.749 '" 190621 N. Am. (ii) Ct' litre of press urt' (h *)

Using eq uation (3.10). we Ita ve 11· '" It

/(

4

l G '" 64 (t '" 64 0) '" 3.976

where

4

III

I, . '" 3.976 x (.8333) x.8333 7.0685 x 2.749

+ 2.749 '" 0.1420 + 2. 749

= 2.89 ] m. AilS.

Problem 3.15 (b)

If in the (lbol'l! problem. Iile gi)'en circu lar plale is hu\'ing a COIla'II/ric circular hole of diameter /.5 m, Ihen calculi"e Ihe lOla/ pressu'e , md positioll of Ihe CClil r /! of pressu'e OIl Oll l! face of the /llare. Solution. Gi ven: IRefer to Fig. 3.20 «(1 )1

d=3.0rn

Dia. of plate. .

1(

,

Area of solid plale '" - d-

4

1t

,

View normal 10 plate

,

= - (3)" = 7.0685 Ill4

Dia. of hole in th e plate. do'" 1.5 III Area of ho le

Area of the g ive n plate. A '" Area of solid plate - Area of '" 7.0685 - 1.767 1 = 5.30 14 rn 1 Distance CD '" 1.5. BE ", 4 rn Distance ofCG. frorn tile

fr~e

Fig. 3.20 (a)

surface.

/'",CD+GCsine

= 1.5+ 1.5sin e Bm

sine = AB",BE - AE = 4 - 1.5=2.5

BCBC

33

25 h:I.5+1.5X 3" = 1.5+ 1.25= 2.75 rn (i) Total press u N fOITt' (F) F: pgA h

'" 1000 x 9.8 1 x 5.3014 x 2.75 '" 143018 N '" 143.018 kN. An s.

I I

Ii

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IL Hydrostatic Forces on Surfaces

91 1

( ii) Pos ition o f centre of press ure (h*)

Usin g equation (3 .1 0 ). we ha ve /' I tr

~4

fc si n2e '"

Ah

1t

_ + 11

444

where I c= 64 [d - d o l = 64 [3 - 1.5 [m

- ' [d' A-

4

sin a ",

2~5

-

d' [ -- ' [3' 0

and

4

-

15'[ . !l1 ,

h = 2.75

+ 2.75

+ 2.75 '" I x 11.25 x 6.25 + 2.75 16x 2.75x9

= 0.]77 + 2 .75 = 2.927 m. An s. Problem 3.16 A circular pll11e 3 metre diameter is submerged ill WaferaJ shown ill Fig. 3.21. Its wealeH and leasl deplhs are be/ow Ihe surfaces being 2 metre and I metre respeClil'eiy. Find: (i) Ihe lolal pressu re 011 f rom face o/rhe plole, amI (ii) Ille position of cemre of pressure. Solution. Gi ve n : Dia. o f plate.

Are a. Distance.

Water surtace

d=3.0 m

"

A = - (3.0)" '" 7 J)6R5 4 DC= 1 m.llE=2 m

III

I ,

<-Jit

,

t

- f-- --',' m ,

All IJE - AE BE - DC 2.0 - 1.0 sin9 = = = AC Be Be 3.0 3 The ce ntre o f gra l'i ty of the plate is at the middle of BC. i.<1 .• at a di stance 1.5 m from C. The d istance o f ce rlt re of gra vit y fro m th e free surface o f th e wat er is give n by ]nMBe,

-

I

" : CD + CG sin e : 1.0 + 1.5 x 3

(":s in O:!)

'" 1.5 Ill. (i) Tota l pressure o n the front face o f the plate is give n by

F: pgA"ii : 1000 x 9.81 x 7.0685 x 1.5" 10401 3 N. An s. ( ii) Let the di stance of th e cerltre of press ure from th e free surface of th e wmc r be II·. T hen usin g equ ation (3. 10). we have

I I

Ii

~ I

IL Fluid Mechanics where

It tt

IG '" 64

It

4

'" 64 OJ . A",

'4It d-,,

-

II '" 1.5

j

.

III

and Sill e '" _

Substituting the values. we get

, d , x (')' 1 - - + 1.5 , , 3 + ! 5= -16x -9xl.5

-

d~

64

-4 "" x 1.5

3'

= ~--"~~+ 1.5 ", J1416 + 1.5", 1.5416 m. Ails . 16x 9x l 5

Problem 3.17 A rec/alls"lar 8m .. 5 In xl m is /I;nged at its bll$e "IIlI indi/led ar 60" 10 IIIe horizon (a/lIS shown i" Fig. 3.22. To keep Ihe 8<1(e in "Mable position. a coulller II'ciglll of 5000 kgfis a ((ached aT Il,e upper elld of 8a /e as sholt'/I il/jigll re. Find Ille depll! ofwa rer {I/ which Ihe gllle begins 10 fall. Neglect Ihe weight of the gtlle and /ricliOIl at rhe Ilinge and /mlley. Solution. Given: Lenglh o f g ~le = 511\ Widlh of gale =2 m

II,,.

""". ,

9", 60 0 Weight.

--~~ -

IV", 5000 kgf '" 5000 x 9.81 N

= 49050

(': I kgf = 9.8 1 N) As the pulle y is fric ti on less. Ihe force actin g at H= 49050 N. First find th e 100ai force F actin g o n Ihe g ate All for a give n de plh o f N

"

"

C A_ -

60

HINGE

Fig. 3.2.2.

wate r. From figure,

,-

,

AE II 5 211 AD", sine '" sin 60o"'""J3i2=T3

211 411 2 Area of gate imme rsed in water. A = AD)( Widt h x ""7:' x 2 == ""7:' m ..;3 ..;3

"

Also deplh of the e.G. o f th e immcrs.::d area'" -h '" - '" 0.5 h 2 . . 411 II 19620 , Tot al force F IS gIven by F = pgA" == 1000 x 9.81 x ""7:' )( ""7:' = ----r.:-- II-N

..;3

,,2

,,3

The centre o f pressure of the immersed surface. h· is give n by 1

lG sin 9 +ii AI,

where

IG'" M.O.1. of Ihe immersed area

=

I I

bx(AD'J

12

2 (2" y

12)( .J3)

Ii

~ I

IL Hydrostatic Forces on Surfaces

~

II '"

J3)' [ --- x + - =- - + - = - + - =- - " 2

9x13

~x~

J3

3h l 18Ir 2

II

4h l

2

II

II

II

h+3/1

211

2

6

2

6

3

93 1

2

2/,

CH= ,,- '" - 3 . LCDH= 60°

Now in the !J.CHD.

CH '" sin 600

CD

CD=.....0:!...-=~= 211 sin 60° si n 60° 3 x .fj AC= AD- CD '"

211

411

2 61! - 4h

2h ---"'7ii" - - '" ./3 3./3 J../3 3./3

Taking the moment s aoou l hinge. we gel

49050)( 5,0 = F x AC =

19620

~

,,3

1

2h

II x-:::-;;:

3,,3

.245250 = 39240 Itl

0'

3x3 III = 9 x 245250 = 56.25 39240 II

= (56.25) 1/3 = 3.S3

m.

Ail S.

Problem 3 .18 A" inclilled re"/{1II8,,/ar J/uice gme All, 1.2 111 by 5 III size as sho wn in f"ig. 3.23 is iustal/ed 10 cOlll'Olllle di.{Clwrge o/wmer. The end A is/lillged. Determitll' Ille force norma/to lite gMf! " f'piied ul 8 Iv open il. Solution. Gi ve n : A = Area o f ga le = 1.2 x 5.0 '" 6.0 111 2

Dt:plli of e.G . of the gale from free surface of the wate r = = DC '" 8C _ BE = 5.0 _ BG si n 45 0

== 5.0 - 0.6 x

h FREE WATER SURFACE 0

C

0

I Ji == 4.576 m

The IOtal pressure furce (1) acting on the gate. F=pgAII == l000x9.81 x6.0x4.576 == 269343 N This force is actin g at H. where the depth of If from

Fig. 3.23

free surface is g iven by

,,* = "'O'-';';'i"_'~. + "AI,

I I

Ii

~ I

IL 194

Fluid Mechanics 1 I G" M.O.!. of gate", _bd _

where

'" 5.0 X 1.2 ) ,,0.72

12

III

12 l

Depth of centre of pressure 11*" n Ul from Fig. 3.23 (a),

",

0.72 x sin 45° "'''''''0:::",'''+ 4.576 " 6x4.576

.013 + 4.5 76 '" 4.589 m

__ = s in 45 0 OH 11*

Oi~t3nce.

4.589

OH= - - ' 1 - " 4.589 sin 45°

x.fi" 6.489 In

T2

Distance.

80 = - '= 5)(.fi" 7.071 sin 45°

Dista nce.

8H = 80 - OH = 7.071 - 6.489 = 0.582

Distance

AH

= AD

- BH

III

= 1.2 - 0.582 = 0.618

III

111

Taking the moments about the hinge A P xA8= F x (AH)

where P is the force norma l to the gate applied a1 B p x 1.2 = 269343 x 0.618

p= 269343)(0.6 18" 1J8708 N. An s.

L2 Problem 3 .19 A gale .\"upporring Waler i.~ JlloWII ill Fig. 3.2-1. Find the fleig/II II afille \Va/a so I/WI the gall' lips abolll Ihe hinge. Take Ille ..... idtl! of fhe gale as unit)'. Solution. Given: e = 60° AC = __" _,, 211 sin 60°

Distance.

J3

where II = Depth of water. The ga te wi ll start tippi ng ahout hinge 8 if the resul tant pressure force acts at B. If the resultant pr.:ssure force passes through a point which is ly ing from /J to C anywhere on th.: gat.:. the gat.: will tip ov.:r the hinge. Hence limi tin g case is when th.: resultant force passes through B. But the resultant force passes through the centre of pressure. Hence for th.: given position. point IJ becomes th e centre of pressure. Hence depth of centre of pressure. I,· '" (II - 3) m fREE WATER SURfACE But h· is also given hy Taking wid th of gat.: unity.

Area.

Th~n

211

-

II

A=ACxl=~xl:II= -

"1/3

2 2/,

Ix ( "0/;' "3 12

I I

)3

Fig.3.24 gil ' 211 , "'12x3xlf,"'9xJJ

Ii

~ I

IL Hydrostatic Forces on Surfaces

95 1

Equa ti ng the two v al ues of 1/*,

2h 3

11 - 3= -

or

11 -

2/,

3

=3

~3 '" .3

or

h=3x3=9m

Height of water for tipping the gate = 9 m . An s. Problem 3 .20 A reCllmglflar sluice gale An. 1 In wide Wid 3 m 10llg is hinged at A as ,~lIo\\'n ill "-ig. 3.25. II is kepI closed by a weiglllfl.I"<,J 10 Ille gare. 11w IOwl weighl ofrlw gale {/tulwl'igll1jixed 10 Ihe gmt' is 3-13350 N. Find Ihe height of tile water 'II' wlliell .... ill jlw eOfue 1111" gale to open. Ti,e anlre olgral'il}, ofrhe ...eight and gale is a/ G.

Solution. Given: Width of g,nc.

Area.

b '" 2 m: Length or gate L= 3 A=2)(3=6m

Weight of gate ami

W.:: 343350 N

Angle of j"clination.

e '"

III

1

45° Let /, is the required height of wale r. Depth of e.G. of the gate and weight = II From Fig. 3.25 (a).

h '" h -

ED '" II - (AD - AE)

'" II - (AB sin

e - EG Ian 9)

AE :. AE = { ... I,m& = EG

EG I~n&l

= II - (3 sin 45° - 0.6 Ian 45° )

::/! - (2.121-0.6)::(IJ-1.521)m The 10lal pressure force. F is given by F:: pgAii :: 1000 x 9.81 x 6 x (II - 1.521)

:58860(h - 1.52 1)N. The IOlal force F is
' cX":::"'c'c4",'o + (II C 1.52 1)

C' I, ' = ~4 6x(1t

I I

cO,.3c7'o'~ + (II -

1.521) = --;

(h

1.521)

<0 I) m 1..,_

Ii

~ I

IL 196

Fluid Mechanics FREE WATER SURFACE}

--.-.-.-.-.-.-.-.--

~

• 1, '".

-

" t-e '" 45"

, "

-_ -_ -. K-. --

e

:: F~ -

(b)

(a)

Fig. J .ZS

Now laking rn om,," \s about hin ge A. We gel 343350xEG",FxAH 343350

X

[

AK

0.6 '" F x - sin 450 rro m AAKH. Fig. 3.25 (b) AK '" AU sin ij ", AH sin

= AK=

Bw Bw

45~ ,', AH '" ~l sin 45 0

58860 (Il - 1.521) x AK

343350 x 0.6 x sin 45" 03535 x 7 '" 58860 (" 1521) (II 1.52 1)

A K = h· - AC =

~c3"7c5= + (II _

... ( i)

1.52 1) - AC

(1i - 1.52 1)

... ( ii)

AC = CD - AD", II - AS si n 45° '" II - 3 X sin 4:;0= 1,- 2. 12 1

Substitut in g th is value in (;,). we get

5 C" +(h - 1.52 1) - (1I - 2.12 1) AK: -:-~J"-7,,, 11-1.521

=

375 -:-'7CC: + 2. 11 1 Ii 1.521

152 J '"

375 -:-'7CC: II 1.521

+ 0.6

.. ,(i ii)

Equating the two va lu es of AK from (i) and (iii)

I I

Ii

~ I

IL Hydrostatic Forces on Surfaces

971

03535 x 7 0375 + 0.6 11 1.521 "1.52 1 0.3S35 x 7" 0.375 + 0.6 (II - 1.521 )" 0.375 + 0.6 II - 0.6 x 1.521 0.611 '" 2.4745 - .375 + 0.6 x 1.521 == 2.0995 + 0.9 126 " 3.0121

01

5. . 2m. Ans. 0.6 Pro blem 3.21 Find Ihe 1010/ pressure and position of celilfe of pressure 0/1 a Irhmgular plale of base 2 m and heiglll J m which is immersed in Waler in SUcil (I way rlta/IIle plane' ofrhe plale makes an angle 0[60" wilh the free su rface of Ihe water. The b{/~'e oftlw plate is parallel to water surface and m {/ depOt of 2.5 III from Wafer surface. FREE WATER SURFACE Solution. Given : - "=--:;.~t - '1 '60~j Base o f plate. b '" 2 III h 25: ]'Ieiglll of plate, II '" 3 III h' ~ 3.0 12 I,= -1 :

I

Ar~a.

A:bx"=2x3=3m2

2

L

(J-1

<'Ill..,..

2

Incl ination.

Dcptll of ce ntre of gravity from free surface of water. h=2.5+AGsin60°

=2.5+

1

3

fi

x3xT

Fig . .1.26

{-: AG '" ±Of hdg ht of triangle }

'" 2.5 + .866 m = 3.366 m (i) T o ta l p ress ure

for~

(1-')

F = pgA/' '" 1000 x 9.8 1 x 3 x 3.366 '" 9906 1.38 N. Am. (ii) Ce nt re o r p ress ure (h "'). Dept h of ce ntre of pres.wrc from free surface of water is gil·en hy le 1,· = Icsin , + II All

where

2x 3J 3 4 = - - = - = 1.5m 36 36 2

bill

Ic = -

l

n f:iJ' I,· " 1.53 xX si3.366 + 3.366" 0.111 + 3.366" 3 .477 m . AIlS.

to- 3.6

CURVED SURFACE SUB -MERGED IN LIQU ID

Consider a curved surface AB, sub-merged in a s tatic nuid as shown in Fig. 3.27. Let dA is the area of a small strip at a depth of II from water surface. Thcn pressure intensity on the area dA is "pgil and pressure forcc. Iff"" P X Area" pgll X dA ... (3 . 11) This force dF acts nomlalto the surface. Hcnce IOlal pressure force on the cu rved surface should be F=

I I

f

pghdA

... {3. 12)

Ii

~ I

IL 198

Fluid Mechanics

WATER SURFACE

,

C

E4~

--------- .....----- --::::=::~:: dF,::--:.,::--,

dA cos {I

AREAdA

(')

(.)

FIg . 3.27

Bul here as Ihe direction of Ihe forces on tbe small areas arc 110t in the same direction. but varies from point 10 ()Oint. Hence in1egralion of equation (3.1 1) for curved surface is irnpos~ible. The problem can. however. be solved by rc!\Olving the force dF in two components dF, and
inlcgraling dF, and lIF,. T hen 100al force on tile curved surface is

2 F=JF, +F'

,

· .. 0 ·13)

;md inclin~tiol1 of rcsuitanl wilh horizonwl is tan

4' ==

F

... (3.14 )

-2:.

F,

Resolving the force dF given by equation (3.1 1) in x and)' directions: dF, = rlF sin e = pgildA sin €I and
and

F, =

J J

dFA

'"

dF, '"

J J

p.!:luiA sin 8 '" pg pgllllA cos 8 "" pg

J J

I': dF=pghdAI

II/fA sin 0

... (3. 15)

IIlJA cos 8

...(3. 16)

Fig. 3.27 (b) shows the cnlarged area dA. From this figure. i. e .• n.EFG. EF=dA FG::dAsin8 EG=dAcos9 Thus in equation (3.15). dA sin 8= FG = Vcrtical projec ti on of the

pg

J

~rea

dA and

h~nce

the expression

repr~scnts the total pressure force 011 the projected area of the curved surface on Ihe

IldA sin e

vertical plane. Thus Fx :: Tutal pressure force on the projected area of the curvcd surface on vertical plane. . .. (3. 17) Also dA l'OS = EG = hurilOntal projecliun of dA and hence /IdA cos is the voluUle o f the liquid

e

e

contained in the elementary arca dA UplO free surface of the liquid. Thus volume contained between the curved surface Hence pg

J

I1dA cos

e~tended

J

/ldA cos

upto free surface.

e is the 10lal weighl suppo l1ed by the curved surface. Thus F. "'pg

JhdAcosO

'" weight uf liquid su pported by the curved surface UplO free surface of liquid.

I I

e is Ihe total

. .. (3. 1&)

Ii

~ I

IL Hydrostatic Forces on Surfaces In Fig. 3.28. the curved s urface AD is not support in g an y !luid. In sucll cases, F, is eq ual to th e weigh t of th e imag inary liquid ~upponcd by AB UplO free su rface of liquid. The direction of F), will be taken in upward dircdion. Problem 3.22 Comp"te 1/11' lroriZO/llal m,,! ('ertical components of the lolal force aCling on a CU')'ed SIlr/llce AB. "'hieh is i'l IiiI' form of l! 'I"ail"m( of a circle a/ radills 2 III 115 sho",,, ill Fig. 3.29. Take IiiI' wiilll, of II!I' gate as IlIIil),.

WATER SURFACE

0 ;- - -

.. . .. ______ _

----------------------------- -_._--------- --------------

A _ _ • __ _

">,,,;;;-;;,,;---~ -- --;;,,; --";,."",,-

Solution. Give n : Widlh of gale Radius of the gale

Fig. 3.28

= 1.0 In

= 2.0 III Di stan<:c AO = OB = 2 rn Horizomal force. Fx cxcT1cd by waler on g ale is given by l 1.5 m equa tio n (3. 17) as F , = Total pressure force on the projccted area of curved surface AB on vcnica l planc == Total pressure force on OU {projected area of curve d surface on vertical plane = OB x I} ==

99 1

o

FREE SUR FACE OF WATER

_ C

-- -- --

, Fig. 3.29

pgAh

== 1000 x 9.8 1 x 2 x I x ( 1.5 +

f)

{ .:

Ii = F, == 9.81 x 2000 x 1.5

Areaof08==A=ROxl==lxl=1.

Depth o f e.G. of 08 from free surface = 1.5 + .,. }

= 49050 N. An s.

The point of application of F, is g il'en by Jr . =

I

-

~ + 11

Ah

l bd lx2 l 2 4 where Ie == M.O.I. of 08 about its e.G. == - - = - - == - 111 12 12 3

2 1,. == __3_ + 2.5 = - '- + 25 III 2x2.5 7.5 == 0.1333 + 2.5 = 2.633 m from free surface. Venical forcc, F y • cxerted by wa ter is given by cquation (3.18)

Fy " Weight of wate r s upponed by AB upto free surface = Weight of portion DABOe

'" Weight of DAOe + Wei gh t of water AOB '" pg [Volume of DAOe + Volume of AOB] == I OOO x 9.8 1 [ ADXAO XI +;(AOfxl ]

I I

Ii

~ I

IL 1100

Fluid Mechani cs = 1000 x 9.81

[L5X2.0 Xl+~X 21 XI]

== 1000 x 9.81 [3.0 + nlN

= 60249.1 N. An s.

Problem 3 .23 "-ig. 3.30 .lhows II gme I1m'ing II qlladroll/ shape of wdilIJ 2 111. "-ind Ihe re.mltal1l fo rce dlle 10 \l'ater per metre lellg lll of Ihe gme. Find also Ihe lingle ar ....hieh Ihe IOtal fo rce \\'ill ae/. Solution. Given: R,I
=2 m = Im

For~

F,

= Force o n the projected area o f Ihe curved surfa,!' on vertical plane == Fo rce on

no = pgAI!

Fi g. 3.30

,I where A =ArcaofB O=2x 1 =2 m , 11= - x 2= 1m: 2 P, = 1000 x9.8 1 )( 2)( 1 == 19620 N This wi ll 3CI al a depth o f

~ 3

)( 2 =

~ 111 froln 3

free surface o f liq uid.

Vertica l Force , " .. F ,":= Weight of water (imagi ned) supported by AB =pgxArea of A OU x 1.0

"lOOOx9.Rl)( !: (2)4 )( 1.0 = )08 19 N 4 This will act a1 a disi:l11cC of 4H = 4 x 2.0 = 0.848 m from OB.

3rt Res ultant force. F is given by

311

F = JCFo,:-+-::F,'-' = .j19620+30819

.j384944400 + 949810761

= 36534.4 N. Ans. The ang le made by the resultant with horizontal is give n by

Fy 30819 Ian 9= - = - - " 1.5708

F.

e " Ian

19620

1.5708 = 57' 3 1'. Ans.

Problem 3 .24 Find tile magnitude and direclion of II,e re.illitwlI force due 10 water acting on (I roller gale of cylindrical form of 4.0 m diameter. wl,ell the gate is placed all tile dam ill such a way that ,,·Clter is just goillg to spill. Toke Ihe lenglh of the gClte os 8 III. Solution. Given: Dia. of gale =4 (11 R =2 rn Rad ius. Len gth of gal~. 1= 8 rn

I I

Ii

~ I

IL Hydrostatic Forces on Surfaces lI o rizont;oI F,

where

ror~,

101 1

F, acting on the gate is

= pgA!J = Force on projected area of curved surface

,,

ACB on vertical plane '" Force on vertica l area AOB A = Area of AOB = 4 .0 x R.O = 32.0 Ill l

;0

T 1

4 .0m

h = Depth of e.G. of AOB = 412 = 2.0 III F,'" 1000x9.81 )(32.0x2.0 = 627840 N. Fig. 3.3 1 Vertical fo ...,e, Fy is given by F ," '" Weight of water enc losed or supponcd (actually or ima ginary) by the curved surface ACB '" pg x Volume of portion AC/J = pg xArcaofAC8x/ IT

,

IT

2

'" l ()()()x9.8 1 x-(R )" x8.0=9810x-(2) x 8.0=493 104 N

2

2

will be acting in the upward direction.

[t

F= JF,~ + F.} - J62784fJ+ 493104 '" 798328 N. AilS.

Rcsulwrn force, o

•

,

•

Dm:clIon ofrcsultarn force ISgIVC" by tan 9=

493 104 f:F, '" 627840

0.7853

e = 3 1" 8', An s. Problem 3.25

F illd the /wriZOlilul and I'f!r/ica/ campanelli of W(ller pressu re (lc/i' ,g all lite face of II rainIer ga/eo/90° ~'eClorofrlldi"s 4 m (l~' si,o ,,,,, in Fig. 3.32. Take widlll ofgllte ullily. SolutIon. Given: Radius of g~te, R",4rn Horizontal cornponenl of force acting on the gale is Fx'" Porce on area of gate projected o n I'enica l plane '" Force on area ADlJ

whe re

= pgA l1 A", A8 x Width of gate ",2xADxl (": AlJ =: 2AD) =: 2 x 4 x sin 45° =: 8 )( .707 =: 5.656 rn ! [':AD=4sin4·n

Fi g . 3.32

-h _ _ _An __ 5,656 _ 2 828 m ~~

2

_

.

2 F, = 1000x9.81 x 5.656 x 2.828 N = 156',)11 N. Ans.

Vertica l co mpone n'

Fy = Weight o f Winer suproncd or endosed by the curved surface '" Weight of water in ronion ACHDA '" pg x Area of ACBDA x Width of gate

'" 1000 x 9.81 x [Area of sector ACBOA - Area of Il.ABO[ x

I I

Ii

~ I

IL 1102

Fluid Mechani cs I ': dAOE is;\ right angled]

'" 98 10 x ['::' 42 _ 4 x 4 ] '" 44796 N. An s.

4

2

Problem 3.26 Ca/cu/rue Iile IlOriZOn/u/ lind "n/icil/ components of Ihe W{lter pressure exerted on r(lilller gale of radius 8 II! (IS shown ill Fig. 3.33. Take ,..idlh of gale "IIily. Solution. Th e hori7.0nlal compone nt of water pressure is gil-e n by

II

F , '" pgA /, '" Fo rce on th e area proj ected on ve ni ca l plane "" Force on the vertical area o f

whe re

BD

WATER SURFACE c

A", IJD x W idth o f gate", 4 .0 x I '" 4.0 III -

'.Lm

I

h=-x4=2m

2

F, = lOOOx9.8 1 x 4 .0x2 .0= 78480N. An s. Fig . l,lJ Vertical rompone nl of Ih e wale r pressu re is give n by F, '" Weight of Wale r su prmncd or enclosed (imagi nary) by c urved

surface

en

'" Weight of water in the portion eBDC '" pg x [Area of portion CBD C ] x Wid t h of gale '" pg x (Area of sector eBO - Area of the trian g le BODI x 1

= 1000 x9 .81 x

[~x 7tH! ~ BDx DO ] 360

2

= 9810 X [ -.!... 7t x 8l ~ ~ 4.~ O ~X~8~.8~OO='=3~O_O] 12

2

1': DO" 80 cos 30° = 8 x cos 30° 1 '" 9810 x 116.755 ~ 13.856] = 28439 N. An s. Problem 3.27 A cylindriClIl gllle of 4 m diameler 2 m long has ...lIIer on ils bolll sides as sllo ...n in Fig. 3.3 4. Delermine Ihe magnilllde. 10culioll wId direClion of Ihe resuillml force e.wrted b)' Ihe ...lIler on Ihe gille. Fillli aiso Ihe /easl "'eiglll of Ille cylinder so Ihlll il may 1101 be lifted 11 ... (1)' f rom Ihe floor. Solution. Given: WATER SURFACE A Dia. of g ale =4m Radius ,,2 In WATER (i) The forces acting on th e left Sidc of the cy linder aTe F 4m _ _ _ _ ~Q ___ _ D SURFACE B c::~~~:::::: The ho ri zontal co mpone nt. F. , x,__~_ ~-where Force of wate r on area projected on ve rtlcal -=-_:_::::::~=~ ~~ _:__ FY, : Fy,

i :t

f"", "

._:_:_:_:._:_-=-_ :._:._-=-_:._.

p l an e

= Force o n ,uea AOC " pgA h " looox9.8 1 x8x2

where

A"ACxW idth ,, 4 x2 1 "" 8 m

'C

-:_

'm

Fx

Fig. 3.34

" 156960 N

I I

Ii

~I

IL Hydrostatic Forces on Surfaces

103 1

F)", '" weight of water e nc losed by AlleOA

'" 1000 x 9.81 x

[~Rl ]

x 2.0 = 9810 x

%x 22 X 2.0 = 1232.76 N.

Ri ghi S id". of Ihe Cy linder

F" = pgA2'1l = Force on vcni ca l area CO

'" l000x9.81

x2x2x~J A,= COXI=2XI=2m l.l;l=~= 1.0) 2\

-

2

= 39240 N F11 = Weight of water enclosed by DOeD

=pgX[~ R l] xWidlhof ga te = 1000 x9.8 1 x ~ x2 1 x2 = 6 1638 N 4

Resulta nt force in tile direction o f x. Fx = F" - F" = 156960 - 39240 = 11 7720 N Resultant force in the direction o f y. F ., =F,• I +F,, = 123276+6 1638= 1S49 14 N (i) Res ulhmt fore", F is give n 3S

F = ~CFC,"+C-;F,~2 _ J(1 17720)l + (1849 14)2 = 2 19206 N. An s. (ii) IJiredlon of res ultant force is given by

tan

F . 184914 e = --.L =

Fx

11 7720

= 1.5707

e = 57° 3 1'. An s. (iii) Location o f Ih e r ..sultan! force rorcc. F" acts

OIl

ac ts at a distance of

a distance of 2; 4 = 2.67 m from the tOp su rface of water o n left side. while F.,

t

x 2 = 1.:\3 III from free surfa<:e on the right side o f the <:y linder. The rcsu lWnt

force F, in the d irection of.r wi ll act at a distance of y fro m the bonom as

F,

x)'

== F" [4 - 2.67 1- F" [2 - 1.33[

117720 x)' = 156960 x 1.33 - 39240 x .67 == 208756.8 - 26290.8 == 182466

Y=

182466 = 1.55 m from the bonom. 117720

Force F a<:ts at a disl
II

Ii

~ I

IL 1104

"'

Fluid Mechani cs

Py )( x = FYI x .8488 - Fy, x .8488 184914 x x == 123276 x .8488 - 61638 x .8488 = .8488 [123276 - 61638 1 = 52318.4 523 18.4

0.2829 m from AOC. 18491 4 (i,-) uasl weight of ,,'Iinder. The resultant force in Ihe up ward direction is Fy = 184914 N Thus the weight of 91indtr should not be less lh:1I1 the upward force F),. Hcnl"t: least weight of cylinder should be at leas!. = 184914 N. A ilS. Problem 3.28 Fig. 3.35 il'llow:; the crU!iJ"J'eCliu n of" /lllikfull of ",,,Ier ""der pre5I>''''''. The /eng/h of lire 'a"k is :I m. Ali emply cylinder lies along the lenglh of Ille [(",k on 01!e of ils corner {IS S"O"'". Find the horiWIl/a/ fwd renic,,/ comflom:nlS of Ihe force oc/i"g Oil 11u: cun'ed iJ'urface ABC of the cylinder. Solutio n. Radius. R==lm Length of lank . 1==2 m -----Pressu re . p '" 0.2 kgf/cm 2 '" 0.2 x 9.8 1 Nkm 2 -----------4 --- - -2 -----'" 1.962 Nlcm = 1.962 x 10 Nlm ~ 4

11= L= 1.962xl0 =2m pg IOOOx9.8 1

Pressure head.

Free surface of water will be at a height uf 2 III from the lOp of the lank. Fig. 3.36 shuws the equi va lent free su rface of water.

----------- ----------------------------------------- ----- ----------------- --- --- --- Fig . 3.35

:::::::::;f~'"

(i) lIurl lo ntnl C umpunent uf Fure..

F, =pgA/'

where A == Area projected on vc.rtical plane = 1.5 x 2.0 = 3.0 1l1 ~

~~~IM~~~~~~ , ' , ' , ' , ' , ' f--G

::::::::-:t-:a

15 "=2+ =2.75

_:_:_:_:1~5 ~ .:

2

F~ =

l000x9.81 x3.0x2.75

= 80932.5 N. An s . (ii) Vt'rtical CO llll)Ollent of Forcc

:0T

:

2.5 m

,- _ ,

~ _Ol..; __

'\

H ,0

:::::}:r:::~

,:,:,:,:,:,:,:,:,:,:,:::~

------------------------------------

F, '" Weigh! of water enclosed or supported a(;lually or imaginary by (;urvcd surface ABC

== Weight of water in the portion CODE ABC = Weight of water ill COOFflC - Weight of water in A£FB But weight of water in CODFBC == Weight of wata in [COB + ODFBO[

= pg

[It:"

+ BOXOD] x 2 '" 1000 x9.81

== 64458.5 N Weight of water in AEFB '" pg [Area of AEFB[ x 2.0

I I

Ii

~ I

IL Hydrostatic Forces on Surfaces

105 1

'" 1000 x 9.81 [Area of (AEFG + AGBII - A ll/I)] x 2.0 AI{ 0.5 sinO ", ==0.5 AO 1.0 EH", 80- HO = LO- AD cos e = 1.0 - J x cos 30°", 0. 134

In MHO.

ABH == Ar;)3 ABO - Area AHO

Area.

30 _ AHXHO 360 2.0

= rtR1x

Weigh! of

WOller

nRl _ 0.5 x .866 = 0.0453 12 2

in AEF8 = 98 10 x [AE x AG + AG x AI/ - 0.0453[ x 2.0

= 9810

x 12.0 x . 134

+ . 134 x.5 - J14 53 1 x2.0

== 98 10 x [.268 + .067 - .04531 x 2.0 = 5684 N Problem 3.29

Fy '" 64458.5 - 5684 '" 58774.5 N. An s. Find file magnitude and direction of the resu/wlil water pressure (Ictillg

,

011

a CllrI'ed

face of (I dam which is shaped uccordili g /0 Ihe f e/Mioll Y = ~ as silow" ill Fig . 3.37. Th e heighT of the 9 "'mer retained by III<' dam is 10 m. Consider 1/11' widlll of the dom (IS Imit)'. Solut ion. Eq uat ion of curve AB is

,

.( 1

y= -

or

,

x' ",9y

9 .I '"

J9Y '" 3JY

Hcig tll of waler.

11= 10m

Width.

b=l m

The hori zonta l compone nt , f'~ is g ive n by Fx '" Pressure due to water on the c urved area '" Pressu re on area BC

' '''J'''''" o n vertical plane

'" pgAIi whe re

A : BCx I: lOx I m 2.1i:! x 10:5 m

F,: l ooox9.8 1 x IOx5= 490500N Vertical component, F,. is given by

Fy = Weight o f water su pported by the c urve AB '" We ig ht of water in th e portion ABC '" pg lArea of ABC ) x Width of dam '" pg

[J~~t X dY ] x

1.0

{Area of strip '" .ldy

1'"

: 1000 x 9.81 x 0 3JY dy =29430

[)~/~'" l" =29430x f[/'2 1~O =19620(I0Jl2( 0

: 19620 x 31.622 '" 620439 N

I I

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IL 1106

Fluid Mechani cs RcsuJtam water prCS5ur~ on dam

F'" JF/ + F/ '" J (490500)2 + (620439)2 '" 790907 N '" 790.907 kN . An s. Directio n of the resultant is given by

F1 620439 = '" 1.265 F, 490500

tan9= -

9 ", 51 ° 40', An s.

Problem 3 .30

A dam has" p"rabolic shapl! y '" Yo

(c<..)2 a~' shown ill Fig. 3.38 below IUn'illg xQ '" 6 m .',

{lnd Yo '" 9 m. The fluid is water with density '" /000 kg/m.!, Compule the "ori2Omal. !'Imical alld Ihe resu/Ullil thrust exerted by ",mer per melre lellglh of the dam. Solullon. Given: Equalion of Ihe curve OA is P)'o

"'

.t

2

(;:r =9(~r

=9 x ;: '"

1

x:

'" 4)'

.J4Y '"

of '" 2ylr.' Width of dam. b: lnl. (i) lIorizontnllhrus t t'xerled by w:ller

F, '" Force c"cned by wate r on venica l surface OB, i.e.. the surfa(;c obtained by projc(;ling the curved s urface 011 vertical plane Fig . 3.38

"' pgA;; '" 1000x9.81 x(9x I)

Xl9

== 397305 N. Ans.

(ii) Vertical thru st exerted by water F1 == Weight or water sup()Orl cd by curved surfa!;e OA upto free surface of water == Weight of water in th~ portion ABO == pg x Area of OAR x Width o f dam

== 1000 x 9.81 x [ f rx == lOOOx9.81

X[I: 2/12 XilY]x 1.0

I

=19620 x [(~:i~ ) = = 19620 x

I I

dY]x 1.0

,

19620 x ~ [9.\12 1

3"2 x 27 == 353HiO N. AilS.

Ii

~ I

IL Hydrostatic Forces on Surfaces 107 1 (iii) Res ult a nt th rus t exe rt ed

b~'

wa te r

F" JF} + F} - .J397305 + 353160 = 53 1574 N. Ans. Direction of resultant is given by tan

e=

F, = 353 160

F,

= 0.888

397305

9 = (an -I 0.888 = 41.6 3°. An s.

Problem 3 .31

A cylinder 3 m ill diameter (lnd., III 10llg retaitl.l Imler 011 one ,I'id,'. Ti,e c)'litlder is

supported as sllo .....tl ill Fig. 3.39. Determine lilt' horizolltal reaclioll III A and rhe I'erlical reaction ar B.

The cylinder weighs 196.2 kN. Ignore friction. Solution. Given : Dia.ofcyli[l(lcr

WATER SURFACE

--~-~ ~- ~-~-~-~-~-= _~_~_~ _~_~_~_:_:

= 3 11)

lA:ngth of cylinder = 4 In Weight of cy linder. IV= 196.2 k.N = 196200N HoriwnwJ fOT..,c exerted by water F , = Force on vcnical area SOC : pgAh

whe re

C

---

----

3m

l

:,

F.

L

'0

~

0" --- + ~ --A

tFy

,

f ig. 3.39

, ~ I A =BOCx/= 3x4= 12m-.iI = - x3= 1.5m

2 F,= l000 x9 _81 x 12x L5= 176580N The ve ni cal force e~ened by water F, = Weight of wate r enclosed in BDC08 =pg><

(~R l) >
1000><9.8 1 ><~ X(1.5)lX4= 138684N

Force F, is acting in th e upward direction. For the equilihrium of cy lind e r Hnri7.onta l reactio n at A = Fx = 176580 N Vertical reac tion at B '" Weight o f cyl inder - F,.

:: 196200 - 1386M::S75 16 N. Ans.

... 3.7 TOTAL PRESSURE AND CENTRE OF PRESSURE ON LOCK GATES Lock gates arc the devices 1I',ed for changing the water level in a cana l or a rive r for na viga ti on. Fig. 3.40 shows plan and elev ati on of a pair o f lock gates. Let All and IlC he the two lock ga tes. Each gate is supponed Orl twO hinges Ii~ed on their top arid ool1om at the ends A and C. In the c losed position. th e gat~s meet at 8. Let F = Resulta nt force due to water on the gate All or IlC acting arc right angles to the gate R = Reaction at the lower and upper hin ge P = Reaction at the oommon oontact surface of the two gates and acti ng perpendic ular to the con tact surface. Let the force P and F meet at O. Then the reactio n R must pass through 0 as the gate All is in the equilib rium under the adion of three forces. Let e is the inclination o f the lock gate with the normal to the side of the lock.

I I

Ii

~ I

IL 1108

Fluid Mechani cs

In LAllO. LOAD", LAlJO '" a. Resolving all forces along the gale All and puning cquallU zero. we get R cos a~ Peose '" Oor R '" I'

... (3. 19)

...- HINGE

~"'~ WATER SURFACE

ELEVATION

DOWNSTREAM SIDE

'CAN

Fig. 3.40

"'

Resolving forces nonnailO the gale All Rsin9+Psin9 - F=O F", R sin €I + P sin 21' sin

e '"

e

p. _ F_

1":R=P)

... 0.20)

2 si n e

T o .,alcul"l ~ P .. nd R In equ at ion (3.20), I' can be calculated if F and e arc I;nown. The value of 9 is calculated fro m Ihe angle between the lock gales. The angle between the two lock gate is equa l to IRO" - 29. Hence 9 ca n be calculated. The value of F is calculated as : II I '" He igh! o f water on Ihe upstream side 11 2 '" Hc igh.t o f water u n tllc downstream side F j '" Water pressure on the gate on upstream side F1 '" Water pressure on the gate on dowlIslream side of the gate J '" Width of gate

Now

F\ '" pgA\h\ H ",pgXH\XJX - ' 2

H'

Similarl y, Resu ltant force

'" pgJ - ' 2 _ H , pgJH ' F1 "'pgA 1 h, = pgX(HlxJ)X 2"'~ F = F\ _ F _ pgIH\' _ pglHi 2-

2

2

Substituting the val ue o f ij and F in equation (3.20). the value of P and R e~n be Reaction s .. t th e tol' li nd bott om hin ges Let R, '" Reaction o f the top hi nge

I I

c~kulated.

Ii

~ I

IL Hydrostatic Forces on Surfaces

109 1

Rb ", Reaction of Ih.e bonum hinge R=R,+R b

Then

The rcsuitam water pressure Fads norillal 10 the g ale. Half of the value of F is resisted by Ihe hi nges of one lock gates and other half will be resisted by the hinges of01h.c r lock gate. Also FI acts at a distance of

~ 3

from bollOOl while F, acts at a di stance o f

-

H~3

from bonoill.

Taking moments about Ihe lower tJingc

R

.

X Sln

I

Fi HI F, H , &x H = - x - - - - x -2 3 2 3

.• ,(1)

where H = DiMallcc between twO hinges Resol ving force s horiZOntally . . fj F, R,sme + Rb slne= - - .....:...

...(ii)

2 2 From equations (0 and (ii). we can find R, and Rho Problem 3.32 Each gare of a lock i,y 6 In high Wilt is supported by two hillges placcd ()II IIIe lap alld bOl/om of lile gille. W/lell Iile gale.! (lfe c/o.!ei/. 111e)' make GIl angle of 110°. The widlll of lack is 5 m. Ifrlie ",/rer (e"e!s lIr,' -I m lind 2 III Oil 1101' upsrream and dOll'lIsrr.-am sides reSl'eC(iI'ely. derermine rlie Imlg,lirl,,!e of rile forces 011 rlie liillges dill' ro lI'arer pressllre. Solution. Given: :: 6m Heig tu of loc k: Width of lock :: 5m Width of each lock gate =A B

HINGE

1'm

r -:-;:;?;-:-

10

"' Angle belwee ll gales

AD cos 30°

't

2.5

0

cos 30°

ELEVATIQ

:: 2.887 m :: 120°

eo

I ,~~' C

~~~ ~

180·- 120'

2

wa1~r

'"

R R

''''.oT- L "'fm

PLAN A

HI = 4m

TOlal

INGE

~

DOWN$TRE

Height of wa ter on upslream side and

''''

-

Hl=2 m press ure on upstream side

30"

Fig . lA I

f") = pgA1ii , . wh~re AI = II I x! '" 4.0 x 2 .887 m ~

(H 4 iii, =-t=2'= 2.0m

= 1000 x 9 .81 x 4 x 2. 887 x 2 .0

I

= 22657 1 N F'orce

F I

will be acting at a distance of

Similarly. to tal

wal~r

fl

4

= - :: 1.33 m fro ln boltom. 3 3

_ I

pressure on th e downstrea m side

-

Fl = pgA~IIl. whe re

A~

= Hl X I = 2 x 2.887 III

= l000x9.81 x2x2.887x 1.0

I I

, II 2 "l = .:..:.l.= _ = 1.0m 2 2

-

Ii

~ I

IL 1110

Fluid Mechani cs

'" 56643 N F, will act a1 a distance of H, = ~ = 0.67 III from bonom . 3 3 Resultant water pressure on each gale F == FI - F, '" 226571 - 56643 '" 169928 N. Le t .r is heig ht of F from the bottom. then taking moments of Fl' F 2 and F about the bollom. we hal'c Fxx= FI X 1.33 - F 2 x 0.67 or 169928 xx: 226571 x 1.33 - 56643 x 0.67

.r =

22657 1 x 1.33 - 56643 x 0.67 169928

301339 - 37950 = 1.55111 169928

From equation (3.20). P ~ _ F_ = 169928 = 169928 N. 2sin e 2 ... i030 prom equation (3.19). R = P == 169928 N. If R T ~ n d R8 arc the rCJ"Iions at the lOp
RTx6.0+RexO=Rx 1.55 RT = [69928 x 1.55 '" 43898 N

60 RIJ=R - R r = 169928 - 43898= 126030 N. Ans.

Problem 3.33 Ti,e end gilles ABC of
180°- 1200 2

'" 30

Q

A

,-~-~-~-~--- L- HINGES

,

'm

•

, (3)

PlAN

(b)

ELEVATION

Fig.J.42

I I

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IL Hydrostatic Forces on Surfaces

111 1

: 10m

W id th of loc I.:

5 " cos 30° or l= S.773 m

Width of e ach lock 0':1'111 of water o n upstre am side.

U I " 8 III

Depth of wate r o n dow nstream side. 112 " 4 III (i) Wa ter pressure on upstrea m s id e

F I '" pgA,ii, w here

-

H

8

A I '" I X H I " 5.773 x 8 " 46. 184 Ill. ' 11 '" _I = - = 4.0 III

2 , F I = 1000 x 9.81 x 46.1 84 x 4.0" 181 2260 N = 18 12.26 k N

Wate r pressure o n dow nstream sid e.

1'1 = pgA1'1! where

A l = I X tl2 = 5.773 x 4 = 23.092

111.

";! ="24 = 2.0

FI = 1000 x 9.81 x 23.092 x 2.0 = 453065 N = 453.065 kN Resulta nt water pressure = " I - F2 = 181 2.26 - 453.065 = 1359. 195 kN ( ti) Reaction be tween the gales All a nd He. The re acti on (P) betwee n th e gmes All and Be is give n by cqumi o n (3.20) as

F F: -. , • ~In

0,

1359.195

.3no = 1359.195kN, An s.

X Sm

v

( iii) Force o n each hin ge. If NT and Nil arc the reactio ns at th e top and bono m hinges th en

Rr + Rs"' R But fro m equation (3.1 9), R = P = 1359.1 95 Rr+ R s '" ])59. 195 , . HI 8 H, 4 The fu rl'c FI IS aetmg at - '" - '" 2.67 III from bonu m and F, at -" '" ~ = 1.33 m fro m bonum . 3 3 3 3 The resuhant fo rce F will act a1 a di stance.r from bo tlo m is give n by Fxx= F I X 2.67 - F z x!.33 0'

.r =

c'c."X"'."67c-,c-F'LX " ,, 1.3,,, 3 F

lS I2. 26x2.67 - 453.065x 1.33 1359.195

4838.734 - 602.576 3 3 '" .1I6= . . lI m 1359.195 Hence R is a lso aC iillg ai a d istance 3, 11 m from bol1om. Tak ing mo ment s o f Rr and R aboll1 the bo no ll1 hinge '"

RTx 16.0 - 1.01= R x (x - 1.0)

RT =

RX(I' - I.O)

1359.195x2.11

5.0

5.0

= 573.58 N

Rs"' R - RT'" 1359.195-573.58 '" 785.6\5 kN. A ilS .

I I

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IL 1112 Fluid Mechani cs .. J.8 PRESSURE DISTRIBUTION IN A LIQUID SUBJECTED TO CONSTANT HORIZONTA LNERTICAL

ACCELERATION

In c hapl ers 2 and 3 . the containers which co ntains liquids. arc assumed to be at resl. Hence the liquids are also at res!. They arc in st:uic cquilibri ulll wilh respecI to con lainas. Bul if Ih e container containing a liq uid is mad e (0 move Wilh a COn S[an l acceleration. the liquid panicles initi all y will mOVe relmive 10 each o lh er and after some tim e. there will nO! be an y relative motion between the liquid part icles and bound ari es o f Ihe conta in er . The liquid w ill tak.e up a new position under th e effect of acceleration imparted to ils con tainer. T he liqu id will come \0 rest in this neW position rda tive 10 \ll<)

container. Th e en li re nu id ma ss moves as a si ng le unit. Since the liqu id aFter anainin g a new position is in sta tic condi tion relative to the cOlllainer. the laws o f hydrostatic can be applied to determine the liquid pressure. As there is no re lat ive motion between the liquid panicles. hence the shear stresses and shear forces be tween liquid panicles will be zero. T he pressure will be normal to th e surface in co ntact with the liqu id. The following are the important c ases under consideration: (i) Liquid (.'Ontainers subject to constant horizontal accele ration. (i,) Liquid cont ain ers subjl'ct to ronstant vertical acceleration. ) .8. I Liquid Containers Subject to Constant Horizonta l Acceleration. Fig. 3.43 (<<) shows a tank conta inin g a liqui d upto a certain depth. The ta nk is s tationary and free surface of liqu id is horizontal. Let this tank is moving with a ronstant accelerat ion 'n' in the horizo nta l direction towards right as shown in Fig. 3.43 (b). The initia l free su rface of liqu id which was horizontal, now takes the "hape as shown in Fig. 3.43 (b). Now AB represcn ts the new free surface of th e liquid. Thus the free surface of liqu id due to horizo ntal acceleration wi ll become a downward slopi ng inclined plan". with the liquid ri sing at the back end. the liqu id fallin g at the front end . The equation for the free liqu id surface c an be derived by considering the equilibrium of a fluid elcmelll C lying on the free su rface. The forces acting o n the clement Care: Free surface of l;quOd Originat liquid sorlsCfl

,

Moving

c

Tank

e

- - - --

horizontat C

mX8

""Rear or

,", end

• lW)

(.,

•

•

•

(0'

(0'

Fig. 3.43 (i) the pressure force P exerted by the s urround ing fluid on the cle melll

C. This force is normal to

the free surface. (it) the weig ht of the fluid cle melll i.e .. m x g acting vertically downward. (iii) accelerating force i.e .. Nt x n aC ling in horizontal direc tion.

I I

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IL Hydrostatic Forces on Surfaces 113 1 Resolving th e forces horizontally, we get Psi n 6+m)(a=O or

Psin8= - mll

... (r)

Resolving Ihe forces vertically. we get Pcos6 - mg=O or

Pcos8=mxg

. ..(ii)

Dividing (I) by (ii), we gel

tan

e" -

~

( or ; NumeriCallY )

... O .20A )

The above cqUillion, gives the slope of the free surface of the liquid which is contained in a lank which is subjec ted 10 horizon tal CO nsta nt acceleration. The [cn n «(jIg) is 11 constalll and hence ta n will be constant. The -ve s ign shows that the free surface of liquid is sloping downw ards. He",:e the free surface is a straighT pla ne indilled down al:1n angle e along the directio n of :Iu;clcrmioll. Now let us fi nd the ex pression for the pressure al any point D in the liq u id mass SUbjected to hori zon tal ac.::clcralion. LeI the point D is at a depth of '''' from the free surface, COl1 sider all e lementary prism DE of height 'Il' and cross·scctional area dA as show n in Fig. 3.44.

e

, , 01 constant /'JI:=:;,. LInes pressure

"

' ~M::l: f -pgh,

Fi g, 3.44

Consider the equilibrium of the eleme nta ry prism DE. The forces acting on this pri~m DE in the ve nil:al direl:tion arc: (i) the atmospheric pressure force (Po x riA) althe top end of the prism ac ting downwards, (ii) tile weigh t of th e clemen t (p x g x" x dA) at the e.G. o f Ihe c le ment acting in Ih e downward d irection. and (iii) tile pressure force (p x
"'

p - Po- pgll=O

or or

or

P=Po +pgll

P - Po =pg/r gauge pressure at point D is given by P

or

=0

pressure head at point D.

= pglr

- I' =/1.

PS

I I

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IL 1114 Fluid Mechani cs From th e above cqual io n, it is c lea r thai press ure head al an y poim in a liqu id subjected to a consta nt llOri zo ntal acce lerat io n is eq ua l (0 th e he ight of th e liqu id co lum n above lli at point. Therefore the pressure distribu tio n in a liquid su bjected to a constant ho rizontal accclc rmi on is sa me as hydrostat ic pressure di stributi on. T he pl a nes of co nst ant pressure arc therefore. para ll e l 10 th e indincd surface as s hown in Fig. 3.44. T hi s fig ure also shows the vari al ioll of pressure o n th e rear and f ront end of Ihe la nk .

II( = Depth of li qu id at Ihe rear end o f Ih e ta nk II~ = De pth o f liquid at the fro nl end of I h~ tank F I = Total pressure force exerted by liq uid 011 the rear sid e of th e tank

If

F z = T ot al pressu re force exert ed by liq uid o n th e fronl FI

th" n

= (A rea of tri ang le

s id ~

of th ~ lank .

AM L) x W idth

I

I

= ( -rxLM X AM Xb)=-rxpglllxIlIXb=

~. b.h ~ 2

F2 = (A rea o f tri a ngle DNO) x Widt h

an d

I

= ( .....•

x BN x NO)

I

= .....•

x It, x pgh, x b

•

-

=

pg .b .hi 2

where b = Widt h of tank perpend ic ular to the plane of th e pa per. T he valu es o f FI and F2 can al so he o btained as [R efer to F ig. 3 .44 (a )l

T-.--------------- 1 t 1 -----------

=P Xg x(hl x b) x

Itl _

=-2I

2

A,

md

pg. b.

= h,

-

", ;[[[g{;~[,,-..... .

"I", I"~

=-=- 2

xh and h,

I"~

~pXgx(",xb) x ......:..

-

I

:.:.:.:.:.:.:.:.: F.-·:·:~'.. ••••••••- •• -:. --.-.---••--••• ~

h

- ,~~~::~~~:::~:

2

,

Fig. 3.44(a)

= "2 pg.bxI12-·

It ca n also he prove d that th e diffe re nce of th ese two forces (i.e .. FI - FJJ is eq ual to the force to acce lerate the lIla~s o f the liq uid co ntained in th e ta nk i.e ..

r~ qui red

f'1 - F 2= M xa whe re M = T o ta l m ass o f the liqu id con tained in the tank II

= Horizon tal

co nstant acceleratio n .

Note : (I) If a tank completely filled with liquid and open at the top is subjected to a COIIstam horirontal acceleration . then some of the liquid will spill out from the tank snd new free surf""e wit h il< slope given by equation tan 9 = _!!. will be de'·eloped.

g

(ii ) If a tanK par1ly filled with liquid and open at the top is subjected to a constant horizontal acce leration.

spilling of the liquid may taKe plae<.: depending upon the magn itude of the acceleration. (iii) If a lan k complelely filled wilh liquid and closed althe top is subjecled 10 a con,tam horizontal accelera tion.lhen the liquid wo uld not spill oul from the tan k all d also there wil l be no adjustment in Ihe surface cle"alion

"g

oflhe liquid. Bullhc eq uation tan e = - (i" ) The example for a ail1'l~n"

I I

during

t~ke

tan~

i< applicable for Ihi' case also.

with liquid .
off.

Ii

~ I

IL Hydrostatic Forces on Surfaces 115 1 Problem 3 .34

A reC/angular /ank i.1 mOI'iIlS IlOrizOIl/(lI/), III lilt! directiOIl of ils '<'ngrll Willi U

CQIIS/atll acceleratio/l of 2.4 m/i. The length. ",idth and deptll of Ille /(IlIk are 6 III. 2.51/1 and 2 respecril'l'iy. If the deplh of !l'mer in III,> /(Ink is I In and {(lilt is open a/ the lOp then calcu/are :

In

(I) tile Iwgle of the \l'ater $Iu jace /0 the IlOrizonial.

(ij) Ihe max im/WI and lIIinilllum flres.lure in/ensilie,1 (j{ Ille bo l/om. (iii) Ihe /0/(1/ forc e due 10 ...tller (lelillg 0/1 eaeil end of flu! tank.

Solution. Gi ve n

F..... surface t'

Constant ;!(;ce lcrmion.

a = 2.4 mls l,

Le ngth" 6 III : Width = 2.5 m and depth '" 2 Ill. 2 Dc ptll o f Wa!U in tank . II = I m (I) The a ngle or the ... aler s urface tn the h o rb;onlul ~

Let 9 ", tile an g le of wa ter surface to th e hori zontal

0.2662 m

Usi ng equ ation (3.20 ). we get Fig. 3.45

" 2.4 tan • " - - ,, - - - " - 0 .2446 g 9.81

(Ihe - ve sign shows that Ihe free surface o f waler is s lo pin g dow nward as sho wn in Fig . 3.45) Jan

e"

0.2 446 (slope downw~rd) fJ " lan- I 0 .2446" 13.7446" or 13° 44.6'. An s .

( if) Th e maximum and minimum press llrt' intens ities a t the bottom of th e tank

From th e Fi g. 3.45. Depth o f water at th e front e nd, irl" 1- 3 t3n fJ " 1- 3 x O.2446" O. 2662 m Depth of wate r at th c re ar e nd. it1 " I + 3 tan fJ " I + 3 x 0 .2446 " 1.7338 111 The pressure inte nsity will be IllJ ximum at the oouo m. where depth o f wat er is max imulll .

Now the maximum pressure inte ns ity m the OOuo m will be at point A and it is g iven by. fI ....," p xgxl,!

" 1000 x 9 .8 1 x 1.7338 N/m 1 " 17008.5 N/m 2 • An s. The minimum pressure intensity at th e OOllo m will be a t point B a nd it is g iven by Pmin" p xg x l'l ,, 1000 x9 .8 1 x O. 266l" 2611.4N/ m 2. Ans. (iii) Thl."" 101,,1 for.:I."" dul."" 10 wull.""r "cling on eu.:h I.""nd of Ihe lunk

Let

F I "to tal force actin g

On

the fmnt side (i.e .• on face BlJ )

F 1 " tota l fo ree acting on the rear side (i.e .. on face A C)

Th en rnd

F I " pgAl h l, where AI " BD x width of lank ""I x 2.5,,0.2662 xl.5

,-11 __ BD

_!!L _

_ _ 0.2662 _ _ 0 . 133 1 m 2 2 2 " 1000 x 9.8 1 x (0.2662 x 2.5) x 0. 133 1

" M8.95 N. Ans.

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IL 1116 Fluid Mechanics F 2 ", p.g.A 2.hl. w herc A2 '" AB x width o f mnk '" 112 X 2.5 '" 1.7338 x 2.5

and

'ii, '" •

AB '" III = 1.7338 '" 0.8669 m

2

2

2

= 1000 x 9.81 x (1.7 338 x 2.5) x 0.8669 = 3686 1.8 N. An s.

Rcsul! ant force'" F, - "'2

c

= 36861 .8 N - 868.95

liquid (water)

= 35992.88 N Not .... The difference of the for~cs acting on the lWO ends of the lan k is equal 10 the force necessary 10 acccicraic the liq u id maS,. This can be provc
F2

__ . '. '••••••••••••••• '••• ".

----'-- ,:::::~:::::::::::::::::::::: ~:: -:-. . D -----------------------------------------------------------------------------------------------------------------------------~I.-------"m ------~'

eration of the liquid.

F ig. 3.46

(F, - f',) 5!>1 XlI

= (p

X

, "

,

volumc of cont rol vol umc) x

II

.. (1000 X Area of tl BDC£x width) x 2.4

5

. 1x2.4 2 xA8xwldth [lOOO x ( AC"D) [-: Area of trapezium = ( AC;

IJI))x tlB]

) x 6 x_. ~5 x •~ .4 .. '000 x (1.7 338 +0.2662 2 = 36000 N

(": AC = h, " 1.7338 Ill. SO = h, = 0.2662 tll. and tlB = 6 Ill. width," 2.5 m) Th e abovc force is "early the same as thc difference of the forces acting on thc two ends of the tank. (i.e,. 35992,88 :< 36000) Problem 3.35 Tile reClang"lar lallk of lite abOl'" probl"m contaill S ...aler 10 a deplll of 1.5 III. Fin d Ille I,orizon/{,l acceleralion wllicll m"y be im(!llfled 10 lite IlI/lk ilt lite direCliOIl of ilJ' lengllt J"O Ilwl (i) Ihe spillillg of ..."ter f rom Ihe 1'lIIk is jllst 011 Ihe !"erge of latill g pl(!ce. (il) Ihe f r0l11 bo/lom co mer of Ihe ,,,"k iJ' j ll S! e_.posed. (iii) lite bOl/om of I/Il' /(IIIk is eXpOJ'ed upla ils mid·poinl. AI,w c,dClllate Ihe tolal force.! exerted by II,e ...mer on each end of lite lank ilt eaclt caJ'e. A 1$0 pro!"e Illat IIII' di/fer('/tce belweelt Ilte,le forces is elflw l to lite fo rce "eCeJsa ,)" 10 accelerale lite mass of ...ater wllk.

Solution. Given: Dimensions of the tank from pre vio us problem. L '" 6

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Ill.

width (b ) '" 2.5 m and depth '" 2 III

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IL Hydrostatic Forces on Surfaces 117 1

/,=

IXplh of water in tan k.

Horizo ntal """elcrnlion Imparted 10 the lank

(0 (iI) IV/len Ille splllwg of'Huer from Ihe tonk IS JUII 0/111/1'

"

,age oftak./Ilg place d'

I

---

l .5m

C

000aco>

I ~:::... HA T__

u_.

_ _ 3",

-, -,,-. ". 8

I

":. 3.. ::._

"-:.::"j,":

:_:::_:::':::':::_:':~_:::_: _. _: ._:

..xl a '" reqUiTe ..omo nla aeee cral10n

When the sp lllm g of water from thc lank IS JUS1 o n th e 15m . ~::: :_:'~. ~:.~: ~:_~~~:- ~:::_ \ ergc of lakmg pi ace. tlie water wou Id rise u PIO the rear ' .' : -,::, .:_ . _ . _.: _' - .:_.:::.:__

tOp corne r of the tank as sliown 111 FIg. 3.47 (a)

E,"- - - - .-,-•. , .-, •. , - ,-,-- - - ,.

I · ---' m --~ ~ ' I

Ian 9= AC '" (2- 1.5) '" 0.5 =0. 1667

3

AO

But frolll equation (3.20) Ian (l

a = -"g

Fig. 3.47

3

F

(J) Spilling of 'IlNlur is just on

tht tJerge of taking place. (Numerica ll y)

= g x tan e = 9.8 1 x 0.1667 = 1.635 m/s 2 • Ans.

(b ) TOIIlI forces e.u:rled by water on each end of IIII' tallk The furce exerted by water un the end CE uf the tank is

FI = pgA ,/',. where A, = CE x width of the tan~ = 2 x 2.5 CE 2 /11 = - = - = 1111

2

2

= 1000 x9.81 x(2 x 2.5) x I = 49050 N. Ans. The force exerted by wa ter un the end PD of the tank is

F1 = pgA 1 x

/'1.

where A1 = FD x width = 1 x 2.5

(.: AC= BD=0.5 111.

= l ooox9.81 x(J x2.5)xO.5

FD=Bf' - BD = 1.5 - 0.5= I ) -

FD

I

I" = - - = - =0.5 III , 2 2

= 12262.5 N. An s. (c ) DijJere1> ce of IIII' forces is eqlla/to the force 1>eCeSI(lry 10 (lcce/emle IIII' mllSS ofw(lier in Ille /(lllk Difference o f the forces = FI - Fl

= 49050 - 12262.5 = 36787.5 N Volume of water in the tank before acceleration is imparted to it '" L x b x dcpth of water =6x2 .5x 1.5 =22.5m 3. The force neces-"llry to accelerate the mass of watrr in th e tank = Mass of water in lank x Acceleration

= (p x volu me o f watrr) x 1.635 = 1000 x 22.5 x 1.635 [Thcre is no spi llin g o f water a nd volu me of wate r = 22.5 mO[

= 36787.5 N

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IL 1118 Fl uid Mechan ics Hence Ihe difference between the forces on the 1WO ends of the tank is equal to the force necessary \0 acccierate tlie mass of water in the tank. Volume of water in Ihe !>mk elll1 also be calculmcd as volume = ( CC+ 2 FD ) x EFx Width. [Refer to

Fig. 3.47 (a) 1

=

F"I'8$urtaoe

(22+1 ) x6x2.5=22.5m.,

Original l"",

aHer ~e<.1ioo sunace

(Ii) (iI) Horizontal acceieral;on ... I,en tile frolll

bOl/om cortler of Ihe Iallk

i,~

just e.rpoud

Refe r 10 Fig. 3.47 (b). In this c ase the free su rface of water in the tank will be along CD. leI 1I '" required hori zontal acceleration. In th is <:ase,

2 6

CE

I 3

,

1309 = - = - = -

ED

BUI from equatioll (J.l7),

tan

•

e = ~ (Numerically)

,

I .---- 'm ------~ " .I Fig. 3.47 (b)

g

a=gxtan8=9.81 xi =3.17m1s!. Ans. (b) Tolal jorces exerted by waler 011 eaell end of Ihe l(llik The force exerted by water on the end CE of the tank is F, "'pg xA , x

whe re

h'

A,=CExw idth= 2x2.S",'; m"

CE 2 : - = lm 2 2

h, = -

= 1{lOOx9.81 x5x 1

= 49050 N. Ans. The force exerted by Waler on the end BD of the tank is zero as there is no water against the face BD

Fl =0 Difference of forces = 49050 - 0 = 49050 N

(e) Difference offorces is eqlw/IO rhe force neeeHary 10 accelerate rhe lIIass ofwarer ill rhe Volume o f water in the tank = Area of CEO x Width of tank = (

CEXED) 2

x 2.5

( ...

/(llik.

Width of tank =2.5 m)

= 2;6 x2.5= 15 m3 Force necessary to accele rate the mass of water in th e tank = Mass of water in tank x Ac(;e lcratiun = (1000 x Volume of water) x 3.27 = 1000 x 15 x3.27 =4,)050 N Difference of two forces is also = 490S0 N

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IL Hydrostatic Forces on Surfaces Hence d ifference be twee n Ihe forces on the two e nds of the ta n~ is eq ual \0 the force necessary to accelerate the mass of wate r in th e tank. (iii) (a) Horizon/al (lac/era/ioll wllell the bOl/om of till' lallk is eXJlosed lip/a its mid-poiru Refer 10 Fig. 3.47 (e). hi thi s casc the free surface of W31cr in the tank w ill be a lo ng mid -po int of ED.

CD'. where D * is the

Let (/ '" required horizontal acce lerat io n from Fig. 3.47 (e), it is clear that

C£

2

ED'

3

Free >urf"""

olle, accew.ralion

119 1

Original I, .. If

i ....... 7. '

r

•

:-:-:-:-:':':':-:-.

lane= -- ~ ­

But from cquiltion (3.20 ) numerically

,

1ane=~

,

a=gxlan 6=9.81 x -

3

,

=6.54 mls. Ail S.

(b) Tolal forces exerted by ,m/er on cod, end of Ihe lank

The force cxcrwd by water on the end C£ of the tant is F I=pxgxAlx/IJ

where

A 1 =CExWidlh=2x2.S=Sm! CE 2 1,,= = - : lm , 2 "lOOOx9.81 x5x 1

'" 49050 N. AilS. The force exen cd by waler on lhe end 8D is zero as lh ere is no waler againsllhe face BD. Fl '" 0 Differe nce o f lhe forces'" FI - Fl = 49050 - 0 = 49050 N (c) Difference of rhe 1»"0 force.! is equtl/ fa rile force neces.lary fa aealerare rile Intls.f of lI"aler reI/wining ill Ihe Iallk Volume o f waler in lhe lank:: Area CED* x Widlh o f tank C£xED* 2x3 2 X2.5" - , - X2.5=7.5 m) Force necessary

10

accele rate th e mass of water in the tank " Mass of water x Acce leration = p x Volume of water x 6.54 = 1000 x 7.5 x 6.54

=49050 N This is th e same force JS the difference of the lWO forces on the lWO ends of the wnk. Problem 3.36 A reclallgular wnk of lellglll 6 m. widlll 2.5 m (md lleighl 2 In is completely filled lI"ilh K·(ller when al resl. The tUIlk. 6· opell 01 Ihe lop. The 1(JIlk is subjected 10 0 hor;zoll/o/ COlis/mi l {illear accelemlioll of 2..1 mli ill Ihe directiOl1 of ils /ellglh. Find Ihe ,·olume of ,,"ala spilled from 1111;0 lallk.

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IL 1120

Fluid Mechani cs

So lution. Given:

L "" 6 Ill . b " 2.5 111 and height, H", 2 m 2 Horizontal acceleration. 1I '" 2.4 rnls , The slope of Ihe free surface of water after the lank is subjected to linear comaa,,1 acceleration is given by equation (3 .20) as Ian

. e = -" (Nullwrlcally) g

'" 2.4 '" 0.2446 9.& 1 From Fig. 3.48,

8e

tan6= -

AB

BC=ABxtan8 =6)(0.2446

I'

(": AB '" Length = 6 III ; Ian

e = 0.2446)

8m

'1

Fig . 3.48

= 1.4676 III Volume of water spilled = Area of ABC x Width of tank

=(+XABxBC)

x 2.5 =+)(6)(1.4676)(2.5

'" i 1.007 nrl,

(': (,:

Width = 2.5 10)

IJC= 1.4676111)

Ail S.

3.8.2 Liquid Container Subjected to Constant Vertical Acceleration . Fig. 3.49 sliows a tank containi ng a liquid and the tank is moving vertically upward wilh a constant 3(;ccicrmion. The liquid in the tank will be su bjected to the same ve nic al accckration. To obtain Ihe expression for the pressure at any point in the liqu id Illass subjec ted to venical upward accele ration. consider a vertical elementary prism of liquid CDFE. Free surface

•

1

I 1--01- ""-1 f-

pgh(1 · t ) - 1

Fig. 3.49 LeI dA '" Cross-sectional area of prism I,,,, Hdghl of prism /10 " Atmospheric pre5.';u re acting on Ihe face CE p " Pn:ssure at a depth II acting on the fat"'! DF

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IL Hydrostatic Forces on Surfaces

121 1

The forces acting on tile clcmcnlary prism arc: (0 Pressu re fOrl:c equa l 10 PI;) x dA acting on the face CE vcnically downward (ii) Pressu re fOrl:c equal !O p x dA acting on the face OF vertically upward (iii) Weight of the prism eq llal to p x g )( dA )( II acting through C.G. of the clement vcnicalJy downward.

According to Newton"s second law of motion. the nel force acting on Ih" clement must be eq ual 10 mass multiplied by accekration in (tIC same di rection. Net force in \'cnically upward di rection'" Mass x accelerat ion p x ciA - Po x
or

P-Po=pgll + plw

= pglt [I ,1

...

+~]

But (p - Po) is the g~uge pressure. l'len<:(; gauge pressure OIl any point in the liquid constam vertical upward ~lcceleration , is given by

m~ ss

P~ ==pgh[I+;]

(3.11)

subjec ted to

... (3.22)

... (3 .22A)

== pgh + plw

where P, == P - 1'0 == gauge pressure In equation (3.22) p. g and a are constant. Hence variation of gu~ge pressure is linear. Also when h == 0.1'8 == O. This means P - Po == 0 or I' = Po' Hence when II == O. the pressure is equa l to atmospheric pressure. Hence free surface of liquid subject~d to constant venical acceleration wi ll be horizon tal. From equation (3.22A) it is also c lear that the pressu re at any poi nt in the liquid mass is greater than the hydrostatic pressure (hydrostatic pressure is = pgil) by an amount of p x II x a. rig. 3.49 shows the variation o f pressure for the liquid mass subjected to a constant vertical upward acceleration. If the tank containing liquid is mo ving vertica ll y downward with a constant acceleration. then the gauge pressure at any point in the liquid at a depth of II from the free surface will be given by

(P-Po)=pgh[ I-~] =pgh-Pha

... (3.23)

,fo

The above equation shows that the pressure at any point in the liquid mass is less than the hydrostatic pressure by an ~mount of plla. Fig. 3.50 shows the variation of pressure for the liquid mass subjected to a constant vertical dow nward acceleration. If the wnk cunt aini ng liquid is moving downw ard with a constant ac(;c!eration equal to g (i.e., when a == g). then equation reduces to P - Po == 0 or P = Po' This means the pgn 9 pressure at an y point in the liquid is equal to surrounding atmospheric pressure. There will be no force o n the Fig. 3.50 walls or on the base of the tank, Note. If a lank containing a liquid is 'Ubjectl'{/ to a constant acceleration in the inclined direction. then the acceleration may be resolved ~long the horiront~l direction aTTd ,"ertkal direclion. Then each of these cases may be separately analysed in accordance with the aoo"e procedure.

-I'"' I N"~-~) I

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IL 1122 Fluid Mechani cs Problem 3.37 A limk cOlllailling ,water upla a dep/II of 500 mm is moring I'erlically upward WillI a conSWlI1 acceleralion of 1.45 mh-, Find IiiI' force exert,'d by Waler 011 Ihe side of II,e wnk. Also mlcu/are II,e force on Ihe side of Ihe /allk wh"11 the wid/II of tank i.! 1 111 and (i) {(1IIi: is mOI·ing I'erlically downward wilh II conSlalit accelermion of 2.45 m/I, and (ii) Ihe limk is IW( moving oj all. Solution. Given: [krIll of waler,

/, = 500

Vcnil:al acce leration.

a = 2.45 fills!

Width of tank.

b= 2m

n11n

= 0.5

III

I

"

To find the force exerted by wate r 011 the side

of the tank when moving vertically upward. Ie! us first tlnd the pressure al the honom o f th e tank. The gauge pressure at the bonolll (i.e .. at point B) for this case is give R by equation as

JIB == pgl!

(I+~)

= IOOOx9.8 1 xO.5

This pressure is re prcscl11cd by line Now the force on the

~idc

Fig. 1.51

2.45) - - ==6 131.25Nfm , (1+9> '

Be.

AB = Area of triangle ABC x Width of tank

==(txA8X8C) Xb = (t x O.5x6131.25) x2

(.: BC=6131.25 and/!= 2 m)

== 3065.6 N. AilS. (r) Fo rce On the s ide or th., honk, wh.,n honk Is m ovi ng nrlio:u ll y do wn ward.

The pressure variation is shown in Fig. 3.52. For this easc. th" pressure m th" bottom of the tank (i.e., at point 8) is given by equmion (3.23) as PB== pgli

,

(1 -~)

2.45) ( ,.,

==1000x9.8IxO.51- - -

"

== 3678.75 Nfn/

This pressure is represented by line

Be.

Now th" force on the side AB == Area of triangle ABC x Width ==(tXAJJxIJC)X b

= (t X 0.5 x 3678.75) x 2

Fig. 3.52 ('c Be. "78.75. b. 2)

== 1839.37 N. Am .

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IL Hydrostatic Forces on Surfaces 123 1 ~'o rc" 0 11 th e s id e or Iht' lank, ... ht'li tunk Is st..lionary. The pressure at point B is given by. fl8'"pgh = lO00x9.81 x05 = 4905 Nlrn 2

(ii)

This pressure is represented by line BD in Fig. 3.52 Force on the side An

:: Area of triangle ABD x Width

=(t XA B xBD)X b

= (i-xO.5x 4905)x2

(': ED :: 4905)

= 2452.5 N. AilS.

For this case. the force on An can also be obtained as "AIl= pgA.;;

wllerc

A:: AB x Widlll = 0.5 x 2 = 1 m 2

AS 0 .5 It:: - : =0.25111 = 1000 x9.81 x 1 xO.25

2

2

= 2452.5 N. AilS. Problem 3.38 A umk coll/oins »'a/er "plo (j depth of 1.5 m, The leng/II amI wid/I, of /lIe lank (lrl! -I m (lIld :z m respec/indy. Th e wnk is mo\'ing lip "" jllc/ined p/(Jlle with (I cons/,w/ (lccderu/iol! of -I ",Ii. The inc/inaliOll of Ihe p/mw will, the /u)riZOJ1la/ is 30° as 11110 ...11 in Fig. 3.53. Find. (i) Ihe ''''gle made hy Ihefree J'u rf"ce ofll'lIler will! Ihe horiWII{(,/. (ii) Ihe pressure III Ihe hOI/om of I/le lallt (1/ Ihe fro n/ ami rear ellds. Solution. Given:

Dcptli of water. 11 == 1.5 10 : Lengt li. L == 4 III and Width. b == 2 Ul

'm

Constant acceleration along th e inclined plane. II '" 4 Ill /5

1

Inclination of plaoe. 0: == 3D"

a

Let == Angle mad~ by the free surfac", of w~ter after tlie acceleration is imparted to the tank p~

== Pressure ,lithe bo1l0111 of the tank :It the from ",nd and Pv == Pressure at the bottom of the taJlk at the reaT end. Fig. 3.53 This proble m can be donr by resolving the give n acceleration along thc horizontal direction and vertical d irection. The n eac h of these c ases Ill ay be separately anal ysed according to the sct procedure. Horizontal and vertical eomponcnts of the acceleration arc: 2 a" '" II cos Cl " 4 cos 30° '" 3.464 lll/s

,

{/ == a sin Cl == 4 sin 30° == 2 m/s 2 When th~ tnnk is station~ry on the inclined plane, free s urfac", o f liquid will be along EF ~s shown in Fig, 3.53, But when til", tank is moving upward alo ng the indi Jled plane th e fr"'e surface of liquid will be along ne. When the tank containing a liquid is moving up an inclined plane with a constant accderation. thc angl e madc by the frce surface of thc liquid with thc horizontal is g iven by

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IL 1124 Fl uid Mechan ics _ "_,_ tan , '" 1I., +g

3.464

2+9.8 1

"" 0.293)

= tan - I 0.2933 '" 16 .346

Ans. "' Now lei us first find the depth of liquid at the front and rear end of the tank. {I

G

IXrlll of liquid at front end", II I '" AB [krill of liquid at Tear ~nd '" 11 2 '" CD From Fig. 3.53, in triangle eOE, tan

a = ~~

CE = EO tan

"'

e = 2 x 0.2933

(,: EO", 2 m. tan

= 0.5866 III CD = 112 = ED + CE = 1.5 + 0.5866 '" 2.0&66

e = 0.2933)

m

hl=AB=AF-BF

Similarly

= 1.5 - 0.5866

(": AF = 1.5. BF = CE = 0.5866)

'" 0.9134 m

The pressure at the bouom of tank 3l Ihe fear end is g iven by.

(I

+ ';)

PD= pghl

= 1000 x9.81 x Ul866 ( , + _ 2_) '" 24642.7 N/m z. Ans. 9$ ' The pressure at the boU01n of lank at the front e nd is given by

Pit '" pgh t ( I + (:')

'" 1000 x9.81 xO.9134 ( , + -'-) '" 10787.2 Ntml, Ans..

9>,

HIGHLIGHTS 1. When the nuid is at rest. Ihe shear stress i. zero.

2. The force exerted by

~

Sl:J.tic fluid on a vertical. horitontal Or an inclined plane immersed surface. /' '" pgAh

where

p • Density of the liquid. A • Area of the immersed surface. and

ii .. Depth of the centre of gravity of the immersed .urface from

free .urface of the liq uid.

J. Centre of pressure is defined as the point of application of the resultant pressure.

4. The depth of centre of pre,sure of an immersed surface from free surface of the liquid. for vertically inlinCrsc
for inclined immcr>ed surface.

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IL Hydrostatic Forces on Surfaces 125 1 S. The centre of pressure for a plane ,'crtical surface III's al a depth of two-third the height of the immersed surface. 6. The t01a1 force on a curved surface is given by F.

where

f~ ",

K

Hnriwntal force on cun"cd surface and is equal to total pressure force on the projected area of Ihe curved surface on the ,-"nieal plane. pgAh

F, '" Vertical force on sub-merged curvcd surface and is equal 10 Ihc weight of liquid actually or imaginary supported by the curved surface.

and

F 7. The inclinalion of Ihc result,ml force on curved surface with horizontal. tan e = .2.... F,

8. The resultant for • .:: on a sluice gate. F = FI - Fl where

F, = Pressure force On (he upstream side o f (he sluice !laIc and

F : '" Pressure force on the downstream side of the sluice gate. 9. Po. a lock gate. the reaction ""'tween the two gates i. equal to the r~action at the

hing~"

R" I'.

Also the reaction between the two gate •• p .. __ F_ 2 sin {I where F .. Resultant water pressure on the lod gate .. F, - F2 and

{I

= Indination of the gale with the nonnal to the side of the lock.

EXERCISE (A) THEORETICAL PROBLEMS I . What do you understand by 'Totall'Tessure' and "Centre of Pressure" ? 2. Derive an expres,ion for the force exerted on a sub·merged vertical plane surface by the static liquid and locate the position of centre of pressure. J . Prove that the centre of press ure of a completely sub·merged plane surface is always ""' low the centre of gravity of the sub·merged surface or at most coincklc with the centre of gravity when the plane surface is horirontal. 4. Prove th~tthe tOl~1 pressure Herted by a static liquid on an inclined plane sub·merged surface is lhe ""me a< the foree exerted on a vertical plane surface as long as the depth of the centre of gravity of the ,urface is unal!ered. 5. Deri~e an expression for thc deplh of centre of pressure from free surface of liquid of an inclined plane surface sub·merged in the liquid. 6. (u) How would you detcnnine the horiwntal and wrtical components of the re,uUant pres,ure on a sub· merged curved surface? (h) Explain the procedure of finding hydrostatic forces on curved surfaces. (f)elhi Un;\·er.
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IL 1126 Fluid Mechani cs !I. De .. ribe briefly with sketches the various methods used fo r measuring pressure excncd by fluid s. 10 . Prove that the "cr1ical component of the resu ltant pressure On a sub-merge
I I . What is the difference between sluice gale atld lock gate?

,.

11. Prove 1ha1 the reaction between the gates of a lock is equal to the reaction at the hinge.

u.

Derive an expression for the reaction between the gates as I' '" _ __

2 sin e

where"" Resultant water pressure on lock gale, (I '" inclination of the gate with nonnal to the side oflhe lock. 1-1 . When will centfe of pressure and centre of gmvity of an immersed plane surface coincide?

IS. Find an expression for the force cxened and ccnlre of pressure for a complclely SUb-merged inclined pi:me surf~ce. Can Ihe same melhod be applied for finding Ihe resull~nl for.:e On a curved surface innners..'
(B) NUMERICAL PROBLEMS I . DClcnninc Ihe 10lal press ure and dcplh of cenlre of pressure on a plane re<:I:,ngulur surface of 1 111 wide and 3 111 deep when ils upper edge is horizonlal and (al coincides wilh Waler surface (b) 2 111 below Ihe [Ans. (a) 44 145 N. 2.0m. (b) 103005 N. 3.71 4 ml free waler surface. 2. Dctennine the IOlal pressure on a cir.:ular plale of diamcler 1.5 m which is placed verlkally in walcr in such a way Ihal ceTllrc of plale is 2 m below Ihe free surface of waler Find Ihe posilion of cenlre of pressure also. IAn< . 34668.54N.2.07rnl 3. A re<;langular sluice gale is situaled on the ,·crlkal wall of a lock. The verlkal side of the sluice is 6 m in lenglh and depth of centroid of area is 8 m below the Water surface. Prove Ihat Ihe depth of cemre of pressure is given by 8.475 m. 4. A circular opening. 3 m diamelcr. in a vertkal side of a tank is closro by a disc of 3 m diameter Which can rotate about a horizontal diameter. Calculate (i) the force on the disc. and (iI) the torque required 10 maintain the di", in equilibrium in the .. ertical position when the head of water abo,·e the hori,ontal IAns. (i)416.05 kN . ( ii) 39005 Nml diameter is 6 m. 5. The pressure at the centre of a pipe of diatneter 3 m is 29.43 Nkml. The pire contains oi l of sp. gr. 0.87 and is filled wilh a gate valve . Find thc force cxene<:l by the oi! on the gate and posilion of centre of pressure. IAns.l.OIl MN ..016 m below cenlre of pipe] 6. Determine the tota l pre.<>ure and ccntre of prc
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IL Hydrostatic Forces on Surfaces 127 1 K. A cOlis>')" for closing thc enlrance [0 a dry dock is of Impel0idal form 16

'I.

10 .

II .

I Z.

13.

14.

In

wide at the lOp and 12 m

wide at the bonolll ,md 8 111 deep. Find Ihc 10lai pressure and cenlre of pressure on Ihc caisson if the water on the outside is 1 m below the lOp le"cl of thc caisson and dod is empty. [Ans. 3.164 MN. 4.56 m below water surface I A .Iiding gale 2 III wide and 1.5 In high lie, in a venical plane and has a co-effocicnt of friction of 0.2 between itself and guides. If Inc gate weighs one tonne. find the vertical force required to raise thc gate if its upper edge is at a depth of 4 m from free surface of wmer. IAns. )7768.5 NI A lank conuin, WOller upto a height o f 1 In above the ha"", An imm;",,;ble liquid of _'po gr. 0.8 is (,lied on the lOp of waler UplD 1.5 m height . Calculale , (il total pre«ure on One side of the lank. (iiI Ihe position of cenlre of pressure for one side of the tank. which is 3 In wide. IAn •. 76518 N. 1686 In from lopl A reclangular tank 4 m long. 1.5 m wide cont";,,s water Uplo ~I height of 2 m. Calculale Ihe force duc 10 waler pressure on Ihe base of Ihe lank. Find also Ihe deplh of cenlre of pres,ure from frcc surface. IAns. 117720 N. 2 m from free surface I A reclangular plane surface I m wi
Fig..H4

Fig.3.SS

15 . An inclined rcct:lngular gale of width 5 m and depth .5 m is installed to comrol the discharge of waler as shown in Fig. 3.55. The end A is hinged. Dcle""inc Ihe force nonnal to Ihe gale applied al H to open it. {An s. 97435.8NI 16. A gale supporting wa(er is shown in Fig. 3.56. F;nd lhe heigh( 'h' of (he wa1er so tha( lhe ga(c begins to (ip aooul 1he hinge. Take the widlh of lhe gale as unilY.

IAns. 3)(./3 "'I

17. Find the tOlal pres,u re and depth of cen1re of pressure on a Iriangular plate of base 3 m and heighl 3 III which is immersed in waler in such a way lhal plane of lhe pla1e makes an angle of 60" Fig. 3.56 wilh the free surface. The ba
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IL 1128 Fluid Mechani cs Ill . Find the horizontal and vertical components of lhe 101al force acting on a curved surface AB. which is

in lhe fonn of a quadrant of a ,-irde of radius 2 m as shown in Fig. 3.57. Take lhe width of Ihe gale 2 111. IAns. F, = 117.72 kN. F, = 140.114tNI WATER SURFACE

-------------

, Fig. 3.S7

Fig. 3.58

19 . fig. 3.58 shows a gale having a quadrant shape of radius of 3 Ill. rind Ihe resultant force due to waler pcr metre length of the gale. rind also the angle at which the total force wil l acl. [Ans. 82.201 kN, 0 = 57° 31 ' I 20 . A roller gate is shown in I:ig . ).59. It is cylindrical fonn of 6.0 In diameter. It is placed on lhe dam. I'ind lhe magnitude and direction of lhe reSUllan! force d ue 10 waler aCling On the gale when the waler is just

going 10 spilL The I

1 of [he gal" is gi"en 10m.

Fig. 3.59

( Ans. 2.245 M N.e~3l!" g·1

Fig. 3.60

21. Find Ihe horizontal and vertical components of Ihe water pressure exerted on a tainter gate of radius 4 rn as shown in Fig. 3.60. Consider widlh of thc gate unity. (Ans. F, _1962 kN. F, _ 7102.44 N I 22. Find thc magnitude and dirtttion of thc resu ltant water pressure act illg on a cun'cd face of a dam which is shaped

",

according to Ihe relalion y '" -

WATER SURFACE

as shown in I' ig. 3.61. The

height of water retained by the dalll is 12 Ill , Take the width IAns. 970 ,74 kN. 43 " 19'] of dalll as un ily 23 . Each gale of a lock is 5 III high and is supporled by two hinges placed on Ihe lOp and bollom of Ihe gale. When Ihe

e ..

Fig. 3.61

gales are closed. Ihcy make an angle of 120". Tn e widTh of The lock is 4 Ill. [f The depths of water on the IWO sides of the gates are 4 m and 3 m respectively. o;ietem,inc , (I) the magnitude of resultant pressure on cach gate. and (ii) magnitude of the hinge reactions. IAns.(i) 79,279 kN. (ii) Rr • 27.924 kN. R~. 51.355 kN] 24 . The end gates ARC of a lock arc 8 m high and when closed make an angle of 120°. The width of lock is 10 m. Each gate is supJXlrlcd by IWO hinge s located at 1 III and 5 m above the bollOIll of the loc\;, The depth of Waler on the upslream ami downstream side, of the lock are 6 m :md 4 HI respccti~ely . Find : (I) Rcsulta1l1 water foree on each gate.

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IL Hydrostatic Forces on Surfaces 129 1 (ii) R""c(ion between the gates AB and BC, and (iii) f orce on each hinge, considering the reaclion of the gate acting in Ihe same horizomal plane 3S

re sultant water pressure. [A ns. 566.33 i:N. (iiI 566.33 kN. and (iii) R, _ ! 73.64 kN. R8 _ 392.69 kNI 25. A hollow circ ular pl~IC of 2 m external and I 111 internal diameter is immer>ed ""l1i'31Iy in waler such thm the cenlre of plate is 4 111 deep from water surface. Find the to(al pressure and depth of centre of pressure. [Ans. 92 .508 kN. 4.078 ml 26 . A rec~~ngular opening 2 111 wide and I 111 deq, in the vertical side of a tanl; is closed by a sluice gate of the same size. The gUll: can turn abom the horirontai ~en\rOidal axis. Dclcrminc : (.) !he lotl! prc,;surc on the slu;"" gale and (il) the torque on the ~Iuice gatc. 11Ie head of wale, alxwe the upper edge of the gale is 1.5 111. IAns. (il 39.24 IN. (il) 1635 Nml

,

27 . Octennine the total force and kx:ation of .'entre of pressure on one face of FREE SURFACE OF L1aU ID the plate shown in Fig. 3.62 immersed in a liquid of specific gravity 0.9. IAn • . 62. 4 kN. 3.04 ml 28 . A circular opening. 3 m diameter. in the "er1kal side of water tank: is cloSC
-

(' ,] X 1.75", 121.J5kN.

F=pgAh '" l000x9.81 X "4 x3 "

h' •

+ 1.75 .. 0.08 + 1.75 .. 1.83 m.1

Fig. 3.63

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IL

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CHAPTER

.. 4. 1 INTRODUCTION In th i~ c hapte r. the equ ilibrium Ml he Oo alin g and sub-me rged bodies will be co nsidered . Thu s Ihe c hapter will inc lude: I. Buoya ncy. 2. Ce ntre of buoy ancy. 3. Mclace ntre. 4. Metace ntri c hf.> ight. 5. Ana lytica l method fo r deta minin g mdacentric heig ht. 6. Condition s of equilibrium o f a fl oatin g and sub -merged bod y. and 7. Expe rim ental met hod fo r metacc ntri c he ight.

.. 4.2

BUOYANCY

Whe n a bod y is imm ersed in a fluid_ an up ward force is exe n cd by the fluid o n the bod y. Th i~ upw ard forec is e qual to the we ight o f the fluid displaced by the body and is c al k d th e force of buoy an cy or s impl y bu oy ancy .

.. 4.3

CENTRE OF BUOYANCY

It is de fin ed as the poi nt. th rough whi ch th e force of buoy ancy is supposed to ac t. As tlw force of

buoyancy is a ve rtica l force and is equal to th e we ight of th e Ou id d isplaced by th e body. the centre of b uoy ancy wi ll be thc ce ntre o f grav it y of th e fluid di splaced. Proble m 4.1 Fittd lite I'Olume of Ille lI'aler displaced alld pwiliOlt of cenlre of buoyancy for a lI'oodell block of wid III 2.5 m and of depth 1.5 m. whell il floals horizollwl/y in Waler. The densilY of woodell block i.f 650 kg/III ) {Illd ils lenglli 6.0 m. Solution . Gi vc n : Width " 2.5 10 Ikplh Le ngth Vol ume of Ihe bl oc k

" 1.5 111 " 6.0 111 = 2.5 x 1.5 x 6.0 = 22.50 ln 3 p = 650 kgfm} Iknsi ty of wood. Wf.> ig ht of bloc k = px g x Volum c = 650 x 9 .81 x 22.50N= 14347 1 N

jw G 2.5 m

Fig. 4.1

131

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IL 1132

Fluid Mechani cs

For equilibrium Ihe wei gh l of water di sp la ced'" We igh.t of wooden block = 14 347 1 N Volum~

o f W31.-:r displaced

Weig ht of wa ter di sp laced We ight de nsit y of water

14347 1 ~~"':'cc= 14.625 ml, AilS. 1{)()() X 9.8 1

(': We ight density of wmcr '" 1000 x 9.81 N/ml) Pos it io n uf Ce ntre of Uuoyancy. Vol ume of woode n b lock in water or

'" Vo lume o f water d ispla(;cd 25 x II x 6.0 '" 14.625 m) , w here /, i s depth of wooden block in W3Ie r II '"

14.625 '" 0 .975 2.5 X 6.0

III

0.975 Cent I"C of Buoyanc y '" - 2- '" 0.4875 m from base.

Ail S.

Problem 4.2 A ...oodell fog 0/0.6 m diameter and 5 m lenglll is j7
w =

,

, o o

px g

Fig. 4.2

=700x9.81 Nfm 3 Find depth of immersion ur II W~igtl! of wooden log

::

'" Weigh! densily x Volum e of log

,

,

=7 00x9.8 1 X 4 (DrXL =700x 9 .81 Fur equi librium. Weigh! of wooden log

X~(.6)2 X5N=989.6X9.81 N

= Weigh! of wa!er displaced

= Weigh! densi!y of water X Volu ,nc o fwatcr di splaced Vo lume u f wa!cr displaced '" 989.6 x 9.8 1 = 0.9896 l000x9.81

m3

(': We igh! densi!y o f wat~r '" 1000 x 9.81 Nfm3) Let" is the dcpth o fi'nmc rs ion Volume of log ins ide wa ter'" Area of ADCA x Len gth

= Area of ADCA x 5.0 But vo lum ~ of log inside water == Vo lum e of water displaced == 0.9896 m}

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IL Buoyancy and Floa tation

133 1

0.9896:: Area of A DCA x 5.0 :. Area of ADCA

__ 0.9896 __0. 1979 5.0

Bm area of ADCA

:: Area o f cu rved su rface ADCOA + Are a of MOe

:: It?

)·-"l [3(j(360"

:: It?

[l - ~l IS(r + , 2 OOS. 9 .sin 9

0. 1979 = It

.00157

or

e-

III '

+

"21 r (;Os e x 2r si n e

(.3)~ [ l- l ~.] + (.3)~cos 9sin e

0. 1979:: .2827 - .{1()157 e + 0.91:0s e si n e .09 cos e sin 9:: .2827 - . 1979 :: 0.0848

, ., .0848 , - -.09 - - cos S10 :: - - .00157 .00 157 e - 57.32 cos e sin e:: 54.01. or e - 57.32 ("(IS e si n e - 54.01 = 0 For 0:: 6(}0, 60 - 57.32 x 0.5 x .S()6 - 54.01'" 60 - 24.81 - 54.01:: - llUn For

e:: 70°,

70 - 57.32 x .342 x 0.9396 - 54.01 '" 70 - 18.4 - 54.01 :: - 2.41

For

e::

72 - 57.32 x.309 x .951 - 54.01 = 72 - 16.&4 - 54.01 = + 1.14

72°,

For e:: 7 1°,

e:: 71.5", T hen

7 1 - 57.32 x .325 x .9455 - 54.01 :: 7 1 - 17.61 - 54.01:: - 0.376

7 1.5- 57.32x .3 173 x .948- 54.0 1 :: 7 1.5- 17.24-54.0 1 = + .248

/,=

r + rl'Os7 15 °

= 0.3 + 0.3 x 0.3173 = 0.395 m . Am. Problem 4 .3 A slOlIe weighs 392.4 N ill air and 196.2 N in .mler. Compule Ihe )"olume of slon e and ils specific grm·ily. Solution . Given: Weigtlt of SlO ne in air Wdglll of Slone in water

= 392.4 N == 196.2 N

ror equilibrium. Weigtll in air - Weight of sto ne in water = Weigh t of w~H er displa<:ed or

392.4 - 196.2 == 196.2 = 1000 x 9.81 x Volume o f wata displaced Volume o f wa ler displaced 196.2 t t = ~~~~== _ rn 3 == - x 1 0~ c rn 3 == 2 x 10' em l . Ans. looox9.8 1 50 50

= Volume of Slo ne Volume o f Slo ne

I I

== 2 x 10· cm 3. An s.

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IL 1134

Fluid Mechani cs

Specific G rav it y or Ston e '" Wcig tll in air '" 392.4 = 40 kg g 9.8 1 Mass in air '" 40.0 kg

Mass o f stone

=

Densi ty o f stone

Vol ume

~rn J

kg

= 40x50=2000 -

rn

l

5()

De nsi ty of Slone

Sp. gr. uf stolle

'"

2000

'"

'" 2.0. AilS. Density o f wak r 1000 Problem 4 .4 A body of dimensions 1.5", x f.O '" x 21M, weighs /962 N in ....afer. Find ils I\"eight in air. IViI,,/ will be ilJ' JJlecific gra,·;ly ? Solution . Gi ve n : Volum e of bod y '" 1.50 x 1.0 x 2 .0 '" 3.0 Ill l Wcig tll of bod y in wa!c r '" 1962 N Vol um e of rhe wa ter di splaced", Vo lume o f the body '" 3.0 Ill]

Weight of water d isplaced = 1000 x 9.8 1 x 3.0 '" 29430 N For th e equilibrium o f rhe hod y Wc igtn o f body in air - We ight of water di sp laced'" Weight in water \¥.. r - 19430= 1962 IV..,'" 29430 + 1962 = 31 392 N

= Weight in air '" 31392 '" )200 kg

Mass of body

9.81

g

=

Den sit y o f rhe body

Sp. grav ity o f th e bod y

=

M ass

Vo lu me 1066.67

'" 3200 '" 1066.67

3.0

'" 1.067. Ans .

1000 Problem 4.5 Find rlre den .~ily of a melilllic hody which /loms at Ihe inlerface of merCllry of Sf'. gr. 13.6 alld Waler silch 1/1(1140% of ils \·olume is sub-merged ill mercllr)" and 60% ill wal"'. Solution . Le t th e vo lume of the body '" V m 3 The n vo lume of body sub-me rged in me rcu ry

40

= - V", 0 .4 Vm 3 100 Vo lum e of body

sub - m c r~ed

in wate r

""

~_ )( V",0.6 Vm 3

I()()

For the equilihrium of the bod y

Fig . 4. 3

To ta l bu oy ant fo r"e (up ward force) '" We ig ht of the body Bu t to tal buoy ant force '" Forc~ of buoy an cy du ~ to wa ter + Force of buoyan cy d ue to merc ury Force of buoyancy d ue to water '" Weig ht o f wa tn d isplaced hy IxHl y '" De ns ity of water )( g )( Vo lume o f wate r d isplaced = 1(00)( g )( Vo lume of body in water

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IL Buoyancy and Floa tation

135 1

= 1000 x g x 0.6 x V N

and Force of buoya llcy du e to mercu ry

= Weig ht of me rcury di splaced by body = g x D ensi ty o f mercury x Vo lume of mercury displ ace d = g x 13.6 x 1000 x Vo lum e o f body in merc ury

We igh t of the bod y wtwre p is the de nsit y of th e bod y For equilibriu m. we ha ve

= Densi ty x

=Ex [) .6x I OOOx O.4 VN g x V olume of body

=Px

To tal buoyant force = Weight o f th e body I OOOxgxO.6xV + 1),6x 1000xg x .4 V=pXg x V or p = 600 + 13600 x.4 = 600 + 54400 De ns it y o f the body '" 6040.00 kg / m ) . An s.

Problem 4 .6

g x V

= 6040.00 kgfm l

A flolli WIll'" regulates I/le flow of oil of sp. gr. 0.8 illio "ciSlem, The JpileriClil float

iJ' 15 em ill diameter. AOlJ is {/ weighlleS!>' link carrying lile floal al anI' <'11
end wllich closes Ille pipe Iltmugll wllich oil flows illla IIII' cis/cm . The /i,,1: i.1 mounled in (I friClionle.H liillge m 0 alld llie (llIgle AOB i.f 135°. Tlit' lellglh ofOA is 20 em, (l1II! Ille disrtillee belweell Ille eelllrt' of Ihe flom mId II,e hillge is 50 em. WIIt'll Ihe flow i.l slOpped AO will be I·ulieal. The \"G/re is 10 be pressed all 10 II,e se(ll WillI a force of 9.81 N 10 co mplerel)' slop Ihe flow of oil inlo Ille cistern. It W(lS obserl'ed IIwl the flow of oil i')' slopped whell Ihe free SlI rfrlCe of ai/ ill Ille cislem is 35 em bela,,· Ille hillge. Delermil1f: Ihe weiglll of Ihe floo/. Solution . Gi ve n : 0"

Sp. gr. of oil

SUPPLY

'" 0.8 Po = 0.8 x 1000

Densit y of oil.

0:

800 kglm'

D o: I5cm

Dia. of flo at.

LAOB = 135 Q OA=20cm

Force .

P=9.8 1 N

O B = 50 cm Find tile weig llt of [h e float. Le t it is eq ual [ 0 W. Wile n til e flow of o il is stopped. the ce ntre of float is s how n in Fig. 4.4 Tlic leve l of o il is al50 sllown. Til e ce ntre of fl oat is be low th e Icve l of o il. by a de pth '/r' .

From tJB OD,

sin 450 0: 00 " OC + CD

OB

OB

35 + II

50

50xsin45° ,,35 + h II = 50 x

I

J'i - 35 '" 35.355 -

35 " 0.355 Ctn " .00355 tn.

Th e weig lu of floa t is ac ting th ro ugh 8, but Ih" upward buoya nt force is acti ng throug h th" CCl)tre o f weight o f o il displa(;ed . Vol um e uf oil disp l;lccd

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IL 1136

Fluid Mechani cs

'" t Buoy ant force

x

It

X {.075)3 + .{)(}355 x

It

x (.075) 2 '" 0.000945 111 3

'" We ig ht of oil di splaced '" Po x g X Volume of oil

: 800x9.8 1 x.000945 = 7.416 N The buoyall1 force ~nd weight of Ihe floal passes through th e sa me vertical li ne. pass ing through B. Let the we ig ht of float is W. Th en net vertical force o n 110.11 '" Buoyalll fo rce - Wc igtu of flo at'" (7.416 - IV) Tak ing mo ments about the hin ge O. we gel

or

Px 20 '" (7 .4 16- IV) x 8D = (7.4 16 - IV) x 50 x \:0545 0 9.8 1 x 20 '" (7 .4 16 - IV) x 35.355

16 - 20x9.8 1"=7.4 T 35.355 1 = 7 .4 16 - 5.55= U!66 N. Ans .

.. 4.4

META -CENTRE

It is defined as Ihe (Xlint about wllich a body Slarts oscillatin g when Ih e body is tilled by a small an gle. The 11lc1a-ccntre may also be de fi ned as the point at which Ihe li ne of act io n of th e force of b uoya ncy will meet th e nonnal axi s of th e body whe n th e body is give n a s mal! angulardisplaccmcnt. Consider a body fi oating in a liq uid as s hown in Fig. 4.5 (a). Let the body is in equilibrium and G is th e ce ntre of grav it y and B the ce ntre o f buoyancy. Fo r equilibrium. both the points li e on the norma l axis. whic h is ve n ic al.

J-... ANGUlAR

NOj MAL AXtS

- "'-!-".

!;

M DISPlACEME

•. . . :.. t... ~

-~

I

,, ,

I"

/

I

-;;'~?';-'

I'.

NORMAL AXIS (b)

(a)

Fig. 4.5

M na-<:f'nlr t

Le t th e body is given a small angular displa~"t!ment in the clockwise di recti on as Show n in Fig. 4.5 (b). The I:entre of buoya ncy. wllich is Ille I:entre of gravity of Ill e di spl al:ed liquid or cenlre uf grav ily uf the ponion of Ihe body sub-11Ie rJ(cd in liquid. will now be Shifted towa rds righl from Ihe nurm al axi s. Lei iI is al B[ as show n in Fig. 4.5 (b ). The line o f action of lhe force of buoyancy in Ih is new posi tion. will in tc["Sectlhe [lOTinal a~is of th e hody at SOllie point say M. This point /If is called Met,,· IT n lre .

.. 4 .S

META -CENTRIC HEIGHT

The d islanc e MG. i.e .. Ih e disl anl:e belween th e mcl a-ce ntre of a n oal in g body and Ille cemre of grav it y o f the body is called IllCla-centric heig ht.

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IL Buoyancy and Floa tation

.. 4.6

137 1

ANALYTICAL METHOD FOR META -CENTRE HEIGHT

Fig. 4.6 (0) shows the position of a noming body in equilibrium. The location of centre of gravity and centre of buoyallcy in Ihis position is al G and 8. The floming body is given a small angu lar displacement in the clocl;wisc direction. Tliis is shown in Fig. 4.6 (b). The new centre of buoyancy is al 8 1" The vcnica l linc through 8 1 cuts Ihe normal ax is at M. Hence M is the meta-centre and GM is mew-centric height.

----.

ANGULAR

M

A

ISPLACEMENT

"'.

S'

S Go S

Gi--(+~I:--"C

,

'. C

T

101

(

l (c)

PLAN OF BODY AT WATER LINE

~.E-". Fig. 4.6

Mela-Cf'nlre height of f/Qdfillg body.

The angular displacement of the body in Ihe cloc kwise direction causcs the wedge-shaped prism B08' un Ihe rigtu of the axis 10 go inside the water while tile idemical wedge-shaped prism reprcscmed by AOA' emerges OuI of tile water un tile len uf tile axis. These wedges represent a gain in buuyam force on the right side and a corresponding loss of buoyant force on the len side. The gain is r~pr~scnted by a vertical force dFJj acti ng th rough the c.G. of the prism BOB' while th e loss is represcnt~d by an equal and op]Xlsite force dFo ac ting Vertically downward through the centroid of AOA'. The coup le due to these buoyant forces
If L is the lenglh of lhe floating body. then Volume of strip

:AreaxL "xx6xLxdx

Weight of stri p := pg X Volu1\le:= pgx fJL dx Similarl y, if a small Slrip of thickn~ss dx at a distance .l from 0 towards the left of the axis is considered, the weight of Slrip will he pg_.fJ L d.L The two weights are acting in the opposite direction and h~nce constitute a coupl~.

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IL 1138

Fluid Mechani cs

Moment of ttJis couple

'" Weight of each stri p x Distance between Ihes<: two weigh ts =pgx fJLdxlx+_t]

= pgx aL dx

x 2x = 2pgxl 9L dx

Moment of the couple for the whole wedge

"J 2pg_~ aL dx

.. ,(4 . 1)

MomcnI of coup!c duc 10 shirting of cemrc of buoyancy from B 10 HI :FyxBBI

= FHXBMxfJ

('0" BB I

"

BMx fJif9is "cry

" WxBMx9

~m alll

I': Fe" WI ...(4.2)

BUl these two l'Oup lcs arc the same. Hence equating equations (4.1) and (4.2), we get

x HM x e" J lpg_'.! e ulx IV x 8M x e" 2pg9 J:h_Ax IV

IV x 8M "" 2pg

J_'~Ldx

Now LAx == Elem en tal area on the wmcr line shown in Fig. 4.6 (c) and == dA

II' x 8M '" 2pg

j .,.2dA .

BUl from Fig. 4.5 (el it is clear lhm 2/ x 2 dA is lh", second moment of area of the plan o f Ihe body at water surface about the axis Y~Y. Therefore 2 Wx JiM", pgl lwhere I '" 2 f x dAl 8M = pgl IV

IV" Weight of the body

""'

= Weight of the fluid d isp laced by the body

= pg x Volume of the fiuid displaced by th e body = pg x Volume of the body sub-merged in water

= pgx V BM = pgx'=i

pgxV

.. .(4 .3)

V

I GM=BM - BG= - - BG Meta-ccntric height

Problem 4.7

I

•

= GM = - -8G.

•

... (4 .4 )

A rec/(Ingu/ar prwlOafl is 5 /II 101lg. J m ...ide and 1.20 m lIigll. The deplll of immerJ'iOll of Ihe pOllloOll is 0.80 m ill se(l waler. If /lle cell/re of grtll"ily is 0.6 '" abo\'e Ihe bollom of (h e pOll/OOll. delermill e lite meia -celllric heighl. The dellJ'ily for Sell waler = 1025 kg/", J.

Solution. Given: Dimension of pontoon !kpth of imm ersion

I I

=5mx3mx l. 20m = 0.8 m

Ii

~ I

IL Buoyancy and Floa tation AG", 0.6

Distance Distance

AB

=

t

1-'-

III

x Depth of imlllersion

'" -t. x

.8=0.4 III Iknsity for se a Waler '" 1025 kg/Ill ' Meta -ce ntre height GM. given by equat ion (4.4) is GM= -

where

f'" M.D.

IlIcni~

I

•

3 m -----.,

",,-r-:t; G ,Tl+. , ...&, .. s ' 4 6m 1mO.6m T , A

T

-BG

of th'" plan of the poll1oon about Y-Y allis

S.Om

45 m~ 12 4 '
'"

139 1

1

~ x 5)( 31 m~ '"

,

=3)(0.8)(5.0= 12.0ml

PLAN AT WATER SURFACE

BG = AG - Ali = 0.6 - 0.4 '" 0.2 111 45

GM = -

4

Fig . 4.7 I 45 x - - - 0.2 '" - - 0.2 '" 0.9375 - 0.2 = 0.7375 m. Ans.

12.0

48

Problem 4 .8 A wllform hody of size 3 m long X 2 m wide x I 111 deep jlOlllS in ImIN. IV/WI is the weighl of I/le botly if depth of immersion is 0.8 //I ? De/ermine the mew-centric fleig/II (I/$(). Solution. (jiven : Dimcnsioll o f body

= 3x2xl

Deptli of imm ersion

= 0.8

-t--. --+-

III

Find (i)Wcigh t of lxKly. IV (ii) Meta-centric lIeiglit. GM (i) Wt:ight of Bod~' , W

j.- 3.0m ~.J

== Wcigllt of water displaced == pg x Volume of water displaced

-~c4'C!tft--~-

== lOOO x 9.81 X Volume o f lxKly in water == H)OOx9.8 1 x J x 2 xO.8 N == 47088 N. AilS.

A ELEVATION

Fig. 4.8

(iI) Ml."h,-cl."ntrk !lelght . GM

Using equation (4.4). we get I GM== - - BG whe re

f

•

= M.O.I about y. y axis of the plan of the body

! 3 ]x2 J ~ = - x]x2 = - - =2.0 m 12 12 V == Volume of body in wmer =3x2xO.8=4.8 m3 BG= AG - AB =

~_ 2

0.8 = 0.5 - 0.4 =0. 1

2

GM = 2.0 _ 0.1 = 0.4167 _ 0.1 = 0.3 167 III. A" s. 4.8

~ I

I~

~ I

IL 1140

Fluid Mechani cs

Problem 4 .9 A block of wood of specific gm!'il}' 0.7 j1oalJ' ill Waler. Determine Ihe mew-celllric hciglll of the blod:. if its size i.! 2 III X I III x O.S m. Solution. Given: =2 xlxO.8 Dimension of block

1

2.0 m

leI depth of immersion

'" II

Sp. gr. of wood

'" 0.7

Wcigllt of wooden piece

'" Weight density of wood· x Volume

In

:

'CAN ,

= 0.7 x IOOOx 9.Rl x 2 x 1 xO.8 N

Wc igllt uf water displ
-

wa ter

I

-'-

-.

~! A

~_8';;- ~ c

1

~

=1000x9.8Ix2xlxhN Fig. 4.9

For eq uilibrium. Wciglll of wooden piece

'" Weighl of watl'r displaced

700 x9.8 1 x 2 x 1 x O.ll '" 1{lOO x9Jll x2 x I x II

/,=

100x9.81x2x l xO.8 :O.7xO.8=O.56m lOOOx9.8 1x2x l

Distance of ccnlre of Buoyancy from 001l001, i.e .• AB = and

I!

0.56

- '" - - '" 0.28 1lI 2 2

AG '" 0.8/2.0 '" 0.4 m UG =AG - AB '" 0.4 - 0.28= 0.12 In

The Inew -ce ntril: height is given by equation (4.4) or GM: -

where

1

\

1

4

f

•

-BG

/: - x2x1.0· : - m 12 6 'I;f:

Volume of wood in wate r

:2x 1 xll",2x I x.56: 112m l I

I

GM: - X - 0.12 '" 0.1 488 - 0.12 '" 0.0288 tn. Ans. 6 1.1 2 Problem 4.10 A solid cylillder of diollleler -1.0 III has (I heighl of 3 melres. Filld Ille melli-centric /!eiglll of Ihe cylillder whell il is floil/illg ill Wil/er wilh ils axis I"eflical. Ti, e sp. gr. of Ihe cylillder '" 0.6. Solution. Given:

Dia. o f cylindcr.

D", 4.0 m

Height of cylinder,

II '" 3.0 III

• Weight density of wood

I I

.. p x g. where p .. density of wood .. 0.7 X 1000 = 700 kg/m ). Hence ... for wood = 700 x 9.8 1 Nlm l.

Ii

~ I

IL Buoyancy and Floa tation Sp. gr.

or cylinder

'" 0.6

j-o-

,

4m -

[kpIII of im mersion of cylinde r

'" 0.6 x 3.0 '" 1.8

1.8 09 AB"'T'" .

141 1

o<

III

III

'CAN

3

AG=2'" 1.5m

I

T

~l, 'r ~

~

BG: AG - All

= 1.5-0,9=0.6111 Now the meta-centric he ight GM is given by equation (4.4) GM= -

"0'

/

~

Fig. 4.10

-nG

"

J = M.O.!. about Y-Y ax is of the plan of Ih.-: bod y

= ~ 0 4 : ~ X (4.0)4 64 64 md

'V '" Volume of cylinder in wata

= .::. D! x Depth of inllllcrsion 4

GM=

.

~x ( 4.0)· 64 ,

_ 0.6

- x (4.0 )" x 1.8 4

I

= 16

4.0'

,

x - - - 0.(, = 1.8

I

1.8

- 0.6 = 0.55 - 0.6 = - 0.05 m. AilS.

- vc sig n 1I\eans 1ha1 IllcI3 -cemrc. (M) is below the c.;,olre of gravity (G).

Problem 4.11 A hotly /W.I lilt! cylintlricalllpper ponion of 3 m diameter alld I.S //I deep. nle lower POrllOlI is a cun'cd one....hich displaces a )'oilllne of 0.6 /IIJ of wa/er. The cenrre of blw)'oncy of ,he cun'ell pOrlion is at a diswnce of 1.95 m below the lop of the cylinder. Th e celilre of gwriry of Ihe whole body is 1.20 m below the lOp of the cylinder. The lolal displacement ofwatl'( is 3.9 1011llt'S. Find Ihe mela--cenlric heiglll of the body. Solution. Given: Dia. of body [kptli of body

" J.O

III

" 1.8 III Volume disp laced by curved portion

= 0.6

1

m o f water.

Let 8 I is tlie cemre of buoyancy o f the curved surf:lce and G is tlie centre of gravity of the whole body.

I I

Ii

~ I

IL 1142 Fluid Mechani cs CB, '" 1.95 III CG:I.20m

Then

Total wCigtll of water displaced by body = 3.9 lonncs == 3.9 x 1000 == 3900 kgf

+-.-+

= 3900 x 9.81 N = 38259 N Find mela- ct'nlric heighl of the body.

Lei tile heig ht of the body abov" th e water s urface x 111. Total weight of water displa\:cd by body = Weight density of water x [Vulume of water displaced I

'" 1000

x 9.81 x IVolu nw uflhe body in water1

= 9810 [Volume o f "ylindrical pan in wmcr + Volume

of curved portion I '" 98 [0

, ELEVATION Fig. 4. 11

[~x D" xDe pth ofcylindrical part in water + Vo lume di spl aced by cur.... ed port ion ]

"'

38259

=98 1 0 [~ (3)l x(1.8-X ) +O.6]

) + 0 .6= 38259 -' 3 () ' xO. 8 -x - =3.9 4 98 10

,

'4

,

x 3 )( (1.8-x)= 3.9 - 0.6

1.8 _.( =

= 3.3

3.3 x 4 '" 0.4668 Itx 3 x 3

x = 1.8 - .4668 = 1.33 m li:l B~ is the centre of buoyancy of cy li ndrical part and B is the centre of buoyancy of thc who le body. Then depth of cylind rical part in water = 1.8 - x = 0.467 111 CH 2 "'.r +

.4~7

= 1.33 + .2335 = 1.5635 m.

Th" d is tanc" of the ccntm o f buoyancy of the who l" body from the top o f the cylindrical part is given as CB", (Vo lume of curved portion x CHI + Vo lume of cylindrical part in water x CB l ) + (TOlal volume of wmcr displaced) =

0.6 )( 1.95 + 3.J)( 1.5635 1.1 7+5. 159 '" 1.623 m. '" (0.6+33) 39

Then BG = C8 - CG= 1.623 - 1.20= .423m. Met a-centric heig ht, GM. is givcn by /

GM= - - BG

•

I I

Ii

~ I

IL Buoyancy and Floa tation whe re

f'" M.O.I. or the plan of llie body al water surface about )'[

';j

4

)'[

143 1

y.y axi s

44

= - xD = - x3 III 64 64 '" Vo lum e ort he body in water '" 3.9 Ill l IT

3'

x - - .423 " 1.0 19 - .423 '" 0.596 m. Ans. 64 3.9

GM'" -

.. 4.7 CONDITIONS Of EQUILIBRIUM Of A fLOATING AND SUB-MERGED BODIES II. sub- Ill crj;cd o r a floating body is said 10 be stable if it w ill es back to its original position afl cr a slig ht disturbance. T he relative position of the ce lllfC of grav ity (G) and cemfC of buoyancy (Ill) of a body de termines the stability of a sub-ilwrgcd hody.

4 . 7 . 1 Stilbility of iI Sub- merged Body. The position of ce lltre of grav ity and centre of buoy ancy in case of a complete ly su b-merged hody arc fixed. Consider a balloon. which is complete ly su bmerged in air. LeI the lower portion of the balloon comai ns heavie r IHalerial, so thaI ils eemfe of gravity is luwe r thnn its wntre of buoyancy ~s shown in Fig. 4.12 «(I). Let the weight uf the balloon is W. The weight W is acting through O. verticall y in the downward direction. while the buoyant force FB is acting vertica ll y up. Ihrough n. For the equilibrium of the balloon IV= Fl!' If Ihe balloon is given an angular di splacement in the clocl.:wisc di r~ction as shown in Fig. 4.1 2 (a) . then IV and FIJ const itute a couple act in g in the ant i-c lockwise direction and brings the balloon in the orig inal pos ition. T hu s the balloon in the position. show n by Fig. 4.12 (a) is in stab le equ ilibrium.

G

, (,)

w

(b)

STABLE EQUILIBRIUM

Fig. 4.12

UNST ABLE EQUILIBRIUM

o «) NEUTRAL EQU ILIBR IUM

Stabiliries of mb-merged bodies.

(a) Sta hl e Equillhrium. When IV = Fe and poin l n is above G. the body is said 10 be in stahk equilibrium. (b) Unstahle Equilibri um . If IV = F o' bUI Ihe ce ntre o f buoyancy (8) is h~low ee nlre of gravity (a), Ihe body is in unstable equilibrium as shown in Fig. 4. 12 (b). A ~light disp lacement 10 Ihe body. in Ihe clockwise direc tion. gives the co upl e due to W and I'IJ also in [he c lock wise dirt."ction. Th us Ih e body docs nOI relurn 10 its origi nal po~ilion and hence Ih e body is in unslable equi lihrium. (c) Ne utml Equi lib rium . If F/I= Wand IJ and G arc at [he same point. as shown in Fig. 4.12 (c). the body is said to be in neutral equili brium.

4 .7.2 Stability of Floilting Body. The sWbility of a nual in g body is determined from [he positiun of Meta-ce ntre (M). In C3SC of fi(Hlting body. the weig ht o f the bod y is equ al to dIe weight of liquid displaced.

I I

Ii

~ I

IL 1144

Fluid Mechani cs

(a) SlabJe EqullibrlullI.

[flh~

Jl'Oilli M is above G. the floalillg bod y will be in stab le equilibrium as

shown in Fig. 4.[3 (a). If a slight angular displacement is given to Ihe flo atin g body in Ihe clockwise direction. the centre of buoyancy shifts from 8 to 8 1 such thm the vcnicallinc through 8 1 cutS al M. Then Ihe buoyant force Felhroug h 8, ,IIlO weight IV lhrough G constitute a couple aCling in the anticlockwise dircclion and thus brillging th e floaling body in the original position. DISTURB ING ----.. COUPLE

, tal Slablc equilibrium /If is above G

Flg. 4.13

(b) UnSlablt equilibrium /If i< below G.

Stability offloating bodin.

(b) Unstab le Equilibrium. If the point hi is below G. the floating body will be in unstable equilib·

rium as shown in Pig. 4.13 (b). The disturbing co upl e is acting in tht clod:wis.: direction. The coup lt due to buoyaru force F/:I and IV is also aCling ill the cloc kwis.: dirc.::tion 311d thus oven urnin g the floating body. (c) Ncutr:ll Equilibrium. Jfthe poim M is at Iliecel1treofgravily oflhc body. 1he floa1ing body will be in neu1ral equilibri um .

Problem 4.12 A solid cy/inda of diameler -1. 0 '" ha~· a lreiglll of -1.0 m. Find Ihe meta-cenlric heigill of Ille cylinder if rile specific gflll"ily of Illl' material of cylinde r'" 0.6 ami ir is floaling in l..arer lI"illl ils a.{is I·urical. Slale II"herher IIIe eqw·librillm is ~·I,jbll' or unswble. SolutIon. Given:

D=4m

Heigh!.

11=4m

Sp. gr. !kptll of cylinder in water

1 4·~ 1

+- .

'" 0.6 =Sp.gr.xh '" 0.6 x 4.0 '" 2.4 m

PLAN

y

Distance of cent re o f buoyancy (B) from A AD=

24 i'" 1.2m

-------:.~:.

4.0m 2.4 m

Distance of centre of gravity (G) from A

"'

AG =

,

~ '" 4.0 =2.0 m 2

2

BG '" AG - AB

, " -".1:

,G

-L 1

Fig . 4 .14

= 2.0 -

1.2

= 0.8 III

Now 1he 111em·ccruric height GM is givcn by I GM"'"r/-BG

I I

Ii

~ I

IL Buoyancy and Floa tation w ltc rc

145 1

f = M ,0.1. o f Ihe plan of tile body abo ut Y- Y ax is

= ~ 04

",

64

2:.

X (4 .0)4

64

'1 " Vo lume o f cyli nder in w ater = ~ x D2 X I)"pl h of cy lind er in wat er '" ~ X 4 2 x 2.4 m 3

4.0

4

~ )(4' ] 4! 1 = ~64>'c-_ _ = ~ x = =0.4167 m 'r:I ;X 4 1 )( 2.4 16 2.4 2.4 I GM = 'r/ - BG '" 0.41 67 - 0. 8 '" - O.3H33 m. Ail s.

- vc sig n me ans thalthc meta-centre (M ) is be low the centre o f grav ity (G). T itus th e c ylind er is in un stabl e equilibrium . Ails. Problem 4.13 A solid cylinder of !O em diwneler alld 40 em /ollg. COIIS;SfS of nm parrs made of differel17 mll/erin/s. Theftrsl pari GIllie base is 1.0 em long and of specific gral'iry '" 6.0. The Oilier pari of Ille cylinder ;s mode of tlie mareria/ IWI'il1g specific gWl'ily 0.6. Siale, if if call flom \'enically in Irma.

Solution. G ive n: Length. Len gth o f I st pa rt. Sp. gr .. Densit y of 1st part. Le ngth o f 2nd pan, Sp. gr .. Dc n~i t y of 2nd part ,

D : Wc m

L ", 40 cm 1.0 u n SI == 6.0 'I:

PI'" 6 x 1000 '" 6000 kg/m , ", 40- 1.0=39.0cm l

l

S2 '" 0.6

Sp.gr. =0.6 -

T

P2 '" 0.6 x 1000 '" 600 kgfml The cy linde r will floal l'c n ic all y in W31c r iF its meta-ce ntric he ight GM is 1.0cm posi tive. To find me ta -cc ntric hc ight. find the location of ccntrc o f gra vit y (G) and ce nt re o f huoy all cy ( 8 ) of th e combined so lid c ylind er. Th e di s tance of the ce nt re of gra vity o f th e so lid cy linder fro m A is given as Sp.gr = 6.0 A G : [(Weight o f 1st pan x Distance ofCG. o f 1st pa rt from A) F ig . 4.15 + (Wcight of 2nd pan of cy lindcr x Di stance uf CG. of 2nd pan fro m A) I + IWc ight uf 1st part + wcig ht of 2nd p~rtl

U.

T LJ.-l ,

==

(~O l

1

X1.0 x 6.0 x 0.5) + (~O l X39.0 x 0.6 x (1.0 x 3912 )) ( : 0 " x 1.0 x 6.0 + : 0 " x 39 x 0.6)

=

1.0 x 6.0 x 0.5 + 39.0 x.6 x (20.5) I.Ox6.0+39.0xO.6

Cance l : Dl in th e Numerator and De nu mina to r ==

I I

3.0 + 479.7 482.7 4 = - - == 16.42. 6.0 +23. 29.4

Ii

~ I

IL 1146

Fluid Mechani cs

To fi nd th e ce lllfC o f bu oyancy of th e co mbined two parts o r of th e cy lind er. de term ine Ihe dcptll of imme rsion of the cyl in der. Let the de pth of imm ersion of the cy linde r is II . T he n Weigh t of the cy lin de r " We ight of wa ter displaced It , 39 .0 11: , 1.0 It , II - )( (. It x - - )(600 )( 9.81 + - Cit x )(6000 )(9,8 1 = - (. 1)" x x 1000 x9.8 1 4 100 4 [00 4 100 I ": h is in cml I:a ncc llin g

~

, cIOOO =:~X~9~.8,,1 th rougho ut. we get ( I) x 100

39.0xO.6+ I.Ox6.0 =11 or 11=23.4+6.0=29.4 Th", d istance of the cen tre of the buoyancy B, of the cylinder from A is 29.4 AB = 1!12 '" - , - = 14.7 BG

= AG - AS '" 16.42 - 14.70 = 1.72

C Ill .

Meta -ce ntric heig ht GM is give n by

GM= ~ - BG wllerc

•

1 = M.O.1. of pla n of the body about Y- Y ax is

'" ~ D~= ~ (lO)~c m ~

64

.64

'It = Volume o f cy linder in water

=

, , 1 '4If D-xh= 4"It (10)-x29.4 111

~ '" ~ (10)4/ .:!.( 1O ! 'V 64 4 )

x 29.4 '"

-.!... x [6

10 ' = -,-'~OO O;;;-c = 0.2 12 29.4 19 x 29.4

GM = 0.2 12 - 1.72 '" - 1.50S e rn As GM is - ve. it mea ns th at the Me ta -ce ntre M is below the cen tre o f gra vi ty (G). T hu s tlw cy linder is in uns tab le eq ui librium and !\O it ca nnot fl oat ven ica ll y in water. An s. Problem 4.14 A rectangillar POII/OOII 10.0/11 long. 7 m broad (lnd 2.5 m deep weighs 686.7 kN. /I carries Oil ils "pper deck (In empty boiler 0/5.0 m di",,,eter weighing 588.6 kN. The celi lre 0/ gw!"il)' o/Ihe boiler alllilile POIi/OOII are (1/ Illei r respectil'e relltres alollg a l"erlicallille. Filld tlte //Iela-celllric height. lI'eighl dellsity a/sea lrater is /0./04 kN1mJ. Solution. Give n : Di me nsiu n u f pon tuon = [0 x 7 x 2 .5

Weig ht o f pont oon. Dia. of boile r. Wr ight o f boile r.

D=5.0 m 11'1 '" 5SS.6 kN

= [0. [04 kN/m 3 \\' fu r sea wate r To find the met a-ce nt ric heig ht. first dc tc nni n ~ the co mmo n ce ntre of gravi ty G and cOl11 mon ce ntre of buoya ncy B of the ooi la and po ntOO n. Let G I nnd G , nrc the ce ntre of grav iti es of pontOOn and boiler r.:sj)Ccli ,·e ly. The n

I I

f

11'1'" 6S6.7 kN

IT

IG, lG

"t

G

B

I

,

A

T

5.0 m

gm ~

1.

1 -7.0m- l Fig. 4.16

Ii

~ I

IL Buoyancy and Floa tation

AG 1 ",

2.5

2 '"

AG1 ", 2.5 +

1.25

147 1

,

III

I

25.0 "" 2.5 +2.5 '" 5.0 III

The distanl'C of co mm on (;cnt rc of gravity G from A is given as

1\1 x AC, + IV, X AG,

AG '" ~---.~o.'--~ \~ + IVz

686.7 x 1.25 + 588.6 x 5.0 (686.7 + 588.6)

to.Om

'" 2.98

111.

Let!J is th e depth of immersion . Then Total weigh t of polUoon and boiler'" Weight of sea water displaced (686.7 + 588.6) = Ii' x Volume o f th e pomoon in water

= 10.104 x I. x h x Deplh of immersion

I·

7.0

~ ------l

J

Fig . 4. 17 Plan of tbe body

a/ water·line

1275.3", 10.104 x 10 x 7 X II

1/: -;-;~~1~27~5~.3~~ : 1.803m 10 x7 x 10.104 The di stance of the com mon cen tre of buoyancy IJ from A is All=

~==

1.803 == .90 15 III 2 2 BG == AG - AB '" 2.98 - .9015 '" 2.0785

Mct3 -ccmric height is given by GM '" where

.!.... V

In -

2.078 111

BG

1== M.O.1. of the pl'lll of the body at th e water leve l along Y-Y

=...:..

x [O.Ox7J: IOx 49x7 m4

[2

';j

12

== V o lume of Ille lxxl y in water == Lxbx II == [0.0 x 7 x 1.857

J : ';j

GM "

IOx49x7 [2x[Ox7xI.857 I

';j -

CC~4~9~ =2.[9&111 12 x 1.857

= ~

BG = 2.198 - 2.078 = 0. 12 m.

Mela·t...,nlric lIeighl o f bolh Ih e ponloon 1lnd boiler == 0.12 Ill. An s. Problem 4.15 A lI"oodell cy/illder of sp. gr. = 0.6 lmd cirCli/ll f ill cron·· seClioll i~· reqllired 10 flo (ll in oil (sp. gf. = 0.90). Fint/ Ihe LID ralio for Ihe cylint/er 10 floal will, ils /Ongilut/inal a.ris ,·erliC(l/ in oil ....IIue L is Ih e Ireiglll of cylinder fwd 0 is ils diam eler. Solution. Gi\"~n : Dia. of cy linder Heighl o f cyli nder Sr . gr. of cylinder.

I I

Ii

~ I

IL 1148 Fluid Mechani cs Sp. gr. of oil 52'" 0.9 Let the depth of cylinder immersed in oi l '" II ror the principle of buoyancy Wcigtlt of q lindc r '" WI. of oil displaced 11:

It

,

C' ~

,

,

- 0- x L x 0.6 x 1000)( 9.81: - 0- x II x 0.9 x 1000 )( 9.81

4

4

h",O.6)(L = ~L

0.9

3 L

,

0

1•

LxO.6 : /1)(0.9

0'

'-----' - "4

lG

, ° Fig . 4.18

1

The distance of centre o f grav it y G from A, AU '" -

2

The dist ance of ccnlT" o f buoyanc y B from A.

The meta -ce ntric height GM is given by

GM"' ~- BG where

I ",""::'"

M

D~ and ';j

•

'"

Volum e of cylinder in oil '" ~ 0 2 X "

4

!....=(~ D./~ D2")=-.!... V

64

4

16

Ol =_ D,,'-;-_ =3Dl II

16

x,

2L

32 L

{ h=H

3Dl L GM: - - - . 32L (, For Mable equilibrium, GM should be +vc or GM>O

3D'

or

32 L

L - >0 6

1

3D L -- > -

0'

32L 6 1 t 18 -
0'

9 16

~ <~ =% IJD < 3/4. A. ns.

Problem 4 .16

511011' /lw / a cy/;ndrit'(I/ buoy of I m diameter Illld 2.0 m heighl \\'eiglling 7.848 kN wil/ 110/ floll i "ulial/ly in sea waler of densily f()JO kglm J• Find Ihe force necessary in a "erlica/ chain a((a("/It:d al Ihe cell ire of base of Ihe buoy Ihal will keep il ,·ulica/.

I I

Ii

~ I

IL Buoyancy and Floa tation

149 1

Solution . Gi ve n : Dia. o f bu oy. 0 '" I III

= 2.0 III

Hciglil.

H

Wci~1I1.

IV = 7.848 kN

Density.

= 7.348 x 1000", 784 8 N p::: 1030 kg/ml

(i) Show th e cylinder will not l1oa1 ve rticall y.

( ii) Find the force in the c hain. Purl I. Th e cy linder will no t fl oat if me ta·centric heig ht is - 'Ie. LeI the d~pll1 of immersio n be II

---,-_. 2.0 h

Then for equilibriuln , Wei ght o f cy linder '" Wcig lll of wa ter displaced

"" Density x g x Volume of cylinder in water

7848 = 1030 )(9.81 x~ Olxh

Fig. 4.19

4

,

, 1-,, -1

,

::: 10 104.3 x - (It)( II 4

4 x 7848 10104.3)(11

The di stance of centre of buo ya ncy

::: 0.989

n from A,

AB::: ~ ::: 0.989 = 0 .494

2

Ill.

Ill.

2

And the distance of ccmre of gra\'ily G. from A is AG :::

22.0 :::

1.0 III

8C::: AG - AB::: 1.0 - .494 ::: .506 m.

Now meta-ce nlri c height GM is given by where and

1= It;f

GM:::'!"" -BG V

~ D~= ~ X(I)~m4

== Volume of cy linde r in wa ter == ..::. D I x " == ..::. I I x .989 , 4 J

~X 1 4

-, x I'

64 V " ~ D 2 xh 4

6'

"::'X I ~ X.989 4

== ~ xtlx _'_ = 16

.989

=0.063 11\ 16x.989

GM = .063 - .506 == - 0.443 m. Am. As the m ~la·ce nlri c hdghl is - ye, Ih e poinl M ties below G and he nce Ihe cy linde r will be in un sta bl e equilibrium :lI1d hence cy linder will nOl float vertica ll y.

I I

Ii

~ I

IL 1150

Fluid Mechani cs

Purl II. Let tlie force applied in a ve rtical chain 3nachcd aI the centre .,.,.,J-t-1~,,-~ of the base of Ihe buoy is T (0 ~ccp the buoy venical. :0 Now find the combined position of ccmn: of grav ity (G') and cen tre of buoyancy (B'). For the combined Centfe uf bUUyJIH':Y. let /J' = depth of immersion when the force T is applied. Then TQlal downward force = Weight of water displaced or (784S + = Density o f wate r x g x Volume of cy lilldcr in water

n

~ D')( Ii' I

= lO30x9.81 x

,

II'

1 [ 7848 + T ]

AB=T="2

7935.9

where II': depth ufimmeThion I

7848 + T

10 I ()..I.3+ T

I Ol04.3X~Xl l

7935.9

m

,

A

Fig. 4.20

7848 + T = 15871.8 m.

The combined cenlTC of gravity (G') duc 10 wcigtn of cy linder and d uc to tension T ill the chain from A is

AG'" IWI. of l:Ylindc r x Distance of CG. of cylinder from A

+ T x Distance of e.G. of T from A] + ]Wcig hl of cylinder + T] " (7848 x B'G'

~ + T x 0) 2

= AG' _ AB' =

+ ]7848 + T] ==

7848 (7848+ T)

The meta-centric height GM is given by GM ==

7848 m 7848 + T

(7848 + T) 15871.8

Vf - B'G'

== ~XD4== ~ X]4=~nl4

where

64 V

= ~ D~

64 X II'

=~

4

64 X ]1

4

x (7848 + T) ==.!:. x 7848 + T 7935.9 4 7935.9

,

f

• " =oi'64~", (7&48+ T ) 4 7935.9 It

= (7848 + T )

GM = -:-C~79,"3:"'.,-9

16

71\48 [ (7848 + T)

T) ]

(7848+ 15871.8

For stable equilibriunl GM shou ld be positi ve GM> 0

7935.9 16 (7848 +T)

I I

[

7848 (7848 + T)

(7848 +

T)] >

15871.8

0

Ii

~ I

IL Buoyancy and Floa tation 7935.9

16 (7848 + T )

151 1

7848 + 7848 + T , 0 (7848 +T) 15871.8

,-79"3;;',.;9,,-~'6;-':..;7~84,,,8 + (7848 + T) 16 (7&48 + T)

>

0

15871.8

~-"I~'7C;6~:n",+ (7848 + T) > 0 J6 (7848+T)

15871.8

(7848 + T)

:>

15871.8

,,-~''~7~63~2"" 16(7848 + T)

(7&48 + n 2:> 117632 x 1587 1.8

0'

16.0 :> 11 6689473.5 :> ( 10802.))!

7848 + T:> 10802.3 T:>

10802.3 - 7848

:>

2954. 3 N. A ns .

Th.e force in lhe chain IllUS! be at least 2954.3 N so that the cy lindric al buoy can be !;cpt in vert ic al position. An s. Pro blem 4.17 A solid coile floats in warer ",itl, irs apex downwards. Determill e Ille lew/ apex ollgle of cone jar .liable equilibrium. Tlw specific 8WI';I)' of lile material of the coile is gil'el l O.S. Solullo n. Given: Spo gr. of cone == 0.8 P == 0.8 x 1000 == SOO kg/m l DensiTy uf I:one.

/

D == Dia. of the cone d == Dia. of cone ,H water leve l

Co<

29

+

,

CAN 0' C ONE AT

= Apex angl e o f cOil e

w" ER LINE

H == Hei ght o f C()nc

D

I'

•I

II = De pth of cone in water

For tllc I:onc. thc

di st~ncc

G'" Centre of gravi ty of the cone B = Centrc uf buoya ncy of Th c conc of ce nTre of gravity from the ~[lcx A is

AC = t height of conc '" tH also

AB", t dcpth of cone in wa1cr '"

Volume of water di splaced '" Volume of cone

= 1-

:. Weight of cone

=800 xgx tx TtR'" xH

Now from JiAEF.

Similarly.

I I

t It? x II X TtR 2 X II

EF

Ian 9= -

t"

"

, Fig. 4.2 1

il

R

=-

EA H R=Hta n 9

r",htan6

Ii

~ I

IL 1152 :.

Fluid Mechani cs Wei~hl

of cOile

t

.. Weight of water displaced", ]()OO x g x

x II'! X II

. .000"""x""' '" lOOOxgx tXI[(htan 9) , X/I= -1 For equilibrium Wcigllt of l"OIlC

3.0

'" Weight of water displaced 3.0

3.0

800xlt=

or

'X'O'ITfX~h" _'""'''_',>.

1000

x

II)

1000 X11.1or!!'" = (1000)'"

H3 =

800

/,

800

For stable cljuilibrium. Meta -centric height GM should be positive. But GM is given by GM= -

/

•

-8G

where 1= M.O.! . of cone al wateT-lil'" = ~ d~ 64 . In, V = Volume of con" In water =- - d" x h

34

/

IT d ' l "3)("4 ' IT d ,- X "

V=64

I x3 d 2 3d' 3 ' 3r' =x - ==x(2, r = - 16

=

II

161,

1611

4 II

3 (/l lan6)1 4

I':

1/

r=I,[an91

=t illa ,,2e and

BG '" AG - AB = fH - til'" f (H - II) GM =

til tanl e -

t(ll - II)

For stable equilibrium GM should be positive or t ht3n!e -1 (1I - 11»O

or

htan 2e>(H _ h)

or

ur

htan!e - (If - h»O htan 2 e + h>H

hltan!a+ 11>11 So<

!!.. = ('OOO)iIJ = 1.077 h

800

, I scc~ e > 1.077 or cos' 0 > - - '" 0.9285

cos

e > 0.9635

1.077

0> 15° 30' or 20>31 ° Apex angle (20) s hould be aileasl 3 1°. Am.

I I

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IL 153 1

Buoyancy and Floatati on Problem 4 .18

di(lmeler

f)

A cOile of Ifleciftc gra\'it)' S. is floating ill wmer willi its apex dowlI.",ard,~. II has a

(md l'erlin,/Iu:ighl H. Show Ilwl jor stah/e equilibrium of the COli I' H <

!.. [V i. 5;;; ]"1, 2 2

S

Solution. Given: Dia. of oone =D Hciglll o f cone'" H Sp. gr. o f cone = S Let G", CemTC of gravity of cone

B '" Ccmrc of

PLAN OF CONE AT WATER LINE

bUOYHIlCy

~ D _I

29 = Arcx angle

T

= Apex of the cone II'" Depth of immersion J = Dia. of cmw at Waler surface

A

Then

"

1

,

AG:2H 4

fig. 4.22

3 A8= - II 4 Also weight of cone =Weigh t of water displaced. I' I , lOOOSxgX,lTWX H : IOOOxgx-rrr'xh

3

or

'2 sn-H=,h

SR!H II = - ,-

B"

R" ,

t~ne = - = -

H

"

N=Hmne.r=ltlan8

SX(Hlan6)l xH II '"

(1/ tan

er,

SX II l xtan 1 exfI SH J " : __ It" tan " e I,,,, (SH1)1fJ = Sill H

I,:

Distancc.

,,1

or h 3 =SH 3 ... (1)

8G '" AG -An

J 4

3 3 (H - b)= -3H(- .1/3 S /I) 4 4 4

=-H - - b= -

'" 2 HI I _SI Il I

Also

... (2)

4 I", M.O. Inenia of the plan of body at wate r surface

'" 3... It 64

.

V", Volume o f cu ne III water'"

I I

, In: , 3In: x 4 x iI + x h '" 34 iI+

IH.S

III

I

Ii

~ I

IL 1154

Fluid Mechani cs

-

J

V

=

~d' 64

.----o"'c----c 1 x

3

It

d

H.S'1l

16.H.slll

4

Now Mcl 0

... (3)

Also we know R = titan

e and R

r

r= h tan

e

H D = - =II

d

If '" DII '" D X HS' fl '" DS' fJ

H H Substituting the value of di n cqum io n (3), we get

3(0 5

1 1))1

3H

~::':;--"i, > -

1.6.H.S'IJ

4

.lt3

(l - S

) or D J .S'/J

H2< 4 ( 1_ 5"1)

"' II>

4.8

EXPERIMENTAL METHOD Of DETERMINATION OF META -CENTRIC HEIGHT

The mcla-cc ntric heigh t of a noati ng vessel can be determined, provided we know th e (emre of grav it y of the Ooaling vc~scl. Lei"", is a known we ig hl placed over the ce ntre of the VCS1;e J as show n in Fig. 4.23 (iI) and the vesse l is floating.

w

(a) Floating body

(b) Tilted body

Fig. 4.23 M l!l d-<:f'rltric bright.

I I

Ii

~ I

IL Buoyancy and Floa tation Let

IV '" Weight of vessel including

155 1

WI

G '" Celltrc of gmvity of the I'essel

B '" Celllrc of buuyancy of the vessel The weight 11" 1 is 1Oo\'Cd across Ihe vessel towards right through a di Sinn/;c x as shown in Fig. 4.23 (b). The vessel will be tilled. The angle of heel is measured by means of a plumblinc and a protractor attached on tlie vesse l. The neW cenlrcof gravity of the VC~!ic J will sh ift to G 1 as the we ighl WI hasbcen moved towards Ihe right Also the centre of buoyancy will change to B] as Ih" vcss<:l ha.~ lih~d. Under equilibrium. the moment caused by the movement of lhe load WI through a distance x must be eq ual to Ihe rnOlTIcl\t caused by the s hifl o f the relltrc of g ravity from G to G I' Thus

e

= GG I X IV= movement of WI = 11'1 X X

The momenl due to change of

The moment duc to

IVx GM tall

G

e

11'1 '\ '" IVGM tan e GM = cc'"~"~'cc ...(4 .5) IV tan Problem 4 .19 A J'l!ip 70 III 10llg {md 10 III bro{{d h{{s a displacemenl of 19610 kN. A weight of 343.35 kN is mOI'l'd across Ille deck: IhrOllgh a tii.,·I{Wce of6 III. The .IMp is lilled Ihrollgll 6 ". The momelll of inertia of Ille ship m wmer-line abOli1 ils fore Wid afl axis is 75% of M.O.I. of I/Il' circlIIIIJ"cribing reClmlg/e. Tlw celllre of buoyancy is 1.15 III belo", ",ala-lille. Find tile meta -cen/rit" IJeigllt lIlIIl PO$ilioli of celilre of gr""ity of J'hip. Specific weight of sell '\"lIler is 10 I 04 Nlm J •

Hence

e

Solution. Given: Length of ship. Bread th of ship.

L=7UIlI b = to m IV= 19620kN

Displacement. Angle of heel. ", .0.1. of ship at water-line ". for sea- water

e" 6° " 75% of M.O,I. of circumscribing rectangle

= 10 I 04 N11lI 1

"

l

10.104 kN/m

w, " 343.35 kN

Movable weight. Distance l\loved by "'I' Centre o f buoyancy Find (I) Meta-ce ntri \: height. GM

.\ =6 m = 2.25 m below Waler surface

(ii) Position of centre of grav ity. G. (i) Meta- cc ntric height . GM is given by equation (4.5)

GM =

"

"'IX IV tan e

343.35 kN x 6.0 19620 kN x lal16°

34335 kN x6.0 19620 kN x .105!

'" 0.999 m. Ans.

(ii) Pos ition of Centre of Gra vit y, G

GM=.!.... - BG

where

~ I

•

1== M.O.l. of the ship al water-line about Y-Y

I~

~ I

IL 1156

Fluid Mechani cs

.-WATER LINE

2.25

~10m ~

Fig. 4.24 '" 75% o f lUlU

-.!.... [2

x 70 X 101

Fig. 4.25 .75 X...!... x 70 X 101

",

12

V" Vu lume of ship in wme r '"

.!... '" 'rJ

",

4375

~h ip

We ight of -:::c-:'c'-'C'':'''-'O''--:-:c: Weight df.>nsily of wate r

4375

'" 2.253

111 4

19620 '" 1941.74 m1

10. 104

111

1941. 74

GM", 2.253 - IJG or .999 '" 2.25] - BG BG

= 2.253 - .999 = 1.254 m.

From Fig. 4. 25. it is c lear lhal Ih e di stan ce of G from free surface o f the wa ter '" dist ance o f B from water surface - BG " 2.25 - 1.254 ", 0.996 111. An s. A pOIl/QOI! of 15696 kN di~placemt'1lI iJ floating in water. A »'<'igh! of 245.25 kN is IIwred 1/"014gh a diSllIllce of 8 m across the deck ojpOIl/OOIl, II'hic// lills the pontoon l!lrOJlgll all angle

Problem 4 .20

·r.

Find mela-anlric height of the pontOOIi. Solution. Given: Weig ht of (lOntoo n = Di splace ment

"' MOl' abl e we ight.

W = 15696 1; N

245,25 kN Distan ce moved by weig ht "'I ' x == 8 m Angle of hee l. 6 == 4 ° Th.;, nwta-ce ntri c heig ht. GM is giv.;, ,, hy eq uati o n (4.5) 11'( '"

GM == ~w",x'-o 1\1 tan 6

"'

245.25 kN x 8 15696 kN x tan 4°

--;:""",,96,,2;.-;;:,;;;;

== -;-: == 1.788 15696 x 0.0699 ~

4.9

III . '\ IIS.

OSCILLATION (ROLLING) OF A FLOATING BODY

Consider a fio at ing body. which is tilted thro ugh an a ng k by a n ol'e nurnin g co uple as s how n in Fig. 4. 26 . Let the o\'Crtu rn ing co upl e is suddenly re moved . The bod y will start osc ill ati ng. T hu s. the

I I

Ii

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IL Buoyancy and Floa tation

157 1

body will be in a Siale of oscillation as if suspended allhe mcta·CCnlrc M. This is similar 10 the case of a pe ndulu lll . The only force acting on the body is due to Ihe restoring couple duc to the weight Wof the body force of buoyancy FR"

,

• Fig. 4.26

Restoring couple

'" \I' x Distance GA

= II'x GM sin 9

. .. ( i)

This couple tries 10 decrease Ihe imgle Angular 3(;celcralion of the body.

(l

d1 S = - - ,- .

d,

- ve sign has been introduced as Ihe restoring couple tries to dcr rcasc the angle B. Torque due \0 inenia = Moment of 11lcrlia about Y-Y x Angular acceleration ,[- -dtlr"J

= I y_y x

IV , f r' r = gK where

IV = Weigh t of body, K = Radius of gyra tion about r·Y _ '" K' ( _

[nenin torque

g

d1,' J__ _~ .,2 dle, g

dt "

"

dl '

... (U)

Equating (I) and (il). we ge l IVxGMsin9=

For sma ll angle

K'

Dividing by -

,

e.

, we gel

_~ K 2 d2? g

dr '

"'

sine - e

d1?+ GM Xfx e ",0 dl-

K-

Th" above equalion is a differenlial equat ion of second deg ree. The solUlion is

. JGM.g ~

e ", C ,S l n

I I

XI + C2 COS

JGM.g xI Kl

...(iii)

Ii

~ I

IL 1158

Fluid Mechani cs

where C t and C2 arc constants of integration. The values o f C 1 and C 1 arc obtained from boundary conditions whic h arc (i) 311==0,6==0 (ii)

all=

~,e =o

where Tis Ihe time period o f one complete o!;(:ilimion. Suhstituting the 1st ooundary condition in (iii), we get

(': si06=0.0:0s6= 11

O=C 1 xO+C1 X1

C1 == 0 Substituting 2nd boundary conditions in (iii), we get

Bm C 1 cannot Ix: equal to sin

1.Cro

and so tile olher ahcmalivc is

JGM.s: T - - - x -=O=sinlt Kl

(.: sinlt=OI

2

T=2n~ Kl

...(4 .6)

GM .g

Time period of oscillation is give n by ~qualion (4.6). Problem 4.21 Til e rat/illS of gy,III;OIl of {/ sllip is S III la/e Ihe lime period of o.\'ciliatiQlI of Ihe Jllip. Solution. Given: Least radiu s of gy ration. K = 8 III Mcta -cc nlric height. GM == 70 em == 0.70 111 The tim e period of oscillation is given by equation (4.6).

"',1.11

T: 2n

j£,'

(/111/

meta-centric height 70 CIII. Caleu-

8x8

- - : 2n , 1 ;cOCC~~ : 19. IH sec. Ans . 0.7x9.81

GM.g

Problem 4 .22 The lime period of rof/illg of a ship of weighI 29430 k.N ill sea waler is 10 seconds. T/l e cellire of buo}'(l/Icy of Ihe ship is /.5 m be/ow II/e cen/re of grlll'it)'. Fifl(llile radius of g)"",liOl/ of Ihe ~IIip if Ihe moment of inerlia af Ihe ship al IIIe lI'aler Urle aboUI fore and aft luis is I()(x) m~. Take j'l'ecijic weiglll af sea Imler as = 10 I (X) Ntm J• Solution. Given: T : 10 s«c Time period. Distance between centre of buoyancy and centre of gravity. BG : I.S 111

Mome nt of Inertia.

I:

10000

m~

WciglH. IV" 29430 kN '" 29430 x 1000 N Let the radi us of gyration: K PiTSt calculate the meta-cenuie height GM. whieh is given as GM: BM - BG '" -

I

•

I I

- BG

Ii

~ I

IL Buoyancy and Floa tation wlterc

.,d

159 1

1 = M.O. llicnia

V = Volume of water displaced Weight o f ship

-

29430 x 1000

- ~~"";;';~' = 2912.6

- Sp. we ight of sea water -

10 104

rn

3

10000

GM = - - - 1.5 = 3.433- I.S = 1.933 29 12.6

Using equation (4.6), we get

T= lit

~ '

GM xg

K'

2,K

1.933x9.81

JI.933 x 9.8 1

10= 211 ,I~~~~

"

Ill.

J~I'~3c3C'C'C',,-1 = 6.93 m . An s . K = clO - c,-,-

"

2,

HIGHLIGHTS 1. The upward fOrl'e cxcncd by a liquid on a body when the body is immersed in the liquid is known 8S buoyancy or force of huoyancy. 2. The point through which force o f buoyancy is _,upposed to act is called centre of buoyancy.

J. The point about which a body slanS oscillating when the body is ti lled is known as mcla-ccntre, 4. The distance between the mela-ccnlre and centre of llravilY is known as mcla-Ccnlric height

5. The meta-ccnlric height (OM) is gi,'en by GII1 = where

I

~

,

V

-IJG

Moment of lnenia of \he floating body (in plan) at water surface nboutthe axis Y-Y

'!j .. Volume of the OOdy sub-merged in wuter

BG '" Distance between

~entre

of grnvity and centre of buoyancy.

b. Conditions of equilibrium of a floating and sub-merged OOdy are :

Equilibrium (i) Stable Equilibrium

(ii) Unstable Equilibrium {iii} Ncutr.. 1 Equilibrium

Flooting Hody

Sub·merged /JOlly

111 is above G M is below G M and G coincide

/J is abo,·" G /J is below G

Band G coincide

' oc

7. The value of meta-centric height GM, experimentally is given as GM "' -;c,' "'"'

Wtan6

where

11',

~

of ..

Movable

w~ight

Distance through which 11'1 is mowd

W .. Weight of lhe ship or flooling body including

1<',

6., Angle through the ship or flo.1ling OOdy is lilted due to the movement of 11',. 8. The time period of o!'Cillation or rolling of a floating OOdy is given by T", 21f ,,'here

v~ c;;;t;g

K .. Radius of gymlion. GM .. Mela -cenlric height

T", Time of one complete o!'Ciliation.

I I

Ii

~ I

IL 1160

Fluid Mechani cs

EXERCISE (A) THEORETICAL PROBLEMS l. Define the lenn, 'buoyanq"

~nd

'cenlre of buoyancy' ,

2 . Explain the terms 'meta-<:entrc' and 'meta-centric heighI'.

3. Derive an expression for the mela-ccmric height of a floating body. 4 . Show thai the distance

bctw~..,n

the mcta-ccntre and centre of buoyancy is giw" by Hili

~

-

,

V

where I • Moment of incnia of the plan of the float ing body at water surface about longitudinal axis. V _ VOlume of the body sub -merged in liq uid. S . What arc the conditions of equilibrium of a floallng I>ody and a sub-merged body ?

6. How will you deten"inc the meta"':c"l,;C height of a floating body experimentally ? Explain wilh ncal sketch. 7. Sclectlhc correct Staiement: (a) The buoyant force for a floating body passes through the (i) centre of gravity of the body ( iii) mcta-.celllre of the body

(ii) centroid of vol ume of the body

(il·) cCllIre of gravity of the suh.merged pan of the body

(,,) ccntroid of thc displaced volume. (b )

A body

sub-mer~~"<.l

in liquid is in equilibrium when;

(i) it' meta-.centre is above the centre of gravity

(I;) its mcta...:cn!rc is above the cenlre of buoyancy

(iii) its centre of gravity is above thc centre of buoyancy

(i" ) its centre of buopn,y is abo,·e the centre of gral·ity

(,,) noneoflhesc. IAn s. 7 (II) (" J. (b ) (i1')1 8. Dcri,·c an cxpression for !he lime pcriO
9. Define the termS meta-<.·entrc. ccntrc of buoyancy. mcla-centric height. ga uge pres,ure and absolute pressure. 10 . What do you understand by the hydroslatic equation ? With the he lpof this equation. derivc the expression for the buoyant force acting on a sub·merged body. II . With neal sketches. explain lhe conditions of equilibrium for fiOal ing and sub-merged bodies. 12. Differenlime between: (i) Dynamic viseosi1y and kinematic ,·iseosity. (ii) Absol ute and ga uge pressure (iii) Simple and differenti al manomelers (i" ) Cent re of gr~vily and cenlrc of buoyancy. (Del/,i Ulli'·t"rsiry. Dec. 10M)

(B) NUMER ICAL PROBLEMS 1. 1\ wooden block of width 2 m, depth 1.5 til and lenglh 4 til floats horizontally in water. f ind the volume

of water displaced and position of centre of buoyancy. Thc specific gravily of thc wooden hloc k is 0.7. IAns. 8.4 til ! . 0.525 m from (he basel

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IL Buoyancy and Floatation

161 1

2 . A wooden log ofO .8m di,,,"c(er and 6 III length is noaling in rivcr Waler. Find the depth of wooden

log in water when the sp. gr. of the wooden log is 0.7. [Ans. O.54m ] 3. A Slone weighs 4SlO.5 N in air and 196.2 N in waler. Determine the volume of Slone and its specific gravity. IAn s. 0.03 Ill J or 3 x W' eml, 1.671 4 . A body of dimen,ions 1.0 III X 1.0 m x 3.0 III weighs 3924 N in water. I:ind its weight in air. W hat will be

its spedfic gravity?

[Ans. 62784 N. 1.06671

5. A metallic body floats at the interface of mercury of sp. gr. 13.6 and water in such a way that 30% of its volume is sub-merged in mercury and 70% in water. I:;nd the den axis vertical. State whether thc equilibrium is Slable Or unstable. IAR •. - 0 ,304 m. Unstable Equilibriuml 13. A solid cylin<.lcr of 15 cm diameter and 60 Crn long . consists of tWO parts rna<.le of different materials. The f,rsl part al the base is 1,20 cm long and of specific gravily '" 5 ,0. The olhcr parts of the cylindcr is made of Ihe material havins specific gravily 0 ,6, Slale. if il can noat \'er(ically in water. (Ans. G/I1 _ - 5.26. Unstable. Equilibriuml 14. A rectangular pontoon 80 to long. 7 m broad and 3.0 m <.leep weiShs 588.6 kN.lt carries on its uppcrdeck nn empty boiler of 4.0 to diameter weigh ing 392.4 kN. Th e centre of gravity of the boiler and the pontoon mc m their respective centres along a vertical line. Find the mcta--ccntric height. Weight density of sea water is 10104 N/ml . (Ans. 0.325 ml I S. A wooden cylinder of sp. gr, 0.6 and circular in cross-secti on is require<.l to nOOt in oil (sp gr. 0 .8 ). Find the /.JO ratio for the cylinder to noot with its longitudinal a~is ,'ertical in oi l where L is the heigh t of cylinder and 0 is it, diameter, IAns. (UO) < 0. 81641 16. Show that a cylindrical buoy of 1.5", di~meter an<.l 3 10 long weighing 2.5 tonne> will not n
I I

Ii

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IL 1162 Fluid Mechani cs walcr line aboUl its fo",,, and aft axis is 75% of mOment of inertia the circumscribing rec(,mgle. The ccnlre of buoyancy is 2.75 Tn below watcr line. Find the mcl~.ccnuic height and position of centre of gnlvity of ship. Take spc<:ific weight of sea walcr = 10104 Nlrn J , IA ns. 1.1145 m. 0.53 m below walcrsurfaccl 19. II pontoon of 1500 lnnnes dispiacClllCnl is !looting in watef. II weight of 20 lonnes is moved through a distance of 6 m across the deck of poilloon. which tilts the pontoon through an angle of 5", Find Illcla-<:cntric height of the pontoon. lA ns. O.9145ml

20 . Find the lime period of rolling of a solid e;"",lm cylinder of radius 2.5 Tn and 5.0 m long. Th,- specific gravity of the cylinder is 0.9 and is floating in water wilh its axis vertical. IAn s, 0.35 secl

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CR~ER

A. KIN EMATICS OF flOW ~

S.I

INTRODUCTION

Kinematics is dctincd as that branch of sc ience whic h deals with motion of particles without co nsidering the forces causing the motion. The ve locit y at any point in a flow field at an y till1~ is studied in this brandt of fluid ll1~chanics. Once the velocity is known. (h e n the pressure distri bution and hence forces ac tin g on the fluid can be dClemlincd. In this <:hapter. the methods of determining velocity and acceleration arc discussed .

.. S .2

METHODS OF DESCRIBING flUID MOTION

The fluid motion is described by two methods. The y arc - (i) Lagrangian Method. and (ii) Eulerian Method. In the Lagrangian method. a sin gle fluid particle is followed during its nlO1ion and its yelocity. acceicration, density. clc .• ilre described . In case of Eulerian melhod. lhe velocily. ilcceleration. prcssure. dens it y etc .. arc described ,,( n point in flow field. The Eulerian method is common ly used in fluid mechanics .

II> S.l

TYPES OF FLUID FLOW

The fluid flow is classified as: (i ) Sleady and unsteady flows: (ii) Un ifo rm and non·uniform flows: (iii ) Lami nar and turbulent flows; (i v) Comp ressible and in compressible flows: (v ) Rotational and irrOlationaJ flows; and (ri ) One. two and thrt,e -dimensional flows.

S.l . I Steady and Un steady FloW5. Steady flow iSdefi lled as thilttypc of flow in which the fluid charac teristics Ii!;e ve loc ity. pressure. density. etc.. at a point do not change with time. T hlls for steady flow, 1I1athematically, we have

163

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IL 1164

Fluid Mechanics

,.

- 0 -P - . ( dill )

•. J.....

~

('P) "" ,.",

=0 -

'at

=0

where (.(0- )'0 - 4 ) is a fixed poim in fluid field. Unsteady flow is that type of now, in which the velocity, pressure or dcnsily ;n a point changes wilh respect to time. Thus. mathematically. for uIl5tcady flow

(~~)

.~o.(~:)

... ·'0· '"

.... y• • ""

~OCIC

.

5. 3. 2 Uniform and Non -uniform Flows. Uniform flow is dcfillcd as lhal lyPl' of flow in which Ih~ velocity at an y given time does nOI cllange with respect 10 space (i.e .• length of di~ct ion of the flow ). Mathematically. for uniform flow

(~~L~I =0 where

av = Change of ve locity

as = Length of flow in the ui rcction S. Non-uniform flow is lhal type of flow in which the veloci ty al any given time ctl311gCS with respect \0 space. Thus, mathematically. for non -uniform flow

('V) as ' _ron"",

.0

.

5. 3. 3 Laminar and Turbulent Flows. Laminar flow is defined as that type of flow in whieh the fluid panicles move along we ll·deflned paths or stream line and all the stream-lines arc straight and parallel. Thus the panicles move in laminas or layers gliding smooth ly over the adj acent layer. This type of Ilow is also called stream-line Ilow or viscous flow. Turbulent now is that type of flow in which the fluid panicles move in a zig-zag way. Due to the movement of fluid particles in a zig-zag way. the eddies fonnation takes place which arc responsible VD for high energy loss. For a pipe flow. the type of flow is determined by a non-dimensional number-

" called the Reynold number. where D = Diamckr of pipe V = Mean vcioci ty of now in pipe and t' = Kinematic viscosity of fluid. If the Reynold number is less than 2000. the now is calkd laminar. If th~ Reynold number is more thaI14000. it is called turbulent now . Irthe Reynold number lies between 2000 and 4()()().thc now may be laminar or turbul~nt. 5.3.4

Compressible and Incompressible Flows. Co mpressible flow is that type of flow in which the density of the fluid c hanges from point \0 point o r in other words the densit y (p) is not constant for the fluid. Thus, math~matically, for compressible flow p ... Constant Incompressible flow is that type of flow in which the density is constant for the fluid flow. Liquids arc generally incompressible while gases arc compressi ble. Mathematically. for incompressible flow p = ConslanL

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IL Kinematics o f Flow and Ideal Flow

165 1

S. 3. S Rotational and Irrotational flows . ROlalional flow is 1lial type of flow in wllich the fluid panicles wll ile flowing along stream-lines. also rolate about thcir own axis. And if the fluid panicles white flowi llg along stream-li nes, do not rotale abou t their own axis then ttJallypc of flow is "ailed irrotational flow.

5.3.6

One- , Two- ilnd Three- Dimensional Flows.

Onc·dirn cnskmlll n ow is thai lyp" of

flow in which Ihe flow parameter such as ve locity is a function of lime a nd One space co-ordinate on ly_ say.T. For a steady one-dimensional flow. the I'doc ity is a func tion of onc-spacc-co-ordinatc only. The variation of veloc itie s in ol her 1WO mutually perpendicular directions is assumed negligible. H ence

mathematically. for one-dimensional flow II =j(x).,·= Oand w= 0 where 11_ V and W
= f,( .... y ). v = f 2(x. y) and w :: O.

Th ~ _dimcn sional fl nw is that Iype of flow in which the veloci ty is a function of time and th ree mutually perpemlkular diredions. BUI for a sle:,dy three-dimensional flow the fluid p
now II::

.. 5.4

fl(x . y . z) . ,' :: fl (. y . z) and

11'::

fix . y. z) .

RATE OF FLOW OR DISCHARGE (Q)

It is delined as the quanti ly of a fluid flowing per second through a seclion of a pipe or a chanrlCl. For an incom pressib le fluid (or liquid) Ihe rale of flow or discharge is expres.scd as the vo lum e of fluid flowing across the s.cclion per second. For compressibl e fluids. Ihe rale o f flow is usually expressed as the we ight of fluid flowing across Ihc section . Thus (i) For liquids Ihe uni ls of Q arc mJfs or iilres/s (ii) For gascs the uni ls of Q is kgffs or Newtonfs Consider a liquid flowing through a pipe in which A :: Cross-sectional area of pipe V:: Average ve loc it y of fluid a(;fOSS the section ...(5.1 ) Q= A xY. Then discharge

.. S .S

CONTINUITY EQUATION

The eq uat ion based on Ihe principle of conservation o f mass is called contin uit y eq uati on. Thus for a fluid flow in g through th e pipe at all the cross-sectiun. the quantity of fluid per secund is constant. Consider IWO ",ross-sections of a pipe as show n in Fig. 5. 1. Let VI:: A vera ge velocity at cross-sedion I- I PI = Density at section I- I A I = Area o f pipe at section I-I

I I

Ii

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IL 1166 and

Fluid Mcchanics V2. P2'

A~


arc correspo ndin g valu es al section. 2-2.

~~

The n rate of now section 1- 1 "" p ,A, VI Rate of flow at "ce uon 2-2 == P ~!Vl

,_",.l~'"'.'"'.'"'.'"".'"'."11.,"".,.,." i ' , 7""

DIRECTl ON~

-

Acco rdin g 10 law o f co nse rva ti on of 1ll,ISS OF FLOW Rate of flow al seclio n I - I == Rate of fl ow al sec lio n 2-2

= PzAIVl

@

i

+.

b"""'" " ',~"",..,.

... (5 .2 ) Eq u:u ion (5.2) is appli ca ble 10 the co mpressi ble as we ll as inCQlIl - Fig . 5.1 Fluid flowi ng through pn: ...s iblc fluids and is callcd Contlnull}' Equation. If the fluid is in a pi~. compressi ble. Ihc n P, '" p, and continuity eq uati on (5.2) redu ces \0 or

p, A I V,

... ( 5.3 )

Problem 5.1

The diameters of a pipe atlhe seclions I (/lId 2 lire /0 em mid 15 em respeclil-ely. Find rhe disc/wrge Ihrough lilt' pipe if the relocily of W
At scc tio n 2.

"'..::.

4

(0 , 2)

:"::'

4

(. I) l " 0 .007854 Ill l _

V, '" 5 m/s. O 2 ,,, 15cIll ", 0.1 5 A2

III

="4" (. 15t =0 .0 1767 m-,

, D,=1Ocm

V," 5m./s.ec Fig. 5.2

(/) DiSCharge through pipe is g iven by equa tion (5. 1)

or

Q=Al x V,

= 0.007854 x 5 = 11.03927 III lls. A/l s. Using equ.u io n (5.3). we ha ve A,V, " A 2 V1 ( ii)

A, V,

,O~.= OO 78~54 ,=-

V 2 = - A, -,- " - 0.0 1767

)( 5 .0 = 2.22 m'-~. Ans.

Problem 5 .2 A 30 em diameler pipe. coIII'eying warer. branches illlo 111'0 pipes of diametl'Ts 20 em and 15 em respecti)'e/y. If IIIe (II'erage ~'e/oeir)' in II,e 30 em diamerl'r pipe is 2.5 mls. find the disc/mrge ill this pipe. Also derermille rhe ),elocity ill 15 em pipe if tile (II'ew ge I'e/oeify ill 20 em diumeler pipe is 2 mls. Solution. G iven :

V," 2.5m/ sec 0, "3Ocm-

(j) Fig. 5.3

I I

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IL Kinematics o f Flow and Ideal Flow

D, '" 3Dcm =0.30 11"

1

167 1

III

11"

,

,

A, '" - V , = - x .3" =0.07068 m " 4 4 V, '" 2.5 In/s

D2 '" 20 ern = 0.20 Al

=~

III

(.2)2 = ~ x .4

= 0.0 3 14 OI l ,

V2 =2 rn /s

DJ = 15cm = O.15m

, II

A):-

Find

,

, If

,

(. Is r =- x 0.22S '" 0.01767 m -

(I) Discharge in pi pe I o r Q , (ii) Veloc ity in pipe of di a. 15 (Ill or V)

Le t QI' Q1 nnd QJ are disch
Q,=Q2 + Q3

.. .(1 )

(,) Th~ dl~c hu rgc Q, in pi pe I is g ive n hy

Q, = A, l', = 0.07068 x 25 lO lls = 0.1767 rn l/s . A n~. ( ii) Value o r V.l

Q l = A1V1 = 0.03 14 )( 2.0 = 0.0628 m3/s Substitutin g th e values of Q, a nd Q1 in cqu~t io n (I) 0.1 767 = 0.0628 + Q3 QJ'" 0 .1 767 - 0.0628 '" 0. 1139 m1ls

Q3 = A1 XV3 =O.0 1767XV)

or

O. I 139=O.O I 767 XV3

0.1 139 = 6.·U m/s. An s. 0.01767 Problem 5 .3 \Yarer flows ["ro ugh II pipe AH 1.2 m diameler a/ j mls and Ihell pt/SIes II!rollgh II pipe BC 1.5 m diameter. AI C, the pipe brallches. Ilrallch CD is 0.8 m it! diameter alld carries Ollel/!ird of Ille flow in All. The flow I"elocity ill brallch CE i.~ 2.5 mls. Pilld Ihe I"olume rale of flow ill AIJ. Ihe )"e/OCily ill BC, Ihe \"e1ocif)' ill CD alld l/ie diameter of CEo V}=

Sol ut io n. G ive n: Dia meter of pipe AB. DAS= I .2 m Veloci ty o f fl ow th rough AB. VAS = 3.0 m/s Di a. of pi pe HC, lJ /iC = 1.5 m Dia. o f branch ed pipe CD. Dcv = O.8m Ve locity o f flow in pi pe CEo Vn · = 2.5 mls Lei lhe fl ow rate in pi pe AB=Q mlts Ve loc ity o f fl ow in pipe BC = VIIC mts Ve loc it y of fl ow in pipe

I I

CD = Ve/J mts

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IL 1168

Fluid Mechanics

,

A

B,

t 1.5

12m

,

m

I

! V>B= 3 m/sec

~

0

~

~

,

VeE" 2.5 m/$<)C

Fig. 5.4

Diameter o f pipe Then now ratc thrQu~h

CE", Da : CD : QI3

CE=Q-QI3= 2Q 3 (i) Now volume flow ralc th roug h AB = Q = V,\~ x Area of An

'IliU now ralc th roug h

= 3.0 x -

1t

2

1t

2

11

2

J

(DAR) = 3.0 x - (1.2) = 3.393 m Is. Ans . 4 4 (rr) App lying co ntinu ity equatio n \0 pipe AB and pipe BC, VAH x Area of pipe AB == VHe x Area o f pipe Be It

,

3.0 x- (D.... /I)" == V/IC x- (Doc) 4 4

"'

3.0 x (1.2)2 == VBe X ( 1.5) 2

V nc

3)( 1.2 2

1.5

l

=

[Di Vide by

~]

1.92 m/s . An s.

(iii) The flow ratc through pipe

CD=QI= Q = 3.393 = 1.13 1 rn J/s 3 3 Q 1 '" VCD x Are a uf pipe CD

. 131", Veo

)(~

(Om)!

, )(0.8-, '" 0.5026 Veo .

)(4

1.1 31 = 2.25 m /s. Ans. 0.5026

V ClJ : - - -

(i,') Flow rate th rough CEo Q 2 == Q - Q 1 = 3.393 - 1.1 3 1 = 2.262 rn J/s

Q 1 = Vet- )( Area of pipe CE

2.263 = 2.5

"'

DeE::

,

= Vc£~ (Dn/

,

)("4 x (DClf

2.2(3)( 4 2.5 x Il

~5'

=,, \.1;).0" 1.0735m

Diame ter of pipe CE:: 1.0735 m. A n s.

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IL Kinematics of Flow and Ideal Flow

169 1

Problem 5.4 A 25 elll diallle/I'r pipe carries oil of sp. gr. 0.9 aI II \'e/ociry of 3 rnls. AI atlOt!!er secriolllire diameter is 20 em. Filld the \'elociry a/ Ihis .Iec/ioll and also mass rate offlow of oil. Solution. Given at section I. v, = 25 em =0.25 m IT

at section

2

It

1

A , : - D , =- x 0 .25 = 0.049 4 4 V,=3 rnls DJ '" 20 ern = 0.2 rn

2.

OJ

1

"

Al : - (D.l t '" 0.0314 111 -, 4 V2 "'? Mass ra tc o f flow o f oi l = ? Apply ing continuity equat io n al sectio ns and 2 . A,V, = AlV l 0'

0.049 x 3.0 = 0.03 14 X Vl 0.049 x 3.0 V. '" = 4.68 IIlls. AilS. 0.03[4 Mass ratc of fl ow of o il = Ma ss de nsity x Q = p X A , )( VI o

Density of o il

Sr . gr. of oil

Densit y of water '" Sp. gr. of oil x Densi ty of' wate r 1 900 ~g = 0 .9 x 1000 kgfm = - ,-

De ns ity of oil

m

Mass ratc of flow = 900 x 0.049 x 3.0 kg/s = 132.23 kg/so AilS. Problem 5.5 A jel ofll'aler from a 15 mm diameter nozzle is directed l'erlically up\>'drds. Anl/ming Illat llie jet remains circular and neglerling any loss of energy, 111(11 lI'ill be llie diameler at II poim 4.5 m abow! Ille IIOU Ie, if Ille "elodly "'ill, II'/licll Ille jel {cares rile 1I0ule is /1 mls.

,

Solution. Give n Dia. o f nozz le.

0, '" 25 mm '" 0.025 rn

Velocity o f jet at nozz le.

VI'" 12 rn/s

Heigllt of poin t A.

II'" 45 III

Lei Ille velocily of Ih e jel at a heig ll t 4.5 rn '" V 2 Consider th e ve rlkal motion of Ihe jel from th e out ld of Ille nozzle to Ille poilU A (neglecting an y loss of e nergy).

Ini lial veiocily.

U=

Final veloc it y.

V = V2 g= - 9.1I 1 m/s 2 and li0: 4.5m

Value o f Using.

V, '" 12m/s

JET Of WATE

NOZZLE

T I

• .• m

J

- 1/ 0: 2g/l . we gel ,-12 , =2)«(-9.81 »( 45

V! V~

Fig . 5.5

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IL 1170 Fluid Mechanics

~1 21

V2

2x 9.8 1:.:4.5

Jl44

88.29

7.46111/s

Now applying continu ity cqu:nion to the ouliet of nouk and at point A. we gel

"

,

A\c VI-X VI Itx (0.025)2 xI2 Al = - ' _I", 4 = 0.0007896 V1 V1 4x 1.46 OJ = Di31nclcr of jd 3\ point A.

Then

.. 5. 6

Al

="2If 02 !

OJ

x4 "'V/0.0007896 It

or 0.0007396

="411

x DJ

2

= 0.03 17 m '" 31.7 mm. A ns .

CONTINUITY EQUATION IN THREE · DIMENSIONS

Consider a Iluid cle ment of Icnglhsdx. tty and dZ in the direct ion o f .(, y and Z. LeI U, I' and II' arc the x.)' and Z directions respectively. Mass of fluid ente ring th e face ABCD

inlet ve loc it y co mponents in per seco nd

= p X Velocity in x-d irection x Area o f ABeD =pxux(dvxdz) Then mass of fluid leavi ng the fa<:c EFGH pe r second = pu dydz

+~ a., (pu dydz ) d...

Gain of mass in x-dircction

= Mass through ABeD - Ma ss through EFGH per sccolld '" pu dplz - pJI dydz - -

=-

=-

a - (pll dydz) dx

o (pu dydz)tlx

ih-

,

ax o - (pu) Ii.( dydz ax

.. dyllz is l:o nstant I

Similarly. the lI et g ain of mass in y-dircctioll

"

ay (pI') d,l.dydz a - (pw) d_l dydz a,

'" - -

and in z- di rcct ion

'" -

Net gaill of masses '"

y

Fig. 5.6

-[~ (plI) + ~ (PI') + ~ (pw)] dxdyd: O.l oy oz

Since the mass is neither created nor destroyed in the fl uid cle ment. the ne t inc rease of mass per unit time in the fluid cle ment must be equal to the rate of increase of mass of fluid in thc clement. But mass

I I

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IL 171 1

Kinematics o f Flow and Ideal Flow of fluid ill the cleme n! is p. d.\". dy. tlz and its ralC of incrca!;C wilh time

is~ a,

(p dx. dy. tlz) or

. dx d)' dz.

Equati ng the two cxprcssio lls.

"'

"'

1

- [-' (pu) + -a (pI') + -a (rw) ax

ay

(Jz

Or . dxdydz

d_w/yilz '" -

dt

ap + ~ (pu)+ ~(pL') + ~ (pw) = 0

at ,h

oy


ICancelling dx.dy.dz from both sides1 ... (5. 3.-\ )

Equation (5.3A) is the continuity cq um iun in I:arlcsian l'O-urdinatcs in its most general rOlln. This equation is upplicablc to : (il Steady and unsteady flow, (ii) U niform and non -uniform flow, and

(iii) Compressible lInd incompressible fluids.

For stead y flow. Up = 0 and hence equation (5.3A) becomes as

a, o , 0 - (pu) +- (pl') + - trw) = 0 ax uy
.. .(5 .3R)

If the fluid is incompress ible, then p is COnSt31l1 and the above eq uat ion becomes as

all

vv vW vy VZ

-+-+-~O

0.(

... ( 5.4 )

Equation (5.4 ) is th~ continuity "quation in three-dimensions. For a two-dimen sional flow. the component w" 0 and hence continui ty equation becomes as

au + a,· ,,0.

... (5 .5)

ax ay

5. 6. 1 Continuity Equation in Cylindrical Polar Co-ordinates. The continu it y equation in cylindrical polar en·ordinates (i.e .• ~. 9. Z co-ordinates) is deri ved hy the procedure given below. Consider a two-dimensional ineompr~ssibl e now field. The two-dimensional polar co-ord inates are, and 9. Consider a fluid ~ · dr element ABCD hctween the radii, and r + dr as s hown in Fig. 5.7. The angle subtendcd by the clement at the centre is d9. Th" componen ts of the velocity V arc "r in the radial direction and ue in the tangential direction. The sides of the c lement ar~ having the lengths as Side All "" rd9.BC "" dr. DC", ('+ Iff) d9. AD" Ifr. The thidncss of the c leme nt perpendicular to the plane o r the paper is assumed to be unity. COllsider the flow in radial direclioll Mass of fluid en tering the fa"e AB per unit tim e "p x Ve locity in , -direction x Area

•

",

,."

Fig. 5.7

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IL 1172

Fluid Mechanics

Mass o f fluid

Ic~ving

"'pxu,x(A LIx I) '" p X II, X (rilO x I ) = p. the fa<:c CD per uni t time = p x Velocity x Arca

= px(u, +

(

=px u, +

(,: Arca '" AB x Thickness = rdO x I ) II," rdO

a;; .df) x(CDx

J:

a..

I)

(,: Area = COx I )

dr X(r+dr)dO

I': CD=(r+tlr)dO I

)

a.., d,+ -a.., (d,')' 1d O aT ar

=p xlI,xr + u,dr+r -

[

'" r{lIrxr +u , xdr+ rJ;; .df] dO [Th e tenn containing (11,)1 is very small and has been ncgk~ t cd l Gain of mass in r-di rect ion per un it time '" (M aS/; through All - Mass through CD) pcr unit time '" p. ",. rdO - P[U,.f + 1I,.dr+ r

a;~ .lIr] dO

a..,

[

1

= p.Il ,.rdO- p.u r·r.de - p u,.dr+r dr,ilf de

''', 1

[

= -p 1I, .dr + r dr,ilT .dO [Tlli s is writlcn in this form because (r. de. dr. I) is eq ual to vo lum e of

ele mentl Now COllsir/er the flow in S-direr/io" Gain in mass in a·direction per uni t time = (Mass through Be - Mass through AD) per unit lime

'" [p x Velocity th rough Be x Area - p x Velocity through AD x Area l

:: [P.ua.ii, x1-p (110 + i.l~~ ,de)xdr xlJ ", - p( ~ .de) drx

(': Area=drx I )

1

aU a r.J6.dr = -P aE! '

,

[Multiplying and div iding by

rl

Total ga in in fluid mass per unit time

u,

(lU 'l . r. dr. dfJ - p-aUa- · rtia,. tir

= - P [ -;:- + -, -,

I I

ae

...(S.SA )

Ii

~ I

IL Kinematics o f Flow and Ideal Flow

173 1

p x Volume of fluid clemen! =px[rd6xdrxll

Bul11lilSS of fluid clement

=:

=px,,'9.dr Rat e o f in creasc of fluid mass in the cle ment with lime

a [p. rde.dr ] = a;' ap rdedr "a;

... (5.58)

(": rde .
1

-p [ -U, + -(111, r.dr.d9-p -dUo rar

"'

Ja

U,

"'

rde. dr

- p[

Ju,]

---; + Jr

[u,

r dUe

-p

Ja

ilp rd9dr dt

= -

r =

dP

[Cancelling rilr . d9 from bolh sides ]

at

Ju,] + p -aile . ~ __ 0

-dp +p - + r dr

...(5.50 r Equation (5.5 C) is Ihe continuity equation in polar co -ordinates for two-dimensional flow.

at

For steady flow

ap :

a,

i.l9

0 and hence equation (5.5 C) reduces to u, au, ] aUe I p[ - + +p - . - :O r

ae

ar

r

~+au,+ aue

.!.:o ,a, ae ', au, aUe u,+ '-a;+ de :0

"'

:, (ru,)

"'

+ dd

O

(un) '" 0

Equation (5.5 /J) represcnts the cuntinuity cquation in incompressible flow,

Problem 5.5A flow?

[.: : , pol~ r

(r. u,) '" r. daU; + u,]

...(5 .5D)

cu-ordin ates for two-di mensional steady

Examine whetllef Ille [o/lQ ..... illg I'e/oeity compollell ts rt'preselll a physically possible

", = r Jill O. Uo '" 2rtos O. u, '" r sin e and ue '" 2r cos 8 For phys ica ll y possible flow. the cont inu ity equation.

Solution . Given:

a

ar (,u,) +

Now Multiplying the

I I

asa (110): 0 should bt'satisfied. e

abov~

",: 'sin equation by,. we gd fII,=?sin8

Ii

~ I

IL 1174 Fluid Mechanics Differentiating Ihe preceding equation W.r. l. r. we get

~ {)T

(furl"

~ (r sin aT

8)

=2rsin8

(":

si n

e is cons tant

w.r.!. r)

Now jje'" 2r cos e Differentiati ng th e nbovc cqumion w.r.l . e. we gel

a

{}9 (" 0) '"

a (2,- cos 9) as (.:

'" 2, (- sin 8)

2,- is conSlalll W.r.I. 9)

= -lrsi n e

i.(nI,) +~ (u e ) = 2r sin

a,

ae

e - 2,- sin e = 0

Hene.: the cominuity eq U
to 5.7

VELOCITY AND ACCELERATION

LeI V is the rcsu hant ve locit y a1 any poim in a fluid flow. LeI 1/. I' and w arc its componclll in .1', y and Z di rections. The velocity co mponents arc functions of space-eo-ordinates and lime. M31llcmal icall y. Ihe ve locit y co m ponents arc given as U=ft(.I',Y·Z,I)

\' "'li·I, y, Z,

I)

w=f,(.r.), . z. t) and

R~~ullaI11

vdoc ily.

V=

IIi + vj+ wI: = Ju l + v l + w l

Lei <1r <1 , and a, arc Ilic lot al acce lerdtlo n in x. y and Z directions respective ly. Tllen by the cliain rulc o f d ifferc ntia tion. we lIave

du au iI.l au dy au dz au a = - = --+--+- .- + , ill dl til til dl

ax

ay

az

dx ely dz - =u. - = I'anO- =w til ill til du til

dll dx

d,'

dl'

dll dy

du dZ

du

a = -;II-+I'-+W-+-

,

dl

a,· d" av ax +v -d), +w -dZ + -(II

a . = - =u ) til

Similarl y.

dw

a.= -

I

If

dW

=U -

ih

dW

+>' -

ay

dW

...(5.6)

dW

+>' - + -

azal

av '" O. wllcrc V is result ant veloc it y a,

('or ste ad y no w. -

I I

Ii

~ I

IL Kinematics of Flow and Ideal Flow

175 1

VII =O,Jv =Oand Jw =0 Jt iJl U/

"' Hence acceleration in

-I',

y and Z directions bci:OIllCS dll

all


(Ix

dl'

o - - =' -

all

+l' -

au

~,

+ Wdz

a,·

dv

dv

dll'

aw

all'

all'

dl

ax

()y

Jz

, -

... (5.7 )

a , = - =II - +I' - +W } dl ax 0;, Jz

" =-=,-+,-+wt

Acceleration vector

A=ai +aj+at:

J ' +a, , +d" , .

=

1

.. .(S .&)

ii,

5. 7. 1 local Acceleration and Convective Acceleration . Local acct'lcrution is defined as the rate o fincrcasc ofvcloci ty with respect to lime at a given poillt in a now field. In the equation given .(JUUI,

by (5.6), the cxprcsslOn -

.-

dt dl

C(l n n:o;Ii,'C acceleration

i~

(JW,

,

or "",,\ IS known as local acceleration. of

defined as th e ratc of change of veloci ty d ue 10 the challge of position of

Ju dv

fluid particles in a fluid now. The expressions other than -

,-

at at

Jw

afld -

at

in cqualioll (5.6) arc known

as convcclive acceleralion. Problem 5 .6

Til e I·elocity ,'eClor ill a fl' lid flo w iJ' gil'en V", 4.';i - IOYyj

+ 21k.

f-illd rile l'e1ocity lIml acre/amioll of a flllid particle m (2, I. 3) aI rim e I '" I. Solution . The velocity componel11s II. v and IV arc II == 4.r'. I' '" - IO.~ y. IV '" 21 For Ihe poinl (2. 1. 3). we have x == 2. ), == I and z = 3 allime I = I. Hence ve loc ily components at (2. I. 3) arc II = 4 X (2» ) == 32 units I' = - 10(2)1(1) = - 40 unils w=2x I =2uni ls Velocily veclor Vat (2. I. 3) = 32; - 40j + 2k Resultant velocity =

Jil"+ ,.1 + ","

==b2l

+(-40)' +2 2 -JI024 + 1600 + 4 = 51.26 units. Ans.

Ac ce leration is given by equation (5.6)

all

all

ih

dY

all all

a =11 -+ 1' -+ 11' - + -

az at av av av av ,. =II -ax + V-dy +W -az + -ar

.(

(l

I I

Ii

~ I

IL 1176 Fluid Mechanics " == ;:

d",

dw Il -

(.Ix

h

-

~.

dw Jz

+W -

d... + -

at

Now from velocity components. we have

au

, all == o. -au

-

= 12.,-, -

ax

dz

dY

Jy = - 20.\)" ~ ax Jy

dW

ax

= 0,

ow

Oil = 0 alld -

at

~

0

= - I O_~. a~ =0('11' = 0 iJz

ow

oy '" 0, az '" 0

iJ!

dW

and

at = 2.1

Su bstitutin g the valu es. the accel erati o n co mpone nts at (2. l. 3 ) at lime I '" I arc

= 4~.3 ( 12x2) + (- 10;')') (0) + 21 x (0) + 0

lIx

'" 4&x~ = 48 )( (2)~ = 48 x 32 = 1536 units

",." 4x3 (- 20xy) + (- IOx!y) (- IOx~) + 21 (0) + 0

=-

4

&ox y + l00x\-

=_80(2)4( 1)+ HlO (2)4 X I = - 1280+ 1600= 32Q unils.
"'

== 4.yl (0) + (- In;)') (0) + (11) (0) + 2.1 == 2 .0 units

II! = 15361 + 320j + 2k. An s.

Accele ration is

A = a,i + (I) +

Resultan t

A =J( 1536)l + (320)1 + (2)l units =J2359296 + 102400 + 4 '" 1568.9 unit s. Ans.

Problem 5.7

Th e [o/lowing CUjCS represent Ihe Iwo ,-e/oe ily compOI.enIS, determine the third componnll of "e/ociry SlIdl flwr riley salisf)' rile COII/in uir), eqllllfio" : (i) 1/ =.y2 + l + z" : "=xl-Y{] +.t)' (Ii) \. = = 2.\)"z.

2/. '"

Solution.

The continuity cllualio n for inco mpressible fluid is ~ivc n by equation (5.4) as

all + 0" + aw ih

a),

=0

ih

a" =2.T a." a, , - =2x)' -z + x

Cuse I.

-

iJy

Substituting the va lu cs of

h+2q- l

I I

all dx

and dl' in continuity equa tio n.

d),

, +x+ aw~o a, ow ,"\ ' - =- 3x - 2~)" + z-orow= (- 3.(- 2ty + z-) dz a,

Ii

~ I

IL Kinematics o f Flow and Ideal Flow Integration of both sidc ~ gives j dw '"

i( - 3..- -

177 1

l .ry + Zl) til

+

W '" ( - 3.>.l - 2.\)"

~: ) + Const31l1 of integration,

w here l.'O ll stam of integratio n canllot Ix: a function of z. BUl it can be a funct ion o f x and y Ilia! is f(x . y). IV

=(

- 3u - 2:
+ f(:<, y). An s.

a,· =4)'

\' '" 2l

Case II.

z:)

-

dy

-

aw = 2.')' a,

-

Su bstitutin g the va lu es of ~ and dW in conti nuity eq uatioll. we get

Jz

dY

dw

-

d.•

+4y+2xy=O

d,

".

-

d.•

" = _ 4.\)' _ 2)' _;1 + f ey,

[nt egrati ng. we get

Problem 5 .8

'" - 4)' - 2.\)' or dlj '" (- 4y - 2ry) dx Z) = _ 4xy _ x 2y

+ fey, z), Am .

A fluid flow field is gi,-<'I/ by V", xl )'; +

/;j _(2.>.)'z + YZ 1)/c

PrOl'e Ilwl il is (I case of possible slnlll), incompreJJiblc fluid jlow. Co/ell/are lire I'e/ociry and aeee/em/ioll at II,e point (2. I. J). Solution.

For the givcll Ouid flow field"

= ..?y

I' :.lz

ow -ax

a,·

-

ay

= lxy :2yz

aw

az-,, - h y - 2yz. ['or a case o f possible stead y inco mpre ssib le fluid flow. th e continuity eqU3lion (5.4 ) should be sm isfied.

i.e .. Substituting the I'alues of dll ,

d.'

dw

,uld - . we ge t

d,

dll dl' dll' - + - + - '" 2xy + 2)"2 dx dy d,

I I

2.1)' - 2)'2 '" 0

Ii

~ I

IL 1178 Fluid Mechanics Hence the veloc it y fi eld

v '" .~}'i + yZzj _ (2.ry< + ),Zl)

is a possible case of fluid fiow. AilS.

j:

Ve locity al (2. 1,3) Substituting the va lues

z= 3

.f '" 2. Y = I and V

=

+

' . [JI

y 1!j

in velocity field, we gel

' k - (1.1)'< + yz')

= 2"x l i+ l'x 3j - (2x2 x I x3+ I x3')k = 4i+3j-21k. An s.

:~4 Z + 3'- + (_ 2 If == J16 + 9 + 441

and Rcsuhant veloc it y

..}466 '" 21.587 unils. AilS.

Accd.,ratkm nt (2, 1,3) The acccJcralio li co mpon ents (Ix' a" and 11, for Slc;\(ly now arc 1I

au

.(

au +W -i}J j

:U - + V -

ih

dy

ilv

dV

(l , =U -

)

+ V-

ax

dy

OZ

ai' dZ

+II' -

all' h -ali' h ' -a!l'

il,= It -

-

ax

2

dz

d)'

dU

all

ai'

dy

au

l

u"'_t)'.-= l.ry. -= x 300 2


th'

dZ

00

rlv,

V=Y Z. ax = 0. {)y =2y:, dZ =y' , i)waw

w= - 21yz-yz·.-:;-= - 2yz. -, oX

y

, aw

= - 2.t: -:' , -,

= - 21)'- 2)': .

Z

Substituting these values in acceleration components, we gel acceleration

~t

(2. 1. 3)

a, == .\1)' (2x)') + iz (xi - (21),z + )'Z') (0)

= l,.-ll + .r"iz =2(2)11'+2'x l ' x3 =2 x8+ 12 == 16+ 12 ==28 units

a,. == x 2)' (0) + iz (2),:) - (21)'z + YZ2) (i ) . == l.l'? - 2xlz - lz2 ==2x 11 X3'-2x2x 11 X3_1 1X3'= 18-12-9 == -3units U;.

= x")' (- 2)"z) + iz (- 2xz- Z2 ) - (ll)': + ),Z2) (- 2.\")" - 2)"z) = - 2Xlit - lxiz' _iZl + [4Jy'z + l.ll:' + 4.,).1: 2 + l izl r == _ 2 X 22 X l ' x 3 _ 2 x 2 X J2 X 3 2 - J2 X 3.1 + [4 X 22 x J2 X 3 + 2 x 2 X J2 X 32 + 4 x 2 X 12 X 3 2 + 2 X 12 x 31r = - 24 - 36 - 27 + 14K + 36 + 72 + 54r = - 24 - 36 - 27 + 48 + 36 + 72 + 54 = 123

Acceleration

I I

=

ai + II) + {If = 281- 3J +

I2Jk. An s.

Ii

~ I

IL Kinematics o f Flow and [deal Flo w or

J28 l + ( 3)1 + 123 2

Resultant acce leration

179 1

../784 + 9 + 15 129

= .J15922 = 126 .1 8 units. Am . Problem 5.9 "-ind Ille conr<'Ctil'l! acceleration allil e middle of a pipe whiel! co/ll'erg<'J uniformly from 0..1 m diameter to 0.2 In diameter OJ'a 2 III lengtll. Tile rale off/o ..... is 20 lith. If Ihe rale of flow c/wnge.! IlIliformly from 20 lis 10 40 lis in 30 seconds. find rhe towl acceleration ill II,e middle of Ihe pipe a/ 151h second. Solut ion. Given: Diameter at section I. D I = 0.4 m ; D2= 0.2 Ill. L = 2 Ill. Q = 20 lis = 0.02 111 3/S a~ one lilre =O.OOlm}= 1000 ern} Find (I) Con>'c<:!;v" ac<:clcration m middle i.e .. at A when Q = 20 lis. (ii) Total acceleration al II when Q c hanges from 20 lis 10 40 lis in 30 seconds. Case I. In this case. tile rate of flow is constant and equal 10 0.02 m}/s. Th" velocity of flow is in _(-d irecti on only. Hence this is one-dimensional flow and velocity componenl~ in y and Z direction s are lcroorl' =O,Z=O.

"'

= I' ay only

Convectivc acce Jcration

.

Lct us find the value o j

I'

all

and -

".,

.. , (i)

_

.

at a dlstance.f from Inlet

(lJ

Th e diameter (D,) at a distance _f from inlet or at section X-X is given by,

0.4 - 0.2 D, =O.4 xx 2 =(0.4-0.1 x)m The area of cross-section (At) at sec tion X-X is givcn by. A

"'~ D 2= ~ (0.4 - 0.1

''' -. T r-+----' ~·T - -' 1 _ x...: x :

O.4m

O.2m

,

:""' lm - ,

;- 2m

X) l

4 ' 4 Ve locity (1') at the section X -X in terms of Q (i.e .. in tenns of rate of flow) .r

I'

=

---.SL '"

~ '"

Ar"a

A,

---.SL=

~D l 4

1.2730 = (0.4 O.I _r)!

"ax

To find -

~

It

Fig. 5.8

4Q Il(OA - O.lx/

'

1.273 0 (0.4 - 0.1

-Irl

fills

... (ii)

. we must diffe re ntiate equation (it) with respcci to x.

au = -:;a [1.2 73 Q (0.4 - 0.1 x)--•[ ax

-

0.1"

= 1.273 Q (- 2) (0.4 - 0.1 .1)- 1x (- 0.1) = 0.2546 Q (0.4 - 0.1 _r) , Substituting the val ue of

uand au ax

[Here Q is co nstant] ... (iii)

in equation (i). we get

Convective accelerat ion = 11.273 Q (0.4 - 0.1 .I;-l ] x [0.2546 0 (0.4 - 0. 1 _rr'] = 1.273 x 0.2546 x Q2 x (0.4 _ 0.1 _IT l

~ I

I~

~ I

IL 1180

Fluid Mcchanics '" 1.27] x 0.2546 x (0.02) 2 X (0.4 - 0.1

:.Convective

acccl~ralion

x»

[-:

Q '" 0.D2 m'/sl

al the middle (where x = I 111) = 1.273 x 0.2546 x (0.02)2 x (0.4 _ 0 .1 x 1)-3 m/s 2

"" 1.273 x 0.2546 x

(o.oll x

(O.3r3 mls 2

= 0. 0048 mls·. An s. Case II . When Q changes from 0.02 m3/s 10 0.04 1I1 3/s in ]0 seconds. find the 10lai acceleration at x'" I 111 and r '" 15 seconds. Total accclcr;l1ion = Convective 'lccclcrmion + Local ac(;clcration ,11 t'" 15 seconds. The rmc of now at I = 15 sccunds is given by

Q'" Q 1 + Qz 3~ Q1 x 15 where Q2 = 0.04 m 31s and Q1 = 0.02 1ll 3,s

=0.02+

The velocity (14) and gradient

:. Convective acceleration",

(0.04 - 0,02)

30

J

x15= O.03mfs

('"a.. ) in terms of Q arc given by equations ,I.

(Ii) and (iii) res[l<-'l:tivcly

a"

"

'" I 1.273 Q (0.4 - 0.1 xr21 x [0.2546 Q (0.4 _ 0.1 = 1.273 x 0.2546 Q2 x (0.4 _ 0.1 )( 1)-3

.(r l [

:. Convective acceleration (w hen Q = 0.03 lO lls nnd x = I rn) : 1.273 x 0.2546 x (0.03)! )( (0.4 - 0. 1 x 1 = 1.273 )( 0.2546 )( (0.03) 2 )( (0.3)-.1 m/s2

rl

=0.0108m/s" Local acceleration:

... (il')

i.l" -a; : -a;i.l [1.273 Q (0.4 - 0.1 .rr ! 1

I .:

II

from equation (ii) is

: 1.273 x (0.4 - 0.1 x)

I .:

1x

II:

1.273 Q (0.4 - 0.1 X)-l ]

,Q

-

a,

Local acceleration is at a point wherc.f is constant but Q is changi ng]

Local acceleration (at x: 1 m)

, aQ : 1.2731'.(0.4 -0.1 x I) 1'.= 1.273 x (0.3)

-2

0.02 x 30

"

[

. .'Q = QrQ I: 0.04 - 0.02: 0.D2 ] 30 30 .. .Iv J

. a,

= 0.00943 m/s 2 Hcnce adding cquations (i,,) and ("). we gc t lOtal acceleration.

:. Tmal
I I

Ii

~ I

IL Kinematics of Flow and Ideal Flow

.. S. 8

VELOCITY POTENTIAL FUNCTION AND STREAM FUNCTION

5.8 . 1

Velocity Potential Function .

181 1

II is defined as a scalar function of space and time such

that its negative derivative w ilh rcspccl l0 any dircrtion g ives the nuid velocity in that direction. It is defined by 0;. (Ph i). Mathematically. Ihe velocity. polcnlial is defined as 0;. '" f(x. y. z) for steady flow sucli that

U=_dQ

d' v'" -

dQ ily

...(5.9)

d,

11'=- -

il,

where II, \' and I\' MC the components of velocity in .l.), and z dircdions respectively. The velocity components in cy lindrical polar co.()rdinales in terms of vctocity potential funct ion
lie =

)

.!. dQ

.. .(5.9A )

'"

whe re

II, '" ve locity component in radial direction (i.c .. in , dircctioll)

and

lie = velocity component in tangential direction (i.e .• in 0 direction)

..

.

.

.

. au + -a" + -a\~ == o. a.T ay at

The COl11lnUlty equation fur an IIKumpresstble steady nuw tS -

Substituting the values of u. \' and I,' from equati on (5.9). we gtl

... (5. 10) Equation (5.10) is a Laplace equation.

alq.

a~9

f'or two -dimension case. equation (5.10) reduces to - , + - , '" o. a.T - ay ' If any valu e of

9 that

. .. (5. 1 I)

s.11isfies Ihe Laplace equation. wilil"()rrespond IU some ,",
P ro lle rties o r th e l'o te nt !:,1 I'- unctio n. The rotational co mponents· arc gil'en by

• Pleasc. refer to equation (5. t 7) on page 192.

I I

Ii

~ I

IL 1182 Fluid Mechanics ro '"

.!. (all _dlY)

==

.!.[aw - ~l 2 ayaZ

,. 2 dZ

U) X

ih·

Substituting the values. of I•. I' and w from cquatioll (5.9) in the above rotational components. we gel

'"' a!q.

(l! Q

a!q.

a"41

(hay

dyi}x

ilzih

dxilz

If 9 is a continuous function. IlIcn - - " - - 0- - " - - : etc. ro~ =w. =wx =O.

When rotational components
c!\iS!~.

the flow should be irrotalional.

2. If veloci ty potential (41) satisfies the Lapla(;c cqualion. it represents the possi b le steady inc01l1pressible irrotational flow. S.S.2 Stream function . It is defined as Ihe scalar fUIl(;tion of space and time. SU(;h. Ihat its partial derivative with rcsp.:ct to any direction gives the veloci ty component at rig ht angles to that direction. It is denoted by IjI (Psi) and defined unly for two·dimc nsional flow. Mathcmati<:alJy. for steady flow it is defined as IjI : / (x. y ) such thut

I

a. Jx ",.

... {5.12)

J IjI : _ "

and The velocity

cOl1lponenl~

'J'

in cy lindrical polar co·ordinates in ICrms of stream function are given as I

aljl

-

-

aljl

... (S .12A)

and "a" - r ar where "r'" mdi al velocity and "9 " tangential velocity II , :

. _.

ae

_ .

' , a" + -(/1' " ax ay

The COntln ully equation for two-dimensional Ilow

I I

IS -

o.

Ii

~ I

IL Kinematics o f Flow and Ideal Flow Substituting the values of

1/

and \' from equation (5.12). we get

a( aya.) a(""J a:;- '"

,h

+ dy

-

183 1

a'. a'.

0 or - a.•ay + (lxdy '" O.

Hence cxistclH:c of \jI means a possible case of fl uid flow. The flow may be rot ational or irrmat ion al.

'(',ax '")

The rolallonal component W, .IS gIven by 00, '" . .

2

Substituting

th~

-

- -

it),

.

va lues of 1/ and I' from cqualio rl (5. 12) in the above rotational component. we gel

For irrotational flow. U1, '" O. Hence above equation becomes

a'

a

as~ + ~ ax- G1y-

=0

which is Lap!iI!;c equ ation for 1jI.

The properties of strea m function (1jI) arc: L If stream function (1jI) exists. it is a possible use of fluid flow whic h may be rotational or inOlational. 2. If stream function (ljI) sat isfi es the Laplace equation. it is a poss ible case of an irrotat ional flow.

5 .8 .3 Equipotential Line . A line a long which the velocity potenti al 9 is cons tant. is called equipotential li ne. For equipotential line
=0

,.ax a.a),


B"

d
'" -

For equipotential line. - (utfx

0'

But 5.8-4

+

an -

IIlf.f -

ay

wi)'

'" - (utf.{ + I'd}'). tf
dy '"

"

tfx

\'

dy = d.,

... (S. 13)

Slop~ of equipoten lial

line.

Line of Constant Stream function ~

'" Constanl

d~:O

Bm

I I

d~ = aljl

ax

d.{ +

aljl a},

dy '" + "dx _ udl'

.

Ii

~ I

IL 1184

Fluid Mechanics

For a line o f constant stream function

'" dljl = 0 or

dy

:

dx

IIdy '" 0

1'1/-'.' -

I'

... (5. 14)

u

But dy is slope of stream line.

d,

From cqU3tions(5.13) and (5.14) it is clear thallhc product Orlhc slope of Ihe equipolenliallinc and the s lope of Ihe stream line a1 the point of inlers.ection is equal to - I. Thus Ih" equipotcnliallines are

othogonal to the stream lines at all points of intersection.

5.8 .5 Flow Net. A grid oblliined by drnwing a scricsof cquipotcntiallincs and stream lines is called a now net The flow n<:1 is an illlporlanilool in analysillg two-dirn"nsional irrOlalional flow problclllS. 5. 8. 6

Relation between Stream function and Velocity Potential function

From equation (5.9),

_df

we have

andv =

"

JIjI dltl and v = ""'\" dy ax

From equation (5.12). we have u = - -

u= _ d¢= _ aIj/ 3Ild,·= _ #

Thus. we h3vc

ax

ilQ

ay

dy

= alj/

ax

'xilQ "''y" :

) -"' " " ",

... (5. 15)

:

Problem 5 .10

The "elocily pOlelllial funclion ($) is gil'en by (III e.>.pressiolJ xy

1

,

.f y

$=---x + -

+y

1

J 3 (i) Fi/ld IIJe "e/oeily compolle/l/s ill x alllll' direclion. (ii) Sholl' 11101 9 represelllS U possible wse offloll'.

The p3nial dcriv31il'cs of

1

3

q.

,

xy

~

3

W.r.l..l and yare

dQ

i

J.l ly

-

ax

=- -

-2 x + - -

... (1 )

d$

3xy".I J , : - - - + - +-y

... (2)

-

ay

I I

,

x)'

9=- - - X + - + ,

Solution. Given:

3 3

3

3

Ii

~ I

IL Kinematics o f Flow and Ideal Flow

185 1

(il The velocity components II and I' arc given by eq uation (5.9)

""d.l : - [y' - 3- 2

11= -

, . 11= "'3

ly ]X X+ -- ]

3

y'

"' )+ 2.r -x ' y

,

+2x -x-y. Ans.

An s . (ii) The given val ue of
a

2

<1>

- , + - , =0 ih · dy · From cqumion s ( 1) and (2). we have

dQ = - J'/3 - 2r+x2y

Now

ax

,f
-, : - 2 + 2.ry

,hO {)¢I -

and

,Xl

= - .ry- + -

(ly

3

+2y

(1 141

= - lx)'+2 dy l

d1¢l ih

J1 dy

- , + - 1 = (- 2 + 2\)') + (- 213' + 2) = 0

Laplace equat ion is sa tisficu and he nce Q represe nt a possible case of flow . Ans. Problem 5 .11 The ndocill' pOlential j,,"(lioll is gil'e" by

dQ '"

ax

lOx

i.lQ

= - lOy.

ay

Bul velocity components" and

I'

i).

arc given by t>{jualion (5.9) as

u:-d1p:-IOx

ax

a. = - (-10y) = lOy

" ,, - -

ay

Th" velocity components at the point (4. 5), i.e .. al x " 4. Y " 5 "" - 10 X 4 " - 40 units. An s. I'" 10 x .5 " SO unit s, A ns.

I I

Ii

~ I

IL 1186 Fluid Mechanics Problem 5.12 A stream junclioll is given by

IjI '" Sol - 6)'. C,,/cu/mf! the ..efocil)' componellts o /l d ,,/so lII(lg1l;llIile (mil direction of Ihe resu/I"/I' I'e/ocity al ,my poill/.

Solution .

1jI =5x - 6y

-"" = 5 and -d. = - 6. ax dY But the velocity l'OmponCllts II and I' in tel ms of stre am function arc g iven by equation (512) as !I=

_ dljl = - (- 6)= 6 uni ts/sec.

d),

a.

1' = -

a.,

AilS .

= 5 un li s/sec. An s.

Rcsul1ant velocity

,. 5 lanS ", - = - =0.833 " 6 == lan- l .833 = 39° 48', An s. Problem 5 .13 If for II two-dimensional potelltial flow, the )'elociTy pOlelitial is gil'ell by 41=..-(2), - 1) delerm;" " Ihe I'docily III Ihe [XliII/ P (4, 5). Dekrm;IIe "bo Ihe WI/Ill.' of slreOIll "melioll IjI (II the poillt P. Direction is give n by .

e

Solution. G i ve n ~=x(2y - l) (i) The vc loc ity componcllIs in the directi o n of x and ), arc

a;

a

= - -ih '" - -ax Ix (2)' - 1)J =- (2)' - I J = I \'= _ i>t", _ ~ Ix(2y - I)] = - [2x] = - 2x

1/

dY

AtlhepointP (4,5 ),i.e.. alx= u= )' = .: VelocitY;11 P

Of

=

Res ul1ant veloc ity at P

2)'

dY

4.y=5 1-2x5 = - 9 unit s/sec - 2 x 4 = - 8 uni lsiscc

- 9i -8j

J9' + 8

1

'"

~81 + 64 =

12.04 units/sec = 12.04 unit,Jse<:. Ans.

(ii) Vallie of S trea m Function a t I' We know that

and

d1f

= - u= - ( 1 - 2)')=2)' - 1

:

=I'= - 2X

d),

Integrat ing equat io n ( I) w.r.t . 'y', we get

j dIy "" j (2y - I) dy o r 1.[1 =

I I

... ( i)

... ( i i)

+2

.1

y + Constall! of integratio n.

Ii

~ I

IL Kinematics o f Flow and Ideal Flow

187 1

The const ant o f integration is not a fUliction of y bUI il can be a fUllcti on of x. Le t the val ue of constant of integration is /e Tllen \V '" l- y + k. . ..(iii) Differentiating the above equation w.r.L 'x', we get

dill '" (J.T But f rom equation (ii),

ak

ax

Oljl = _ 2x

0,

dljl

dk

ox

oX

Equating the value of -,;;--. we get -:;- '" - 2.1. 1ntcgratlng . th " ' n. we get k IS equalto

= f - 2xd x = - -2x~ , - '" - X-.,

Substituting this value of k in equation (iii), we gel It' '" y1 _ , . _ ,,1, AilS .

Stream fun(;tion If a1 I' (4. 5) '" 5' - 5 - 4' '" 25 - 5 - 16", 4 IInits. Problem 5.14

711,. stream /1I1II:tioll for

II

two-dimensional flow is git'en by \jI = 2xy. (aleulMe Ille

\'cloeily at IIII' point P (2. 3). Find lile I'elocity pOfelllillljullc/ion

Solution. GiVt'1l : The ve locity componen ts

IjI 1/

1/=

At

th~

point P (2. 3).

w<"

get

$.

= 2 ly

and I' in tcrms of Itf

I'

d.

ar~

a

- - = - - (21}')= - 2x

ay

= -"" = -a ih ax

ay

(2.1)')

= 2y.

= - 2 x 2 = - 4 units/sec v = 2 x 3 '" 6 units/sec

1/

/, , /, , 1"'=

Resultant vcloci ty at

p=" .. + 1' _ ", 4 +6 = ,,1 6 + 36 Vel ocity l'olenli,.1 Fu n ction ¢I We know

A ilS.

~

.J5i = 7.21

"' -Il = -(-2x) =lx

d.jl = -v=-2y

units/sec,

...( i) ... (i i)

Oy

Int egrating equation (i). we get

!d$=!2xdx

"'

,

2x' 1 ., +C= x + C

... (iii)

where C is a constant which is indepe ndent of x bu t can be a function of y. Differentiating equation (iii) w. r.t.

I I

y. we ge t

ay oc oy

Oljl '"

Ii

~ I

IL 1188

Fluid Mcchanics

But fro m (ii),

~

0,

= - 2y

'0,C "' - 2.1'

-

Integrating lliis equatio n. we get C ==

2 '

J- 2y dy =- ~ . == - l

Substituting this val ue of C in eq uati on ( iii). we get q. == x~ _ yl, Problem 5.15 Skelch Ihe s/re(lfll lines repreSented hy W == .; + Also Jim{ QuI Ihe I'e/ocil)' lind ilJ' dir ediOIl al poin/

Solution. Given \j1 '" ,,) + The ve locity compon ents II and I' arc

AilS .

i.

(I. 1 ).

I

,

_ iA¥ =_i. (..-2 + I)=-2y dy

I'

==

Al tlie poi nt ( I, 2). tlie velocity 11=

()y

,

oW = i. (.(2 + l ) = 2.1'

,h

ax

compon~nL~

4 UNITS/SEC

Fig. 5.9

arc

- 2 x 2 '" - 4 units/sec

y

\' = 2 x I = 2 uni ts/sec

Resul tant ve loci ty

==

~u! + vl_~(_4)1 +2 '

=.fiO lane=

oed

I'

== 4.47 uniHi/so.-"C

2

,

-HH--- +--+-+-t--~

I

-=-:II

4

e = t all

I .

2 5 = 26° 34'

Res ultant veloc ity makes an an gle o f 26" 34' wilh .f- axis.

Sketch of Stream li nes

w=x2+l Ld Then We ha ve

1jI =

Fig. 5. 10

1.2.3 and so on.

1 =->.,l + l

2=-I) + l

3=·,J+i and so 011. Each equa ti on is a equm io ll of a ci rcl e. T hus we s hall ge t concemric circles o f differcm diam ete rs as show n in Fig. 5.10. Problem 5 .16 Th e l'e!oCily components in (I Il!"o -dimensioll(l/ flow field fo r (In incompressible fluid (Ire liS follows: u=

LJ

+ 2-1· _ -Ily {lnd

obtain {Ill expressioll for rhe Slream fUllc/io n

I I

I'

= -1/ _ 2y -

7!/3

v.

Ii

~ I

IL Kinematics o f Flow and Ideal Flow

189 1

u " if] + 2\" _ _(2y = .1)'2 _ 2y _ x 1/3 .

Solution. Given:

I'

The velocity components in terms of stream function arc

a; = 1' = .l:l -2y - ~f3 ihv OJ'

Int egrating (il w.r.l, x. we gd

···(il

= - 11= - 113 - 2\"+.,-2,)' IjI

... (; i)

= f (xi- 2y - ..:"3) dx ~

1

~

q - -2xy- -"- + ' . .: 2 4 x3

. .. (iii)

where k is a COnSl3J11 o f integrat ion which is independent of x but can be a function of y. Differentiating equation (iii) w.r.l. y. we gd l

Oljl == 2.t y _ 21" + ~

dljl

Bul from (ii),

-

ay

at = .f \' _2.1 + ~

2

~

J

1

~

= - yf3 - 2 l +XY

, dljl'
,

Comparmg the value of -

oy

oy

dk= _ 113

oy

k =

Integrating. we gel

, f (- 113)

,

-=.L = -y

dy =

4 x3

12

Substituting this value in Uii). we get 1

1

••

. : -11: ·V- - 2xv - -" V - -· - . Ans. 2

Problem 5.17

.

12

12

III a rWQ·dimensional incompressible j/ow. ,lie fluid reiociry compoll ents are gircn by II = .l - 4YG/ld I' = - y - 4.1".

SilO»" Ihat reiocil)' pOlelilial e.(i.~IS allli de/ermine ils form . Find ol.fo lite slrcom filliCliotl . Solution. Gi ve n:

II" .l - 4y and I'" - )' - 4x

a" :

-

a.<

1

VI'

-= - 1

ay

~+,)I'= I - I =O ih

d)' Hence flow is continuous and velocity potent ial exists. 41 = Velocity potential. Let

I I

Ii

~ I

IL 1190

Fluid Mechanics

Let veloc ity componcnls in

t~rrns

a.;. ax ,"d

of velocity potential is given by

= - Il= - (.r - 4v)=-x+ 4y

. .. ( i)

d.jl = - I' =-(-y - 4x)=y + 4x

... (i I)

.

dy

"2

Int egrating eq uation (il, we ge t ¢ '" - -

+ 4xy + C

where C is a co nstant of integration. which is independent o f x. This conSla nt can be a function of y. Differentiating Ih e aoov" equatio n, i.e .• equation (iii) wi lh respect 10

..

-

dy

... (i i i)

y, we ge l

de

=0+4x + -

dy

But from equation (iii), we have

at i)y

== y + 4.•'

Equati llg Ihe two valu es of d¢ ,we get

oy

de

de

4.l+ - =y+4x or dy Integrating the above eq uation. we get

c= L

where C 1 is a CO ll Man !

=y OJ'

,

+ C1 2 of integration. which is independent o f x and y.

Tak ing it equa l to zero . we gel C == yl .

2 Substituting trw va lu e o f C in equation (iii), we ge t -1, 2


=- -

2

)'1

+ 4 .'1+ -

2

. Ans.

Value of Stream functions u:t \II = Stream function The veloci ty com ponell!s in terms of SCream function arc

d. d, d. -

- = I ' = -y- 4x

and

=-,,=-(.(- 4y)=-.\"+4y

... {;,,) ... ( v)

OJ' Integrating eq uati on (1"1')

W.f.!.

\II

x. we get 4.r 1

= - )'.\"- - -

H

2 where k is a cOnStalll of illlegration which is independent of x but can be a function o f y.

I I

Ii

~ I

IL Kinematics o f Flow and Ideal Flow

191 1

Ok Diffacnliating equ ation (ri) w.r.1. y. We gel QIjI", _ x _ o+

oy

il)'

a.

But from cqu3lion (I'). we Itave

-

ay

"oy

Equatillg the twO values of d\jJ ,we gel

-x+ -

"

= - .t+4,}'

'y"

=-x+4y

-

=4y

Integrating the above equation. we get SultS\ilulinlllhc va lu e o f k in equat ion (I'i). we gel

IV =

... 5.9

-)'x -

2.\"2 + 2/ . AilS•

TYPES OF MOTION

A fluid pani cle whi le moving may undergo anyollc or combination of follow ing four ly pes of disp lacements: (i) Linear Translation or Pure Tran slation. (il) Linear Dcformmion. (iii) Angular Dcfornwtion. and (iJ') R01alion. S. 9. 1 Lineilr Translation. It is defined as the movement of a fluid c lement in such a way thai it moves bodily from olle position 10 another posit ion and the twO axes ab and cd rcprcsemcd ill new positions by (I'b' and c'd' lire parallel as shown in Fig. 5.11 (a). 5 .9.2 linear Defo rmati o n . II i~ddlned as the defonnalion of a fluid clemenl in lineardireclion when lhe eleme nl moves. The axes of lhe elemenl in lhe deformed POSilion and un-deformed posilion are parallel. bUI lheir lenglhs change as shown in Fig. 5 II (b).

,

,

.' .'

,

eI ~+~

• -t 0

0

(0) LINEAR TRANSLATtON ~

0

0

(b) LI NEAR DEFORMATION

,"

'" , .

t ,'

,

• , • ,,' " , -~-

,

, '!~ '" " .-"'

o.

,

(e) ANGULAR DEFORMATION

Fig . 5.11.

~ I

_4 d'

0

o

,

,

(eI) PURE ROTATION

DiJplaument of a fluid deme"f.

I~

~ I

IL 1192

Fluid Mcchanics

5.9.3

Angu lar Deformation or She af Deformation.

It is defi ned as the average change in lhe angle conta in ed by two adjacent s id es. LCl ll.9 1 and "'6 2 is the change in angle ~ twccn two adjacent sides of a fluid c lement as shown in Fig. 5.11 (c) . then angular deformation or s hear strain rale

.... 6 1 :

Nuw

Angular d erorma tion '"

dV AI" ':h· -x- · -

ax ~

S h c ll r slru]n r>lt" '" -I

ax

Ih

and l!.6,=

-

a"

6)' all . oy ll.y dy' -

[d6 1 + d B! 1

[a" + -a,,] -

2 i,h

... ( 5. 16)

(ly

5 .9.4 Rotation . II is ddincd as Ihe mOVellll'nl of a flui d element in sucll a way Ihat both of it~ a)les (horizontal as well as ... crtical) rolah! in Ihe same dir"ction as s hown il\ Fig. 5.1 1 (d). II is equal

ox Jill oy for a two-dimensional clemen t in x-y p lane. The rotational compon ents are

to~(~2

W. '

-

~(a"

2i1x

-,,,] Jy

.. ,( 5.17)

W •

J'

~(a"

_awl

2ozox

Vorticity. It is defined as the vai lic twicc o f the rotation and h ~ncc it is given as 200. Problem S.18 Aflilidflow is gi),en by V '" 8fi _ iOAJyj. Find Ihe ~'he(lr slmin rale lind ~'/('Ie whelher Ihe floK' is rola/iuII(I/ or irrol(lliOllll/. 5 .9 .5

Solution. Given

V", S..-li - I Q~yj

II'"

\' '" - lOx·y. ~ h car

'"

24.,.2.

~

'" 0

a,·

,

20.1)" (ly = - lOx"

strain ratc is givcn by equation (5.16) as

=

I I

~:

,av a.t = -

and (i)

8..-1.

.!..(av + au]",..!. (-20.•y +O) = - 10sv. An~. 2axd}'2 .

Ii

~ I

IL Kinematics of Flo w and Ideal Flow (ii) Rotation in

,f -

193 1

Y plane is given by equation (5.17) or 00, ==

-.!. [~- d

)= ~ (- lOX)'. _ 0) == _ tnt)'

U

2ihdyl

As rotation w, ~ O. Hence flow is rolmional. Ans. Problem 5.19 The "e/OCil)' com/l(ments in 1I (wo -dim,," siOll,,/ flow lIre ' , - 2y - x ' I J. ""'),'I3 + 2x - x)'lmd,'=xY SllOw flwt Ille.le compolI<'lIrJ represelll a po.l.fible case of lIIl irrOlalional flow .

Solution. Given

II"

if3 + 2,1' - _~y

0" -=2 -2xy 0,<

a"3y ' , , ,

-

oy = -3- - .C=Y--.I\' == xy2 _ 2)' -

:(31)

0' -=2.l'y - 2

oy

dV

1

3 .\'!

-== y - --=y

ax

(i)

3

" 1 0 ow. continu " ity r'o r a two-dmlcnSlona

Sub~liluling

au

the value of -

<'h

2

1 - ,f.

. .IS -audv cqu3110n + - == O

a\ ay

dl-

and -

oy. we gel

au ai'

- + - =2 -2->')'+ 2-')'-2 ax Jy

==

0

It is a possible case of fluid now.

,

..

(ii) ROla1101l. 00, IS given by w, == -

[al' au)

I -

- -

2a.ray

1 , , 1 ' == - [(y- - .c ) - (y - x-)] .: 0

2

Rotation is zao. which means it is case of irrotational flow . AilS.

.. 5.10

VORTEX FLOW

Vortcx flow is defined as thc flow of a fluid along a curvcd path or the flow of a rotating mass of fluid is knowll a . VOrtell Flow'. Thc VOrtCll flow is of twO types namely: I, ForlCcd vOrtell flow, ;lnd 2. Free vonell flow .

5 . 10. 1 Forced Vortex Flow. Forced vo rtex flow is defined as that type of vortex flow. in which some ellternal torque is requircd to rotate the fluid mass. T he fluid mass in thi s type of flow. rotates at constant angular velocity. (0. Thc tangential vcloc ity of any fluid particle is given by I'= rox r ... (5.18)

I I

Ii

~ I

IL 1194

Fluid ML>chanics

where, '" Radius of fluid particle from tlie axis of rotation. I CENTRAL AXIS

V

I

_::1'__ _ =~=~=-{--::::~ VERT iCAL .... CYLI NDER

LIQUID

:::::=t::-:: _-_-_1--_-_-

::=:::f:::::

:-:-:-£cccc

_-:-:-:t-:-:-

-=-=-}-:-=-=

•

L1

(a) CYLINDER IS STATIONARY (b) CYLINDER IS ROTATING

Fig. S.12

Forced 'OOrrex floUi.

Hence angular I'clocil y ro is g iven by 1-

,

... (5.19)

m" - '" Consta nt.

Examp les of forced \'OnCll arc: L A vert ica l cylinder containing liquid wll ietl is rOlaled abollt its centra l axis with a constant ang ul ar veloc ity ro, as sliow n in Fig. 5. 12. 2. Fl ow of liquid inside the im peller of a cen trifugal pump. ] . Fl ow of waier through the runner of a lUrbinc. S. IO.2 free Vortex flow. When no cx tcmal torq ue is required to rotate the fluid mass. that type of flow is call ed free vortex flow. Thus the liquid in case of free vonex is rotating due to the rotation which is imparted to the fluid prev iously. Examp les of the free \'onex flow are : I. Flow of liquid through a ho le provided at the OOllom of a contai ner. 2 . Fl ow of liquid around a c ircular bend in a pipe. 3. A whirlpool in a river. 4. Fl ow of fluid in a cent rifuga l pump casing. The relation between velocity and radius. in free vort ex is obtai ned by putting the va lu e uf extern,1I torque equal to zero, or. the ti1l1e Tate of change of angu],lr momenlum. i.e .. 1I10melll of m01l1emum must be zero. Consider a fluid panicle uf mass 'm" at a radial distance, from the axis of rotation. havin g a tangential velocity I'. Then " Mass)( Velocity" m )( v Angular momemum Morucm of mOnlemum " Momenlum x r'" /II X I' X , Time rate o f change of angular momentum '"

~ a,

(lin',)

a (ml") '" 0

For free vonex Int egratin g. we get

I I

"'

1111"

'"

Constan t or (" '"

Constan t m

'" Constant

... (5.20)

Ii

~ I

IL Kinematics o f Flow and Ideal Flow

195 1

S. 10.3 Equiltion of Motion for Vortex flow . Consider a fluid clemen! ABeD (shown shaded) in Fig. 'i. 13 rotating at a uniform velocity in a horizontal plane about an ax is perpendicular to Ihe plane of paper and passing through O. Let r= Radius of the element from O.
+ ~~

tJ.r) M

on llie face CD.

, (iii) Centrifugal force, till" acting in the dirc(;tion away

,

Fig . 5.13

from the cemfC. O. Now. the mass of Ihe clement'" M ass density x Volume

=pxMx!:J.r

",

Centrifugal force

'" piMAr - ,

Equali ng the forces in the rad ial direc tion, we get

l

ap) ,,' a, ty L\A - pL\A = pL\A6r - , ap v a, 6r L\A = pL\A6r - , ,

p+-

l

dp

Cancelling 6r)( L\A from both sides. we get -

a,

,,1

= p -

...(5.2 t)

,

Equation (5.21) gives the pressure variation along the r:,dii.1 di rection for a forced or free von cJ\ flow in

J

horiwllml plJnc. The cJ\pression dp is called pressure gradient in the radial direct ion. As ap

ar

ar

is posilive. hence pressure increases with the increase of radius ' r' , The pres~ure variation in the "enical plane is given by Ihe hydrostatic law, i.e. ,

ap

az-= - pg

... (5 .22)

In equation (5.22). z is measured venically in the upward direction. The pressure, p varies with respeci to rand z or pis J fune lion of rand Z and hence total derivative of {! is

ap dr+ -ap dz . or aZ

dp: -

ap a,

Substituting the values of -

I I

frOnl eljuatioll (5.21) and

ap aZ

front cljumion (5.22), we gel

Ii

~ I

IL 1196 Fluid ML>chanics dp '"

P -" dr - pgd:

,

...(5 .23)

Equation (5.23) gives the variation of pressure of a rmating fluid in any plane.

S. IO.4

Equation of Forced Vortex Flow.

For th., forced vorlex flow. from equation (5. 18).

Wi' have

.'=",xr where ill '" Angular ve locity'" Constant. Substituti ng lhe value of,' in equat ion (5.23), we gel (!) ~ r l

dp '"

,

P x- - dr - pg dz.

Consider two points 1 and 2 in Ihe fluid ha vin g forced von e,; flow as shown in Fig. 5.14. Int egrat in g the abo ve equation for poilUs I and 2. we gel

J,~p

'" f pm!

rdr -

I~gdZ

"'

"' p '1 l' ='2(00-'1 - oor n - pg lz2- zd

= £b}-I',ll- pgI Zl-z dJ': V1=rof1 l

1

2

v,"' (!)r, f

Fig. 5.1 4

If Ihe points 1 and 2 lie on the free surfal'C of the liquid. then PI = /11 and hence above equation bc(;OUlCS

0= ~ [1'/ - ",21 _ pg Izl-zil pg

[Z1 -

zd '"

f 11'/ _

[zl-z d= -

I

2,

----...r: _:_:...;z, __ :..r_

",2(

-:-:1:-- -

[ "/ - "I~l.

[f thc point I lies on the axis of rotation, then \., "ro x 'I "ro x 0" O. Thc abovc equati on becomes as

Ld

I I

I',·,

l , - l , = Z. thcn wc ha vc Z= - '-

-

2g

=

ro

,

'

Xr,-

28

Fig. 5.15

•

...(5 .2 4 )

Ii

~ I

IL Kinematics of Flo w and Ideal Flow

197 1

Thus Z var ies wilh th e square o f r. He nce equation (5 .24) is an equatio n of para bo la. This means Ih e free surface of th e liquid is a paraboloid. Problem 5 .20 Pro,'/< Ilwl ;1' case of fo rced I'orlex, l/i e riSt: of liquid {el'e!
.. U) Lcl1lie cylinder is rmated al consta nt angular ve locity w. T he liq uid will risc at Ihe e nds and will fall ,H th e ce ntrc. Let y = Rise of liquid at the e nds from 0·0 x = Fall of liquid <11 the (e mf" fronl 0-0. The n volum e of liqu id '" IVolume o f qtl inder UplO level 8 -81 - IVolume of pJraboloidl '" InR ! j( He ig h1 of liquid up10 level 8-81

,

- [2' H'

j(

He ight o f paraboloid

,

1

""'

, ~

'" nR-, x ( II + X + Y) - - , - x (.( + Y)

AXtS OF ROTATION

Fig_ 5.16

, , nR l " nR- x II + nR - (x + Y) x (x + Y)

2

, nRl " nR - x II +- - (x + y)

... (ii)

2

Equ atin g (I) J nd (ii). we get ,

rtR- (II

1

nR" 2

+ .f )" nR x It +- - (.l + y)

", nR ' nRl - - .f= - - y

or

2 Fall of liquid at ce ntre = Rise of liquid at th e e nds.

2

or

.f" Y

Problem 5 .21 All ope" firc"fa r la"t of 20 cm d;wl1t:ler 'IIId 100 CIII fOllg COllla;" 5 ...aler "1"0 a lreiglll of 60 CIII. Tile /'IIIt is rolated {,bOUI ils I'erlical llXiJ' at 300 r.p.m., fiml Ille deplll of (lflfabola forllled a/ lite f ree Sllrface of "'I'/er. Solution. Given ,,20 em Diame ter o f cy linder

Rad iu s.

I I

2. =!Oem

'. -

2

Ii

~ I

IL 1198

Fluid Mechanics

Hcigllt of liq uid.

II : 6Q em

Speed.

N", )00 r.p.m. 21f.N 2x

An~ular

ve locity,

Ld the depth of parabol a

w= -- =

1'()(

60

300

60

,,3 1.41 rad/scl:.

=z

Usin g eq uation (5.24),

w'R' (31.41)' x (tOi == "'''''"-c''CCC'"-= 50.28 e m . An s. 2g 2x981 Problem 5.22 An open circular cyiindt'f of 15 em diameter and 100 em long conwin.i water JjfllO a lleight of 80 em. Find ti,e maximum sp<'cd 0 1 ....lliell tile cylinder is to be rOIa/ed a/W ill liS I'ertical axis so thm 110 ",mer spills. Solution. Given Diamclcr of cy linder = 15 em :. Radiu s . Leng th o f cy li nd er. Initial he ight of water

15 R =-=7.5cm

2

L= lOQcm

= 80 CIlI. Let the cyli nder is TOMkd at an angular sP<"cd of ill radfscc , wh en Ihe water is i!hou llO spi ll. Then

usi ng. Ri se o f liquid at e nds BIll rise o f liqu id at ends

Fa!! of liqu id at ce ntre He ight of parabo la

'" Fall of liquid at centre '" Le nglh - Inilial he ight : IOO - 80:20cm '" 20 cm ",20 + 20",40em Z= 40em I!J ~ R ~

Usin g Ih e rci:lt ion.

Z= - - , we gel 40 '"

2,

w1 ",

4Ox 2x98 [ 7.5 x 7.5

1!J ~{7 .5 ).!

2 x981

'" 1395.2

w", J1395.2 : 37.35 radls Speed. N is give n by

2>
60

x00 '" -"60~X"3"7".3"5 '" 356.66 r.p.m. An s. 60N= 2n 2 x It

Prob lem 5.23 A cylindrical \·<'.uei 11 cm in diameter and 30 cm deep is filled wirll water apro Ille lOp. The renei is opell at Ihe lOp. Filld Ihe qualilily of liquid left in Ihe I'e.uel. wilen il is rotated aboul ils rerlical axis wilh a speed of (a) J()(X) r.p.m .. and (b) 600 r.p.m. Solution. Given: Diumclcr of cy linder '" [ 2 em Radiu ~. R '" 6 em Initial height of water '" 30 em

I I

Ii

~ I

IL Kinematics of Flow and Ideal Flow Initial vo lume o f water

199 1

'" Area x Initial he ight of wate r

:: ~ X 12l X 30 em) = 3392.9 ern ) 4

(II) Speed,

N = 300 cp.m.

,"" 60

21t x 300 =31.4 1 rad/s 60

W= - - =

(jJ l Rl (3 1.4 1)" X6 1 Hcigtu of parabola is g iven by Z= - - = = 18.10 em. 2g 2x98 1 As vessel is initially ful! of water. water will be spilled if it is rOlll1ed. Volume of wmeT spilled is equal 10 the volume of paraboloid. But volume of paraboloi d = li\rea of cross-section x Height of parabola I + 2

,

It

Z

'" - V - x-

If

424

Volume of w ater left

(b) Speed.

,

= - x [2-

18.10

x- - :: 1023.53 ern 2

l

= Initial vo lume - Volume o f water spilled :: 3392,9 - 1023.53:: 2369.37 em l, Ans. N:: 600 r.p.m.

,"" 60

21t x 600 60 '" 62.82 radls

W=-- =

Height of parabola. As the Itcight o f parabola is more than the Itcight of cy linder th e shape of imaginary parabola wi ll be

as showil ill Fig. 5.11. Lei r =' Radius of the parahQla at the boltom of the vessel. Hei glll of imaginary parahQla == 12.40 - 30 == 42.40 em. Volume of wa lU left in Ihe vessel = Volume o f water in portions ABC and OEF '" Initial vo lume o f wala - Volume of parabo loid AOF + Volume of paraboloid COD. I" \ 4cm Now vol ume of paraboloid 42. l \

ill~ I "'tmSIj~ :~ :E 72.4cm T

AOF ='

~

x Oi.

X

4

It

I I

Height of panlbola

72.4:

'" - x 12" x - - == 4094. 12 em 4 , For Ihe imaginary parabola (COD). w == 62.82 rad/sec

3

I' I

I

I

I

I

I I I

1

1GL,_jt>

I

IMAGINARY CYLINDER

IMAGINARY PARABOLA

I

,

\

I. \

.

\ I '\

' _IH

FIg. S.17

Z == 42.4 ern r == Radius ill Ihe bottom of vessc l

I I

Ii

~ I

IL 1200

Fluid Mechanics

Using Ihe rdation

ro' "

Z = - -.

2,

r ==

get 42.4 '"

We

62.822

X,'

2 x981

2 )(981 x 42.40 '" 2 1.079

62.82 X62.82 r '" J2 1.079

==

4.59 em

Volume ofparabo luid COD

,

==

~ x Area at llic lOp of the imaginary parabola x Heig lll of parabola

==

"21 x It? x 42.4 "'21 x 1t x 459 2 x 42.4 '"

1403.89 eml

Volume of wate r len '" 3392.9 _ 4094.12 + 1403.89", 702.67 em J • Ans.

Problem 5 .24

An open circular (flllll/er of 15 em diameter and 100 em 1001g ( 0111<1;11$ waler 14plO (1/ "'hiel, rile cylinda iJ /0 b,' row/ed abou i its l'erlicaiaxi.l . .\0 IlulI

II ',eigh l of 70 em. f 'jm!l/Je .~peed the {lXial deplll becomes zero.

Solution. Given

Diameter of cy linder Radius.

u,ngth of cylinder Initial height o f wa ler

==

15

I T"'"

em

15 R = -=7.5cm == ==

2 100 ~m 70 em.

om

Tl

When axial depth is zero. the depth of paraboloid = 100 e m.

Usin g the

......... 15cm ~

rcl~tion.

Fig. 5.18

w =.~I00 ~X~2~X~9.~8}1 " 442.92 "59.05 radls 75x75

.". Speed. N is given by

0'

7.5

',H

00 = - -

60

,v __

6Q x

21t

w _

,60~X;,'"9c·O",

21t

" 563.88 r.p.m . An s.

Problem 5.25 For tile problelll (5.24), find tile difference in IOtal preHlfreforce (i) at tile boltOIll of cylinder. {lnd (ii) at tile sides of the cylinder due to row tion.

I I

Ii

~ I

IL Kinematics of Flo w and Ideal Flow

201 1

Solution. (i) T he data is given in Problem 5.24. The difference in \olal pressure force al the bonom of cy lind er is obtained by finding 100ai hydrostatic force at tbe bottom before rotation and aflcr row lion. Before rotation.

force'" pgAil /f

1

l

it

22 -

1000 kg/m-. A '" Area of bonum = - D : - x (0. 15) m . II '" 70 em '" 0.70 III

where

3

rorcc= 1000x9.81

x~

4

X (0.15)1 xO.7 N '" 121.35 N

After rotation. the depth of water at the bottom is nut cunstant and hell,",c pn:ssurc force duc 10 the height of w~ tcr. will 1I0t be l'OIISlanL Consider a circular ring of radiu s r and width tiT as shown in Fig. 5.19. Lee Ihe height of water from tlie boUOIll of the lank uplo free ~urface of water at a radius

fil','

r=Z= - - . 2, Hydrostatic force on ring a1 Ihe bonum.
1

'" WOO x 9.81

TOlal

pr~ssun: forc~

al

tll~

bollom

o/r'

R

J io

'" dF'"

r·

071

'"

o

d.

9810x2xltrx--d,

2,

00 '

19620XItX -

2,

/

J

r dr

Fig . 5.19

I"'rom PrOblem 5.24.

ro '" 59.05 radls R", 7.5 em '" .075 nl. Substituling Ihese values. we ~cI lOla I pressur~ force 19620 x rc. x (59.05) '

2x9.8 1 '"

19620 x It x (59.05)1

2 x9.8 1

x

(.07 5)'

4

",86.62N

Dill"ercnce in pre~ure forces al the bottom 121.35 - 86.62", 34.73 N. An s. (;i) Forces Orl Ihe sides of the c ylind er

Before rOlalion where

~ I

'" pgA/' A '" Surface area of Ihe sides of Ihe cy linder UplO heighl of water '" itO X Heigl\! o f waler '" It x .15 x 0.70 1112 '" 0.33 m 1

I~

~ I

IL 1202 Fluid ML>chanics

h '" e.G. '"

~ 2

of the wcucd area of the sides

x height of wate r = 0.70 '" 0.35 m

2

!'oree on Ihe sides before rotation '" I{X)() x 9.S1 x 0.33 x 0.35 '" 1133 N After rotation, the water is UplO llie lOp o f Ihe cylinde r and lienee force on the sides '" 1000 x 9.81 x Wcucd area of the sides x

= 9810 x rrD x 1.0 x

~

x 1.0", 9810 x

11:

~

x Height of water

x .15 x

~ '" 2311.43 N

Difference in pressure On Ih e sides 2311.43 - 1133 ", 11 7H.4J N. Ans. S.IO.S Clos ed Cy lind rical V ess els. If a cy lindrical vesse l is closed allhe lOp, which con tains some liquid, lhe shape of paraboloid formed duc to rotation of the I'essel will be as shown in Fig. 5.20 for different s~d of rotations. Fig. 5.20 (a) shows Ihe initial siage of the cylinder. when it is nO! rotated. Fig. 5.20 (b) sliows Ihe shape of Ihe paraboloid fonned when the speed of rotat ion is WI' If the speed is increased funher say w1• the shape of paraboloid formed will be as s hown in Fig. 5.20 (c). In this case the radius of the parabola m the tOp of the vessel is unknown. Also the height of the parJboloid fonned corresponding to angular S[)t....,d 001 is unknown . Thus to solve the twO unknown. we should hav~ twO equations. One equation is

z=

lI) ;r1

2, The second cquation is obtaincd from the fact that ror closed I'essel. I'olume of air berore rotation is equal to thc volume of air after rotation. Volume of air before rotation'" Volume of closed vcsscl- Volume of liquid in vessel /tr l x Z Volume of air after rotation'" Volu mc o f paraboloid formed '" "'-~-"

2

----y'l'--= -- -

-- -

=:=:=t=::-:-]-:---i"

~ : --' -_-_t::-_c

-:-1-:-:---::E --- -: i'l

i<1

fig. 5.20

Pro blem 5.26 II "en'd, c)'/illlhiw/ ill shape (11111 cloud at Ihe lop (llId bollom, eOlllaills .wler IIplo a "eighl of 80 em. Th e diameter of rhe ressei is 20 em (lIId lellglh of rnse/ is 120 cm. The "esse/ is rOlmed (II (I speed of -I()() r.p.m. ahOll1 ils "erli~'(li luis. Filld Ihe heighl of pllwh%id formed.

I I

Ii

~ I

IL Kinematics of Flo w and Ideal Flow Solution. Given: Initial height of water Diameter of vesse l :. Radius,

,,80

em

= 20 em R = 10 em = 120 em N = 400 T.p.m.

Length of V('sscl

Speed.

w~

2.N

-60

~

120

60

'" 41.88 radls

When tlie vcs.sc l is rotated. let Z '" Hei ght of paraboloid formed

11

==,=

r '" Radius of paraboloid at the top of Ihe vesse l This is lhe ca'ie of closed vesseL Volume o f air !x·forc rol31ion '" Volume of air MIN rotation tt

2

Jt

2

,

T , 1

TT

2 11 x4()()

203 1

Fig. S.21

Z

- 0 xL - - D x80=/(,· x -

"'

442

where Z = He ight of paraboloid. r = Radius of parabola.

"' It

11

,

,

,Z

- xV- x ( 120 - 00)= - 0-)(40= It,· X -

"'

4

4

2

~X 202X40= 4000x1t=1f.rx ~

"'

,

,·xZ= Using relation

4000xllx2

•

= 8000

... ( i)

.4C ""'';-'CX ,-,-,'_ = -.4C"C"8~'Ox",_'

w' , ! ,wcgc\Z= Z~ -2g

28

2 x98 1

=0.894 ,2

? ~ --'--

0.894 Substituting this value of? in (I). we get

"

_ Z _ xZ= 8000 0.89 4

Z2 = 8000 x 0.894 = 7 152 Z"" J7152 "" H4.56 em . Ans. lind Method Lc( Z, = Hcighl of paraboloid, if (hc vcssel wo uld nOi havc bcen closed at (hc top. corrcspond ing 10 speed, N"" 400 r.p.lll. or ro "" 41.88 radfs Then

I I

ro

2

R 2

" I,

"1

I,

•

r - -,--,

.4~L8~8~';'~L~0-_'

Z, "" - - "" 2g 2x98 1

TT 11 120

I,

:--:I- :

:-:-r-:-:

T " 1

Fig. 5.22

"" 89.34 Clll.

Ii

~ I

IL 1204

Flui d Mechanics

Half of Z, will be below the initial hei ght of water in the vessel

i.e..

AO

= 2, '" 89.34 '" 44.67 em

2

2

But heig ht of pa raboloid for dosed vessel '" CO '" CA + AO= ( 120 - 80) +44.67 em '" 40 + 44.6 7 '" 84.67 em . An s. Problem 5.27 For Ihe dll/a gin'lI in Problem 5.26. find Iile speed of rota/ion of Ihe ,'essel. when tHill1 depth of Waler is zero. Solution. Given Diameter of vessel '" 20 em :. Radius. R '" 10 em Initial height uf wa ter '" 80 (;111 u:ngth o f vessel 120 = 120cnI LeI (oj is Ihe angular speed. when axial depth is zero. When axial dcplh. is zero. the heig ht of paraboloid is 120 em and radius of the parabola at the tOP of the vessel is r.

,,

:. Using tlie relation.

IT '11"

, ,

Z:~ or l20=ooxr 2g 2 x980 (fi ,J '" 2 x 980 x 120" 235200

''''

1

Fig. 5.23 ... (i)

Vulume of air before rOl
=

II(1)(

Z

2

"'

r

1 1 = 11 )(10 )(40)(2 ~

""-"'::'C;;;"'-'"

=

r

11)(120 Substituting the value of? in equat ion (i). we get {Ii )( 66.67 " 235200 to

Speed N is given hy

(oj

35200

66.67

= 2rr.N 60 60 )( 1,1

N 0 ~~211

"'

8000 0

-

120

= 66.67

== 59.4 radls

60)( 59.4

0

"'~"'":: == 567.22 r . p,m . Ans .

211

Problem 5 .28

The q/i",hk(.Il'(:n'c/ of IllI! problem 5.26 is r Olflled {II 700 ,.p.m. ahO"1 ils }'erlieal luis. Find Ihe area ,mCOl'ued {II Iile bOl/om of Ille lank.

Solullon. Given Initial height of water Diameter of vessel Rad ius. u:ngth of vessel

I I

" KO cm " 20 cm R" 10 em = 120cnI

Ii

~ I

IL Kinematics of Flow and Ideal Flow

205 1

N", 700 T.p.lll.

Speed,

2rr.N

w ~ - - '" 60

2x

It

X 700

60

'" 73.30 rad/s.

If the tank is nOI closed allhc lOp and also is ve ry long. then th e height of to w = 7].3 will be 73.3 l

2Xg

X

IO ~

,'--- 1----\, ------. ""

I

':

Froll) Fig. 5.24.

correspo nding

I .

"

== 274.12 ern

2 x 980

par~bol;\

'.

"" I- r, --.t: '

A K

Of

I

r/

l

'"

x l +x 2 =174.12 - 120= 154.1 2c m . .. {i) From the parabola. KOM. we ha ve 73.3~ x 2 )(980

I

x,

... (ii)

['or the parabola. LON. We have 73.3~ x rl!

... (iii)

2 )( 980

Fig. 5.24

Now. vo lume o f air before rotation = Volume o f air after rotation Volume of air before rotation = rrRl x ( 120 - 80) = It X 102 X 40 = 12566.3 em J Volume of air afta rotation = Volume of parabolo id KOM - I'olu me o f parabol oi d LON = ltr,

Equating (il') and

(1') .

l

x

(120+ x, ) l

-1I:r,

.

2

Xi 1

x-

..• (v)

we gel

Substituling th e va lue of 'l2 from (ii) in (I'i) . we get 12566.3 = /t x ( 120 + XI ) x, 2 x 980 x (120 + 2 73,3-

1 .:

xd - -'"O,,'- ,,+;.c:"L'

.

2

1

From (U)"I '"

2X980X ( 12o + x (733)"

'

)1

12566.3 = 0.573 ( 120 + .l l)" _ Substituting th e va lue of XI from (iii) ill the above eq uati o n

_

21

2

2

_ 0.573 ( 120 + 2,74 '1) - 4,3 x '1 X'l '" 0.573 [1 20 ' + 2,74 ' ,,4 ,: 1 _ 4.3 . + 2 x 120 x 2.74 ,,2 . .

I I

Ii

~ I

IL 1206

Fluid ML>chanics

'" 0.573 114400 + 7.506 ,:4 + 657.6 r/ I - 4.3,:4

~5~66~.3o '" 2 1930 = 14400 + 7.506'24 + 657.6 -~J20.573

r/ (7.506 -

'l

~

-

4.3 ' 14

r/ + 14400 - 2 1930= 0 r/ + 657.6 r/ - 7530", 0

4.3) + 657.6

3.206

"'

-657.6 ± J657.6 1 - 4 X(- 7530) x (3. 206) 2 x 3.206

=_ -~6.~ S7~.6~±~J~4~3~2~ 43~7~.7~6~+~9~6~ 564 =.7,, 2 6.4 12

= - 657.6 ± 727.32 '" _ 21 5.98 or 10. 87 6.412

Neg at ive va lu e is not possi ble

r/ '" Area uncovere d at [h e base '"

10.87 ern

Itr/ '"

If

2 x 10.87'" 34.149 <.: rn 2• An s.

Problem 5.29 A closed cylindrical I'esse! of diameter 30 0/1 and II
(l/

the edge.

Solution. G iven: Diameter of vesse l :. Radiu s .

T

:: 30 em R :: 15 em

Initial heig ht of wale r. H = 80cm Le ngth of cy linder, L :: 100 elll Pressure of a ir above water :: 5.886 N/cm!

"

1

P '" 5.886 x

"' H ead du e 10

prc~s u rc.

II

= plpg ::

S(lecd .

"'"

5.886 x 10" IOOOx9.8J

l - 30cm- l

=6 m ofw ater

Fig. 5.25

N = 250 T. p. m.

21tN

w ~ - - :: 60

2/t x 250

60

:: 26. 18 radJs

Le t xI:: He ig h! of parabol o id fo rm ed. if the vesse l is assum ed ope n at the to p and il is vc ry long. Then we ha ve

'

w-R': : -.26.".1",8-,'"x ", " :...' XI:: 2g

2x 981

= 71\.60cl11

... (i)

leI 'I is the radiu s of llie actua l para bo la of hc ig hl .T2

I I

Ii

~ I

IL Kinematics of Flo w and Ideal Flow Then

207 1 ... (ii)

The volume of air before rotat ion

= rcR" (100 - 80) '"

1[

x 152 X 20 = 14 1]7 em}

Volume o f air after rotation'" Volum e of paraboloid EOF I

= '2

, x It"

X Xl

But volu me of air before and after rolmion is sa me. I , 14137 = zxltr , XX l Btu from (ii),

1 xl '" 0.35 '1 I , , 14137 = 2" x 1If, x 0.35 " ,,4 = 2 x 14137 = 257 14 Il

x 0.35

" '" (25714)"4 '" 12.66 ern Substituting Ihe value of " in (ii), we gel Xl = 0.35 X 12.66 Pressun' head at th e bottom o f Ih e H'SS e!

1

",

56. 1 em

(a) At Ihc centre. The pressure head at the centre. i .e.. aI II = Press ure head duc [0 air + 011

[ ": OH=LH - LOI

=6.0+ (HL - LO)

HL = IOOcm =l rn LO = Xl = 56. 1em = .561 m

= 6.0 + (1.0 - 0561)

(b) Mille edge. i.e .. at

l

= 6.439 m or wu le r . An s. a = Pressu re head due to air + height of water above G

= 6.0 + AG = 6.0 + (GM + MA) = 6.0 + HO + 0,786 = 6.0 + 0.439 + 0.786

== 6.0 + (HO + x,)

l' (

.f ,

= 78.0 cm = 0.786 m}

': HO : Uf -LO : lOO -56.ll = 43.9 cm = 0,4 39111

== 7.225 In or waleT, AilS. A closed cylillder of wdills R alld lIeiglll H is compte/ely filled lI"illl lI"alU. II is rolated abolll its rerlical a.lis lI"ilh a speed ofro radialls/s. Determille tile t01il1 pressllre exerted by "'iller on Ihe rap (/111/ bottom of IIIe cyti"der.

Problem 5.30

Solution, Given Radiu s of cyl inder Heiglll of cylindcr Angular sp.:cd

I I

Ii

~ I

IL 1208 Fluid ML>chanics As llie cylinder is c losed and completely Illl ed with water. Ihe ri sc o f wate r level a1 the ends and depression of wate r at the ce ntre due \0 rotation of llie vessel. will be preve nted. Thus th e water will e llen force o n the complete top of Ihe vessel. Also the pressure will be cxcncd a1 th e oonom of Ihe cylinde r. Total )' ressure exerted on Ihe lop or cylinder. T he top o f cyl inde r is in con tact w ith. wata and is in hori zoma l plane. Thc pressure var iation a1 an y radius in horizontal plane is g iven by eq uation (5.2 1)

1

" , l up PI'-' POO", , - = - - = - - " pro r [o.' 1'=CllXrl

dr

r

J

r

Integrating. we gel 0'

Consider an elementary t: ircular rin g of rlldiu s r and width dr 0111he lOp of Ihe cy linder as shown in Fig. 5.26. Are a of circul ar ring = 2rt rdr Force 011 Ihe demclllary ri ng = Intensity o f pressure x Area of ring "P X 2nrdr

-

"" e . II ' ·

-

Fig. 5.26

" x 2nrdr. '" '2p orr' Total force on the top o f th e e ylinda is obtainL..:l by ink grat in g the above equat ion be tween the lim its and R.

°

.

i"

i"'

Total force or FT ", -(f) r dr P -r" x2 n rdr= -(f)P ' x2 n o 2 2 0

... {S.2S) Total pressure foree on the bollom of cy linder. F /I = Weight of water in cylinder + total force on the top of cylinder . .. (5.26)

p" Density of wate r. Problem 5.31 A closed cylillder of diameter 200 1/11/1 alld hdglll 150 1/11/1 is comple/ely filled WillI Wilier. C{I/culme rhe 101111 pre.!Sl jre force exerted by Wilier is rormed aboul irs I'utical axis (1/ 200 r.p.m. Solut ion. G iven Dia. of cy linder " 200 mm = 0.20 m Radius. R"O.llIl He igh t of cylinder. H" ISO mm = O. IS III Speed. N" 2oo r.p.m. Angular spL'"Cd.

I I

21tN

w = -- = 60

OIl

rhe rap {llId bO/fom of tile cylillder. if it

21t x 200 " 20.94 rad/s 60

Ii

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IL Kinematics o f Flow and Ideal Flow

209 1

TOlal pressure force on Ih e top of Ihe cy linder is g iven by equ ation (5.25)

p , 4 100 0 , 4 J'[ x R '" - - )(20.94")( It)( (0.1) '" 34.44 N. An s. 4 4 Now total pressure force on the bottom of the cylinder is given by equation (5.26) as F T ", -)( ro-)(

F B ", pg)( nR!)( H + Fr

= 1000 x 9.81 )( It x (0.1) 1 x 0.15 + 34.44 '" 46.22 + 34.44 '" 80.66 N. Ail s. 5 . 10.6

Equation of Free Vortex Flow.

For the free vo rtex, from eq uation (5.20), we have \' x r'" Constanl '" say c

,

"'

\'

"'-,

Substituting the value of \' in equation (5.23), we gCI dp =

t'

P-

2

r

ilr -

C

1

pg liZ = p)( - , r

X

1

ilr -

r

pg liZ = P X

C - )

r

ilr -

pg dZ

Consider two poims I and 2 in the fluid having radiu s 'I and'2 from the centml axis respectively as shown in rig. 5.27. The heights of tlie poill1s from bonom o f the vessel is 'I and Z2' Integrating the above equation for the points 1 and 2. we get , , pel 1

Z 1,r-dp"' J I- - , ) d'- Jpgd I

[ -"']',

, ,

"'PC" - -2

pc,

-pg [z l - zd " ' -

- 2

"' -fH - vn - pg[Zl - ZI]

'" f [VIl - I"ij- pg [Zl - ZI] Dividing by

pg. we gel

I

Pl - P,

L

pg

"'

/ +ZI .E.!... + _I pg 2g

I,

...(5.27)

Equation (5.27) is Bernoulli' s equation. Helice in case of free vortex flow. Berlloul[i"s equation is applicable. Fi g. 5.27

I I

Ii

~ I

IL 1210 Fluid ML>chanics Problem 5.32 In a free cylindrical !'Or/ex jlow. at a point ill 111<' fluid at a radii/.\' of 200 ml/) and at (/ height of 100 mm. rhe \'e1ocil)' and pre.l.mres are JO mh alld 117.72 kNlm' absolute. Find Ille pressure at a radi/IS of 400 mm alld III a IIdg/1I of 200 min. Ti,e fluid is air IIa)'ing dell.!!t)' eqllal to

1.24 k,g/m J • Solution. Al Poin! I : Give n : RJdiu s. 'I'" 200 mill '" 0.20 m Heigh t. ZI'" 100 mm '" 0. 10 III Velocity. \', '" 10 m/s Pressu re. P, '" 117.72 kN/rn ! '" 11 7.72 x 10 1 N/m" At Puint 2 : ' 2 = 400 10m '" 0.4 111 Zl=

200 10m '" 0.2 III

1'1 = press ure at point 2 p '" 1.24 I; g/ml For the frce vortex from equation (5.20), we have \' X

r = constant or

1'1', '" I'l'l

X ' I lOxO.2 v1 = - - = '" 5 IIl/s \' 1

Now usin g equation (5.27) , we ge t

117.72 X [o j 1.24 x9.8 1

"'

+

10 2 2x9.8 1

0.4

"

+ 0. 1= !!1.. +

pg

f!.l.= pg

"

2 x9.81

+0.2

1 117.72x IOl + 10 + 0.152 - 0.2 1.24x9,S I 2x9.SI 2x9.8 1

"" 9677.4 + 5.096 + 0. 1 - 1.274 - 0.2 '" 9676.22 Pl "" 9676.22 x pg = 9676.22 x 1.24 x 9.8 1

= 117705 Nfm 2 = 117.705 x 10) Nfl1l 2 = 117,705 kNfml (abs.) = 117.705 kNfm l. " " s. (8) IDEAL FLOW (POTENTIAL flOW)

to .s . 11

INTRODUCTION

Ideal fluid is a fluid which is incom pressible and inv iscid. Incompressible fluid is 3 fluid for whi ch density (p) re main s conSlanl. ln viscid fluid is a fluid for which viscosity (1-1) is zero. Hence a fluid for whic h densilY is constanT and viscos it y is zero. is kn own as an ideal fluid. The shear s tress is give n by. t = 1-1 dll . Hel1l'e for ideal flu id th e shear Slress will be zero as 1-1 "" 0 dy for ideal fluid. Also the shear force (w hi ch is equal to s hear stress multiplied by area) will be ze ro in

I I

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IL Kinematics of Flo w and Ideal Flow

211 1

case of ideal or polclUial flow. The ideal fluids wi ll be mo ving with uniform ve locit y. Alillic fluid panicles wi ll be movi ng w ith llie s ame veloc ity. The l'Onccpt of ide al fluid si mplifies the typical mat hem ati ca l anal ysis. £lluids such as Waler and air hav e low visl:os it y. Alw when the speed of air is apprec iabl y lower than that of sou nd in it. the rom prcssibili ty is so low Ihat air is assumed to be incompress ibl e. Helice und er certain co ndit ions. c~rtain rea l fluid S such as wa ler and aiT may bc t,-.,atcd like idea l f lu ids .

.. S. 12

IMPORTANT CASES Of POTENTIAL FLOW

The following arc th e import ant cases of pot ential flow: (i) Uni form flow, (ii) Source flow, (il·) Free-vo rtex flow.

(iii) S in k flow. ( v) Superimposed flow .

.. 5. 1)

UNIFORM flOW

In a uniform fl ow. th e velocit y remains eonstall1. All the fluid panicles are movi ng with th e same veloci ty. The uniform flow may be : (i) Par
Fig. 5.27 (a)

• • • • •

U" Ve locit y which is uniform o r constall1 along x-axis II and V" Co mpone nts of uniform ve loc ity U along x and y-ax is. For the uni form flow. parallel w x- axis, thc ve locit y components II and v ~ rc givcn ~s I I " U and ,. " 0 ... (5.28) But th~ ve loci ty II in term~ of st rea m functio n is give n by. Let

1/"

o~

Oy

and intcTms o f ve locit y potcntial thc vc locit y II is givcn by.

,= -a, a.,
Similarly, il ca n be shown

lh~t

v=_

... (5.29)

o~ "" o¢

,h

oy

.. .(5,29A)

But II = U fro m equatio n (5.28). Substituting u = U in equation (5.29). we hav e

I I

Ii

~ I

IL 1212

Fluid Mechanics

u '"

dljl ""
u= -a.

...(5 .30)

#

ity

a nd also U= -

a.I'

Pirs1 pa rI g ives tI\¥ '" U tly whereas seco nd pan gives dip '" U d.\" . Jm cgrati on of th ese pans gives as \11 == Uy + C 1 and 9= Ux + C2 where C ! and C 1 arc constant of integrat io n. Now leI us plm the stream lin es and p01 cn lial lin es fo r un ifonn flo w pilrallc l lo x- ax is. Pl otti ng or Stream li JlL'!j. For stre am Ji nes, the eq uati on is Ij/ =Uxy +C t le I IjI '" O. w h ~ rc Y '" O. Substitutin g lh es.e values in [h,:, above equatio n, we gel O=U x O+Ctor Ct",O Hence the equati on of st ream lin es occo mcs as Ij/ :U. )' ... (5.1 I) Stream tines The stre am lines arc straight lines paral le l \0 x-axis and at a di Sla nce y from th e .I'-axis as show n in Fig . 5 .28. In equat io n (5. 3 1), U. Y re prese nts the vo lum e flow ra te (i.e., m3/ s) be~ U x4 ~4U twee n x-a xis and th at stream line al a d islance y. 3 N ull'. The thickness of the fluid stream perpcndkulm to the y" 3 '1';1. mU x 3 ~ 3U plnne is assumed to be unity. Thcn y x I or}" represenls thc area of y~2 ~ U x 2~2U flow . And U. )' represents tne product of velocity and area. Hence p ' ... a U X \EU U. )" rcpresenls the volume flow rate.

,

,., Ii

, , , ,

•

"

y~lJ

.,

•

,

Plotti ng lJf potcntiH I lin es. For pote nti a l lines, th ~ ~quat i on is Fig. 5.28 1\1 == U . x + C2 ... (5 .32) Lei ~ == O. where.f '" O. Substituting these va lu es in the above equ alion. we gel C2 == O. Hence equat io n of po te llli a l lines becomes as ~ = U . .t The above equat ion shows that polc ntialli nes arc slrai ghtlin es parali c I to y· axis and at a di slance o f .f from y- axis as s how n in r ig . 5.29. Fi g. 5.30 shows th c plol of strea m lines and polenliallines for uniform flow para li c I to x-ax is. The Slrca m li nes a nd pote llli a l lin es interseci each other 31 ri g hT angles.

,

,

"• •• ~

"'

•

••

• • "• ~ • • • •

0

,

~

-

V

,

PolenUat lines

,"

5' • • • •" •"

o

3

Po1ential -?1 ~"'"

;; ,

Fig. 5.21

I I

o

•••

•

tream lines

•

e-- 1/, ,. II

,

V

"

,." ,-,

,

Fig. 5.30

Ii

~ I

IL Kinematics o f Flow and Idea l Flo w

213 1

S. I 3.2 Uniform Potential flow Puallel to y· Axis. Fig. 5.31 shows the uniform potclUial flow parallel to y- axis in which U is the uniform velocity along y- ax is.

Fig . 5.3 1 Th" velocity components u, y a long ,(-axis and )~axis are given by Il=Oand,' =U ... (5.33) These velocity components in Icnns of strcmn functi on (1jI) and ve locit y polcmial function (9) arc given as

ow

II" -

.(5 .34 )

,, = _ dl¥=Jq. Jx dy

>0' But from equation (5.33),

ilQ

0 -

ay ax

"==

... (5.3 5)

U. Subst ituting

U=_ dlj/ = dQ dY

ax

I ' '"

or

U in eq uati on (5.35), we gel

U=_dW 3ndaL<;oU=

ax

at iJy

First p~rl gives iI'lI = - U dx whe reas scl:o nd pan giv~s dq. = U d),. Inte gration of Ihese pans gives as 1( = - U . x + C 1 and 9 = U. y + C 2 •.. (5.36) where C 1 and C 2 are constant of integration. Lei us now plot tlte stream lines and potential lin es. Pl utt lng or S lreum lin es. For s tream lines. tlt e ~q uatjon is \jI '" U .. I' + C 1 Let \jI '" 0. wlt ere x '" 0 , Tlten C 1 '" 0. Hence tlt~ equatio n of stre am lin es becomes as \jI '" - U.X ... (5 .37) Tlte allo\'~ equation sltows that stream lines are slTaigltt lines paralle l to y~axis and at a distance of x from tlte )~ ax i s as shown in Fig. 5.32. Tlte -ve sig n sltows that tlte stream lines are in tlte downward direction.

,

Stream lines o

•

>

o

., . , , ,. , ,. , ,,, , >

;'

;' 5

potenUat

, ,

N

"• "

0

. /,

I "• •

•

"•

I

•

•

" "-

-

Fig. 5.32

I I

~""S

' -0

•

Ii

~ I

IL 1214

Fluid Mechanics

Pl otting of Potential lin es. For potentia! lines. the equation is
to 5 . 14 SOURCE FLOW The source flow is the now coming from a point (source) and mov ing uut r:ldial1y in all dir~tions of a plane at uniform Tmc. Fig. 5.33 shows a sour(.'C flow in whi<:h the point a is the source from which the fluid moves radially outward. Tlw strength of a source is defined as the volume flow rate per unit depth. The unit of strength of soun;" is m 2ls. It is rcprcsemcd by q. Let u , = radial veloci ty of flow at a radius r from the source 0 If = volume flow rat¢ per unit d¢pth r = radius The radial velocity II, at any radius (is gil'¢n by.

"

.. . (5.39)

", = 21((

o

Fig.S.33

SoIlrr:t{lIJW(Flvwa'way

fro m wll rct")

The above equation shows that with th¢ increase of r. the radial velocity dec re3S<:s. And at a large distance away from the source. the veloc it y will be approximately equal to zero. The flow is in radial direction. hence the tangenTial velocity,,& = O. li:t us now find the equation of stream function and velocity potenTi,tl function for the source flow . As in this case. /Ie '" O. the equation of stream funct ion and velocity potential function will be obtained from 11,. Equation of Stream Function

By definition. the radial velocity and tangential velocity compone nt s in tenns of stream function arc given by /I

"m

I aW =- andll e = -ihv -

ar

' rae

(See equation (j. llA)] (Sec equation (5.39) ]

u, = 2!r

_ "" = --.!L r dO

21(r

d'V= r. -

'/

21((

.d6= -

q

2 11

de

InTegrating the above equatiun w.r.1. 6. we get Ill '"

..!L x e +

2, IjI = O. when 6 = O. then C t = O.

Ct. where C t is constant of integration.

Let Hence the equation of stream function becomes as

I I

Ii

~ I

IL Kinematics of Flow and Ideal Flow

1jI",.iL. e

215 1

.. . (5.40)

2,

In the above equation. q is constant. The above cqumion shows Ihal stream fUllction is a function of a. For 11 given v~luc of a. the stream function IjI will be l'()l1stant. And this will be a radial line. The stream lines \:an be ploued by hav ing different values of Here is taken in rndians.

e.

't'-t

e

Plotting of 5tream lines

"••,

When 9 =O, IjI = O

, 9=45° = _ 4

9=90Q=~ ra d "I:ms. 2 31'[. 135" = -

"

.~;'/>_, Stream lines

units

4

IjI '" -

q

21t

radlans,ljI =

-

.

!!..",!1 2

4

q3Jt3q

21t

.- = 4

o· ,

unil~

8

are radial

S 9_0

.

units

The stream lines will be radial lines as shown in ri g. 5.34.

'I' - ~

Equation of Potential Function

Fig . 5.3~

By de fi nition. the radial and tangential components in terms of velocity funct ion arc given by

source flow .

a. andua = -I ·-a.

It,= But from

~quation (5.39). ", =

Equati ng

th~

[See equation (."i.9A)[

, iJO

"2!r

Strea m lin e for

two values of "" we get

dQ = ...!L or
2nr

2nr


Integrating the above equation. we get

Jd9 = J...!L.dr 2" 2 f -;-dr q

1jl=

=

n

q

2,

Potential tir>8S are cirele q

[ : 2n

is a eonstant term]

,. 2

log,

r

,. ,

... (5.4 1)

In the above equation. q is constant. The above equation shows. thatth~ vclocily potential function is a function of r. For a given value of r. the velocity function 9 will be C()nstant. H~nce it will he a circle with origin at the source. The velocity potential lines will bc circles wilh origin at the sourc" as shown in Fig. 5.35. Let us now find an expression for the pressure in terms of radius.

I I

~I _"'"''''>o\:.t o. / -v

s

Fig. S.l S

'0

Pottwtiallinn for souru.

Ii

~ I

IL 1216

Fluid Mechanics

Pressure distribution in a plant source flow The pressure distri bution in n pl~nc source flow can be obtained with the tion. Let us assume that the plane of the now is horiwnt,d. In that case the for two poin ts of now. LeI p" prc~sure at a (lOint I which is at a radius r from Ihe source al II,: velocity al point I Po" pressure al point 2, whirl! is al a large dist ance away from Ihe be zero at point 2. [Refer \0 equation (5.39)] Apply ing Bernoulli's equation. we get

(p- Po) pg

, p .u,

help uf Bernoulli's equadmum head will be S<11llC point 1

source. The velocity will

=_ u; 2g

({) - I'o)= - - -

0'

2

But from cqu ation(5.39),

II

q

,

=-2 ltr

Substituting the value of II, in the .(bove equation. we get

=-

..

8~(1

(5.4 2)

In the above cqumion. p and II arc constants. The abovc cqualiOll shows lhat the pressure is im'ersdy pro)Xlrt iooal to the square of Ihe radius fro m the source .

... S.15

SINK FLOW

The s i o~ flow is the flow in which fluid mov es rad ially inwards tow ards a )Xlint where it disappears at a consta nt ratc. This flow is JUSt opposite tu the source flow. Fig. 5.36 shows a si nk flow in which the fluid moves radially inwilrds towards point 0, where it disappea rs at a constant rate. The pattern of strealll lines and equipotential lines of a sill k flow is the sallie as thai of a source flow. All the equations derived for a '-Ource flow shall hold to good for sin~ flow also except that in sin~ flow equations. q is 10 be replaced by (- q). Problem S.33

Plor Ihe srream lilies for

II

o

uniform flo ... of:

(i) 5 IIIls pllralld to Ihe positi!"e direcIioli of the x-a.>.is alld Fig . 5.36 Sillk flow (ii) 10 IIIls pUTIJlIeito Ille pO~·ili>·e direCI;OIl of the }"-(lxis. (Flow toward centre) Solution . (i) The stream func tion for a uniform flow parallel to the positive direction of the

.. - a~is is givell by equation (5.31) as If== Ux)' The above equation shows that stream lines are straight li nes p
I I

x-a~is

al a distance )'

Ii

~ I

IL Kinematics of Flow and Id eal Flow IjI stream function IjI For), '" 0.2. stream function IjI For y'" 0.4. S!Tcam function IjI

["or )' '" 0,

217 1

'" 5)' '" 0 '" 5 x 0.2 = I unit

'" :; x 0.4 '" 2 unil The other values of S1rcam function can be obtain ed by substituting the different values of y . The Slrcam lillcs arc horizontal as shown in Fig. 5.36 «(I).

,

,.",

'+' '" 4

Y E 0.8

,.

...., \II '" 3

Y E 0.6 y

,.,

z

0.4

,

y: 0.2

,

,.,

p'

,

Fig. 5.36(11)

(ii) The stream funcTion for a uniform flow parallclto the positive direction of the y- axis is given by equatiun (5.37) as 1jI = -UX .l

The above equation shows thal stream lines arc straight lines parallcl 10 the y·axis .11 a distance from the y-axis. Here U", 10 mts and hence the ahove equation hecornes as

.l

1jI= - lOxx

The For For For For The stream

negative sign shows Ihm the stream lines arc in .l" O. the stream fU"':tlon '¥" 0 .I = 0.1. the stream function '¥" - I 0 x 0.1 x = 0.2. the stream (un ction '¥ = - lOx 0.2 .' = 0.3. the stream function '¥ = - lOx 0.3 oth~r valu~s o f stream function can be obt ain ed l i n ~s arc vertical as shown in Fig. 5.36 (h).

,

•

•

..•

•

the downward di roxtion.

= - 1.0 unit

= - 2.0 unit = - 3.0 unit by substitutin g the different va lues of .1". The

7

..

•;0

M

o

• o

O.t

o

..

.. •

..

.. •

• 0.2

0.3

Fig. 5.36 (b)

~ I

I~

~ I

IL 1218 Fluid Mechanics Problem 5.34

D elerllline (lie \'clocily a/Jlow

(1/

radii 0/0.11/1,0.4 1/1 {lin/ 0.8 m. Id,ell Ille water

II source (1/ a slrength of 12 1/1"/.1. Solution. Given Strength of so urce. q '" 12 111 '/s The rad ia l veloci ty II, at any radius r is give n by equa tio n (5 .39) as

is flowing radially QIII\\'ard in a /wrizonlal pia/Ie from

" , '" .!!...2 1(, Wh en r =

O.l m.

=

12 = 'J,SS m/s. An s. 21txO.2

II, '"

12 '" 4.77 m/s. An s. lit x 0.4

II,

Wh en r = 0.4 m.

12 = 2.31\ mls. Ail S. 2n x O.8 Problem 5 .35 T wo di,KS (lrc placed in II hQriZ(lnld/ plane, onc OI'a lire OIlier. Tile lI'aler elllUS M tile centre of the lower disc oml flows radially oil/ward from a source of sirength 0.628 1/1"/.1. The Wh en r= 0.8 !l1 .

II,

=

pressure, at a radius 50 mm. is ZOO kN/m' . Find: 0) pre;'sure in tN/III' lIl" ",dillS of 500 mm mil/ (ii) .{{rcwo JuneliOIl (1/ anglCl" of 30" tlllt! 60" if \I' = 0 at 0 = 0". Solution . Gi ve n So urce Mre ng th . q = 0.628 m 21s Pre~surc a t radius 50 mm . P, '" 200 kN/m2 '" 200 x 10) N/m2 ( i) Prcs!iIIre III (I radil/J 500 mIn p, = pressure at rad ius 500 111m

Let

(ur), '" \'eloci ty at radiu s 50 nlill

(u r), '" velocity at radiu s 500 mm The rad ial ve loci ty at any radius r is give n hy equat ion (5.39) as II, '"

....!1....-

2"

When

r = 50 mill '" 0.05 111.

(Ur), =

When

r = 500 111111 = 0.5 Ill.

(urh '"

0.628

= 1.998 IIl/s '" 2 IIl /s

21t X 0.05 0.628 21t xO.5

'" 0.2 IIl/s

Apply ing Berno ull i's equati on at radiu s 0.05 III and at radiu s 0.5 III.

P,+(u,)~ P' +(u,): - - = _. -_ . pg 2g pg 2g P,

-

I I

(u,)~ _

p,

(u,):

+ ---- + -p 2 P 2

Ii

~ I

IL Kinematics of Flow and Id eal Flow

=2()()'7,;'~'~0,-' + _2 1

'"

2

]()oo

21 9 1

_1'_'_ + _0,_2_' 1000

2

p,

200 + 2= - '- +0.Q2 1000

1 ~ '" 202 - 0.02 = 201.98 p, = 201.98 x 1000 N/ml (ii) StrewlI fUllctiol1,~ at

= 201.98 kN/m l . Ans.

0", 30" Gm! 0", 60"

For the SOUT(;C flow. the equation of stream functi on is given by equation (5.40) as \11 '"

.1....0. where e is in 2,

radians

0.628 30)( .' --,-lit 180

e '" 300 '" ]0)( It

it

radians)

ISO

'" 0.052J m l/s. AilS.

0.628

6011

21l

180

- - , - - '" 0. 1046

.. 5. 16

1 111

Is. Am .

fREE · VORTEX flOW

Frcc·vo ... c~ flow is a <:ircul:llory flow of a fluid suc h Ihm its stream lines arc conccmric <:ircles. For a free-vortex flow, u& x r '" constalll (say C) Also. circulation around a Slrealll Ii"c of an irrolalion vonc~ is

r",2rrrxue=lnxC

(":rx"e=C)

where ue '" tangential velocity at any radius r from the centre.

r

jje=-27fr

The circulation r is takcn positivc if the frec vortex is anticlockwisc. ror a frcc -vortex flow. thc vclocity cumponents are

r

lie = - 2"

and

1/,

=0

Equation of Stream Function By definitiun. the stream function is given by

-a.

"e=-a,

and

I ", " , =de --

(See equ ati on (5. I2AlJ

In casc of freC-I'ortel flow_the radi al velocity (u,) is zero. Hence equation of s tream fum:tion will be obtained from t:lngential velocity. I/e' The value of lie is given by

r

I/e = - -

2"

Equating the two values of "e- we get

I I

Ii

~ I

IL 1220 Fluid Mechanics

d. r - -- = d, 2 lt r

or

d'¥ = - -r- d, 2"

Integrating the above equation. we get

-

"'

r .IS a consta nt (e nn )

2,

The above equat ion shows 1li al stream function is a fUliction o f radius. For a given value of r. the stream function is constant. Hence the stream lines arc concentric circles as shown in Fig. 5.37. Equ a llon of p otential fun ction. By definition. the potent ial function is given by. I

u~=

Here

II,

=

d,

--

d e

0 and

and

... (5.·B )

, -"'" Stream lin
[Sec equation (S.9A)[

r

ue = - - , Hence. the cquntion of polcnlial 2"

Fig. 5.37

function will be obwincd from ue' Equaling Ihe two values of us- we gel

~ d$ = r

d~

ae

r 2 tt r Integrating the above equation. we get

fd¢= f;ll d9

or

= r.

r 2"

r

.<19= 2,

de

!II = ~ J d9 = ~. e 2n

21l

... (5.4.f)

The aho ve equatio n shows thai velocity potenti al function is a function of 9. for a g iven value or9. potential function is a constant. Hence equipotenti al lines are radial as shown in Fig. 5.38. Veloci!y pote!1t~t tines

~

• ••

•

Y-

o '

,., • Fig. 5.38

~ I

POlenlialli"~1 ar~

radial.

I~

~ I

IL Kinematics of Flow and Id eal Flow

.. 5. 11

221

I

SUPER-IMPOSED FLOW

The flow pancms due to uniform flow. a source tlow. a sink. flow and a free vortex flow can be

supe r-imposed in any lin ear combination to ge\ a resultant flow whic h close ly resembles th e fl ow around bodies. The rcsultan! flow will st ill be potential and ideal. The following arc the importan t supcr-im(X>scd flow: (i) Source and sin k pai r (ii) Doublet (specia l case of source and sin k combination) (iii) A pl;lnc source in a uniform flow (flow past a half body) (i,') A soun:c and sink pair in a unifonn Flow (I') A doublet in a uniform flow.

5.1 7. 1 Source and Sink Pair. Fig. 5.39 shows a source and a s ink of s tre ngt h q and (- q) placed at A and lJ respectively a1 equa l distance from tile point 0 on the x-axis. TtIUS tile source and s ink are plal'Cd symmetrically o n the x-axis. The sou rce of s tre ngth q is placed at A and sin k of strength (- q) is placed al 8. The combination of the source and the si nk would result in a !lownet whe re stream lin es will be circ ular arcs staM ing from point A and end ing at poi nt IJ as s ho wn in Pig. 5.40.

A

o

/1-1-'--- . ------II-- ' ------I Source (q)

Fig. 5.39

Sou ~u alld sillk pair.

Stream

line"

,

~-~ Sink

Fig. 5.40

StrMm lillo for lvuree-lillk pair.

Equation of stream function and po tential fun ction Ld P be any point in the resultant !lownci of source and sink as shown in Fig. 5.41.

~ I

I~

~ I

IL 1222

Fluid Mechanics

e

= Cylindri cal co-ordinates of point P wilh rcs(lCct 10 origi n 0 x.Y'" Corresponding co·ordinates of point P ',_ 8 , " Position of point P with respect to sourcr placed at A '2_ 6 2 '" Position of point P Wilh respect to sink placed a1 n a", Angle subtcndcd at P by the join of source and sink i.e .. angle APE. LeI us find th e equation for the resultant stream function and velocity potential function. The

leI

r,

equation for stream function due \0 source is given by equation (5.40) as III , '" If· 8 , wllcrcas due 10

2,

sink it is given by IV, '"

-

(- qe 2 l. The equation for resultant stream function 2,

(\II) will be the sum of

these twO stream function.

. U

~

[

...

- q.U

... (5 ...1 5)

The eq ua tion for potential function due 10 source is given hy equation (5.41) as 4>,

'" .!L log " , and 2,

due to sin~ il is g iven as 4', = - q log,r,. The equation for re~ultanl potential function (9) will be the - l /t sum of these two potential function. 4'=4'1+4',

(-q)

=.!L log, r l + log, r 2 l/t l /t

I I

Ii

~ I

IL Kinematics of Flow and Ideal Flow

'" ..!L 21(

110&,' 1 - log,'21""

.!L log ,

(.'L)

223 1

...(5 .46)

21 t ,l.

To pron' that n'Sult lml strt'!lnl lines will be circular arc pass ing through source and s ink The resultant stre am fUllcti o n is give n by equ ati on (5.45) as

-q.a .~ -­

"

For a given stream lin e l¥ '" co nstant. In the above eq uati on tile te rm .!L is al so constant. T liis

"

mea ns th ai (9 2 - e,l Of ang le 0: will also be constant fo r vario us positio ns o f P in the pl ane. To s,uisfy ttJi s. th e locus of P mu st be a (;ircJc with AB as chord . hav ing its ce ntre on y- ax is. as shown in Fi g. 5 .40. Consider th e cquali on (5.45) ag ain as

- q 21\:

-q (EI1 - EI,) 2/f

. ~ - "~ -

v

= ...2...(6 1 - O

(e, -a!)=

"

2 lTIjI

q Tak ing tange nt 10 both s ides. we ge t

1= lan (2:1tf )

HIn (9 1 - 9 2

B",

tan 9 1 '"

-

and

'-

x +a

Suhstilut ing Ih e values o f Ian 9 1 and Ian

- )-' - - - '(x+ a)

(.I· - a)

= Ian

1+ -' - .--'(.1+ (1) (X- II )

9~

tan 9,

or

-

tan

0;

,".( i)

1+l an9 , .1an8.

ta n 9 , = - Y• x- a

... (S A M )

in equ al ion (i).

(2'.)

,

q

y (x-o))' (.1'+0) = Ian ( " . ) , l x'

a + .vl

, ,: I

"

0'

,"0 ' I-- , x·, - a"" +

I I

y- = -

2a)" COl

q (".)

---t=.:..:'-=1,__ F ig. 5.41 (d )

Ii

~ I

IL 1224

Fluid Mechanics

, "

.\--a + y" + 2(1},l:ol

.l!+ /

(2"' q 1')

=0

COl (2:W) -(/ = 0 ,[2+ / + 2ay COl(2:'V ) + al COIl (2:"') _al +

2a)'

CO,2

(2:1JI ) _al= 0

[ Adding and subt ract ing I' ! cot! (2:1¥)]

... (5.4 7)

The above is the eq uat ion of a cin:lc· with centre on y·axis M a distance of± a COl (2:"' ) from Ih e origi n. The radius of the cin:lc will be a cusc<: (2:'41 ). Sim ilarly, il Can be show n th al Ihe poten tial lin es for Ihe sou rc e-si nk pair will bc eccentric nonimcrse(;,i ng cirdcs with th eir centres 011 the x- axis as shuw n in Fig. 5.41 (b). PoIent;alline. Sink

Fig. SAl (b)

Polentia/linn for SQuru sink pair (potentia/lina are «untric non-inlerlt'Ct ing circlN with Ibrir t:t'ntrn on x-axis).

°The equation.C + y' ~

I I

(II

is the equation of a c;rcle wilh centre a1 origin and of radius ',,-,

Ii

~ I

IL Kinematics of Flow and Id eal Flow

2251

Problem 5.36 A source atilt II sillk of slrcng lll 4 m!ls and 8 m!h are IOC{/led at (- I, 0) lind ( I. 0) respeC/ire/y. D<'lermillt' tile I'e/aeily and stream fllllNion (I{ II paiM P ( I. I j which is lying 0/1 Ihe jlulI'llei of Ille resu l/mil stream lille.

Solution. Gil'en Source Sl re ng lh.

'I,

= 4 In l{s

'I, = 8 m 2{s Distance of Ihe source and sink from origin. (I = I unit. The position o f Ihe source. s ink and point P in the flow field is s hown in Fig. 5 .42. From Fig. S.42. ;1 is dear that angle 6 2 will be 900 and angle 8 1 can be ca kulatcd fro m right angled tri:lIlglc ABP. Sink strength.

The eq ua tion for stream func tion duc to sou rce is g ive n by equ ati o n (5.40) as . k· w h.:reas ...' lIe to Sin ' II ·IS g .i ven by 1jI ,

-=

-'I'- x e,• . The rcsultam

,-,

Mrc~ m

fUllction

IjI

1jI,

= ", x 0 , .

is givell

2,

~s

, p (• . y) (1 , 1)

,

Fig. 5.42

+(- l/l X6 l )= l/, X6, 21t

21t

... ( i)

Le i us find llie values o f 6 , and 6 2 in radian s. Fro m th e geomelry. it is cl ear tlim llie traingle ABP is a rig hl angl ed Irian gle wilh angle 0, '" 90° '" 90 x • 180 AI<;o

It"

~ 2

radian s.

liP 1 Ian Ell :: - : : - " 0.5

2

AB

,

8 1 :: tan

,

0.5" 26.56° " 26.56 x 180 radians" 0..463

Subslituting Ih ese values in eq uati on (i).

1jI "~ x0.463 - {Il x~ 21t 21t 2

I I

Ii

~ I

IL 1226

Fluid Mechanics

8 , xO.463 - - x l it In 2

' -

,

= 0.294 - 2.0 = - 1.706 m !/s. Ans. To lind the veloc it y m th e poin t P. let us tirst fi nd the stream function in terms of o rd inates. TIle stream function in terms of 9 1 and 9 2 is giv,, " by equ ati on (i) above as

.l

and y

(.X).

1f= rJ,x9 1 _(h X9 l

211

21'(

The values of 9 1 and 9 2 in terms of x. y and a arc give n by equation (5 .4611) as tan 9 1 =

- '-

9 1 ""

13n-

"' Substituting th ese

va lu c~

1

- Y-

(.[ -a )

and

H'

- ,-

9 2 ", tan

(x - a )

of 9 1 and 9 2 in eq uat ion (iI, we gel

, _ Y_ _ q1 [an 1_ '_'_

1jI =.!l.!.. t:l n

2,

The velocity com ponen t

Ian 9 2 = - y-

and

x +a

,, -a.0, /I

=

\" + a ~nd

lit

x- a

I' = _d\v .

"

d",

0, =

1an ~ or [!!J... 21l

q

'"

2~ x

, _ ,__ (h tan x + a 21(

1_'_] .t -

{l

q.

I

x +a

Ifl

I

[

(Y J' x -( _,- +-,-) - -2-', x 1+ (Y J' x -(,- - -,-) 1+ - -x -a

(x + ,, )2

I

if'

(x -a )!

, -,-, '(.-, '+-0")"+ ~,,, x -(_'-+-0- ) - -,-; x 7(_,~-oC),C+~y2'X -(,- _- ,-,)

(x+a) (h (.t -a ) = 2; (.\ +(1)1 + / - 21l (x_all + vl q,

At th e poilll P( 1. I). the componelll "is obtained by substituting .f = I and )' = I in th e above equa ti on. Th e value of" is also ,;,qual to o ne.

q,

u= -

I+I

2 lt(1 + 1) ;: +1 ~

I I

(12

(I - I )

2ft(l _ 1)"+ l l

Ii

~ I

IL Kinematics o f Flo w and Ideal Flow

Now

I' '" _

2:1.7 1

iilv

ax

'" - -

, [q, -

;:h 21t

=_ [!!J..

- - - - " lall

21t

x +a

1

y

--

-I'-a

, x y(- I), Xl _iL X

I

l it 1+

q, -,

_I)'

""

(_Y_)"

(.I + af

2 1t

x+a

, x Y(- I), XI ] I + (~ )- (.\ - a)" I

x

a

(12

Y 2 n (.f _a)l + / At th e point P( I , I).

'I, I til ~_',-_~ I' " -2n x (I + III + I! - -21t x (I _ 1)1 + ]~

(': a= I)

'I' I = -q, x -J - -" x-

2n

5

l it

1

= _, _, x .!.. - _q_l = _4_ x .!. _ _8_ = 0.1272 _ 1.272 = _ L 145 mlsl 21t 5 2n 21t 5 21t T he n:sulta n! vc IOl.'it y, V =

Ju 1 + v" _ ~rOc2c5c44:c;,-+-(C_-IC,C4C,::T)1 = I 174 m/ s. AilS.

For Ihe abo,'" problem, de/ami" " Ihe preswre al P( I, I) if Ihe pressure al infinity is zero Wid dellsily o/fluid is 1000 kg/m J.

Problem 5.37

Solut ion. Given: Pressure at in fini ty, Po = 0 Dcn~ily of fluid , p = 1000 kgfm 1 The ve loc ity· o f fl uid aI infi nity w ill be zero, If Vo = ve loci ty at infi ni t y. thcn Vo = 0, The result ant veloc it y of fluid at P( I , I) = 1.174 mfs (c alc ul akd above) or V= 1. 174 m/s. LCl p = pressure al P( I . I ) A pp ly in g Be rno ull i's tlleorem at point m infi ni ty and m poin t P, we get

Po Vo' P V" -+= -+pg 2g pg 2g

0' 0'

V'

0 + 0= L+ _ pg

2g

or

~= _ V' =_ 1.174 ' P

2

-------'--

p

V '-

pg

2g

0= - + -

or

p

V' 2

0= - + p

(':

V", 1.1 74 mls)

2

* From equation (5,39), the velocity at a distance 'r' from sourcc or sink is g;"cn by u," ~ , At infinity, 2lfr

r is very very !:\fge hene" velocity is Zero,

I I

Ii

~ I

IL 1228

Fluid Mechanics 1.1 74'

1.174 l x 1000

= _ 689. 14 N/ml. A il S. 2 S. 17 .2 Doublet. It is a spec ial case of a source anti sin k pair (bo th of them arc o f equ al strength) when the 1WO approach each other in suc h a way that the "islancc 2a between them approac hes ze ro and the product 2a . II remains constant. This product 2(1 . q is kno wn as doubl e! strength and is

P= - - ,-

deno ted by

x p =-

1-1.

Doublet strength. Jl '" 2a . " ... {5 .4 8) Let q and (- q) may be th e slrenglh of the source and th e si nk respectively as show n ill Fig. 5.43. leI 2(1 be the distance between 1hel11 aud P be any poin t in Ihe coulbincd field of so urce and sin k.

Fig . 5A)

Let e is the angle made by P al A whereas (9 + Now the stream funct ion at P.

Ifa

Be) is the angle at B.

q If (9 + 00) ': - - 39 ... ( 5 .4 9) 21t 21t 21t From B. draw Be J. on AP. Le I AC '" or. CP = rand AP = r + or. Also angle HPC: 09. Tile angl~ is very sma ll. Tile distance BC can Ix: taken equal to r x 00. In triangle ABC. angle BCA = 90 0 and h~nce distanc~ BC is al so equal to 2<1 . sin G. Eq u ~ting the two values of Be. we get • = -

- -

oa

fxOO=2<1.s inG

SG= 2a.si nG

,

Substituting the va lue of

oa in equation (5.49). we get q l it

21t sin

J.l

si n

2,

,

W': - - x

e

r

e

= -- x --

[.,'

2n. q = J.l

from equa tio n (5.48)1 ... (5.50)

In Fi g. 5A3, when 2(1 -t O. th e a ngle 09 sub tendcd by point P with A and B becomes very small . Also or --+ 0 and AP hecornes ~qua l to r. Then s inG: PD =2'. AP , Apl '" A02 + P02 or r2", .( 2 + Also Substituting the va lue of sin G in equation (550). we get

I I

)'2

Ii

~ I

IL Kinematics of Flow and Ideal Flow \jI :

~

)'

I

I-iy

--X-X- = - - ,= 21t r r 21lr-

I-'Y "

2rr (_t- + y'

229 1

) ... (S.SOA)

.~+l=-....!2....

2,.

The above equati on can be written as

.r2+i+2X YX-"+[-")' -(_ , _)2=0 41t1jl 41t'l' 41f\jl or

.t

[Addi ng and subtra<:ting

1+ (Y + - "-)' = ( - "- )' 41t1t/

(4~\I r] ... (5,51)

4ltlj1

The above is Ihe equmion of a circle with centre (0. - " - ) ,md radius - "- . The centre o f Ihe

41t1jJ

circle lkson

41t'll

y -a~is at a distance of - " - from x-ax is. As Ihe radius of the circle is also equal 10 - "- . 41l1j1

41l0/

hence the circk will be tangent to the x-axis. Hence str.:am lines of Ihe doublet will be the fam ily of circles tangen l to (he x- axis as shown in Fig. 5.44. Stream

~"""

are circles

tangent 10 . -aX;S wittl

centre on y-axiS

Stream lines y lines

Fig. 5.44 Potent i ~1

fu" c:t ion at P

Refer to Fig. 5.43.

Th ~

potential function at P is give n by $=

I I

SITI'am /inff for II doub/fl .

;~ IOg , (r+or)+(-;~) log ,r

[Refer to equation (5. 41 )]

Ii

~ I

IL 1230

Fluid Mechanics

q Iog '= q '0, : - q Jog (r +or) - l it

'

l it

'

:'L[" +("]' X~+ ...l 211

r

",..!L

or

2,

l it

'

(<+,,]

q Iog (1+ "] -- : -' r

l it

r

2

r

[ "

.

("]'

As ---; IS a sllIall quamu y. Hence ---;

,

'

BUI in fi g. 5.43. from triangle ABC, we gel

or",

Or '" cos e

2"

e

2" cos Substituting lhe value of 15r, we gel

. 1

becomes neglig ible

OP '" ...!L )( 2" cos 9 2,

,

Jl cos e : - X- 11l r

1-: 2{/

x q '" Jl from equation (ill ... (5.52)

oe

In Fig. 5.43. when 2a ~ O. Ihe angle becomes very sma ll. Also or ....... 0 and AP becomes equal to r. Then AD

.r

AI'

r

cos9= - - : -

l

Also Ap l '" AD2 + pD l or ? '" _~ + Substituting [he valu e o f cos 9 in equatio n (5.52), we gel

41= L x (~]x!=J:..x -~ 21[ r r 2)[ , -

.~ +l", Lx~

lit 0;. The above equation c;m be wrinen ~

~- ilt i+( 4~r -(4~~r +l= o or

(.I" - 4~4>

r

+i =

(4~9

r

The above is Ihe equation of a cin:]" wilh cenlre circle lies on x-axis at a diM3nce of

l

4,.

(-"-.0] 4mp

[Addingand subtracting(4~9r]

.

(5.53)

and radius ( -'-'- ]. T he centre of the

41'(4)

frotH y-a xis. As the radius of the circic is equal to the

distanw of the cent re of the circle from the y-axis, henc e the cirdc will be tangent to the y-axis.

*

I I

Expansion of log, ( I + xl '" x +

2""" " +,

Ii

~ I

IL Kinematics of Flow and Ideal Flow

231

I

Hence tile polclUiallincs of a doub let wi ll be a family of ci rcles tange nt to the y- axis with their centres on the x-axis as sliown in Fig. 5.4'5. Potential lio&S

POlentia l lines

Of

, lines are ardes w ith centre on x_axis

£.::::~Q'-'"" :; 1 13og,,01 10 y-axis

Problem 5.38

Fig . 5. 45 Potenlial linn fo r a doublet. A paim P(O.5, I) is SilUaled ill II,e flow field of

II doubler of streng/II 5 mlls. CII/cu/ale IIII' velocity al Iilis point and "/50 Ihe "olue of Ihe slreom fu nction . Solution. Given: Point 1'(0.5. 1). This means of = 0.5 ,md y == 1.0 Strength of doublet. J.I " 5 111 2/s (i) VelOcify at point P The velocity at the given point Can be oblain"d if we know the stream funclion (1jI). But stream function is givell by equation (5.50A) as

1jI=- ~ X

21l The velocity componc llts

1/

y (Xl + /)

and I' arc obtained from the st ream funct ion as

d,, [-2;X(X2+l) " ,]

u=ay-=(1),

"-:, ;y[(>'~r')]

~ is a consw nt ternl) 2,

"-;, [(:,':)~>l [ ... ~'[!' ('<'+r') }y[-Jj[.<'+"'l ' [2y] +(>' +,.' ) _ 2),2

(.ol +l) ,

I I

.I

1

+ (X-'

+)' ') -

Ii

~ I

IL 1232

Fluid Mechanics

= irr ;, [i" ~ y' )1=irr [ix;:~;)' 1 Su bstitutin g the va lu es o f )l = 5 m'ls, x" 0 .5 and y" 1.0, we ge t tile veloc ity co mpone nts as

u= _L[ (XlXl +-if i'l = _ ~[ O.5l -ll'1= _ 2- 0.7~ + IT 2 J't

and

2 1t (0.5'

= - 0.382

2It 1.25"

\'" ;n [(x::"/ )!]'" n[(O~5~ :'~l)l[ ] '" ~[~] ,,- 0.509 5 2

Res ultant ve loc ity,

V=

Ju

1

- ~(- 0.382 i

+ v"

+ ( - O.509 )" = O.63fi m/s. ADS.

( ii) Value of strewn lime/ion m poilll P ,

I

l it

1.25

-- x == _ 0 .636 m1/s. A DS.

Solution in polar co-ordinates

e

The above qu estion ca n also be done in r. (i.e.. polar ) co-ord inalCs. The s tream func tion in r. 9 co-ord inates is give n by equatio n (550) as )l si n 9 1jI =

-- x --

2rr

--.(i)

,

and vel oc it y contlx mc nt s in radial and tangC llli a l di rec ti o ns arc given as

" : ~ x ,)IjI= ~ ~ [_ LSi lle] ,

ae rae

r

'" ..!. x r

,,'

21t

r

(_L] x ..!.~(sin 9) J9 lit

r

[ .: L2rr

is a constant te rm and also r is conSlanl w. r. t. e]

=- -x...., " I cos e 21t r·

... (Ii)

"" '[" ';0 ']

ue : - - : - - - - - ar ar 21t r = _ ( _ _11

21t

sin e]~ [~]= ~ sine (-I ) . -

ar '

21t

, !

[.: ~

sin e

21t

,.

= - - ' -,-

I I

~~i;eiSaCOnSlanl w.r. I. '] ... (iii)

Ii

~ I

IL Kinematics of Flow and Ideal Flow

233 1

1.",5 ~.f " + Y _ ~0.5 " +1 _","" y ' 894 and cos , x O.5 sin , =-:~=O. =-:~= 0 .44 7

Now

r=

r

Substituti ng the values of

r,

r

,,1.25

"1.25

a and cos e in above cqumions (i), (ii) and (iii), we gel J.I sin e 5 0.894 z ' l I = - - - - = - - x r.= =- O.636 m Is. Ans. sin

21t

r

J.I 21t

I

21t

If,'" - - x ----,- x cos ,"

,, 1.25

e '" - - 5

I x - - x 0.447 = - 0.2845 IlI/s 21t (1.25)

J.I sin 9 5 0.894 lie'" - - x -,- =- - x - - = - 0.569 rnls

lUlU

21t

r

211

1.25

Resu ltant velocity.

= J(-Q.2845)! + (-0.569)2 = 0.636 S. 11.3

m'.~.

AilS.

A Plane Source in a Uniform flow (flow Past a Half-Body), Fig. 5.46 (d) shows

a unifonn flow of velocity U and Fig. 5.46 (b) shows a source fluw of strength q. When this unifonn flow is flowing over the sourc<; flow , a rcsuhant flow will b<: obtained as shown in Fig. 5.46. This resultant flow is also known as the fluw past a Ilalf-body. Let the source is placed on The ori~in o. Consider II point P(.\". y) lying in The resultant now field with pol~r co-ordin
,

,

+

• (al Uniform !low

'01 (b) Soufce now Point P(x . y) (f. &)

Hall body

,

"

fG ~

\

( \&

0

,'

",

~

," , 0

, . 0

1--'. _ Fig.5.4-6

I I

FlO'W pall"n reuliling from Iht combinalir)n of a uniform flow and a $Our("t.

Ii

~ I

IL 1234

Fluid Mechanics

Th e stream fUllction (\V) and pot ential fllllCiion (IJI) for the rc suhant flo w arc obtained as give n belo w : It' " Stream func ti on du c \0 uniform flow + stream func tio n du c to source

= u . y + .1... e 2,

... ( 5 .54 )

=U. r sin9+!Le

(":

2,

and

9" Vdocity po te ntial

y =

r

sin Ell ... (554A)

function due to unifonn flow + Ve locity potcr)[i al function due to source

= U . .1" + - '- Iog , r= U . rcos 9 +.!L log , '

21(

... (5.5-1.8 )

211

The follow in g arc Ihe importa nt po ints for tile resultant flow panc rn : (i) SraglUuiQlI poi/I I. On th e lefl side of the source. at the point S lying o n the x- axis , th e ve loc ity o f

unifo rn! flow and Iha! due to source are equal and opposite 10 each othe r.

H~ncc

the ne t ve locit y o f the

combined fl ow fi eld is ze ro . Thi s point i s know n as sta gnation point and is dcnOlcd by S. The polar co-

ord inat es o f the stagnatio n point S arc 's and It, where 's is radial diSl31lCC of point S from O. The nd veloc it y (or resultant veloc ity) is zero at the stag nation point S. ", =

=

(v. .!La) +.!L] v. ...!i.....-

..!. altl =.!. ~ r aa

..!.

[v.r

r

r

ae

sin a +

21t

= cos a + 21t 2 J(r At the s tag nati on point. f) = II radian s ( 180G ) and f = fsand net ve locit y is zero. This mean s u, = 0 and 1'& = O. Substituting these values ill the abo ve equati on. we get

r

cos a

0= V. cos 180° + - ' 2J(f S

=- v+ - ' -

"'

2J(rs

q

's=2,U

...(5 .55)

From th e abo ve equation it is d e ar that positio n of st agnati on point depends upon th e f ree stream ve loci ty V and source s trength 1/. Al th e Slngnatio n point. the value o f strea m fundi on is obtained fr011l equat ion (5.54A) a~

If= V .fsina+ .!L. 8

2,

For th e stagnatio n point. the above equation becomes as

ItI s = V .T, si n ] 80 0 ( .: . 0,1 2 = 12

+.!L2,

x 8

At sta gnation point. 8 '"

It

radian s = 180 0 and r '" fs l

... ( <<6) ~ ..'

The abo ve re lat io ll g ives the equatio n o f stre am line passin g throu gh sta gnation point. We kn ow that no fluid mass crosses a stream line . He nce a stre am line is a I'irlua l solid surface.

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IL Kinematics of Flow and Ideal Flow

235 1

(ii) Shape of resul/atll flow. 1\( tlie stagnation point S. the net ve loci ty is zero. The fluid panic les lhal isslie from tlie source ca nn ot proceed further to the left of stagnation poin!. They arc carried along the contour BSB' Ihm separates the source flow from uniform flo w. The curve BSB' ca n be regarded as ttte so lid bound a r y or a rou nd nosed body suc h. as a bridge pie r arou nd wll ic h th e

uniform flow i ~ forced to pass. The contour RSB' is ca ll ed the half body. bcc auS<) il ha s o nl y th e leading point. it trails [() infinity al down "tream e nd. The value of stream fUll ction of the strc~m line passing through stilgll,Hion poin t Sand p
!l. 2

Thus the composite fluw cunsists of : (1) flow over a plalle half-body (i.e., flow over curve BSB')

out~idc

IjI

= '1 alld 2

(2) source flow wi thill the plane half-body.

The p lane half·body is described by the dividillg stream lille,

IjI

=

%.

But the stream function at allY poi nt ill the combined flow field is !!ivell by equatioll (5.54) as IjI=U.y+..!L{}

If we take

IjI

=

"

%in th e above equation. we will!!et the equalioll of the divid illg stream line.

EquatiOIl o f the dividing stream line (i.e .. equation of curve 8SB') will be

'1 = U 2

. Y+

.!L . {} or 2lt

U . Y = 'i 2

_.!L 2lt

6

=

" "--(I 2U -Ill

"'

... (5.57)

It

From tlie above eq uation, tlie main dimensions of tlie plane lialf·body may be obt ain ed. From tliis eq uation. it is clear that y is maximu m. when 6", O. Hence At 6 '" O.

Y is maximum and y ...., '" 2(~ y=

At6=lt,

At

6

3lt

=2'

y

2(~ ( 1 -

= -.!L 2U

T' ~) '" 4~

(I- ~l '" 0 It

)' = 2(~ (I- ~: ) =- 4~

... tlie maximum ordinate

... tlie ordi nate above tlie orig in

... tlie !cadiug po int of the half-bod y ... the ordinate be low the origill.

The main dimensions arc shown ill Fig. 5.47. (iii) Re.m ltll/!/ !"dodt)' III Ilny poi/lt Tlie vclocity components at any point in the flow field are givcll by iI,=

I I

~ dljl = ~ ~[u.rsine +.!Lel r de

r dfJ

2lt

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IL 1236

Fluid Mechani cs ==

.!.[u.rcos a +.2....] '" V.cos e +..!!...r

2lt

Leading peOn!

2M

-"-

Y_ -- 21)

-"-

'"

Origin

Fig. SA7

The above equati on gives the radial veloc it y at any point in the now field. Tllis radi al ve locity is due

,,,

tn uniform flow and due \0 sou rce. Due \0 source the radial veloc ity is~. lience the velocity due 10 source dimillishcs wilh im:rease in radial distance from the source. At lar~c distance from the source the contribution of soure", is negligible ,md hence free stream uniform flow is nOl influenced by Ihe prescncc of source.

'" - IV. sin Rcsul1aol velocity,

V

e + 01 '" -

U si n

.2.... e is cO lman! w. f . \.

e

2.

r]

= ~u; + uJ

(il') Locatioll of swgnafion poilll

At the stagnati on point. the velocity compone nt s are lero. Hence equating the radial and tan gent illl velocily componen ts to zero, we gel

or

11,=0

,.U

0'

rl:os e = - - q-

,.,

ucuse + ~=o BU1

0'

fl'OSe=X

x= - -q2nU Wh~n

"6= 0 or e = 0 or

-Usine=O

Hence Stagna tion point is at

I I

or nBu1 y=rsine

(- ~. 0). the

,.U

e :: 0 :. y'" 0

si n

as U canno t be zero

leading point of the hal f-body.

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IL Kinematics of Flow and Ideal Flow

237 1

(I') Pressure ar (IllY point ill flow field Let Po" prc~so re at infinity where velocity is U p == pressore at any point P in the fiow field, where velocity is V Now applying the Bernoulli>s equation at a point at infinity and at a point P in the now field. we gel 1

Po

U

P8

2g

- +-

V~ 28

P

: - +pg

"'

_v _' _ _V_1 =.l!....._!!..2...", P-Po pg 2, 2, pg pg

The press ure l"()-cm"icnt is defined as c=P-Po

,

1

"2

PU

,

V' V'] [ : ----'T"--:c'''-'' P8

-

Zg

2g

!pu 1 2

... (:'i.58)

Problem 5.39 A III1i/orm flow .... irh a I'e/ocity of J mh is flowing ora a p/(lIIe source of slrell8111 30 m!k Til<' uniform jlow lind $Ollrce flow are in IIIC sallie plwle. A palm P is siwaled i/I the flow field. The dimmee of Ihe /loim P from Ihe source i.! 0.5 III ond il is III a/l angle of 30 0 10 Ille ulliform flow. De/ermine: (i) Slreamj,m CliOIl a/ point P, (ii) resuil'IIIll'docil), of jIow
2,

at poi nt p, r = 0.5 m and (} = 30° or

30

180

x 1t radians.

,

, ----l

1.59m

I""~­

Fig. 5.48

Stream runction at point P. 1jI=3x0.5xsin30o + 30 x 21t

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(30 X1t) 180

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IL 1238

Fluid Mechanics '" 0.75 + 2.5 = 3.25 m I,s. Ans.

(ii) Resu/I(lI'/l'e/ocity II/ P

The velocity componen ts anywhere in the flow arc given by

"r=..!. dljl :.!.~[u.rSin 9+ ..2.... 9] r ae rae 2rr '" ..!.[u.rcos 9 r

+.!L] = V.cos e + l 2 1t 2rc r

.-::3~O~ :]xcos300 + -::;21txO.5 '" 2.59& + 9.55 '" 12. 14

,II1U

lie =

-:,ljI = - ~: [u.rSine + ;~. 6]

= - U sin

e

+ 0 '" - U sin = - ]xsin30 = - 1.5

e

Q

Rcs ulwm velocity.

v=

Ju; + u~

= J12.l 4 1 +(- LSl '" 12.24 m/s . An s. (iii) LoctUiOlI of stagnation poilll

The horizontal uistancc of the

st:'~l1alion

poilll 5 from tlie source is given by equation (5.55) as

's= -q- = - ]0 -- =

1.59 m . An s.

lit x 3 The stagnation point wi ll be at a distance o f 159 III to tlie lert side o f Ihe source on the x- axis. Problem 5 .40 A wli/o'm jlOII' Willi" ,'eiOCil)' of 20 mls is ]/0"" " 8 orer " J'ou,ce of Slrength 10 111 21$. Tlu~ uniform flow Glllt -Wlm; e flow are in Ille same pili/Ie. Oblain III" eqllMi011 of Ille dil'idil1g Siream /in e ""'/ skelCh Ihe flow p"l/em. Solution. Given: Un iform 1'~ loc ity. U :: 20 m/s : Source strength. q :: 10 m 2/s (i) £,/lIl1lion of Ille dil'iding SIri'lim line The st re31l1 function a1 any point in th~ combined now field is given by equation (5.54A) 2/tU

IjI

= U. rsi n e+..2.... 2rr

e

::20x rsine+..!.Q.e 2rr

Th" value of th" stream function for the d ividing stream line is 1jI::

2.. He nce substituti ng IjI '" 2. 2

2

in

the above equation. we get the equation of the dividing stream line.

1. :: 20r sin e + ..!.Q. e 2

"'

I I

2rr

10 10 - =20rsin6+ - 9 2

2rr

c·

q:: 10)

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IL Kinematics of Flow and Ideal Flow

. 10 10 5 = 20rslll 9+ - 9 :20y + - , 2n lit

"'

(":

239 1

rsin8= y)

10

20)'=5 - - ' 2,

"'

e

e

5 10 - - x - = 0,25- 20 l it 20 4n The above relation gives the equation of the di viding stream line. From llie above equmion. for d iffe re nt values o f e the value of}' is obtained as : y= -

Value of 9

0

, 2

,

3,

-

2

"

Value of y from 0)

0.25

... ( i)

Remarks

Max. ha lf width of body

III

0.125 m

T he + vc ordin ate above the orig in

0

T he leading point

- 0. 125 m

The - vc ordinalc below the origin

- 0.25

The max. - \'C ordin ale

III

(ii) 5kelci! of flow pal/ern

For ske tc hin g Ihe flow pattern. Jet us first find the location of the stagnation poilll. The horizomal distance o f the stagnation p
', = -q- :

10

'" 0.0795 III

2rr.U lit x 20 Hen,;:" Ihe s tagnat ion poin t lies on the x- a~is a1 a di~lanc", o f 0.0795 m o r 79.5 mm frorn Ih'" source tow'lrds left of th e so urce. The flow p'l\1Crn is shown in Fig. 5.49.

-~§:,,~ . ";~ J} 79 .5 mm

Fig. 5.49

Problem 5.41 A uniform flow wi,1I a I'e/ocity of 2 m/s is flowing ol'er a source placed III ,lie origin. Tile s/agl1a /ioll poill' occIlrs (II ( - 0.398. 0). De/ermille " (i) Slrellg/II of Ille source. (ii) Maximum widlll of Rallkine half-body allli (iii) OIlier pril1cipal dimellsiolls of Ihe Rallkine Imlf-body. Solution. Gi ve n Uniforrn ve locity. u= 2 rnls

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IL 1240

Fluid Mechani cs

Co-aminalcs o( stagnation point'" (- 0.398. 0) This means,. " 0.398 and stagnation point lies on x-axis al a distance 0(0.398 IIl lowards left of origin. The soum: is placed al urigin. (i) SI":tIgll! of tile J'ource Let q " strength of tlie source We know thm

or

r == ~ ,

2 lf U

q= 2ltUx ' . " 2ltx 2 x 0.398 = 5.0014 m1/s '" 5 m l/s. Ans. (it) Maximum Iddtl' of Ril/lid/le half-body The main dimensions of the Rankine half-body arc obtained from equation (5.57) as y=

.!L(l-!) 2U ,

...

(1)

The value o f )' is maximum. when G = O. Y""" '"

2~ ( I - ~) '" 2(~ '" 2 : 2 '" 1.25

III

Maxim um width of Rankine body '" 2 )( Y ...... '" 2 )( 1.25 = 2.5 Ill. AilS. dillWII$iOIlS of Rallkin<' lIal/-body Using equation (5.57), we gel

(iii) Olher Principal

y = -"2U

y

(I --"-),

-,,- [Ji)1= -"-[1_-'-]=-"- =- '

=2V

It

2V

2

4U

4x2

=0.625 m

The above value gives the upper ordinate at the origin. w here source is placed. Width of body at origin == 2 x 0.625 == 1.25 III At the stagnation point. th e width of the body is zero.

0.625m

5 0.398 Origin

Stagnation point

(Soon;
Rankine ha ll·body

Fig. 5.511

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IL Kinematics o f Flo w and Ideal Flow

241 I

.5 . 11 .4 A SOUTce and Sink Pair in a Uniform Flow (Flow Past a Rankine Ova l Body). Fig. 5.51 «I) shows a unifonn now of velocit y U and r ig. 5.51 (b) shows a source sink pair of equal strength. When this I,mifoTm flow is flowing over the source sink pair. a resultant now wil l he obta ined as shown in Fig. 5.51 (e). This resultant now is also known as the now past a Rankine oval body. Sink

"'""'

·;te'

•

Source

(a) Unif"",, flow

•

Sink

,

I-

~{'

•

0

I·

' --I

(b) Source and sink pair

"

,

Rank;"" Oval8od~ Pix.

~)

"' ,

-

,'

'.

0

• •

•

"

"

•

• x,

••

,,'

Fig. 5.51 U == Velocity of uniform flow along .I-axis If" Strength of source (- q)" Strc ngt li of si nk 20" DiSlancc between source and sink whid is alun g x- axis. The origin 0 of the .(-y co-ordinates is mid -w ay between source and sin k. Consider a poilU P(x. y) lying in the resu ltant flow field. The stream function (III) and veloc it y potential function (q.) for the resultant now fie ld are obta ined as given below: III == Stream function due to uniform flow + stream function due to source + stream fu nction due to sin k

Let

== 1jI..;{orm now + 1jI><>I/fCe + 1jI..""

q

(- q) 2lT

=Uxy+-',+--xe,

2n (where 9 1 is the ungle made by P with source along x- a;o;is ilnd 9! with sin k)

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IL 1242 Fluid Mechanics

" u x y + (Jf.l , _ qfJ) "u)( y + !L 21t

2rr

(G, _ G,)

21t·

(... )' =, "0 9) ... (5.59) .p" (I'Otcll1iaJ

and

fUllction duc to uniform flow

+ potential fUllction duc 10

source + potential function duc co sink

'" U x r cos

e + !L [log, '1 2,

" U )( r cus

e+

;' _It

log,'21

(.:

[10g,.2..]

-,""reos6) ... (5 .60)

')

The following arc the importam poil11s for the resultant flow p.dtcrn : (a) The re will be two stagnati on points 5, and 52' one to the left o f the source and other to the right of the sink. /\11hc stagnat ion points. llie resu ltant ve locity (i.e .• velocity duc 10 uniform flow. ve locity duc to sou rce and ve locity due to s ink) wi ll he zero. The stagnation point 51 is to the left o f th e source and st agnat ion point S2 will be to tlie rig ht of th e sink on the x-ax is.

Let ).'s = Distancc of tile slaguation points from origin 0 along .r-axis. Let us ca lcu lat<) this distane<) " 5' For the stag nat ion poin! 51' (i) Velocity due to uniform flow = U (ii) Velocity due 10 source

The ve locily a( any radius due 10 source == ~l

= cc,,~q,-, 2n (xs a)

r"Or .':II' the radius from source == ("'s

-"

(iii) Velocily due 10 sink '" ~,-'!.,--,

Al 5 1'

2n ("s + a)

th~

- a)

2nr

radius from sink'" ("'5 + a))

At poi nt 51' the velocity d ue to uniform flow is in the positive .l -direction whereas due 10 source and sink are in Ihe -ve x-direction. The resultant ve locity at 5 1 = U - cc-,' ::; '' --, 2n ( x.~ a)

(- q)

Butthc resu ltan t velocit y at stagnation point 51 shou ld be zero.

uU=

II

2n (x s -a) q

2n(" 5-a)

I I

-

+

q

=0

21l" (xs+ll) q

2n(xs +a)

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IL Kinematics of Flow and Ideal Flow

q [ '" l it

(xs

243 1

I

"'

(1+:~U )

... (5.61 )

Th" above eq uatio n gives th e locati o n of the stagnation point on the .(· axis. (b) The s tream line passing thmugh th e stagnation points is having zero ve loc it y and hence ca n be re p laced by a so lid body. This so lid body is having a s hape of 0'131 as show n in Fig. 5 ,5 1. T here wilt be two flow fields. o ne wit hin the ova l contour and the Olh er outside the solid bod y. TIle flow fi el d

within Ihe ova l co ntour w ill be due 10 source and s ink whereas tilt! flow field outside the body w ill be

du e to unifonn flow onl y. The shape of so lid body is obtained from Ih e s tream lin e ha vin g stream function eq ual 10 zero. But th e stream function is g iv en by equat io n as IjI

= U x r sin

a + .!L (9 , _ 0,) 2]'( •

For th e shape of so lid body.\jI = 0

0 = U x r sin a +

"'

U)( rSin

.!L (9 1 _ e ,) 21'f

•

e ", _ .!L (ti l _ e,)",.!L (e, - ( 1) 2rr

•

21t

•

(el - 6, )

if

'" -2rr

... (5 .62 )

Usi n 6

Prom th e abo ve equation. th e distances of the s urface o f th e so lid body from the origin call be obcai ncd or th e shape of th e so lid body can be obtained. The maxi mum width of the body (Y""",) will be equal to OM as shown in Fig. 5.52. M

" s,

A

"

]<- '

, 0

,

- +- .----<

s,

F ig. 3.32

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IL 1244

Fluid Mechanics

From triang le AOM. we have

OM

tane = - -

,

AO

OM == AO tan 6 1 "

"'

Y......

"' LeI us lind th e va lu e of 6 1,

Whe n lhe point Plies 011 M. then r == OM.

and

{/

tan 6 1

= a tan 6 1

(.,. OM " Y,,..,) ...(5.63)

e" 90

Q

== ;

6 2 = 180 0 -6[ =1t-6 1

I":

AM=BM

[R efer \0 Fig. 5.521 Ang le ARM" Ang le BAM == 6 1J

Substituting th ese va lu es in equatio n (5.62), we gel OM

~(I_ '_ - '-",,)-,-',,-,)

= _" -" It

U

.

Sill

I'f

2 Iwhere OM "

"' or 2rr. UYm, ==

It _

y"",, 1

2 (1 1

q " _ _ I -

"'

It

_ l ItU)'.,,,

q

"'

Substituting thi s value of 6 1 in cqualiOIl (5 .63). we get Y.m == a Ian

[% _rru;m>x ] " a WI [ 1t U~;m~ ]

... ( 5 .6 4 )

From th e above cqumion.lhc valu e ofy ..., is obtained by hit and triallll cth od till L.H.S." R.H.S . In this equation

( J'[ u~..., ) is in radian s.

The kngl h and w idth of the Rank i ne oval is obtained as :

Length.

L=2

X .l S

"2xo and Width.

B=2 x =

(,+"1 ,0u

y"""

2a (nu;-. )cot

... ( :'i .66)

Prob lem 5.42 A Iwifarm flo ... of I'eloeity 6 mls is flo ..... ing {liong x-{l.fi,~ 01'(' , {I SOilree {lnd a sink ...hicil are silllmed along .(-axis. The strengtli of source and sink is 15 m"/s alld they are m a distallce of 1. 5 III alwrt. Determille :

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IL Kinematics o f Flo w a nd Ideal Flow

245 1

(i) wCaiiOIl of srtlgnali(1II points, (ii) unglh lind width of Ihe R(mkil1c oml (iii) Eqllation of profile of Ihe Rankine body.

Solulion. Given: Uniform fl ow velocity. U'" 6 In/s Strength of source and sin k. q'" 15 rn ~15 Distance bt,',twccn source and sink, 2,,: 1.5 III II'"

1.5 ""2

= 0.75

III

(Refer 10 Fig. 5.51) For Finding tlw localion of the Slagll31iOH points. Ilic equation (5.61) is used.

(I) wnllioll of sl(lgnmioJi poillls

x. =a

J(I+rwUq )=0.75 [1+ IlxO.75x6 15 1'" 1.076rn

Ttw above cqu ali oll gives lhe di,13nce of Ihe stagnation points frolll the origin. There will be two stagnation poi nts. The diswnce ofslllgoation poims from the source and sink'" x, - II = 1.076 - 0.75 = 0.326 m. Ans. (Ii) Lenglh olld ",idlll 0/ fhe Runkine 0>'01 Length. L = 2 x x, = 2 x 1.076 = 2.152 nl. Width. B '" 2 x y""" ... (i) Let us now lind Ihe value of Y"",. Usin g equ3lion (5.64). we gel

y

Y"",,"'
= 0.75 COl

1

=0.75col

(" 6X" ) 15'"''

= 0.75 COl (O.4Il Y....,)

(0.411 Y..., x I~}

[ .; (OAIl y ..... ) is in rad ians an d hence (OAII Yo.,,) x

I~

will be in degrees]

'" 0.75 cot (72 )( y ..... )O The above equation will be sol\'ed by hit and trial method. The value of x, = 1.076. But x, is equal 10 length of major axis of Rankine body and y""", is Ihe length of minor axis of Ihe Rankine body. The length of minor axis will be less Ihan length of major axi s. Le i us flrsl assume Yma.\ = 0.8 ]I). Then

y"""

[" H.s.

R.H.S.

0.75 COl (72 x 0.8)° = 0.75 COl 5l N '" 0.475 07 07 0.75 cot (72 x 0.7)° '" 0.75 cot 50.4" = 0.577 06 06 0.75 cot (72 x 0.6)" '" 0.75 cot 43.2" '" 0.798 0.65 0.65 0. 75 cot (72 x 0.65)" = 0.75 cot 46.8° '" 0.704 0.67 0.67 0.75 COt (72 x 0.67)° = 0.75 cot 48.24° = 0.669 '" 0.67 f.r0111 above it is clear that. when You, '" 0.67. then L.H.S. '" R.H.S. 0.8

0.8

y ..... '" 0.67 II) Substituting this value in equ3lion (i). we get Width. 8", 2 x Yn"" = 2 x 0.67 '" 1.34 rn . Ans. (iii) Eqlwlion of profile of Ille Ronkilte body The equation uf profile of the Rank ine body is give n by eq uJ tion (5.62) as

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IL 1246

Fluid Mechani cs

,= -q la, -a,) 211: U si n e 5. 17.5

A Doublet in

ill

15 (8 2 - 9,) 0.398 (9 2 - 6, ) = . A ilS . 21l 6 x si n 9

Uniform Flow (Flow Past

ill

sine

Circular Cylinder).

Fig. 5.53

(a)

shows a unifonn flow of velocity U in the positive .H lircction and Fig. 5.53 (b) shows a doublet at the origin. Doublet is a spcdal case of a source and a sink combination in which bolh of equal strength approad each olher such that distance between them tends to be zero. When the unifonn flow is flowi ng over tlie doublet. a resultant flow will be obtained as shown ill Fig. 5.53 (c). This res ultan! now is kn ow n as th e n ow llas l a Rankine ova l or equ;'tl :,xes o r n ow p as t a circular cylinde r .

/

H f--' (a) Unif"",, flow

\\

"

(b) Dout»el

Poten~al l >nes

Stream lines

.., ~~,m ., s,

$,

,,' Fig. 5.53

The stream function ( 1jI) and velocity potenti al functioo ($) for the resul tant flow is obtained as given below: I¥ " SITe,lm function due 10 uniform flow + stream fUlH:lion due to doublet

,,, a)

"uxy+(-J.l

si n

[ Stream function due to doublet is given by equation (5.50) as'" -

~ I

,,,

L

sin

e]

I~

~ I

IL Kinematics of Flow and Id eal Flow

L

=Ux rxsin6 -

(";

sin6

2rcr

247 1

y=rsin6)

~(u xr-L) sino

... (5.6 7)

2"

(jI '" Pote ntial func tion due to unifo rm flow

=UX .l+ -

).1

2,

~"()s

+ potential fU llction due to do ubl et

e

x --

,

[ From

COS6] eq uation (5.52). potent ial function due to do ublet ", - ).l x 21t

II

cos

e

'" Uxrcos 6+ - x - 2, ,

r

(':x=rcos 6)

~(U x,+L)cos e

...(5.68)

2"

Shape of Rankine oval of equal axes

To get the profile of th e Rankine ov al of equal axes. the stream lin e IjI is taken as zero. He nce substitutin g IjI = 0 in equa tion (5.67), we ge l

O~ (u xr- L)Sine 2"

This mean s either

. e =0 or UX·, _ ..1:...-_o

sm

e

(i) If si n = O. [h en 9 = 0 and hori zontal line is the x-ax is.

(ii) If

Uxr - L 2n:r

2" ± 1'( i.e.• a tlOri zonl al tine th ro ug h lh e origi " of the douhlet Th is

=0, then

Vxr= Lor ?", - "2ltr 211:U

,-- J).1 = a constant 2n:U

as).l and U arc constant.

Let 1his eo nswm is equal 10 R. r=

~2~tU

=R

This givcs that the close d body profile is a circular e ylimler of T:ldius R wi1h ce ntre on double t. The di vidin g strcam linc corrcsponds to 'V = O. Th is strcam linc is a circlc of radius H. The stream line'll = 0 has twO stagnat ion poims SI and S2. At SI . the un iform fl ow sp li ts into two stream s that !low along the c ircle with radius R "

~ 2n:U P . the twO branches mect again at the stagnation poin! S, and th e flow -

continucs in the downward diredion. The uniform !low occurs outs ide the circle whereas the fl ow fi eld du e to doublet lies entirel y wi thin the ci rclc. Th e strcam function for the co mposite flow is g ivc R by equation (5.67) as

~ I

I~

~ I

IL 1248

Fluid Mechanics

.' (u xr -l-it,"- )Sin =U

9= U ( , - - "- ) 2ltU,

R' ]sine (r--;-

sin e - P- , R

2,U

2) ... (5.69)

V .... oclt y Co m po ne nts (u , a n d ( 9)

Th" I'elocity components at any point in Ihe flow fie ld are given by.

R'].sinS1", -I U ( , - -R 'l cuse

","'-,IJ. -ae =- [U( , - - r ," (}9

r

r

... (5. 70)

wd

11 0

a.(lr

a [U (, - -R '].sin e1=-u (1+-, R'].sma

=--= - -

(Jr

f

, "

.. .(5 .7 t)

. .. ( 5.72)

Resu ltanl veloc ity.

On IIII'

surfac~

of the cy linder. r

=R

I l r=U[I - :~ ] cos e

1-"

In equal ion (5.70),r=R]

,0

[ R'l

"0'

u9= - U 1+ R'

sinfl=-2Usinfl

... (5 .73)

- VI' sign s liows the cloc kwise direction of tangential ve locity at that point. The value of lie i~ maximum. when e '" 9(10 ,md 270°. At e = 0 " or 180".lhc value o f lie = O. Hence on lhe surface of the cylinder.l hc resultant ve locity is zero, when 0", 0" or ]lIDo, These two points on the surface of cyli nder li. e .. at 0 '" 0 0 and 1&0°1 whae resultant velocity is ze ro. are known as stagn3lion points. Th~y ar~ denoted by SI and S2' Stagnation poin t SI corresponds to e "" 180° and S1 corresponds to e '" 0°.

Pressure dis tribution on the surface of the cyli nder Let

I I

Po =

pr~ssure at a point in the unifonn flow far away frolll Ihe cy linder and towards the left of Ihe cylinder [i.e" approaching uniform flow ] U '" velocity of uniform flow 31 that point /I = pressure at a point on the surfa<.:e of the <.:ylindc r V '" resu ltant vc locity at th at point on the surface of the cy linder. T his velocity is equ al to lie as ", is zero on the surfacc of th~ cylinder.

Ii

~ I

IL Kinematics of Flow and Id eal Flow

249 1

V=u a = - 2Usin9 Applying Bernoulli's equatio n at the above two points.

Po

-

pg

UI

+-

II

VI

: - +28 pg 28

~ + ~ == ..E... + [-

"'

pg

28

2Us in

pg

at

[-:

2g

v=

09=

- 2U sin 91

-p" + -U" = -P +-,4cUC'...,'C"C'_'C p

"'

2

P

2

p

But P - Po is a dim,,"sionle_~s term and is kn ow n as dimensionless pr~ssure co-effi cie nt and is 1 U'

"

denoted by

Cpo Cp = ~-p~ = 1_4 sin 2 e

,

- pU'

Va lue of preuure co-efficient for di fferent values of Vulue of B

o

e

Value of Cp 1 _ 4sin 2 1

e

= 1- 0 = 1

GY =1-~ :

30·

1 _ 4 sin 30o : 1 - 4x

90·

1 _ 4 sin 1 90o = 1 - 4x I == 1- 4= - 3

I - I =0

I - I =0 1 _ 4 sin 1 ISOo = 1- 0= I

AI Al Al The sh own

I I

9 = 0 a nd 180°, th ere arc stag nati on points 52 and 51 respectivel y. 30 Q and 150 0 , the pressure co-efficient is zero. 9 = 90¢, lhc prcs~urc cO-1>'ffici<' nl is - 3 (i.e.,leaM pressure) va riation o f press ure co-efficie nt along th" surface of th ~ c ylind er for diff~renl valu<,s o f in Fig. 5.54.

e=

e are

Ii

~ I

IL 1250

Fluid Mecha nics

The positi ve pressure is actin g normal 10 th e surface and IOwards the surface o f th e cy l inder

whe reas th e neg at ive pressure is acting normal to tile surface and away from th e surface o f the cylinder as sllow n in Fi g. 5.55.

' f'

.i\

o· ",.

w·

I-

\

-,

,

,

00"

-

"-

V

. / 150~180· 0

/

F ig. 5.54 ~

/ --,r--<:- - pressu r"

,",~- pressure ".

Fig. 5.55

Problem 5.43 A IIl1iform jlow of 11 m/s is flowillg ora a doublet of strengtll II:! is in Ihe line of the ulliform flow. Determine: (i) slrope of Ihe Rankine oml (iij radius of Ihe Rmlkine circle

•

I/I"/S.

'I'Iledoublel

(iii) )'(1111<' of .l'lrealll litle fimcliOl1 (1/ Rankine circle (il'J re~·"II<"'1 re/ocily at (I poinl 011 Ille Rankine circle a/ lm angle of 30° from x·,uis ( I' j \'{jille of II/{uimum ,·e/oi.'iry 0' 1 the Ran/due circle (lnd lQ(;lIliOlI of Ihe poilll wllere "e[o<:iry is maT. Solution. Given U = 12 m/s: J.I = 18 1I1 2/s (i ) Shape of Ihe Rankine 0,",,/ When a uniform no w is Flowing over a do ubl et and doublet and uni foml n ow are in lin e. then the

sh ape of the R ankine ova l w ill be a circle of radius =

I I

J2,U I-l •

A DS.

Ii

~ I

IL Kinematics of Flow and Ideal Flow

251

I

(ii) RIUIiII$ of Ille Rallkille circle

R=r=

J)l 2nU

=

~

V~

= 0.488 m. AilS.

(iii) Value of stream lille /,melion al /I,e Rallkin e ('irc/e

The va lue o f strea m lin e function (\11) at th e Ranki ne c ircle is zero i.e .. \jJ = O. = JO On the s urface of the cyl inder, the radial velocity (u,) is lero. The langenlin] veloci ty ("a) is give n by equation (5.73) as ua=-2Usin 6=- 2 x 12 x sin 300 =- l2m/s. AilS. (;v) Rend/mil ndocily 011 the sur/ace of 1/"" rir dl!. ""hell

e

Q

- ve sign shows the clockwise direction of tangential velocity a1 thai poi n\.

Rcsul1alll ve locity. (,,) Maximum "e1ocity and ils foell/ioll

The rcslllwnt ve locity at an y point 011 Ih e surface of the I:ylindcr is equal to I'a" But "6 is give n by, " a=-2UsiIl6 This velocity wi ll be maximum. when = 90°.

e

Max. ve locity " ~ 2U" - 2 x 12 = ~ 24 m/s . Ans. Problem 5.44 A uniform flow 0/ JO /Ills i$ flowillg OI'er a doublel of j'lrenglll 15 m 2/$. Tile doublel i.~ ill IIII' lille of Ihl' ulliformJIow. 1"11<' polar co-ordilWles of a poim P ill IIII' flow field are 0.9 III and 30°. Find: (i) Slrealll line/ul1Cliol1 lIml (ii) /Ile "'j'ullllll/ "docil)" a/lire poini. Solution. Given: U = 10 I11 ls: ~ = 15 111 2/s: r = 0.9111 and = 30°. Let us fi rst find the radius (R) of the Rankine c ircle. This is given by

e

R=

~

Jl5

P = =0.488m 21!U V~

The polar co-ordinates of the point /' are 0.9 111 and 30°. Hence r '" 0.9 111 and 30°. As the va lu e of r is more than th e radius of the Rankine circle. hence point P lies outside cylinder. (i) Vallte 0/ Siream line/wne/ioll III III<' poil1( P The st ream line function for the cOllljXlsite flow al any point is given by equa ti on (5.69) as

e '"

th ~

,

P{O .9 m, 30' )

-

Fig. 5.56

I I

Ii

~ I

IL 1252 Fluid Mechanics

,= 0.9 Ill. R = 0.488 and 9 '" 30") I

,

= 10(0.9 - 0.2646) x - '" 3. 177 m Is. An s. 2

(ii) He,mllalll \'e1ociry al lilt! poilll P

The radial ve loc ity and tangential veloc ity at any point in tlie flow field arc g iven by equations (S.70) and (5.7 1) respectively.

[ 7")

"r=U 1-

(0488')

cos9= 10 1-

~.9 l

CQs300 =6 11 m/s

+vc sign s how s th e radial \'clocit y is o utward. ue =- U

and

[1+ 7")

sin9=-10

[ 0488') I +~

si n 30 0

= - 6.47

mf~

-vc sign shows the clockwise direction of tangential ve locit y. R csul1~ nt velocity,

= J6.l 12 + (- 6,47)' = .)3733 + 44,86 = 8.89 m /s. A ns . HIGHLIGHTS 1. If the fluid characteristics like velocity. pres.ure. density etc. do not change at 3 point with ",spc<;t to lime. the tluid flow is called steady flow. If they change w,r.l. time . the fluid now is called unsteady

now.

(~;)

= 0 for sleady flow

and

(~,)"" 0 for un.leady flow.

2. If lhe velocily in a fluid flow does nm change with re'pee! to space (length of direction of flow). the flow is said unifoml otherwise non-unifonn. Thus.

(~) '" 0 for unifoml flow and (~;) c#- 0 for non-unifonn flow. J. If the Reynolds number in a pipe is less than 2000. the flow is said to be laminar and if Reynold number is more Ihun 4000. the flow is said 10 be turbulent. 4. For compressible flow. p"" constant For incompressible flow. p '" constant. 5. Rate of discharge for incompressible fluid (liquid). Q ~ A x I·. 6. Continuity equation is wrillen a, A,", '" A,,·).

A,'·. '"

I I

Ii

~ I

IL 253 1

Kinematics of Flow and Ideal Flow 7. Continuity C{Juation in di ffcrcnliul fonn. (Ju

(n'

ilx + dy +

flu


a( + iJy

.. 0 for two-dimensional flow.

II. The components of acceleration in x. )' and Z direction are (,

au

'

dU au iJX{)yi.lZ(J1 ()u

",u-+,·~+

.. '~ + -

;n, ;),0 i),. a,· + " - + II' - + axiJyilz dt

II : U-

'

al<'

all'

: ,h

a", a...

" " - H -()y +W -at ' -i)I 9. The components of

~elocit y

u__

in x. )' and

l dir~""lion

'

in Icnns of velocity polential

(~)

are

"*." K_'*ily and", _ _ ()q>dZ . d.t

10. T he stream funclion (1jI) is defmed only for Iwo -<;l imensional flow. The ""Ioc i!y CQln ponems in

y direClions in {CnTIS of stream function are u • _

a'll ;ry

and """

~ ~

J

and

.

II. Angular defonnation or shear strain rate is given as Shear strai n rale _

~ [().. + au] 2 ax

ily

12. Rolational components of a fluid par1icie are

13. Vonicity is two times the value of rotation . 14. flow of a f1uid along a curved p;tth is known as vortcx f1ow . If the part icles afl' moving round in curved p;tth with Ihe help of some external torque the f10w is called forced '"Ortex f1ow. And if no external torque is req ll ircd to rotate the nuid p;tnicles. the f10w is called free -vortex f1ow. IS. T he rclation beNeen tangential velocity and radi us: for forced vortex, \. ,. to X r. for free vortex, I' X r = con,tall!. 16. The pressure variation along the radial direction for ,-ortex now along a horizont:ll plane. and pressure variation in the venical plane ,,2

For the where

I I

W 2,l

iJp _ _ P8.

a,

wl R l

for~e<.l

vortex now, z,. - ' " - - . - 28 28 28 Z .. height of p;trJboloid fanned w = angular velocity.

Ii

~ I

IL 1254

Fl uid Mechanics

18. For D forced vonex flow in a ope" mol. Fall of liquid level at cemre Rise of liquid level at the ends. %

19 . In case of clo"'" cylinder. thc volume of air before rotalion is equal to Ihc volume of ai r after rotation. 20. If a close cylindrical vessel completely filled wi th waler is rolated about its "cMical axis. the IOU!

pressure forces acting on lhe top and OO1(om arc Fr"

and where

FT= F 8 .. 1Il .. R=

..e. oillK

4 F 1/"' FT + weight of wale' in cylinder Pressure force on lOp of cylinder Pressure force on the bonom of cylinder Angular vclocity Radius of the vessel

w p .. Density of fluid,. - , g

21. For a free vonex flow the equation is

.£L -+- ,·f pg

2g

~

+ ZI .. .!!.l... + -+pg 2g

Zl'

EXERCISE (Al THEORETICAL PROBLEMS 1. What arc the methods of describing fluid flow ?

2. bplain thc lcnns: (i) Path line,

(iii) Stream line, and

( ii) Streak line, (i") Stream tube.

3. Distinguish between (i) Steady flow and un-steady flow. (ii) Unifonn and non-uniform flow, (iii ) Compre,siblc and incompressible flow. (i") Rotati onal and irrOlational flow. (v) Laminar and turbulent flow, ~ . Define the following and give one pmctical example for each (I) Laminar flow. (ii) Turbulcnl flow. (iii) Steady flow. and (i") Unifonn flow. S. Define the equation of continuity. Obtain an expression for continuity equation for a three-
I I

Ii

~ I

IL Kinematics of Flow and Ideal Flow

255 1

12 . Dcrive an e~pres,ion for the depth of paraboloid formed hy the surface of a liquid cOnlaincd in a cylindrical lank which is rotated at a constant angular vel ocity (II ahou! ils venical axis. U . Derive an expression for the difference of pres,ure between two points in ~ frce "orln flow. Does the difference of pressure sat isfy Ucmoullrs equation? Can Bemoulli"s equation t>c applied to a forced ,'one" flow 'I 14 . Dc,;"c. from first principles. the condition for irrola(ional flow. I'row that. for potential flow. OOlh the stream fu nction and velocity potential function ",-[isfy the Laplace equation. 15 . Define "elClCity potential function and slream funnion. 16 . Under Whal conditions Call one trcat real fluid flow as irrolationa] (as an approximation).

17. Define the following: (I) Steady now. (Ii) Non -unifomt now. (iii) Laminar now. and (i") Two·dimcnsiona l now. Ill . (a) Disting uish between rotational flow and irrotational flow. Give One example of e.leh (b) Cite two examples of unsteady . non-uniform flow, How ~"n thc Unsteady flow be transformed to steady flow ? /J.N.T. Vlli,·asil),. S 10(1) 19. Explain uni fonn flow with souree and sink. Obtain expressions for stream atld velocity potential functions. 20 . II poi1ll source is a point where an incompressible fluid is imagined 10 he created and senl OUI e"enly in al l directions . Detenn;ne its velocity potential and Slream function. 21 . (I) Explain doublet and define the strength (If the doublet (ii) Di stingui'h between a source and a sink. 22 . SketCh the fl(lw pattern of an ideal fluid flow past a cylinder with circulation. 23 . Show that in case (If forced vortex flow. the rise of liquid IC"e! at the ends is "'Iual to thc fall of liquid le"el at thc axL. of rotati(ln, 2~ . Differentiate between (i) Stream function and "elocity potential function (ii) Stream line and streak line and (iii) Rotalional and irrolational fl(lws.

(B) NUMERICAL PROBLEMS 1.

2.

3. 4.

S.

~ I

The diameler:s of a pipe .11 the seclions I and 2 me 15 cm and 20 cm respecti,·e ]y. Find the discharge through the pipe if "ciocily of water at section I is 4 mIs, Dctemline also the ve locity at section 2. IAns. 0.07068 m'/s, 2.25 m/sl A 40 cm diameter pipe. conveying water. branches into two pipes of diameters 30 cm and 20 cm respectively. If the averagc veloci ty in the 40 em diameter pipe is 3 m/s. Find tne discharge in this pipe. lliso detennine the velocity in 20 em pipe if the average velocity in 30 em diameter pipe is 2 m/s. IAns. 0.3769 ml/s . 7.5 mlsl II 30 em diameter pipe carries oil of sp. gr. 0.8 at a velocity of 2 m/s. lit another section the diameter is 20 Cm. Find the velocity at this >cetion and also mass rat~ of flow of oil. IAn s. 4.5 m/s. 113 kg/sl The velocity "cctor in a nuid now is given by V = 2~ 1 - 5.rYJ + 4/k. Find the velo<'ity and accelemtion of a fluid panicle at ( I. 2. 3) at time, / ~ I. (Ans. 10.95 units. 16.12 unitsl The following case .. represent the two velocily components. delenninc the thin! component of velocity such that they sa(isfy (he continuity equat ion

I~

~ I

IL 1256

Fluid Mechanics (Il u., 4,,-2, \' = 4.I)'Z

(ii) u = 4Xl + 3.,)" ", ..

[An•. (i) ... _ - 8
t - 4.')' -

2y< _

2xz1 + I( x. y)

(ii) I' = -!UY

>" + 3}"l '' + I(x, t)] - "2

Calculate the unknown "clocily components so (hm they satisfy the following cqualions (II i t . 2.-2, v. 29'<.'" _? (ii) ,,_ 2K + 2sy. 1<'. - 4xz + 2yz .•'.? [AilS. (ill<' _ - LIz

r

6. A fluid now is given by : II =

7.

H. 9.

10 . 11 .

_ x'zi

xy'i - 2yz'j _ ( 2),' - 2t) k.

Prove that it is a ca5e of possible steady incompressible fluid f1ow_ Cakulatc the velocity and acceleration at the point [1. 2. 3]. [A ns. 36.7 units. 874.50 unils] Find the convective accclcr.tlion al1hc middle of a pipe which co,,,'crges unifonn ly from 0.6 III diameter to 0.3 m diameter over 3 111 length. The Me of flow is 40 litis. If the rate of flow changes unironn l), from 40 liUs to 80 hils in 40 second,. find Ihe IOlal a"elcmlion al the middle of the pipe al 20lh second. lAos• .04<)<) tnls' : .1 IS74 mis' I T he velocily potential function. , . is given by, '" x' - /. Find the "cloc ity components in x and y direction . AI"" show thaI, rep rcscms a possible case of fluid flow. IAns. u = 2x and I' ~ - 2yl f or Ihe "elod ty polential function. " fi nd the velocily components at the point (4. 5). [Ans . u = S. ,. = - 10 IInitsl II stream function is given by: 'I' _ 2r - 5)". Calculate the "elocity components and also magni1Ude and direction of the res ultanl velocity at any point. [Ans. u .. 5. 1'" 2. Resultant .. 5.384 and e .. 21 0 48'1 If for a two-dimen,ional potential now. thc velocity potential is given by : , = 4.t(3)' - 4). deteonine the velocity at the point (2. 3). Dctennine also the value of stream fun ction 'I' al Ihe poinl (2. 3).

.r -/'

[ Ans. 40 units. '1''' 6.r' - 4

(f)'l - 4 Y). - IS]

12 . The stream function for a Iwo-
1 S. II fluid now is gh'cn by : V _ I o..-Ji - 8.1'\;' f ind the shear strain rale and state whether the flow is rotational or irrotational. [AilS. - &ry. rotational I 16 . T he vc!ocity components in a two-
u ..

&r), - ~ }'l and v .. _ &xl + ~ r.

J 3 Show that these veloci ty components represent a possible case of an irrotational flow . [

An s.

au + a,· 0=0. ro . 0=0] ax ay .

17 . An open circular cylinder of 20 em diameter and 100 cm long contains water upto a hcight of 80 em. It is rolated about it.. ,"enieal axis. I' ind the .• peed of rotation when (I) no Water spills. (iil uial depth is zero. [AilS. (I) 267.51 r.p.111 .• (ii) 422.98 r.p.m.] 18 . A cylindrical \"Cs~1 15 cm in diametcr and 40 em long is complelely filled with watcr. The ves~el is open at the top. Find the quantity of watcr left in the ve,,,,l, when il is rotated about ils ,'ertieal axis wilh a [Ans.4566.3cm ' ] speed of 300 r.p.m.

~I

I~

~ I

IL Kinematics of Flow and Ideal Flow

257 1

19 . An open circular cylinder of 20 em diameter and 120 em long contains water upto a height of W em. It is rotated about its vertical a~is at 400 r.p.m. Find the difference in lotal pressure force (il at the OOno111 of the cylinder. and (Ii) al the sides of the cylinder dilC 10 rotation. [Ans. (i) 14 .52 N. ( if) 2465,45 N I 20 . A c1o<;ed cylindrical \'e,sel of diameter 15 em and length 100 em contains water urlO a he ight of 80 em. The \'esse l is rotated at a speed of 500 Lp,m, about ils wrtical axis. Find the height of paraboloid formed. [Ans. 5606 em] 2 1. For the data gi"cn in question 20. find the speed of rotation of the vessel. when axial depth is zero. [Ans. 891.7 r,p .m.1 22 . If the cylindrical vessel of question 20. is rot ~t a point al a rad ius of ] 50 nun the velocity and pressure arc 5 mls and 14.7 15 Nkm l , Find the pressure at a radius of 300 nnn , [Ans. 15,65 Niem I ] 26. 1)0 the follow ing "elocity components rcpre""nt physically possible flows ? II .. .r' + l' + 5. ,... / + z'. 1<' .. 4.tyZ. [Ans. No.[ 27. State if the flow represented by" .. 3x + 4y ana \' .. 2r - 3)' is rotational or irrotational. [Ans. Rotational[ 28. A Yes.e!. cylinarical in shape ana dosea at the top and bOllom. <"ontain. water upto a height of 700 mm. T he di~meter of the vessel is 200 mm and length of vessel is 1.1 m. Find the speed of rotation of the ves",] if the axial depth of water is '.ero. 29 . Define rotational ana irrolational flow. The stream fUn <'I;on ~nd veloc;ty potenti al for a flow are given by, 'lI 3 2 (),.¢/ 3 X< - j . Show that th~ cond il;on' of continuity and irrotational flow arc sati.fied.

30 . f or Ihe steady incompres .• ihlc flow. are the following val ues of u and .' possible ? (I) " z 4.ry+ / .,· .. 6.ry+3xand (ii) u=2.C+ / ••'=-4.l)·. [Ans. (i) No. (ii) Yes[ J] . Dcfinc two ·dimensional stream fun ction and velocity potential. Show that followi ng stream function 'lI .. 6x - 4y+7xy + 9 represents an irrOiational now, I;ind its velocity potential. (A n.•. 41 = 4.< + 6y - 3.5.<1 + 3.5/ + C] 32 . Check if oil '" .•' - y' + y r{'presents the >'C locity potential for 2--dimensional irrot:ltional flow . If it docs. thcn dctcmline the stream function 1(1. [Ans. Yes. 1(1 = - ] .ry + x l 3J . If stream function for steady now is given by 1(1 _ li - ...-l). dctenninc whether thc flow is rouuional or [Ans. IrrotJtional. 41" - 21y + CI irrotational. Th en detcrmine the velocily potential 41. .l4 . A pipe ( I ) 4SO mm in diameter branc.hcs into two pipes (2 ) and (3 ) of diameters 300 mm and 200 mm respectiYely as shown in Fig. 5,57. If the 3YCrage .'elocity in 450 mm diameter pipe is 3 mls. fin,,,!. S 200]) [Hint. Given:

~ I

tI , =4SOmm=0.45m.tI,,,,300mm=O.Jm tI J .. 200 mm = 0.2 m. VI = 3 mIs, V.! .. 2.5 Illis

I~

~ I

IL 1258

Fluid Mechanics (i)

It

~

J

0 , "A,V, "'"4 (0.45)x 3 '" 0.'177 m Is.

(j)

-

d ,_ 450mm

Fig. 5.57 _ ; (.3 1 ) x 25., 0. 176 rrhs

(ii)

OJ •

IJ ut

0 , '" OJ + QJ

Also

O J "'A l X V1

A l V:

QJ '" 0 , - Ol = 0 477 -0.176= 0.301

" (0.2-), X V1 "' "4

Q,

V, • - _.-

~(O_ 2~ 4

I I

D.lO l

• -0.0314

~

11.6 m/s. ]

Ii

.. 6. 1

INTRODUCTION

In the previous c hapter. we stud ied the ve locity and acceleration 31 a point in a fluid flow. without tak ing into consideration the forces causing the flow. Th is chapter includes the study o f forces causing fluid Flow. Thu s d ynamics of fluid flow is thc st udy of fluid motion with the forces causing flow. The dy namic behaviour of thc fluid flow is anal ysed by Ihe Newlon'S second law of molion, which relates Ihe accele ralion wilh Ihe forces. The tluid is assumed to be incompressib le and non- viscous .

.. 6.2

EQUATIONS OF MOTION

According 10 Newlon's seco nd law of motion. Ihc nel force F , acti ng On a fluid elemenl in the direClion of X is eq ual 10 mass In o f the fluid clement multiplied by Ih e acceleration (/. in Ihe x-direction. Th us mathemati cally. ...(6.1 ) F, = m.(/, In Ihe fluid flow. Ihe following forces are pre.'lCnt: (i) F g• gravity force, (ii) Fr Ihe pressure force. (iii) F .. force due 10 vis.:osi ty. (iI') F,. force due 10 lurbulence. (I') Fe' force due 10 compre"sibilily. Th us in equa tion (6. 1). the net force F, = (F, )" + (Fp) , + (F) , + (F,)x + (Fe),' (0 If the force due to compressibility. Fe is neg ligible, the resulting net force F,

= (F, )" + (Fp) , + (F,.) , + ( F,),

an,j equ31ion of motions arc called Reynold' s ell uation s of motion . (ii) For flow, where (F,) is negligible. th e resulting equatio ns of 1Il0t ion are known as Na"ier-Stokes Equation. (iii) If the flow is aSl;umcd \0 be ideal. viSl:ous force (F.) is zero and equation of 1Il0tions arc known as Eule r 's elillation of mo tion.

259

I I

Ii

~ I

IL 1260

Fluid Mechanics

EU LER'S EQUATION OF MOTION

... 6.l

This is equation of motion in which the forces due to gravity and press ure arc take n il1m consideration. This is derived by considering the motion of a fluid cle ment along a s trea m- line as : Consider a strcam-lin
cylindrical clcrncllt of cross-section
(p + ~ dS) dA opposite 10 Ihe direction of flow.

3. Weight of clement pgdAds. Let

e is the angle between the d irect ion of flow and the line of action of the weig ht of clement.

The resultant force on the flu id c lement in the direction of clement X ae.::cleralion in the dire.::tion s.

pdA - [p +

= pdAds )( where

ap

~'

must be equal to the mass of fiuid

dS) dA _ pgdAds cos

d,I'

s

a

... (6 .2)


is the acceleration in the direction of s.


til'

Now

" . '" - . where v is a function of sa nd r.

,"

=

av ds + av a,~ dt at

I'dl' + ai' (.; a~· at

=

Id'" '"

I')

a,· = 0

",

If the flow is steady. -

I~I'

= --

it

,

"'

Substituting the value of II, in equa tion (6.2) and simpl ify· ing the equalioll. we gCl

ap

- -

as

pd,j'dA. -

pas

Forcl!S on

(b)
fluid l'Ieml'rll.

ap

ill·

pas

as

- - + gcos 9+1' - =0

a= ~ d,

lip
I I

Fig . 6 .1

"""',

a~·

But from Fig. 6.1 (b). We have eos

0'

(a )

a = pdA
dsdA - pg dAds '::OS

Dividing by

0'

d,'

or

lip

-

p

+ gdz + \"lit, = 0

dp - +gdz+"d,,=O p Equmioll (6.3) is known as EuJc(s equmion of mot ion.

.. .(6.3)

Ii

~ I

IL Dynamics of Fluid Flow

.. 6.4

261 1

BERNOULLI'S EQUATION fROM EULER'S EQUATION

Bernoulli's eq uati on is obta ined by integrating tlie Euler's equatio n of motion (6.3) as

f d; + Jgdz + J

I'dl' '"

constant

If flow is incompressible. p is oon,13n1 and I

\ .:

p

2

L + gz + )

I ,l

-'- + <+ P8 2g

'" constant

'" constant

l

-P + -v +z=constant

"'

pg

...(6--')

28

Equation (6.4) is a Bernoulli's equation in which

~ '" pressure energy per unit we ight of fluid o r pressure he~d. pg

,?-ng '" kinetic ene rgy per unit weig ht u r kinetic head. l '"

.. 6 . .5

p01cmial energy per unit weight or potential head .

ASSUMPTIONS

The following arc the assumptions made in th e dcri vati on o f Bernoulli' s equat io n : (il Th e fluid is ideal. i.e., viscos ity is zero (U) Tile flow is Meady (iiI) The flow is incompressible (iv) Tile fl ow is irrOlalional. Problem 6 .1 Water is flowing Ihrough a pipt! of 5 cm diameler under 0 pressllre of 29.43 Nkm ? (gouge) lind lI'ilh lI!ellll I'e/oeily of2.0 mA Filld Ille 100a/lleml o r rollIl ellergy per utilI weiglll oflllt! water m a cross-seclioll. which is 5 m IIbol't! Ihe dmum lille. Solution. Given: Diameler of pipe = 5 el11 = 0.5 m 2 Pressu re. p '" 29.43 Nkm '" 29.43 x v=2.0mfs VelocilY. Datum head. z =5 m Total head '" pressure head + kille ti c head + datum head Pressure head

:

~

:

pg 1,2

Kindic head

Total head

:

2,

:

29.43 x 10" =30 m 1000 x 9.81

2x2 2x9.81

{ p for wa ter =

IOO(L.~~} l m

= 0.204 III

: -p + -" + Z'" 30 + 0.204 + 5 '" 35.204 m. Ail S. pg 2g

Prob lem 6 .2 A pipe. 111f00'gh ... "ieh Wilie r isj/owill8. is /1(II'illg diameters. 20 cm (IIul10 cm allhe cross-seCliOIl S I alld 2 respeClil"e/y. The "elOCily of waler al secliull 1 i~' givell 4.0 mls. Filld Ihe veloCily IWlld II/ secliolls I alld 2 IIl1d also rale of discharge.

~ I

I~

~ I

IL 1262 Fluid Mechanics Solution. Given

D,,,, 20cm=O.2 m /flit A,,,,D, = - (.2) ! =O.031 4 m1

Area.

4 VI = 4.0 111/5

4

_

D,~20cm

V, ., 40 mlsec

D1 =0. 1 III

r

A l '" ..::. (.1)1 '" .00785 m 1

4

Fig. 6.2

(i) Vclocily head at section 1

=~= 4.0x 4.0 '" 0.815 m . Ans. 2g

2 x 9.8 1

(iiI Velocity hC~ld at .s«tion 2 = V/12g To rind VI_ apply cont inuity ~-quntion at 1 and 2 VI

VI

Velocity head at SCi:lion 2 '" ---.L = 2g

A,V,

.03 14

Al

.00785

= -- = ---

16.0 x 16.0 2x9.81

x 4 .0

= 16.0 m/s

'" 83.047 m. A il S.

:: A ,V, or A 1V2 '" 0.03 14 x 4.0 = 0. 1256 ml/s = 12S.6I1tNs/s. r\,,-~ . Problem 6.3 SIMI' Bem ol/IIi 's tileorem for steady flow of "n ;nco mpTHsib/e fluid. Deril'C (In c.
Pressure energy"

.1!.... pg

, ,,'

Kindie energy" -

2,

Datum e nergy" Z Thus mathematically, Bernuulli's theorem is wnnen as p

"t

- + - + z " Constant. pg 21( Derh'ution or lh-rn oulll 's theONIII. ror derivation o f Bernoull i's thL'Orem, Aniclcs 6.1 and 6.4 should bt;: written. Ass umptlons arc given in Article 6.5.

I I

Ii

~ I

IL Dynamics of Fluid Flow

263 1

Problem 6.4 Ti,e WIlier i,~ flowing (I/fougll a pipe IlaI'ing dhmreters 20 em and 10 em (1/ seCliolis f alld 2 respec /il'e/y. Tile rate of flow IhrOi,gh pipe h 35 Ii/fo/s. The seC!iOlI I is 6 m "bol'<' dlllum alii! iJ'eclioli 2 is./ In abo,'e dll/Ilm . Iftl", prnyu'e III sec~ lion I is 39.24 Nkm 1, find the ill tensify of preSSllre (II iJ'eclioli 2.

Solution. G i ven: A I sec/io n I.

D 1 ",20clll =O.2 m

Fig. 6.3

"

A, '" - (.2) '" .03 14111-' 4

P, '" 39.24 Nk m! '" 39.24 x 10 4 N/m '

z,=6.0 m AI secti on 2.

D2

:=

0. 10

111

"

4 (0. 1)" '" .00785

A, '" -

,

III

,

Zl=4 m po'" '1

Rat e o f n ow .

Now

Q" 35 li tis'"

.l2... = .035 m 3/s

1000 Q = A1V I = A1V,

v,=iL= A,

.035 = 1.114mls .031 4

V, = -Q.Q35 = - - - : 4.456 mIs A! .00785

and

Apply ing Bc mu ull i's equ ation al sc(;l ions I and 2. we gel

4

oc

39.24x 10 + (1.1 14 )1 + 6.0 ",,,,,;"PL',,,,, + (4.456)1 +4.0 IOOOx9.81 2 x9.81 l 000x9.81 2 x9.81

40+0 . 063 + 6.0=~+ 1.0 12+ 4.0 9810

oc

46.063 =....!!L + 5.0 12 9810

~ = 46.063 - 5.0 12 '" 41 .051 9810

Pl ", 41 .051 x98 1ONlm 1 '"

I I

4 1.05 1x9810

10'

Nlc m 1 '" 40.27 Nlcm 1. Ans.

Ii

~ I

IL 1264

Fluid Mechanics

Problem 6 .5 Warer i,~ flowing Illrough II pipe iIarillg diameter 300 mill alii/ 200 111m III Ille boftom and upper <'lid rcspecli\'e/y. The illtensi1), of pressure ar Ihe halTom t1nd i.! 24.525 Nkml and flw pressure a/ Ihe upper end is 9.81 Nlcm l. D elumine Ihe differe' lce in darum /wad if Ille mIl! of flow through pipe is 40 lilk

Solution.

Gi\'~n

:

Sect io n 1,

DI = 300 mm = 0.3 111 P I'" 24.525 Nlc rn ! '" 24.525 x 104 Nlm!

Sect io n 2,

D1 '" 200 111111 '" 0.2 m ,

40

004

'I

"'

m

P2 "9.81 Nlcnl

\ ill

Z,

'" 4 0 litis

Q " lOOCl " ' .

0 ," 200 mm

1

4'

Pl=9.8 1 Nkm-=9.81 x 10 Nfm-

Rate of n ow

I

D,"'300mm

.-

",

.__ .

2

P1 '" 24.525 Nlcm

DATUM LINE

S

Fig. 6.4

Now .04

.04

,,0.5658 mfs

V,"'-" - - " ~ D' 4 ' 0.566 m/s

AI

0::

v,,, _"'_ " ~.04",-_ " -

Al

..!:. (D,)l

-;,0",.04,-- _ 1.274 III/s ~(O.2)2

4 ' 4 Applying Bernou lli 's equation al secti ons ( I) and (2). we gel ~l

V!

EL+-'- +z, '" P! + ---.L+ z, pg2g

24.525 X 10

4

l000 x 9.81

+

.566x.566 2x9.81

pg 2 g "

+ ZI =

9.8 I xI0 · rOOOx9.81

+

{1.274)1 2x9.81

25 + .32 + ZI = 10 + 1.623 + Z2 25.32 + Zl = 11.623 + Zl Zl - Zl '" 25.32 - r 1.623 '" 13.697", 13.70 111 Differe nce in datu111 head = Z2 - ZI = 13.70 m . Ans. Problem 6 .6 The water is jlowillg Illrough a taper pipe of /e/l glh IOQ 1/1 IUII'ins diometers 60Q mm (lllhe upper elld ,md JOQ mm (II Ihe 10l<'er end. allile role of 50 lilres/s. The pipe lUIS a slope of I in 30. 2 Find Ihe press lire 01 Ihe lower elld if Ihe pressure III Ihe higller lerel is 19.62 Nkm • Solution. Given: u,ng th o f pipe, L= 100m Dia. at the upper e nd. D I '" (lOO 111111 '" 0.6 111

"' "'

Arc~.

1t

2

1t

,

AI= - DI = - x ( .6) -

4

4

= 0 .2827 111 2 c'" PI'" pressure at upper end 0 ~.yP./= 19.62 N/cI11-,

'

I I

Fig . 6.5

Ii

~ I

IL 265 1

Dynamics of Fluid Flow 4 '" 19.62 X 10 N/m2

Di a. at lo we r e nd ,

D2 '" 300 111m '" 0.3 111

: . Arca.

A , '" - D,' '" - (.3) - '" 0 .07068 • 4 • 4

/t

,

I(

,

Q = ralC o f now = 50 lit rcsls = Ld

lh ~

datum Ii Jlc

III

..22.... '" 0 .05 m31s 1()()()

th ro ugh (h e ce nl re o f the lowe r end.

p a~s

T hen

Zl = 0

As slope is 1 in 30 mean s

ZI'" -

Also we know

Q=A 1V1=A1V1

1

30

10

)( HlO = -

3

III

VI = Q = 0.05 = 0 . 176& m/sec = 0. 177 Ill/s A .2827

V, '" .Q. =

ami

+

Al

~ .07068

= 0.707 4 m/sec = 0.70 7 Ill/s

Apply ing Be rno ulli 's eq uati on a1 scc li o lls ( 1) and (2) , we gel

Pl

-

,

V,

+-

pg2g

"' 0'

,

P' V, +ZI = - ' + - '- + ,. pg

2g

+

19.62 )(10' .177l 10 Pl .707 2 +0 + + - "' - + looox9.8 1 2 x 9.81 3 pg 2x9.81 20 + 0.00 1596 + 3.334 '" Pl + 0.0254 pg 23.335 - 0,0254 = -;;:"" -;- PL'"'"

looox9.8 1 Ill '" 23.3 x 98 10 Nlm 1 '" 228573 N/m 1 = 22.857 N/cm 1. Ans .

or

... 6 .6

BERNOULLI 'S EQUATION FOR REAL FLUID

The Be rno ulli 's equat io n was de ri ved o n th e assumpt io n thm fluid is invis.:id (non-\' is,",ous) and the refore fricti o nl ess. But a ll the rea ll1uids Jre viscous and he n,",e o ffer resistan ce to now. Thus there arc alwa>'s so mc losses in fl uid flows and hence in the applic ati o n of Bc rno ulli' s equati o n, thcse losses hJ\'e to be ta ~ e n into conside ration. Thus the Bc rn oulli 's equ ati o n (o r re al flu ids betwee n points I and

2 is g il'c n 3S

,

,

..b.+~ + Z l = P2 + ~ pg 2g pg 2g

where

I I

+Z2 + !JL

.,.(6.5)

"L is loss of ene rgy betwee n points I and 2,

Ii

~ I

IL 1266

Fluid Mechanics

Problem 6.7 A pipe of diameter 400,mm carries water,III (/ I'elocity of 25 mls. The pre.~,~ure.f lU Ihe points A and B are gil'ell as 29.43 Nkm- "lilt 21.563 Nk",- respec/iI-ely ...hile IIII' da/lun /,e(l(/ {/( A and B "re 28 m lind 30 m. Find I/,e IOH' of head ber".een A lind B. Solution. Given:

Dia. of pipe.

D = 400 mm '" 0.4

Vclocily.

V=25mfs

At point A ,

111

4

<'~

I'll '" I' '"

25

,

~9""'~

ZB

_ _ _ _D o ATuM l LINE

=

fig. 6.6

29.43 x [ O~ 25' + + 28 IOOOx 9.81 2x9.8 1

'" 30 + 3 1.85 + 28 '" 89.85 111 PH'" 22.563 Nlcm' '" 22563 x 104 Nlm l zo",30m \'/1 '" I ' '"

Total Crlcrgy at B,

~?l

""?~-~

Ill/s

Total energy at A.

At Ilo ini B .

•

"" = 29.43 Nlcm 2 :: 29.43 X 10 N/m 2 ~ z" = 28 III 1:P ((\(I'I~\C~

V" '" 25 mls

P

1,2

P8

2g

Eo= ....!. + ....!. + ZB 4

'" 22.563xt0 + 25' +30=23 + 3 1.85+30=84.85m JOOOx9.8 1 2x9.8 1 Loss of e nergy

= Ell - EIJ= 89.85 - 84.85 '" 5.0 m. Ans.

Problem 6.8 A conical lube of lengrh 2.0 m is fixed I'erlicallv wirh ils smaller end upwards. The ~'elociry offlow ar fhe smaller elld is 5 mls wllile at rhe lower end if is 2 mls. The pressure head ar fhe l'mllller<'IId is 2.5111 ofliqw·d. The loss of liead ill file IIIbe

i,~ 0.35(\,/ -i'll: . wllere 1'1 is file I'docify at 2,

Ille SlIIlIlIer end lIlid v: {II lile lower end relpeCfil·dy. D ef<'fmim: Iile pfeSJ'llfe ilead af Iile lo ..... er end. Flow lakes pl(lct' in lile dOIl'nW(lrd direcfion.

Solution. LCllhc smalle r clld is rcprcsclllcd by (I) alld lowcrcild by (2) Givell : U:llgth of lube. L=2.0m 1' 1" 5 nils P1/pg = 2.5 m of liquid \'1 '" 2 mls Loss of head

'------+1-----' ® Fi g. 6.7

I I

Ii

~ I

IL Dynamics of Fluid Flow

0.35 ]5 - 2]'

"

28

0.35 x 9

'"

2 x9.8 1

267 1

:=U. 16 m

{/, "'?

Pressure he ad.

PH App ly ing Be rno ulli 's equ ation at secti ons ( I) and (2). we gel

", lg

,.2 pg 2 g - -

12. + _1 + z, ,,, Pl + -'- + z, + h/

pg

Le t the datu m li ne passes th ro ugh see ti o ll (2). Theil Z2 '" 0 , Zj '" 2.0

5'

2.5+~',-,,, + 2.0= 2 x 9.8 1

2.5 + 1.27 + 2.0 '"

Problem 6.9

P1

pg

2'

+~o",, + O+O. 1 6 2)( 9.81

l!.l.. + 0.203 + .16 pg

p, - - '" (2 .5 + 1.27 + 2.0) - (.203 + . 16) pg '" 5.77 - .363 '" 5.407 III of nuid . Ans. A flipclille ca r,-yillg oil of .~pedjic gral'iry 0.87, chatlge,~ ill iliameler from zoo mm

dioll!efer (I{ (/ 11O.lllioll A /0 500 mil! diameter at a po.filioll B which is 4 melres at a hig/IeT lel'el. If lile fI' f'Swres at A all
Solution . Disc hargc, Sp. gr. of oil

Q = 200 li tis = 0.2 m 31s = 0.87 kg

p for oil

=.87x 1000= 870 - ,

T

m

=200 111111 =0.2 m

Givcn: AI s l'(! lion A .

DII

Area.

All =

'm

~ (D~ ) 2 = ~ (.2)2 4

4

= 0.03 14 m 2

PA = 9.S1 Nlcm" =9.Sl x 1 0~N/m l Fig. 6.8

If d atum li ne is passi ng throu gh A, th en ZII = 0 VII =

JL = ~ = 6 .369 Illis AA

0.031 4

At s rtliun fl ,

DIJ = 500 mm = 0.50

Arc ~ .

AB= "4DB ="4 ( .5>-=0.1 963 111 -

If

l

/!

III

'

,

PB = 5.886 Nlcm-, = 5.886 x 10" Nlrn -

I I

Ii

~ I

IL 1268 Fluid Mechanics 2/1 '" 4.0

III

Q

02

Arc ~

.1963

vB = - - " - - = 1.018111/5 V' =E~=&+.....:1.. + Z... pg 28

Total e nergy al A

'"

Total e ne rgy at B

9.8 1)( 10 '

870x9.8 1

+

(6.369)1

2x9.8 1

p~

vi

pg

2g

+ 0= [1.49 + 2.067: 13.S57 m

=E8 = - + - + z,

=

5.886 x 10"

+

(1.0 18):

+ 4.0 = 6.896 + 0.052 + 4.0 = 10.948 m 870x9.81 2x9.8 1 (i) Directio n a rn ow. As Ell is more Ihan E8 and hence flow is taking place from A to 8. An .•. (ii) Losso fh ead =IIL=E,, - E 8 = 13.557 - IO.943 = 2.609rn . An s•

.. 6.7

PRACTICAL APPLICATIONS OF BERNOULLI 'S EQUATION

Bernoulli' s equatio n is appli ed in all problems of incomprcssi bk fluid now where ene rgy cons;(I erations are in vo lve d. But we s hall consider its applicat ion 10 the follow ing measuring devices: l. Vemurimctcr. 2. Orifice meter. 3 . Pilot-lube. 6 . 7 . 1 Venturi meter. /I. vc nturimctcr is a device used for measuring the rate of a flow o f a fluid flowing through a pipe. It co nsists of three pans: (i) A s hort converging parI. (ii) Throat. and ( iii) Di ve rging pan. II is hascd on the Principle of Bernoulli' s eq uati on.

Expression for ute of flow through ve nturi meter Consider ~ vc murim eter fllted in a hori zoillal pipe through whil:h a fluid is flowing (say water). show n in Fig. 6.9. Let til = diame ter at in le t or at SCl:tion ( I). PI = pressure at section ( I) 1'1 = ve locit y of fluid at sec tion ( 1).


1

_ T

- '.

" "I

. a == area at section (I ) == 1. 4 and tl2 _ P 2' "2' £l2 arc correspondin g va lu es at secti on (2). Apply ing Berno ulli' s equation at sect ions ( I ) and (2). we get

,

HROAT

,

F Ig. 6.9

Venlur;meUr.

.£!.. + s... + ZI== P2 + ~+Zl pg

pg

2g

As pipe is horizonwi. he nt"e ZI == PI

2 1'1

2g

Zl

Pl

vi

- +- " - +pg 2g pg 2g

I I

or

2 P I ~ PI _ I'; _ 1'1

.

-

pg

2g

28

Ii

~ I

IL Dynamics of Fluid Flow BUI

PI-P2 pg

is Ihe difference of prc:;sure head s m scc~ iulls I and 2 and it is equal 10 " or

269 1

Pl - P2 pg

== II

S ub~ti!uling Ihi~ val ue of PI - PI in Ih e abo ve equation. we gel pg

,,2

/

2g

28

11= -'- -'

..,(6.6)

Now appl ying cont inuit y equatio n al sec tion s I and 2 (I,V

I

vl= ~

Substituting this value of 1'1 in equatio n (6.6)

" [";:'r

" = -' 2g

.[ '] vi'["d,· -, (/; ]

"i "--';f::"-28 2g

1 _ (j~ (/,-

=

28

(Ii

x

~2 gh

"'

Discharge.

Q '" ==

" 2" 2 <1 2

II I

Ja~ ai

)(

~2gh"

(I I":

Ja~

a;

...(6.7)

Equation (6.7 ) gives lh e discharge ullder ideal condition s and is ca ll ed . theoretic al disdtargc. Act ual discharge will be less than th eoretical discharge.

Q= Cx "'"

where

d

"I'"

I'

'lU i

7

,

ai

x

2,,, .J2ih

...(6.8)

Cd == Co -efficie nt o f vc nlurimdcr and irs va lu e is less than I.

Value of ' h' ,iven by differential U· tube milnometer Case I. Let the differential throu gh thc pipe. Lct

manom~ te r

contai ns a liquid which is heavier than

th~

liquid flowing

Sp. gra vi ty of thc ht.> a l·ier liquid So '" Sp. gravi ty of the liquid flowing throu gh pipe .\" '" Differe nce o f th e hcavier liquid column in U-tubc

S h :=

T hen

"=.,[,,s" -,]

...(6 .9)

Cllse II. If the dilTerc nt ial manometer contai ns a liquid which is lig hter than th e liquid nowing through the pipe. th e value of II is give n by

~ I

I~

~ I

IL 1270 Fluid Mechanics ... (6. 10) whe re

5, = Sr. gr. of lighter liquid in U·1Ube

So = Sp. gr. of fluid flowi ng through pipe .r = D ifference o f Ihe lighter liquid columns in U -lube. Case III . Incl ined Ventu r lrn eler wi th mrr.,renilu l U. lu be mu norn" le r. The above two cas.cs are given for a horizunlal \'CnmrirnClcr. This case is rdated \1.1 inclined vcnturimClcr having d iffcrcmial

V-lube Inanometer. LClth., diffcrcIltialllHlrlOIllCler contains heavier liquid [hell" is given as

,,= (;~ + ll) - (~ + Zl)

= .r

[~: -I]

... (6.11)

Case IV. Similarly. for inclined \,cnlurimeli:r in which differential IllallomCler contains a liquid which is lighter Ihan lhe liquid flowing through Ihe pipe. lite value of II is given as

II = (;~ +Z, ) - (~; + Zl) =x [I-

:J

... (6. 12)

Problem 6.10 A horizolltal relllllrimelf'r wilh inlet and Il"oat diameters 30 em and 15 em respectinny is I/Sed to mellsure Ihe flow of 'WlIer. Ti,e rellding of differenti,,' manomeler conllecled to II,e inlet and Ihe rhrom ;.1 20 em of mercury. Dererlllille rile flUe of flow. Ta/(e Cd '" 0.98. S olution. Given: Dia. at inlet. Area at inlet,

a,

, = ~/f
4 d 2 = 15cm

Dia. at throat.

4

,

,

(30)" = 706.85 em-

a,,,,~xI51"'176.7cm2

.

4

Cd '" 0.98 Reading of different ial manometer'" of '" 20 cm of mcrcury. Difference of pressure head is given by (6.9)

where Sh = Sp. gral'ity of me rcu ry'" 13.6. So '" Sp. gral'ity o f water =

1

=20 [ -136 , - -I

=20 x 12.6cm = 252.0 em ofwatcr.

The discha rge through \'cnturimcter is givcn by cqn. (6.8)

Q=C

II , a,

, J'a ,

'" 0.98 x

-

,x

.Jfiii 28/'

"2

706.85 x 176.7

x ~2 x 9.& 1 x 252

J(706.85)2 _( 176.7)2

~ I

I~

~ I

IL Dynamics of Fluid Flow &606759336

=~~~= '/499636.9 3 1222.9 '" 125756

C1Il

3

/s '"

1~:6

271 1

86067 593.36

684.4

litIs", 125.756 litis. An s.

Problem 6.11 An oil of sp. gr. 0.8 is flowing !hrougll (I l'en7llrillleler /1(I)';n8 illier diameter 20 em and IhrOal diameter /0 em. Th e oi/·mere",)' differential mOllomeler sllOWS ( 1 reading of 25 em. CalcuIme lite discharge of ai/through IIII' horizol1lal 1'l'lIlUrimeler. Toke Cd '" 0.98. Solution. Given: Sr . gr. of oil. So '" 0.8 s~ " 13.6 R eadin g of differential manometer. x:= 25 em

Sp. gr. of merc ury.

DifJcrcncc of pressure head. II"

.f [

~:

I]

-

1

13.6 :25 [ - -I cm o f o il =25 [! 7 - 1J = 400c lll ofoi l. 0.8 Dia. nt inl et.

rr

,

,

"4 x 20" '" 31 4 .16 c m d 2 '" IQem

"1 '" ~ X 102 ", 78.54 cm 1 4 CJ = 0.98 Th e discharg.-:

Q is g i wn by Cq U31io rl (6.8) Q=C

(/ 1(/,

dJ

2 1 " I - U,

'" 0.9& x

~I

Xv"'!;"

314.16 x 78.54

Jc.~ 14.1 6)1

2 1421375.68

J98696

6 168

x

J2 X 98 1 x 400

(785 4 )1 1/

2 1421375.68 "

3().1

(; 111

S

" 70465 CI111/S = 70.465 1itn' 'iI.~. An s. Problem 6.12 A horiZOl1tlll l"el1turimeter .... ilh il1lel di","eler 20 e ll! allli Ihroat diameter 10 em is used to ",e"sure Ihe flow of oil of Sf!. gr. 0.8. TIl e discharge of oi/through "e1lluri",eter is 60 litres/s. Fil1d 'he re(U/ili g of Ille oil·ml'rCliry diJferellfill1 mO/Jometer. Tuke CJ = 0.98. Solution . Gi vc n :
rr

,

,

", = - 20- = 31 4.1 6 cm4
I I

Ii

~ I

IL 1272 Fluid Mechanics a, '" ~ x 102 ", 78.54 em! •

4

Cd == 0.98 Q = 60 IilTCsls = 60

x

IQOO .::rn 1ls

Using the equati on (6.8).

60

"'

314.16x78.54 x 1000 = 9 .81 x c.;;;;;;:;:,;~~..". x .J2 x 981 x II

J(314.l6)~ _(78.54)2

=

I07 1068.78Jh

304

IT 304 x 60000 "h = 107 1068.78 = 17.029

"'

II '" (17.029)" = 289.98 nn of oil

B~

II= '{~: -I]

wh ere ::i/o = Sp. gr. of me rcury = 13.6 50 = Sr. gr. o f oil = 0.8 .f == Reading of manometer 239.9S=x [ 13.6_ 1] = 16.<

0.8

289,9 8 X= - - -

16

= 18. 12 em.

Rcadin g o f oil-mercury differential manom eter", 18.12 em . Ans. Problem 6.13 A IlOriWl!lai I'cl1lUrimelcr )I"rlll ililet diamerer 20 em alld Ihro(l/ diameter 10 em is u!led 10 measure 'he jlow of w(I/er. The pressure (1/ ;,,/el is /7.658 Niem i alld the "('euum pressu re (1/ the IhrO(l/ is JO em a/mercury_ Fi" d the disclwrge ojlt/(lier through ",mlllri",cler. Take Cd '" 0.98. Solution. Gh'cn : Dia. :II inlet. al Dia. at throat.

,

'" -

.

,

x (20t '" 3 14. 16 em"

4 d1", IO em

.

"

a, '" - x 10' '" 7&.74 em ' . 4

PI'" 17.6$8 Nfem z '" 17.658 x 104 Nfm 1 p for wate r

'" 10
[.;: ~

m

and

p

17.658 X 10·

pg

9.81 x 1000

_, '"

'" 18 1ll of wata

p, - - '" - 30 cm of mercury pg '" - 0.30 m of lI1~rcu ry '" - 0.30 )( 13.6'" - 4.08 m of

I I

wat~r

Ii

~ I

IL Dynamics of Fluid Flow

= II = Ji _ P, '" 18-(-4.08) pg pg '" 18 + 4.08 '" 22.08 III of w ater

DilTcrc nt ial head

'"

2208

273 1

em o f wat<'r

The di!\Chargc Q is given by eq uation (6.8)

'" 0.98 x

3 14. 16 x 78.54 x .jr"C,"9ii'''IC,C2n2'''0'' J(3 14.1 6) 1 - (78.74)'

'" 50328837.2 1 x 165555 cm 3/s

=:

165.555 litis. Ans .

304 Problem 6.14

The ;II/el 'IIld I/"oal diameters of II horiWllla/ l'"n(urimt'ler are 30 em ,,,,d 10 em respeClil'('ly. Tile liquid flowing l11rollgll Ihe meIer is water. The flrCn"llft' i,llen)"ily at i" lel i.~

f 3.734 Nkm l wllile I/'e mel'"m pnrimeler.

Solution. Given: Dia. at in let.

"

", '" - (30r = 706.85 cm-, 4

Dia. at throat.

d 2 =lOcm

"

"1 =="4 (10) == 78.54 eln-' Pressure. Pressure head.

P, = 13.734 Nfcm 1 = 13.714 x 104 N/m 2 p --.!..

pg

p,

~

pg

==

13.714 x 10' 'C2;:':'ii'c = 14 m of water IOOOx9.8 1

,,- 37 em o f mercury = -37 x 13.6 100

Differentia l head.

In

of water" - 5. 032 m of water

II '" !'I/pg - pipg = 14.0 - (- 5.032) " 14.0 + 5.032

== 19.032 m uf w ater = 1903.2 em Head lost.

4 1I/ =4%ofI1= 100 x 19.032=0.1613 m

Cd "

~h-hl

"

I I

==

19.032

.76 13

19.032

== 0.98

Ii

~ I

IL 1274

Fluid Mechanics

Discharge

'" 0.98 x 706.85 x 78.54 x ./2 x 98 1x 1903.2 J(706.85) ' -(78.54f [05[32247.8 3 " ",;::~"'~~'" '" 149692.8 cm /s '" 0.1 4969 mJ/s. ,\IIS. ,1499636.9 6168

PROBLEMS ON INCLINED VENTURIMETER Problem 6 .15 A 30 em x 15 em )'emurimeler is iliseY/cd i/l {l >'crlic,,/ pipe nlfryillg "'(lIer, flo K'illg in Ihe upward direc/iOlL A different;,,/ mI'TCIl'Y mU/lomder C01mccll'd /0 'he inlel (md IhTool girl'S II Teadillg 0[ 20 em. Find Ihe discharge. Take Cd := 0.98. Solution. Given: Di3. at inlet. d,=30cm

{/, = !:. (30)2 = 706.85 Dia. at th roat.

cm 2

4 d l = 15cm

"

a,= - ( 15)-= 176.7cl11

-

4

,

[.s~ -1]= 20 [13.6 _LO] = 20x 12.6 = 252 .0 em of wate r

II "' .r

So

1.0

Cd = 0.98 Discharge,

,

Q= C

(1 , (/, /,

"a

l

, x ,/28/' {/ l

706.85 x 176.7

= 0.98 x [:~'§¥~~~ x ./2 x 981 x 252 ~(706.85)1 (176.7 )1 8606759336

" rtiiii~~ ./499636.3 31222.9 "

86067593.36 684.4

= 125756 cm 3 1s = 125.756 litis. Am. Problem 6.16 A 20 CIII X 10 cm renlurimel/!f is i"sel·'ed in a rerlin,l pipe earning oil ofsp. gr. 0.8. Ib e flo .... of oil is in "fI'nm/ direClion. Tbe tlijJerenn! of /el"e/s hel"·een Ille II'TOal and in/el J·eclion iJ· 50 cm . The oil merc"ry differenlial manometer Ril"es a reading of )0 em of mercury. Find li'I! tlisc/wrge of oil. Neglectlon·es. Solution. Dia. at inlet. d , = 20 em

"

a , = - (20) - = 3 14. 16cm Dia. at throat .

I I

4 d z =lOem

,

Ii

~ I

IL 275 1

Dynamics of Fluid Flow {/! '"

Sp. gr. of oil. Sp. gr. o f mercury.

'4" (10)' '" 7854 em ,

So '" 0.8

S~ = 13.6 Differential mano met er reading. x '" 30 e rn

=30

13.6 [ ~0.8

1

1

=30(17 - 1]= 30x 16 =480crn of oil

T he discharge.

'" '"

1.0 x 3 14.1 6 x 78.54

)( J2 x 98 1 )(480 c m3/s

J(3 14.16)1 - (7854 )1

23932630.7 J • = 78725.75 em Is", 78.725 llt resls. Ans. 304

Problem 6.17 I" (I "erlic,,/ pipe cOlll'eying oil of specific 8m)'il), 0.8, Iwo pressure gllllge:; hU\"I: been illS/ailed (II A and n where 1/", diameters are 16 ClIIllnd 8 em respectire/y. A is 2 melres abo",: 8. The preSJure glwge readings 1/(/\'/: s/loll'n Illallhe pres.illTe al B is greater t!ian III A by 0.98/ Nlem", Neglectillg all losses. cu/cu/a/e I/!e flow f(!le. Iflhe glluges (II A alUi B aTe rep/aced by lUbes filled wilh Ihe Jllllle liquid (//1(1 collnec/ed 10 a lj·lllbe COIIWilling mercllry. ClI/cli/aie Ih,. difference lIf lel'eI of mercllry ill Ihe two limbs oflhe U-lIIb,.. Solution. Given: Sp. gr. of oil. S,, '" O.S

Density.

P '" O.S x 1000 '" SOO DA

Dia. at A.

",

k.~

16cm

m

16 cm",0. 16m

-'

®

AI=~(.16)''''0.020Im'

Area at A. Dia. at 8.

D IJ '"

8 e m'" 0.08

(i) Differe nce of pressures. PR - p", '" O.9S1 Nlc m'

'" 0.98 1 x

Ht

9SlON Nlm ' '" - ,-

®

f-t-LG\ , ~ _I

m

Fig. 6.9 (,,)

Differe nce of pressure head

p

p 8 -

pg

I I

'm

m

A, '" -" (.OS) - '" 0.005026 m , • 4

Area at 8.

t--+----f'-"':)

A '"

98 10 ;;;:""~7 '" 1.25 800 x 9.8 1

Ii

~ I

IL 1276 Fluid Mechanics Applying Bernoull i's th eore m at A and B and takin g the refere nce line passing through sec tion B. we gel

PA pg

PB. P,

- - - -ZA~Z8=

P"'p~PB ( ]

vi v;

- - 2g 2g

v'v'

+2.0-0.0= 2; - 2;

v"

Vl

2g

2g

-1.25+2.0:....!..- --
... ( i)

Now appl ying conti nuity equation at A and B. we gel

VA xAI = V8 xA!

L

! ®

0'

/

, Substituting the va lu e o f VI! in ctjualion (i). we gel

®

0.75 = L6V,;' _ V; " 15V;

28 V" '" Rate of nuw.

28

2g

O.75X2X9.8 1

J

15

_ '" 0.99 m/s

,

J

,

T

.L

Fig. 6.9 (h)

Q=V... XA 1

'" 0.99 xO.OlOI '" 0.01 989 m'/s. Ail S. (ii) DifFcrcn;;c of le ve l of mercury in the U-lu be. u,1 II "" Differe nce of mercury level.

h=X(~:-I)

Then

where

""

"'"

(~; + Z... J- (~; + 2 8) = ,p"'Cp-C8'P~8 + Z... -

Z/I

=-1.25+2.0-0

PB - P... ==

" 0.75

pg

1

1.25]

136 - - 1 =xx 16 0.75=_1' [ -

0'

x'" 0.75 '" O.(l4687

III

= 4.687 e m . Ail S.

16

I I

Ii

~ I

IL Dynamics of Fluid Flow

277 1

Problem 6.18 Find Ihe diRharge of wmer flowing II/rough a pipe 30 em diameter placed in WI inclilled POSilioll where a )'{mll/rime/a is itlserled. /wI'ing a th roat diameter of 15 CIII. Ti,e difference of pressure belween lilt: main mid Ihroar is mcas" red by Ii liqu it! of sp. gr. 0.6 ill an jm'a/ed U -/" be which girl'S {/ , e(liIing of 30 Clil. The lo.H a/head be/ween Ihe lII{1illlllU/IIIfQa/ is 0.1 limes Ille killerle head of Ih e pipe.

Solution.

Dia. at ink1.

til'" 30 I:In

" (30)

(/1'" -

4 Dia. al throat.

= 706.85 em

,

II

til'" 15cm

"

a,= - { 15t= 176.7cl11',

,

4

Rc~ding of differenti al manometer. x = 30 em Difkrcnce o f pressure head. II is give" by

(~~ + Zl ) - ( ~; +Zl)=11 A lso

where 51 " 0.6 and So " 1.0

= 30 [ 1- 0.6 ] = JOx.4 = 12.0c1I1 ofwa!cr 1.0

Fig. 6.10

"

Loss of head. ilL = 0.2 x kinetic head of pipe = 0.2 x - '

2g

Now app lying Bernoulli' s equation at sections (I) and (2). we get

(:~ +'1) - (~ + Zl) +;: -;~ 0"

(:~ + Zl ) - (~ + Zl) =11= 12.0cm o f wmcr

and \'1"

12.0+ -

28

12.0 + 0.8

~2g

I I

, .;

- -'

2g

1

=0.2 X VI

,.; .0 2g

28 ... (1 )

Ii

~ I

IL 1278 Fluid Mech a nics Apply ing continuit y eq uation a1 sect ions (I) and (2), we gel (1\,,\ '"

al". tt

,

4 (15) ~l If

4

=~ 4

(3of

Substituting tliis valu e of 1', in eq uation (I ) , we gel

1'; ["' 16- 11== 0

v; '" 0 28 4 - 2~

12.0 + 0.8 (",)"

or 12.0 + 2~

-";'- [.05 - 1[ =- 12.0 or 0.951'i == 12.0

2,

2,

2x98 1x12.0

0.95 Disclia rgc

'" 157.4 emfs

= a 2 "l '" 176.7 x 157.4 cm 31S:o 27800 cml/s

= 27.8 lil rt."Sls. Ail S. Problem 6.19 A 301.'111 x 15 em \'tmmrimeler is pro l'ided in a )'eflim/ pipe lille carrying oil of specific grlll·ily 0.9, Ihe flow being upwurds. The difference in eiel'alion of the I/Iroa/ sec/ion and e/!lrance sectioll of the I'emurimeler is 30 1.'111. The differential U-lIIbe mercury mallometer shows a gauge deflectioll of 25 cm. Calculate: (i) Ille di.~c1/arge of oil. lind (ii) the pressure differellce between rile elllrance secriol1 and rile rlima! secriol1. Take rlie cQ-efficiellr of discliarge lIS 0.98 (mu ~·pecific gra,·iry of mercury as 13.6. Solution. G iven: Dia. at inl et. ti, = 30 em

Area. Dia. at throat. Area.

a , = ..:: (30)2 = 706.85 c m 2

4

d 2 = 15cm

tt

,

a,= - (15) = 176.7 em 2

,

4

u.,t section ( I) rcprescil is inle t and sect io n (2) represents throat. Then Z2 - Z, = 30 cm

S" = 0.9 S~ = 13.6 Read ing of diff. manometer. .l = 25 cm The differen tial head. II is g ivcn by

Sr· gr. of oil. Sr. g r. o f mercury.

/,=

(.E.!. +ZI) - (!!.pgJ. . +Z1) pg

=X

[' 1 [".6 1 ~- 1

So

I I

=25 - - - I = 352.77 Clll o f oi l 0.9

Ii

~ I

IL 279 1

Dynamics of Fluid Flow (i) Th e discharge. Q of oil

0.98)(706.85)( 176.7 12 98 3' '" ~(706.85) =" x Ix 5_.77 1 _ (176.7)l

'" 101832219.9", 148790.5 cl11 3/s 684.4

= 148.79 lilrl's/s. AilS. (ii) Pressure diffucnce betwee n en trance and throat seeli on

I,: (ll.+ ZI)-(l!l.pg +Zl) =352.77 pg

1

!/

, ,

!
"'

"m

r"O

li _ Pl)_ 30 '" 352.77 ( pg pg

, ~

~

ttf Fig. 6.11

.Ii _.!!.J... '" 352.77 + 30 '" 382.77 pg

pg

(PI - Pl) '" 3.8277 )(

ern of oi l = 3.8277 m or oil. AilS.

pg

= Sp. gr. o f o il x 1000 kg/Ill l

But dt nsi ty of o il

'" 0.9 )( 1000 = 900 kg/em' N (PI - p,) '" 3.8277 )( 900 x 9.8 1 - , -

111 -

33795 1 1 : -, - Nlc m '" 3.J7~ Ntcm . AilS. 10 Problem 6.20 Crude oil of specific gral-it)' 0 .85 flo .... s up ..... ards ar a roll/me rale offloll' 0[60 lilre per second II"Of/gil a I'err;cal I'ellwrimeter .... illI lUi inlel diameler of zoo mm and a Ihroal diameter of 100 mm. Tile co-efficielll of di.~c1w'ge of Ille rel/llfrimeler is 0.98. The I'ulical di.ltance belween Ihe pre.~.mre tappings is J(}() 111m. (i) If Iwo preJ'sure gauges are cunnee led 01 Ihe lappings sIIch Ilwl liley are posilioned 01 'h e (el'els of Iheir corre,"pondillg lapp;'lg poinls, determine Ihe difference of readings in Nkm 1 of Ihe two pfeSJ'ure glmges. Iii) If a men""y differelilial m(lIiomeler is COlinected. in fllace uf preSJ'ure gauges. 10 tile tappingJ' J'uch Ihal Iile cOli/leCling lube IIfllO merc'lf)" are filled wilh oil. delermine Ihe dif/erenCl' iii 111<' level of 'he mercury CO/Willi. Solution. Gi ven: Specific gra vi ty of oil.

I I

Ii

~ I

IL 1280

Fluid Mechanics

Densit y. Discha rg",

p - 0.85 x 1000 850 ~ g/Ill J Q - 60 litrels 60 , · - - =O.06 m /s 1000

Inle t dia.

tI, '" 200

a, "' "4" (.2)

Arca. Throat d ia .•

0.2

lllill '"

I ~

IJ

III

'" O.03 14m

,

til'" 100 mill '" a. lm

4"

a, = - (0. 1) '" 0.00785

Area.

•

Ill ''

,

JOO mm

..L

Val ue of Cd

'" 0.98 Let section ( 1) represents in let and sect ion (2) re presents throa t. Then

I (j)

!

(il Difference of readings in Nfc ml of tile /h'O prl'SSlIre gauges

The discharge Q is given by.

,~

(/,(/,. '

(I,

0'

-

, (11

x

.Jfih 2,1,

0.06 = 0.98 x 0.Q314 x 0.00785 x

~O.03 1 4 z _ 0.00785 '

'" 0.98

\=

I, 1

Zl- Z, = 300 111111 = 0.3 In

Q=C

--

~I

!

T, ~

...

F ig. 6. 11 (a)

..j'2c,",".8"'c,c,',

x 0.00024649 x 4.429 ,fh 0.0304

.[h =

O.U6 x 0.0304 = 1.705 0.98 x 0.00024649 x 4.429

II'" 1.7051

BIH for a venica ] vC llturimc tcr. II =

2.908",

(;~

= 2.908 m

+:,) -(;;

+ l, )

(li zl)_(fi Zl) . (l!J.. _li) pg

+

pg

+

pg

pg

+ z, -

Zl

= 3.208 III of oil p, - Pl =pgx3.208 '" 350 x 9.8 1 x 3.208 N /m ~ '" 850 x 9.8~ x 3.208 N/c m ~

10

I I

Ii

~ I

IL Dynamics of Fluid Flow (ii) Difference

ill

rhe /el'el.l of mereul)'

The v alue of 10 is g i ven by.

COIIIIllIl ,\'

281 1

(i.e .. x)

iI=_r [~: -I]

2.908:x[ [3.6 - I] :x I1 6 - 1)= I S.l 0.85

2.908

.r '" - - '" 0. 1938 m '" 19.JS em of 011. Ans.

IS

Problem 6.21 III II 100 mm diameter iJoriZO/lfa/ pipe (/ l'elllUrimeler 0/0.5 cOlltractioll ratio /1
Solution. Gi ven Dia. of pi pe.

il, '" 100 mm '" 10 em

Area, Dia. at throat.

il1 '" 0.5 X il, '" 0.5)( 10 '" S em

Area,

"

(/, '" - (S) '" 19.635 em

•

4

,

H ead of water for no flow

'" i!..L '" 3 m (ga uge ) '" 3 + 10.3", 13.3 In (abs.) p,

Throat pressure he ad

p, = - ' = 2 III of wa~er absolute. pg

-. P, p, DI !fere nec of press ure lIe ad. II '" - - - ' : 13.3 - 2.0: 1 1.3 m : 1130 e m

pg

Ra~cofflow.

pg

Q is~iven by Q= Cd :0.97)(

78.54 )(1 9.635

xJ2x98 1x ll 30

J(7&54)' - (19.635)' :

22273 18.17 76

3

•

: 29306.8 cm Is : 29.3Ofi hlres/s. Ans.

6. 7.2 Orifice Meter or Orifice Plate. It is a device u.',cd for measuring thc rak of flow of a flu id tllrough a pipe. It is a cheaper devicc as co mpared to I'cnturimd cr. It also worb on the Same pri ncip le as that of vcnturimetc r. It co nsists of a flat circular plate which has a ci rcular sharp edged hole ca ll ed orifice. which is coocen tri c with the pipe. Tile orifice diamder is kept ge nerally 0.5 times the diameter of Ihe pipe. though il Ill~y vary from 0.4 10 0.8 tim es th e pipe dimncter. A differemial manomete r is connected Jt stttion (I). which is at a distan"" of aboul 1.5 to 2.0 tim<,s the pipe diameter upstream from the orifl"e plate. and at section (2). whid is at a distance of about half the diameter of Ihe orifl"e on the downstream side from Ihe orifice plate.

I I

Ii

~ I

IL 1282 Fluid Mechanics Let PI" pressure at section (I). VI" velocity at section (I). U("

area of pi pc al section ( II, and


PIPE

OR IFICE METER

12>

--

DIRECTION OF FLOW

.i

,

.-- D IFFERENTIALMANOMETER

T

Fig. 6.12. Orifiu m('ler. Pl>

)'1' "1 arc corresponding values al section (2). Applying Bernoulli 's equation al sections (I) and

{21, we gel 1 PIl'I

- +pg 28

'I

pg

28

+',

(li+ZI)-(li+Zl) 28"i _ ~28 (Plpg ll) '" h" ( lpgi + pg

3m

+

!

p, ", == -" + -"-

==

pg

Zl)-

+

Differential head

Now section (2) is al the vcna·co nlracta and the area of orifice then. we have

<12

represents [he area al the vena -cont racta. If "0 is

",

Cc = -

".

where Cc == Co-effident of contraction "l="OXCc

... (ii)

By colltinuity equation. we lIavc

...(iii)

I I

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IL Dynamics of Fluid Flow

,

283 1

(""J'

\', =2gl1+ ~

\', ==

J2gl1

'-[::1' c',

Substituting this va lue o f C< in equatio n (II'). we get

Q="o XC d

c', '-[""1' ",

x

C; ,-["" ",]' ,-[::]' CJuoat M

,- [",]' ",

~2g"

Ja~ -l/&

... (6. 13)

where Cd == Co-efficient of discharge for orifice meter. The co-efficicill of discharg e for orifice mClCf is Illuc h smaller than that for a venturi meter. Problem 6.22 All orifice lIIeler will! orifice diameter {O em is iliseT/cd ill a pipe of 20 em diameler. The pressure gauges fifled upstream wId down.nream of Ihe orifice meIer gil'e.! readillgs of 19.62 N/cm 2 alld 9.81 Nlem z respeC/iI'ely. Co-efficietll of discharge for Ille orifice meier is gil'ell {IS 0.6. Filld Ille di,~clllI'ge of Waler IflrOllgll pipe.

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IL 1284

Fluid Mechanics

Solution. Given

Dia. of orifice.

do'"

Area.

aO

] 0 em

"' "4" (10) '

'" 78.54 em

,

til'" 20 em

Dia. of pipe.

Area,

(11=

~ (20)1= 314.16c01 '

1 4 PI'" 19.62 N/(;m = 19.62)( 10 N/rn'

19.62)(104

P

---.!. '"

pg

-:-;O:::-""' C- = 20 III of waIi: r 1000)(9.81 9.8 1x I0· "" 10 m of water 1000 x 9.8 1

Similarly

II =

~ - p~ '" 20.0 - 10.0 = 10 m of water = 1000 em of wate r pg pg

Cd = 0.6 The discharge, Q is given by equalion (6.13)

'" 0.6 x

c~7~8".5"4",,,.;,3,,14~.l~6~ x .}2 x 981 x 1000 J (314. 16)' -(78.54)1

'" 20736838.09 '" 68213.28 cm 3 /s = 68.Z1 liires/s. Ans. 304 Problem 6.23 All orifice meier wilh orifice diameter 15 em is inserted ;11 IIp'pe of 30 elll tlir,meler. Tile pressure diffeTe"c/! meaSllred hy II mercury oil differential m(mameleT Oil lire /"'0 sideJ' of lile orij"'" meier gi.'es" reading of 50 of mercury. Find Ihe rMe offlow vf oil of sp. gr. 0.9 .,.hen the co efJicil'1lt of disch"rge of the orifice mcter = 0.64. Solution. Giwn : Oia. of orifice, " 0 =15cm

"m

"

a o = - ( 15)"= 176.7 cm-, 4 til = 30 cm

Area. Oia. of pipe,

"

(30) = 706.85 cm 4 Sp. gr. of oil. So = 0.9 cm of mercury Reading of diff. ll1anOlneter. x =

Area.

Oiffcr~l)l i al

I I

III

head,

=-

,

50 h= x [S,So - I] = 50 [13.6 -I] cm of oil 0 .9

Ii

~ I

IL 285 1

Dynamics of Fluid Flow '" 50 x 14.11 '" 705.5 em o f o il

Cd '" 0.64 Th e rat e o rlh e fl ow. Q is given by eq uation (6. 13)

",0.64 x

176.7 x 706.85

xJ2x981x 705.5

~(7 06.85)2 _ (176.7)2

'" 94046317.78 '" 137414.25 cm 3/s= 137.414I11resJs. Ans. 684.4 6 . 7 .3 Pitot- tube. 11 is a d~vicc uscd for measuring the velocity of now al all Y poil1l i n a pipe or a channel. It is baso:d on th e principle Ihm i f the ve locit y of flow m a poill1 becomes le ro. th e pressur.: there is in<:rcased due w the (;unvcrsion of til..., kine!i" energy into pressun: energy. In its simp iesl ronn, Ihe pilot-mix: consists of a g lass lube. bent at right ang les as shown in rig. 6. 13.

T

" :-C-,

~:":":":":":":":""""""""""":__;:-""

H

--q>---------------------.------------ -"l'1..-=,..!! .i

The lower end. wlli ch is bent th rough 90 Q i s directed in [h e up -

--------------------------- . - '

stre3111 direction as show n in Fig. 6. [3. The liquid rises up in the ---------------------------------tu~ du~ to th ~ co n v~ rsion of kinetic e nergy into pressure energy. The veloc it y is determ ined by meas urin g the risc o f liquid in the lU be. Fig. 6. IJ PiIQt-tl/be. Consider two po ints (I) and (2) mt he sa me level in suc h a way that poim (2) is j ust as the inl et of th e pitot-tube and point ( I) is far away f m m th e tube. 1'1: intensity of pressure at point
).

,-"+"

v, - ' - + '1: - ' pg2g pg 2 g p,

But ZI =

Z2

",

p,

as points (I) and (2) arc On th e sa me lin c and

.!i '" pressun: he ad at pg

1'2

= O.

(I ) '" H

Po : pressure head at (2) : (II + Il) pg Substituting these values. we get 1,2

H , -'- =(h +/f)

2,

1,1

I,,, _ l 2,

or

1'1" J2g/r

This is th eore tica l ve loc ity. Actual veloc ity is given by

I I

Ii

~ I

IL 1286

Fluid Mechanics (I',) "" '" C. J2g/r

where C , = Co-efficient of pilot-lube

Ve locity al an y point

... (6 . 14 )

"'" C• .j2gh

Velud ty of n ow in a Pi llt' by pitot-tu be. For findin g the veloc ity J i an y point ill a pipe by piwltube. the following arrangements arc adopted: I. Pilot-lube al ong wilh a >'c l1ira] pie zo mder lube as sllown in Fi g. 6 .14. 2. Pitol-Iulk connected wilh pic;o;omc!er tube as shown in Fi g. 6.]5.

3. Pilot-tu be a nd vertica l pi ezo me ter \tJhc conn ected wilh a diffe re nt ial V- tube man o meter as sh own in Fig. 6. 16. PIEZOMETER .....

TUBE

1

"

if

V-PITOT.TUBE ~

-_

f~:: ~ =::=::=:=:=:=:=:=:=:=:=:

------------

= __ _-----------_ ::."'" ----.

IT ... _-_._--

.. ., _- ......___--.. .... <

::::~ -:-:-:-:-:-:-:-:-:-:-:

':..-'10

-----------

---------------------

Fig. 6.16

4 . Pi lOt-stat ic lube. wllich co nsists o f two circu la r concel11ric lubeS o ne inside the ol her wilh some annular space in between as show n in Fig. 6.1 7. The outlel of lh ese lwo lubes are co nn ec ted 10 the diffe rcl11 i;ll mano meTe r whe re Th e d iffe rence of pressure head 'h' is measured by kn ow in g the

difference o f Th e levels o f Th e mallo mcter liquid sa y.\". Thell h

"'.f

[~:

-I}

Pro blem 6 .24 A pilOt-.I/atic lUbe placed in the cemre of a 300 mm pipe lille has one orifice pointing upslream mid olher perpelli/icu/ilr 10 il. Ti,e meall I'e/oeily in II,e pipe is 0.80 of the cell/TtlI I·e/ocil)'. "-ind II,e discharge Ihrougll the pipe If the pre.l$llre difference bet .....een the 11>'0 orifices i.l 60 mm af ..... ater. Take the co-effieielll of pitot IIIbe as C. '" 0.98. Solu t io n. G iven: Dia. of pipe. d", 300 mm '" 0.30 III Di ff. o f pr~ssure head. I,,,, 60 mm o f wa ter '" .06 m of wate r C. '" 0.98 Mean veloc iT y. V", 0.80 x Central ve loci ty Central ve locity is give n by eq uatio ll (6. 14) '" C • .J2g11 = 0.98 x .J2 x 9.8 1 x.06 '" 1.063 m/s

I I

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IL Dynamics of Fluid Flow

287 1

V "" 0.80 x 1.063 = 0.8504 m/s

Discharge.

Q = Ar~a of pipe x V j'(

,_

== 4d- x V ==

1t

"4 (.30)

1

.

J

x 0.8504 == 0.06 m Is. Ans.

Problem 6.25 Find the I'e/oeify ofl/I(' flow of GIl oil Ihrough
II = x

[~:

-I]

=. 1

[_'g_.:__ I]

""

].6 III of oil

'" C. J2gl1 = 0.98 .J2 x 9.81 x 1.6 '" 5.49 m/s. Ans.

Problem 6.26 A pilol -Slalic (ubI' is u~'ed /0 ",ea~'ure Ihe ..docily oflt'(l/er in a pipe. The ~'Illg'illlio" pressure head is 6 II! (I"d Sialic pressure head is 5 m. Ca/culllle Ihe "e/OCily of flow aSJ"umillg Ihe co I'jficil'nl of 'uhe equulla 0.98. Solullon. Given: Stagnation pressul\! liead, 11, =6111 SIalic pressure head. ",=5m 11=6 - 5= I m

Velocity of now.

V = C. JZg" = 0.98 JZ x 9.81 x I = 4.34

III/S.

Ans.

Problem 6.27 A sub-marille mOI'es IlOrizollilllly ill seo Imd I/Os ils oxis /5 m bdow 1/1/' sllrface of "·Iller. A pilo/./ube properly placed jusl ill frOlll of Ihe suh -marille olld (l/oll g ils axis is cOimecled 10 Ihe IWO limbs of II U-lIIbe COIlIO;Ilillg mercury. The differellce of mercury lel'el is found ro be 170 mm. Filld l/ie speed of Ihe SUb'lIlllrille kllowin8 1//01 Ihe sp. gr. of mercury is 13.6 and 1//01 of sea· Willer is 1.026 willI re.lpect offre.lh Willer. Solulion. Given: Diff. of lIIercury level, .r = 170 111111 = 0. 17 11\ Sp. gr. of mercury , Sj= 13.6 Sp. gr. of !.ea-watcr. SQ= 1.026

[S'l

lI=x -

So

['3.6 1

- I =0. 17 - - -J 1.026

=2.083411\

V = J2gh = J2 x 9.81 x 2.0834 = 6.393 Ill/s =

6.393 x 60 x 60

1000

kmlhr = 23.01 km/hr.

An.~ .

Problem 6.28 A pilOI-lube iJ' ill$er/t"d ill a pipe of 300 IIlm diameler. The J'Ia/ie pressu re ill pipe is tOO 111m of mercury (mCllllm). The slllgllaliOli pressure aI Ihe e<:llire of IIIe pipe, reco rded by Ihe

I I

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IL 1288

Fluid Mechanics

pilOI-lUbe is 0.981 N/cm<. ea/clliare Ille rale of flow of Wilier II/rough pipe, if Ille meatl \'elocity of flow is 0.85 limes rhe cemral 1't1/ociry. Toke C. '" 0.98. Solutio n. Given: Dia. of pipe. d" 300 mm '" 0.30 III

Area. SIalic pressure head

a = ~ <1 2 '" ~ (.3) 2 ", 0.07068 Ill l 4 4 " I()() nUll of mercury (vac uum)

100 = - - - x 13.6= - 1.36 III of wate r

1000

Stagnation prc,,-w rc Stagnation pressure head

= .981 Nfcm l = .981 x 10' Nfrn l =

.98 I x IO· = .98 1 X 10' = I m WOO x 9.8 1 pg

II " Stagnation pressure head - Slatic pressure head = 1.0 - (- 1.36) = 1.0 + 1.36 = 2.36 III of water Veloci ty at centre

'" C. J2g/1 = 0.98 x ...;RjC,c9'g"C'''2.'36'' 6.668 tnls

Mean velocity. Rate of fl ow of water

~

6.8

V = 0.85 x 6 .668 = 5.6678 Illis

= V x area of pipe

= 5.6678 x 0.07068 rn l/s = 0.4006 1Il 3/s. An s.

THE MOMENTUM EQUATION

It is based on the law of conservation of momentum or on the mOllie ilium principle. which states th~t the net force "cling on ~ fiu id mass is cqual to Ihe ch"ngc in moment um of now p.:r unit time in that dircction. The force acting on a fillid mass '/11' is givcn by thc Newton's second law of motion. F:/II:>
where a is the acce leration acting in the sallie direction as force F.

8", d,' F= m -

," ,"

'" d(m •. )

F ==

d(IIII') d'

!m

is constant and can be taken inside the diffe re lll ial } ... (6. 15)

Equation (6.15) is known as the momentum principle. ... {6 . 16) Equation (6.15) can be written as FAt", d(ml') which is known as the illlpu{se'lIIomenllllll equation and states that the im pulse o f a force F ac ting on a fiuid of mass /II in a shon interval ofti mc dt is equal to the ch~nge of momentu m d(mv) in thc direction of force.

I I

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IL Dynamics of Fluid Flow

Force exerted by

~

fl owing fluid on

OJ

289 1

pipe bend

The impulse-momentum l"
flowing fluid on a pipe be nd. Consider two sections (I) and (2). as shown in Fig. 6. 11\. leI VI'" ve locity of flow at seclion (I), PI'" pressure intensity at section (1).

AI'" area of cross-section of pipe al sedion (1) and 1'2- Pl- Al '" corresponding va lues of velocity. pressure and area al section (2). Lei F. and F, be Ihe components o f tile forces exerted hy [lie flowing fluid o n the bend in x-and y-dircctions r..:spcclivcly. Then Ihe force cxcn cd by the bend on the fluid in Ihe directions of x and y will be equa l \0 F. aut! Fy bul in the opposi te di rections. Hence component of lhe force exe rted by bend on the fluid in the ,[·direction = - F, and in the direction of y '" - Fr The other externa l forces actin g on the fiuid arc PIAl and py\~ on the sections (I) and (2) respectively. Then momentum equation in x-di rection is give n by

"• P,A"_-ll-_ (0)

(0)

Fig. 6. 111

For~

011 bend.

Net force acting o n fiui d in the direction of x = Rate of change of momentum in x-direction PIA I - Pt\~ cos 0 (Mass per sec) (change of ve locit y) '" pQ (Final velocity in the direction of x - Initial ve loc ity in the direction of .f) ...(6. 17) =pQ(V2CO~O-VI) .(6. 18) F , '" pQ (VI - V2 cos 0) + PIAl - PY\l cos Similarl y the momentum equation in y-dircction gives ... (6. 19) Pt\~ sin F. = pQ (V2 sin 0) ... (6.20) ,..,. '" pQ (- Vl sin 0) - p0~ sin 0 Now lhe resultam force (FR ) 3<:ling on lhe bend

"'x'"

e

o-

e-

e-

"

'" 'Fz +Fl.

, And the angle made by lhe resultant force with horizontal direction is given by

,.

tanO"'~

F,

... (6.2 1)

... (6. 22 )

Problem 6 .29 A 45° reducillg belld is connecled ill a pipe line. ,be diamerers ar 'he inlel and OIl/lei of,he bend being 600 mm Imd 300 mm respec/ire/y. Filld lire force e:terled b), warer 011 II,e bend if/be in/ellsil), of pressure ar illiel 10 belld is 8.829 Nkm! lind rale of flo ..... of ..... arer is 600 lirres/s.

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IL 1290

Fluid Mech a nics

Solution. Given:

9 = 45"

Angle of bend. Dia. at inl c!'

DI '" 6CKl1ll1ll = 0.6 III f[

It

:

2

Area.

A1='4D1 ='4(.6)

Dia. al outle!,

= 0,2827 m2 Dl = 300 mill '" 0.30

ill

,

- - ""'1----/ / _P2~ ___ ~ ~'

III

Fig. 6.19

:. Area.

"

, 8.829 x 10 N/rn" PI'" 8.829 Nlcm-::: Q = 600 litis", 0.6 mJ/ s

Pressure at inlet.

Q 0.6 VI"" - = - - =2. 122 m/s AI

.2827

Q 0.6 = - - - '" 8.488 mls. Al .07068

V, '" +

Apply ing Bernoulli's equation a1 scc ti o lls ( 1) and (2) , we gel

,

.£.!..+~+Zl= pg

B",

2g

V'

II) + ....L+zz PI! 28

11'" 12

,

,

l!J...+~ = III + V1 pg 28 pg 28

or

8.829 X 10

4

1000x9,81

+

2.1 22" /I, 8.488' =~ + 2 x9.8 1 pg 2 )(9.81

9 + .2295'" PtPS + 3.672 P! '" 9.2295

~ 3.672 '" 5.5575 m of wmcr

"

li z = 5.5575 x 1000 x 9.81 Nfm 1 '" 5.45)( 104 Nlm 1 Forces on lht: bend in x- and y-dircclions arc given by equ ation s (6. 18) and (6.20)

a~

F, '" pQ IVI - V1 cos 91 + PI A l - P0 2 cos a = 1000 x 0.6 [2.122 - 8.488 cus 45°1 + 8.829 x 10' x .2827 - 5.45 x 104 x .07068 x cos 45"

'" - 2327.9 + 24959.6 - 2720.3 '" 24959.6 - 5048.2 '" 19911.4 N

,..,. '" pQ 1- V1 si n

md

- vc sign means

r,.

sin

e

. '" 1000 x 0.6 [- 8.488 si n 45° [_ 5.45 x 104 x .07068 x sin 45° '" - 3601.1 - 272 1.1 ", - 6322.2 N is acting in tile down ward direction

Res ultant force.

I I

e] - PY\ l

,.. I? '"

Jr} + F/

Ii

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IL Dynamics of Fluid Flow ==

291 1

~(1991 1 .4)Z + ( _6 322.2)2

= 20890.9 N. An ~. The ang le made by resultan! force with x-ax is is given by equa tion (6.22) or F,

Fi g . 6.20

= 0.3175

tan9= -'- = 6322.2

199 11.4

F,

9= tan - I .3 175 == 17° 36' , An s. Problem 6.30 250 IilTes/s of ,Wiler is jlOWillg ill a pipe Illn'itlg a di"meler of 300 mill. If tile pipe is bellI by 135 0 (IIml i.1 c/range from illilial 10 filial "iree/iOIl i,1 I J 5 °J. find I/Ie mag/lilude amI
Solution. Givcn : Pressu re. 1'1 = 1'2 = 39.24 Nlcrn 1 = 39.24 x 104 N/m 2 Disch~rgc. Q == 250 litrcsls == 0.25 m3/s Dia. of bend at inlet nod o ullet. VI = D1 = 300 111m = 0.3 m l'I:

Area. Veloc it y of

l

ll

l l

A 1=A!=4"D 1 =4X.3 =0.07068 111

wat~r at S~"(:lioJ\s (1) and (2), V = VI = V, = ~ :

V2 sin 45'

~

.1'

p;/"

~

cos 45'

"- ",

I")

-

V

0.25 : 3.537 m/s. .07068

Area

o

,

p' $ -----IDI

~

V, cos 4 5'

''I

135'

Lx

Fig. 6.21

Force along .(·axis

: F, : pQ[V I,- V~,J + PI..A I + P;o..Al initia l ".. Iocily in th .. direction of.>:: 3.537 m/s V~, " final ve loci ty ill the dircctioll of x = - V~ cos 4 5° = - 3.537 x .707 1 PI.< : pressure at s,:,ction ( I) in x-dir.. c ti o n : 39.24 N/cm ~ " 39.24 x 104 Nlm 2 Pl, : pressure at section (2) in x-dircction : p~ cos 45°: 39.24 X 104 X .70 71 F,= 1000 X .25[3.537 - (- 3.537 X .7071)[ + 39.24 X 104 X .07068 + 39.24 X 104 X .07068 X .707 1 : 1000 X .25[3.537 + 3.537 X .7071 [+ 39.24 X 104 X .07068 II + .70711

V lx :

I I

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IL 1292 Fluid Mechanics " 1509.4 + 47346 = 48855.4 N '" Fy " pQIYI , - V!yJ + (PIA 1),_+ (P:01) ..

= init ial velocity in y-d ireClion = 0 V 1., = final velocity in y-dircclion = - V! si11 45" = 3.537 x .707] (PIA,), = pressure force in y-dirccli on = 0

whe re

VI ."

(P02) , = pressure force at (2) in y-dircction - = - P0 2 si n 45 6 = - 39.24 X [0 4 X .07068 x .7071 F, = 1000 x .2510 - 3.537 x .70711 + 0 + (- 39.24 x IO~ x .07068 x .7071) = - 625.2 - 19611.1 = - 20236.3 N - vc sign means "'., is acting in the downward direction

Rcsuhant force.

,

,

F = J F,1 + Fl F,=4980.1

• "'CJi-='------, °

= 528110.6 N . Ans . Th" direction of the resultant force F H' wi th the x-axis is given as "

~

Ian

e = F,. = Fs

LL" " t _______________ _

20236.3 = 0.4 142 48855.4

Fig. 6.22

e" 22" 30 ' , Ans. Problem 6.31 A 300 "'''' diameter pipe C(lrries W(1/er under a Ilead of 20 me lres will, a ndociry of 3.5 ",/~·. if Iile a.tis of Iile pipe lu rtlS Illrougil 45 °, find Ille mag" il"de and dire/;Iioll of Iile rcsu/lllJIl farce allll<' bend. Solu t ion. Given: Dia. of bend. D = DI = D z = 300 mm = 0.30 m 1t

Area.

A=A , =A 2 = -

4

Velocity.

2

1t

D = -

4

2

x.) =0.07068m

2

V=V , =V1 =]·5m/s

e = 45° Q = A x V = 0.07068 x 3.5 = 0.2475 m 3/s

Discharge.

= 20 m of water

Pressure head

or

...E.... = 20

pg

III

of water

p=20xpg=20x IOOOx9.81 N/m 2 = 196200Nlm 1 Pressure intensity.

Now

p = P, = P2 '" 196200 N/m!

= 3.5 1l1/S, VZr = V1 oos 4 5 0 = 3.5 x .707 J V ly = O. V 2y = V2 sin 45" = 3.5 x .7071 V"

(PIA I )~ '" PIA I = 196200 x .07068. (PIA I),. = 0 (P;oAl)~ '" - p0!

Forte along

I I

.{-{I.ri.l.

cos 45°, (P;oAl),' '" - p0! sin 45° pQl VI., - V2,1 + (PI A ,)~ + (P:0- 2)~ = 1000 x .247513.5 - 3.5 x .70711 + 196200 x .07068 -

F~ '"

p;oA~

x tOS 4SO

Ii

~ I

IL Dynamics of Fluid Flow

293 1

'" 253.6& + 196200 x .07068 - 196200 x .07068 x 0.7071 '" 253.68 + 13871.34 - 9808.()..I = 4316.98 N

F,,: pQ [ VIY - Vy'l + (PI A I)\' + (1']"\2),. '" 1000 x .247510 - 3.5 x .707 11 + 0 + [- Pfo2 si n 45°] '" - 612.44 - 196200 x .07068 x .7071

Force along Y'{/XiI,

'" - 612.44 - 9808 '" - 10420.44 N

F II '"

Resultant fore\'

JF.1 + F/ '" J( 4316.98) 2 + (10420.44 )2 '" 11279 N. ,\IIS.

"~ A'""''' .'

V.sin45"

q'

I7A,cos 45"

V, 45· V.COS 45·

, /

- - ~- -

(j)

L

'.

Fig. 6.23

The ang le made by F/I with .r-axis

lane", F, '" 10420.44 = 2,411 4316.98 F, tan - I 2.411 '" ii7 c 28'. An s. belli! {/ rcell/llgular (lir duel of f m 2 cross-secliQIWlllrell iI grll
a",

Problem 6.32

In {/

-15°

reduced 10 0.5 m l area. Find the magllirude oml direc/ioll of the jorce required /0 Iwld Ihe duel ill position iflhe I'e/Orit)' offlow ai/he I m ! seClioll is /0 mIs, and I'rej'sure is 2.9-/3 N/cml . Take dellj'ily of "ir as /./6 kg/mJ. Solution. Give n : Arca at section ( I). Area at section (2) . Velocily at section (1 ), Pressure at section ( I ). Density of air,

A I" lm

1

A 1 " 0.5 m

1

VI "lOm/s PI" 2.943 Nfcm 2 '" 2.943 x 104 N/m 1 " 29430 N/m 1

P '" 1.16kgfm 1

Apply in g continuity equ ati on at sections (I) and (2) A1 V I ",A 1 V1 AI VI 1 V1 ", - - o - x 10" 20 m/s

Al

~ I

0.5

I~

~ I

IL 1294

Fluid Mechanics

Discharge V, sin 4 5 '

V, cos 45"

/ )\f7L

p,A, cos 45'

". Fig. 6.24

Apply ing BCrrloulli' s cqualioll al sections (I) and (1)

P,

V,~

P!

¥l!

-+-=-+pg 2g pg 2g 2.943 x 10' + -;;,-J~ O '= "" P! +

"'

1.1 6x9.8 1

pg

2x9.8 1

20'

P2 = 2.943 X 10' +

pg

1.16x9.81

2x9.8 1

2 x9.8]

"" 2586.2 + 5.0968 - 20.387 '" 2570.90 m p, '" 2570.90 x 1.l6 x 9.81 '" 29255.8 N

Force along .I·axis. whe re

A I,

'"

F~

= pQ IV" - Yhl + (p,A,). + (PY\l),

10 mls. Vb"" VI (;OS 45° '" 20 x .707 1.

(PIA, ), '" PIA , == 29430 x I '" 29430 N

and

(1'02), = - P02 cos 4.5° = - 29255.8 x 0.5 x .7071 F, = 1.16 x 10110 - 20 x .7011 J + 29430 x 1 - 29255.8 x .5 x .707 1

'" - 48.04 + 29430 - 10343.:n '" 0 - 19038.59 N Similarly furce (1/0118 )"-luis. F. := pQ[V h, - V2,.1 + V',A,)y + (p;Y\:J ), wh~ re V,), = 0, V2r = V1 si n 45 ° = 20 x .7071 = 14.142 (PIA,),_= 0 and (pyl.!»), = - PY\l sin 45° = - 29255.8 x.5 x ,707 1 : - 10343.37 F,, : 1.16 x 10 [0 - 14.142 [ + 0 - 10343.37 : - 164.05 - 10343.37: - 10507.42 N R,,~ullanl

force,

FK: ~F} + F/ : ~( 1 9038.6)~ + (10507.42 )"

: 21746.6 N. An s.

The direclion of FII Wilh x- ax is is givcn as lanG:

10507.42 :0.5519 19038.6 t 8 : tan - .5519: 28 c 53'. Ans. Fy

:

Fx

I I

Ii

~ I

IL Dynamics of Fluid Flow

295 1

PI< is (h e force exerted on bend. Hence th c fo rce required 10 hold th e duct in position i s eq ual lO

2 1746.6 N but it is actin g in th e opposite direction of FR' AD S. Problem 6.33 A pipe of 300 mm diameter con "eyillg 0.30 ", JIJ' of ,mler hus a righlllllgied be.rd in II horiZOliful pilll/c. Find Ihe rC$u/tallf jorce e.fer/ed on Ille bend If IIIC pro.fllre at i, 111'1 lind QU lld of Ille

bend are 24.525 N/cll/ ami 23.544 Nkm 1.

Solution. Given: 0;3. of bend.

D '" 300 Imn '" 0.3 111

"

Area.

A=A 1 =A 2 = - ( .3) =0.0706810 4

DiM:hargc.

Q" 0.30 rn J/s

Ve locity.

v=

VI" V2 ",

~

==

.O~~8

~-

/

p,A,

=4.244 Ill is

e - - <2!

v

(j)

v,

,

y

L,

Fig. 6.25

a == 90° PI = 24525 Nfcm 1 = 24.525 x 10· Nlm 2 = 245250 Nf m l 1'2'" 23.544 Nfcm! " 23.544 x 104 Nlm 2 " 235440 N/m!

Angle of bend.

['orce whe re

DO

Fx '" pQ lVI" - V2, 1 + (PIAl)" + <,,02), p = I(X)(}. VI> = VI = 4.244 mls. V~, = 0

bend along x-axis

(PIAl), = PI A l = 245250 x .07008 (P:A2) ,

=0 ,..~

= 1000 x 0.30 [4.244 - 0] + 245 250 x .07068 + 0 = 1273.2 + 17334.3 = 18607.5 N

Foree on bend alon g y-axis. Fy = pQ lVI , - V2)· [ + (PI A l), + (Pi'-2)y where VI , = O. V2, = V2 = 4 .244 Ill is (PIAl),. = O. (P:A:U,. = - P0-2 = - 235440 x .07068 = - 16640.9

F,. '" 1000 x 0.30]0 - 4. 244] + 0 - 16640.9

= -1 273.2 -16640.9= -1 79 14.1 N Resullanl force. and

F,,=

~F,' +F,'

F, lana : - . 17914.1 = 0.9627 18607.5 F,

a = 43

I I

= J(l8&J7.5)' +(179 14 .1 )' = - 25829.3 N

0

54'. Ans.

Ii

~ I

IL 1296 Fluid Mechanics Problem 6 .34 A tlOU!<- of diameter 20 mm i.1 jilled 10 a pipe of diameter ~O mill. "';lId lite force "'.I'erled by Ille noule 011 the water ..... hich is flowing 1III0llgii I/Ie pipe m lilt! rale of 1.2 111 3/lIIlllIIle. Solution. Giv~n : Dia. of pipe. 0 , = 40 mm = 40 x 10-1 m ", .04 m

Area.

Dia. of nozzk.

VI '" 20

0.02

m ill ",

111

Area. Discharge.

i

(j) Fig. 6.26

Apply ing continuity equalion at scrtions (I) and (2 ). A,V,=A 1Y2=Q VI =

~I

=

.,d

'(Xl~'~56

= 15.92 rnls

0.2

'" 63.69 rn ls

.0003 14

Applying Bernoulli's equ ation


sec tions (1) and (2). we gel

V'

Ie '

1:l + _'_ + z, '" ~ + - '- ' " pg

2g

Now

pg

28

Z, '" Zl' 1'1 '"

pg

mmosphcriC pressure", 0

-PI + -VI' : -V} pg

28

28

1:l =

(1S.92') V,' _ V,' = (63.69') ""~~ = 206.749-12.917

pg

28

2g

2 x9.8 1

2 x9.81

= 193.83 m of wate r

N

N

In '

rn '

p, = 193.83 x 1000 x 9 .81 - , = 1901472 - ,

I I

Ii

~ I

IL Dynamics of Fluid Flow

297 1

Let the force exerted by Ihe nozz le on water'" F, .l '" r~tc

Net force in the di rection of PI A l - P;tAl

of change of momentum in the direction of x

+ F, '" pQWl -

VI)

where 1'1 '" at mosp heric pressure", 0 and p '" 1000 1901472 )( .00 1256 - 0 + F, '" 1000 x 0.02(63.69 - 15.92) or 2388.24 + Fx'" 916.15 F , '" - 2388.24 + 916. 15 '" - 1472.09. AilS. - VI.' s ign indicates lhal the force exe rted by Ihe noa k on waler is acting froln right to left. Problem 6.35 The diameter v/a pipe grodual/y ~educeJfro'" I III /0 0.7 III G J' sho ..." ill Fig. 6. 27. Tlw pressure ;nlensily II/Ihe C<'IIlre-/i",: of I III sec/ion 7.848 kNlm! (1111/ Tale offlow a/woler Illrougil th e pipe is 600 li/res/s. Fill d Ihe ;n'CIIsity of pressure a l Ihe cen/re·/ille of0.7 m seclion, Also delermin/!

rhe force au/ed by flowing Imrer 011 rratlJirion of rile pipe. Solution. Given: Di~ . of pipe at scl:lion 1. D I '" I In

"

Are a.

A I ",-(I),,,,0.7854m

4

, ~

GJ I

3r'f '+ '"

;_~ll""'""v;_-"A,!""---1 1

A

CD

@

Fi g. 6.27 Dia. of pipe at scction 2.

,

"

A, '" - (0.7)' '" 0.3848 III • 4 1 2 PI'" 7.848 kNlm '" 7848 N/m

Are a. Pressure at scction I.

Q'" 600 litres/s '" 600 '" 0.6 rn'ls

Discharge.

1000

Apply ing continuity equation. A I V I ", A ZV2 ",

Q Al

VI'" -

~

Q

O.

- - - '" 0.764 IlIls

0.7854

V,,,,.Q.= ~ ,,,

1.55m/s • A! 3854 Apply ing Bernoulli' s equation at scl:tiuns (1) and (2). PI

-

pg

7848

1000 x 9.81

I I

VI!

+-

2g

~ P2 V} - - +pg

2g

!...

pipe is horizontal.

+ (.764)2 '" Ii + ",(Ic:'",S);C' 2 x9.8 1 pg 2x9.81

Ii

~ I

IL 1298 Fluid Mechanics ",

,( . ,,,,64 ,,,Ic,-' 2 x9.8 1

-'-"- == 0.8 + 7"

pg

(J.55)' 2x9.8 1

== 0.8 + 0.0297 - 0.122 == 0.7077 111 of water 1', "" 0,7077

x 9,81 x

1000

== 6942.54 N/ml or 6.942 kN/ml, Ans.

Let

F~

== the force cxcrlcd by pi pe lransition on the

flowin~

water in the dircdion of flow

Then net force in the direction of flow == ralC of change of momentum in Ihe dirc<:!ion of flow

= p(V2 - VI) 7848 x .7854 - 6942.54 x .3848 + F, = 1000 x 0.6]155 - .764] 6163.8 - 2671.5 + F, = 471.56 F, "" 47156 - 6163.8 + 2671.5" - 3020.74 N PIAl - P~l + F,

or

The force exu[cd by water on pipe transition = - F, " - (- 3020.74) == 3020.74 N. AilS.

to 6. 9

MOMENT OF MOMENTUM EQUATION

MomcnI o f rnomclltum equation is derived from moment of 1ll0lllCnlum prillciplc which states that the resulting torque acting on a rotating fluid is equal 10 the rate of change of moment of momentum. Let

VI == velocity of fluid at section I. '1 == radius of curvature at section I. Q == rate of flow of fluid. P == density of fl uid.

and

V z and '1 = velocity and

radiu~

of curvature nt sec tion 2

Momentum of fluid at section 1 = mass x velocity == pQ x Vl/,f MOlllent of 1Il0tllcntUill per second nt sectiOn I. =PQXV1x'l Similarly moment o f momentum per second of fluid at section 2

=pQxV!x,! Rate of change of moment of momentum

= pQV!'2 - pQVl'1 = pOlY!'! - VI'tI According to moment of momentum principle Resultant torque == rate of chan ge of moment of momentum ... (6.23)

Equation (6.23) is k.now n as moment of momentum equation. This eq uati on is applied : I. ror analysis of flow problems in turb ines and centrifugal pumps. 2. ror fmding torque exert ed by water on sprink.ler.

Problem 6 .36 A lawn Jprlllkler II'Ilh MO nou{ej of diamele," mm each Is cotlnecteil aCfOSJ a lap of Wilier as shown ill Fig. 6.28. The 1I0zzies are at a distance of 30 cm alld 20 Cllifrolll Ihe cellrre of IIIe tap. Tile rale of flow of waler Ihrollgll lap is 120 cmJ/.i. The 1I0zzies discll
I I

Ii

~ I

IL Dynamics of Fluid Flow

r-

Solution. Given Dia. of nozzles A and B. D- D~ - D8

A",

Area.

(.004)2 ",

"I

_ /0 ____

30

cm---J

ro t

III

.00001256 111 2

4

F Ig . 1>.28

Q'" 120 em}'s

Discharge Assuming

2:

4mm -.OO4

20cm

299 1

(h~

di.<;ehargc 10 be equally divided bclWc.:1l the two nozz les. we have

Q 120 3 6 ) Q,,=QI/= 2 =T=60c11l Is=60X IO 11I Is Velocity of

Wal~r

at the ou tld of each nonle.

Q

...

whe re

fA'"

8

=_

6OxlO --<>

A '" = 4.777 m/s. A .0000 1256 The jet of water coming oul from nozzics A aJld 8 is havi ng veloc ity 4.777 mls. These jets of water will exc n force in llle opposite direction, i.e .. force exerted hy Ihe jets will be in Ihe upward direction. The torque exerted will also be ;lIlhe opposite directiun. Hence IOrquc at B wi II be in the ami-clockwise direction imd al A in the dockwis.: di rection. But torque at B is 1001\: (him the torque at A ;md hence sprin kle. if free. wi ll rot ate in the anti·clockwise di rection as s hown in rig. 6.28. lA:t (0 '" angular veloci ty o f the sprin kl er. Then absol ute velocity of wata at A.

V '" V

VI = VA + (Ox 'A distance of nozzle A from the centre of tap

" 20

elll

= 0.2

IW x

III

'A=

tangential ve loc it y d ue to rota tion I

VI = (4.777 + 00 x 0.2) lilts Here 00 x ' A is added to VA as VA and tangential ve locity d ue to rotation (00 x fA) are in the sallie dire<.:tion as shown in Fig. 6.28. Similarly. absolute velocity of wate r at B. V 1 ", VB - wngential velocity due to rotation =4.777 - wx

(where 'B= 30 em = 0.3 III)

'IJ

'" (4.777 - 00 x 0.3) Now applying eq uation (6.23), we get

I Here,z" 'B"I='A

T" pQ(VZ'2 - VI'I( '" pQAIV 1f B - Vlr,, 1

= 1000 x 60 x 10

6

Q'" QA '" Q R

((4.777 x 0.3 (0) x .3 - (4.777 + 0.2

(0)

x .21

The moment of momell1 UIII of the fluid entering sp rinkl er is given zero and also there is no external torque applied 0 11 the sp rinkler. Hence resultant external torque is zero. i.e.. T", 0 1000 x 60 x 10 6 [(4.777 - 0.3 (0) x.3 - (4.777 + 0.2 (0) x .21 = 0 (4.777 - 0.3 (0) x 0.3 - (4.777 + 0.2 (0) x.2 '" 0 4.777 x.3 - .09 00 - 4.777 x.2 - .04 (1)= 0 0. 1 x 4.777 '" (.09 + .4777

w" - -

0.13

I I

.(4)(0 '"

. 13

OJ

= 3.6746 rod/s. An s.

Ii

~ I

IL 1300

Fluid Mecha nics

Problem 6.37 A h,WII sfJrillkler shown ill Pig. 6.29 lIa.1 O.S elll diameter 'lOuie ar lile end of a fOlilIing arlll alllt discharges Wafer ill the rate of to mls \'e/ocily. Defemlim' Ille torque required /0 hold Ihe rofaling "rm s/alio,wry. Also determine the conJ'tan' spud of rOI,,'iOIl of Ille If free 10

'If"'.

t10

rO/lIIe.

Solution. Dia. of each nOlzle = 0.8 elll = .008 III

20 em

Are a of each nozzle

Velocity of !low al each nozzic = IU m/s. Discharge through each noak.

I

25 em

'9

II'j

1

mise.;

liA

Q '" Area x Velocity 10 mlsec Fig . 6.29 '" JXlOO5026 x 10 = .0005026 rn 3/s Torque cxc-fled by wate r coming through nozzle A on the sprinkler = moment o f momentum of

water through A

=

f" X P x Q x V~ '" 0.25 x 1000 x .0005026 x 10 clockwise

Torque exerted by water coming through nozzle B on the spr inkler

= 'II X P x Q X VB'" 0.20 x 1000 x .0005026 x 10 clockwise Tot al torque exerted by wmcr on sprinkler

Torque required

10

'" .25 x 1000 x .0005026 )( 10+ .20 x 1000 x .0005026 x 10 '" 1.2565 + 1.0052 '" 2.26 Nm ho ld the rotating arm stmionary '" Torque exerted by water on sprinkler '" 2.26 Nm. Ans.

Speed of rotation of arm , if fret' to rot:.t e Let 00 '" speed of rotation of the sprinkle r The absolute velocity of flow of wate r at the nozzles A and Bare VI'" 10.0 ~ 0.25 x 00 and V! '" 10.0 Torq ue exerted by water coming out at A. on sprin k ler

~

0.20 x 00

= FAX P x Q X VI '" 0.25 x 1000 x .0005026 x (10 - 0.2500) '" 0.12565 (10 ~ 0.25 (0) Torque exerted by water coming out at B. on sprin k ler X P x Q X V! '" 0.20 x 1000 x .0005026 x (10.0 '" 0.10052 (10.0 ~ 0.2 (0) Total torque exerted by water: 0. 12565 (10.0 - 0.25 (0) + 0.10052 (10.0 - 0.2 (0) :

Tg

~

0.200)

Since moment of momentum o f the flow entering is zero and no external torque is applied on sprinkler. so the result ant torq ue on the sprinkler must be ze ro. 0.12565 ( 10.0 - 0.25 (0) + 0.10052(10.0 - 0.2 (0) '" 0 1.2565 - 0.0314 00 + 1.0052 - 0.0201 00: 0 1.2565 + 1.0052 '" 00 (0.Q3 14 + 0.020 I) 2.26 17: 0.0515 00 2.2617

w: - - - '" 43.9 md /s . A n s. 0.051 5

I I

Ii

~ I

IL 301 1

Dynamics of Fluid Flow 60 x 43.9 ~-';::?'"- h

and

... 6. 10

'" 419.2 r,ll.tn. Ails .

FREE LIQUID JETS

Free liquid jet is defined a ~ th e jd of water corning oul from the noule in atmosphere. The palh travelled hy the frcejet is parabolic. Consider a jet coming from the nozzle as shown in Fig. 6.30. LeI the jet at A. nwkcs an ilngle with th e horizontal dirc~lion. If U is the velocity or jet o f waler, Ih ,," the horizontal component and vcrticnl ~u11lponcm of Ihis velOCiTy al A arc U cos e and U sin 9. Consider another point P(x. y) on the centre line of Ihe jeT. The co-ordinates of P from A arc x and y. Let the ve locity ofjct at P in the of- and y,d ircctiolls arc u and v. LeI a liquid particle tak es lime '( to reac h from A to P. Tllen the horizontal ami vert ical distances tra velled by thc liquid panicle in lime 'r' are;

e

V

TRAJECTORY

J":;;::::1- -..(/

.,

u

"

PATH

"

: -;:--·L f

;; ;.., \c':-:-: ' Ucos9 . --1

NOZZLE Fru liquid il'l .

Fig. 6.3O

x'" velocity l"t>mpone nt in x-direction x

1

.. ( i)

=Ucos €lx t

and

y = (I'enica l component in y-direction

= Usin 9XI ,.

±

g1

X

tim., _

.!. g(2) 2

2

... (,. i)

Horizontal componen! of velocity is consta n! whi le th e venka l distance is affected by gravity I

From eq uation (i). the valu e of ( is given as I = --,-"' -; C·'" U cos Substituting this val ue in equation (il)

a

y=UsinOx

=.Ttan 9 _

x UeosO

g.<

I

(

x

- xgx 2 Ucos9

1J.:

)'

. sma =x - - -

cos 9

gx

2U

2

cos'e

e) . .

sec' a - '-,- = sec! (6.24 ) 2Ucos-O Equation (6.24) gives the varia tion of y with thc squarc of.T. Hence thi s is the equatio n of a parabola. Thus the path travelled by the free jet in atmosphere is parabo lic.

I I

Ii

~ I

IL 1302 Fluid Mechanics (I) l\ la xl mum h eig ht uttnlntod b~' th e Jet . Using the relation V/ - VI! '" - 2g5. we get in this case VI '" 0 at the highest point VI = Ill ilial vertical component =UsinG -vc sign on right tHmd side is taken as g is acting in the downw:ml dircdion but panicles is moving up.

0- (U sin e)~ '" - 2g x S where S is Ihe Illnimurn vertical heig h! attained by the particle. or - U 1 sin1e == - 2gS

s=

Ulsin " e

... (6.25)

2,

(ii) Tim e of fli ght. It is the time taken by Ihe fluid particle in reaching fro m A to IJ as shown in Fig. 6.30. Let T is the time of fligh!.

Using equation (iiI, we have y = U sin

when the panicle reaches at B.

e x /_ ..!. gIl 2

y'" 0 and I'" T

Above equation becomes asO '" U sin exT -

..!.g x r2

O=Usin9 - ..!.g'f

"'

1Cancelling T)

2

,

T = .2~U~si~"" ,

... {6.26)

(iii) T im e 10 rea l' h hIghes t po int. The tillle to rcaeli liigliest point is halftlie tillle of niglit. Let P is tlie tillle to reacli liigliest point. tlien

P= T = 2Usin8 = Usi n8

... (6.27)

2

gx2 g (iv ) Horl:ton tul run g" of th " j cl. Th~ tOlal horizontal distauce travelled hy til<: fluid particle i~ called horizontal range of the jet, i.e .. the horizo ntal distance AB in Fig. 6.30 is c alled horizontal range of the jet. LeI this range is deooted by x·. Then x· = velocily .::omponent in .l-dire'::l ion x time taken by IiiI' panicle 10 rea.::h from A 10 B = U ,::os 8 x Time uf flight =U\'os8x 2Usin 8

g U

1

=-

e sin 8 = -U

1

.,,(6.28) sin 28 g g (v) Valu e of 8 fo r lll" x illlUnl range. The range x· will be maximum for a given velocity of projection (U). when sin 28 is maximulll or when sin 28 = 1 or si n 28 '" sin 90° = 28=900or8=45°

Then maximum range.

I I

2 \'Os

U" . '

.l *rn'" '" -5111- 8

g

U = g

l

{ '.'

,;" 90" = 1{ ... {6.29)

Ii

~ I

IL 303 1

Dynamics of Fluid Flow

Problem 6 .38 A I'ulica/wall is of 8 In ;" Iwighr. A jel of warer is cOllling QUI from a !lOule WillI a 1't!locily of 20 mls. The noule i.! silllated ar (I diswllce 0[20 //I from Ill<' I'ulieal wall. Find the {/lIgle of projection of lire llOule to Ihe //OriZOllwl 50 Illar the jet of \\"(l/er jusl clears rhe lOp of Ihe wall. Solution. Given : 8m Hci gllt of wall

Velocil y of jd. Dislancc of jet from wall. l;:t Ihe required angle

T 'm

U=20 mfs .P'

20

WALL:

In

=e

~i 'f---20 m------t

Using equation (6.24), we ha ve y=x tan

e-

1,'.(

,

1

,

-"- , sec" a

Fig . 6.31

2U·

where y = 8 m. x = 20 Ill. U = 20 Illis 8 = 201:tnfl-

"' "'

9.8lx20 2 , 2 x 20-

scc 2

e

'" 20 tan

e - 4 .905 scr 2 e

'" 20 (an '" 20 (,III

e - 4 .905 [I + (an ! 81 e - 4.905 - 4.905 1:1112 e

e - 20 Ian e + 8 + 4.905 '" 0 4.905 ta n' e - 20 Ian e + 12.905'" 0 4.905 (an'

tan e ""

""

20 ±

J20

l

-

4 x 12.9{lS x 4.905

2x 4.905 20±JI46.81 9.81

=

20 ± Jr.400M-2~5~1C;19"

"" ",ccce,;;,:,;.",,,,,,-

9.8 1

20±12.116 32. 116 7.889 = or 9.81 9.8 1 9.81

= 3.273 or 0.8036 9 = 73 9 0.8' or 38 9 37'. An s. Problem 6.39 A fire-brigade Imm iI holding II fire sIred'" nozzle of 50 ",m did",eler aI shown in Fig. 6.32. Thejel isslles 0141 wilh a reloClr)' of JJ ",h and strik.es rile willdow. Find Ihe angle orallgles of in dill ali 011 will! ...!!ieh Ihe jel issues from lite nozzle. Wlwi ... il/ be rhe WIIO/Ull of ",,,'er [dl/illg 011 Ihe willdow ? Solution. Givell : WINDOW d,=50mm,=.05m Di3. of nozzle.

Area.

A

= 2:(.05)! = 0.001963

10

I I

f

2

4 Veloc ity of jet. U,= 13ml'i. The jet is corning out from n07,7. le at A. It strikes the window and let the angle made by the jet at A wit h hori zo ntal is eq ual to 9. The co-ordinates of win dow. with respect to origin at A. .\" '= 5 Ill. ), '= 7.5 - 1.5,= 6.0 m The equation of the jet is give n by (6.24) as

,

7.5m WALL

1

1-- 5 m---.1 Fig. 6.32

Ii

~ I

IL 1304

Fluid Mechanics

,

'"

,a

y=xI1ln9- --, sec 2U"

9.S ! )(5 , , 11 + tan-91 2 x 13 -

6.0= 5 X (3n 9 6.0", .5

0 _ .7256 (1

(an

+

(a1l

2 EI)

e-

'" 5 tan .7256 - .7256 tan l e l 0.7256 lan e - 5 Ian 0 + 6 + .1256'" 0 0.7256 tan 2 5 tan + 6.7256 '" 0

e-

This is a quadratic equation in tan

e. Hence solution is 5±

I an

e '"

e

JSl - 4 x .7256 )(6.7256

"-''-'''---"""""",,,,,7-'''':'='' 2x.7256

'" s± J25

19.52 '"

~5;,:+,,2.,,34;C'

1.4 512

1.45 12

'" 5.058 or 1.8322

9= lan - I 5.058 or tan- I 1.8322", 78.8 0 o r 61.3r. AilS. Amoun t of W31t'T f alling on window = Discharge from nou ie '" Area of nozzle )( Velocity of jet at nozzle = 0.001963 x U = 0.001963 x 13.0 = 0.0255 m J/s. AilS. Problem 6.40 A "ozzle is J'il""kd at" diSlan",' of f '" "bOI"e the gmllml !e"e/ a",/ is inclined III an angle of 45 0 10 file Ilorizonlal. The diameler of Ihe "oule is 50 111m (llid Ihe jet V/Warer from Ihe ,'ouie s/rike5 the grOimd (1/ (l "oriZOlll(l/ diII(mce of.J Ill. Find Ihe Tille of flow ofw(IIer. Solution. Given: m Distance of noule abo ve grou nd '" I III Anglc o f inclination. 6", 45° Dia. of nozzle. d", 50 mill '" .05 III

1 _________ '

J~ , m

•

Area.

B

Thc horizontal distance .I" '" 4 III Fig. 6.33 The co·ordinates of the point B. which is on the ccmre·linc of the jet o f wate r and is situatcd on thc ground. with respect to A (ori gin ) are .1": 4 m 3ndy "'- I.O m (From A. poim B isvcnically down by I III I

,

Thc eq ua tion of the jet is given by (6.24) a.~)' "" x tan 9 _ ~s"c2e

2U

Substituling the known values as

-1.0=4tan450 -

9.8l~4 1

xSC(:145 0

2U

"" 4 _

7~~S x (J2y

{sec 45° "" COSI450

"" - : - ""

J2}

.fi

~ I

I~

~ I

IL Dynamics of Fluid Flow - 10=4 - 78.48x2

.

U'

"'

305 1

78.4&, x 2 '" + 4.0 + 1.0 " 5.0

U·

U1" 78.48 x 2.0 ,,31.39 5.0 u '" .J31.19 '" 5.60 111/s = Are a x Velocity of jet '" A xU ", .001963 x 5.6 1l) 3 lscc '" 0.0 1099 .. .011 nlls. AD S. Problem 6.41 A window. ill II ,'erlieal wal/. is a/ a disrance of 30 III abo!,t! rllt' grOlllld lel'el. A jet of "'(!fer. issuing/rom (l /louie of diameter 50 mil! is 10 slrike Ihe wi"do .... The rille offlow o/,,"(lier IhrOl'gil the lIou/e is 3.5 ",J/minute (lnd /lou./e is si/IUlled "/ (I dis/lweI! of I /II abo)'e ground le"f:/. Find th e greate:;1 /Iorizonlili di~'I{mce fro m th e .ml/ of IIII' /lou.le so l/ul/ jel afwaler MrikeJ' the window. Solution. Given: :,;.W INDOW Di stance of window from ground level = 30 m Dia. of nozzle. d=50mm=O.05m

Now the rntc of flow of fluid

, T

" , , A", "4 (.05 t = 0.001 963 In-

Are a

30 em

3

Q = 3.5 m /mi nute

The discharge.

x

= 3.5 = 0.0583 m3/s 6Q

Fig. 6.34 Distance of nozz le from ground '" I n1. Let the greatest horizonlal di51ance of the nozz le from the wa ll = x and le t an gle of inclination = O. If the jet reaches the window. the n the poinl B on th e window is on the centre-l in e of the jel. The co-ordinates of B with respect 10 A arc .r=x.)'=30 - 1.0=29rn The velocity of jcc.

U=

Dis<:ha r~e

Q .0583 '" - = = 29.69 m/kc

Are a A .00 1963 Using the equal ion (6.34). which is Ihe <"quation of jeC )'

gx

,

' sec- e = x lane - ~-

2V'

0",,' e

9.8h· 2 1 "" 2 x (29.69) = .f tan 0 _ 0.0055 seclO x x l .0055.1. 1 =.r lan 0 , cos- 0

29 .0 =.r lan 9 -

.. (i)

or tan e - .0055 .l leos'e - 29 '" 0 The

ma~imurn

value of or with res[)Cct to 0 is obtained, by diffcremi
dx o and Subsliluling the value of ~ [

I I

.r sec l 9 + tan 0 x

dxl

"9

"9

'" O. Hence differe miating the eq uation (il W. r.t. O. we have

[ , ((-2»).

- 0.0055 x x

I

d' ]

~-,- (-sIl10)+ ~-,- x h~

cos 0

cos 9

dO

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IL 1306

Fluid Mechanics

x tanB) = x scc 1J.: !!.-( d9 x sec

2

2

e + tan e dx and .!:!.- [~l = x 2 .!!...(_I_)+ _ 1_, --"-Ix'l) de de cos· e d(J COO l e cos · e de

2.r dx ] ,_.t·sln ' · 0 - .0055 [ oos J 0 + ----rcos O-'( 0 = 0

dx

e + tan 9 -IIG

For max imum val ue of .T. w.r.L

e,

we have d.l '" 0 dO Substituting this val ue in the above equmion. we hav e

, ' 0 - .00 55 .I SCC

[2X1sinej_ J _ 0 ,~

.OO55 X2x l Si nB _

x

\:05"9 -

"' "'

0

cool e

-

.\" - .011 X" iII" 9 = 0 or 1- .O ll .r lan .I"

o

orx -.011

.1sin8 __0 cosS

X .I

e=0

I Ian 9 = - - = 90.9

.,.( i i)

.011

90.9

... (i i i)

.r : - tan {I

Substituting this value of .r in equIItion (i), we ge l

90.9 lane

- - x tan

0

- .0055 x

(90.9)" I \ x ----;-:- - 29 '" 0 tan - a cos ' s

909 - 45.445 - 29=0 or 6 1.9 _ 45.~45 = 0 . si n ' e sin " a 45.445 . 10 = 45.445 61.9 = -,- or sin - - = 0 .7 341 si n "

e

61.90

sin e '" JO.7341 '" 0.8568

9= Substitutin g this va lue of

lan- I

e in equation (iii). we gel 90.9 Ian 9

x= - = 54.76

I I

.8568 = 58° 57.8'

~

n1.

90.9 _~9~O~.9= 90.9 = =- =54.759 In Ian 58° 57.8' Ian 58.95 1.66 An s.

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IL Dynamics of Fluid Flow

307 1

HIGHLIGHTS I. The .tudy of fluid motiun with the forces causing flow is called dynamics of fluid flow, which i, analysed by the Newton"s second law of motion. 2. Ilemoulli"s equation is obtained by integrating the Euler"s equation of motion. Bernoulli's equation Slales "Fo r a steady. ideal flow of an incompressible fluid. the lotal energy which consist, of pressure energy, kinetic energy and datum energy, at any point of Ihc fluid is ron,l;ml' . Mathematically •

..El.+ ,} 28

pg where

li

+t

_!!.l.+ ' - PB

,·i

28

+" '-

= pressure energy per unit weight = pressur<: head

"

~

'"

"

~inctic

energy pcr unit weight., kinetic head

z, • datum energy pcr unit

weighl. da!iJ111 head.

J. Bernoulli s equation for real fluids 'il + z : [11 + "i + z, +h L pg2g'pg2g"

..El.+ where

hL

K

loss of energy Detween section> I

~nd

2,

4. The discharge. Q. through a venturi meter Or an orifke meter is given by _C

Q-

°to;:

" J'lit - (/;:,

x J2gh

where {II '" area at the inlet of "enlUrimetcr. 0, z area at the throat of ,·cniurimelcr. Cd ~ co-efficienl of \'enturimeter, h '" difference o f pressure head in tcnns of fluid head flowing through venturimclcr. 5. The value of his gi\'cn by diffcrential U-tUDe manometer

h~ .{ [Sh

S"

-I]

h .. .,{ I - ::]

h ..

(~+
.

(when differentia l man01Heier contains heavier liquid )

... (when differential munometer comu;ns lightcr liquid )

... (for inclined ve1l1urimetcr in which diffcre1l1ial manometer contains

hC~Lvier

liquid )

... (for inclined "eolurimeter in which differential man0111elCr contains lighter liquid ) where x. difference in the readings of differe1l1ial manometer. S. '" .p. gr. of heavier liquid S• •• p. gr. o f fluid flowing through velllurimetcr S, .. sp. gr. o f lighler liquid.

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IL 1308 Fluid Mechanics 6 . PilOt-lube is used 10 fl nd the ,-c!ocity of a flowing fluid a1 any point in a pip" or a channel. The velocity is given by Ihc relation \I ~ C, J2gh where C, z co-efficknt of PiloHuhe h '" rise of li'luid in thc lube ~bo,'e free surface of liquid

. x[ ~: -1] (for pipes or channels). 7. The 1110menlum equalion Slales thaI Ihe nct force acting on a fluid tum

p<."'

second in that dire.:lion. This i, gh'en HS F

M

ma~S

is equal 10 Ihc change in

mom~n­

,"

.!!..(I/II·j

The impulse -momentum equation is given by F . dl "d{m"j. II. The force excned by a fluid On a pi"" bend in the directions of ..- and y arc given by

mass

F , '" - -

=

(Initial velocity in the direction of.r - Final wlocily in x-direction )

.;. In itial pressure force in .T-direclion + Final pressure force in - pQlV" - V.:.. J + (p ,A,), + (1'0.:) , Ii, = pQ I v" - V"J + (p ,A,)" + (p,A,)r

""'

Ii" _

Resultant force.

.r-dir~"::lion

JI-~l + 1-~1

and Ihe direction of Ihe rcsullalll wilh horizonl:!1 is Ian

a.

F ---.!....

F,

9. The forcc exerted by (he nozzle on (he walcr is givcn by F, " pQW.:.. -V" l and force exerted by the watcr on Ihe noUle is ~ - Fx ~ pQI V,-, - V.:..l. 10. Moment of momenlum equation Slales lhalthe resullanl lorque Hcting on a rotating nuid i, equal to lhe rate of change of 1HOlnent of momenlum. Ma1hema1ically.itisgivenbyT .. pQIV••::_V•• ll. II. f ree liquid jet is Ihe jet of water i"suing from a noule in atmosphere. The palh lravelled by Ihe free jel is parabolic. The eq uation of lhejel is given by

grl

Y "'x Ian 6 - ~sccl{l 2U

where x, y .. oo-ordinales of any point on jet w.r .I. 10 Ihe nozzle U .. velocily of je( of waler issuing from nozzle 6 • inclination of jet issuing from nozzle wi1h horizontal. 1 1 U sin 6 ll. (i) Maximum heighl attained by jel " "--"'-'~

" g

(iii) Time 10 reach highesl poinl, .,... •

!.. • 2

U sin {I

18

1

(il') 1·lmimm:!1 range of Ihe jet ..• _ '"

U sin 26 g

(.,) Value of {I for maximum range. {I. 45· ~ U1'g.

(..,1 Maximum range, x-__

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IL Dynamics of Fluid Flow

309 1

EXERCISE ( A) THEORETICAL PROBLEM S l. Name the di ffer':,I! forces present in a fl uid flo ..... , For the Eu ler"s equation of mOlion. which forces arc

taken in10 con,ideralion. 2. What is Eulcr"s equation of mOl ion ? How will you obtnin llemuull;'s equation from it? J . Derive Bernoulli's eq uation for the flow of an incompressible f.iclionlcss fluid from consideration of momentum. 4. State Bemoulli's theorem for steady flow of an incompressible fluid. Derive an expression for Bernoulli s theorem from first principle and Slale Ihe assumptions made for such a Jcrivalion.

5. What is a "cnwrimetcr ? Derive an expression for the discharge through a "cnlUrimetcr. 6. Explain the principle of "cnwrimet.:. with a oem sKelch. Derive the cxprcSSiOl1 for the rate of flow of fluid through it 7. Discuss the re lativc mer;ls and demerits of "c1l1urimeter with respect to oriflce -mctcr. (Delhi Ulliwrsily. Dec. 2(02) II . Dcfine an orifice ·melcL l>rovc that the discharge Ihrough an orificc-mc ter is given by the relation Q~C"

"aa,

r:>7 x ,,2gh

J",l - ,<5

'" ~ area of pipe in which orifice -muter is fitl~-d " 0 • area of orifice (Technical Uni.-ersity of M.P.. S 2(02) 9. What is a piloc -Iube? How will you dctennine the velocity at any point wilh the help ofpilOl-tube? (Dd"i UlliWfSily. Dec. 2(02) Ill. What is the difference between pitot-tube and pitot _static tube ? II . Siale the momcntum C4uation . How will you apply mOmentum equation for dClennin ing thc fo.-.:c cxcrted by a flow ing liquid on a pipe bend ? 12. Whal is Ihe difference belween momentum equmion and impulse momentum equation. 13. Define momenl of momentum equation. Whe.-.: Ihis equation is used. l~ . What is a free jet of liquid? Deri"e an e ~prcssion for thc path tra"clled by free jet i,suing from a nOl7.le. I S. Prove Ihatlhe equalion of Ihe free jet of liquid is given by the expression. where

l

gx l 2Uwhere x. y • eO-umptions made. How is it modified while applying in practice? List OUI its engineering applications. Define continuity equalion and Bernoulli's equation.

y ,. x Un 0 - -::--:-:osee 0

16 .

17 . Ill. 19 .

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IL 1310 Fluid Mechanics 20 . What arc the different forms of energy in a flowing fluid? Represent s<;hematically the Il ernoulli s equation for flow through a tapering pipe and show Ihe position of total energy line and the datum line. 21 . Write Eulers equation of mmion a long a stream line and integrate it to obtain BernouUi"" equation.

Stale all ass umptions {"ade 22. Describe wi th the he lp of sketch the construction. operation and u>c of 1';I01-slalic lube .

ll. Staning wilh Eu lds equation ormation along a stream line. Obtain Bernou lli's equation by its in!Cgralion. List al11he a"umplions made.

24. Slate Ihe different devices that one "In use 10 meaSure Ihe discharge through a pipe and also through an open channel. D,'scribe one of such devices with a Ilcal sketch and nplain how one can obuin the actual discharge with its help' 25. Derive Bemoullfs equation from fundamentals.

(8) NUMERICAL PROBLEMS 1. Water is flowing through a pipe of 100 mm diameter under a pressure of 19.62 Niem I (gauge) and with mean ~elocity of 3.0 mls. Find the total hend of the water at a cross-section. which is 8 m above the datum line. [Ans. 28.458 ml 2. A pipe. through " 'hich water is flowing is having diameters 40 cm and 20 Cm at the cross-sections I and 2 respectively. The velocity of wmer at sect ion I is given 5.0 mls. fi nd the velocity head m the scrtions I IAns. 1,274 m : 20.387 m; 0.628 mJlsl and 2 and also rate of discharge , J . The water is flowing through a pipe having diameters 20 cm and 15 cm at sections I and 2 respectively. The rate of flow through pipe is 40 litresls. The 'lCCtion I is 6 m above datum line and section 2 is 3 m above lhe datum. If the pressure at section 1 is 29, 43 Nlcm l . find the intensi ty of pressure at section 2, [Ans. 32.19 Nlcm' l 4. Water is flowing through a pipe having diameters JO cm and 15 em at the bottom and upper end respec tive ly. The inten~ity of pre .• sure at the bottom end is 29,43 Nlcm l and the pres,ure at the upper end i~ 14.715 Nlem1. Delennine Ihe difference in dalum head if the rale of flow through pipe is 50 Hlis [,\ns. 14.618 Illi 5. The waler is flowing through a taper pipe of length 50 m having diameters 40 em at the upper end and 20 emallhelowerend.allhe rate of 60 litresls. The pipe has a slope of I in 40. Find the pressure at the [Ans. 25.58 Nlem' l lower end iflhe pr~ss ure al lhe higher le~eI is 24.525 Nkm', 6. II pipe of diameter 30 em carries water at a "elocity of 20 mlscr. The pressures al Ihe points A and Bare given as 34,335 Nlem' and 29.43 Nlem' respecti,·ely. while the datum head at A and II arc 25 m and 28m. Find lhe loss of head belweeTT A and B, [Ans . 2 ml 7. A conical tube of length J.O m is fixed \'enically WiTh i" smaller end upwards. The velocity of flow at the smaller end is 4 mls while al the lower end it is 2 mis, The pressure head al the smallcr end is 2,0 m of liquid. The loss of head in Ihe tube is 0.95 \" 1 - ,·,)'ng. where "J is the velocity at the smaller end and ", at the 10"'cr end respectively, DClennine the pressure head at the lower end. Flow takes place in downward di re<:tion , [Ans. 5.56 m of fluid I II . A pipe line carrying oil of specific gravily 0.8. changes in diameter from 300 mm al a position A 10 500 mm diameter 10 a position B which is 5 m at a higher level. If the pre"ures at A and Bare 19.62 Nlem ' aT,d 14.91 N!cm ' respective ly. and Ihc discharge is 150 Htresls. detennine the loss of head and [Ans. J .45 m. I-l ow takes place from A to 8 1 direction of 110w, 9 . A horizontal venturimeler with in let and throat diamelers 30 em and J5 em respecl ivcly is used 10 measure the flow of water. The reading of differential manometer connected 10 inlet and throat is 10 cm of mel"(:ury . Detennine lhe rme of flow. Take CJ • 0,98 IAn s. 88.92 litreslsl

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Ii

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IL Dynamics of Fluid Flow

311 1

10 . An oil of ~p _ gr. 0.9 is flowing through a "cOlurimeter having inlet diameter 20 em and lhroat diameter 10 em. T he oil-mercury differential manometer shows a rcading of 20 em. Calculate the discharge of oil through the horizontal venturimcl"r. Ta ke Cd " 0.98. IAn s. 59.15 lilrcslsl II . A horizontal venturiml-('" Wilh inlet diameter 30 em and throat diameter 15 em;, used to measure the flow of oil of sp . gr. 0 .8. Th e discharge of oilthroullh venturim,,!cr is 50 litTCsls. find the reading of the oil mercury differential manometer. Take Cd ~ 0.98. IAns. 2.4!!9 eml 12. A horiwntal vcnturimeter Wilh inlet diameter 20 em and throat diameter 10 em j. u-"cd 10 meas ure Ihe flow of water. The pressure at inle! is 14. 715 Nlcm 1 and Va~Uum pressure 31 the Ihroal is 40 em of mercury . Fin
C".

21. Find the vclocily of flow of nn oillhrough a pipe. when the d ifference of mercury level in a differenti al U-tube mUnomeler ,:onneeted to the two tappings of the pilOt -tube is 15 crn. Taite sp. gr. of oil ~ 0.8 and co-effle ien l of pitot_tube as O.9 R. [An •. 6.72 m/sl 22. A sub -marine mOveS horizontally in sea and has ils axis 20 m below the surface of waler. A pitot -st:lIic lUbe placed in front of ,ub· marine and along it' axis. is connecled to the two limbs of aU -tube containi ng mereury . Th c difference of mercury IC"el is found to be 20 em. Find Ihe speed of sub-marine. Ta ke sp. gr. of mercury 13.6 and of sea-wate r 1.026. [Ans. 24.95l\ kmlhr.1 23 . A 45 " reducing bend is connected in a pipe line, the diameters at the in lel and outlet of the bend being 40 C111 and 20 em respectively. Find the force exerted by water on the bend if the intensity of pressure at inlet of bend is 2 1.58 N!cm'. The rate of flow of water is 500 litres/s. [.... ns. 22696.5 N; 20" 3.5'1

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IL 1312 Fluid Mechanics 24 . The discharge of water through a pipe of diameter 40 em

i~

4()() ]ilres/s. If the pipe is bend by I J5°, find

the magnitude and direction of the re
29.43 N/c", l, [Ans. 7063 .2 N. 8 _ 22 ° 29.9' wilh x-axis clockwise[ 25 . A 30 em diameter pipe carries watcr under a hca
[Ans. 11717.5 N. 8 _ 67" 30' 1

",'!"""

26. A pipe of 20 em diameter conveying 0 .20 of water has a right angled bend in a horizontal plane. Find (he rcsull:mt force exerted on (he bend if thc pressure at inlet and outlet of the bend are 22.563 Nlcm'! Dnd 21.582 NiemI resf>C<;ti\'cly. I An~. 116047 N. 0 " 43" 54.2']

27. A n07.zle of diameter 30 "nn is fitted 10 a pipe of60 nun diameter, Find lhe force exerted by the nOlzle on the water which is now;ng through the pipe at the rate of 4.0 mJ!minute. [Ans, 7057.7 NI 28. A lawn sprinkler with two nozz les of diameters 3 mm each is connlXted across a tap of water. The nozzles are at a distance of 40 em and 30 em from the centre of the tap. The rate of water through tap is 100 em 3!s. Th e nozzle discharges water in the downward directions. Detennine the angular speed at which the sprinkler will rot"te free . [Ans, 2.83 ""tlsi 29 , A lawn sprinkler has two 'tozzles of diameters 8 mm each at the end of a rotating ann and the "ciocity of now of water from each nOI.l.le is 12 m/s. Onc noule discharges water in the downward dirlXtion. while the other nozzle discharges water ~ertically up. The nozzles arc at a distance of 40 em from the centre of the rotating ann. Detennine lhe torque require{! to hold the rotating ann stationary. Also detennine the constant .
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IL Dynamics of Fluid Flow

313 1

(I) the difference in readings of Ihe Iwo pressure gauges. which are connected to the two pressure

tappings, and (iil Ihe difference in the level of the mercury columns of the differential manometer which is connected to the tappings. in place of pressure gauges . [An s. (i)42.928I
.

II

l

•

_

11,= _(. 1 ) =O.Q07854m ",UsmsQ=Cd

-

.

,

,

m. II," _ (.2") ",0,0314 m" 4

4


n, •

Ja,2- ai

M-::i:

x .. 2gh

Find 'h , This "3lue of

h '" 5.17 Ill.

Now use

, • [.!!L!i) .;. (ZI _ '2)' where p .. 800 kg/m!, find {PI - I'll. P8 pg

Now use the fonnub

where h" 5 . 17 m, S," 13.6 and Sf" 0.8. Find the ~alue of x which will be 32.3 em. I JII. A venturimetcr is installed in a 300 mm diameter horizontal pipe line. The throat pipe rates is 1/3. Water flows through the installation. The pressure in the pipe line is 13783 Nkml (gauge) and vacuum in the throat is 37.5 em of mercury. Negl"'ting head los< in the vemurimcter. delennine the rate of flow in thc pipe linc . [Ans. O. 153 m'/sec[

[llin t. ill ,. 300 mm .. 0.3 m. ill " Hence

Value of

"3I x 300 ,. 100 mm = 0.1

• ~ . m. P t ., 1].783 Nlcm " .. 13.783 x Iv Nhn -.

p,lp x g ~ (3 .783 x 10'/1000 x 9.81 '" 14.05 m. P/P8 '" - 37.5 em of Il g = - 0.375 x 13.6 'n of water .. - 5.1 In of water. Hence h .. 14.05 - (- 5.1) '" 19.15 m of water. CJ

,.

1.0. Now usc the fonnula Q w CJ

J
39. The ma~imum flow through a 300 mm diameter horiwntal main pipe line is 18200 litre/minute. A vcnturimcter;< inlroduce
fond the "alue of " , ' Thcn " , '"

.

,

"4 d i

and find d , .]

40. The following arc the data given of a change in diameter effected in laying a water .• upply pipe. The change in diameter is grddual from 20 ern at A to 50 ern at H. l'ressures at A and Hare 7.848 Niem I and 5.886 Nkm ' respccliYcly with the end B being 3 m higher than A. If the flow in the pipe line is 200 litre/s. find : (i) direction of flow, (ii) the head lost in friction I:>elwecn A and 11. [Ans. (i) From A to 8. (ii) 1.015 111 ] [Hin t. D A .. 20 em = 0.2 m. DB = 50 em '" 0.5 m. PA = 7.84 8 Niem I = 7.848 x 10"' NIm" fiB = 5.886 Nlem 1 = 5.886 x 10" Nlrnl. ZA '" O. ZB = 3 m. Q '" 0.2 m' ,.

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IL 1314 Fluid Mechanics

v'

E~ .. (r~/p x g ) .;. ....d... .;. Z~ .. (7_848)( 10' /1000 x 9 _81) ';' (6.369'12 x 9.81) + 0 .. 10.067 m

"v'

EB .. (I'i/r )( g) + ~ + ZH"' (5.886 x 10'/1000 x 9.81) + (l.DlS"12 x 9.81 ) .;. 3 .. 9.052 ml 2, 41 . II ,"cnt urimeter of inlet diameter 300 mm and throat diameter 150 mm is fi ~ ed in a vertical pipe li ne. 1\ liquid of 'po 8f. 0. 11 is flowing upward th rQugh the pipe line. II differential manometer containing mercury Siy"," a reading of 100 mm when connected at in let and throat . The vertical difference belween inlet and throat is 500 nllll. If Cd " 0.98. then find : (r) Me of flow of liq uid in lilre pcr seeo",' and (i') differcn.:c of pressure between in lel and throat in Nlm ' [AIL<. (i) 100 litre/s. (i') 15980 N/ml ] 4 2. 1\ "colurimeter with a throat diameter o f 7.5 CHI is installed in a IS em diameter pipe . The pressure al the entrance 10 Ihe meIer is 70 kPa (gauge) and il is desired Ihallhe pressure al any poinl sho uld nOI fa ll below 2.5 m nf absolute water. Detenn ine the ma~imUln flow rate of water through the meter. Take Cd " 0 .97 and atmo spheric pressure as 100 )':Pa. (J.N.T V .. Ifyliaab(l(/ S 2(02 ) IHln t. The press ure at the throat will be minimum . Hence ~ '" 2.5 m (abs .)

.,

pg , tI , '" 15cm:. A , '" 4" ( 15-) '" 176.7 em

G iven:

"

til "' 7.5 em ;. Al ", - (7 .5 ) ", 44. 175 em -' 4 p, : 70 kPa : 70x 10) Nhn l(gauj;ej,p.... '" 100 k Pa '" loox 10J Nhn l PI (abs. ) .. 70 x 101 + 100 X 10J .. 170 X IO J Nhn l (abs .)

P Pi:

LL ..

h ..

170 x 103 .. 17.33 m of water (abs. ) )OOOx 9 .81

.!i_ ~ pg

Q~

Now

PS

C"AI A1

z 17.33 _ 2.5 .. 14.83 m o f waler .. 1483 em o f water

JA,!- Ai

r:>7 O.97 xI 76.7 x44 .175x.J2x 98IXI483 x,,2Xh '"

J176. 71 -44 .175!

z 7S.41UI litre/s.] 43. Find the discharge of waler flowing through a pipe 20 em diameter placed in an inclined position. where a vcnturimCler is inscrted, having a thro at diametcr of 10 cm. The difference o f pressure between the main and Ihroat is measured by a liquid of specific gravity OA in an inverted V-tube . which gi"cs a reading of 30 CI1I . The loss of head Octween the main and throal is 0.2 timcs the kinetic he~d of pipe. (i>elh; VIII"t.",sily. Dec. 2(02)

II " II " Illin t_ G ivcn : ii, '" 20 em :. " I '" -(20' ) = 100 II em' ; tI,_ = 10 em :. a,. '" -( 4 4 10-) ~ 25 II em-.

[ S"s, ] ( 01.0.4 )

x .. 30 em.h ..... 1- ----;- .. 30 1- -

Il UI h is

I I

~1 8 cm

.. O.1 8 1TI

al
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IL Dynamics of Fluid Flow

315 1

"'28

h L ",O.2x..:.L

Ii

From Bernoulli's C
PI(

m

1':"

-+- - '-

28

f' V" +" .. ~ -+- ---.L -I- z: + hL

pg

21(

(l!L+ZI)-(!i+ Zl) + .1.._~ = ilL pg pg 28 28 "

O. 18 +~ - ~ .. 0.2 "1 28

28

, [

28

.. (12 + pg

ZI)-(rlpg Zl) .. o.18m and -+-

hL ..

0.2 v,l) 21(

v,z

\til V,l 0.2 0.8 \li l V,l 0.18 + - - - ' - - - - = 0 orOIS + - - - - " - - = 0 28 28 28 28 2g

From continuity l"lUalion.

Now

(I,

V, = ",V,

O.I!! + 0.8 V,l 28

~4V,

or

_:1 .. 0 or 0.1 8 + 0.8 V,! _ (4\)2 28

0. 18 -+- 0.8 \<]1 _ 16"1 28 28

1

28

~0

28

= 0 or 0.1 8 = 16111 _ 0.8 V,l = 15.2\11

2$

21(

21(

~O:II~'~'~'~'~9~'~1 = 0.48 mls = 48 crnls

V, = ,/

IS.2

,

,

, Q = A , V, = "4 (20" ) x 48 = 15140 em Is = 15. 14 li tre/ • . 1

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.. 7. 1

INTRODUCTION

Orifice is a small opening of any cross-section (~uch a~ circular. triangular. rectangulardC.) on the side or al Ihe bonom of a tan k. throug h which a fluid is flow ing. A mouthpiece is a short lenglh of a pipe which is IWO to three times its diameter in lenglh. fllted in a tan k or vesse l conlaining Ihe fluid. Orifices as we ll as mouthpieces arc used for measuring the rale of fl ow of fluid .

.. 7.2

CLASSIFICATIONS OF ORIFICES

The orifices arc classified on the basis of their size, shape, nature of discharge and shape of the upSlrealn edge. The follow ing are Ihe important classifiea t io n ~: I. The orifices arc class ifi ed as s ma ll orifice or la rgt: o rifice depending upon the size of orifice and head o f liquid from the centre of Ihe orifice. If the head of liquid from the centre of orifice is more than five times the depth of orifice, Ihe orifice is c all ed small orifice. And iflhe head of liquids is less Ihan five limes Ihe dept h of orifice. il is known as large orifice. 2. The orifices arc classified as (,) Circular ori fi ce. (i,) Triangular orifit""', (ii,) Rectang ula r orifice and (iv) Sq uare orifice depending upon their cross-secliollal areas. 3. The orifices arc c l a~sified as (i) Sharp-cdged oriflce and (ii) Bell-mouthed orifice depending upon Ihe shape of upslream edge of Ih.:: orifices. 4. The orifices are c lassifi.::d as (i) Free diseharging orifices and (i,) Drow ned or sub-merged orifices depending upon Ihe nalure o f discharge. The sub-merged orifices arc further classified as (al r ully sub· merged orifices and (b) Parti ally sub-merged orifices .

.. 7.3

FLOW THROUGH AN ORIF ICE

Consider a tank filled wilh a ci rcular orifice in one of ils sides as shown in Fig. 7. 1. Let I I be the head of the liqu id above the centre of the orifice. The liquid flowing through the orifice fonns II jet of liquid whose area of cross-section is less than that of orifice. The area of jet of fluid goes on deneasing and at II section C·C, the area is minimum. This section is ~pproxim~tdy al a distan ce of half of dia meter of the orifice. At this section, the streamli nes arc straighl and parallel to each other and perpendicular 10 the

317

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IL 1318 Fluid Mechanics plane of Ihe orifice. This sec ti on is ca ll ed Venu-contr"cla. Beyond thi s section. the jet diverges and is anrac lcd in the downward direction by the gravity. Consider twO poinTS 1 and 2 as shown in Fig. 7. 1. Point 1 is inside 111e tank and point 2 31 the vcn J-contracla. LC11hc now is stead y a nd at a constant head H. Applying Bernoulli's equation at points I and 2.

,

",i!!+I'i+ z, pg 2g -

""'

Fig. 7.1 Tank with " 11 orific~.

ZI '" Z2

,

'

P! -PI + -VI = +I'i -

pg

2g

!i

Now

pg

2g

=H

pg

P! '" 0 (atmosp heric pressure) pg I' , is very s mall ill comparison to v 2 as area of tank is very large as compared 10 the area of the jet of

.,,

liquid.

f/+O=O+ - ' 2g

.J2gll This is th eoretical ve loc ity. Actual veloci ty will be Jess th a n this va lu e. \'2 '"

to 7.4

...(7.1 )

HYDRA U LIC CQ -EFFICIENTS

The hyd raulic co-cfflcients are I . Co-efficient of velocity. C. 2. Co·efflcielll uf cumractiun. Co ]. Cu-efflcien! uf discharge. Cd' 7 .4 . 1 Co-e fficient of Velocity (C y ) ' It is defined as the ratio between the actual velocity of a jet of liquid at vena-con tracta and the theoretical velocity of jet. It is denoted hy C. and mathemati cally. C. is given a.s

C. = =

Actual velocit y of jet at vena -cont raeta Tlleoretieal velocit y

v

r:>::U '

v2gJ1

whe re V = actual veloc ity . .)2gH '" Theoret ica l velocit y

... (7. 2 )

The value of C,. varies from 0.95 10 0.99 for different orifices. de(l<:nding On the ~ha pc. size of the orifice and on the twad under which flow takes p lace. Generally the value of C. = 0.98 is taken for sharp-edged orifices.

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IL Orifices and Mouthpieces 319 1

7.4 .2 Co-efficient of Contraction (C e) . h is defined as the rmia of Ih e area of the jet at vena-contracta to the area o f th e orifice. It is denoted by Ce . (I '" area of orifice and G" = area of jet al vcna -co n lr~C l a.

C,, =

T hen

area o f jet at \'cna-con tracta area o f orifice

'" G o

...(7 .3)

"

The va lu e of Cc varies from 0.61 100.69 dep"nding on slla pe and size of [h e orifice and tlead of liquid unda w hich flow takes p lace. In ge nera l. the va lu e of Cc m~y be taken as 0.64. 7.4. 3 Co-efficient of Discharge (Cd)' It is defined as th e mtio of the actua l discharge from:m ori fice to the th eoretical discharge from the oriti<:c. It isdc nolcd by CJ' If Q is actual di scharge i1 nd Q,. is Ih" th~"(Irc tical discharge then mathematically. Cd is give n as

Cd'"

JL", Q,h

Actual ve locil y x Actual area Theoretical ve locit y x Th eoretica l area

I : ",~A" ':"C'CI:'O'e""oc"'";'~Yc:: x 'CCA"'C""'O ' C,~''C'C' ::Theore tica l ve loc ity T heoretical area ... ( 7.4)

Cd",C"x C c

The value of Cd varies from 0 .6 1 to 0.65. For genera l purpose the va lu e of Cd is take n as 0.62. Problem 7 .1 The head of W(lter o\"er (m orifice of diameter -10 mm is 10 m. Find the actlla/ dischargc ""d "elua/"e/oeily of thc jet al ,·c"a-contracta. Take Cd '" 0.6 ,md C. '" 0.98. Solution. Give n: H~nd. H = 10cm Dia. of ori fi ce. 40 mill 0.04 III

,,=

Area,

a

= 4,

= . =.001256 ,

~ (.04 )-

111

Cd = 0.6

C. '" 0,98

-c.C , A"':"e'C'cdC;"~""h,",~gc'=

(i)

Th eore tical di sc harge

BUI

: 0 ,6

Theoretical discharge = V'h x Area of orifice

V/h '" Theoretical veloc it y. where V'h '" J2g H = J2 x 9.S ! x JO =14 m/s

m2

,

T heoretic al discharge = 14 x .00 1256 = 0.0175SAc tua l dL.;;charge

= 0.6 x Theoretica l di scharge = 0.6 x .0 175S = 0.01054 mlfs. Ans .

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IL 1

320 Fluid Mechanics

-o;CA"' :; C"'C"C"C'CICOC';"""Y;:::

(ii )

Tht:orctical \'d ocit y

'" c. = 0.98

Actual ve locity '" 0.98 x Thcorctical ve locity '" 0.98 x 14", 13.72 m/s. Ails. Problem 7.2 Th e head of \I"o/er o"er tile Cenlre of (ill orifice of diameter 20 """ is I m. Th e {lelul' / d;~'c1wrge Ihrough IIII' Qrifice is 0.85 lilre/s. Find Ihe co -effieien, of discharge. Solullon. Given: Dia. of orifice. d = 20 mill = 0.02 III Area.

l/=

~ (O.02 ) 2 '" 0.000314111 2 4

Head.

H ", I

Actua l discharge,

Q = 0.85 litr.. /s = 0.00085 1I1 3/s

Theoretical ve loci ty.

III

V'h = J2g H = J2 X 9.8 1 X I = 4.429 Ill is

Theoretical discharge. Q,h = V,h x Area of orifice = 4.4 29 x 0.000314 = OJ,OI39 ml/s .. Co-effici e nt of discharge =

... 7.S

Actual discharge 0.00085 = '" 0.61. AilS. Theoret ic al discharge 0.00 139

EXPERIMENTAL DETERMINATION Of HYDRAULIC CO - EFFICIENTS

7 .S . 1 Determination of Co-e.fficie.nt of Discharge (Cd)' T ile waTer is allowed to flow through an orifice filled to a tank under a constant head. H as s hown in rig. 7.2. The water is collecTed in a measuring Tank for a known time. /. The heighT uf water in the measuring Tank is noted dow n. Then acwal discharge through orifice.

Q" ,A"C'c.,",r"'"'''''="=","c,~,,=,,'k~Xc;H='=i'.'hC'CoCf~w,'=',,·'=i="~m='c·c',""=·,".,,',"='k= Time (/)

and

thcOrc1ical disch~rge = area of orifice x ~2gH

S~Y~ WATER

I

I

'!

b"",',h, ,r :,.·l

1

Fig. 7.2

Valu~ofCd.

. .. (7 .5 )

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IL Orifices and Mouthpieces 321 1

7.S .2 Determiniltion of Co- efficient of Velocity (C v ) ' Let C-C represents the ve narontracta of ,( je t of W3lCr com ing out from an orifice under constant heal! H as s ho wn in Fig. 7.2. Consider a liquid particle which is at vc na-co ntraCIa at a ny time and takes Ih e position al P alon g the j et intim c '(. Let .1 '" horizontal distance tra ve ll ed by the part id c in tim e ' r' )' '" vertic al di stance betwee n P and C-C V", actual velocit y of j et al vc na-colltracta. The n ho rizomal di stance , x'" V X I . ..( i) I , ...(ii) and vertic al di stance. y= - gr 2

Prom equatio n (i) ,

"

1= -

V Substituting thi s valu e of 'r' in (iil, we get

.( 1 1 Y= - g x - ,

2

V-

BU11hcorcti cai ve loc it y.

:. Co-cfficic nl of I'c loc it y. C. '"

~ = ~gXl V", 2y

X

~ = J4yH x'

"Z g H

"

= J4 yH '

7 . S. l Determination of Co -efficient of Contraction (C e) . is dcle nnilled from the equat ion (7.4) as

...(7 .6 )

Th e co-cfnci ent of contracti on

Cd=Cv xCe C = Cd c C,'

... (7.7)

Problem 7 .3 A jel of !l'afer. issuing frolll a sllarp-edgell rerrical orifice under a conSlall1 head of 10.0 cm. af a cerlain poim. liaS 'he horizon/aland !'erricai co-ordinates measllred fro/ll IIII' l·nla·comraC/II (IS 20.0 C/ll Imd /0.5 elli resper/ire/y. Fi" d 'he mille of C,.. A IIO find /111' I'lllue ofCe if Cd = 0.60. Solution. Give n :

Head. Horizo ntal di s tance. Vc n ic al dis tan ce .

I I

II = 10.0 ern

x = 20. 0 c rn y = 10.5 ern Cd = 0 .6

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IL 1322

Fluid Mechanics

The value of C,. is given by equat ion (7.6) as C = _ .'_ " .)4)'/1

'"

J4

20.0 x [0.5 x 10.0

20 - - '" 0.9759 == 0.976. An s. 20.493

The value of Cc is given by equation (7.7) as

CJ 0.6 C, = - = - - == 0.6147 '" 0.6 15. Ans. C" 0.976 Problem 7.4 The head o/wafer ol'a (lfl orifice of diamerer 100 mm is 10 m. The waler coming Oll[ from orifice is collected in a circu/a, /mlk of diameter 1.5 m. The rist> of water iel'e/ in Il,is lank is 1.0/11 in 25 seconds. Also 1/11.' co-ordinates of (l poinT 011 1/,1.' jet. measured'rom rena-COll/wel« ar/! 4.3 In /Iorizon/ll/ (lnd 0.5 lit l"I:rlica/. FillJ tile co-efficients, Cd> C. (mlf Cr' Solution. Given: H~ad. H == 10 III Dia. of orifice.

Area ofoJificc. Dia. of Illca5uring lank. Area.

Rise of wate r. Time. Horizontal distance. Venical distance.

d= lOOrnrn:O. lm (I'"

,

,

,

- (.1)' '" 0.00785) m ' 4

D: 1.5rn 2

It

A= - (1.5) =1.767 m

2

4

11= 1111

t = 25 seco nds x = 4.3 III y = 05 m

Now theoretical velocity. V,h: .J2g H : .J2 x9.81 x 10 : 14.0 m/s Theoretical disch~rge, Q'A : V,h x Area of orifice: 14.0 x 0,()()7854 = 0.1099 m 3/s Actu al disc harge.

Q: II

XII :

1.767 x 1.0 = 0.07068

,

Q

25 0.07068

Cd: = : 0.643. Ans. 0.1099 Q'h

The value of C,. is given by equation (7.6) as

c= _ .' - = ,.

.J4yH

4.3 .J4 x 0.5 x [0

43 4.4 72

= 0.96. Ans.

C.. is give n by eljuation (7.7) as C( '" CJ = 0.643 = 0.669. AilS. C, 0.96 Problem 7 .5 1V(ller discllarge (II Ihe mle of 98.2 lilresls IlmlUglIll 120 mm di(lmeler ..erliCCII J'/r"rp -edged or iJke placed II/1du 1I COIIS(lIIll head of 10 melres. II pOilll. Oil Ille jel. meaSll r ed from II,,! I'el!(l-COII/mcw of II,e jel I!(IS co-ordillmes 4.5 melres horiwllwl and 0.54 melres I'erli[(ll. Filld Ihe co -eJJrcielll C,. Cc «lid Cd of Ihe orifice,

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IL Orifices and Mouthpieces 323 1 Solution. Given: Discharge. Dia. of orifice.

Q'" 98.2 litIs" 0.0982 ml/s d= 1201l11n=O. 12m

:. Area of orifice,

O= ~ (O. 1 2)!=O.Dl13 1

ml

fI:lOm Head. Horizontal distance of a point o n the jet from vc na-contracta . .\' '" 4.5 m Jnu vertical distance. }, = 0.54 In Now theoretical velocity.

Theoretical

disc h~rgc.

V'h = J2g x H '" J2 )(9.8 1 x 10 = 14.0 m/s Q'h'" V'h )( Area of orifice 3 '" 14.0 x 0.QI131 '" 0. 1583 m fs

. . Actual discharge Q 0 ,0982 The val ue o f C, IS g ive n by. Cd = . '" = - - - = 0.62. ADS. , TheorclJcal discharge Q'h 0.1583 The value of Cc is given by equation (7.6). C"'x= 4.5 '" 0.968. Ans. '" J4yH J 4x O.54xlO

The value of Cr is g iven by equation (7.7) as

Cd

C< '" -

C,

0.62

= - - = 0.64. Ans.

0.968

Problem 7.6 A 25 mm diameter nozzle di.~c1/{/rge$ 0.76 mJ of water fler minule when Ihe Ilead is 60 III. The diallleter of IIIe j,'1 i.1 22.5 111111. Delerlllin e : (i) Ihe I'allle$ of co-efflciet!/$ C". C. alld Cd ali
Solution. Givcn : Dia. of nou:lc. Actual discllargc.

~~3a.d~fjCt.

D=25mm",,0.025m Qoct " 0.76 m]/minut c =

0~6

,,0.0 1267

t I3 N~"" I ~;'~~ --=~3:.= ==3)-

~:~.~nmm"0.0225m.

(i) VoilleJ ofco-eJftcicllI.l: Co-cfficicm o f contraction (C~) is givcn by.

C _ C -

111]/S

-

.

Area of jcl Area of nozzle

!!. d' = _4__ " !!. Dl

Fig . 7.3

~= 0.0225l l D

0.025

1

,, 0.81. Ans .

4 Co-efficient of discharge (Cd) is givcn by. C _ d -

II

Actu al dischargc Th core ti<.:al disc hargc

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IL 1324 Fluid Mechanics 0.01267

===~"'=== Theo retical vel oc ity x Area of non le =

0.0 1267 0.01267 --="';;---:J2gH X~D " J2 x 9.81 x60 x 4 (0.025)' 4 It

'" 0.752. An s.

Co-efficient of velocity (C.) is

~ivcn

C. ""

by.

Cd

0.752

Co

0.81

- = - - '" 0 .928. An s.

(ii) Loss of ilead due to fluid feJ"iJ'/allce :

Apply ing Be rnoulli' s equation 31thc oU llet of nozz le and \0 thc j cl of water. we gel ~"

l!!.. + _,_ + z, '" pg

28

V

1

Pl + -L + II + Loss of head pg 28

~ '" PI", Atmosphe ric pressu re head pg pg

z, '" l l ' VI '" .)2811 . Vl '" Actual ve locity of jet'" C. J2g H

(.fiiii)'

=

(C, ~2,H)'

2,

+ Loss o f head

H:C;xH+Lossofhcad

0.

=1I -C;x l/ : I/( [ _ C,2) '" 60(1 - 0 .928 2) '" 60 x 0.1388 '" 8.328

:. Loss o f head

m. Am .

Problem 7.7 A pipe. 100 ",m in diumeter, Iws (l /JOule aI/ached to il at rhe discharge elllf. til e diame ler of Ihe IIoufe is 50 "'nt. Tlu" Tole of discharge oj"'(1/er Ihrough Ihe IIou/e is 20 lilres/s ,/lid Ihe I'fes~'''re al Ihe base oj the nou/e is 5.886 Nlem l. Ca/cu/ale Ihe ('o -efficient oj disclwrge. An"me Ilwl Ihe ba~'e oj Ihe nozzle ,/li d OIl1lel oj Ihe nou /e a fe at Ihe same e/el'lliion. Solution. Given: D= 100111111 =0. 1 m

Dia. of pipe.

"

AI= 4 (.0- = .007854 111-, Dia. of nozz le.

d = 50

Al =

111111 '"

4" (.05)

0.05

G)

NOZZLE

c_~,,~=t:/0~'~.tT

~ipe = .001963 III , III

,

BASE OF NOZZLE

AClUal discharge.

Q = 20 litis = 0.02 1I1 3/S

Pressure at the ha st:.

, , N PI'" 5.886 N/cill '" 5.886 x 10 - ,

PI = 5.686 Nlcm Fig . 7.4

m'

.007854 VI'" .001963 V1

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IL Orifices and Mouthpieces 325 1

_ V1 4 where VI and V1 arc theoretical ve locit y at se ctio ns ( I ) and (2 ). App ly ing Bcrno ulli 's eq ua ti on at sec tio ns ( 1) and (2) , we gel V

I963Vl ,-_ .OO.007854

,l

p,

V,1

-PI + -V = - ' + -'pg 2g pg 28 V, )'

4 5.8S6x 10 + ( 4"

I OOOx9.8 1

'" 0 +

v/

1.. l!.l. '" Atmosph eric press ure =01 1 pg

2g

28 V1

V1

.,-"''cc, = - '2g x l 6 2g

6 .0 + ,

2~ V " [1- 16' ] = 6.0

v,'

or 2~

[15] 16 = 6.0

V1 ", J 6.0 X 2X9.8 1X :: "" 11.205m/scc Tlicoretic al d ischarge

'" V 1 X A l = 11.205 x .00 1963 '" 0.022 111 3,s _ C,_

Problem 7 .8

Actua l diiiCh argc _ 0.02 _ o . ~ _ - - _ .7"", Ans. Th eo re tical d isc harge 0.022

A I(mk has Iwo ,dell/ie,,/ orifices VII one of ;I~' l'erliCa/

be/ow Ihe ,..aler $lIr!
~·ide5.

Sur/IK e.

x = horizo ntal di stance of P )"1 '" ve rtica l di stance o f P Fro m o rifi ce ( I ) )'1 '" ve rt ic al di stance of P From o rifice (2) Th en )'1"')'2 + (5 - 3) ==.\'2 + 2 III The value o f C. is given by e quat ion (7.6) as

I I

The upper orifice is J m

If the 'Oll/ue of C./or cad,

Fi g. 7.5

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IL 1326 Fluid Mechanics For orifice (I ).

c

For orifice (2),

C '"

",

=

.r

.,)4v,

H,

., J4y.r fl l

= =

l

"

.. ,(1)

.)4)', x 3.0

., .j4 X)"

.•.( i i)

x 5.0

As both the orifices are identical C" '" C"

"' B",

Y'''')'2 + 2 .O 30'2 + 2.0) '" 5)'2 2)" '" 6.0 y, '" 3.0

From (iI),

"'

0.96 '"

::'~' ~~ J4)( 3.0 x 5.0

.r ", 0.96 x ,)4 x 3.0 x 5.0 = 7.436 nl , An s. Problem 7 .9 A closed I'essei cOllwil lS 'WI/er "(I/O a IJeigill of 1. 5 m 01111 o\'er II,e \l'llter surftlce IllerI' is air /wl ling pressure 7.848 Nkm' (0.8 kgf/cm") lIbOl'e llimospheric f!"'smre. Alllie bOffom of Ihe ,'ej'se! Ihere is at, orifice of diameter I(){) mm. Fi" d Ihe rule of flow of ",,,'er /rOlll orifice. T"ke Cd'" 0.6. Solullon. Gil'en : Dia. of orifice.

If '" 100 mm '" 0. 1 III

p " 7.848 Nlcm

Cd'" 0.6 Hdglll of water.

If '" 15 III

Ai r pressure. p '" 7.848 Nfcm 2 '" 7.848)( 104 N/m' Apply ing Bernoulli's eq uati on 3l sections ( I) (water surface) and

(2), we gd

II

v: 1

P

Vl

_' + _1 +ZL"'.:...l.+..:.L+Z, pg2g pg2g ' Taking datum linc passing throug h (2) whic h is very clo.'iC to the bonom surface o f the tank. Then Zl = 0, tl = 1.5 III Also

"od

,

............. -. --.- ---. _ .. •. -.-. -. '

I I L---1 !,t----'

H

T ~1.5

m

1

Fi g . 7.6

l!1.. = 0 (allnospheric pressure) pg

..!!J.. pg

= 7.848 x 10' 1000)( 9.& 1

v;

8+0+1.5=0+ ---+0

,

2,

=8

III

of W3!cr I Vj is negligible}

9.5 = V2 2g

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IL Orifices and Mouthpieces 327 1

v!

J2x9.8lx9.5 _13.652m/s Rate of now of wawr = Cd x a 1 x V ,

n

~

3

1

:0.6x "4 (. 1) xlJ.652m/s: O.064J m /s. An s.

Problem 7.10 A closed {(10k partially filled willi wat'" UpfO (j heighl of 0.9 III hm'ing lIIl orifice of diwlleler 15 mm a/ the bOl/om oflhe wllk. The air is pumpl!(/ inlO Ihe ljpper part of the Iilllk. De/ermine tI!I' preJJure reqllired for a disc/mrge of 1.5 lilres/s Ihrough Ihe orifice. Takf' Cd = 0.62. Solution. Given: Heiglll of wa ter above orifice, If = 0.9 In Dia.uforifice. d= 15mm=O.O I5m

'4'[']" t/" = '4 (.015t = 0.0001767 In-,

:. Area.

(/ =

Discharge.

Q = 1.5 litrcsls = .0015 mlls

Cd = 0.62 Let p is intensity of press ure required above water surface in Nlcm 1. p x 10" lOp - - III of W3Icr. = "'~"''"' pg lOOOx9.8 1 9.81 p

Then pressure head of ail'= -

If V, is the v,",ocity at outle t of orifice, then

2 x 9.81(0.9+ lOP) 9.81

Q==CJxax J2g(H+p/pg)

Discharge

.0015 == 0.6 x .0001767 x ~r.2-X~9~.S~I" (O~.9~+c,,~1p - ,') r-~---co"

2X9.81(0.9+ lOP) = .0015 = 14.1 48 9.81 0.6x.0001767 or

2x9.8 1 ( 0.9+

IO P

9.81

)= 14. 148 XI 4.148

to II == -,14.,IC40SC'CIC4CIC4,S 0.9 = 10.202 - 0.9 = 9.302 9.81 2 x9.81 9302 x9.81

10 ... 7.6

= 9. 125 Nfcm!. An s.

FLOW THROUGH LARGE ORIFICES

If the head of liquid is less than 5 times the depth of the orifice. the orifice is called large orifice. In case of sm all orifice. the ve locity in tho.) entire cross-section of tho.) jet is consi dered to be const ant and discharge can be calculated by Q == Cd x a x J2gh. But in case of a large orifice. th e ve locit y is not conStant over the en tire cross-section of the jet nnll hence Q cannOl be calc ulated by Q = CJ x a x J2gh.

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IL 1328 Fluid Mechanics 7 .6. 1

Disch.ug~

Through L.uge Rectangular Orifice . Consider a large rectangular orifice

in one side of the tank disch argi ng freely InlU atmosphe re under a constant head. H as sliow n in Fig. 7.7. HI " height of liquid above top edge of o rifi ce H2 = height of liquid above txltlorn edge o f orifice b = breadth of orifice

"',

If = depth of orifi<:c = H2 - HI Cd = co-efficient of discharge.

Consider an ele mentary horizontal strip of d,'pl h . dll' 31 a depth of liquid in the tan~ as show n in Fig. 7.7 (b).

'/1'

below the free surface of the

-

-

---- --

~~-.---

,.) Fig. 7.1

'01

LarKt uctangu!ar ori/ict.

Area of strip = b x dl! and lheordical veloc it y of watCT through strip = ~2gl!.

Oi scharge tllrough elemen tary dQ

~lrip

is give n

= Cd x Area o f strip x Velocity

= CJ x bxdl/x J2gh :Cd bx,j2gh dh By int~grating the above equation between tlw lim its HI and H!, tile tOlal dischar£e tlirough the whole onli<.:e is obtained

,: Cd )( b x

.J2iJI/' Jh dlt ,: C~ x b x.J2i -3f2 "-'" ]'" Ii,

[

1/

,

... (7 .8 )

Problem 7 .11 Find Ihe diIcharge Ihrough a reclangular orifice 2.0 m ...ide and 1.5 m deep filled 10 a w"lu lallk. The "'aler IeI'd in Ihe lallk is 3.0 III abo}'e Ihe lOp edge of Ihe orifire. Take Cd = 0.62. Solution. Given: Width of ori li<.:e, b = 2.0 III !kpth of ori fice, "=1.50t Hei gh t of water ahove top edge of the orifice, H I = 3 III

I I

Ii

~ I

IL Orifices and Mouthpieces 329 1 He igh! of wmc r above bo tto m edge of 111e o rifi ce. H1 ", II I + <1= 3 + 1.5 = 4.5

In

Cd == 0.62 Disc harge Q is give n by equati o n (7.8) as

,

2 ~ 31' 3f1 Q= - Cd xb x,, 2g 11f 2 - - 1f 1 I ==

~ 3

x 0 .62 x 2.0 x

~2 + 9.S! [4.51.,

_ 3 U I mlfs

"3.66[9.545 - S.1961m 3/s '" 15.9 17 molls. '\I1 S . Problem 7 . 12 A r('c{(//Igulor orifice. 1.5 II! wide and 1.0 III deep is discharging Wall'r frolll a wnk. If Ihe \l'ater 11'1'1'/ in rhe /lUI " is 3.0 II! above Ihe lOp edg e of the orifice, filld Ihe discharge through Ih e orifice. T(l ke the co-effrciem of disc/wrgil18 fo r Ihe orifice " 0.6.

Solullon. Give n : b == 15 In d == 1.0 III H 1 =3.0 rn Hl = H I + d = 3.0 + 1.0 = 4.0 111 Cd = 0 .6 Disc harge. Q is giw n by lh e equati o n (7.8) as Width of ori fice. Depth of orifice.

2 ~ Q= ] X Cd XbX ..; 2g

,x 2

= -

IH;3f1 -HIlt2 1

0.6 x 1.5 x .)2 + 9.8114.0 u _ 3.0 u [ rn 1/s

'" 2.657 [8.0 - 5. 196] ml/s = 7.45 rn l/s. ADS. Problem 7 .13 A recumglilar orifice 0.9 m wide alld 1.2 III deep is dischargillg water from a \"essel. The lop edge of tile orifia is 0.6 m be/ow the waler sll rfila in the ,·esse!. Ca/cll/me (he discharge Ih rOllgh the orifice if C J = 0.6 Imd perce ntage error if the orifice iI treall'd as a small orifice. Solution. Givc n : Width of uri flcc, Dcptli uf orificc.

b=O.9rn d= 1.2 rn Hl = O.6 rn 111 = II I +d=0.6 + 1.2= 1.8 rn Cd = 0 .6

Disc li argc Q is g ive n as

Q

="32 X Cd x b x J2i =

X

IlI lll- H i~1l 1

"32 x 0.6 x 2.9 x .)2 x 9.81

11.8.lJ! - 0.63/1] mJ/s

= 1.5946 12.4149 - .4647 J = 3. 1097 III l/s. Ans. Disc liarg in g for a s mall orifi ce QI= C J xa x.j2gh II

where II = II I + "2 =0.6 +

~ I

21.2

= 1.2 m and a= b x 11=0.9 x 1.2

I~

~ I

IL 1330 Fluid Mechanics O.6x.9x1.2xJ2x9.8 I x l.2 = 3.1442 m3/s

Q1

'l> error = OJ - Q = 3.1442 - 3.1 097 = 0.01109 or 1. 109%. Ail S.

Q ~

3. [097

7. 7 DISCHARGE THROUGH FULLY SUB - MERGED ORIFICE

Fu lly sub-maged orifice is one which has its who]" of lhe outlet side sub-merged under liquid so that it d ischarges ajcl of liquid into the liquid of the same kind. It is also called totally drowned orifice. Fig. 7.8 shows the fully sub-merged orifice. Consider two points (I) and (2). point I being in the reservoir on Ihe upstream si de of the ori fice and

point 2 being at the vcna-contracca as shown in Fig. 7.8. Let III '" Height of water above the top of Ihe orifice on the upstream side. H2 '" H eigh! of water above lhe 00[[0111 of the orifice. fI '" Difference in water level. b = Width of orifice. Cd := Co-efficient of discharge. Height of water above the ce ntre of orifice on upstream side

Fig. 7.8

= ", + Hl - H, := H, + Hl 2 2 Hdght of water above the ccntre of orifice on downstream side

= H , + Hl _ H 2 Apply ing Bernoulli·sequalion at ( I) and (2). we get p,

-

pg

Now

li", pg

V,'

+-

2g

p.

V/

pg

2g

... ( 1)

... (2)

: - ' +-

H , + Hl , Pl :

2

Fully sub.merged ori[ir:t'.

- H and V, is n~gligible

pg

H , + Hl + 0 '" H , + Ih 2 2

V,'

H+~'~

2,

V1' = H

2, V2 '" .J2g H =l>x(H1 -H,) Areil of orifice Oischarge through orifice = Cd x Area x Velocity :=

Cdx I> (1/ 2 - H,) x .J2g11

Q:= Cdx I>(H 1 -

I I

H,»( .J2gH.

... (7 .9)

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IL Orifices and Mouthpieces

331 1

Problem 7.14 find r/i,> di.lclwrge rlrmugll a fully Sllb-merged orifice of\\'itllil 2 II! if Ille diliaellc,of water lel'els OIl bOIl! sides of Ihe orifice be 50 em . Ti,e Ileighr of water from lOp and boltom of the orifice are 2.5 III and 2.75 /II r('speCfil'ely. Take Cd " 0.6. Solution. Give n : Widlh o f orifice,

b=2 m

Diffe re nce of wate r leve l. H : JO c m '" 0 .5 Hcigtn of water from top of orifice. II , = 2.5 111 I-Ie ight uf wate r frum bottom o f o rifice. H! '" 2.5 III

III

CJ = 0 .6 Disd,argc lhrough full y suh-merge d orifice is given by equat io n (7.9)

Q = C J Xb X (1I2 - lI ,)X .J2g H

'" 0.6 x 2.0 x (2. 75 _ 2.5) x .J"2cxC9".8",cxc O".<5 m3/s

= 0.93% m 3/s. AilS. Problem 7 .15 Find Ihe di~'ch(lrge I/l rough (l lolul/)' droll"lIed orifice 2.0 m wide (lnd I difference of ...."ler /elds on both Ibe J'ides oflbe orifice be 3 m. TlIke Cd = 0.62. Solut ion. Given: Width of orifice. b = 2.0 m D<: pth of orifice. d = I m. Differe nce o f water leve l on both th e sides H=3m Cd == 0.62

/II

deep. if IllI!

Disc harge throu gh o rifice is Q "" Cd x Area x ,j2g11 "" 0 ,62 x b x If x ,j2gl/

= 0.62 x 2.0 x 1.0 x

,12 x 9.81 x 3

111 3/S = 9.5 13 mJ/s . Ans .

.. 7.8 DISCHARGE THROUGH PARTIALLY SUB · MERGED ORifiCE Paninll y sub· me rged orifi ce is o ne whic h has its Outlet side pnn iall y sub· me rged und er liquid liS show n in Fig. 7.9 . It is nlso kno wn as pnrtia ll y dro wned orifice. Thus th e pnrt iall y sub·m erged orifice has two pon ions. The upper po rtion be haves as an orifi ce d isc harg in g free wh ile th e lower porti on hc ha ves as a s ub· merge d orifice. Onl y a large or ifi ce can be ha ve as a partiall y sub· me rged o rifice. The t0 1a1 di sc harge Q th roug h partiall y sub· merge d o rifice is e qu al to the d isc harges throug h Fig. 7.9 free and the sub · merged porti ons. Disc harge throu gh th e s ub·m e rged portion is give n by equation (7.9)

I I

Partially IIIb.muged orifice.

Ii

~ I

IL 1332

Fluid Mechanics

Disctmrgc through Ihe free ponion is given by equation (7 .8) as Q ~:

2 M-:: 3f1 312 ']C d Xbx v 2g If/; - HI I

Q = Q 1 + Q1

Tutal u ischarge

'" Cd X b X (11 2 - I I) x .J2g11

+f

Problem 7.16

A r{'(/lUlgl/lar orifice 0/2

In ..... id/I!

Cd

X

b x

/ii

[H /,,2 - H

3I2

1. ...(7. 10)

IInif 1.2 m deep isjifleil ill 0I1e J'ide o/a large

Wllk. The ",aler lel'e! 011 Olle side of Ihe orifice is 3 III abo!'e rhe lOp edge of the orifice, II'Mle 011 Ille OJ/WT side of lire orifice. Ihe ...lIla /e"ei is 0.5 '" be/ow its lop edge. Calculate Ihe diiicilarge Ihroug" Ihe orifice if Cd '" 0.64. Sol ulion. Given: Width of orifi~. /,

=2

rn

Deplll of orifice. If = 1.2 111 Height

or w~tcr from

tOP edge of orifi<:c, H I '" ) 111

DiffeT"",:c of water leve l on both sides. H", 3 + O.S '" 3.5 111 Hcigtll o f Waler from the bottom edge of orific,,-

Ifl

= HI + d = 3 + 1,2 = 4.2 HI

Th" oritice is partially sub·maged. The discharg" throug h ,ob-maged ponion. Q I = Cdx bx (H 2 - H)x .J2gH

= 0 .64 x 2.0 x (4. 2 _ 3.5) x

..j",C,c9".oS"I<=3.<5 '"

7.4249 mJ/s

Ttw discharge through free portion is

Q2 =

t

Cd

=

~

x 0.64 x 2.0 x .J2 x 9.81 [3.5"\12 - 3.0.112]

X

bx ,fii[H3J2 -

HIJ/2]

= 3.779 [6.5479 - 5. 1961] = 5.108 mJ/s Total discharge through the orific" i, Q = Q I + Q1 = 7.4249 + 5.108 = 12.532.9 m l/s. Am . ... 7.9 TIME OF EMPTYING A TANK THROUGH AN ORIFICE AT ITS BOTTOM Consider a tank containing SOme liquid upto a hei gh t of HI. Let an orifice is fined at the bollOIll of the tank. 11 is required to find the tillle for the liquid surface to faJi from the height II I to a heigh t H 2. Let A = Area of the lank II = Area of the orifice Ill'" [n itial height oflhe liquid fl 2 = Fi nal height of th~ liquid T", Time in sL'Co nds for llie liqu id to fall from fli to H,. Let at any time, the height of liquid from orifice is II and IClt lie liquid surface fall by a sl1\all h"ighl till in lime tiT. Ttwn Volume of liquid I"aving lhe tank in tim", tiT: A x rill Also lhe theorctkal vcJOCily through orifice, V = .j2g11

I I

Ii

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IL Orifices and Mouthpieces

333 1

Disctmrgc throu gh orifice/scc. dQ = Cd x Area of orifice x Theoretical veloci ty" Cd'

(l •

,J2gl1

Discharge tltrough ori fi ce in time interva l

dT =C,j.a . .j2gll. dT As lh e vo lum e o f liquid

I ~aving

Ihe tank i s equal 10

Ih~

vo lume of liquid Flow ing throug h orifice in

lime dT. we have A (- dll)=C,/.a . .j2gh .dT

- vc s ign is inscn cd becau se with Ihe increase of lim e. head on o rifice decreases. f'>::l:

- Adil=C,j.a. ,,2gll.dTordT=

By integrating the aoove equati on betwee n llie

r'"

T

Iill1it~

CJ .a .,,2gh

H I and H, othe total time. T is obtai ned as

dh -A J ill )1' h -· .lI . .j2i - C" . a . .,ffi II,

_ All -I I!

jo d7'-- j II,, C

J

-AdII F>7.

~

[ru ruj -_ 2A [JH, -= ,lH,j ,H, _,H,

-2 A Fe Cd .(1.,,2g·

Cd . a. ,,2g

... (7 . 1 I)

For emp tyi ng th e lank comp letely. Hl == 0 and hence

lA - Cd

T-

.[ii;

.a.J2i"

... (7. 12)

Problem 7 .17 A circu/
A=

~ (4)! = 12.566 Ill ! 4

d=0.5 111

"

(1= - (. 5) =0. 1963 111"' 4

1-1 ,=5 111 I-I! = 2 III (ii) H ! = 0 fll =2 111

Using eq uation (7.1 1). we have T =

I I

2A

Jii {.{ii; ~.JH;""]

Cd . ( I . 2g

Ii

~ I

IL 1334

Fluid Mechanics

""

2 x 12.566 0.6)( .1963 x ../2)( 9.81

[r. ,,5- ....=1 2,0 ~conds

20,653 '" 0.5217 '" 39.58 lIt'm nd s. An ~ .

Senmd

ellS'"

When H2 = 0

2A

T",

Cd .a .

.Jii

~ '" ;;c:~2~X~I~2.~566""x~,f5",~ 0.6x.1963x..)2)(9.8 1

= 107.7 seco nd s . Ans . Problem 7.18 A circu/M /{mk of
Solution. Given: Dia. of lank.

D:l.25m

:. Arc",

Dia. of orifice.

d=50mm=.05m

:. Area,

(I

"

= - (.05)- = .001963

,

01-

4

Initial height of w ater.

Time in seconds.

",=5111 T= 1.5x60=90seoonds

Lei the height of wa ter after 90 seconds = Hl

Usi ng equat ion (7.11 ) , we have T =

JH: -JH:]

2A

- -:;'---'---ic--'-" Cd .a . 2g

JH:] J' =455.2 15 [2.236-~1 0.62 x 0.00 1963 x 2x9.8 1 . 2 x 1m[,f5 -

90:

,[iT;. '" 2.236 -

90 455.2 15

Hl '" 2.0383)( 2.0383

I I

'" 236 - 0.1977 '" 2.0383

= 4.154 m. An s.

Ii

~ I

IL Orifices and Mouthpieces

.. 7. 10

335 1

TIME OF IMPTYING A HEMISPHERICAL TANK

Consider a Ilcrnisphtrical lanl.: of radius R filled with an orifice of area '(( al its bouo m as shown in Fig. 7. 10. Th.e lank cOnlains some liquid whose initial height is H I and in lillie T. the hciglit of liqu id falls to H! . It is required tn find the time T. U:i a1 any instant o f time. the head of liquid over the orifice is hand
ORIFICE

and theoretical ve locity of liquid '" J2gh.

Fig. 7.10

lI~milpbmcaf

tallk.

Let th.: liquid level falls down by an amo unt of dh in time dT. Volume of liquid leaving lank in lim", tiT = A x III!

= 1t1.2 x till Also

volurn~

•.. ( i)

o f liquid nowing through orifice

= Cd x area of orifice x velocity

= Cd.a . ./2811 second

Vulume or liquid fluwing through orifke in time liT ... (ii)

: Ca.a. J2gh )( dT

From equations (i) and (ii). we get 2 7U ( - dll) '" Cd.a .

..j2gh . tiT

-ve sign is introduced. because wilh the increase or T. h will - ltr!

de<.:Te~se

... (iii)

dl,: CJ.a . J2gh . dT

But from Fig. 7.10. for ""'OCD. we hal'c OC '" R DO=R-II OJ",.( == .\J

Substituting

~OCl _ OD 1 _ ~Rl _(R_h)l

== R'- _ (R _ It)" == R" _ ( R" + h'- _ 2Rh) '" 2RII _ Il"

x"in cquation (iii). we get - n(2RII - IIl)dll == Cd.li. J2gh . dT

"'

dT:

1 - n (2 Rh - h )dll",

CJ .a.J2gh

The total tim e Treq uired 10 bring th e liquid lel'd from HI to H2 is obtained by integrating the above equation between the limits H, and H2. T-

f'-;c---, -c'-".- (2Rll ln - "Jnjdh JI, Ca ·a.j2i -, fll, (2 Rh In- - II -jdll ~

In

Ca .a.,,2g II,

I I

Ii

~ I

IL 1336

Fluid Mechanics

'"

[' R-,,-,,_,._I - "lll C+] ]"'

-It

~+1

cJ xax fii

C_'3 -+ ,C 2

2

=

[:: R ( H~i2 _ lIi'1) _ ~ (1I11) - 111m )]

- It

3

Cd X II X,fii

:

[.:!. R(H,l!! - H;'2 ) _ ~ (Htl) - Hi/C )] ... (7.13)

'

For comp lete ly emptyi ng lh e lank . HI = 0 and

Problem 7.19

5

Cd x a x ,ffi 3

T=

It

H,

5

h c ~ cc

fii[:: RHi'! - ~ H~12] .

Cd ,a . 2/1

3

... (7.1 4 )

5

A hemispher jw/ IW11:: of dill/nerer .,

III

CQl lfains waler lip/a

II

height of 1.5 m. All

orifice of diameter 50 111111 is prol'ided allile bOl/olll. Pilld IIII' lillie required by W
:. Radiu s. Dia. o f or ifice.

R ", 2.0 In II" 50 mm = 0.05

:. Area.

(1

III

"

= - (.05) = O'()O I963

'

111 -

4

Ini tial heig h! of wate r.

111= 1.5 111

Cd '" 0.6 Fi rs l Case. H ~ = 1.0 Time T is give n by equatio n (7.13)

T-= -=

[~ R ( H:'l _ H~ll )_ '-( HI)ll _ H;ll )]

II

Cd XIIX jii 3

5

X [~X 2.0 (1.5)12 - I.O lll ) O.6x.OO I96 3xJ2X9 .8 1 3 II

-

~ (1.5 512 - I .O~ll)] 5

-= 602. 189 [2.2323 - 0 .70221 " 921.4 second = 15 min 21.4 sec. Ans. SeCQnd Case. H2 = 0 and he nce ti m", T is give n by equ at ion (7. 14)

'r __

' Cd . a .

:

I I

.J2i

[4 H lil 2H $!l] 3 5 - R

I

- -

I

' [~X2.0X 0.6x.OOI963J2x9.81 3

I .5Jll _ ~ X I..5m] .5

Ii

~ I

IL Orifices a nd Mo uthp ieces

337 1

'" 602. 189 [4.8989 - 1 10221 sec '" 2286.:n sec '" 3H min 6.33 s.,.", An s.

Problem 7 .20

A hemis{!/laical cis/em of 6 III radius iii full of Waler. II is filled willi a 75 mm dh'meler sllllrp edged orifice III "ie bollom. CalCl'/ale (I,e lime required 10 lower the le"e/ in 1/"" cislem by :z melres. An'ume co-effie;..'", of discharge Jor lire orifice is 0.6.

Solullon. (jive n : R~d ius

uf

hc rni ~phc rica l

ciste rn. R = 6

11\

",=6 m

Initial he ig h! of w ater.

Dia. of ori fi ce.

d = 75 mill = 0.075 III

:. Area.

0=

! (.075)l", .00441 8 m 2 4

Fall of height of wa ter :2 m (' i nal hei ght o f wa in, 111 == 6 - 2 == 4 III Cd == 0.6 The ti me T is given by equat ion (7.3 1)

T= :

[~R ( 1IIm _H~I1) _~ ( llt'2 _Hi'2)]

II

Cd x a x,fii 3

:;

,

~=;';;-c-~~ 0.6 x .Q0..\4IS x x9.81

J2 x

JI [~x , 6 (6.0 ! -

4.0 lll ) -

~ s (6.0$12 -

4.0" 2)]

'" 267.56 [8( 14.6969 - 8.0) - 0 .4 (88. 18 - 32.0) ) :: 267.56 153.575 - 22.472 ) sec :: 832 1.9 sec '" 2hrs 18 min 42 sec. Ans. Problem 7.21 A cylilldriw/ tllllk is IUII-ing a /utm isplwrical bllst'. Tilt' heig/II of cy/ilidriCi// portion orifice of diameler 200 mm is jilled. Find the is 5 m alld diameter is -.I m. All/Ie bottom of Ihis 10111.

,III

lim;~:~~'i ~~~ ~i~~,7~llele/y emplying

Ihe

/(111/;.

T(I/;e Cd :: 0.6.

T -:

_\~~~~{~~~~;~~~(~:_; -

Heig ht o f cy lind ri cal port ion (II ) = 5 m =4.0 m Di a. of lank

- :.:.:.:.:.:. _.

A :: ~ (4)!:: 12.566 m 2

Area.

"

4

tI---- 4.0 m -


Di a. of orifi ce. Area.

(I'" -

" {.2t '" 0 .03 14 m-,

4

- {1

z.o m

Cd:: 0 .6 ORIFICE

7

Fig. 7. 11

I I

Ii

~ I

IL 1338 Fluid Mechanics The tan~ is splittcd in two ponions. Fi rst portion is a hemispherical tank ami second portion is cylindrical tank. LeI T, '" time for e mptying hemispherical portion I. T) == time for emptying cylindrical portion II. Then lowl time T= 1'1 + T1" For Portio n I. H I == 2.0 Ill. H 2 = O. Then TI isgi\'cn by equalion (7 .14) as

T, = =

It

[ :.

Cd xax,fii 3

RH~l _2. Hi'l ] 5

' [~X2.0X2.0JI2 O.6x.031 4 xJ2x9.8 1 3

_ 2.X2.0)ll] 5

== 37.646 [75424 - 2.2621 sec == 198.78 sec. For Port io n II. H I == 2.0 + 5.0", 7.0 Ill. 112 '" 2.0. Then 1'2 is given by equation (7.11) as

2A[,fH; -,IH,]

Tl = "

Cd X{lX,ffi

2X12.566[J7-J2:0] O.6x.oJ1 4 XJ2x9.81

To: 7', + 1'2"" 198.78 + 370.92

:. TOlallimc.

== 9 min 29

.... 7. 11

==

S L"<:,

== 569.7

sec = 370.92 sec

SL'i:

Ans .

TIME OF EMPTYING A CIRCULAR HORIZONTAL TANK

Consider a circular horizontal tank of length L and radius R, comaining liquid upto a height of H I' Let 3n orifice of ~rea ' Q' is fillcd at the bottom of the tank. Then the time requi red to bring the liquid level from fli to 112 is obtaincd ~s : Lc t at any time. tlie licigli t of liquid over orifice is '/1' and in time dT. Jet tlie hciglit falls by an licight of 'dh·. Le t at tliis time. the width of liquid surface == AC as sliown in Fig. 7.12.

I_

L

_I

Fi g . 7.12

SurFace area of liquid = L )( A C

Bo<

I

AC = 2 xAD = 2[JAO' - OB' =

~ ~H' - (R -

")'

J

= Z~R l _ (R' + Il " _ ZRII) = ZJZRII _ Il "

I I

Ii

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IL Orifices and Mouthpieces

339 1

Surface area. Vo lum e of liquid lea vin g tank in til llC tiT ==

A xdh == 2L J2RII - II " xdll

... (1)

A lso th e vo lum e o f liqu id fl ow in g th ro ugh orifice in time dT ==

Cdx Are a of o rifkc x Vclm: ity x tiT

Bm the veloc ity of liquid at the l im e cons ide red == .J2gh

Vo lum e o f liq uid flowing th rough urifi ce in tim e tiT

= CJ xax ../2g11 xlrr

... ( i i)

Equatin g (il and ( ii). we get

2L JZRh

_ 1, "

x (- dll) = Cd X II

X

,JZgh x tiT

- vc s ign is introd uced as w illi Ihe increase of

r. the

he ight II dccrcascs,

- lL.j (2R

C~

III dll

Xi/ Xlii ITakin g

", I

.fh COl11lll on]

",

- 2L (2R - h) - dll

To ta l tim e. T=

II,

Jii CJ xax 2g

==.:-,,..1 x ax,f2i - 2L

=

==

II,

Cd

II,

- 2L

III

[2K - I,]

dh

[12K - h)'" '' X(-I)] "' ~+I

C,s x a x ,ffi

2

= =

2L

CJ x a x ./2i

",

x ~ X[(2R - h)lI1 J"'

4L 3Cd x a x ..{ii

3

H,

[(2R_H1)JIl_(2 R _ Ht )Jil ]

... O . IS)

For l'On1pl ctc ly em pt ying th e tank. Hl == 0 a nd hence

!].

J1 4L [(2R/ 12 - (2R- Hd ... (7 . 16) 3C" x " x Problem 7,2.2 A/I orifice of diameler f()() mm iJ filled III rhe hOllom of a boiler drllm of fe/lglfl 5 1M GIld of diameler 2 m. The drllm is IlOfizollla{ (IIld half full of warer. Filld Ille rime required 10 empl}' rhe boiler. gil'ell Ihe mlue of Cd " 0.6.

T=

I I

J2i

Ii

~ I

IL 1340

Fluid Mechanics

Solution. Gi ven: Di3. of orifice.

d '" 100 !lllll '" 0.1 III

"

a'" - (. n- '" .007854 4

Area. Length.

L =S Ill

Dia. of drum.

D= 2 m

Radiu s.

,

nJ '

R=lm

Initial he ig ht of wa ter.

11 1= 1111

Final height o f wa t~ r.

Ill'" 0

Cd'" 0.6 f'orcompletely emptyin g the tank. Tis give n by equati o n (7. 16)

'"

4 x5.0

[(2 x])Y.! _ (2x l _ l)3r2]

3 x.06 x .007854 x ./2 x 9.81 '" 319.39 12.8284 - 1.01 = 583.98 sec = 9 min 44 sec. Ans. An orifice of diameter 150 mm iJ filled at lile' bOl/om of II boiler ilrllm of lenglll 8 m and of diameter J metres. 1'I,e drum ;.1 horizontal (md cOll/ain.! "'
Problem 7 .23

Area.

(I

"

= "4 (. 15)- = 0.01767

Lenglh.

L =8.0 rn

Dia. of boiler,

D=3.0m

:. Radius,

R= 1.5 [1)

Ini,ial height of water.

H)=2.4m

rind he ighT o f wa Ter.

Hl '" 0

,

Ill "

Cd'" 0.6. For co mple tel y empl ying the lanl;. T is given by equ ation (7. 16) as

4 x &.0 3 x.6 x .0 1767 x

J2 x9.8 1

[(2 X I.S).lIl _ (2 x I.S _ 2 .4)"~n)

:227. 14[5.196 - 0.4647): 1074.66 sec : 17 min 54.66 sec. An s.

I I

Ii

~ I

IL Orifices and Mouthpieces

.. 1. 12

341 1

CLASSIFICATION Of MOUTHPIECES

I , The mouthpieces arc classified as (i) External mouthpkcc o r {iil Inlc mal mouthpiece depend·

in g upon their position wi th respect 10 the tank or vesse l to which th.ey are fined. 2. The mouthpiece arc c lass ified as (I) Cy lind rical Illoulllpiccc o r ( ii) Con ve rge nt mouthpiece or (iii) Convergent -divergent mouthpiece depending upon their shapes. 3. The mouthpi eces arc d a.%ifi ed ,IS (i) Mo uthp ieces running full or (ii) MUUThpieces running free. depending upon the nmure of discharge ;u the outlet of th e mo uthpiece. T his d assificatiun is o nl y for internal rnoulhpi c<:cs which arc known Borda's or Rc-cnlran t mouthpieces. A mouthpiece is said to bo: rurming free if th e ~ l of liqu id after contraction docs no t touch the si des o f the rl1outhpiece. But if the jet after contraction expa nds and fill s the whole mouthpiece it is known as run ning full. .. 7 _13

FLOW THROUGH AN EXTERNAL CYLINDRICAL MOUTHPIECE

A mouthpi«:e is a short le ngth of a pipe whi ch is two o r three times its d iameter in leng th. If thi s pipe is fitted e xte rnall y to the ori fice. the mouthpiece is called ex ternal cyli ndrical mouthpi ece and the discharge thro ugh orifice increases. Conside r a tank ha vi ng an exlernal cylindrical mouthpi ece o f cross-sectional area (i l ' allached 10 one of it s s ides as s how n in Fig. 7. 13. The jet o f liquid e ntering th e mo uthpiece co ntracts to form a ve na-co ntracta al a section C-c. Beyond this section. the jet agai n expands and fiB the mouthpiece comp lete ly. Let II = Height of liquid above the centre of mouthpiece Fig. 7_13 v,. = Velocity of liqu id at C-C set: li on a,. = Area o f fl ow at ve na-contraCla "t = Velocity of liqu id al o utl el at = Area o f mouthpicce at outlet C,. = Co-e ffic ie nt of contraction. Appl ying continuity eq uation at C-C and (1) -(1)_ we get

ExuTnal cylindrical mOlflbpiecN .

ae x 1',. = arv r " == al"1 == - "-' -

..

B",

!!.c.

",

a,.

lIJar

= C, = Co-effici ent of contraction

Taking C,. = 0 .62 . we get.:!..r.. = 0 .62

",

'" 0.62

" = --

,.

,

The jet of liljuid from section C-C s udde nly e nlarges at sec tion (1 )-( 1). Duc to s udd en e nlarg eme nt.

(",. -vd Ihe re will be a loss of head. IlL· which is gi ve n as hi = "''-;;-''"-2,

•

I I

Please refer An. 11.4. I for loss of head duc to sudden enlargement

Ii

~ I

IL 1342

Fluid Mechanics

( -0'.6' 2

But " '" - -' 'c 0.62

-"1)' =;~ [O.~2- 1 j'- = 0.375 vi

2g

2,

Apply ing Bernoulli' s equati on \0 point A and ( I H I)

v1

{!

.....::!.. + ~ + Z~

pg

wher" Zot

= ZI' l'A is

2g

P

1,1

= - ' + - '- +ZI +h L pg 28

neg li gi ble.

~ = atmospheric pressure = 0 pg

, ,1

If '" 1. 375 - '

2g

"g il = 0.855 J2g H 1.J75

-- -

Theoretical veloc ity of liquid at outlet is

V'h

= JZglI

Co -efficient o f ve locity for mOUlhpkcc

J2iii

0.855 C '" Actua l ve locity '" J2gH '" Th eoretical velocit y

= 0 .855.

Cc for mouthpiece ", I as Ihe are a o f jet of liqu id at o l,lliet is equa l to the area of mouthpiece at o Ull et.

T hu s Cd = Cc x C. = 1.0 x .855 = 0.855 Thus Ih e va lu e of Cd for mouthpiece is more Ihan the va lue of Cd for o rifice. and so di ;;c hargc throu gh mouthpi c<:c will be more. Problem 7.24 Find the discharge from II 100 mm diameler c.u emal mouthpiece. filled to a side of (l /tlfge \'esse/ if the head ol'a the mouthpiece is../ metres. Solution. Given: Dia. of mouthpiece = 100 11\ = 0. 1 In Area.

Hc.:u./, Cd ror mo uthpiece Disc harge

(l

= ~ (0. 1 ) 2 =0.007854 11\ 2 4

H = 4.0 11\ = 0.855 = Cd x Area x Veloc ity = 0.855 X li X.)2g H = .855 x .007854 x .)2 x9.81 x4.0 = .05948 IIl l /s.

AilS.

Problem 7.25 An eX le",,,1 cylindriCll/ moulhpiece of diame/a 150 mm is discllllrging "."Ier u",fer" conJHml llead of6 m. Delermine rhe disclwrge "nd "bw/"re pressure head of ,.."Ia 1I1 n'nll·conlraC/". TlIke Cd = 0.855 alld C,. fo r \'ellll'COlllfllCW = 0.62 . Allllospilaic pressure head = 10.3 til of Wafer.

I I

Ii

~ I

IL Orifices a nd Mo uthp ieces Solution. Given: Dia. of mouthpiece.

d= 150 Imn:Q. 15cllI

:. Area.

a '" "::(.15) 2 '" 0.0 1767 rn 2

343 1

4

H= 6.0 III

Head,

Cd = 0.855 C~

al vcna-contracta '" 0.62 Atmospheric pressure head.

H~

Discharge

= [0.3

III

=CJ xaxJ2g H '" 0.855 X .01767 X

.J2 X 9.8 1 X 6.0 '" 0.1639 ",lis. An s.

I' ress UN Head a t Vena-co ntractu

Apply ing Bernoulli's equation at A and C-G. we get

------. _-------. ............. .........

p I '! p~! ~+.....1..+ZA=_ c +-'-Hc

pg

B",

28

pg

2g

.!!..t,.. = II. + II. pg

I'A '"

"

O. A

I

Fig . 7.14

Bm

,~ =-0".62'1I,, =Hu+lI- ( - " ) ' X - I = "'"+ II - -,,''

.62

B",

H", 1.315

2g

28

x -I-, (.62)

I'I!

2,

Vil = ~= 0.7272 II 28 1.375 I

He'" fl. + f/ - .7272 Hx - ,

(.62)'

'" fl. + f/- 1.89 H= Ha - ,89 H = 10.3 - .89 X 6.0 {-: Ha '" 10.3 and 11= 6.0) = [0.3 - 5.34 = 4.96 m (Absolute). An s.

I I

Ii

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IL 1344

Fluid Mechanics

.. 7. 14

flOW THROUGH A CONVERGENT- DIVERGENT MOUTHPIECE

If 3 mouthpiece converges UplO v"na-conlraCla and then di verges as s hown in Fi g. 1 15 tlle n that type of mouthpiece is ca ll ed COllvcrgcm-Divcrgcnt " 1outhp iecc. As in this lIl oulhpicr c the re is no s udde n clllargc men! o f llie jet. th e loss of ene rgy du e to sudd en en large me nt is elim inated. The coefficie nt of discharge for this mouth piece is unit y. Let H is th e head of liqu id over llie mouthpiece. Applying Bernoulli's equation to the free SUrf,I CC of water in tank and sect io n C-c. we hav e

-

II~'

pg

,

+-

2g

Pc

,

Vc

+ z= - + - +," P8 2g

Taking datum pass ing through (he centre of orifice. we ge l

... ( i) ,. ~

~ : Il +/f - I '

28

~

I'e '"

... (ii)

c

~2g( H. + II

Conv ergentdivergent mQllt hpiece.

Fig. 7.15

Hr)

Now a ppl ying Bernoulli' s equatio n at sect ions C·C and (I )·(l)

,

,

I'" -P' ' + ~'~+ Z =' + _1 pgZg '

pgZg

+Z I

Bm 1,2

1,2

H,. + ~'~ =Hd + -'

Also from (;) .

Zg Zg Hc + v/ IZg=II + H" lid + v/IZg = II + lid

... ( iii)

"I=.j2g H Now by co ntinuity equatio n.

tlc v e

= "I X (I I

!2. _ 3:..= JZg( H. + H He ) lI,

-

VI

.jZgH

1 + lId - l-l~ II

The discharge. Q is given as Q = a, x .jZgH where

I I

li e

Hd

lie

H

H

- +l~ ­

...(7 . 17) ... (7 . 18)

= area at vena-conlraC la.

Ii

~ I

IL Orifices a nd Mo uthp ieces

345 1

Problem 7 .26 A com 'ergel1 /·dil'ergelll mOlllhpiece /ral'ing llirom diameter of 4.0 em is discharg ing Waler IIIlder a cOlIswlllllead of 2.0 m. deTermine II,e maximum OIlier dimneler for maximum di.IClwrge. "-ind maximum discharge also. Take Ha " /0.3 III afwater and H ",p ,,2.5 III a/waler (absolule).

Solution. Given : Dia. of throat. Area.

Il c "

" (4t

= 12.566 ern-,

-

4

fI '" 2.0 III

ConSlant head.

Find max. dia. at o Ull e!, d l and

Om.,

H. '" 10.3

H"p '" 2.5

In

In

(a bso lute)

The di scharge, Q in converge nt.diverge nl mouthpiece depends 0 1> lh e are a al Ihroat. Qm>., '" a, x ,)28 1/ '" 12.566 x .J2 x 9.& 1 x 200 '" 7H7 1.5 CIlIJ/S. Ails. Now ratio of are as at oullet and throat is give n by eq uati on (7 .17) as

~

oc

'" I]+ 11" V

He"

11

II + 010=.3" ,,=2.='

V

2.0

=2.2 135

,

,/'

- lr

4

I

- d

4

, '" 2.2 135 or

c

r::r

= 2. 2 1JS

ill '" 1.4877 x d~ "" 1.4877 x 4 .0 "" 5.95 em. A ns .

Problem 7.27 Ti,e tl/rom Illld exil diameters of conl'ergenl·d;"ergelll mouthpiece are 5 em Illld 10 CIIl respec/il'e/y. { I isfilled 10 Ihe l'erlielll side olillollk. cOIII(lilli"g Inlier. Find 'he lIliuim ,u" helld of II water lor steady floK'. The maxim"m "lIcuum preS~'ure is 8 m of \l'lIler alltl lake lIlmospheric pressure"" 10.3 m I\·(l/er.

S o lution. Give n : Dia. al throal, Dia. a[ exi t. A [mo~ph e ric p re~sure

de "" 5 em

", "" 10 cm head. H. "" 10.3 m

The maximum vacuum pressur.:: will be at a throat o nl y Pressure h.::ad at [hroal " 8 m ( vacuum ) He "" HQ - 8.0 (ab,'IO IUle)

" 10.3 - R,O " 2.3 m (abs.) Let maximum head o f wate r ove r mouthpi ece "" H m o f water. The ratio of areas al out le t and th roat o f a convc rge lll-di l'erge lll moulhpiece is given by equat ion (7. I 7).

I I

Ii

~ I

IL 1346

Fluid Mechanics

~1 + 1O.3

23

H

Hi

to ! -,-: 4= ,-

0'

If '"

8 or 15", -8

1+ - or 16= 1+ H H

H

.! '" 0.5333 m of water 15

Maximum head of water'" 0 .533 m. An s.

Problem 7.28 A con",:'gf!n/-t/ij'ergenl moull/piece is filled /0 Ihe side of (I Illllk. Th e discharge through muulilpier( "",Ier (I COll:!/,ml lier,d of /.5 In is 5 liITes/s. Th e helld loss ill Ihe dil'erg<',,1 portion is 0.10 limes Ihe killetic IW{ld at oliliel. Find the ,"roalllnd ".til diameters. if 5cl'(I"'liOlI pressure is 2.5 In w/(I a/mo,~pheric pressure llead '" 10.3 III ojW(,ler. Solullon. Given: Constant head. 1/ '" 1.5 III Discharge. Q '" 5 rilres '" .005 111 3,s ilL or H ead loss in divergent '" 0.1 x k i netic head at oUllet

lie or H"p = 2.5 (abs.) If" "" 10.3 m of wmcr Find (i) Dia. a1 th roat. d< (ii) Dia. at oUIIc!. til (r) Ilia. a l th roa t (d <). Applying Bernoulli' s equation to the free water surface and throat section. we get (See Fig. 7.15).

-

P

pg

I,l

+-

2g

+ z==

P

I,l

pg

2g

- ' +-'- +<,

Taking th e centre line of mouthpiece as datum. we get

,

"

H" + O+H ==H,. + -

2,

1,2

-'-== Ha+H-Hc== 10.3+ 1.5-J.5 ==9.3 morwater

2,

I'e ==

J2 )( 9.81 )( 9.3 == 13.508 IlIfs

Now

.005 x 4 It X

13.508

" J.0CXJ47 == .0217 m

==

2. 17

CIII .

AilS.

(ii) Dia. a \ oullet (d,). App lying Bernoulli's eq uation to the free water surface and ou tlet of mouth· piece (See !'ig. 7.15). we get

I I

Ii

~ I

IL Orifices and Mouthpieces

347 1

1,2

II~ + 0+ 1/ : H" + - ' +0 + 0.1

2,

.,2

",l

28

28

H = - ' + .I x 1'1

=

"J

= .1 - '

28

J2gH = ,p2~X~9~.8~'GX~1.5~= 5. 1724 l.l

l.l

Now 4 x .005 11: X I',

II> 7 . 1S

=

,1:04::X7:.00 ",5~ '" 0.035 III = 3.5 .:m. Ans. 1f

X 5. 1724

FLOW THROUGH INTERNAL OR RE- ENT RANT ON BORDA 'S MOUTH PIECE

A shon cy lind ric al tu be anached to an orifice ill s uc h a way Ilial th e lube projects in wardl y \0 a tank . is c alled an inlc rn al mo uthpiece. II is also ca lled Rc-c mraru or Borda 's mouthpi ece. If the length of the tu be is equ al to its di ame te r. the jet of liqu id co mes Oll t fro m mouthpi ece with out to uc hin g the s idc50ftbc mbe as ~h ow n in ri ~. 7.16. T he moU!hpiccc is kn ow n as fUlI/ling/ree. But if the leng th of the tu be is about 3 li mes its d iameter. th e jet comes out wi th its diameter equal to th e d iamete r o f mouthpi ece a t o ut let as s how n in Fi g . 7. 17. T he llI o uth piece is said to be fUlUlingfull. (i) Bo ni n's !\Io ut hpi l'C" Run n in g Fre.. , Fi g. 7. 16 s hows the Bonia's llI o uthpiece runnin g free. Let H = heig ht o f liquid above th e Ill outhp iece, a = area of Illouthpiece, " e " area of contracted jet in th e mo uthpi ece, ve " veloc ity th ro ugh mouthpiece .

RUN NING FREE

Fig. 7.16

RU NNt NG FULL

Fig. 7.17

Thc n ow of fluid th ro ugh mo uth p iece is takin g place d ue to th e press ure force exc n ed by the flu id o n th e entrance SC1:tion of the mo uth p iece. As the area of the mOllthpk'i:e is "a' hence total pressure force o n entrance "pg.a . 11 where II = di stance of e.G. of area ',,' from free s urface = H. , ..(i) =pg.a. H

I I

Ii

~ I

IL 1348

Fluid Mechanics

According 10 Newton's second law of motion. the net force is equal lO the ralC of change of mOlllen ttl 111.

Now mass of liquid flowing/sec '" p x (I e )( I'e The liquid is initially at rcst and hence initial velociTy is zero bUI final veloci ty of fluid is I'c' Riue of change of lllomClllum = mass of liquid flowing/sec x [tinal velocity - initial velocityj = pac x 1',lv, - 0] = pac v/

... (Ii)

Eq uating (I) and (Ii). we get

pg ,a.H. = pac' \./

... (i ii)

Apply ing Be rnoulli's equation to free su rface of liquid and section (1) -( 1) of Fig. 7.16

Taking the centre line of mouthpiece as datum. we have

z=

H.

z, '"

p

0, pg

v= 0

'" 0,

,

0+0+11=0 +-'- +0 or H= ~

"

2g

28

Substituting the valu e of

Vc

in (iii), we gel

pg .
a,_ _ _I __O .5 " 2

l!_~a c or -

.

.

U

Cu-cfficlcill of cun tTacUOIl. C<" ...L '" 0.5

"

Since there is no loss of h~ad . co-effident of velocity. C, '" 1.0 ..

Co-effident of discharge. Cd" C< X C. " 0.5)( 1.0" 0.5 Discharge

Q '" Cd a.}2gff

...(7. 19)

'" O.S )( aJ2gH (ii) Bord a's l\l oulhp iece RUllnill1l: l-' ull . rig. 7.17 shows Borda"s mouthpic(;c rUllning full. Let H"" height of liquid above the mouthpic(;~. "I = velocity UI outlet or lit (1).(1) of mouth[licc~ , (l = ar~a of mouthpiece, a < = area of the flow at C· C, V C = velocity of liquid at \,ena·contrac ta or at C·c. The jet of liqu id after passing through C·C, suddrnly enlargrs at sc£tion (1 H I). Thus therr will be a loss of head due to ~uddc n cn l arg~1llen\. It L

I I

(v<_1'd

'"

1

'-'-c;-"2,

... ( i)

Ii

~ I

IL Orifices and Mouthpieces Now f rom continuit y . we have a,. x I'c '"

v~

0'

Substituting this va lu e of

I ' ..

~ a~

Vi'

349 1

= a t x VI

X I' I

==

--"'- = ~ = ~ aC I {' I

CC

0.5

= 21"1

in (i). we ge t hI."

(2" 1-1(1)1

"-",,...c,,-2s

_

1',' 28

Apply ing Berno ulli 's equation to free surface of water in tan k and section (1).( I). we ge t p

1, 2

pg

lg

-+ T~kil1g

PI

+z= -

pg

VI'

+-

2g

+ z, + il l

.

dmulll line passing through th e CClllre line of mo uthpi ece I' 1

V !

O+O+ H =O + - '- + O+ -'-

28

28

2

~

2g

28

l

H=~+~: !L g

== .JgH He re \', is act ual velocity as losses ha ve b<:c n taken illlo consider-Hion. \'1

But th eoretical ve loc ity.

.

.

\' ,h

==

,JZgH

l'l = J,[iii. =

I M= 0.707 2811 ,,2 As th e are a of th e jet at o utle t is equal to the are a of the mouthp iece. hence co-effi cient of :. Co-efficlen l of ve loc1t y, C, = -

V'h

contracti on ", I

Discharge.

Cd '" Cc X C,.= 1.0 x .707" 0.707 Q" Cd X (/ x J2 gH "" 0 .707 x a x J2gH

Problem 7 .29 An in tern(ll mouthpiece of 80 mm diameter iJ" di~"C!lIIrging ....aler unde r head of 8 metres. Find Ih e discharg e through mouthpiece .....1"'" (i) Ti,e mouthpiece is runnillg free . alld (ii ) The mo uthpiece is rI/IIlling JIlII. Solullon. Give n : Dia. of mou thpiece. d = 80 10m = 0.08 III

Area. Constant head.

... (7.20) (I

constant

a = ~ (.08 ) 2 = .005026 m 2 4

11 = 4 m.

(i) Mouthp;"ce runu;n g f ree . The di sc harge. Q is g iven by equat io n (7.19) as

Q=0.5x(lx J 2gH = 0 .5

x .005026 x J2 x 9.8 1 x 4.0

= 0.02226 m 3 /s "" 22.26 lit res/s. Aos . (ii) Mouthpiece ruuning full . The discharge. Q is given by l'q uation (7.20) as

I I

Ii

~ I

IL 1350

Fluid Mechanics

Q = 0.707 x a x J2gl/ '" 0.707 x .005026 x

J2

x9.81 x4.Q == 0.03147 m 3fs '" 31.47 1I1rt'ls. A ns .

HIGHLIGHTS 1. Orifice is a small opening on (he side or al the bottom of a lank while mouthpiece is a shon length of pipe which is 1WO or three limes its diameter in length. 2. Orifices as well as moulhpic~es arc used for measuring (he rJle of flow of liquid. J. Theoretical "clocity of jet of waler from orifice iii giwn by V ~ J 2gH • where If _ Height of Water from the cenlre of orifice. 4. There arc three hydraulic co-efricients namely: c

(a) Co-efficicnl of ,-elocity.

= Actual ve loc ity at vcna-contmcla =

x

Theoretical velocity

J4 yll

•

(b) Co-efficicnl of contraction.

k ) Co-efficicnt of discharge.

C • Area of jet at vena - COIHracta

Area of o rifice

<

c = d

Actual disdarge ... C x C Theoretical disc harge ' ,

where.t and yare the co-<;>rdinate .• uf any point of jct of walcr from ,'ena-«lnlracta. S. A large orifice is one , where the head of liquid above the centre of orifice is Ie,s thun 5 times the depth of orifice. The discharge thro ugh a large rectangular orifice is

,

Q=~ where

C" xb X.J2iI H,JI1 _ H ,JI:~J

b ~ Width ofurifice. Cd = Co-cfficient of discharge for orifice.

II J s Height of liq uid abo,'e top edge of ori fice. und Hl = Il eight of liquid above bollom edge of orifice. 6. The discharge through fully sub-merged urifice. Q C" X b X (H l - H,) X .j2gH where b ~ Width of orifice. Cd .. Co-cfficienl of discharge for orifice. Hl '" Height of liquid ,tbovc OOtlOrll edge of orifice on upstream side. II , ~ Height of liquid aoow top edge of orifice On upstrcum side. H = Difference of liquid h,,'els on ooth sides of the orifice. 7. Discharge through panially sub-merged ori f,ce. K

Q-Q , + Ql D

C,I> (H, _ H)

X.J2gH

+ 2/3 CJ>

X.J2i W ,n _ H, Jn ]

where b .. Width of orifice C", H,. H, and H are having their usual meaning . 8. Time of emptying a lank through un orifice at its OOtlOrll is given by.

where H , = In itial heigh! of liquid in tank. H, ~ I:inal he igh! of liquid in tank ,

I I

Ii

~ I

IL Orifices and Mouthpieces

351 1

A _ Area of lant, a = Arca of orifice. Cd" Co-efficicnt of discharge. If the tan); is to be completely emptied. then time T.

T

2A.[ii .. Cd.ll,jf;'

9. Time of emptying a hemispherical

taTI);

by an orifice fiued at its bonom.

[:. R( 111311 - Hil!) - ~ (ffl~i 1 - lIil!)]

,

Cd.a./ii)

and for completely emptying the where

tan~.

T=

5

It

.{I ,j2i

CJ

[~ RH~12 - ~ 1I1~12] 3

5

R .. Radius of the hemispherical {anl. If I

..

Initial height of liquid,

" l " Final heigh! of liquid, II .. Area o f orifice, and CJ .. Co-efficicnl of discharge. 10. Time o f e mptying a circu lar horizontal tan k by an orifice at thc bouom of the tan k,

T_

4C }CJ .(/ . ..[fi

](2R_ H,r_(2R_ H, )Jr. 1

where L .. Length of horizonta l lant. II . Co-efficicnl of dischmgc for. E ~lcmal mouthpiece, Cd" 0.855 Internal moul hpicre. f\lnning fuli, Cd" 0.707 Interna l mouthpiece . running free. Cd ~O ..so Convergent or con"crgent -
(i) (ii) (iii) (i" )

H, .. Hu -O .89H where Ha "atmospheric pressure hcad " 10.3 m of water H '" head of liquid above the mouthpiece. 13 . For a cOl\ vcrgcnt·di,·crgcnt mouthpiece, the ratio of areas at outlet and at vena ·contracta is

'"

J

- : 1+ H. - H, ", H where il , '" Area of mouthpiece at outlet il, " Area of mouthpiece at vcna-contracta H• .. At mosp heric pressure head If, "Absolute pressure head at vcna-contmcta If " Hcight of liquid above mouthpiece. 14. In case of internal mouthpieces. if the jet of liquid cOmeS out from mo uthpiece without touching its sidcs it is known as running free. But if the jetlOuchc, the sides of the mouthpiece. it is known as runn ing full.

~ I

I~

~ I

IL 1352 Fluid Mechanics EXERCISE (A) THEORETICAL PROBLEM S I. Defi ne an orifice and a mouthpiece. Wh at i, the difference between the two ?

2. h plain the classification of orifices and mout hpi«cs b;tscd on thelr shape. size and sharpness ? J. W hat are hydraulic co..,ffk icnl. ? Name them. 4. Defme the following "o -dficienls (i) Co -... rr.cicnl of vciocity. ( ii) Co-efficient of contraction ~nd (iii) Co...,fficient of discharge. S. Derive the c~ p rcssion Cd . C. xC,. 6 . Define Ycna--<:onlracta. 7. D ifferemiatc between a large and a small ori fice. Obtai n an cxp",,,;on for d;",hargc through a b rge rectangular orifice. 8. What <10 you understand by Ihc [cnllS wholly sub-merged orifice and panially sub-merged orifIce?

9. Prove that the expression for discharge through an external mouthpiece is gi"cn by Q .. .855 x <1)<"

10. II . 12.

13. 14.

where a .. Area of mo ut hpiece m outlet nnd •• • Velocity of jet of water at outlet. Distinguish bet ween : (i) External mo ut hpiece and internal mo uthpiece. (ii) Mouthpiece runn ing free and momhpie<:e running full. Obt:,in an expression for absolute pressure head at vcna-cuntmcta for an cxtcm~ 1 mouthpiece . Whm is a conwrFnt-olutc pre"ure head at Vcnu COnlracta. head of liqu id above mout hpiece and atmospheric pres.''''c head, T hc length of the divcrgeOl outlet part in a venlurimeter is usually made longer compared with that of the converging in let part. Why? Justify the statemcnt . '" In a con"ergcnt-
(B) NUMERICAL PROBLEMS I . T hc head of water over an orifice of diameter 50 10m is 12 m. Find the :tctual discharge and actual velocity of jet at vena-COntr~cta. Take Cd ~ 0.6:md C, ~ 0.98. IAns • .018 m'/s ; 15.04 mlsi 2. The head of water o,'cr the centre of an orifice of diametc r 30 10m is 1.5 m. The actual discharge through the orifice is 2.35 litrcs/sec. Fi nd the co-cfficient of discharge. IAn s. 0.613] 3. A jet of W:ltcr. iss uing from a sharp cdged "crtical orifice under a constant hend of 60 em. has the horizon tal and "enical co-ordinalcs mcasured from lhc vena-conlmcta at a cenain point as 10,0 cm and 0.45 cm respectively , Find thc "alu~ of C •. Also find the value of C, if Cd -O.60_ [Ans. 0_962. 0 _6231 4. The head of "':tIer over an orifice of diameter 100 mm is 5 111. The waler coming 011\ from ori fice is COlleCICd in a circular tank of diameter 2 m. The rise of ",ater level in circ ular tank is .45 111 in 30 seconds. Also the co-ordinales of a cenain point On the jet. measur~..J from \'ena
I I

Ii

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IL Orifices and Mouthpieces 7. A closed laok partially filled with water Upl0 a heig ht of I

In,

353 1

having an orifice of diameter 20"'111 althe

bonom of thc wnk. Dctcmlinc thc pressure required for a discharge o f 3.0 litresls through thc orifice . Take

C" .,0.62.

IAns. 1088 N/cm' j

8. Find (he dischnrge through a rccwngular orifice 3.0 m wide and 2 In deep fined to a water tank. The watcr Ic,-eI in the tank is4 m nixwc thc lOp edge of thc orifice. Ta~c Cd _ 0.62 IA ns. 36.77 mJ/sl

9. A rectangular orifice. 2.0 In wide and 1.5 III deep is discharging water from a lanio;. . l rlhe water level in thc tun\: is 3.0 m above the tOP edge of the orifice. find the discharge through Ihc orifice. Take Cd 0.6. IA ns. 15.40 Inl/sl 10. A rectangular orifice . 1.0 m wide and 1.5111 deep is dischargiTlg water from a vessel, The lOp edge of the orifice is 0.8 m below the water surface in Ihe vessel. Calculate the discharge through the orifice if C J ", 0.6. At"" calculale the percemage error if the orifice is treated as a small orifice. [Ans. 1.0511%[ 11 . Find Ihe discharge through a fully sub· merged orifice of width 2 m if the difference of water levels on both the sides of the orifice t>e 8(X) I1U11. The height o f water from top and bonom of the orifice are 2 .5 m and 3 m rcspc.;tively. TaKe Cd ~ 0.6. [Ans. 2.377 ", l/sl 12. Find the discharge through a toully drowned o rifice 1.5 III wide and I m deep. if the difference of "'ater le"els on both the side", of the orifice t>e 2.5 m. Take Cd - 0 .62 . IAns. 6.513 ml/sl 13. A rectangular orifkc of 1.5 01 wide and 1.2 rn deep is fined in one side of a large tank. The water le\"el On One side of the orifi~e is 2 m abo,·c the top edge of the orifice. While On the other side of the o rifice. the water IO"el is 0.4 m be low its top edge. Calculate the discharge th rough the orifice if Cd ~ 0.62. IA ns. 7.549 mlls l 14. A circular tank of diameter 3 m conta ins water upto a height of 4 m. The tank is provided with an orifICe of diameter 0.4 m at the oollom. Find the time taken by water: (i) 10 fall from 4 m 102m and (ii) for completely emptying the tank. Take C" '" 0.6. IAn •. (i) 24.8 s. (ii) 84.7 sl 15 . A circular lank of diameter 1.5 m contains water UplO a height of 4 m. An orifice of 40 mm diameter is provided at its bollom. If Cd '" 0.62 . find the heighl of w"lcr abov~ the orifice after 10 minules. IA ns. 2 m I 16 . A hemispherical tank of diameter 4 m co~tains ",ater upto a hei ght of 2 .0 m. An orifice of diameter 50 mm is provided al the bollom . Find the time required by water (/) 10 fall from 2 .0 III to 1.0 III (ii) for complelely emptying the unk. Take Cd" 0.6 I"' ns. (I) 30 min 14.34 s. (ii) 52 min 59 51 11. A hemispherical ci'lern of 4 III radius is full of"'''ler. It is filled with a 60 mOl diameler sharp edged orifice al Ihc bollom. Calculate the time required to lower the level in the cistern by 2 metrn. Take CJ ,", 0.6. [Ans. I hr 58 min 45.9 s l IS. A cylindrical tall. is having a hemispherical base. The height of cylindrical portion is 4 111 and diameter is 3 m. At the bottom of th i, lank 3n orifice o f diameter 300 mm is fitted. Find thc time required to completely cmplying Ihe tank . Take Co z 0.6. [Ans. 2 min 7.37 ,[ I II . An orifice of diameter 200 l1un is fitted at Ihe bottom of a boiler drum of length 6 III and of diamete r 2 m. The drum is hori7.0l1lal and half full of water. Find the time required to empty the boiler. given the "alue of Cd - 0.6 [Ans. 2 min 55.20 51 20. An orifl,e of diameter 150 mm is filled at Ihe boltom o f a boiler drum of length 6 m and of diameter 2 m. The drum is horizontal "nd contain, water upto a height of 1.8 01. Find the time required to empty the boiler. Take Cd '" 0 .6 . IA ns. 7 min 46.64 s l 2 1. Find the discharge from a 80 mm diameler extern"l mouthpiece. filled to" 'ide of a large vessel if the head O,'er Ihc moulhpk'<;e is 6 m. [Ans. OJ)466 10)/5 1 22 . An external cylindrical mOUlhpiCO"e of diameter 100 mm is discharging water under a con .• tal1l head o f 8 m. Detemline the discharge and absolute pressure head of water at vena-eontracta. Take CJ • 0.855 and Co for vena-eont racta .. 0 .62. Take almospheric pressure head", 10.3 m of water. IA n,. 0.084m3,s; 3.18 ml 23 . A eonvergent-
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IL 1354

Fluid Mechanics

24 . The throat and ~ ~ it diameter of a convergem-d;\,ergcnt lHouthpiece are 40 mm and 80 ["In re~peclively. It is fitled to thc ""Mical side of a ImIO: . containing water. Find the maximum head of water for steady flow. The maximum vacUum pressure is 8

In

of Wa ler. T ake atmospheric pressure head = 10.3 In of "",,(eT.

[Ans. 0.533 ml 25. The discharge through a (:{)nvergcnl-divergcnt mouthpiece titled [0 the side of a (anK under a constant head of2111 is 7litrcsls. The head loss in the divergent portion is 0.10 limes the kinelic head at OUtlet Find the throat and exit diameters. if separation pressure head ~ 2.5 m and atmospheric pressure head 10.3 111 of waler. [Ans. 25.3 nnn ; 38.6 mml 26. An i1l1cnml mouthpiece of 100 mTll diameter is discharging Water Ilnder a COIlstant head of 5 m. Find the E

discharge through mouthpiece. when (i) the mouthpiece is running free. ,md (ii) the mouthpiece is running full. I Ans. (,) 38.8 lines/s. (ii) 54 .86 l itre",1

~ I

I~

.. 8. 1

INTRODUCTION

A notch is a device used for measuring Ihe rate of flow of a liquid through a small chan nel or a tank. It may be del1ned as an opening in the side of a tank or a sm all ehannel in sueh a way that the liquid surface in the tan k or channel is below the tOp edge of the opening. A weir is a concrete or m3!\onary structure, placed in an open channel over which the flow occurs. It is generally in the form of ven ic~1 wall. with a sharp edge at the tOp. running all the way across the open dannel. The notch is of small size while the weir isof a bigger size. The notch is generally made of metallic plate whi Ie wei r is Illade of concrete or masonary structure. I. Na p pe o r Vein. The sheet of water flowing through a notch or over a weir is called Nappe or Vein. 2. C res t or Sill. The bottom edge of a notch or a top of a weir over which the water flows. is known as the sill or crest.

.. 8 .2

CLASSifiCATION Of NOTCHES AND WEIRS

The notches arc classified as : I. According to the shape of the opening: (a) Rc.:: langular notch. (b) Triangular notc h. (e) Trape7,Oidai notc h. and (d) Stepped notch. 2. According to the effect of the sides on Ihe nappe: (a) Notch with end conlraction. (b) Notch without end contraction or suppressed notch.

Weirs arc classified according to the shape of the opening. the sh ape of the crest. the effect of the sides on the nappe and n~ture of d ischarge. The following arc importal11 classificmions. (a) According 10 the shape of the opening: (I) Rc.:: langular weir.

(ii) Triangular weir. and

(iii) Trapezoidal weir (C ipollcui weir) (b) According to the shape of the crest:

(i) Sharp-crested weir.

(iii) Narrow-crested weir. and

(ii) Broad-crested weir_ (iv) Ogee-shaped weir.

355

I I

Ii

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IL 1356

Fluid Mechanics

(e) According to the effect of sides on the emerging nappe : (i) Wei r with end contraction. and (il) WeiT without end contraction.

to 8 .3

DISCHARGE OVER A RECTANGULAR NOTCH OR WEIR

The expression for discharge over a rcctallgular notch or weir is the same. NAPPE

~~ I 7>,-'

i

'''''''j'''fJQ TI CREST ' / OR SILL

,' ~

1

I-- L

IT'"

----..j

(e) SECTION AT

CREST (al RECTANGUlAR NOTCH

NAPPE

c:!,~r\

(b) RECTANGULAR WEIR

Fig. 8.1 R~ctangular notch alld 'Il;rir.

Consider a rectangular notch or weir provided in a channel carrying water as shown in Fig. 8.1. Le! H " Head of water over Ihe nest L " Length of the notch or weir For finding the discharge of water flowing over the weir or notch, cOlIsidcT an elementary horiwntal strip of water of thickl1es~ dll alld length L at a depth" fronl the free surface of wakr as shown in Fig. 8.1 (c). The area of strip "Lxdh and theore tical velocity of water flowing through strip" J2gl! The dischargc dO. through strip is dQ'" Cd X Arca of Strip x Th~'Orctical velocity ... (1)

where Cd '" Co·cffleient of dischargc. The total discharge. Q. for the whole notch or weir is dctcnn ined by imcgrating equation (I) bctwCt:n the limits 0 and If.

=C, 'L'.j2i -'1111" ,

[2 + [

1' =C, 'L'.j2i [h- )I1]"

2 ' " IHI ~. ="3Cd X/.X.y2g

3/2

0

...(8. 1)

Problem 8. 1 Find rhe discharg e of Wilier flowillg o ra a reclangulor norell of 2 m /engrll Wllell ,he COIISIaIll head ora rile lIorch is 300 mm. Take Cd " 0.60. Solu l io n. Given: Length of the notch.

I I

L=2.0m

Ii

~ I

IL Notches and Weirs Head over no tch.

II '"

300 m '" 0.30

357 1

III

Cd '" 0 .60 Di sc harge,

Q=

~ Cd XL x.J2g" [HJIl ] 1

2 = - xO.6 x 2.0 x J2x9.S r x [O.3011,s m J {s 3 '" 3.5435 x 0.1643 '" O.SSZ m' /s. ADS. Problem 8 .2

D e/ami/Ie I/Ie /leighl of II reC/(lIIgu/M weir of I
Solution. Given: Length of wt'ir. !krIll o f wate r. DiSl:hargc.

L=6 m H I= 1.8rn

Q '" 2000 litIs", 2 m 3/s

Cd '" 0.6 Let II is he ig ht o f Wale r above the crest of wei r. and Ill'" he ight o f we ir (rig. 8.2) The d isch:lrgc uve r Ihe we ir is gi ve n by the equatio n (8 . 1) as Q '"

2.0 '"

"'

"32 Cd X L x./2i

,,312

~ x 0.6 x 6.0 x ,J"2~<~9~.8"1

'" 10.623 H

X 11 312

Jn

1l3l2 = ~

Fig. 8.2

10.623

II =

He ight of we ir.

'- -0 )'" '" 0.328 m ( -10.623

H1 ", H , -H '" De pth. of water on upstream side - /I

= 1.8 - .328 = 1.472 m. An~. Problem 8.3 The he(ld of ,,'«Ier 01'1." ( I recl""glll(lr nolch ;s 900 mm. Tile disclwrge is 300 lilres/s. Find Ihe /eng lh oflht, nO ld,. wilen Cd '" 0.62 . Solution. Give n : H~ad

over notc h.

Disc harge.

/I = 90 cm '" 0 .9 m

Q =300 litIs = 0.3 m 3 /s Cd'" 0.62

lA:t le ngth. of nutch Using equatio n (8 . 1), we have

"L

2 Q'" 1

I I

X

Cd X l.

x./fi X /1 )12

Ii

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IL 1358

Fluid Mechanics

0.3 =

~ )( 0.62 x L x

..}2 x 9.8 1 x (0.9) .112

= 1.83 x L x 0.8538 L:

... 8.4

7"CCO"·" '''''O = .192 1.83 x .8538

III

= 192

mm.

Ans.

DISCHARGE OVER Po TRIANGULAR NOTCH OR WEIR

The expression for the discharge over
e = -AC :

t;lI1 -

2

AC ;-;:C"' " ( H II )

OC

AC=(If - II) tan

,

,

,

~ r

-z

•

Width of strip

=AB=2AC=2(H - l!)tan-

:. Area of Mrip

=2( H-h) ian 2"xdh

,

.'\t'o--'-

,.,

<0,

Fig. 8.3

2

Th~

triangular notch.

The theoretical velocity of water through strip '" ../2gh Oischarg~ ,

tllrough the "trip.

,

dQ = Cd x Area ofslrip x Velocity (theoretical)

'" Cd

X

2 (H - II) tnn

'" 2C,, ( N - II) tan

:. Total di scharge,

,

"2 x till X

"2 x ./2gh

./2gll

X dll

jo" 2Cd (H - II) tall -,2 x ./2gh x dll =2C , xwn -' x,,2g (H - /J)II '"-till re j" , 2 " == Cd x tan -, x J2i j" (HI,! n

Q=

2 X

_11

2

==

2 x C x tan -, d

I I

2

312

"

Hhlll " SI! x,fig [ -- -3/2

5/2

)

dll

r

Ii

~ I

IL Notches and Weirs

=2 x C" xtan

'"

!XJ2i[~ H3Ill 2

Q=.!

Problem 8 .4

,'2 '"

xO.6x I x .}2 )(9.8 1 x

15 '" 1.4 17 H5I2 ,

Frill! t/ie discharge oreT

...(8.2)

2

tan

Discharge,

15

~ Cd x Ian! x J2i x Hsn.

15 For 3 right -allglcd V-nOich. if Cd '" 0.6

II

359 1

wn

... (8.3)

Iriatlgl4/ar nOlcll of /wgle 60" ..... hell the Ilead Mer Ille

V-nore/! is 0 .3111. Anume Cd '" 0.6. Solution. Gi ve n :

Angle of V-notch .

e '" 60"

Head over nOlch.

H =O.3 m

Cd '" 0.6 Disc harge, Q over a V-notch is give n by equation (8.2)

8 e Q = _ xC" xtan _ x.j'fi x HSfl 15

8

2

.

60'

'" IS X 0.6 tan 2)(

.}2 x 9.8 1 x (O.3)Y.!

'" 0,8 182 x 0.0493 = 0.040 mJ/s. Ans. Problem 8 .5 Warer flows 0I'e( II rec/allgular weir I III wide m a deptll of 150 mm and aftenI'Grds pancs {llrol jgh a rriangular righi-angled weir. Taking C,,/or II,e rec/angulM (Uullrhlllgular weir as 0.62 (l1Ii1 0.59 resper/iI-e/y. find the deprh Ol'/'f rhe triangular ,,·{'ir. Solu t Ion. G ive n: For rectangular weir. length ,

t:o I m

Depth of water.

H = 150 111m = 0. 15 111

Cd = 0.62

e == 90°

For trian g ular weir.

Cd == 0.59 lA:t depth uvcr The

I I

disch~rl(e

tri angul~r

ove r the

weir

:0

reet~ n gular

HI

weir is given by equat ion (8. 1) as

Ii

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IL 1360

Fluid Mechanics

2 Q= - xCdXLX.Jfi xHIf!

3

'" '32 x 0.62 x

1.0 x .)2 x 9.8 1 x (.15)">11 mJ/s '" 0.10635 ml/s

The same dischargc rasS<.'s through the tria ngular Tight -angled weir. But discharge, Q, is giver> by equation (8.2) for a triangular weir as

8 15

Q= -

X

e

Cd x tan - x.[fi X H~f!

2

8 90' 0. 10635 == - x .59 x tan ""2 x 15

8

= -

15

H

Y1.

FE x H/'/2

x .59 x I x 4.429 X H

IY2=

1.3936

HI 512

= 0.10635 = 0.07631 1.3936

I

HI == (.07631)11.4 = 0.3572 m. Ans.

Problem a.SA W,,[er flows 111f0Ug/1 II (rfangulllT rig/u-ongled weir firJ"' amllhen Ol'('( {/ r<,ClanguIlIr weir of I In widill. The di.~cha'ge cQ-ef!iciellfJ of Ihe Iriangul
=

e 2g xl-/ w Q= -8 xCd xlan -xv 15 2

8

== - xO.6x tan 15

(90°) -

2

xJ2X9.81 x (0.36)w= 0.11 02 1ll 3/s

The same discharge is passing through the rcct.lngular wei r. But disc harge for a I"\:l:tangul ar weir is given by equation (IU) as

2 Q="3 xCd xLx 0.1 102 ==

"'

I I

~ 3

Iii xfl·Vl.

xO.7 x I x J2 x9.8 1 x fI·Vl. == 2.067 H )J2

H312 '" 0.1 102 '" 0.05:n 2.067 1-/ = (0.0533) U3 '" 0.1 41 5 m '" 141.5 mm. Ans.

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IL Notches and Weirs

361 1

Problem 8 .6 A f<'c/{wgular c1wlllwl 2.0 m wide ilaJ a discharge of 250 litres per .>'ecalili. whiel! is measured by (I righl·ongled V·notell weir. Find the positioll oflhe apex of tile nOlel, from lilt' bed of the dUlI/nel if maximum depth afwiller is no/ ro exceed 1.3 m. Td/(e Cd '" 0.62 . Solution. Given: Width of rccwngular chnno.:!. L = 2.0!l1 Dischilrgc. Q = 250 litis = 0.25 Ol lis [)cplll of wa te r in channel = 1.3 In Lei lhe height of wakr ove r V-notch = H

The rate of n ow throug h V- notch is given by equation (8.2) as

8 e xcdx.[fi xtan- xll 5t2 15 2

Q= -

90· 8 X.62x.j2x9. Rl x tan 2 15

Q= -

"' "'

xH~

0.25 '" 18 x .62 x 4.429 x 1 x HSi'l 5 .25 x 15 H Y! = ;;-c~=~~ "" 0. 1707 R x .62 x 4.429 H = (.1707):!I~ '" (. [ 707)°·4 '" 0.493 III Position of apex of the notch from the bed of channe l = de pth of waler in ch
.. 8.5

ADVANTAGES Of TRIANGULAR NOTCH OR WE IR OVER RECTANGULAR NOTCH OR WEIR

II. triangular notch or weir is preferred to a rectangular weir or notch due to following reasons: I. The expression for discharge for a right-angled V-notch or weir is very simple. 2. For rnc~suring low disdargc. a tri;mgu)ar notch givcs 1110re a<.:curate results th~n a rectangular notch. 3. In case oftriangulur notch. only one reading. i.e., H is required for the computation of dischnrge. 4. Ventilation of a triangular notch is not necessary .

.. 8.6

DISCHARGE OVER A TRAPEZOIDAL NOTCH OR WEIR

II.s s hown in Fig. 8.4. a trapezoidal notd or wci r is a combinati on of a rectangular and triangular notch or weir. Thus the total di scharge will be eq ual 10 the sum of discharge through a rectangular weir or notch and discharge through a triangular notch or weir. Let H = Height of water o\"er the notch L = Length of the crest of the notch

F

------

e

\l\H~r D

E

C

I - - , --.,

F ig. 8.4 The t~(lpezQida/ ,,(Itch.

I I

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IL 1362 Fluid Mechanics c~,

= Co-efficient of discharge for rectangu lar portion ABeD of Fig. &.4.

Cd, " Co-cffkicnt of discharge for Triangular portion IFAD and BeE1 The diSCharge through rectangu lar portion ABeD is given by (8.1)

,

2 Q, '" - x CJ

'

)(

I'>::

1I2

L )( ,,28 )( H -

The discharge through twO lri~ngular notches FDA and BeE is equal 10 Ihe discharge through a single triangular notch of angle 9 and it is given by equation (8.2) as

a

8 ~ Y1 Q1 = - XCd X I
2 3

"" -

Cd L I

'28 x fI -1I2 + -158 Cd' )( Ian en)( v"X f2ii x H SIl.

v -'-I;

...(8 .4)

Pro blem 8 .7

Find Ihe discharge through (l Ir(lpezoiJal II oleil "'hid, is I m "'ide al the lOp Gild 0..10 m til the bulIO'" wu/ is 30 em jll height. The /W{Id of ",Mer 011 the li DICh is 20 em. A ssume C,,Jor reCllmg"{"f portio" '" 0.62 ,,"/life/Of lTitmg,,/ar portioll '" 0.60.

, ,

Solution. Giv",n : Top width.

liE = I III

Base width.

CD", L = 0.4

Head o f wate r.

H", 0.20

III

:........ , ,

III

-

Cd, = 0.62

,

For triangular portion.

Cd," 0.60

f0- e

WI'

-

~:-"i!i"-

For r"C!angular portion.

From l!.ABC.

, ,

--- - --

0

T , 1

-i

have Fig. 8.5

tan ~ " _A_8 : c( A~E~---;; C~ D~ )I~2 2 Be 11

=

( 1.0 - 0.4 )f2

0.3

0.6/2

0.3

0.3

0.3

; --: -

:1

OiSl:liarge through trapezoidal notd is given by eq uati on (8.4)

Q= -2 Cd 3 '

X

L x.ffi x~ +

' -8 Cd x tan - x../2i x 15

2 3

'

2

H '>r.

8 x .60 x I x ./2 x 9.81 x (0.2)Y.! 15

= - x 0.02 x 0.4 x ./2 x 9.81 x (0.2 l Y.! + -

= 0.06549 + 0.02535 '" 0.09084 m3fs " 90.84 Ill res/s. An s .

... 8.7

DISCHARGE OVER A STEPPED NOTCH

A stepped notch is a combination of rectangu lar notclies. The di .>eharge through equal to the su m of the discliarges through tlie different rectangular notclies.

I I

st~ppcd

notch is

Ii

~ I

IL 363 1

Notches and Weirs Co nsider a ste ppe d no tc h as show n in Fig. 8 .6.

Lei H I" Height o f water abo ve the ercS( of nOlc h !. Ll

"

Length of notch I.

H1. L2 and H3_ LJ arc corre sponding v:tlucs for notdlcS 2 and 3 respect ive ly.

f-- L' --1

Cd " Co-effi cie nt of di sc harge fo r all no tc hes Total di Sl: hargc Q '" Q 1 + Q2 + QJ

r-- l , I•

2

r;;: 31'1 .Yl Q= "] xCd x LI x ,,2g IHI -Hl 1

_I

L,

•

I

Thr Itrpped notch .

Fig. 8.6

... (8 .5)

Problem 8 .8 = 0.62.

Fig. 8.7 sho ws

II

stepped lIolc/'. Find lilt' discharg/' l"rOJ/gll Ihe notch

il e d/o r 1111

sec/ion

Solution. Given:

T L I = 40 em. ~" 80 em.

LJ = 120 Clll

"I

= 50 + 30 + 15 =95 em.

Hl '" 80 em. HJ

fl,

50cm

= SO em.

Cd = 0.62

-~ Iscm

T

_ ,.

Total di scharge. Q '" Q, + Ql + QJ where

~40cm""'1

80cm _

120 cm

. ,

Fig. 8.7

2 r->:: QI ="3 X C" X L1x .;2g

J.i2

IHI

- H!

.lI2

1

= ~ x 0.62 x 40 )( J2 x 98 1 x [95 31l - 80Jl'1 1 3

= 732.261925.94 - 7 1'i.54 1 == 154067 cm3/s '" 154.067 li tis f'>:: 1I1 1I1 2 Q2 ""3xCJ xL2 X,,2 g xlH2 - HJ 1

" ~XO.62X80 X J2X98 1 3

x [80 312 _ 50 312 1

== 1464.52(7 15.54 - 353.55( cm3/s= 5301 41 cl1I 3ls" 530.144 li tis

"

~ 3

x 0.62 x 120 x

J2 x 98 1 x 50ll.l " 77677 1 cm 3,s" 776 .77 1 litis

Q = Q 1 + Q 2 + Q 3 = 154.067 + 530 . 144 + 776.77 1 = 1460.98 ]iUs. ADS.

I I

Ii

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IL 1364

Fluid Mechanics

.. B.8

EffECT ON DISCHARGE OVER A NOTCH OR WEIR DUE TO ERROR IN THE MEASUREMENT OF HEAD

For all accurate value of the discharge over a weir or notc h. an accurate measurement o f he ad ove r the weir or notch is ve ry esse ntial as the disch arg e ove r a triangular !lotch is proportional 10 HYl and in case of rectang ular no tch i1 is proportional \0 HY2 . A s mall error in thc measurement o f head. will affect lhe di.<;chargc co ns iderabl y. The fo llowing c ases of error in Ihe rnCa5 urcmcnt of head will be considered : ( i) For Rectangular Weir or No tch. (ii) For Trian gular Weir or Notclt.

8 . 8 . 1 for Rectangular Weir or Notch . by equation (8.1) as

Th e disc harge for a rectangular weir or notch is g iven

=

~ Q= -2 xCJ xLx,,2g xH3 == KH1I2

where K

2

= "3 Cd X L )(

...( i)

.Jfi

Differe ntiating th e above equation. we get

3

I12 dQ- Kx -2 H dH

K x ~xH"2dH

dQ '" -~2~;n- '" 3 dH Q KH JI2 2 H

Di vidin g (ii) by (i).

...(8.6 )

Equ ati o n (1\.6) shows that an error o f 1% in mcasurin g fJ will produce 5<;<, c rror in di scharge ovcr a rectang u lar weir or notch.

8. 8.2 For Triangulilr Weir or Notch. by equation (8.2) as

Q'"

The discharge uver a triangular weir or nutch is g ivcn

8 e r->: 15 C d' tan T,,2 g

'" KH

xH

YJ.

SIZ

... (iii)

8 e Cd' tan - ,J2i IS 2

where K '" -

Differe nli~ting equ~tion

(iii). we gel

dQ= K~H3I2 XdH

.. .( i,')

2

K~

1/ 311

dll

Di viding (il') by (iii ). we get dQ '" _.2co",,~JIl Q K fi Equation (8.7) show s thaI :111 error of 1% in a triangul~r weir or notch.

I I

5 dH 2 H

III c ~ suring

H will produce 2.5% e rror in

...(8.7) disch~rg c

over

Ii

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IL Notches and Weirs

365 1

Problem 8 .9 A recrangular 1I(1Icii 40 em hlllg is IIsed for lIIea~'llfing a dj.~c1wrge of 30 Iilre.1 per second. All error of 1.5 mm was made. wllile medsurillg the head ol'a Ihe notcll. Calcula te IIII' pefcelilage error in II,e discharge. Take Cd '" 0.60. Solu tion. Given: Length of notch. L=40cm Dischilrgc. Q = 30 litis = 30000 c11l 3fs Error in head. dH= l.5mm = O.IScm Cd = 0.60 Let the hdght of water over rectangu lar notch = Ii Ttw discharge through a r;:c(angular notch is given by (8.1 )

2

Q=")XCd XLx.[fiXH 3I!

or

,

30000

"'

=..:. x 0.60 x 40 x .J2 x 98 1 x 11Y2 3

H3Il =

3 x 300Xl 2 x.6O x 40x

.J2 x 98 1

= 42.33

II : (42 .33)Z13 = 12.16cm

Usi ng eq uatio n (8.6). we gel dQ

Q

= ~dH =~X~ =O.OI85= 2 H

2

1.85%. A ns.

12.16

Problem 8 .10 A righl ·angled V.notch is IHed Jor me(lSllring (/ di~Tll(lrge of JO li/res/s. All e rror of /.5 "'''' "'a5 ",ade "'bi/e ",,,aJ'uring Ill" ile"d Ol"er IIIe n"'cll. C,,/culale Ille percell/age eTTor in lile disdwrge. Take Cd '" 0.61. Soluti on. Give n : Angle of V-no tc h. e '" 900 Dis.:harg'" Error in head.

Q '" 30 litis'" 30000 cm 1/s dH:: l.5mm",0. 15cm

Cd:: 0.62 Let the he~d over the V- notch:: H The d ischarge Q through a triangu lar notch is given by eq uation (8.2)

8

e

fi":"

Q'" - Cd ' tan - x ,,2g x H 15 2

30000 ::

Y.!

.! x 0.62 x tan ( _90_°) x , "lex"9","1 x HYl. 15

2

:: -8 X .62 x 1 x 44.29 X H Y2 15 30000 x 15 :: 2048.44 8 x .62 x 44.29 H '" (2048 .44):!I·~:: 2 1. 1 1 ern

I I

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IL 1366

Fluid Mechanics

Usi ng equatio n (8 .7). we get dQ '"

Q

~ dH 2 11

'" 2.5 x 0.15 '" 0 .0 1776 '" 1.77 % . 21.11

AD S.

Problem 8.11 Tire /!"lItl ojwMer Ol'e, II Irilmgu/ar II oleil of lingle 60' is 50 em ami co -effieien, of discharge is 0.62. Ti,e flow measured by il is 10 be wililill all accuracy of 1.5% up or dowl!. Find Ihe limiting )'a/ues of the Iwail. So lution. Gi ve n : Angle o f V-no tch , H~ad

o f waler,

9 = 60° H "'SO cm Cd'" 0 .62

dQ '"

±

1.5'll '" ± 0 .0 15

Q The di sc harge Q over a triang ular no tc h is

8 Q'" -15

8

=-

15

c d .,f2i Carl x 0.62 x

, -::;- H Y2

_

Jz x 981

60· x tan x (50)Y.! 2

'" 14.64 x 0.5773 x 17677.67 '" 149405.86 C111 3fs Now appl yi ng eq uati on (8.7). we get dH dQ = 5 dH ur :t.O I 5=2.5 Q 2 1/ if

.015 2.5 T h", lim iting va lues of th e he ad

" '

tlH /I

-

.015 :± - -

2.5

.015 2.5

dH ::!: - - x H =::J::. - - x50=:!:0.3

" 1/ ± till " 50 ± 0.3 '" 50.3 e m. 49.7 elll = 50.3 e m und 49.7 e m. AilS.

.. 8 .9. (a)

TIME REQUIRED TO EMPTY A RESERVOIR OR A TANK WITH A RECTANGULAR WEIR OR NOTCH

Consid er a rese rvo ir o r tank o f uniform cross·section al are a A. A rec ta ng ul ar weir or notch is prov ided in one of its si des. L = Length of crest of the wei r o r notc h Cd = Co·efficien t of d i><: harge H , = In it ial height o f liqu id ahove the crest of notc h H2 = Fin al hei g ht of liqu id above th e crest o f no tc h T = Ti me req ui red in seconds to lowe r the heig h! of liqui d from " , 10 H1 • Le t at an y instant. the heig ht of liquid surface a bove the cre"t of wei r or notch be /, and in a small time JT. let the li!.juid surfa\:e fall s by ·d''-. T hc n.

Let

- Adh=Q x dT - ve s ig n is tak c n. as with th e im;rease of T. /, dec re ases.

I I

Ii

~ I

IL 367 1

Notches and Weirs

2 Q= - Cd x LX.fii

""'

3

X

h 3ll

, "" ~ __--=-~A~dl~,_ __ -Adh=iCJ XLx,,2g .11 xdTordT= Z "3 Cd X LX/fiXhli! Th" [01.11 lime Tis oblai",:,d by integrating Ih" :thovc equalion between the limits H, and H!"

lo'

,Fr-

fHl

-

If,

-Adh

~Cd XLx.fiixllm 3

0'

... (8.8)

(b)

TIME REQUIRED TO EMPTY A RESERVOIR OR A TANK WITH A TRIANGULAR WEIR OR NOTCH

Consider II reservoir or tank of unifonn '-Toss-sectional area A. one of ilS sides. LeI a", Angle of the notch

havin~

11 triangular weir o r nOleh in

Cd == Co-emden! of discharge HI'" Initial height of liquid above the apex of 1I00(h H) '" Final heig ht of liquid above the apex of nOH;h

T = Time required in seconds. 10 lower the heighl frolll HI to II! above the apex o f the 1I0\cll. LeI at any instant. Ihe height of liquid surface above the apex of weir or notc h be II and in a small timc dT, let th~ liquid surface falls by 'dll', Then - Adll=QxdT - ve sign is taken, as wit h the increase of T. II decreases, And Q for a triangular notch is

=

8 , Q= - xCd xtan - .. 2g xlrV2 15 2 8 e"-::~, , - Adll= - xCd xtan -x .. 2g xII -xdT 15 2

I I

Ii

~ I

IL 1368 Fluid Mechanics tiT",

Adlt

8 e r;:>;: J/' - xCd xtan - x,,2g xll 15 2

The total lime .,. is obtained by integrating the above equation between th e lim its III and 11 2, T dT- J, -Adh

r

1.

- JH, ~ C 15

Ian

d

~ vJ2; II SI ! 2 "'-e;

f"'

"'

II

Si'2

dh

0,

'"

~15Ae

8xCd xtan

=

2

X

x..{fi

5A a

(-')[Ji" , ]'"

3'"

4xCd xtan - x.j2i

[H Ill' l

II,

1-

2 Problem 8 .12 "-ind the lime "~quired la low", the Waler Jerel from J III 10 2 dimension 80 In x80 m. by a reclangl jlar norch of len gIll 1.5 m. Take C'I '" 0.62. Solution. Given: Initial height of water.

HI=3 m

Final heigh! of wmcr.

" 1=2111

Dime nsion of reservo ir'" 80 or Are a. Length of notCh.

In

...(8.9)

- Hl1l']'

-

III i'l

a

re,~erl'Oir

of

X 80 1ll

A=80x80=6400ml L= 1.5m,C
Using Ih e relal ion given by the equ ation (8.8) T

3A [ , '" Cd X LX .j2i :.{Ji;

, ]

- J'H,

3x6400 [ ' '] = O.62xI.5xJ2 x9.81 J2 - Jj '" 4661.35 [0.7071 - 0.5773J seconds == 605.04 secunds == 10 min 5 sec. AilS.

Problem 8.13 If ill problem 8.12. illSI<'ad of a rectallgular lIo/ch. a righl·allgled V'lIotch is used. filld /I'e lime requiTed. r"ke (11/ o/l,eT da/" same.

I I

Ii

~ I

IL Notches and Weirs Solution. Gi ven: Angle of nOlch. Ini tial height of wate r. Final he ight o f walcr. Ar~a of rese rvoir.

369 1

H, =2 111 A= 80 x 80 = 6400 ml Cd = 0.62

Using th e relation give n by equa ti o n (8.9)

5xMOO

[ "

'" 4 x .62 x Ian 90"x .)2 x 9.81 2

2" -]iT

J

'" 2913.34 x [ -'-

- - ' -] 2.8284 5.1961 '" 29 13.34 [0.3535 - 0. [924J seconds = 469. 33 seco nds = 7 min 49.33 sec. Ail s. Problem 8.1 4 A right-angled V-lIo/eh is illsert,'d ill Ihe side of a tallk of leng'" 4 m mId .....idth 2.5 m. Initial heigh! o[ ...
e

T=

SA 9

[~ - ~l H

4x Cd xlan l x.[fi Hl "

180 =

4x.60xlan

I -

'l

5xlO [ , (90' ) ---y;y - ( )'" 2 xJ2x9.81 Hl 0 .3

' ] = 4 x .60 x501 x 4.429 [ H/' '1 - (0.3)liZ 0'

I I

180 x 4 x 0.60 x 4.429 = 38.266. 50

Ii

~ I

IL 1370

Fluid Mechanics I "I'T - 6.0858 '" 38.266

or

H,

I

-,~ fI ;!

'" 38.266 + 6 .0858 '" 44.35 or H,

I.~

•

1 '" - - '" 0.0225 44.35

112 ", (0.0225)"1.5", (0.0 225 ),6667 = 0.0822

...

m = 8.22 COl, Ans .

VELOCITY OF APPROACH

8 . 10

Velocity of approach is defined as the veloci ty with which the wate r approaches or rc~chcs the weir or nOld before it flows over it . Thus if Va is til.: velocity or 3pproach. then an additi onal head /10 cq uallo

V'

.....!L

2,

duc lU veloc ity of .tpproach. is acting on the water flowing over the notch. Then initial

heigh! of water over the nmelt bc\;OIllCS ( 11+ Il o) and I1nal hei ght becomes equal 10 11 0 , Then all the

formulae arc changed taking into co nsideration of ve loc ity of approach. The ve loc it y of approach. Va is determined by finding the discharg e ove r the nOlch o r weir neglecting veloci ty of appro ach. Then di viding th e discharg e by the cros.'H;•."Ctional area o f the channel on the upstream sidc of the weir or notch. the ve locity of approach is obt ained. Mathe matically.

Q

Vo = CC---C'c----C Arca of ch annel

This vel ocity o f approach is used to find an additional head

(II.

=

~.:). Again the discharge is

l:aJculatcd ,lI1d above process is repeated for l11or~ accurah: discharge. Discharge over a rectan gular weir. wi th velocity of approach 2 X Cd X L x .,ffi [(HI + h o )Y2 - h,:n [

'" "3

.. .(K.10)

Problem 8.15 Wulu is flo,..illg ill II rec/Illlgu/w c/UWIlef of I III ,..ide "JIl/ 0.75 III deep. Filld 11,1' disch"rge o,·u (I reCl,mgu/ar ,.."ir of aes//ellgll, 60c IIl. iflhe head "fwaler v,·er Ihe creSI of weir is 20 cm "",/ ,..(ller from dlm,"el flvws m ·er Ihe weir. Tilke Cd == 0.62. NeglecI elld COll/ractiOllS. Tilke ,·elOeily of approach illlO COllsidew/im, . Solution. Givcn : A", Width x d~pth '" 1.0 x 0.75 '" 0.75 m l Are a o f channel. L=60 cm=0.6m Length of we ir. ~lcadofwmer. H 1 =20cm =0.2 m Cd = 0.62 Discharge over a rectangular weir without velocity of approach is given by

Q==

~ 3

2

== -

3

I I

xCd xLx

J2i

x 0.62 x 0.6 x

Y2 xH 1

J2 x 9.81

x (0. 2)312 111]/5

Ii

~ I

IL Notches and Weirs

371 1

'" 1.098 x 0.0894 '" 0.0982 1I1'/s

Q

.0982

Velocity of approach.

11. =-.:\ ",

:. Additional head.

"" '" - "- '" (. 1309)212 x 9.81'" .0008733 m

0.75 =0.1309l11/s

v'

2,

Then di scharge with velocity of approacl"t

Q '"

~

j~

X Cd X L

given by L-q ualion (8.10)

x

J2i

2 3

'" - x 0.62 x 0.6 x

[(If,

+ lI o)lIl_ 1I.3J2]

J2 x 9.8 1 ](0.2 + .OOO87)Jf2 -

(.OOO87) lI2 ]

= 1.098 [0.09002- .00002566] = 1.098

Problem 8.16

x 0.09017 '" .09811 1 m3,s. An s.

Fil1d III" discharge o,"a u rec/(wgu/ur weir of /e"gll! 100 III. Th e head of "'{ller VI'er

Ille weir is 1. 5 m. The \'e1ocily ofaJlJlrOllcll is ghell {/,~ 0.5 mls. Take Cd '" 0.60.

Solution. Given: L.:ngth of weir. Head o f water. Velocity of approach.

L = HlOm 1I ,= 1.5rn V. '" 05 Illfs

Cd '" 0.60 11 2

If : - "-=

U5xQ5

=0.0 [27 111 2 x9.8 1 The diSl: hargc. Q over a rectangular weir due \0 velocity o f approac h is givc n by equat ion (R. I 0) :. Additional head,

"

2g

Q=

3.. X Cd x Lx 3

J2i I(H I + h.):sn. _h. m ]

2

'" - x 0.6 x 100 x J2 x 9.111 ]( 1.5 + .0 I 27):sn. - .01 27 3/2 ] 3 '" 177. 16] 1.5127·V2 _ .0127m ]

m',s.

= 177. 16 ]1.8605 - .00 143 J = 3 29.35 Ans. Problem 8.17 A rectangular weir of cre),tlengtll 50 cm is used to measure tile rate offlow ofll"ater in a rectangular cliallnel of 80 elll ....ide and 70 em deep . Dererm ille rile discliarge in lire challnel ifrlle warer lel'el i.f 80 111111 IlbOl'e rhe cresr of weir. Take I"elocity of approach illlo consideration and mlue of Cd '" 0.61 . Solution. Given: 1. =50cm =0.5 11) Length of we ir. Ar~a o f channel, A = Wi d th x de pth = 80 CIII X 70 Cill == 0.80 x 0.70 == 0.56 m< H~ad

over weir.

Il '" 80 111m '" 0.08 m

Cd == 0.62 The discharge ove r a rectangular weir without ve locity o f approach is g ive n by equation (8.1)

I I

Ii

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IL 1372 Fluid Mechanics Q '"

~ 3

x

cJ )( L)( Jii )( H lI!

2 3 = 0.9153)( .0226 = .0207 m 3fs

'" - x 0.62 x 0.5 x .}2 x 9.81 )( (D.08)3/"! m 3 1s

Velocity of approach.

Head due to Va>

Q ,0207 Vo '" - = - - =.0369mfs A 056 Ir

"

2 (.0369)" = V;t2g = = .0000697 III 2 )(9.81

Discharge wilh velocity of approach is

,

=::. X 0.62 X 0.5 X .)2 )(9.81 [C08 + .0000(97)312 - .0000697~nl 3 '" 0.9 [53

X

[.0R00697I.S - .00006971.~J

3 '" .9153[.02265 - .0CKl00(582) '" 0.2073 111 /5. AIlS . A suppre.ued reClimgu/ar weir I.! COIIJlrUCled acroJ.l a ellalilie/ of 0.77

Problem 8.18 II! widtll Willi a head of 0.39 //I and Ihe crest 0.6 III (lbOl'1" the bed of the chaliliel. Eslimale the discharge oru ir. Con sider \'e/ociry of approach and aSSU/II(' CJ '" 0.623. Solution. Given; Widlh of channel. b = 0.77 111 Head over wdr. H = 0.39 m Hcigtll of ere,! from bed of channel", 0.6 m Deplh of channel '" 0.6 + 0.39 == 0.99 Value of SuppresSI.""d weir me an s that the width of channel is equal to width of weir i.e .. there is no end contraction. Width of channel" Width of weir" 0.77 m Now area of channel, A" Width of chan nel x D<:pth of ch ann el " 0.77 x 0.99 The diSCharge ove r a rectangular weir without velocity of approach is given by equation (8.1).

Q==~ 3

xC" Xbx.j2iXH Y2

" "32 x 0.623 x 0.77 x J2 x 9.81 Now velocity of approach.

I I

v-

Q

• - Area of chan nel

0.345 ~'"C'!c~ 0.77 x 0.99

(,:

Here b " L)

x O.W Y2 '" 0.34 5 m'ls

" 0.4526 lIlls

Ii

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IL Notches and Weirs

373 1

Head due 10 velocity of approach. 2

1

V 0.4526 hu = - "- = =0.0[04111 28 2 )( 9.81

Now th e discharge wil h velocity of approach is given by. Q:

i

x

C" x b x ..fii

1(1/

+ 11~)·112 -

1l/ J 2

2

'" JX 0.623)( 0.77 x ./2 )(9.81 1(0.39 + 0.0 1(4)312 - (0.0 1(4)3n 1

2

'" 3" x 0.623 x 0.77 x 4.43 [0.2533 -

0.001061

'" 0.3573 mJ/s. An s. Problem 8.19 A slwrp crested rectangular weir of I In heighl e.tfe/ld.i across a rec/(l/lgular c/wrme/ of 3 m ",id,h. If lilt! head afwaler 0\'''' I/Ie weir is 0.45 Ill. co/cu/m(! the discharge. COllsida I-e/oeily ojupproac/I (1111/ IlHume Cd '" 0.623. Solu ti o n. Gi ven: Width o f channel . h=3 m Heigh t of weir H~ad of water over weir.

=I m

H '" 0.45 III Depth of channe l '" Heig ht of we ir + Head of water ove r we ir '" I + 0.45= 1.45m Value of Cd '" 0.623 The discharge ove r a rectan gular we ir with out ve locity o f approac h is given by

2 Q'" 3

X Cd X b

x

(8. 1) as

,ffi X H lI!

'" 3. x 0.623 x 3 x ~2 x 9.81

3 Now ve loc il y of approach is given by

v-

~ qual io n

x O.45·V2

:=

1.665 1I1 3/s

Q

• - Area of c hanne l

= ~~~~~L~~S~~~~ Widlh of cha nn el x {kplh of channel

-"L66~SO ~ '" 0.382 3 x 1.45

1I1/s

Head due 10 ve locily of approac h is give n by.

V l 0382' h = - "- '" = 0.0074 " 2g 2x9.8 1

11\

Now the discharge Wilh ve locity of approach is give n by.

2 Q'" - X Cd X b x Jii I(H + 11)311 _ Ol).ln l 3 2 '" - x 0.623 x 3 x ~ 2 x 9.81 [(0.45 + 0.OO74)3ll _ (0.0074)311 1 3

= 1.703 m 3 /s. An~ .

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IL 1374

Fluid Mechanics

.. 8. 11

EMPIRICAL FORMULAE FOR DISCHARGE OVER RECTANGULAR WEIR

The dis.chargc over a rectan gu lar wc ir is given by Q ==

, i Cd J2i X L x If/JIl l without velocity of approach

= 3: Cd Jfi X LX 3

[(f! + " a)3Il.

~ h/I':!t

... (il

wilh ve locit y of approach ...( ; i)

Equations (i) and (it) arc applicable to the wcir or notch for w hich the c rest length is equal lO Ihe width of the channel. This type o f weir is ca ll ed Suppressed '...·;r. But if Ih" wdr is not suppressed, the <'ffect of end contraction wi ll be tahn in to account. ";,,; .;.;.;.; .; .;.;.; .;.;,,;, , (a) Fr"ncis' s Fo nnuh., Francis on the basis of hi s experiments estab'" "" """"""""""",,

li shed thai end cOlllmcli on decre ases Ihe effective length of the crest of

weir and hence dcucascs the discharge . Each e nd contraction reduces Ihe n cstlcngth by 0. 1 x H. where H is the he ad uver the weir. For a rectangu lar weir Ihere are two e nd cont ra<:1io ns onl y and he nce clfcctivc Icngth L =(L - O.2H) O. lH

..

'"",',',',', ', ',',',, ,', "

:::::::::::::::: ::::::::::::!:: ~ ,~ :~'

2

Q = 3xCdX [L - 0.2XIl1x-!fi flYI

', ',','," ,', ','" " I

',', ',',','

"' ·I:'.:.'O."i~: ' , ',' ,',', "',',', ','

r-O.l H

Fig. 8.8

Cd = 0.623 . g = 9 .8 1 m/s1. the n

If

2 Q = - x .623 x -J2 x 9.8 1 x [L - 0.2 x HI x 113f1 3 = 1.&4 IL - 0.2 x 1I1f1·>l2

•.. (8. 1 I)

If cnd contractions arc s upprcssed . tllc n H = 1.&4 LIlY!.

... (8 . 12)

If vclocity of approacll is considered. tllen

Q = 1.84 L [(fI + 11~»)12 _ h}i21 ... (8 . 13) (b) Bazin 's Fo rmula . On tllc basis of rcsults of a series of exper im cnts. Bazin's proposed thc fOllowing formula for til e discharge ovcr a rectangular wcir as Q =m xLx-!fixH l12 whc r~ /Il

.. .(8 . 14 )

2 .003 = - X Cd = 0.405 + - 3

H

II = hei ght of water OVe r Ihe weir

If veloc ity of approach is considered. the n Q =/Il , XLX,ffi [(H+holl12[

where 111 , = 0.405 +

.. .( 8. 15 )

.003 ( H + II.l

Problem 8.20 The helld of "'(1ler ora (1 ~e("l{mg"'{lT ...e;r is 40 em. The lellglll of Ihe aes l of Ille "'e;r \\";Ih end eo nlmelit", suppren'ed;s 1.5 III. F;lId Ihe d;seharge JM;lIg Ille Jollo ... ;lIg Jormuille: (i) F mnei s' s F ormuli, Will (ii) BlIzin '.1 F orlllu/a.

II

Ii

~ I

IL Notches and Weirs

375 1

Solution. Given: ~lcad of wate r.

H:40cnl=OAOm L=I.5m (I) Fr,mcis's Porm ula for e nd col11raclion suppressed is given by equation (8. 12), u,ngth of weir.

Q'" 1.84 L x H3I! '" 1.84 x 1.5 x (.40)31:'

'" 0.6982 m .l/s (Ii) Ba 7. in 's Formula is given by equat io n (8.1 4)

Q=mxLx,ffi xH J12 where

til '"

0.405 + .003 ", 0.405 + .003 '" 0.41 25 H .40 Q'" .41 25 x 1.5 x

.J2 x9.8 1 x (.4),\12

'" 0.6932 ",.lIs. AIl-"

Problem 8.21

A weir

36 melfes /ollg is dil'ided inlO 11 eqlUll bays by !'<,nical pO,~I.f. eael! 60 em

wide. Determine Ille di,Kllarge o\'er Ille weir approach is 2 metres per second. Solution. Given: Lengt h of wc ir. L[ '" 36 10

if

Ihe head OI'er Ihe crest is 1.20

//I

allli \'e/OCify of

Numbcrofbays. = 12 For 12 bays, no. of vertical post = I I

Width of c;l<:h POSt

Effective length. Head On wdr,

Velocit y of approach.

=60cm=O.6m L= H

LI

- 11 xO.6'" 36 - 6.6 '" 29.4111

= 1.20

Vo :

III

2 mls V2

Head due to Va' Number of e nd con trac tio n,

11 := - "-

"

2g

2l :=

:=

0.2038

III

2x9.81

= 2 x 12 {Each bay has twO e nd co ntrac tion s I = 24 Di scharge by Francis Formula wi th end co ntracti o n and ve locit y of approach is Q'" 1.84 [L - 0. 1 x n( H + Ilalll(H + /t)312 _ 1I/"2 J = 1.84[29.4 - 0.1 x 24( 1.20 + .2038)J x [(1.2 + .2038/~ - .2OJ8L~ 1 II

: 1.841 29.4 - 3.369111.663 - .092 1 : 75.246 m J/s . Am . A discharge of 2000 mJ/s is 10 pdS.1 OI'U a rectangular weir. Tile weir i.1 dil'ided imo a III/Illbu of openings each of SpOIl /0 III. If IIII' I"eloeily of approach is 4 tills. find Ihe number of 0plmi,tgs needed in order l/te heml of ""diU ol"er lite cresl is ltol 10 exceed 2 m. SolutIon. Given:

Problem 8.22

Total

disch~rge.

Leng th of each opening.

I I

Ii

~ I

IL 1376 Fluid Mechanics V" ", 4rnls H=2m

Velocity of approach. Head over weir.

:N lei number of openings Head duc to velocity of approach. Ir: V} ", 4X4 '" 0.8155 " 2g 2x9.81

In

For each opcnillg. number of end contractions arc two. Hence discharge for each opening consideri ng velocity of approad is given by Francis formul a i.e.. Q'" I.MIL - 0. 1 x 2 x (I{ + I.a)][(/l + ha)3f1. - /'a3f1. ) '" 1.84110.0 - 0.2 x {l + .8155)1I2.81551~ _ .81SS u l '" 17.3631 4 .7242 - 0.73641 '" 69.24 m]/s 2{J(J()

Total discharge

: =~='=== Discharge for 0"" opening

Number of opening

69.24

"" lS.88 (say 29) '" 29. Am .

.. 8. 12

CIPOLLETTI WEIR OR NOTCH

Cipollclli weir is a trapezoida l wcir. which has side slopes of horizontal to 4 vertica l as ., hown in Fig. 8.9. Thus in 6ABC .

~ '1- r

' .- ..... -.., , f f

e

AB H f 4 I tan - '" - - : - - : 2 Be H 4

r :=::::: -, .

,

"

I

_ = tan _ = 14 ° 2'. 2 4

"

1

1

By giving this s lope to the sides. an increase in discharge through the triangular portions ABC and DEF of the we ir is obtained. If this slope is not provided the weir would be a rectang ular one. and duc to end Fig. 8.9 contraction. thc di scharge would decrease. T hu s in Ca.>C of cipolletti weir. the factor of end contractio n is IIOt required which is shown be low. The discha rge through a rectangular weir with twO end con tra ct ions is

Q=

~ 1

X

Cd x (L-O.2

If).ffi xH

The cipollelli u·ejr.

Jl2

=~ xCd XLx.fii H3fJ. _ ~XCd X.fii xH sn 3

IS

Thus due to end cOnlr.lction. the discharge discharge

Cilll

decre~scs

2 by 15 x Cd x

.ffii

x

,;noThis decrease in

be compe nsated by giving such a slope to the sides that the di scharge th rough twO

~

x Cd x..ffi x HV1. . Let the slope is given Ily 9/2. Tile discharg" IS through a V·notch of angle 9 is give" by triangular portions is equal to

8

=-

15

I I

X

r:c

Cd X ,,2g

X

9

tan - Hm 2

Ii

~ I

IL Notches and Weirs

Thus

-

8

X

IS

a2

e = -2 x15 [ - =-

Ian -

Thus

disch~rgc

2 15 8 through cipolieui weir is

'

Xlan _ H5I2= ~ xC x

CdX .fii

or

4

15

d

377 1

~xH'>r. v~~

e/2 = t3n- 1 ~ = 14 ° 2'. 4

Q=~xCd)(Lx.fiiH3I2

... {8. 16)

3

If ve locity o f approach, Va is to be taken into consideration.

Q '"

Problem 8.23

~ 3

X

Cd

X

Lx

J2i lUI + Ila)3I1 _ ha1l2

Find rhe diKilarge OI'er a cipo/klli wl'ir aile/18th 2.0

... (8 . 17)

1

III lI'ilell

Ihe he{jd orn tile weir

is /111. Take CJ = 0.62. Solullon. Given: li:ngth of we ir. Head o\'er we ir.

L =20 m 1/ : 1.0111 CJ = 0.62 Using equation (8.16), the disc harge is given as

Q=~XCd XLX.ffi 3

x HJJ2

= -2 x 0.62 x 2.0 x

..j2 x 9.8 1 x ( 1)312 = 3 .66 1 m)/s. Ans. 3 Problem 8.24 A cipol/elli weir of cre.n lenglh 60 em diJClwrge.1 Wafa. The head afwaler ora IIII' ,,'eif is 360 mm. Find the disc/wrge ol'a I/Ie weir if Ihe ellmmel is 80 em ...ide and 50 em deep. Take CJ = 0.60. Solut ion. Given: Cd '" 0.60 Length of weir. L=6Ocm=0.6Om Head o f water. 1f =36O mm =0.36m Channel width =oocm =0.80m Channel depth = 50cm= 0.50 m A = uoss-scClion,ll area of dwnnel = 0.8 x 0.5 = 0.4 m1 To find velocity of appro ach. First ddermine discharge over the weir as 2

Q= - xCd xLx.fiixHln. 3

The velocity of approach, Vo = Q A

,

Q '" ~ x 0.60 x 0.60 x

J2 x 9.8 1 x (0.36)ln. m /s '" 0.2296 m /s 3

3

.2296 0.40

V" = _ _ = 0.514 m/s

I I

Ii

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IL 1378

Fluid Mechanics

Head duc to ve loc it y o f approach, h '" V "/2

" Thus th e

d isc h~rgc

"

is give n by

g

'" (0.574 )2 '" 0.0 168 m

2x9.81

(8 ,17) as

cqu~tio ll

, Q= i xCd xL x J2i I (H +hj -~ - h/~ l =~XO .60 X .6X J2x9.8 1

[(.36 + .0 168)I.S _ ( .OI 68)I.S 1

'" 1.06296 x [.23 13 - .002 177 J '" 0. 2435 mJ/s. A" s. to 8. 1J

DISCHARGE OVER A BROAD- CRESTED WEIR

A wei r ha vin g a w id ~ crest is kn ow n a~ broad-cres ted wei r. Let If = height o f wate r aoove th e c r"st L '" length o f th e creS!

ru=

7::,,:"::,,:~::~:~ : !r, :"::":-".:?:~~-r~~

,,,,,,0 "'i''' O

~ ,

F;g . 8.10

----<

B~Qad<{;Tf'Jted wt:i~.

If 2L > H. tile we ir is ca ll ed broad ,uc~lcd weir If 2L < H. the wei r is c alled a narrow-crested we ir Fig. 8.10 s hows a broad-crested weir. Let II '" head of water m the midd le of we ir w liich is CO nMan! \' = ve locit y of flow over the we ir Appl ying Bernoulli 's equatio n to th e still wate r surface o n the ups tre am side and runnin g water at th e e nd of wei r.

1,2

-

28

Th~

= H - II

1. = J2 g ( H /r) discharge Oller weir Q = Cd x Are a of n ow x Ve loc it y =Cd x Lxll x J2g (H

II)

l J =C d XLXJ2 g( Hll - II )

I I

... (8. 18)

Ii

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IL Notches and Weirs

379 1

The di scharge will be maximum, if (1/111_ lll) is maximum

d (HII-, - II 3 )=Oor 2hxIJ - 3h-=Oor ' 2H= 311 dh

-

/,= -32 If Q""" will be obtained by substitut in g this value of II in equation (IU8) as

'" Cd X L x

J2i J2~ H l

'" Cd X L x

.J2i

x 0.3849 X 11J.r.

"" .3849 x J2x9.8 1 x Cd x Lx HlI2 = 1.7047 x CJx t X 11lll. '" 1.705xCJ xLxH

to 8. 14

JI2

. . . (8.19)

.

DISCHARGE OVER A NARROW-CRESTED WEIR

For a narrow -crested weir. 2L < H. It is simi lar lO a rectangular we ir or notc h hence. Q is given by

2 Q'" -

3

.. 8. 1S

X Cd X L

x

.fii

X

n ln

...(8.20)

DISCHARGE OVER AN aGEE WE IR

Fig. 8.1 1 shows an Ogee weir. in wh ich the crcS! of Ihe weir rises upto maximum heigh! of 0. 11 5 x H (w here H is [he heigh! o f

Wale r above inlet of the weir) and then fall s as shown in f'ig. 8.11 . The disc harge for an Ogee weir is the same as that of a rectangular weir. and it is given by

2 Q= 3 ... 8 . 16

X

Cd X f. x

.fii

x

If'!

,~e

CREST

.. ,(8 .21 ) Fig. S.1l

An Ogff 'Wt'ir .

DISCHARGE OVER SUB·MERGED OR DROWNED WEIR

When the water level on the downstream side of a weir is above the crest o f the weir. the n the wei r is called to be a sub-mcrged or drowned weir. Fig. IU2 shows a sub-mc rged weir. T he tOial discharge. ove r the we ir is obtained by di viding the we ir into two parts. T he port io n between upstream and downstream water surface may be treated as free weir and ponion between dow nstream water surface and erest of weir as a drowned wei r.

I I

Ii

~ I

IL 1380 Fluid Mech a nics

Fig. 8.12

Slth-merged 'U:tir.

Lei H '" hciglil of water on the upstrea m side of the we ir II '" height uf wate r on the dow nstream side of the we ir Then

QI '" discharge ol'e r upper porti o n

,

"'~X Cd '

xL x fii IH- h l)f!

Q 2 '" di scharge thro ugh drowned porti on

'" Cd, x Are a o f flow x Ve locity of fl ow '" Cd, xLxil x J2g( H - h) .. TOI al di Sl' tJ argc .

Q'"

0 1 + Q2 2

'" J ed,

XLx,ffi 11f _ II I)I2 + Cd, x L xhx ~2 g( H - h) . ... (R.22)

Problem 6.25 (a) A broad-crested wei, of 50 m lengtll. /1<105 50 em heigh! O/lI'ater aborc i/$ eres l. "-illd Ihe maximum discharge. T(lke Cd '" 0.60. Neg/eel I'C/Oeil)' of approach. (b) If Ihe I'e/oeil)' of approach is /0 be taken ilifO cDllsidermioll.jilld the mlUimulII disclwrge when Ille challllel liaS a crosssecliOlw/ area of 50 "1 2 011 Ihe IIpstrcam side. Solullon. Give n : Le ngth of we ir. L =50111 Head o f wata. H ::50cm =0.5 m Cd:: 0.60 (i) Nl'"glecting "elocily o f ap proach. Max imum di sc harge is givc n by equatio n (8 . 19) as

1.705 X Cd X L X H3f2 :: 1.705 X 0 .60 X 50 X (.5)'V2 = 18.084 mJ/s. AilS.

On",,::

(ii) T"kin g nloclly o r "ppn)!lch intn consldcnllion

Area of Channel .

A = 50 m 1

Velocity of appro ac h.

18.084 V:: - = - - = 0 .36 m/s " A 50

: . Head du e toY".

h = V} = O.36x .36 =.0066 m

o

"

2g

2x9.8 1

Ma xi mum d isc harge , 0 ....., is give n hy Qn",:: 1.705 X Cd X L x [(f/ + 11~) .\12 _ 11 • .\12 [ " 1.705 X 0 .6 X 50 x [(.50 + .(066)'~ - (.0066>'~ 1 = 5 1.1 5 [0 .3605 - .0005361 = 18.412 m l/s. AilS.

I I

Ii

~ I

IL Notches and Weirs Problem 8.26

All Ogee weir 5 metres 101ig lIa,\" a Ilead of 40 em of wilier.

If Cd '"

381 1

0.6, filld Ille

disellarge ora Ille weir.

Solution. Given:

L=51O

Len gth of we ir. H~ad

o f wala.

H= 4Qc rn '" DAD 111

Cd '" 0.6 Disdlargc over Ogee we ir is g iv en by equation (8.2 1) as

,

Q= ~ 3

X Cd X L

x

.ffi X HY!.

'" ~ x 0.60 x 5.0 x

J2 x 9.81

x (O.4)3i'l '" 2..2409 nh~. A n s.

Problem 8.27 Tlte /JeiglllS o/ . . . aler 011 I/le IIps/reom (mil downs/rem" side of (l sllb·merged we ir of J", /<'/lgl/' are 20 em ,md /0 em rnpeC/ire/y. If Cd /or free ,md dro,.."ed pOr/iolls (Ire 0.6 (lnd 0.8 r espec/il'ldy.

fill"

llie diJ ch(4rge on" Ihe weir.

Solution. Give n : Height or waler on upslream s ide, H '" 20 em = 0.20 III ~I c ight of water 011 downstream side. II '" I0 e m '" 0.10 u,n g[h of weir . L= 3 m

In

Cd, '" 0.6 0.8

Cd: '"

Total di scharge Q is the s um of discharge through free (>Onion and discharge through the drowned [lonion. Th is is g ivcn by equation (8.22) as

,

Q==~X CJ, XLX,ffi ==

~

II2 IH _ II I· + Cd, xLxllx

xO.6 x3 x ,)2 x9.8 1

J2g (H II)

1.20 - .10(~+ 0.8 x3x . IOx

J2 x9.81 (.2 - .1 )

== 0.168 + 0.336 == 0.504 nI'/s. Ans .

HIGHLIGHTS 1. A nouh is a dcvice u.'iCd for measuring [hc rotc of flow of a liquid through a small channel. A weir is

a cOllcrete or masonary ,truewre placed ill the open channel over which Ihe flow occurs. 2. The discharge Ihrough 3 rectangular notch or weir i. gi,·en by

2

Q"'"3CJ XLXHJfl

whue C d '" Co-efticient of
5fl

where e = \o[al angle of triangular notch.

I I

Ii

~ I

IL 1382 Fluid Mechanics 4. The discharge through a trapezoidal notch or weir is equal to the sum of discharge through a rectangular nOlCh and the discharge through a triangular notch. It is given as

2

Q""3 where

Cd ,

Cd, X Lx

Jfi

8

X H II2 + T5CJ, x tan

a ,,28 r::o:: "2)( x It!'!

~ co-efficient (If discharge for rectangular notch.

Cd, ., co-efficient of discharge for triangular notch. 612 ., slope o f the side of lmpezoidal notch. 5. The error in discharge due \0 the errOr in the meaSurement of head OVer a rectangular and triangular notch or weir is given by

dQ = 3dH 2 H Q

...

r~r

5 tlH .--2 H

where

a rC<:tnngular weir or notch

For a triangular weir or notch

Q" discharge through rectangular or trinngular notch or weir H .. head over the notch or weir.

6. The lime required \0 empty a reservoir or a tank by a rectangular or a triangular notch is given by ... By a

r~ctangular

n()(ch

... By a triangular notch

A. cross·sc.:tional area of a lank or a reservoir

where

1/, _ initial height of liquid above the crest or apex of notch

111 - final height of liquid abo,·e. the. crest or apex of notch. 7. The wlocity with which the water approache, the weir or notch is called the velocity of approach. It is denoted by V. and is gi"en by V• •

Di seharge Over the notch or weir Cross -sec tional area of channel

H. The head due to velocity of approach is gi"en by h.

~

v'

- '- .

2g

9. Discharge over a reClJngular weir. with velocity of approach. Q10.

Fr~nci,s

%C,J- J2i Hil t + h)'" - h}I.'I.

Formula for a rectangular weir is given by

Q. 1.84[1. _ 0.2 II I IIlIl .. 1.84 L llJi1. .. 1.8..1 L [(II + h.,)lrl_ h}'2 J where I. .. length of weir. H ., height of water above the crest of the weir. h • • head duc to "clocity of approach. 11. Ilazin's Fonnula for discharge over a rectangular weir.

~ I

... For tWO end conlraclions ... If end contractions are suppressed ... If "elocity of approach is considered

I~

~ I

IL Notches and Weirs Q .. In L .j2;HJI2

... without velocity of approach

~ m L.f2i [(H + hdlrll where

2

mK_

3

... with velocity of approach

cd"O. 405+ -~ .003

... without velocity of approach

H

~O.405+(

.003 H+ha

383 1

I

... wilh velocity of apP!U.lch.

12 . A trapezoidal weir. with side slope or I horizontal 10 4 vertical. is called Clpollelti weir. The dis.::harge through Cipollctti

w~ir

is given by o

Q = ~ Cd X L X '"

J2i HJn.

.. , wilhout velocity of approach

~ CdxLx J2i [(H + h)lfl_ h}1:.'1

... with velocity of appro.1ch.

13 . The discharge over a broad-crestcd wcir is given by,

where

H = height of waler alx,,"e the

()r

cre~l

h ~ head water at the middle of the wcir which is consl:m! L .. length of the wcir.

,

14 . The condilion for maximum discharge ow. a broad-cres\cd weir is h. ~ H and maximum diSCharge is given by Q_

K

1.705 Cd L H J".

2 IS. The discharge O\'er an Ogee weir i. gi"en by Q '" - C)..)(

3

.fii )( HlfI..

16. The discharge OYer sub-merged Or drowned weir is given by

Q "" discharge Over upper portion -+ discharge lhrough downed portion

.. %Cdl LX.[fi (H - h)lf! + Cd, UI where

X

J2g(1f - h)

H _ heighl of waler On [he upstream side of lhe weir. h: height of waler on lhe downstream side of [he weir.

EXERCISE (A) THEORETICAL PROBLEMS I. Deline lhe [cnns : notch. weir. nappe and crcs!. 2. How are the weirs and nolche< cla"ifjed 1 J. r ind an expression for the discharge oVer a rectangular weir in termS of head of waler Over the cre,l of lhe weir. 4. Prove lha[ the discharge through a [riangular notch or weir is g;"en by

8 Q- - Cd xw.n 15 where 1/ .. he,[d of waler ovcr lhc n01Ch or 8 • angle of notch or weir.

I I

8 M- lll -x,, 2g H 2 weir

Ii

~ I

IL 1384

Fluid Mechanics

5. What are the advantages of triangular notch or weir o,"cr ",,,[ang ular notch? 6. Pro"e that the error in discharge due 10 the error in the mea'DWlnen! of head over a rectangular notch is given by dQ

Q

:3 dJ/

."2"

where Q ~ discharge through rectangular notch H ~ head O\'er the .cclangula. notch. lind 7. Find an c~ pression for the time required to empty II tank of area of cross-section A. wilh a rcclnngular notch. II . What do you understand by 'Velocity of Approach " ? Find an e ~ prcssion for the discharge over a rectan gular weir with ~eloci ly of approa,h .

9. Define 'end conlrncl;on' of a weir . What is the effect of end contrnc(;on on the discharge through a weir ? Ill . What is a Cipoliclti We ir 'I Prove thaI the discharge through CipolleU; we ir is given by Q'"

%C.,I.. J2i H JIl.

wbere L '" length of weir. and H '" head of water o,'cr weir. II . Differentiate betw""n Il roa,h,rested weir and Narrow-cre,ted weir. Fi nd the condition for maximum d is-

12. 13. 14.

15 . 16.

charge m'er a Ilroad-crested weir and hence deri,'e an expression for maximum di>eharge m'er a broad crested weir. What do you mean by a drowned weir ? How will you detenninc the di>eharge for the downed weir? Discuss 'cnd contraction' of a weir. State the different devices that Can be used to meaSUre the discharge through a pipe also through an open chan nel. Describe one of such devices with a ti Cat sketch and explain how one Can obtain the actual discharge with its hclp , What is the difference between a notch and a weir? Define vc!ocity of approach . How does the "elocity of approach aff(X:t the di><;harge over a weir?

(B) NUMERICAL PROBLEMS I. Find the discharge of water flowing oVcr rcctangular notch of 3 m length when the conStant head of water OVer tbe noteb is 40 ern. Take Cd'" 0.6. [Ans . 1.344 ml/sl 2. Detenni"e the height of a rKtangular weir of Icngth 5 m to be built aCross a rectangular channe l. The

J. 4.

S.

6.

7.

I I

maximum depth of water on the upstream side of the weir is 1.5 m and dischargc is 1.5 m' per second. Take C. ~ 0.6 and neglect end contractions, [Ans. 1.194 ml fi nd the d ischarge o\'er a triangular notch of angle 60° when the head over the triangular notch is 0.20 m. Take C" '" 0.6. [An s. 0,0164 111 l /s l A rectangular channel 1,5 m wide has a di".:harge of 200 litres per second. which is measured by a right angled V-nolch weir. Find the position o f the apex of thc notch from the be
Ii

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IL Notches and Weirs

385 1

8 . fi nd the time required 10 lower the water level from 3 III to 1.5 m in a reservoi' of d imension 70 III x 70 m. by a 'lX'wngular notch of length 2.0 m . Take Cd~ 0.60. IAns. 11 min I 51 \I . If in the problem 8. instead of a rectangu lar notch. a right angled V-notch i~ used. find the lime required. Take all other data -,arne. IAns. I J min 31 s[ Ill . Water is flowing in a rectangular ~hannel of 1.2 rn wide and 0.8 III deep. Find the discharge over a feClangular weir of crest length 70 e1l1 if the head of water over the crest of weir is 25 em and Water from channel flows o,'cr thc weir. Take CJ 0.60. Neglect end contractions bul con,ider veled is 1.4 m. Find the discharge using following fommlae : (i) Franciss Fonnula and (ii) Hazin·. formula. IAns. (i) 0.91 m'/s. (ii) .901 m' lsl U . A di>charge of 1500 ml/s is to pass o,·cr a rectangu lar weir. The weir is dividcd into a number of opening. each of span 7.5 m. If the velocity of approa~h is 3 m/s. fmd the number of openings nceded in order the head nf water ovcr the crest is not to exceed 1.8. IAIls. 37.5 s.ay J81 14. Find the discharge Over a cipolleui weir of length 1.8 m when the head oVCr the weir is 1.2 m. Take Cd = 0.62 IAns. 4.331 m' lsl 15. (a) A broad ·nested weir of length 40 m. has 400 mm height of water aoo\"C its crest. Find the maximum dischargc. Take Cd " 0.6. Neglect ,·elocity of approach IAn •. 10.352 mllsl (1)) If the ,·clocity o f approach is to be taken into consideration. find the ma~imum discharge when the channel has a cross-sectional area of 40 m' On the upstrcam side. IAns. 10.475 1111/sl 16. An Ogee weir 4 m long has a head of 500 mm of water. If Cd = 0.6. fi nd the discharge ovcr the wcir. IAn s. 2.505 ml/sl 17. The heights of water On the upstrcam and downstream side of a sub-merged weir of length 3.5 mare 300 mm lind 150 mm rcspecti'·cly. If Cd for frcc and drowned portion arc 0.6 and 0. 8 rcspectively. fmd the discharge o'·cr the weir. IAns. 1.0807 ml/sl E

~ I

I~

CR~EK

.. 9. 1 INTRODUCTION This chapler deals with the flow of fluids whic h arc viM:ulIS and flowing at very low veloc ity. At low ve locit y lh e fluid moves in laye rs. Each layer of fluid slidcsovert hc adjaccnllaYCf. Duc 10 relath'" .

du

velocity betwee n two laye rs the velocity grad ient - exists and hence a shear stress 1 '"
1111

j.l -

aCL~

On


I. Flow of vi scous flui d through ci rcular pipe. 2. Flow of vi<;C{}us fluid between 1WO parallel plates.

3. K inctic energy torrcction and momentum correctioll f;!ClOTS. 4 . Power absorbed in viscous flow through (lI ) Journal bearings. (h ) Foot-step bearings. and (el Collar bearings.

to 9 .2

FLOW OF VISCOUS FLUID THROUGH CIRCULAR PIPE

ror the flow or viscous fluid through circular pipe, the velocity distribution across a secti on. the ratio of maximum velocity to average veloci ty. the shear stre.'\S di~tribution and drop of prcs~ure for a given length is to be dete nnined. The fluw throug h the circul~r pipe will be viscous ur I~min~r. if the Reynolds number (R,·) is less thun 2000. The expression for Reynold number is given by

R '" p VD

. "

where p '" De nsity of fluid fl owing through pipe V '" Aver~ge ve locit y of fluid D", Diameter of pipe and )1 '" ViM:Osity of fl uid . DIRECTtON OF FLOW

,

'" VisWUI flow throllgh a pipf'. (_)

Fig. 9.1

(b)

• For derivation, please refer 10 An . 12.8.1.

387

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IL

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IL 1388

Fluid Mechanics

Consider 11 horizontal pipe of radius R, The "iSl."Qus fluid is flowing from len 10 right in the pipe as shown in Pig. 9.1 (a). Cons ider a flui d clcmcllt of radi us r. sliding in a cylindrical fluid clemen! of

radius (r + dr). Let the length of fluid clemen! be Ax. If 'p ' is the intensity of pressure on the face All, Ihen the intensi ty of pressure on face CD will be

(p + ~ a.x). Then the forces acting On the fluid

clcmcru ,m: : L. The pressun: force, p X It,l on face AB. 2. The pressure fOrl:c.

(p +~ AX) rtr2 0n face CD.

3. The shear force. t X 2nrA. on the surface of fluid clement. As there summation of all forces in til.: direction of flow must be zero i.e"

i~

no acceleration. hence lhe

' ra"J ' ax

/lltr-

p+- lh

_ap

ax

1tr""-tX2ItrXal=O

dl7l,l - fx2n:rx6.t'=O

_ ilp .r - 21=0

ax

ap -,

... (9 . 1)

1: - -

(Ix 2

al' across a section is constant. Hence shear ax

The shear stress t across a section varies with " as stress distribution across a section is linear

Ill;

shown in Fig. 9.2 (11).

SHEAR STRESS DISTRIBU,)ON

VELOCITY / DISTRIBUTION

?' ) Fig. 9.2

~

{.,

---, C-

'I

Sbt'ar urt'u and vl'iociry distribution across a st'Ction.

(i) Ve locity Distribution . To obtain the velocity distribution across a section. the value of shear

stress t =

~

du is substituted in equation (9.1). dy

But in the relation t =

~

du ,y is m.::asured [rom the pipe wall. Hence dy

y=R -r and i/u - i/r Substituting this value in (9.1), we get

dy=-dr

i/u dr

t=~ - = - ~ -

I I

Ii

~ I

IL Viscous Flow dll

-, -

up , ax 2

= - --

dr

or

du

-

389 1

1 dp

"' - - , dr 2)1 a.r

Int egrati ng this above equation w,r.t ' , . we gel

,= _'_ dP,2+C

... (9 .2)

ax

4)1

where C is the constant of inwg ration and its va lue is obtained from the boundary conditio n that at r= H. u=O.

o= - '_ QP n2+c

ax

4/1

c :- - '_ QP R2 4).1 ,h Substituting this va lu e o f C in e quation (9.2) , we get 1 dP 2 l iJP 2 ,= - , -4).1- R 4)1 fh ax I

op

2

2

= - - - [R - r l 41J

ax

In ' ljuation (9.3).

v~lucs of ).I. ~~

and R :Irc

... (9.3)

const~nt, which means the ve locit y, II va ries with the

squ are of T, T hus eq uati o n (9.3) is a equ ation o f parabola, This shows Ihm the vciocity di stribution across Ih e sect io n of a pipe is parabolic. Th is ve loc ity distribution is sliow n in r ig. 9.2 (b). (ii) Ra l lo uf !\Iud mum Veloc ll ,f lu AHrage Vel ocl t y. The velocity is maxi mu m. whe n f ' " 0 in eq uation (9.3). Thus maximum veloci ty. Umo, is o btained as

u .....

= __,_ op R2 4J.l ,ll-

...(9 .4)

The average ve loc ity. II. is ob tained by dividing the di sc harge of the nuid across th e seetion by the area of the pipe (rrR\ The di...:hargc (Q) across the sectio n is Obtained by considerin g the now through a circular ring ,dement of radius r and thickn ess dr a~ ~ho w n in Fig. 9 .1 (b ). Th e fiu id fiowing per s
ap

2

2

=- - - [R - r 1x2rrrdr 4J.l ax

Q=

I I

lOp , _ (R 2 _ r'")x2rrrdr J,3' dQ= J,'3 _ _ 4).lax

Ii

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IL 1390

Fluid Mechanics

[-a")X2' [!!'..C_~]' : _1 [ - ''') x 2'['" _"'] ax 2 4 0 4 pfJx 2 4

: -1

4~

: _I [-ap)X2 1'tX~ : ~ [-aap) R' .•' 4)1,Jr

4

-,,-(-a 8j.I ax p

-

,:

Average ",",ocity.

Q --:

8!.1

) R'

,

rrR -

Area

u'" ~1 (-a~:) Rl

..,(9.5)

Dividing equation (9.4) by eq ua lion (9.5) ,

Ra1io of maximum velocity to :IVnagc velocity'" 2.0. II pipe

(iii) Dro p of Prcs~ u ~ fo r a g j,'cn Len gth (L ) of

From equation (9.5). we have

(-,p) "' [-a ):: 8R-1l:-; _ ax ax

" : _I

R'

8p.

p

-

Integrating (he abov e equat ion w. r.t. x. we gel

-

J' dp: fl l

8)1 ,-1/ R"

l

d,

-

-

8~lU

8pu - [PI- Pll = - ,- {XI - x, 1or (PI - p,) = -,- [x, - x d R -R--

8~lii

: -

" (p _ I

Loss of pressure head

R'

L

I":

)."2 - ,( I '"

F ig . ';1.3

L from Fi g. 9.3)

8p.uL (D /2)2

) _ 321.l11L P! D' .

where PI - P! is Ihe drop o f pressure.

= P,-1'2 pg :UpuL '" Ilf = pg D '

...(9.6)

Equation (9.6) is called Jl a gen Po ise uiJI e Fo rmu h. ,

I I

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IL Viscous Flow

391 1

Problem 9.1 A crude oil of \'i.~co.~iry 0.97 poi,le 01111 relalil'e delLlity 0.9 iJ flowing "uougll a lIorizonra/ circular pipe of diameter tOO mm and of /eng/h 10 m. Calculate 1111' difference of pressure alille 1"'0 "lIds a/ lire pipe. if 100 /(g of IIIe oil is collected in 1I lunk in 30 seconds. Solution. Given:

I-l = 0.97 rois.c == 0.97 '" 0.097 Ns/m 1 10

RcI
'" 0.9 '" 0.9 x 1000 ", 900 kg!nr' Po. or dt'llsity. Dia. of pipe. D= l00mm=O. 1 111 L= 10m Mass of oil collected. M=l00kg Time. I'" 30 seconds CaJculalC difference of pressure or (P, - PI)' The differcm:c of press ure (P, - Il l) for viscous or laminar flow is given by

321-l uL Q PI - P2 '" - -,- , where II '" average ve locit y == - D' Area

'00

= -

Now. mass of oil/sec

kg/s

30 =P o xQ:900xQ

(.:

Po= 9(0)

100 =900xQ 30

Q = 100 x-'30

ii == ~= Area

==

0.0037 mJ{s

900 .0037 = .1)037 = 0.471 m/s. ~D1

~(.I) 1

4 4 For laminar or viscous n ow, the Reynolds nUlnocr (R, ) is less than 2000. Let us calcuJacc the Reynolds numbe r for this problem.

Reynolds numbe r, where

pVD R, * = -

"

p=Po= 9oo. V= j; =0.471.D=0.1 m.~1=0.097 R = 900 x .471 xO.1 = 436.91

,

0.097

As Reynolds number is less than 2000. the flow is laminar.

32xO.097)(.47 1)(1 0 NI ' ,

(.q-

m

= 1462.28 Nlm 1 = 1462.28 x 1O~ 4 N/cm 1 = 0.1462 Nlcm!. Ans . • For derivation, plea"" refer to An . 12.8.1

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IL 1392 Fluid Mechanics Problem 9 .2 An oil of I'i.{cosily 0.1 Nsfm " 0111/ felMil'e detlJ'if)' 0.9 is fl o wing Ilrrougll a circular pipe of diameter 50 mill and of /englll 300 m. The rale of jlQW of fluid [I"ough Ihe pipe is 3.5 li/reY! . Find Ihe pressu,e drop in" lenglll of 300 m lind also Iile l'lle", SlreJ'$lIIlire pipe ...al/. Solution. Given: Viscosity.).I '" 0.1 Nslm! = 0.9 Relati ve dcr>sil y Po or densi ty of oil = 0.9 x 1000 = 900 kglm ) D = 50 nlln = .05 III L =300 Ul

(,:

De nsity of waler = 1000 kg/ml)

Q = 3.5 Iitrc sJs = 3.5 = .0035 m1ts 1000 Find (0 Pressure d rop. P, ~ P l (ii) SlIcar stress at piP<-' wall, t o

=

32).111L ---vr-

where

L782 m/s

,

The Rey nolds num ber (R, ) is given by, R, = pVD

, where p = 900 kg/m". V ", average velocit y = II = 1.732 mls R =900x 1.782:-: .05 =801.9 ,

OJ

As Reynolds number is less than 2000, the flow is viscous or lam inar

32 xO. 1x 1.782 x 3000 PI - P2 =

(.05)'

= 6&4288 Nfm" = 6&428 x 10- 4 Nfcm ' = 68.43 Nfc m 2. AilS. (ii) Shea r Stress a t tile pipe w:11I (1:0) The shear stre ss at any rad ius r is give n by the equation (9. l)

dp' dol 2

i.e ..

T'" - - -

Shear stress at pipe wall. where r '" R is gil'e n by To =

Now

-ap

R 2

-~~

ax

~ -a~ p = --, - (,-, p,~--,-p,,,) d.T

PI -Pl", f! I -P l

Xl - XI

Xl

XI

L

= 6&4288 Nfm l '" 2280.96 Nlm3 300 III and

D

.05

R ="2=T= ·025m t o = 2280 .9 6x

I I

.025 N

-~--, Ill ·

2

,

= 28.512N/m ·. Ans.

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IL Viscous Flow Problem 9.3

393 1

A lamillar flow is raking place in a pipe of diameter 200 mm. Ti,e maximum I'elociry (If wl/ieb Illis OCCllrs. Also calculme Ih e I'e/Oeity at

is 1.5 mls. Find the mean \'elocil), olll/lile Tilt/ius oJ em from Ihe wall of II,e pipe.

Solution. Given: Di a. of pipc. D ~ 200 mill = 0.20 In U""" '" 15 m/s

find ( il Mean ve locity, u ( ii) Radius at w hi ch

U OCC lirs

(ii,) V~ locil y al 4 em frolll the wa ll.

(i) l\ lc3n n,' ]oc ily,

ii

u~, '"

Ratio of

or

2.0

.!J. = 2.0

II

( ii) Ra diu s a t wh ich

1.5 II '" = 0.75 m/s . Ails. 2.0

II

II (X;CU P.;

The ve loci ly. u, at an y radius 'r' is g ive n by (9 .3)

ap ax

I - IR2 - r 2 1= u= - 4 ).1

- 1 -ap H1 [ 1- -" , 4).1

ax

1

N-

BUI from equation (9.4) U""" is give n by

u

= __,_ aPR!

Now. [h e radiu s

r

al which

u

.•. ( 1 )

4 ).1 (J.r

"""

ii

==

= 0.75 m's

~: [",J0.5 ~2

0.1

f =

0.1

x.J5 '" 0. 1 x .707 '" .0707 III

'" 70.7 mm. Ans. (iii) Vclocil f HI 4 em rro m Ihe wull

r= R - 4.0= 10 - 4,0 = 6.0 em = 0.06

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IL 1394

I

Fluid Mechanics The velocity al a radius", 0.06 III

or

I

4 em from pipe wall is given by equation (I)

__

R~ r .1__ . _ _

"U-,HJl="HO,6),] -

,.,,=

== 1.511.0 - .361 == 1.5 x .64 '" 0.96 m/s. Ans.

F ig . 9.4

Problem 9 .4 Crllde oil of ~ == /.5 poise and re/mil'l! densit), 0.9 flows Ihrougll a 20 mm diameter I'crlical pipe. Tile pressure gaugesji.\ed 20 In apart read 58.86 Nkm : milt 19.62 Nkm ]
I.S 1 10 == O. J 5 Ns.im

Solution. Given:

)l"

Relative de nsity Density of oi l Dia. of pipe.

=0.9 '" 0.9 x 1000 '" 900 k. g/m) D = 20 mill = 0.02 111

1.5 poisc "

L" 20 III PIl == 58.86 Nfcm-, = 58.86 x 10 Nfrn" Ps = 19.62 Nfcm 2 = 19.62 x 104 N/ml.

"

Find (il Di rect ion of fluw (ii) Rate of flow. (I) Dirulion a rnow. To find th e diw:tion or flow, the IOwl energy ( L +

pg

~ + z ) allhe lower end 2g

A :tnd Jt the uppe r end B is to be cakulmed. The direction uf fluw wil l be give n from th e higher ene rgy to the lower ene rgy. As here the diJmeter of the pipe is same and hence kin etic energy at A Jnd B will

Ix: same. Hence to find Ihe d irec tion of flow , ca lcu la te

Taking Ihe leve l al A as dal uill. The value o f A

(~ +

z)

at A and 8.

('.'g' + zJ "

= l!..4. + Z~ 6x l0 4 x9.81 900x9.8 1

B- -

(:g+ Z)


58.86 N/cm

,

I--!- 20mm Fig . 9.5

= As lh e value of

2 19.62 Nlcm

+0 1·: r=900kg/cm~1

= 66.67 III

The value of

-

>Om

.g

=

l-

2 xIO· x9.81 900x9.8 1

+ 20 = 22.22 + 20 = 42.22 m

(:~ + z Jis higher at A and Ilene" flow lakes place from A 10 B. An s.

(ii) Ral., o f flow. T he Joss of press ure IIcad for viscous flow lh rougll circular pipe is g iven by

I I

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IL Viscous Flow For a vertical pi pe

395 1

/1/ '" Loss o f pc izo mclric head '" ( : ; +ZA

24.45 "

u ==

"'

J-(:;

+28) '" 66.67 -42.22 '" 24.45 III

32 xO.l5xu x20.0 i

900 x 9.8 [ x (.02)" 24.45 x 900 x 9.81 x.()()()..I

32xO.l Sx20.0

== 0.889 ,. 0.9 mls.

The Rey no lds number stlOuld be calcul ated. If Reyno ld s number is less llian 2000. the flow w i ll be

laminar and th e above ex pression for loss of pressure head for laminar flow can be used. Now Reynolds number where

~

p" 900 k g/Ill) and V",

pVD

--

"

w

=900x O.9x.02 = 108 0.15 As Rey nolds number is less than 2000. th e flow is laminar. " average velocity x area Rate of flow Rey nolds number

It "II X "4

,

1t

D-=O.9X "4

X

'3

(.02)- m Is: 2.827 x 10-

43

10

Is

'" 0.2827 lil res/s. Ans. Problem 9 .5 A fluid of "iscosily 0.7 Nsf",' (lnd specific grlll';/Y 1.3 is flo .... ing through" cireulM pipe of diameter 100 "'III, The maximum silear Siress 01 Iile pipe wall is gi,'en OJ' 196.2 N/", 2, find (i) Iile pre~'suf<' gmdie,lI, (ii) Iile ,,,'erage "'!/Ocily. alld (iii) Reyn old.f II I,,"ber of Ille flail".

N,

I.l = 0.7-,

Solution. Given: Density Dia. of pipe. Shear Slres.~, Find

(i) Pressure

m

Sp. gr. = 1.3 "1 .3 x 1000= 1300kglm 3 D = IOOll1m =O. llIl t o = [96.2 NflT/

gr~dienl ,

dp

d.,

(ii) Ave ra ge ve locit y. II (iii) Reynolds numbe r. R,

dp

(/) P ress ure g rad ie nt , r/x

The maximum shear stress (to) is gi ven by

dp f) dp 0. 1 dp H to=- - - O< [96.2= - - x - = - - x d.( 4 4 dX 2

,I,(

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IL 1396

Fluid Mechanics

op

196.2 x 4

, '" -7848 Nlm' perm 0.1 Pressure Grndicnt:: - 7848 N/m l P"I' m. An s.

-

ax -- -

(ii) A \,.,rHge ""]
~ _I ,(~aPlo' - &!-t d.f]' 1

,

:: 8xO.7 x (7848) x (.OSt

:: 3.50 Illfs (iii) Rernold s numbe r , R,

N, = uxD ", ,,xD ",pxllx D

\'

III P

I.!

'" 1300 x 3.50xO.l '" 650.00. Ans .

07

Problem 9 .6 IV/uu power i,~ required per kilomdre of (I {rIle /0 OI'ercome (lie d.~co,,¥ re,~i.~/a'ice /0 the flow of glycerine lilrollgil II iloriZOIl/{11 pipe of diwlle1er 100 mm m (he rale of 10 Iilre,lIs ? Take ).l '" 8 puise fwd kinematic I'isn!Sily (I') '" 6.0 slakes. Solution. Given: Length of pipe.

L '" 1 km '" 1000

Dia. of pipe.

D= IOOlllm =O. l m

Di~hargc.

Q = 10 litrcsls '"

Viscosity.

. 8 Ns 1 1-l=8POISC=- - l ::O.8Ns/1ll 10 m

Kincr11mic Viscosity.

v= 6.0 Sluk.cS

III

~ 1000

m3 /s:: .01 m 3/s

IpoiSC=-.!....NS/m l )

10

:: 6.0 en/Is = 6.0 x 10- 4 m"ls Loss of pressure head is given by equ ation (9.6) as lit "

32'W~ pgD-

Power required == IV x I,/ watts where IV = weight of oil flowing poer sec = pg x Q SUbstituting the va lu es of Wand II/ in equation (i). Power required

"" I I

== (pg x Q) x

(32 "L) , wmts" pg D-

...( i)

Q x 32 pUL , D-

(can(;elling pg)

u == ~= ~ = .Olx4 ==1.273 m!s Area

!!..Dl 4

Itx (.l)'

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IL Viscous Flow

397 1

.0 1 x 32 x 0.8 x 1.273)( ]()OO

Power required

(.I )' '" 32588.8 W" 32.5!SH kW. Ails.

~

9. 3

FLOW OF VISCOUS FLUID BETWEEN TWO PARALLEl PLATES

10 Illis ca se also, Ihe sllcar stress distribution, Ih c vdocily di stribut ion across a sec tion; Ihe ratio of maximum ve locity \0 average ve locit y and diffcrcnc~ of pressure head for a give n length of parallel plates, aTC to be calculated.

,

PARALLEL PlATE

-I-----~-"";...---~

DIRECTION OF FLOW

-

1

, Fig. 9.6

T,

I--<

PARALLEL PLATE

"

ViscOUJ flow bU'U'f't'n two paralld plall'S.

Consider two parallel fixed plates kepI at a distance '( apan as s ho wn in Fig. 9 .6. A viscous fluid is flowing between these two plates from lcft to ri ght. Consider a fluid cJcmC(1{ of length AI" and thick ness Ily at a diMancc), from the lower fixed plale. If p is the intcnsily uf pressure on lhe f
(p + ~ I1r). Let 1: is the shear suess

acting on the face BC then the shear stress on the face AD will be

(1: + ~ Il.Y}

If lhe wid th of lhe

e lement in the d irection perpendic ular to the paper is unily then the forces aCling on the fluid element arc: 1. The pressure force. P x Il.y x I 011 face AD.

2. The pressure force.

(p+ ~ Il....)

fly x 1 011 facc CD.

J. The shear force. 1: x Il.x x I on face Be. 4. The shear force. [r + dr Il.yj AI" X I 011 face AD.

a,

For steady and uniform flow. there is no acceleration and hence th e resultant force in the direc.tion of flow is zero. pll.yX I - (P + QP Ar)Il.YX l - tArx 1 + It+ dr ll.yjAxx I =0 (hay _ dP Il.xll.,. +

dX

~ ll.yAr = 0

ax

Dividing by rull.y. we gel _ dP+at",Oor

a.1" ay

I I

... (9 .7)

Ii

~ I

IL 1398 Fluid Mechanics (0

Ve lodt~'

""

stress T" fl -

"y

Integrating

Dist ri bution. To obtain the ve locity distribution across a section. th e value of s hear

from Newton's la w of viscosity for laminar flow is substi tuted in equation (9.7).

th~

above equation w. r.l. y, we gel

au I af> - "' - - y + C fJy )l ax I

f.,. ap isco nSlant) 1 ax

ap

I )'~ u= --_·_ +C,)' + C2 ).l ill 2 .

lmcgmting again

...(9.8)

where C, and C1 arc constants of integration. Their values ;lrc obtained from th e twO boundary condili ons lhJt is (0 at)' = O. 1/ = 0 (ii) at y '" I. II = O. The substitu tion of y = O. II = 0 in equ ation (9.8) gives The substi tution of

0= 0 + C 1 X 0 + C2or C! " 0 Y = I. I. = 0 in equation (9.8) gives I iJp [ 1 0= - - - + C, x 1+0 ).l

ax

2

___'_ "P,

C _ _ ~ap ~ -

,-

f'Qx2XI

2)JJx

Substituting the va lu es of C, and C 2 in eq uat ion (9.8) 11= _'_ iJp 21-1 (Ix

I

II

l +y[ __'_'r,] ax 2ft

ap , a( -

= - - - ltv - y-1 2)1

...(9.9)

'""'

In the aoove equation. )1.. -::;- and t are constant. It means Ij varies with the square of y. Hence equation (9.9) is a equation of a paraoola. Hence velocity distribution across a section o f the parallcl p late is paraoolic. This \'clocity distribution is show n in Fig. 9.1 (a).

1- to -I

-

- -..

Fig. 9.7

(&)

,0)

Vl'locity diuribution and Jbl'ar Jtrf'JJ diJtriblltion a Jl'ction of parallel plates.

acrOIJ

~ I

I~

~ I

IL Viscous Flow 399 1 (ii) Ratio of Maximum Vl'loc ily to An' rnge Ve locit y. The velocity is maximum. when y '" rIl. Substituting this value in equation (9.9). wc gel

+- ap [I x+-('-]'] =_ 2~ ~~ [I; _t: ]=_ 2~ ~ r: =_ ~ ~/;

U""" '" -

d.(

. )1

_

2

,2

... (9. 10)

The average ve locity, ~. is obtained by dividing the d ischarge (Q) across th e section hy Ihe area of the section (/ x I). And the d ischa rge Q is obtained by considering the ralC of flow of fluid through Ih~ strip of thickness Ify and inlcgrating it. The ralC of flow through strip is dQ'" Velocity

al

a distance), x Area of strip

ap

J 2 = - - - [l), - y 1xdyx 1

2).1 ,h

_

-

Q

"=-~ = -

dP.1 J

12jl ih

Area

Ixl

I dp 1 =- ----, [2)t ax

... (9. 1 I )

Dividing equation (9.10) by cqu;uion (9. 11 ). we get

I all 1 ----,

U'!'." '" II

_''1"'-' 'ap, ,',:- =8='2 12 3 1

... (9. 12)

- - -r 12j.l

ax

( iii) [)ro p uf Press ure h"Hd for" gh'e" Le";;t h . From equation (9. 1 I). we have -

1

ap

u = - ---, 1211 Inte grating this equation w.r.1- x. we get

1

ax

I' 1

I I

ilp=

II 1

12)Ju

- , ~ dx I

Ii

~ I

IL 1400

Fluid Mech a nics (!J

12~IUL

PI - P2 '" - ,- ,-

.

!J '" PI - 1';

... (9 . 13 )

pg

(it·) S hear S tress Distribut io n . II is obtained by su bstituting

•

"

•

•

•

<

.

«

,

•

•

•

~<

••

Ihe va lu e of u from cljuat ion (9.9) int o

. '. - -

d ~ l~

"

,

~-"

C_

'

I_

"J""

•

<

.

<

El.:

If ht is Ihe drop of pressure head. th e n I

,

•. ' --

'

,

"

- -_ , ,

Fig. 9.8

1:11 -

t=11

dU

dy I

ap

2).1

a.1

2).1

ax

.,.(9. 14)

,lIld, ,m: l"OnSlnnt. I'len<:e t

\'~rics Iincarly willi y. The shear S1ress diSlri bU1ion

2

~:

()y

11-2)'J

l=--~

In eq uation (9.1 4),

__

= fl~ [ l dP(ly _y1l]."[ __1 OP(/ _ 2Yl ]

ax

-

is show n in Fig. 9.7 (b). S hear stress is n1a!limul11. w hen y = 0 or I m the w all s uflhe plates, Shc~r stress is zero , w hen y '" 1/2 Ihal is al the centre line between Ihe (WO plates. Max. s hear Slress (t o) is given by l op t o "' - - - t. . .. (9 . 15)

ax

2 Calcu/me: (i) tile pressure grailielll along flo ..... (ii) l/ie a\'erage !'e{oeity. (wd (iii) Ille discharge for Oil oil of !'iseasil), 0.02 Nslm ! j/O"'ing belween 1....0 SlatiQnary ptlralld plates I m wide m(II' II"i»ed 10 mm tlfwrt. Tile !"docily mid ....ay bel,..enl l/ie plates is 2 mA

Problem 9.7

SolutIon. G ive n IJ == .02 N sJm1 Viscosity. Width. b = I 111 Distance between plates. 1= 10 111111 = .0 I 111 Velocity midway be tween the plates. U mo , = 2 mrs. (,) Press ure g rad;"'n!

(dd.,P]

or 2.0 = -

Usi ng eq uati on (9.10).

_'_(dP]

(.01)1

8x.02


.2C·O:cx,8~XCC'.O.. 2 == - 3200 N/ m , per m. An s . .01 x.OI

tip == d., (ii) Ave rage "cloelt)" (;;:)

Usin g eq uati on (9.12),

U ... , --. -3

" (iii) Disc harge (Q )

I I

2

-

2U",,, 3

2x2

3

u == - - - . - - == 1.3 m/s. An s. -

3

-

,

= Area of n ow Xu = b XIX U == I X .01 X 1.33= .0133 m Is. An •.

Ii

1396 Fluid Mechanks Proble~

Dt tennine (a) the prtssurt gradient, (b) the shear stress at the two horiumtal parallel platts and (c) the discharge per metre width for the laminar flow of oil with a maximum Vt!locity of 2 mls betwu n two horiumtal porallel fued plates which are l oo mm apart. Given II = 2.4525 N s/m2• Solution . Given : U-.A = 2 mis, t = 100 mm = 0. 1 m. ~ = 2.4525 N/ml Find

9.8

(i) Pressure gradient .. :

(il) Shear stres at the wall . to (iii) Discharge per metre width. Q. (i) Pressure gradient. dp

d.. Maximum ve locity. Um».' is given by equation (9. 10) U

= __ 1 OPt2

ox

mu

~

2.0= -

p I x aa x(. I )' 8)( 2.4525 x

Substituting the values

2 0 x8 x 2.4 525 = -ap ax =- .I x .l

39"~ NI m• perm. Aos.

(ii) Shear stress at tbe wall, to

to is given by equation (9. 15) as to = - 1. op 2 ax

X

t=-

1. (- 3924) x 0.1 = 1.96.2 N/m1, Am. 2

(iii) Dlscbarge per metre wkllh . Q

= Mean velocity x Area

= ~ U_)( (I x 1) = ~ x 2.0 )( 0. 1 x 1 = 0.133 ml/s. ADS.

) ) An oil of viscosity 10 poise flows betwu n two paraUe/fixed plales which are kept at a distance of 50 mm apart. Find the rate of flow of oil betwu n the plates if the drop of pressure in a length of / .2 m In 0.3 NIe,",. The width o/ the plates is 200 mm. Solution. Give n : ~ = 10 poise Problem 9.9

=..!..2. N s/m2 = I Nslm1

10

t=50mm=0.05m PI - PI = 0.3 N/m2 = 0.3 x l et N/ml L = 1.20 m B = 200 mm = 0.20 m.

Width. Find Q. rate of now The difference of pressure is given by equation (9. 13)

IN') . ( .: I IXHse= iO m 1

~ I

IL 1402 Fluid Mechanics 1 2 ~wL

PI - P2 == - ,- ,-

Substitutin g th e values,

w" gel

0.3 x 104

:

11 =

12x l.Ox

ux 1.20 .05 x .05

03 x 10 ' x 1.0 x .05 x .0 5 == 0.52 rn ls 12 x 1.20

= ii x Are a == 0.52 x (8 x /) = 0.52 x 0.20 x .05 111 3/s = .0052

Rate of fl ow

1I1

3/s

1

== 0.0052 x 10 litreJs " 5.2 lit re's. A ns. Problem 9.10 Wa/a at ISoC flows belweell two l(!fge parillie/ plales ill (l dis/arlce of 1.6 mm oparl. D t!lermille (iJ ,lie mluimll'" !'e/ocir)' (ii) lile pressure drop per Illii/lenglil and (iii) Ihe shear slress (l/ Ihe "'(I /Is of Ihe p/lUes if Ihe Il\"emge veiocily is 0.2 rnA Tile "iscm-;Iy of water (II Ij OC is gi"en as 0.01 poiJe. 3 Solution. Gi\'cn : t= 1.6 mm = 1.6 )(1 0 m = 0.00 16 III

ii

.

== 0.2 m/sc\: , )l == .0 1 PO ISC

=

.OJ ru

==

0.00 1

(i) Max imum ,·clocity. Un", is give n by cqu ,ni on (9. 12)

i.e.,

U" ....

3== "2" = 1.5)( 0.2 = 0.3 m/s. Ans.

( ii) The prt."Ss ure dn.p . (P, - Pl) i ~ given by t'qu3t io n (9. 13)

12 ~I UL

PI - Pl= ~

or

prc~sllre

L211U

drop per un it le ngth = - ,-

r

' " - I 2 x .01 x -;-~0~.2~ = 937 .'N 44 Im ' pcrm . -10 (.eX) 16i

ox -

01

(iii) Shea r s tress :.tthe walls is g ive n by equati on (9.I S)

I op I , t o=- - - X I = - x 937.44 x .00 16 = 0.749 N/m ". Ans .

ox

2 2 Problem 9.11 TilerI' is a ilorilOllla{ c rack 40 111m ... ide and 2.5 mm deep ill a ...al/ of Illickllf'ss 100 mill. W
Width o f crac k. D~plh of crac k. Le nglh o f crack.

I I

b = 40 mm == 0.04 m 1= 2.5 nu n = .0025 111 l. = 100 111m = 0. 1 m

Ii

~ I

IL Viscous Flow

403 1

/II - 1'2 '" 0.0294) Nlc m 2 '" 0.02943 x 104 N/m 2 '" 294.3 Nlm! J..l

.0 1 Ns 10 Ill ·

= .0 1 poise = - ,

Find rate of kak agc (Q) (PI - (2) is g iven by equa tion (9. 13) as

121l;;L or 294.3= 12x :Q!x u xOJ JO (.0025)(.0025)

PI - /l2= - , -, -

294.3 x 10 x .0025 )( .0025 II '"

Rate of lea kage'"

'"

12 ><.0 I x O.l

ii

1.5328

ill 'S

x area o f cross-section of crack

= 1.538x(bx/) '" 1.538 x.04 x JXl25 rn J/s = 1.538 x [0- 4

111 3/s

= 1.538 x 10-4 x 10) litre's = 0.1 538 litre/s. Ans. Problem 9 .12 The rIU1,,,{ c/ermmce be/wee" (I flyi/mulie plullger (md 'he cylinder walls iso.l mm: the lenglh of lilt! plunger is 300 mm (lml diameter 100 111m. Fi"d the I'e/oeiry of leakage and mil.' of iellkage plm the plunger at GIl ;IISWIlI wilell Ihe difference of Ihe press lire be/ween the IWO ellds of Ihe plunger is 9 lIT o/lI'aler. Take /.I = 0.0127 poise. Solution. G iven: The flow through the clearance area will be the same as the flow between two parallel su rfaces. I" 0. 1 mm ,,0.0001 m L,,300mm:O.3m Diameter. D" IOOmm=O. lm "

Differe nce of pressure

PI - P,

- " 9 mof waler

pg

PI - P2 : 9 x 1000 x 9.8 1 Nfm 2 " 88290 Nfm! Viscosity. Find

' . _ .0127 Ns Il "'.0 1'7 _ po ise - - - , 10 III

(i) Veloc ity of leakage. i,e .• mean ve locity (ii) Rate of lea kage. Q

(i) Velocit y o f leaka ge

u

(i/). The average ve loc ity (i/) is given by equati o n (9. 11) -

I()p ,

u: -----c 121.1 dol

I PI-P, x (.000 I ) x (.000 1) = -~", 01 27 12x· L 10

-

=

I

12 x.0127

88290

x - - x(.OOOI)x(.OOOI)

OJ

= .193 mfs == 19.3 cm/s. Ans.

I I

Ii

~ I

IL 1404

Fluid Mechanics

(ii) R:tte of leak;lge. Q

Q " ~ x area o f flow ,,0. 193 x nD x I m3/s= 0. 193 x II x.I x .0001 1I1 31s =6.06x I O-6 m l fs = 6J)6 x 10-6)( IO] li1r(~ls = 6.06 x 10-.1 lit,..,/s . A n s.

to 9.4 KINETIC ENERGY CORRECTION AND MOMENT UM CORRECTION FACTORS Kinetic energy correct ion factor is defined as the rat io of the kinetic energy of the flow pe r second bascl! on acmal velocity across a sectiun to the kinetic energy of the flow per second based 011 avef'Jgc velocity across the same sec tion. II is denOTed by 0.. HelKe nwthcnwtically, "~

K.E.lscc oo based on actual ve locity :eO'C''C '=.CCCCCCOOCC'7'C'K. E'/scc based o n average ve locity

... (9. 16)

Momen tum Cu rredlo n I' Hct Ot . It is defined as the rati o of momentum of th e flow per second based 011 actual vcloc il y \0 the momentum of Ihe flow per second based 011 average veloc it y across u section. It is denoted by ~. Henc~ mathematically. ~

= Momen tum per second based 011 actual vd ocity . Momen tum pe r second based o n avera ge velocit y

. .. (9. 17)

Pro b le m 9.13 Show I/",//he momentum co rrection jador ""'{ energy co rrection judor jllr [llminar flow /hrough" cirClll'" pipe lire -II) WId 2.0 respec/h·e/y. Sol u llo n. (!) I\ lomc ntum Correction Factor or ~ The vciocity distribution through a circ ul ar pipe for laminar now at an y radius r is given by equation (9.3)

"~ _' (_dP) 4~

(R2_?)

d.l

... ( I )

Consider an elementary area dA in the form of a ring at a radius r and of width dr. then rlA = 21(r dr

-Fig. 9.9

Rate of fluid flow inll th rough Ih", ri ng = dQ = velocity x area of ri ng clement =ux2ltrdr Momentum o f Ihe fluiu Ihrough ring per second = mass x velocity = p xdQ x I j = P x 2ltrdrx II x II = 2ltp u 2rdr TOlal aClual momentum o f the fluid per second across the secti on =

I I

Jr' 21(p 11 ,r tiT o

Ii

~ I

IL Viscous Flow

405 1

Substituting Ihe value of u from (I)

=

"',(")'[R'" +~_ 2R"']' ('I')! +
2

a.T

'" IlP, 81-1"

6R

6

6

4

2R6 _ 6R

= PI',

0

('I' )'[~+~ _ 2R' ] 2 6 4

81l";h

6

12

- PI' ( ' , )" - Sil l ax

x ~-....!2...(,p )' 6 - 4&/1 1 ,h

R'

.. .(2)

Mome ntum of Ih e fluid per second based on ave rage veloc ity

=

mass of fluid

, x ,we rag c ve locit y

~,

'" pAu

xii", pAUl

where A= Area uf cross-section = 7tR". Ii

'" average

velucity '" Umo,

2

--' (-"ax )R' - RI-l

Momentum/sec based 0 11 average ve loc it y

'" p

x 7tHl x [ _ ' 811

(al') R, ] l '" P X 7tHl ax

x 64' , (_ ~1

~p)l R4

(IX

... (3)

p=

I I

MomCllIum I sec based on actual ve locity Momentulll I sec based on average velocity

Ii

Viscous Flow

=

is?(:r,R' (a

= --ax !tp

p)

64~

R

= 64 , 48

401 1

=~.A"'. 3

(ii) Energy CorrtCtJon Factor. (L Kinetic energy of the fluid flowing through the elementary ring of radius',' and of width · dr' per sec

=.!.. x mass x u2 =.! xpdQx'; 2

2

=.1.. x p x (u x 27fr dr) x ,i = ..!..p x 21tru ] dr = 1tpru1 dr 2

..

2

Total actual kinetic energy of flow per second

=S: "p",'dr= S: "pr [4~ (-:)(R'- r')r dr =!tp x

=

[4~ (- ::))' S: [R'- ?J' rdr

...!2...(_ ap )' [R' _R' _3R' + 3R'] 64~' ax 2 8 4 6

=...!2...(_ap )' R,[ 12-3-18+12] 64 .... 3

=

ax

...!2... (_ ap )' R' 64~l

ax

24

8 Kinetic energy of the flow based on average velocity I -z I - -2 1 -J =-xmassxu =-xpAuxu = -x pAu 2 2 2 A = (.if) . Substituting the value of and

...(4)

1402

."-

Fluid Mechanics Kinetic energy of the now/sec

=

.,

.

i x p x rul'X[8~ (- ~) R'r

I = -x p x ~x

2

I

64 x 1Ij1'

(il)' _...1!... x H' ilx

(ilp)'x R' ax

_ p. - 128 x8~ 3 a=

K. EJsec based on actual velocity Equation (4) = K.EJsec based on average velocity Equatiofl (5)

...!!£... (_ ='

64~1

p q8x8iJ l ~

9.S

... (5)

ilp )' x R' ax

J8

P) II' (d - ax x

= 128 x8= 2.0.Aru. 64x8

POWER ABSORBED IN VISCOUS FLOW

For the lubrication of the machine pans. an oil is used. A ow of oil in bearings is an example of viscous now. u a highly viscous ~i1 is used for lubrication of bearings. it will offer great resistance and thus a grealCr power loss will take place. But if a light oil is used, a required film between the rotating pan and stationary metal surface will not be possible. Hence. the wear of the two surface will take place. Hence an oil of correct viscosity should be used for lubrication. The power required 10 overcome the viscous resistance in the following cases wiU be detennincd : I . Viscous resistance of Journal Bearings. 2. Viscous resistance ofFoot·step Bearings. 3. Viscous resistance of Collar Bearings. 9.5.1 Viscous Resistance of Journal Bearings. Consider a shaft of diameter D rolating in a journal bearing. The clearance between the shaft andjoumal bearing is filled with a viscous oil. The oil film in contact with the shaft rotates as the same speed as that of shaft while the oil film in contact with journal bearing is stationary. Thus the viscous resistance will be offered by the oil to the rotating shaft. Let N = speed of shaft in r.p.m. t = thickness of oil film L = length of oil film

21CN

..

Angular speed of the shaft, ro = 00

..

Tangenllai speed of the sbaft= m x R or V = - - x - = - 60 2 60

,

' ••• 'I ' ' b d. The ••••• ~ ....... stress In UI'C 0 1 IS given y;t = ~ dy

21CN

D

.DN

Viscous Flow

OIL

403 1

......-1 .

Fig. 9.10 JOllnlal btttrilig. As the thickness of oil film is very small . the velocity distribution in the oil film can be assumed as linear. du V -0 V 1fDN

Hence

dy

= - - = - = 60 X I 1tDN

t=IJ.

60 X1

Shear force o r viscous res istance = t

X

Area of surface of shaft

~1tDN Illt' D' NL = - - x rcDL = =~= 60t 601 Torque required to overcome the viscous resistance.

. D · T = YISCOUS resIStance x 2 = JUt D l NL x D = !:~::",.;D:,J.:.: N:::L 60t 2 t20, 1

Power absorbed in overcoming the viscous resistance .p =

'E:!!!... = 2rcN X 1J.1[1 DJ NL 60

.

60

1201

JUtlDJN1L = 60 60 watts. Ans.

x

xI

... (9. 18)

Problem 9.14 A shaft having D diam~l~r of 50 mm rotates centrally in ajoumol bearing having a diamelt!r of 50. / 5 mm alld length tOO mm. The angular spaCt! ~lWun the shaft 011(1 the bearing is filled with oil having viscosity 0/0.9 poiu. Dcruminc the powt!r absor~d in the inoring when the spud of rotation is 60 T.p.m. Solution . Given : D =50 mmor .05 m Dia. of shaft. 0 1 =50.15 mm o r 0.0501 5 m Dia. of bearing. L = 100 mm o rO. 1 m Length. 2xN 2ttNT 21fNT ·Power P""T XOl "" T x - - - - - watlS: - - kW. . 60 60 60,000

~ I

IL Visco us Flow

409 1

09 poise= -O.9Ns -, 10

~oo f il =.

m-

N = 600 r.p.m. Power= ?

Thickness of oi l film .

, ~ _D_, -_D _ ~

2 ~

Tange nti a l speed of sli afl. Shear stress Shear force (F)

"'0".,,":-2-,,',,0

0.15

- , - = 0.015 mm = 0.075 x 10

V = nON '" c'cXcO~.O~5,Xc600=

6IJ

6IJ

d,

t =Il - ~"

dy

Powe r

= 0.5 x IT

III

Illfs

;;-;;O~5~X::..;;'",

0.9

-; = -10- x 0.075 x 1O ~_1

= 1883.52 N/m2

= t X Ar~a = 1883.52 x itO X L = 1883.52 x

Resistance torque

V

3

T= F x

7[

x .05 x 0.1 = 29.5&6 N

~ = 29586 x .~5

= 0.7387 Nrn

= 2 /tNT '" 21£ X 600 x 0.7387 = 46.4 1 W. An s. 6IJ 6IJ

Problem 9 .15 A silaft of /00 111m. diameter rQlaleI m60 r./I.m. in a 200 mm 10llg bearing. Taking IIwl IIII' two surftlces are uniformly separated by a distance of 0.5 111m alld laking linear \'e/ocit)' tfislribulioll ill I/le {'tbr icalillg oil IIm'illg dynamic ('; scosiry of -I cell(ipoiu s, find Ille power ab wrbed ill tile beoril lg.

Solution. Giv~n : Dia. of shaft. Length of bearing,

D = I()() mm = O. l m L = 200 unll = 0.2 11\

I=O.5mm=.5x 10-3 m

.

0.04 Ns

1.1 = 4 ce ntipoisc = .04 potse = - - --,

10

Ill '

N = 60 r. p.m.

Find power aowrbed

Using equation (9. 18), .04

~ -x

It l X(.I») X (60)1 xO.2

10

6Ox60x0.5xlO ~

J

l : 4 .% 1 )( 10- \V. ADS.

Problem 9.16 A ~'''afl of diameter 0.35 III rotales 1II 200 T.p."" ;"side a s/ee,'e 100 Ill'" [ollg. Th e d)'llllmic ",'scosily of [llbricalitig oil ill Ihe 2 mm gap bellt'<'lm J'lee,'e ami J'/u'f/ is 8 poi,\'es. ell/cIl/are Ille {lower [OJ'/ ill Ille beMing. Solution. Gi ven: Dia. of shaft.

I I

D = 0.35

111

Ii

~ I

IL 1410 Fluid Mechanics N", 200 T.p.lll. Speed of shafl, L= l00l111n :O. 1 111 Length of s lct:\'c. Distance between sleeve and shaft. I = 2 mill '" 2 x I 0- 3 III

Viscosity.

11 '" 8 poise '"

..!. N~ 10 m"

The power lost in [he bcaring is given by "qU31ion (9.18) as

P=

Illt l 0 ) N ~ L

60x60xl

8 to

walts

rr3 x (.35)) X(200)2 xO. 1

= -x

6Ox60x2xlO

= 5'X).!1 W = 0.59 kW. A ns.

1

"'II!. is n",n;ng {II !Zoo f.p.m., is

Problem 9 .17 A slee,'c, in which (I sh aft of di(llnder 75 radill/ clearance 0/0.1 mm. Calculate Ihe torque resi~'I"'lce splJCe is filled wilh oil of d)'llllmic I'iscosily 0.96 poise. Solution. Given: Dia. of shaft. 0= 75 mm = 0.075 m N = 1200 T.p.lll. /=0.1 mm =O. l x [O-3 m L = 100 Ilun = 0. 1111 Le ngth of sleeve.

11

nON

/t x .075x 1200 60 = 4.71 2 Illls

~

Shear stress.

!l - == -

Torque resis tan ce

iI(ll'ilig (I

length of ~'Ieel'e is 100 mm ami Ihe

' 0.96 Ns =.096 poise = - - --, 10 m "

Tangential ve locity of shaft. v=

Shear force,

if Ihe

t =

'"

V

.96

I

10

x

4.712 .lx lO J

, 4523.5 Nlm -

F =t xnDf. = 4523.5xnx.075x.I '" 106.575N = Fx 0

2 = 106.575 x .0;5 = 3.996 Nm. Ans .

Problem 9.18 A slrap of / 00 mm dim"eleT r""$ in a beMin!; of lenglll 200 mm willi II radial clear· ance ofO.o25 mill al 30 r.p.lII. Filld Ille "elocily of IIIe oil, if ille power reqllired 10 overcome I/Ie l'iSCOliS ,e.\·i.~tallce is 183.94 'i'tlll.~. Solullon. Gi ven: D" loornm =O.1 111 L " 200 mm " 0,2 m 3 I " .025 111111 " 0.025 x 10 m N" 30 r.p.rn . : H.P . = 0.25 Find viSl:osity ofoil.).t. The h.p. is given by equat ion (9.18) as

I I

Ii

~ I

IL Visco us Flow

"'

18 3.')4 ==

).In

, x( J ), x(30 )', xO.2

60 x60 xO.025x 10

411 1

J

133.94 x60x60x.025xlO- J Ns It

=2.96

J

x .001 x 900 x 0.2

N~

m-

=2.96x 10= 29.6 po iw. Ans.

9 .S.2 Viscous Resistance of foot -Step Bearing. Fig. 9.11 shows Ihe foot-step hearing. in which a vertical shaft is rotating. An o il film between the oonorn surface of the shaft and bearing is provi ded, 10 reduce tile wear and lear. The viscous resistance is offered by lh" oil 10 [he shaft In this case Ihe rad ius of the surface of the shaft in conlact willi oil is not conSlant as in the case of the journal bearing. Hen!;c. viSl.,()us resistance in foot-step bearing is calculated by consid ering ~n eicmentary circular ring of radius r and lhickne~s dr as shown in Fig. 9.1 I. LeI N", speed of the shaft I = thickness ufuil film R '" radius of the shaft Are a of the ele mentary ring '" 2rrrdr

,

Now shear stress is given by

d,

V

d)'

I

I = IJ-=).I~

t-*=1~;;;;~;;;;:;;;K

where V is Ihe tangential vclocily uf shan al radius r and is equallU 2~

Oil

WXr== - - H 6()

Shear force on the ring == dF == t

X

area of elementary ring

2 ttN r ).l n l Nr l = J.1 x - - x - X 2nr
60

15,

I

Fig. '1.11

Torque required to ovacolllc the viscous resistance. dT=dFxr

Foot-step beari llg .

ft n-N," )1' 3 " dr x r = n-Nr dr 151 151 Tot allUTque required to overcome the viscous resistance. T=

fR1fT", j RL

Jo

0

n2N?dr

151

R '" L n2N r ? dr= L n2N 15/ Jo 15/

'" L

6(),

Power absorbed.

[~lR = LnlN~ 4 151 4

n2NR 4

0

...(9. 19A)

'«NT

p" - - - watts 60

2nN 2 ~ = '"o ' " - ' - J.1 nNR - '"'"N",'R;:,. 60 601 60 X 301

I I

... (9. 19)

... (9.20)

Ii

~ I

IL 1412 Fluid Mechanics Problem 9.19 find the torque required IQ rOlale II l'utieai J'lwft of diameler 100 mil! af 750 r.p.m. TIlt' lower ''fId of IIII' shaft reSIS ill a F;ot-slep bearing. The end of 1/11' slwfl and sur/oce of the bearing are holh flat al1
Dia. of Shaft.

D = l00 l111n =O. 1 111 R = D : Q:!. = O.05 m

2

2 N", 750 r.p.m. Thickness of oil film ,

I = 05 mnl = 0.0005 m

1-1= The to rqu e required isgivcn by T=

1.5 Ns

c.

1 " P0l5C= - - , 10 m " cqu~lioJl

L

60,

(9.19) as

rc 1NH4 Nm

1.5

= - X 10

IT ' X750x (.05)4 = 0.2305 Nm . A ns. 60 x .0005

Problem 9.20 Find Ille po",er required /0 rO/ale a cireu/a, dis{' of diameler 200 mm at 1000 f.p.m, Th e c ircIII", disc I/(Is " ckaran ce 0/0.01 mm from Ihe bollom f1" / plale Olllilile c/ea"mce co nlainJ' oil of\,iscosily / .05 poise. Solution. Given: Dia. of di sc, D = 200 mm = 0.2 m

Th ick ness of oil mm.

R = D '" 0.2 '" 0 .1 m 2 2 N ", 1000 r.p.m. I '" 0.4 mm = 0.0Cl04 m C' 1.05 _,2 I! '" I .. O J poise = - - N ", m

10

The

pow~r

required to mtale the disc is given by equation (9.20) as 1'=

I!n 1N" R4 6Ox30 xI

1.05 10

= -- x

wa llS

n 1 x WOO" x (0.1 / 60 x 30 x .0004

= 452. 1 \V. A n s.

9. 5. 3 Viscous Resistance of Collar Bearing. Fig. 9. 12 shows th e coll ar bearing. where the face of the collar is separated fro m bearing surface by an oil film of uniform thic kness. N '" Specd of the s haft in r.p.m. Let /(1 '" Internal radius of the collar

I I

Ii

~ I

IL Visco us Flow

413 1

R 2 '" Ex ternal radius of the co ll ar I '" Th ickn ess o f o il film.

-
L

",

N,3 df

It!

TOlall0rquc, required 10 overcome Ih e viscous reSiSl
JR,Iff= JR , J:.. H, 15r

It!

Nf)

H,

= - "-

151 x 4

tiT: L151411, nlN [~l"

rr:!N IR,4 -R ~ 1 = L -

I

601

rr: 2N rR ,~ - R"1 -

I

.. ,(9 .21)

Power absorbed in overcoming v isco us reSistance

p _ 2rr.NT _ 21(N 11 2N [R 4 R " [ - ( j ( ) - 6(j X 1l 2- I 601

=

)JI[ l N

2

60 x 301

•

•

IR2 - HI 1 wallS.

... (9 _22)

Problem 9 .21 A co ltor iJeMing 11I1I'ing e.lIemu/ (llIIl ;Ii(em(l/ dillmelers 150 /Il'" 'lnd 100 m", respeclhdy iJ' uJ'ed 10 lake Ihe IlnuSI of II slillfl. An oil film of Ihicknen' 0 .25 """ is m(linlllined beMet'1i IIII' col/ar surface and rhe bellrillg. Filld II,e po ...er 1051 in ol'ercomillg Ihe I'iscous resislallce ...hell the shaft rO/(ltes til 300 r,p.lII. Take jJ = 0.91 poise, Solution. G iven: EXlemal Dia , of co llar, Dl = 150 111m = 0. 15 m

In lerna l Dia. of co llar,

Thickn ess of oil 1111\1,

D, .15 K,,,, - "' - ,,,0.0751ll 2 2 • D , '" loo mm '" 0.1 III D, 0 .1 K, ,,,-=-=0.05m 2 2 I '" 0.25 1\11\1 '" 0.00025 111 N = 3oo r.p.m.

0.9 1 Ns I! = 0.91 poise = - - --, 10 III

I I

Ii

~ I

IL 1414

Fluid Mechanics

The power required is given by eq uation (9.22) or p.

I-lIT1N

1

-'+::';;C-- IR 6O x 30xI

0.9 1

' -- x

4 _

41

R

1

I

I'l l )(300 1 )([.075" ~.054 l

10 60 x 30 x .00025 '" 56431 4 1.00003164 - .()()()()()625] '" 564314 x .00002539 '" 14.327 W. Ans. Problem 9.22 Th e a/ernul (md in(emu/ diameten of 1I CO/lOT heaTing (lTe 200 mm (lnd f 50 mm respedi"d)'. Be/ween IIII' COIlM surface (llid Ihe hearing. an oil film oJ/hickness 0.25 mm alld of I'isco:;;'), 0.9 poise. is main wined. Find Ihe IOrf/oe 01111 l/ie power /OJ/ in orereoming Ihe ,·ismus resist,mc/' of Ihe oil when IiiI' shuft is "",,,ing 01 250 T.p.m. Solution. Given: D l '" 200 mm '" 0.2 m

N.

1

", .!2", 0.2 2

= 0.1

m

2

D ,,,, 150mm=O. 15m o 0.15 RI

= - ' . - - = .075 III

2

2

1 = 0.25 111111

== .(0)25 III

I-l == 0.9 poisc = 0.9 N~ 10 m-

Torque required is given by equmioll (9.21)

. ~( , 4 4 0.9 1t~ X250[O. I ·-.075·] , . /t"N [R , - H,I = x Nm 601 • 10 60 x 0.00025 = 14804.4 [ .000 I - .000031641 = 1.0114 Nm. An s. Power lost in visco us resistance 2nNT 21t x 250 x 1.01 14 • --~= '" 26.48 W . An s. 6Q

... 9 .6

6Q

LOSS OF HEAD DUE TO FRICTION IN VISCOUS FLOW

The loss of pressure

h c ~d.

II, in a pipe of diameter D, in which a viscous Ouid of viscosity

).I

is

Ilowing with a velocity jj is given by Hagcn Poiscuille formula i.c .. by cq uation (9.6) as

II = 32).111L , pgO l where L = length of pipe The loss of head due to friction· is given by

,,_ 4 .j. L . V!:. 4 .j. L. ";/ ,D x2g Dx2g

... (ii)

[.: ve locity in pipe is always averag e I'elocity - For derivmion.

I I

ple~sc

r<:fcr to Art. 10.3.1.

Ii

~ I

IL Visco us Flow

415 1

where!'" co-cfri cicllt of friction between the pipe and fluid .

32~IUL

Equating (i) and (il). we gel

4./ . L.~ ~ D x2g

pg0 2 '"

16,

3211;;£ x D x 28 4 .L.u' . pg. V '

" II .p. D

= 16X - ' - = 16 x....!.... pVD

where

fl

I

pVD

R,

--: -

R,

pVD

and N., == Rey nolds nurnb.:r '" - ).l

f=~.

. .. (9.23)

R,

Problem 9.23

Waler is flowing {lnOllS/' (I 200 111m diameter pipe ... ilh coefficient of friClion 2 (j/ a point 40 mm from Ihe pipe a.tis is O.()()981 Nkm . Calcuhlle II,e shear stress allhe pipe wo/l. Solution. G iven: Dia. of pipe. D '" 200 nlln '" 0.20 m Co-efficient of Frict ion. f'" 0.04 Shearslrcssat r= 40m m, t=0.00981 Nfcm ~ Let the shear stress at pipe wall '" t or irs! find w heth er the flow is viscousor not. The flow will be viscous if Rey no ld s number R, is less (hall 2000.

/=

O.().I. The sileaf Siren'

Using equ ation (9.23). we get/= R

,

~ R,

0'

""~=400 .04

Th is tneans flow is viscous. The s hear s tress in case of viscous flow tll ro ugh a pi"" is give n by tlw equa ti on (9.1) as

ap , ax 2

t = - --

But

~~

is constant across 11 secti on. Across a sectio n. there is no

variation of p.

\'ari~tion of x

and there is no

,- ,

At the pi pe wa ll. radius", 100 nun and shear stress is to

0'(X1981 '" ~ 100 40 100 ]()() x 0.009&1

~ '" ..!.!L. ,

"'

40

I I

'" 0.0245 Nfcm

2

,

A n s.

Ii

~ I

IL 1416

Fluid Mechanics

Prob lem 9.24 A pipe of diameter 20 em Oll
Solution. Given: Dia. of pipe.

D=2Qcm=2.50rn

Length of pipe.

L = IOOOOm

Slope o f pipe.

i= I in 200:

Sp. gr. of oil.

S= 0.9

De nsity of oil.

2~

p" 0.9 x 1000 "" 900 kg/ml 1.5 Ns -, 10 m'

Viscosity of oil.

)l" 1.5 poise == -

Di sc harge.

Q = 20 litre/s = 0.02 m )'.~

Velocity of flow.

ii

== ~ = 0.020 = 0.020 ,,0.6366 m/s Are a ~ 02 ~ ( .2 )2

4

4

R, = Reynolds number = pVD = 900 x 0.6366 x .2 )l 1.5

10 =

900 x .6366 x.2 x 10

{": V:ii:O.6366 1

15

= 763.&9 As lhe Rey nold s number is less llian 2000. the fl ow is visco us. Th e co-efficien t of frict ion for \'iscous fl ow is g ive n by equat io n (9.23) as

/==

~== _'6_ == 0.02094 R,

763.89

-,

Head lost due to frict ion,iI, ==

==

4 ./. L .u D x 2g

4 X .02094 x I0000 x (.6366)' 0.2x2x9.8 1

m " 86.50 m. An s.

Due to slo pe of pipe I in 200. th e height throu gh wh ich oil is to be raised by pump " S lope x Le ngth of pipe =;x L= - ' - x 10000= 5010 200 Total head against which pUlnp is 10 work, H"'lj +iXL"S6.50 +50 = 136.50 m Power required to pump the oil

I I

Ii

~ I

IL Visco us Flow

417 1

'" PI:. Q .11 '" 900 x 9.8 1 x 0.20 x 136.50 = 24. 1 kW. Ail S. 1000 1000

.. 9.7

MOVEMENT OF PISTON IN DASH·POT

Consider a piston moving in a vc nical dash-pot containing oil as shown in f ig. 9. 13.

Let

D" Diame ter of pisto n, L " Le ngth o f piMon.

IV

T

= Weight o f piston.

,

)..l :: Vi scosit y of o il.

V

= Velocity o f piston.

1

ii

= A I'c ragc veloci ty of oil in the clearance. I = Clearance between the dash-pot and piscon.

to.p '"

Difference o f pressure imcilsitics between th e two end s

of the piston. The flo w of oil throug h c leara nce is s imilar 10 th e viscous fl ow bet ween two paralle l plates. Th e differenc e o f pressure for parallel plates for kngth .L' is giv"' ll by

.6.

p

~

12)tilL

Fig. 9.13 ...( i)

- ,-,-

Also th e diffe rence of pressure a1 th e two e nds of piston is given by_ I'J.p =

Weight of pis ton IV 4 1\1 -Area of pisto n - ~ D2 - rr.D ~ 4

... (ii)

... 12)lilL 4 1V Equaling (I) and (u). We get --,- ~ - , r TtD " _

Tt O-

1V/ 1

I~

41V

II '" - ,

x --.

,

...(iii)

12)1L 3n)1 L D " V is lh e vdocily of piston or the veloc ity of o il in dash-poI in con tact wilh piston. The rate of fl ow ofoi l in dash-pot '" veloci ty x a rea o f dash-pot = V x ~ D ! 4 Rat e of fl ow throu gh c learan ce = veloc ity throu gh cle aran ce x area of clearance = iI x nO x I Due to continu ity equati on. ratc of flow through clearance must be equ al to rate of flow throu gh das h-pot.

Ii x nD xl=Vx ~02 4

_ II

Tt

",Vx - D 4

2

[

VD

TtDxl

41

x --- ~ -

... (1"1')

Equating the value o f Ii from
VD

" I I

Ii

~ I

IL 1418 Fluid Mechanics ... (9 .24 )

Problem 9 .25 All oil dash-flol COIIS;S/.! of II pislOn mOl'ing ill a cylinder IWI'iIiS oil. TI,;s arrangemelll is Iised 10 damp out the I'ibratiolls. The pislOn falls willi uniform speed (!lId co!'ers 5 em in foo seconds. If on uddiliOlwll>'eighl of 1.36 N is placed 011 the top oflhe pis/oil. itfalls Ihrough 5 em in 86 ~'ecollds wilh uniform speed, The diameter oflhe pisloll is 7,5 em lind its /ell gl/J is 10 em. Th e clearance between IiiI' piS/Oil ,md Ihe cy/illder is 0./2 em which is uniform Iinoaghoul. Find the \'iscosily of oil.

Solullon. Given: Distance covered by piston due to self we ight, = 5 em Time laken. '" 100 sec Additiunal weight. '" 1.36 N Time taken to cover 5 elll due to additiona l we ight. '" 86 sec Dia. of pisto n. D '" 7.5 em '" 0.075 111 Length of piston. 1_= 10cm=0.1 m Clearance. 1=0.12cm:0.0012m Let the viscos ity of oil IV = Weight of piston. V = Velocity of piston without addit ional weight. ¥" = Velocity of piston with addi tional weight. Using equation (9.24). we have

=,

4 IVt J

_

3ltV J

].\ -

IV

"'

v

=

_

LV -

4[ W + 1.36] IJ

31tD ) LV *

IV + 136 V'

V

IV

V·

1V + 1.36

(

41V, ' ) Cancelling - -,-

31!D L

...(i)

V= Velocity of pis ton due to self weight of piston

""

Distance cove red Time taken

5

- - em/s

100

¥" = Distalll'e cove red due to self weight + additional weight

Similarl y.

Time taken 5

= 86 em/s V 5 86 - = - x -=0.86 V· (00 5

Equating (i) and (ii). we get

"' "'

I I

IV

W

+ 1.36

... (ii)

'" 0.86

1V =0.86W+.86xl .36 IV- 0.86 IV = 0. 14 W = .86 x 1.36

Ii

~ I

IL Visco us Flow

419 1

IV", 0.86 x 1.36 "" 8.354 N 0.1 4 41VI

Using equation (9.24 ), we gel).l:=

J

'3C"~D'i'LCV~ 4x8J54 x(.OO I2f

{

'" JIl x (O.07Sl x.1O x (-'- x _ '_ )

100 :=

.. 9.8

5

5')

": V ,. 100 emls = 100 x 100 mfs

100

0.29 N s/m 2 = 0.29 x 10 poise = 2.9 poi~.,. Ail s .

METHODS OF DETERMINATION OF CO - EfFICIENT OF VISCOSITY

The following arc the cx pcrirn cmal mctllods of dClcrmining llie co-efficient of viscosity of a liquid: I, Capillary lube method. 2. Fa llin g sphere rcsis\JllcC met hod. 3. By rmating cylinder method. and 4. Orifice type viscometer. T he apP;lralUS uscd for determining the viscosity of a liquid is called viscometer. 9 . S. 1 Capillary Tube Me thod . In capillary lube method. Ihe viscosity of a liquid is calculated by measuring th e pressure difference for a given length of th e capillary lube. The Hagen I-'o isc uill e law is used for "3lculatin g viscosity.

~~~~~-~~~ ~~ ~~ ~~

-

--

~

~ ~~

~.

-

-- -

~ ~ ~~ ~ ~

~_

T 1\

- :- : - :- : - :- j )' CONSTANT HEAD ,~,

Fig.9.14

jD

,' JI"\\J

ME~!~~ING --:=:::-' Capillary tllbe Vi5CQmeter.

Fig. 9.14 shows lh~ capillary luhc viscomel,;,r. Th~ liquid whose viscosily is to be del,;,rmin,;,d is fill,;,d in a consta nt head lank. The liquid is mainlained at constanl I,;,mpermure and is allowed 10 pass through Ihe capilluy tube froU1the constant head lunk. Then, Ih~ liquid is collected in a measuring tank for a given lime. Then Ihe rale of liquid collecled in Ih e tank IX'r second is determined. The pressure head "II' is measured at a poim far away from the tank as shown in Fig. 9.14. Then 1/ = Difference of pressure head for length L. The pressure at oUl lel is atmospheric. Lei D = Diameter of (;api lla ry lUbe. L = Length of tube for whic h difference o f pressure head is known. p = Dt:nsity of flu id.

I I

Ii

~ I

IL 1420 Fluid Mechanics ~

and

'" Co-cfllciclU of viscosity.

Using Hagen Poiscuillc's Formula. II = 32 11 ii,L pg D '

: -Q- = -QArea I( D~ 4 where Q is ralc of liquid flowing through lube. So<

II

32f.l x

Q ~D2

xL

_ /, = _ _4" ,,_ - '"_. 128).l Q. I. pgD2

).1=

1tpgD"

npghD' 128Q.L

.. .(9 ,25)

Measurement of D should be done very accurately.

9. 8 .2 Fall ing Sphe re Resis tance Method . Theo r y. Th is method is based on Stoke's Jaw. according to which the drag force. F Oil a small sphere moving with a l'OIiStam velocity_ V tli rough a viscous fluid of ViSl:osity. f.l for viscous conditions is given by ... ( i) F = 31t~IUd d = diameter of sphere U = velocity of sphere. When the sphere attains a conSlant velocity U. Ihe drag forc.: is Ihe difference between the weight of sphere and

where

and

~'~f ~ i.:o 0: ~ ~

~

~

- -U - I - - -

-L- -

~

~I ~

SPHERE

buoyanl force acting 011 it.

Lei

CONSTANT TEMPERATUR BATH

FIXED MARK

L '" dislalH:e Iravelled by sphere in visco us fluid. I'" lime wken by sphere 10 cover distancc I. p, '" dcnsity of sphere. Pf '" density of fluid. IY '" weight of sphere. Fig. 9.15 Falling 5phaf: rf:5i5tam:f: method. F(l '" buoyant force acting on sphere. L

.

,

Then constant vclocny of sphere. U "'Weighl of spher~. IV", volume x dellsity of sphere x g

,

"'6d

;

Xp, Xg

f:

vol ume o f sphere ,. ~dJ}

and buoyant force. F 8'" weighl of fluid displaced '" volumc of liquid displaced x dellsily of flui d x g

,

;

'" - d x Pf X g

6

~ I

I volume of liquid

dL~placed

'" volume of sphere J

I~

~ I

IL Visco us Flow

421 1

For equilibrium.

Drag force'" Weigh! of sphe re - buoyam force

F=W - F8

or

Substituting the val ues of F. Wand P s . wc gel IT

311~Ud= - d

)

1t)

It\

Xp, xg - - d xp, xg= - d- xgIP,- Pjl

6 6 6

11 dl x g[P, _ p, ] gd 1 ~:6 3nUd "'ISU ]P,- P/ ]

... {9 .16)

where PI '" Density of liquid Hence in eq uation (9.26).lhc values of d. U. p, and PJ aTe ~nown and hence the viscosity of liquid can be determined. Mclhod. Thus this method consists of a tall vertical transparent cy lindrical wnk, which is filled wilh. the liquid whose visco~ity is 10 be dctcrrllincd. This lank is surrounded by 31l011wr transparent tank to keep the temperatu re of Ihc liquid in the cylindricallJnk lU be constant. A spherical ball ofsrnall di~mclcr 'd' is r!;lccd un the surfa!;c of liquid. Provision is made to rele~sc this ball. After a shon distance of travel. the ball attains a constant velocity. The time to travel a known vertical distancc betwecn two fixed marks on the cylindrical tank is noted to calculate the constant velocity U of the ball. Theil with the known values of d. p,. PI the I'iscosity tJ of the nuid is calculated by using equation (9.26).

Rotating Cyli nder Method. This method 9.S.3 TORStONAL .- SPRtNG consists of two ,-,oncentri\: cylinders of r~dii R, and R2 as show n in Fig. 9.16. The narrow space between the two ""~b,.,j§i,,,,;;'POINTER cylinders is filled with the liquid whose viscosity is to be determ incd. The inner cy linder is he Id stationary by means of a tors ional spring while outer cylinder is rotaled at conVISCOUS stant angular speed (I). The torque T acting on lhe in ner LIQUID (>I) ,-,ylinder is measured by 'he torsional spring. The torque on the inner cylinder must be equal and opposite to the torque app lied on the outer cylinde r. The to rque applied on the outer ,-,ylinder is due to viscous resistance provided by liquid in the annular space w and at the bottom of the inner cylinder. Let OJ = angular speed of outer ,-,ylinder. Fig. 9.16 Rotl1ting cylinder viscometer. Tangential (peripheral) speed of outer cylinder =OJxR 1 Tangential velocity of liquid layer in ,-,onwct with outer cylinder will be equ al to the tange ntial velocity of outer cy linder. Velocity of liquid layer with outer cylinder = OJ x R1 Velocity of liquid layer with inner cylinder = 0 !. Inner cylinder is stationary I Velocity gradient over the radial distance (Rl - R I) =du",wRl - O

Shear stress (t)

I I

Ii

~ I

IL 1422

Fluid Mechanics Shear force (P)

'" she ar stress x are a of surface '" t X l rr N I H I' shea r stress is acting on surface area '" I n NIx N I

mil, 2 ' ) XllR IH ( R2 - HI The torqu e TI

o lllh ~

inn er c ylinder duc 10 shearin g ac ti on of the liqu id in th e annulM space is '1'1'" shear fo l"('c x radiu s

=, ( ') wR,

HI

_ 21fllwll

-

HI

x2n RI H xRI

H I2 R I

... ( i)

(R2 - Rd appli~"d

If the gap betwee n th e bonom oflh e two cy linders is '/1 ' , th en Ihe torqu e is give n by equatio n (9 .19A) as

on inn er cy li nde r

( T 2)

But he re

11

R =R 1· t = h thcn T1 ", 6011

It

2

N fl l

4

w'" 2/tN or N '" 6O{o 6il 2, Il

'1'1= - - "

2

6000

4

1tJ.1 111

x - - xRI = - -

lit T Ol al \orquc T ac ti ng on Ihe inn cr cy linder i s fijI!

2//

HI

4

... (ii)

T : T 1 + T1 =

11 "

21'qlmHR,~ R2 (

R 1 - R,

)

IlJH])

4 _2

+ -- HI -

211

,[

IfI..lR I

2(R1 - RI)'!T rrR,2w [41f1IR) + Hi

(Rl - HI)]

R

R1H + RI2]

1

R

I

4 11

xw ... (9.27)

where

T " IOrqu e meas ured by th e strain of the tors ional s prin g . N I • N2 " radii of inner and o uter cy linder. II " clearan ce at the botto m o f cy linders. H " he ight o f liq uid in annul ar space. ~ := co-efficie nt of vi.<;cosity to be de te nnined . Hence. th e va lue of ~ (;an be calc ul ~ltcd fro m equat io n (9.27).

9 .8.4 Orifice Type Viscometer. In thi s meth od. th e time taken by a ce na in quant ity o f the liquid whose vis.;:osily is to be detc nnin cd , 10 fl ow thro ugh a s hort c ap illary lube is no ted down. The co -efficie nt o f viscusit y is the n obwined by comparing with the co -efficient of viscosit y of a liqu id whose viSl.'Os it y is kno wn or by th e usc convcrsio n fa ctors. Visco melc rs s uch as Sa ybo lt. Red wood or En g ler arc usua ll y use d . Th e princ ipl e for ali lhe th ree visco meter is same. In th e United Kin gdom. Red wood visco mete r is used whil e in U.S.A .. Sayboh visco mda is co mmonl y used.

I I

Ii

~ I

IL Visco us Flow

Pig. 9.17 shows thal

~aybo lt

viscometer. whi ch consists of a

tan~

at th e bonom o f which a sllon capillary tube is fitted. [n this tan k the

liquid w hose viscosity is to be determined is filled. This tan l.: is su r-

rounded by another tank. called constant te mperature bath. The liq uid ;, allowed now through capillary tube standard

'0

"

,

tcmpcraturc. The time laken by 60 c.c. of Ih" liqu id [0 now through the ell pi liar y tube is noted down. The in itial hci g'ht of Ii uid in the tank is previously adjusted to a standard height. From Ih" time measure· ment. Ihe kinematic vis<'"Qsity of liquid is known from the relation. V=

B AI - -

,

Fcc

ree

· c '~&" " :-: ; 'LI aiD -

::;

423 1

0 .

M

: 0-_ ,~ ~9NSTAN T

. ;:-

,O ~; 0,

TEMPERATUR

,

BATH

-.~~ ~.-_. ._. - --- - .

---:: .' .-=- -

III

l l J - .MEASURING CYLINDER

Saybolt v i5Cometer. where A = 0.24. B = 190. I = time noted in seconds. v = kinematic vis.;osily in slol; es. Problem 9 .26 Tile \'is"osil), 0/ (In oil 0/ ~p. gr. 0.9 is mea~''' reJ by a capillary I"be 0/ diamela 50 mm. Th e differe1l ce o/press"re head belwe,," 111'0 poinlS Z m aparl is 0.5 m a/water. Tile mass 0/ oil collected ill tI metlsllri1lg t(/II1r, is 60 kg ill /00 secollds. Find Ihe I'ismsily 0/ oil. Solu t ion. G iven Sp. gr. of oil =0.9 Dia. of capillary tube. D = 50 mill = 5 cm = 0.05 III Length of tube. L = 2 III Difference of pressure h~ad. 11 = 0.5 m Mass o f oil. At = 60 kg Time. 1= 100 s Mass of oi l per second [knsity of oi l. Di scharge. Using

equ~tion

Fig . 9.17

60 100

= - - = 0.6 kg/s

I' = sp. gr. of oil x 1000 = 0.9 x 1000 = 900 kg/m 3 0.6 Q= Massofo i! Is = - m 3/s= 0000667 . III ' Is De nsity 900 (9.25). we get viscosity. 1tpgll D

4

[here II = 111 = 0.5 J

1'=128Q . L

It X 900 x9.81 x 0.5 x (.05)' ='Ci'C':~:;;;;;;C~~"'- = 0.5075 (SI Units) N shn! 128 x 0 .000667 x 2.0

= 0.5075 x 10 poise = 5.075 polsi.'". Ans. Problem 9 .27 A ClIpillllr)' /lIbe 0/ iliameler Z mm amllellglh {OO mm is used/or mea.mrillg \'iscosit), 0/ tlliquid. nle difference o/pre.iSlIre ber ...een rile IlI"a enils o/rlle wbe is 0.6867 Nlcm ! ""'/Ihe \'iswsily 0/ liquid is 0 .25 poiJ'e. Find IIII! rale 0/ flow 0/ liquid Ihrougll Ihe lube. Solullon. Give n : Dia. of cap illary lube. D=2m=2xIO l 1l1 Length of IUoc. L= 100111111 = IOc1l1=0. 11I1 flp = 0.6867 Nkm2 = 0.6867 x 10" Nfrnl Difference of pressure.

Difference of pressure head.

II

=_' _11 =~O~.6~86~7~x~I~0_· 1'8

I I

I'g

Ii

~ I

IL 1424

Fluid Mechanics ~

Viscosity.

'" 0.25 poise

= 0.25 Ns/ml 10

Ld the rmc of flow of liquid ", Q

0.6867 X 10 ' x (2 x IO -l)' n p)ll1l)4

Using equation (9.25), we ge t 1-1 '"

128 . Q. L

= npg x

pg --""~c-;;CCC;C--128)(Q)(O.l

,

0.25 10

It

=

Q=

"'

X 0.6867 x 104 X (2 X 10- 3) 128 x Q)( 0. 1

It

x 0 .6867 x [0' x 24 X 10- \1 x ]O

In ~

118 x 0.1x 0.25

'" 107.86 x 10 ~ m .lls '" 107.86 x [0 ~ x I06 crn 3/s '" 107.86 x IQ-! C11I 3/S '" 1.078 cm l/s. AilS. Problem 9.28 A sphere of diameter 2 mill falls 150 mm ill 20 J'eCOlll/S in a 1';5CO" 5 /iq"id. Til e density of tile sphere iJ' 7500 kg!", J ' Illd of'iq"ill is 900 kgl", J. Find Ihe cQ-'1ficienl of I-isCQsily of Ih e liql/id.

Solution. Given: Dia.ofsphere.

<1= 2 mlll :2 x 10 .! m

Distance travelled by sphere

Time taken.

Veloc ity of s ph ere. Density of s phere. Densit y of liqu id.

'" 150 !TIm '" 0.15 III

I", 20 seconds

u '"

Dist ance Time

0.1 5 '" .0075 mfs 20

p, '" 7500 kg/m ,l PI = 900 kg/ m! d1

Usin g re lation (9.26). we get!!= ~ Ip, - Pj l =

[8U

9.81 X[2 x 10

Jf

18 x 0 .0075

17500 - 9001

= 9.8 1x 4 xlO -li x6600 = 1.9 [ 7 Ns 18xO.0075 m1

'" 1.9 17 x 10 '" [9. [ 7 poise. AilS. Problem 9.29 Find rhe I'iscosiry of" liquid of sp. gr. O.B. whe" " g"s bubble of diamerer 10 "'''' rise,~ UNUfi/y rhrougl, rhe lilfllid 111" I'e/odty of 1.2 cm!!l. Neglect rhe weight of the bubble. Solution. Given: Sp. gr. of liquid = 0.8 De nsit y of liqu id, PI = 0.8 x 1000 = 800 kgtlll 3 Dia. o f gas bubble. D = 10 nlln = I elll = 0.01 III Ve[ocily of bubble. U = 1.2 emts = .01 2 I1ItS As weight of bubble is neglected ,md density o f bubb le

I I

Ii

~ I

IL Visco us Flow

425 1

d'

II '" ~ [p, - PI! which is for a falling sphere. I8U ['or a rising bubble, the relalion will become ~s Now using the rcl31ion.

j.I '" : : :

Substituting lh..,

valuc~.

[Pr p,1

9.8 I x.otx.O I Ns Ns 1800 - 01 ~ = 3.6]"2 18 x,0 12 III 111

we gel

'" 3.63 )( 10 '" 36,3 pil lse. Ail s. Problem 9.30 Th e I'iscosily of II liq!lit/ is dc/ermined by rolaling cylinde r mel/wd. ill ...llieh c(lse the i""er cylinde r of diameter 20 em is Sia/iunary. The OilIer cylinde r of diameter 20.5 em, C011/llillS rhe liquid upro II heighl of 30 em. The clearance allile bOl/om of IIII! /h'O cylinders is 0.5 em. The ouler cylinder is rO/(l/ed a/ 4()() r.p.lII. The /Orque registered 011 IIII! lorsioll meier (I//(Iched /0 rill! illller

cylinder is 5.886 NIII. Filld IIII! I'iseasily of fluid. Solution. Given: Dia. of inner I:ylindcr. DI '" 20 ern Radius o f inne r cylinder.NI = 10 em = 0.1 m

Dia. of outer cylinder, Radius of oute r eylinder.N~ =

2~.5

= 10.25 e m = ,1025

In

Heigllt of liquid from bott om of outer cylinder = 30 em Clearance at tile bonom of two cylinders. !J '" 0.5 CIlI '" .005 m Height of inner cylinder immersed in liquid

= 30 -II = 30 - 0.5 '" 29.5 m H",

0'

Speed of ou ter cy linde r.

29.5

CIlI '"

.295 m

N = 400 r.p.m.

= 2nN = 2)(/tx400 = 4 1.88 6
III

Torque measured.

=

_-;-;-:,---=2"(1'.I0 :"2='"--;0".I)"X ".00 ,,,,'::X:::'.:8116 -,,,:==--,.,; It

x (.1 )1 x 4 1.88

[4 x .295 x .005 x .1025 + .1 1 (.1025 -. 1)]

2 x .0025 x .005 x 5.886

= -,-:;=~=~=::; /tx.O I x4 1.88 [.0006047 .000025] =0. 19286 Ns/rn 2 = 0.19286 x 10 = 1.9286 po ise. An s.

I I

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IL 1426

Fluid Mechanics

Problem 9.31 A sphere of diamerer I mill /alh 11"014811 335 III ill 100 secQlIl1$ ill (j l'isCOIH fluid. ,e/alil'e den.!ilies of Ihe sphere (Illd Ihe liquid are 7.0 (lIId 0.96 respec/il'ely. de/ermine Ihe

If Ihe

dJllllmic l'isCQsily of Iile liquid.

Solution. Given: Di3. of sphere.

d" 1 mm " 0.00 I m

Distance travelled by sphere = 335 mm == 0.335 Time taken. I" 100 seconds

Distance 0335 - - = 0.00335 m/.<.cc I ()() Time =7 p, == 7 x 1000 == 7000 kg/m l = 0.96 P, = 0.96 x 1000 = 960 kglm 3

U"~==

Veloci ty of sphere. Relative densi ty of sphere Density of sphl're. Relative density of liquid Densi ty of liquid.

Using the relation (9.26).

t11

gd 1

We gC11J

= 18U

9.8 1 x 0.00 I'

Jr, - Pj l '" 18 X 0.00335 17{)()() - 9601

= 0.0000098 1 x 6040 [8xO.00335

= 0.98 1 Ns/rn 2

== 0.98 1 X 10", 9.81 pulse. Am . Problem 9 .32 Delermille Ille fall I'e/ocily of 0.06 mm sand parlide (Ipecijic gradl)' '" 2.65) in ...mer at 20"C. /(Ike j1 '" 10 -J kg/mI. Solu t Ion. Gi\'en : d", 0.06 111111 '" 0.06 X [0- 3 m Dia. of sand particle. Specific gravity of sand '" 2.65 Density of sand. p, '" 2.65 x 1000 kglm l (": P for wat<'r in S .I . uni t '" 1000 kg/rn l) '" 2650 kg/Ill '

Viscosity of water.

)1. '"

10- 1 kg/ms '" 10-3 Nslrn 2

D.:nsity of water. PI "" 1000 kg/Ill ' Sand pan il:1c isj uslli ke a sphere. For equilibrium of sand p~rtic[e. Drag force"" Weigl\! of sa[ld particle - buoyant force F n ", W - F8 0' F n '" Jltj.l x U x d. where U", Velocity of panil: lc '" 31t x 10-' )( U x 0.06 X 10-1 N IV"" Weight of sand particle

"m

""

~ 6

Xtl'

...( i)

x P, Xg '" ~ x (0.06 X 10-,)1 x 2650 x9.81 N

6 P/J = Buoyant force = Weight of waler displaced

Hence vi""nsily '" (1 kg ~ 1 m ) )( --;- '" kglms. Hence kglms '" S

I I

111

_N_~ . m-

Ii

~ I

IL Viscous Flow

427 1

"'~ xtFxp,xg= ~ x (0.06)( IO-')'x 1000)(9.81 N

6 6 Substituti ng the above va lu es in eq uat ion (i). we get 311 X 10- 3 X U x 0.06 X 10- 3 '"

~ x (0.06 X 10-")3 X 2650 x9.8 1 - ~ x (0.06 X 10-3)3 X 1000 x 9.8 1 6 6

Cancelling (/I x OJk) X 10-·)2 throughollt. we get 3x U

=~

X

6

'"

~

0.06 2 X 10 .l x 2650 )(9.81 -

~ 6

X 0.06 2 X lO .l x IOOO x9.81

X 0.(6 2 )( 10-J X 9.81 (2650 - 10(0)

6

= ~ x 0.0036

X

10-J x 9.81 x 1650 = 0.0097 12

U '" 0.009712/3 '" 0.00323 nt/sec. An s.

HIGHLIGHTS I . A flow is said to be viscous if the Reynolds number is less

th~n

2000. Or the fluid flows in layers.

2. For the viscous flow through circular pipes. (I)

Op ,

af 2"

Shear Siress ...... t - -

(i/) Velocity ...... u _ - -'4jl

(iii) Ratio of velocities U : " .. 2.0

where

~

Loss of pressure head. Ill '"

"

32 IiL

~ pgD -

r", radiu.' al any poim.

'" pressure gradient.

u.... ~ maximum ,docity or velocity at,

H '" radius of the pipe.

Ii

~P [,r - ,zJ

OX

~ a,'crage ,'eloxily ~ ~. •R

jJ ~

m

O.

co-cffident of viscoSity .

D '" diameter of the pipe. J . For Ihe ,·iscou.' flow between two parallel plates. I ilr l " " - - - ( 1 ) " - )") ... Vciocity distribution 2jJ

ar

Um>' .. 1.5

"

h

12j.luL

J .~

, 'p

l"-"2a.. 1t - 2y ] where

I I

... Ratio of rna.imutn and avemge velocity ... Loss of pressure head Shear stress dislributioll

I = thi~k"ess or distance between two plates. y. distance in the "cMical direction from the lower plale. 1 '" sbear stress m any point in flow.

Ii

~ I

IL 1428 Fluid Mechanics ~.

The kinetic energy correction faclor a is gi.-cn as

o·

K.E.pcr second based on actual velocity K. E. per second ba>Cd on average "clocity

~ 2.0 ... for a circular pipe. 5. Momentum correction factor. ~ is given by

Ii..

Momentulll per second based on actmt! velocity Momentum pcr second based on average velocily

.. 3"4 ,-, for a circular pipe. 6 . For the viseou ••e,i.lance of Journal Hearing _

V",nDN dw=~=ItDN 60 'dy I 601

Torque.

where L= lenglh of bearing, N .. speed of shaft I" clearance between the shaft and bearing. 7. For the Foot -Step Bearing. the shem force. torque and h,p. absorbed are gj"cn as:

Shear force.

2 J fl rr. N R F. - - - 15 I J

Torque.

T .. ~ l(lNK'

""

,. flIlJNl R' 6OX)OXI

where R_ radius of the shaft. N .. speed of the shaft 8. For the collar bearing the torque and power absorbed are given as

T _ _II

60,

, .. leNIR,

.

" rrJNl p", ..

- R,• I.

6Ox301

where H, .. internal radius oflhe collar. I .. Ihickncss of oil film. 9. For the viscous flow the

IR,~ - R'l

•

,

Rl .. external radius of (he collar. P = power in wallS.

co~fficicn( of friction

is given by.! ""

~ ~

pVD VD where R, . 1he Reynolds number .. - - . - .

"

'

10. The co-efficienl of viscosily is detennined by dash-pol

where IV "" weight of (he piston. L", length of 'he piston. \' "" "cioci,y of the piston .

I I

~rrungcl1lcn(

4IV/)

as)1

E

3/tLO JV

r = dcaran~c between dash -pot and piston. 0 .. diameter of the piston.

Ii

~I

IL Visco us Flow II. The co-efficienl of viscosity o f a liquid is also delcnnined

c~perjlnentally

429 1

hy the following method

/lpgI/O'

(i) Capillary wbe method. ~,. 128QL

(ii) Falling 'phc.c method. II '"

where

>I'

'" spe~ific

,,"1,,-,,1 lSU

L '" length of the lUbe, Q '" rate of flow of fluid through capillary tube. P, _ density of sphere.

weight of fluid.

D = diameter of the capillary lube. d _ diameter of the sphere.

PI '" densi ty of fluid. NI ~ rad ius of outer rotatin g cylinder. T ~ torque.

U", "doci!y of sphere. RI '" radius of inner stal;o"a,y cylinder.

EXERCISE (A) THEORETICAL PROBLEMS I . Define the terms: Vi scosity.

kincmati~

viscosity. '-ciOC;I}, gradient and pressure gradient

2. What do YOIl mcan by 'Vi>eou,; Flow"!

3. Dcrive an expression fOf the velocity distrib ution for "iscous flow through a tircular pipe. Also skc!ch th~ "docity distribution and shear stress distribution across a section of the pipe. 4. 1'<0"<" that the ma.~ imum vel","·ity in a ci rcular pipe for vis.:ou< flow is eq ual to two times the a'·erage Velocity of th~ flow . (Delili Unilusily. Deamber 1(01) 5. find an e ~pre"ion for the loss of head o f a "i",ous flu id flowing through a circular pipe. 6. What is Hagcn Poiseuillc· s fonnula ? Dcrivc an e ~ pression for Hagen Poiscuillc·s Fonnula. 7. Prove that the velocity distribution for viscous flow between two paral lel plates when bOlh plates arc fixed across a .<;cClion i., parabolic in nature. AI"'.! prove that maximum ,·docity is equal to one and a half timcs the a'·emgc vc locity. 8. Show that thc differencc of prcssurc hcad for a givcn length of the two para llel plates which are fi~cd and through which viscous fluid is flowing is givcn by l2fliiL J P81l whcrell = Visco~ity of fluid. Ii ~ Avcrage velocity. 1 ~ Distance between thc two pamllcl plates. L ~ length of the plates. 9. Define the tenns , Kinetic energy correction factor and momentum correction factor. HI . Prove that for viscous flow throuSh a circular pipe the kinetic enersy correction factor is equal to 2 while 4 momentum corrcction factor ~ J.

,. --

II . A shaft is rotal ins in a journal bearing. The clearance belween Ihe shaft and the bearing is filled wilh a ,·iscous oil. Find an cxpression for the power absorbed in overcoming vi>eous resistance. 12 . Prove Ihal power absorbed in o,·ereoming viscous resistance in foot-Slep bearing is given by

P.

""~60'~'xN~'301 ~.r;-

where N ,. Radius of Ihe shaft, t ~ Clearance between shafl and fool-slep bearing.

II

N ,. Speed of 1he shaft.

II ,.,. Visco,ity of fl uid.

Ii

~ I

IL 1430 Fluid Mecha nics U . Show that the value of lhe

~o-efficien t

f .. ~,

where

of friction for v;>
R," Reynolds number.

14. I'rove that the co-efficient of viscosity by the da.;h-pOi arrange",,'"l is g;,'en hy. 4 (I'l l where

).I '"' 3ltLD'V IV ., Weight of Ihe piston.

L '" Length of piston.

I = Clearance between dash -pot and piston. D '" Diamet.:, of piston.

V = Velocity of piston. 15. What arc the different methods o f dClcnnining the co-cFflcicnt of viscosity of a liquid? Describe any two

method in details. 16. Prow that the 10 •• of pre.sure head for lhe viscous flow through a circular pipe is giv<'n by h

f =

32).1 u L

p:;;;iT""

where U ~ Avcrage velocity. w'" Sp<"ific weighl. 17. for a lami nar steady flow. prove thaI the pressure gradient in a direction of motion is equal to the shear gradient normal to the di'e<;tion of motion. 18 . Describe Reynolds e~periments to demon>trate the two types of flow. I~. For the laminar flow through a circular pipe. prove that' (i) the shear stress variation acroSS the seclion of the pipe is linear and (ii) thc vciocity variation is parabolic.

(B) NUMERICAL PROBLEMS I . A crude oil of viscosity 0.9 poise and sp. gf. 0.8 is flowing through a horizontal circular pipe of diameter 80 111m and of length 15 m. Calculate Ihe difference of pre,sure at the two ends of the pipe. if 50 kg uf the

[Ans.0.559 N/cm' l oil is collectcd in a tanK in 15 seconds. 2. A viscous flow is laking place in a pipe of diameter 100 mill. The maximum .-eiocity is 2 m/s. Find the mean velocity and the radius at which this occurs. Al so calculate the "eiocity at 30 mm from the wall of tbe pipe. IAn s. I m/s. ,~ 35.35 mm. u ~ 1.68 mlsi J . A fluid of viscosity 0.5 poise and specific gravity 1.20 is fluwing tbrougb :, circular pipe of diameter 100 mm. The maximum shcar stress at the pipe wall is given as 147.15 N/m ' . fmd (0) thc presslIre gradient. (b) the average velocity. and (e) the Reynolds number of the flow. I/\ n<. (a)- 64746 Nitn 1 per m. (b) 3.678m/s. (e) 882.721 4 . Detemline (II) the pressure gradiem. (b) the shear stress at the two horizontal parallel plates and (c) the discharge per metre width for the laminar flow of oil with a maximum "elocity of 1.5 m/s between two hor;~omal

• parallel "xed plates which arc 80 mm apart. T ake vi.cosity of oil as 1.961,Ns. lIl'

IAn•• (a) - 3678. 7 N/m' per m. (b) 147.15 N/m'. (c) .08 m'lsl 5.. Water is flowing between two large parallel plates which arc 2.0 mm apart. o,,!enninc : {oj maximum vciocity, (bj the pressure drop per unit lenglh and (c) Ihe shear stress a! walls of the piate if the awrage velocity is 0.4 m/s. Take viscosity of water a.< O.QI poi~. IAn s. (a) 0 .6 m/s. (b) 1199.7 N/ttt l per m. (e) 1.199 N/m! 1 6. There is a horizontal crac~ 50 tttm wide and 3 mm deep in a wall of thidness 150 mm. Water leaks through the crack. find the rate of leakage of water through the crack if the difference of pressure between the two ends of the crac~ is 245.25 N/ml. Take the viscosity of water 3> 0.01 poi.,e. [An.•. 183.9 em'lsl

I I

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IL Visco us Flow

431 1

7. A shaft h""in£ ~ diameter of 10 em ro(ate<; centrally in ajoumal bearing ha ving a diameter of 10,02 em and length 20 em. The annular space belw~n the shaft and the bearing is fillexJ with oil having viscosity of 0.8 poise. Dclcnninc the power absorbed in the t.earing when the speed of rotation is 500 r.p.m. [An s. 343.6 WI K. A shaft 150 mill diameter runs in a hearing of length 300 mill, with a radial clearance of 0.04 mm m 40 [ ,p_m. Find the viscosity of the oi l . if the power l"C'quired to o'-ercomc the viscous resistance is 220.725 W . jAn s. 6.32 poise[ 9. Find the torque required to rotalc a ...,"ical shaft of diameter 8 em at 800 r.p.m. The lower end of the shaft rcs!s in a fool -.tep bearing. The end of the shaft and surface of the bearing are both fin! and are sCp.:1r:llcd by an oil film of thickness 0.075 em . The viscosity of the oil is given as 1.2 poise. [Ans . 0.0538 Nrnl Ill . A wllar bearing having external ~nd intern:,l dimnclers 20 em and to ~'TTI respectively is used to take Ihc thrust o f a shafl. An oil film of thickness 0.03 em is maintained between Ihe collar surface ~nd the bearing. Find the power lost in overcoming the viscous resi.tance when the shaft rolates Ul 250 r.p.m. T~ke II s 0.9 poi,.,. IAns. 30.165 WI II . WateT is flowing through a ISO rmn diameler pipe with a co-efficienl of Frictionj ...05. The shear Slress al a point 40 mm from the pipe wall is 0.01962 Nkm l. Calculate the shear Slress allhe pipe wall. IAns. 0.0419 8 Nlem11 12. An oil dash-pot consisls of a piston moving in a cylinder having oil. The pislon falls wilh unifonn spe~'"d and wve" 4.5 em in go seconds. If an addilional weight of 1.5 N is plated On the top of the piston . it failS through 4.5 em in 70 seconds wilh uniform S]><.~d. The diameler of the piston is 1O em and ils length is 15 cm. The clearance between the pi,ton and the cylinder is 0.15 cm. which is unironn throughout. I'ind the viscosity of oil. IAns. 0.177 poisel 13 . The viscosity of oil of sp. gr. 08 is measured by a capillary lUbe of diameter 40 mm. The difference of pressure head between two poinls 1.5 m apart is 0.3 m of water. The maSS of oil ~'Olleeled in a measuring tank is40 kg in 120 seconds. Find the viscosity of the oi l. [Ans. 2.36 poisel 14. A capillary tube of diameter 4 mm and length 150 mm is used for measuring viscosity of a liquid. The difference of pressure between Ihe two ends of the lUbe is 0.7848 N/cm 1 and Ihe "iscosity of Ihe liquid is 0.2 poise. Find the rate of flow of liquid through the tube. IA lls. 16.43 conJlsl 15. A sphere of diametu 3 mm falls 100 mm in 1.5 seconds in a viseous liquid. The density of the sphere is 7000 kg/m' and of liqyid is 800 k£hnJ. Find Ihe eo-efficient of viS<.'Osity of Ihe liquid. [Ans. 45 .61 poise] 16. The viscosity of a liquid is determined by rotming cylinder method. in which case the inner ,"ylinder of diameter 25 ern is slalionary. The outer cylinder of diameter 25.5 em comains the liquid UplO a height of 40 em . The clearance at the bouom of the two cylinders is 0.6 cm. The ouler cylinder i, rotated at 300 r.p.m. The torque regislered on Ihe torsion melre attached to Ihe inner cyli nder is 4. 905 Nm. Find Ihe IAII • . .77 poiscl viscosity of liquid 17. Calculate: (a) the pres.ure gradient along the flow. (1)) the average ,·elocity. and (el the discharge for an oil of viscosity 0.02 Nslm l flowing between IWO stalionary parallel piates I m wide mainlained 10 mm apart. The "e\ocity midway between the plates is 2.5 ml,. [,\ns. (a) - 4000 Nlrn2 per m. (b) 1.667 m/s. (el .01667 m'lsl 18 . Calculale: (I) the pressure gradienl along the flow. (ii) the average velocity. und (iii) the discharge for an oil of viscosity 0.03 N slm~ /lowing between two stationary plates which arc par~lIc1 and arc at 10 mm ~p:1rt. Width of plates is 2 m. "The "elocity midway between the pl~tes is 2.0 mfs. 19. II cylinder of 100 mm diameter. 0.15 m length and weighing 10 N slides a~ially in a "enical pipe of 104 mm dia. If the space belween cylinder surface and pipe wall is filled with liquid of viscosily II and the cylinder slides downwards at a velocity of 0.45 m/, . detennine II . [Hlnt. /J ~ 100 111m _ 0. 1. L _ 0.15 m. LV _ 10 N. Dp _ 1.4 111m _ 0.104 m. V = 0 .45 mi •. Hence t = (0.104 - 0.1)12 = 0.002 m.

I I

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IL 1432 Fluid Mechanics p20 . II liquid is pumped through a 15 em diameter and 300 111 long pipe al the rale o f 20 tonne5 pcr hour. The density of liquid is 910 k glm) and "incm~lic viscosi ty ~ 0.002 m )!s . Dclennine the po" 'cr required and show thaI the flow is viscous . [Hi nt. D '" 15 em _0.15 Ill, L '" 300 m. IV", 20 tormeslhr .. 20 x 1000 kgf/60 x 60 sec • 5.555 kgOsec • 5.555 x 9.81 Nts .

8C' '" 0.0061 111 , Is. Q '" ~= ~5~.55~5~'~9~. , P8 910 )( 9 &1

v_ Q = II

0.0061 ~ (.l51)

~ O.l45 m/s. ,. _ O'()()2 m l!s.

R = pVD

Now

'

VxD

0 .345 x 0.15

,

0.002

P

= 25.87

which is less than 2000. Hence flow is viscous

I"" 32 I-ILVlpgD 1• where v ..

.e. :. !!" v)( p .. 0.002 x 910 .. 1.82 P

•

Hence.

f

32 x 1.82 x 300 x 0.345

•

(91O X 9.8 1 x 0.15 1)

•

30

I' '" pg.Q.hJ'IOOO '" 9 10 x 9 .81 x 0.0061 x 30!l{)()() '" I.6JJ kW. 21. An oil of specifIC gravity 0.9 and viscosity 10 poise is flowing through a pipe of diamdcr 110 mrn , The velocity at the centre is 2 m/s. find: (I) pressure gradie nt in the direction of now. (il) shear mess at the pipe wall: (iii) Reynolds number. and (il') vel ocily al a di,tance of 30 mm from the wall. IUln t.p =900 kglmJ; IJ = 10 poise", IN s/m 2 ; 0= 110 mm =0.11 m,

U.... ~ 2 m/s; ii

(;) (-tlP)= ,Ix (ii)

~ I mi• . V""" ~ - '-

41J

41J X V m ... Rl

( iii ) R =

'

2

pxiixO IJ

"'

tlx

R'

4xlx2 , 0.0551 = 2644.6 Niln

_ (-- dPx ) -Rz 2644 .x 6 -0.055 ' otlx

( - 'p )

2

7' 7' NIm'.

=~.~

9OQ xlx O.ll .,99 ' and I .

r)

" = _,_ (- dP ) (If _ = - ' - (2 644, 6) (0.055 2 _ 0.025 2) = 1.586 m/s.1 4)1
....

I tip, " - - -I' 81J dx

I tip (Ii) t o. - - 2 fix

X ,. -

dp

-

I

2

N shIll

-8).1 U"", -8x2.Sx2 = _1777.77NJml. 11 O.lS! I

(- 1777.77) x 0.15 _ 133.33 N/m .

(iii ) Q = Mean velocity X Area.,

I I

-

dx -

(Delhi VII""",-ily. Drcember 20(1)

= 2.S

.... )

(~V

X (/ X I) '"

(~x

2)

X (0.15 X I)" 0.2 TIl lIs. I

Ii

CRM'J'RK

.. 10. 1

INTRODUCTION

The laminar now has been discussed in chapter 9. In laminar flow the fluid panicles move along straight parallel palh in layers or lam in ae. such that the paths of individual fluid panicles do nO! cross those of neighbouring panicles. Laminar flow is possible only at low velocities and when the fluid is highly viscous. But when the velocity is increased or nuid is less viscous. the fluid panicles do nO! move in straight paths. The fluid panicles move in random manner resuhing in genera l mixing of the part icles. This type of flow is called IlImu Jcm flow. 1\ laminar flow changes!O turbulent flow when (I) veloc ity is increased Of UI) diameter of a pipe is inc reased or (iii) the vi_'\Cosily of fluid is decrea_"",d. O. Reynold was fiut to demon~trate that the transition from laminar to IlImuJcnt depends nOt on ly on the mean velocity but on the quantity pVD . This quanlity p m is a

dimen~ionlessquantity

and is called

Reynold~ number (R, ).

" In ca.<;£! of circular

"

pip'" if R, < 2000 the now is said to be laminar and if R, > 4000. the flow is said to be turbule nt. If

H, lies betwccn 2000 10 4000. the flow changes from la minar 10 tu rbulen t .

.. 10. 2

REYNOLDS EXPERIMENT

pV xd The type of flow is ddermined from Ihe Reynolds number i.e" ~~~. This was demonstrated by

O. Reynold in 1883. His apparatus is

~hown

in Fig. 10.1.

"

Fig. 10.1 R ~nQ/d apparatus.

433

I I

Ii

~ I

IL 1434

Fluid Mechanics

The apparatus consists of : ( i) A tank co ntaining water at constant head.

(il) A s mail tank containing so me dye. (iii) A g lass tube ha ving a be ll -mouthed en tr~ncc at one end and a reg ulatin g va lu e at other ends. The water from the tank was allowed 10 flow through the glass tube. The velocity of flow was varied by [he regulating valve. A liq uid dye having sa llI e spo:ciflc weig ht as waler was ill lroduccd into the g lass lubc as shown in Fig. 10.1. D" The following observati o ns were made by Re yno ld: (i) Wh en tlie veloci ty o f flow was low. llie dye filamen t in the g lass tube was in Ihe (onn of a s trai ght line. This straight li ne uf dye filame nt was parallel to the ~ WAVY g lass tube. whkh was the case of laminar fl ow as show n r~---:C--=--= 7"l FILAMENT in Fig. 10.2 «(I). (i/) With the increase of velocity of flow, the dyeTrans~ion filament was no longer a strai ght-line but it became a W3VY o ne as sho wn in Fig. 10.2 (b) . This sho ws that flow is no longer laminar. (e) Tu.wlent flow (iil) With funh e r incre ase of ve loci ty o f flow. the wavy dye- filam e m broke-up and finall y diffu sed in Fig. 10.2 Diffe rent stages of filame nt. wat er as Shown in Fig. 10.2 (c). Th is mea ns that the

[r;;=;;;=.;=====~;' r ~'3IIlAMENT (b)

•

mFDIIlFAFUSED MENT

F ~ ~<~?,;*J

fluid panicles of the d ye at this hi ghe r veloc ity are mo vin g in random fashion. which shows th e case of turbulent fl ow. Thus in c ase of turbulent fl ow th ~ mixing of dye- filame nt and water is intense and flow is irregular, random and disorderly. [n case o f laminar fl ow. th e loss of pressure head was found to b.: proponiona l to the ve locity but in case of turbulent flow. Reynol d o bserved that loss of he ad is approximate ly proporti ona l to the square of veloc ity. More exactly th e loss of head. 111 "" V · , where It varies from 1.75 to 2.0

to 10, l

FRICTIONAL LOSS IN PIPE flOW

Whc n a liquid is flowing throu gh a pipe, the veloc ity of th e liqu id layer adjaccm to th e pipe wall is zero. The ve loc ity of liq uid goes on in creas in g from the wall and thu s ve[ocit y gradiem and hence shea r Stre~s arc prodm;ed in th e whol e liquid due to vi><:osit y. This viscous action c au ses loss of energy whkh is usually kn own as fric tionn[ loss. On the bas is of his ex perime nts, Wi[lium Froude gave the following la ws of fluid fraction for turbulent flow. The frictional resistance for turhulent fl ow is : (i) proportional to 11". whl,re " var ies from 1.5 to 2.0. (ii) proportion al to thc de ns it y of fluid. (iii) proport ion al to the area o f s urface in cOntact. (i,') independe nt of pressure. (v) depende nt on the nature of th e su rface in contact.

10.l . 1 Expression for Loss of Head Due to Friction in Pipes, Co nsider a uniform hori w mal pipe. havin g stead y fl ow as shown in Fig. 10 .3. Let I - I and 2-2 arc tWO sc<:lio ns of pi pe. Let (It" pressure inte nsity 31 section I-I. V, " ve locit y of flow at secti o n I - I,

~ I

I~

~ I

IL Turbulent Flow

435 1

L = length of the pipe between sections I-I and 2-2.

d = di;ullclc r of pipe.

t' = frictional

resistance per Ulli! wetted area per unit ve loc ity.

Il, " loss of head due 10 friction. and Pl' V! = are va lues of pressure intensity and veloci ty at section 2-2.

--

T,

"

-- "-1-, ------

Fig. 10.3

Uniform horizonr«l pipe.

Apply ing Bernoulli's eq uation s between sect io ll s 1- 1 and 2-2. Total head 31 1- 1 '" Total head at 2-2

+ Loss of head due \0 friction between 1- 1 and 2-2

"'

~+ -'- +Zl = 1'1+V/+ Z2+I! pg2g ! pg 2g Z, "" Z2 as pipe is horizontal

VI'" V 2 as dill. of pipe is same at I I and 2-2

~ '" Pl + II or " = ~ _ 112 pg

pg

f

f

...(i)

pgpg

But lit iS the head lost due \0 frictioll and hence inte ns it y of pressure wil l be reduced in the direc tion of flow by frictional rcsi~tancc. Now frictio nal resistance == fri cti on,1i rcs iswnce per unit wetted area per unit ve locit y x wetted area x velocit/ or F 1 ==l'xrr.dLxV 1 [-: wettedarea=rr.dxL, vc locit y= V== VI = V![ ==!,XPXLXV2 [. ltd = Perimeter = PI ... (ii) The forces acting on th e fluid between sections 1· 1 and 2·2 arc: 1. pressure force at section 1· 1 = PI x A where A = Area o f pipe 2. pressure force at sect ion 2·2 = P! x A ]. frictional force FI as shown in Fig. 10.]_ Resolving all forces in the hori zontal direction. we have PI A

- p0 -

... ( IO. J)

F I=O

(P I - Pl)A

= FI =1'

PI - P2 =

2

x Px L X V

r·

2

From (iil. FI =I'PLV [

I'XPXLXV 1 A

Btu from equat ion (il. PI - 112 = pgli,

I I

Ii

~ I

IL 1436 Fluid Mechanics Equating th e va lue of (PI - P2)' we ge l ,'X P XLXV A

P8 11f = Ilf '"

f'

p

- xpg A

X

P

Wetted perimeter = ~=",c:==c A Area

2

, L x V-

... (iii)

--= -

In equation (iii). -

f' 4 2 f' 4 L V! " r = - X - XLXV = - x - pg d pg d Puning

L= f, wherefis know n as co-effi cient of fricti o n. p

2

Equation (ii') . becomes as

II = _4 ·_f

,

_LV_2 = c4,; f Cc ·L 'o."V_'

28

d

... (1 0.2)

dx 2g

Equ ation (10.2) is know n as Darc y- W c isbad\ cquatio ll. This equation is commonly used for findin g loss of head d ue to friction in pipes. Someti mes equati on (10. 2) is wrintl! as

' f=

j .L.V

1

...(10 .2A)

dx2g

Then / is know n as fric tion facto r.

10.3.2

Expression for Co-efficient of friction in Terms of Shear Stress. The equation (10.1 )

g ives the forces acting on a fluid be tw een sectio ns 1- 1 and 2-2 o f r ig. 10.3 in ho ri zontal diredion as PIA -

pz'l -

FI

(PI - pzli\

=0

= FI = Force due 10 shear siress t o = shcar stress x surface arca =tox ndx L

Cancelling nd frolll both s ides. we hav e d (PI -P 2l 4" = l O X L (PI-Pi)=

Equation (10. 2) can be written

~ I

4to X L d

a;;

II{ =

... (1 0.3)

PI - P2 __ 4/ . L. V

~cc:~

pg

1

dx2g

I~

~ I

IL Turbulent Flow (PI - P2l=

4j.L.V'" xpg d x2g

437 1

... (10 .4)

Equating [he value of (PI - fl 2l in equations (10.3) and (10.4), 41: o XL 4/.L. V 'd '" dx2g xpg

t o :fV1 xpg 2g pv!

0'

TO=[ -

jV 1 xpg 2g ... (10.5)

2

J= 2t o,.

. ..(1 0.6)

pV·

... 10.4

SHEAR STRESS IN TURBULENT flOW

The sllenr stress in viscous flow is given oy Newton ' s law of viscosity as

d,

where t " '" shear stress due to viscosity.

t,. '" j.I - , tly

Similar to the expression for viscous slieaT. J. Boussincsq expressed the tu rbulent shear ill math ematical fonn as
... (10.7)

t , = shea r stress due to turbulence '1 = eddy viscosi ty Ii = average veloc ity at a dislunce y from boundary. The ratio of '1 (edd y viscosity) and p (ma ss den sity) is known as kinematic eddy viscosity :md is denOlcd by {: (c psiloll). M alhemal icaliy it is wriucn as

where

..!l

... ( IO.S) p If the shear stress due 10 viscous now is al so considered. then the IOlal shear stress becomes as £ '"

du t"'t ,. +t,=Il - " dy

du dy

... (10.9)

The value of T] = 0 for laminar now . For other 1'1ISCS tbe value of T] may be scveral thousand times the value of ).I. To find shear stress in turbulent flow , ~"
I I

Ii

~ I

IL 1438 Fluid Mechanics ...(1 0 . 11 ) The turbuicnt shear stress given by equation (10.11) is known as Reynold stress.

10.4 .2 Pr;lndtl Mixing Length Theory for Turbulent Shear Stress . In equation (10. 11). the turbulent shear Slresscan only he calculated if the value of 11'1" is known. But it is very difficult to measure ,i'v'. To overcome Ihis difficulty, L Prandtl in 1925. presented a millin~ length hYP01hcsis which can be uscd 10 express turbulent shear stress in Icons of measurable quamilics. According to PraHdtl, the mix ing length I. is lhal distance bo:lwccn two layers ill the transverse direction sucli lhm the lumps of fluid particles from OnC layer could reach lhe other layer and the

particles are mixed in lhe other layer in such a way that lhe momentum o f the particles in the direction of.( is same. Hc also assumed that thc velocity fluctuation in th e .(-direction ,,' is related (0 the mixing lengt h I as II'

=1 du dy

and I"~ • the fluctuation componenl of velocity in y-dircction is of the same order of magnitude as ,,' and hence ,,' =1 -d" II)"

Now u'x,( becomes as

II',.'

=

(I IIU]X (I dU]=/l (d"dy ]' d)' tly

Substituting the value of II' ,,' in equation (10.11). we get the ex pression for shear stress in turbulent flow due 10 Prandtl as

( ]'

_T = pl-,d" -

... { 10.12)

dy

Thus the total shcar stress at any point in turbulent flow is the sum o f shear stress due to viscous shear and turbulent shear and can be written as

-T=].I -+p <1" /,(<1"]' 'd)'

... ( 10. 13)

d)'

But the vis<:uus shear stress is negligible except ncar the boundary. Eq uatiun (10.13) is used fur must of turbulent fluid flow problems for dctennining shc~r stress in turbulent flow.

to 10 .S VELOCITY DISTRIBUTION IN TURBULENT FLOW IN PIPES In case of turbulent flow, the (Otal shear stress at .my point is the sum uf viscuus shear suess and turbulent shear stress. Also the viscous shear stress is negligible cxcept ncar the boundary. Hence it can be assumed that the shear stress in turbulent flow is given by equa lion ( 10.12). From this equation. the ve locit y distribution can be obtained if the relation between I, the mixing length and ), is ~nown. Prandtl assumed that the mixing length. I is a linear functiun of the di stance y from the pipe wall i.e.. 1= ky. where k is a t"Onstalll. knuwn as Kannan consta nt and = 0.4. Substituting the value of I in equntion (10.12), we get

I I

Ii

~ I

IL Turbulent Flow

439 1

d" ' ( dy J

_ , t o rt :p x (ky),x

...( 10. 14)

For small values of y that is very close to th e boundary of Ihe pipe, Pral1 (.hl assum ed shear stress T to ~ COnstant and app rox imakl y equal to t o wlli ch prese nts Ihe turbulen t shear Slress at the pip.: boundary. Substitu ti ng t "" t o in equ ati on ( 10 . 14), we get dd"Y

In c"uation ( 10.1 5), 'i

fi

=

,'yVp ~ p"

...( 10. 15)

fi

!IIL lr l ML '

pO ha s th e dim ensio ns

~Ll _ _L

L r ) - T " BUI -T is ve locit y and he nce

has th e d i mension of ve lo ci ty. whi c h i s kno wn as shear ve loc ity and is deno ted by" •.

fi n

I

dll

'

= u•. th en cqumlOn ( 10. 15) !J;.,comcs - = - II •. P dy ky For a give n case of lurhul cnl now. u. is conSlant. Hence integ rating ahiwc equ ati on. we gel

Thu s

-

",

... (10. 16)

u =T1og,.y + C

where C = co nstant o f integra tio n. Eq uati o n ( 10 .1 6) shows [hat in lurbule nt flow. lhe vel oc it y varies di reclly with lhe log arilhm of lhe di slance fro m lhe boundary o r in other words the veloc ity distributio n in turbulent fl ow is log arithmic in natu re . To determin e the co nstant of integ ration. C the boundary condition that at)' '" R (rad ius o f pipe ). U '" II """ is s ubstituted in equat io n ([ 0. [6). lin,", ""

k",

",

,

log, R + C

C=U m,,-, - - log R

'

Substitutin g the va[u e o f C in equ ati on ( 10 . [6). we get II.

II '"

k

[og, ), +

"" "mlU + -0",4 '" limo,

+ 2.5

u""" -

II .

k

log, R "" II,,,,,, +

log, (yI R) II .

log, (y IR)

II.

k

(log, ), - log, R)

[ .: t. = 0.4 "" Karm an con, tan t] ... (10. 17)

Equ ati o n ( 10. 17) is ca ll ed ' Prand tr s un il'cT.'ia l vcloc ity di stri but ion equati o n for turbul ent flow in p ipes. Thi s equati on is appli cabl e to smooth as we ll as roug h pipe boundari es. Equation ( 10. 17) is al so wrine n as

I I

Ii

~ I

IL 1440

Fluid Mechanics

"m., -

Dividing by

u., we gel

U '" -

2.5 u. log, (yIR) '" 2.5 u. log, (Rly)

'" 2.5 log.. (Rly) '" 2.5 x 2.3 Iog lo (Rly) (.: log.. (RIy) '" 2.3 Iog lo (IVy)] " .."" - u '" 5.75 10gIO (Rly)

.. .( 10. 18)

,.

In equation (10. 18), the difference b<:lwccn the mal
IOCil1

velocity

1/

al any

10.5 . 1 Hydrodynamiully Smooth and Rough Boundaries.

Let k is Ihe average height of the irregularities projec ting from the surface o f a boundary as shown in Fig. lOA. If Ih<: value o f k is IM!;c for a boundary then the boundary is called roug h ooulldary and if the value of k is less, Ihen boundary is known as smooth boundary. in general. This is the classilkalioll of rough and smoot h boundary based on boundary \.Characteristics. BUI for proper ciassifkation, the flow and fluid daracteristics arc also to be co nsidered. LAM INAR SUBLAYER

T ---------- - - -- -

'L ~ ' (a) Smooth bounda
Fig. 10.4

LAMI NAR SUBLAYER

II

J _ " i~

t (b) Rough

___ _

bounda
Smootb and rougb boundariel.

For turbulent flow analysis along a boundary, the flow is divided in two runions. The first rurtion consists of a thin layer of fluid in th e itnm",diate ndghbourhood of the boundary. where viscous shear stress predominates whi l", the shear st ress due to turbulence is negligihle. This run ion is known as laminar sub- la yer. The height upto which the effect of viscosity predominates in this zone is denot~d by /)'. T he second portion of flow. where shear stress due 10 turbulence arc large as compared to viscous stress is ~nown as turbulent zone. If the average height k of the irregularities. projecting from the surface of a boundary is much less than 0'. the thic kness of laminar sub -layer as shown in Fig. 10.4 (a). the boundary is "ailed smoot h boundary. This is because. outside the laminar sub-layer the flow is turbulent a nd eddies of various size present in turbulent flow try to penetrate the lamimlr sub-layer and reach the su rfac e of Ihe boundary. BUI due to greal thickn ess o f lamin ar sub-laye r Ihe eddies arc unable to reach the s urfa"e irregularities and hence the boundary behaves as a smoot h boundary. This type of boundary is called hydrodyn:llni"ally smoot h boundary. Now, if the Reynolds number of the flow is increased the n Ihe Ihidnes~ of laminar sub-la yer will d"'creasc. If Iht: thiCkn ess of laminar suh-Iayer bcco ln"'s much sl)1all er Ihan the avcragt: height k of irregularities of Ih e surface as show n in Fig. 10.4 (h). the boundary wi ll aCI as rough boundary. This is because the irr"'gu larities of Ihe surface arc above th '" laminar sub- laye r and the eddies prt'scnt in lurbulent zone will come in contact with th'" irreg ularities of the surface and lot of cnergy will be 1051. Such a boundary is ca ll ed hydrodynamically rough bou ndary . From Nikuradsc's experim ent : k k I. If /)' is less than 0.25 or S' < 0.25. the boundary iscallcd smooth bounda ry.

I I

Ii

~ I

IL Turbulent Flow

4411

2. If :' is greater (han 6.0, the bounda ry is rough.

3. If 0.25 < ( ;, ) < 6.0. the boundary is in transition. In terms of roughness Reynolds number -"k ,

,

h I. If -lI,k, < . boun d ary"IS l'Onsldcrcd smoot.

"

Ilk

,

2. If - '- lies betwee n 4 and 100. boundary is in transition stage. and

3. If lI.k > 100. the boundary is rough.

10.5.2

"

Velocity Distribution for Turbulent flow in Smooth Pipes . The velocity distribution for turbulem now in smoOl h urruugh pipe is given by cqumion (10.16) as 11=

"

- Iog,y + C k

It may be sccnlhal at )' '' 0, the velocity /I at wall is - ... This means that velocity II is positive at some distalll:e far away from the wall and-"" (min us infinity);n the wall. Hc",:e at >01Il1' finite distance from wall, the ve locity w ill be equal lO zero. Lei (his distance from pipe wall is y'. Now the eonstanl C is determined from the boundary condition i.e .• at y '" y' . II '" O. Hence above equation becomes as

.. ,

O= T log,

f

11"

+corc=-Tlog,y

Substituting the value of C in the above equation. we get II

=

II . T

log, y -

II. T

, I I.

log, Y =

T

,

log, (yl)')

Substituting the val ue of" '" 0.4, we get II

= ~ log, (yly') = 2.5 0.4

.!!.... '" 2.5 x 2.3 Jog 1o (yly') ,

.

"'

.!!.... = 5.75 logl o (yly')

",

fl.

log, (yly')

I': log, (yly') '" 2.3 loglo (yly'») ... ( 10 . 19)

For thc smooth boundary. there cxists a laminar sub·layer as shown in Fig. 10.4 (a). The velocity distribution in the laminar sub·layer is parabolic in nature. Thus in the laminar sub-layer. logarithmic velocily d iMribulion does nOI hold good. Thus il can be a~sumcd (hal y' is proportional 10 0', whe re 0' is the lhickness of lam inar sub· layer. From Ni kurad sc's e)(P<"rimenl lhe va lue of / is give n a~

I I

Ii

~ I

IL 1442

Fluid Mechanics

where S' = I L6v . where

".

It

= kinemmic viscosity of fluid. ,

11.6v

J

u,

107

)' = - - x -

O.108v = --u.

SUbsti tut ing this value of )" io equation ( 10.19), we obtain

I~' '" 5.75 loglo [.]({sv] ". y '" 5.7510&10 (~) = 5.7510gJO (u. X9.259 ) .I OS\' \' II ·Y

'" 5.75 10&1 0 -

,

!l.y

"" 5.75 loglo -

10.S.3

,

+ 5.75 10&10 9.259

[.: 0.i08 =9.259]

+ 5.55

.. .( 10.20)

Velocity Distribution for Tu rbulent Flow in Rough Pipes .

In case of rough

bound aries. lite thic kness of laminar sub- layer is I'cry sma ll as shown in Fig. 10.4 (b). The surface irregularit ies arc abol'c the lam in ar sub- layer and hence the laminar s ub-layer is com p letely destroyed. Thus y' ca ll be considered proport io nal 10 th e Ilcighl of protru sions k. Nikuradse 's e~pcrimc l11 ~hows the va lu e o f y' for p ipes COaled wit h unifonn sand (rough p ipes) as y' '" -

k

30

.

Substituting this va lu e o f y' in equation ( 10. 19). we gct

~ :: 5.7510g lo 14.

[-y-):: I.:I?IJ

5.75 [loglO (yll.:) x 30]

'" 5.75 logl o ()"II.:) + 5.75 101:10 (30.0) '" 5.75 log 10 0-11.:) + 8.5 ... (10.21) Problem 10 .1 A pipe-lille carryillg water has orerage height of irregularilies projecling from Ihe sllrface of Ihe bOlilidory of Ille pipe as 0.15 mm. W/lal Iype of bOlllldary is il ? The slteM Slress dln'eloped is 4.9 Nlm'. Tile kill ematic riscosit)' of Wilier is .01 stol.:es. Solution. Givcn : Average height of irreg u larities. 1.::: 0. 15 111m '" 0.15 x 10- 3 m Sh~a r stress d~veloped. to'" 4.9 Nlm2 v'" 0.01 stok~s '" .01 cm1ls", .01 x 10-1 m 21s Kin~m:nic viscosity. [knsity of water. p '" WOO kg/J11 3 S h ~ar

I I

vdocity,

".:: Jr: o I p '" J[000 4.9 "" JO.0049 :: 0.07 tnls

Ii

~ I

IL Turbulent Flow u,k

= -

Roughness Reynold number

"

=

0.07 x 0.15 x IO-J

443 1

'" 10.5.

,

Since - '- lies between 4 and 100 and hence pipe s urface behaves as in transilio rl .

"

Problem 10.2

A rOI/gl! pip" is of diomeler 8.0 em. Tile \'('IQcily
~ '" 5.75 loglo (ylk) + 8.5, where k'" height of rough ness.

".

For a poilU. 1 elll from pipe wall. we have

,

-

".

= 5.75 log 10 (I.OIk) + 8.5

... ( i)

For a point. 3 ern from pipe wall. velocity is I,) " and hence

1.311 '" 5.75 log,o (3.0/k) + 8.5

... (ii)

".

3

Di viding (ii) by (i). we ge l 1, =

c',,"'~5c'O:"~'""(i3'cOc'-i'C)+",,"C ' 5.75 log ,o(J 1 k) + 8.5

or

1.3[5.75Iog lO (Ilk) + 8.5[ = 5.75Iog I0 0.0Ik) + 8.5

or

7.475 logl o ( I/k) + 11.05", 5.75 lug lO (3.0/k) + 8.5 7.475 log lO ( l /k) - 5.75 log 10 O/k) '" 8.5 - 11.05 = - 2.55

Of

or or

7.475 [log lo 1.0 -Ioglo k[ - 5.75 [Iogl o 3.0 - log lo k[ = - 2.55 7.475 [0 - loglo kI - 5.75 [.4 771 - 10); 10 kJ '" - 2.55

or

- 7.475 log lo k - 2.7433 + 5.75 log 10 k '" - 2.55

or

- 1.725 logl o k '" 2.7433 - 2.55 '" 0.1933

"'

0. 1933 log w k = - 1.725 = - 0.1120= 1.888 k", .772fi

COl.

An s.

Problem 10.3 A .\1/100111 pip" of diameter 80 mill atlll 800 //I Io/Ig carri,'.1 Wafer af II,e rale of 0,480 m J/lllinule. Calculate Ille 10.1$ of head. wall .Ihearillg .lIre.lS, celltre lill<' 1·e/OCily. ["elocil), alld she!" SlreSJ (1/ 30 mm from pipe wa/l. Also ca/Culmt' II,e II,icklless of lamillar .Illb-layer. Take kill emU/ic ["iscosit)' of water (jJ 0,015 stokes, Take the mlile of co-f'jficiellt offriclioll ,from Ihe re/alioll girt''' (IS

.0791 f= ( R,)'I~' where

I I

R, =

ReYliolds IIllmber.

Ii

~ I

IL 1444

Fluid Mechanics

Solut ion. Given: Dia. of smooth pipe. Leng th of p ipe.

L=800m

Discharge.

0048 Q =.

Kinematic vi!\Cosily.

Density of wata.

v :: .015 sto kes", .0 15 X 10-4 ml{s P'" 1000 kg/Ill )

Mean veloc ity.

V -_

d =RO nun =.ORm

In

3Iminute '" 0.48 60 = .0081031s

-.iL '" _° =8 "=.00 7 Area

It

(.08) 2

= 1.59 1 mls

4 R '" Vxd = 1.591 xO.08 :8.485 x 104

Rey nolds numbe r.

,

V

.015 x I0.....

As the Rey nolds number is more IhJIl 4000. th e flow is turbulent. Now th e va lu e of l' is give n by /=

.079 1

.079 1

------;{.I=

(8.485 x 10

R,

4

1/4

= ,004636

)

(i) Head lost is given by equ atio n (10.2) as

4. f. L. Vl 4 x .tXl4636 x 800 x 1.591 l II, '" d x 2g '" .08x2x9.8 1 : 23.42 m. Ans. (ii) Wall s hearing stress. to is give n by eq uat ion (10.5) as

to= (i ii ) Ce ntre- line ve locity.

U""'-,

7 '" Jj V !

lI. y

1/

,. = 5.75 login - ,

II.

=

is s henr veloci ty and

x 1.591 1 = 5.866 N/m l, An s.

for smoo th pipe is given by eq uation ( 10.20) as

-

where

21000

.004636 x

+ 5.55

~ = J~:

... ( i)

= 0.0765 ntis

The veloci ty will be maximum whe n ), = !!.- = .08 = .04 nt ,

2

Hence

31 )'

= .04

2

Substituting these values in (0, we get

In . II '" II ...."

1/ "",

.0765

'0;

= 5.75 logl a 0.0765)( ,0 15)(10

+ 5.55

= 5.75 logw 2040 + 5.55

(i\') The shear stress, !

"" 5.75)( 3 .309 + 5.55 = 19.03 + 5.55 = 24.58 """" = .0765 )( 24.58 = 1.88 mls. An s, at an y point is give n by

a" ax 2

~ 1=- - -

I I

... (A)

Ii

~ I

IL Turbulent Flow

445 1

where r = distance from ce mre of pipe and hence slica r stress at pipe wall where r = R is

up R to" - - 2 Dividing equation (A) by equa tion (IJ). we gct

... (8)

ax

, ,

'. Slicar stress

R

,. -"R '

A poin t 30 mm fmlll pipe wa ll is having r

= .0 1 Ill) =

t o x .01

=4 -

3 '" I ern

= .01

5.866

III

z

- - = 1.4665 Nfm . A ns • .'" 4 Velodly at a point ) em from pipe wa ll meanS y = 3 ern :: .03 III ! a1 (r

U

and is given by equat ion ( 10.20) as -

u y '" 5.75 loglo - ' - + 5.55, where u. = .0765. y:: .03

II,

V

u .0765 x .03 .0765 '" 5.75 log w .015 x 10 ..... + 5.55

:: 5.75 JOS 10 1530 + 5.55 = 23.86 0.0765 x 23.86:: 1.825 OIls. ADS. (v) Thickness of lamin ar s ub-layer is given by 1/::

1)'= 1J.6xv

11.6)(.0 15xlO--I =2.274xIO-lm .0765 1 :: 2.274 X 10- em = .02274 em. A ns . Prob lem 10 .4 De/ermine Ihe wall siJe(lTing J'ln:ss in (l pipe of diameter 100 111m which ~'lIrries ",,,Ier. The ,-e/OCilie;- al Ihe pipe cenlre and 30 mmfrom Ihe pipe ,'enlre are 2 ",Is ""'/ /.5 mls respeCI;'·e/y. Th e flow in pipe is gi"ell (IS lurbulent, Soluti o n. Given D", 100 mm '" 0. 10 m Dia. of pipe.

Radius,

R=

o~o

= 0.05 m

Veloc ity at ce ntre. " ..... '" 2 mls Velocity at 30 111m or 0.03 m from centre'" I.S m/s Velocity (at r = 0.03 m). II = 1.5 mls Let the wall shearin g stress =lu ['or tu rbulent now. th e ve locit y distrihution in terms of centre tine ve tocity ( un",,) is given by equa tion (to. t8) as

-::.' =.515 -"'".~u_ ,. loglQ ,. y

(K)

where

I I

II '"

[ .S IIl/s at y = (R - r) = 0.05 - 0.03 = .02 III

Ii

~ I

IL 1446

Fluid Mechanics 2.0 -1 5_ _5751 . og, & .05_")288 __ . ur 0.5_':>288 __ .

.02

u.

U . '"

Using the relation

u. '"

0.2185=

2~~8

II.

= 0.2185

mfs

FoTP. where p fur water'" 1000 kglm l Jl~or 1~ =O.2185~=O.0477

t o == 0.0477 x 1000", 47.676 Nfml, An s.

or

10.5.4

Velocity Distribution for Turbulent Flow in Terms of Average Velocity. The

average veloci ty U . through the pipe is obtained by first finding the 100ai discharge

Q and then dividing

the total disdJ.arge by the area of the pipe. ELEMENTARY C IRCULAR RING

/

- - I---FLOW

.

- r --

'

~

,,

A vtTagl' velocity for IUrb"lt'nf flow.

Fig. 10.5

Consider an elementary circular ring of rad ius 'r' and thickn ess dr as shown in Fig. 10.5. The distance ofille ring from pipe wal l is y '" (N - r), where R '" radius o f pipe. Then the discharge. dQ. through the ring is given by dQ = area of ring X veloci ty = 2rrrdr X II = II X 2rrrdr H

TOlal discharge.

Q= f dQ= f o

..

... ( I0.22)

x2rrrdr

(a) for sm oo th pip es. For smooth pipes. the velocity diSlribulion is given by cqu;uion ( 10.20) as

u

-

II.

"'

11=

Bm

u.),

= 5.75 Jog lo + 5.5 v

,.y 1xu. [5.75Jog lO --;-+5.5

)' =(R - r)

Substituting the va lu e o f " in equation ( 10.22). we get Q=

I I

r'[

Ja

5.75 1og lo , .( Hv ,)

1

+ 5.5 II. X 2tt rdr

Ii

~ I

IL Turbulent Flow

447 1

Average veloc ity. '"

~ ( It [5.75 Jog lo11, ( R rrR

Ja

r) +

v

5.5]11. 2Tt rdr

Integ ration of the above eq uation and subscqucllt simp lification g ives th e ave rage veloci ty for turbu Jcm flow in smoo th pipes as U II . R - '" 5.75 10lll(l + 1.75 II. v

...( 10 .23)

(b ) For ro u gh pip"s. For rou gh pipes. t he vdoc ily at any poinl in lurbulenl fl ow is gi ve n by equation (10.2 1) as

" '" 5.75 101110 (y /t) + 8.5 ".

-

"m

y=(R - r)

" '" 5.75 loglo (H -o,) - + 8.5

-

k

u,

"'

Average vcioc it y,

u- = - Q =

1tR ~

'R'

Int eg rati on o f the abo ve eq ualion and sub.""qu ell[ simplifi cat io n wi ll g ive Ih e following relation for 'lVcragc velocity.

fJ

for turbul ent flow in rough pipe as

fJ

-

".

R '" 5.75 10gIO - + 4.75

,

... ( 10.24 )

(el Dirrc~nce of the \' ~Iodty at any point and an' rage ve locity for s mooth and rough Ilipes. The veloc it y at any poin t for turbul ent n ow for s mooth pipes is g ive n by equat io n ( 10.20) as

u

u.(R - r) +5.5 II. v and Ihe average vdocity is given by equation (10.23) as - ~5.75Jogl o

U II.

~

u. R 5.75 Jog lo + 1 75 I'

Differe nce of ve loc ily u and U for

I I

[·: y=R - r [

~1ll00111

pip.: is oblained

a~

Ii

~ I

IL 1448

Fluid Mechanics

, -iJ : -

[ 5.75 10g lO

,.IN-')+ 5.5] - [5.75 101,: .0-,.R+ 1.751 v

II.U,

II-V [ U.(R-r) - - '" 5.75 10g.o:::CC--'"

0'

,

'.

:5 .75 10g lO

[

v

,.N]+

log lO~

5.5 - 1.75

,.R]

,.IRv- ,) + ~

+3.75

') + 3.75 R ,- '" 5.75 login ( '" 5.75 101:'0 (yIR) + 3.75

...( 10.25) 1'; R-r= y ]

S imi larl y the veloc ity. II at any poillt for rough pipe is g ive n by equati on ( 10.2 1) as

,

-

,. '" 5.75 log 10 {ylk ) + 8.5

and ave rage ve locity is g ive n by equat io n (10.24) as

iJ

-

,. '" 5.75 logwCRfk) + 4.75

Diffcrc llcc of ve loc ity II and U for ro ugh pipe is given by

-

,

iJ

- -

II.

'" [5.75 101:'0 (ylk) + 8.5] - [5.75 log 10 (Rlk) + 4.75 1

II.

'" 5.75 101:)0 [(ylk ) + (Rlk)] + 8.5 - 4.75

,-u

- - '" 5.75 log 10 CyfR) + 3.75

,.

...( I 0.26)

Equ ati ons (10.25) and (10.26) arc the Same. This shows that Ih e difference of ve locit y al any point and th e average velocity will be the sam" in case of smoo th as we ll as ro ugh pipes. Problem 10.5 Detert/lint' IIIe diJ'/(/lice frolll Ille pipe ....all al .... hicll Ille local ("dodl)' is equal to Ihe al"erage I"e/adl)' for turbulent fla .... in pipes. Solution. Given: Local ve locity a l a point = ave rage velocilY

u= U For a smooth or rough pipe. Ihe di fference o f ve locit y at any point a nd average I'elocity is gil'e n hy equa li on ( 10.25) o r eq uation (10.26) as II-U

- - = 5.75 loglo (yIR) + 3.75

,.

Substit utin g th e given co nditi on i.e .. u =

fJ -

fJ . we

gel

U

- - = 0 = 5.75 logl o (yIR) + 3.75 or 5.75 IOIlIO (yIR) = 3.75

,.

~ I

I~

~ I

IL Turbulent Flow logl o (yfR)= - 3.75 = - 0.6521 5.75

"'

449 1

=_ 1.3479

ylR", 0.22279 "" 0.2228 or y '" .2228 R. Ans. Problem 10.6 For /Il,hu/,,,,/ flo", ill II pipe of dium eter 300 III"'. Jim/ Ille di sellarge "'/1(:11 Ihe CIOn/reline ,-e/OCily iJ' ].0 ",Is 'lIId Ihe ,-e/Oeily 01 a poi,,/ 100 mm from Ihe centre
Velocity at cenlre,

03 Ro - :O.15m 2 u nWl '" 2.0 m/s

Velocity

(atr= IOOI111l1 =O. lm).u= 1.6mfs

Now

y '" R - r= 0.15 - 0.10 '" 0.05 m (a l r=O.1 10 or aty = 0.05 In)." = 1.6 OIls

Rad ius.

Velocity

The veloci ty in terms of cent re-li ne veloc ily is given by equation (10. 18) as

-"'.·,.,c-:.::' = 5.75 10&10 (Rly )

,.

~ Substilulin g Ihe va lu es. we gel 2.0 - 1.6 - 5751 . og lO II . .05

", YO ,05",] [ R =O.15rn

'" 5.75 logl o 3.0 = 2.7434

0.4

-

'. u.

= 2,7434

OA

= - - - '" 0.1 458 m/s

...(i)

2.74 34

Using equalion (10.26) whicll gives relat ion between veloci ty al any point and average vdocily, we have II - U

- - '" 5.75 log 10 (yIR) + 3.75

,.

y'" H, ve locity II becomes'" lin""

"'

-"'._••~---"Uc '" 5.75 loglo (RIR) + 3.75 '" 5.75 X 0 + 3.75 '" 3.75

,.

"0<

II ....... : 2.0 and u. from (I) '" 0.14511

2.0

- U '" 3.75

0.1 458

fJ : Di scharge.

I I

2.0 - .1458 x 3.75 '" 2.0 - 0.5467 '" 1.4533 m/s Q'" Area )( average veloc ity

Ii

~ I

IL 1450

Fluid Mechanics ==

10.5.5

'41t

~

_

D )( U

"'41t

J

1

(0.3) x 1.4533" 0.1027 m Is . '\n s.

Velocity Distribution for Turbulent Flow in Smooth Pipes by Power Law. The

ve loc ity distribution for tur bulent flow as give n by equations ([0. 18). (10.20) and ( 10.21) arc logarithmic in nature. T hese equatio ns arc nO! cOIlVcniclll 10 usc. Nikuradsc carried OUl experiments for difFerent Reynolds number 10 determine the vd oc it y dis tribu tion law i n smoot h pipes. H e cxprcsscd

the ve loc ity distribution in exponenti al form as

-"- = (yIR) lin

•.,( 10 .27)

" ~,

I where cx ponelll - depends on Reynol d s number

" T he val ue of (~) decreases. with incrca..;;ing Reynolds number. - =-

Por

"

6 I

- =-

I

I

"

7

"

7

/I

10

Thus if - = - . the vel oci ty distribution law becomes as

u:.

=

( ~yn

...(10.28)

Equiltion ( 10.28) is kflown as I n ih power law of velocity disnibution for s mooth pip.:s.

to 10. 6

RES ISTANCE OF SMOOTH AND ROUGH PIPES

The loss of h.-:ad. due to fri cti on in pipes is given by equation (10.2) as II = I

~4~.!~.~L~.~V_' rlx2g

In this equation. the val ue of co-"fficie nt of friction. f should h,:, known accurately for predi cting th.-: loss of head due to friction in pipes. On the basis of dimensional analysis. it can h,:, shown that the pressure Joss in a stra ight pipe of d iarne ta D , length L. roughness k. average velocity o f now fJ. viscosity and density o f nuid J.l and p is

L\P = p~2$ [R,,~.~]

or

p&l=Q [R,.~.~] 2

E~perirnenla lJy it was found that

~ I

pressure drop is a fUIll:tion of

~

to the first power and hence

I~

~ I

IL Turbulent Flow

pi;l

=

~ 9 [n,.~J

451 1

or

2 The te rm of lh e rig ht hand side is called co"c fficic nl offri cli on f. Thus/ = 1\1

[R" ~]

Th is equati on shows that friction co-e m de n! is a functio n of Rey no lds num ber a nd kID rati o . where k is the ave rage hei ght of pi pe wall roughn ess protrusio ns. (a) Variation of 'I' for Laminar Flow. In vi sco us flow chapte r. it is show n that co-effic ie nt of fric tion "f fo r la min ar fl ow in pipes is g ive n by

f= ~

... (10.29)

R,

Thus friction co-e ffi c ient is olll y a funcli OI) o f Rey no lds number in case of lam inar n ow. It is indepe ndent of (kiD ) rat io. (b) Variation uf l' fu r Turbul~nl Fl ow. For turhule nt flow . the co-dfi c ient of fri ct io n is a funct ion of R, and kID ratio. For rel at i ve rou ghness (kiD), in th e Itlrbulc nt !lo w Ihe boundary may be smooth or

rough and hence the va lue of J ' wi ll be diffcrCIII for thcse bound ari es. (,) '/' rur s muuth pip es. For turbul ent n ow in smooth pipes. co -e Ffici~nt o f frictio n is a func tion of Rey no lds num ber o nl y. The value of laminar sub- laye r in case of s mooth pipe is large as compared to the a vc ra ge hc ight of surface roughn ess k. The va lu c of 't' fo r s mooth pi pe for Rcy no ld s num be r varying from 4000 to [00000 is give n by the re lati on

/=

.0791

(R,t

...(10.30 )

4

The equati on (1 0 .30) is give n by Blasiu s. The valucof J ' for H, :> JO~ is obta ined from equati on ( [0. 23) whi c h gives til<' vel ocit y di stribu tion for smooth pipe in te nn s o f avc ra~e ve locity

fJ

-

!t.

From equat iOll (10. 6), we

(U)

as

("oR)

= 5.75 10g lo - -+ 1.75 v

have/= ~ , where pV'

...( [0 .3 1)

V= ave ra!!:", velocit y

... ( 10 .3 IA )

"' Su bstitutin g the valu e o f u. in equatio n ( 10.3 1). we get

fj

U J/ 12

I I

=5.75JOg 1o [ J /1 2 )R +t.75 v

Ii

~ I

IL Fluid Mechanics

~

vf Taking R

'" 5.75 10g( D

12

[fIRv ..JTT2) + 1.75

= on and simplifying, tile above equation is wriucn as I :2.03 Jog 1o r<-.:

v4!

But U D = R, and hence above

"

~

l~uillion

[UD - ,ffJ ) - 0.9 1 v

is writ1cn as

= 2.03 Jog lo (R,..[4j) - 0.91

, 4/

... (10.32)

Equ ation (10.32) is valid UplO R, = 4 x 1O~ Nikuradse's experimental resuit for t UTbtlk"! now in SlllOOlh pipe for 'ris

I

fA7 '"

,4 /

2.0 ]ogl o (R,

J41) - 0.8

... (10.33)

7 This is applicahk uPIO R, = 4 X 10 . But the equation ( 10.33) is solved by hit and tria l method. The val ue of'j ' (i.e .. co -c fficicnl of friction) can alternate ly be obtained as

000 /=.8+

.05525

... (10.34)

(N,) 0'\1 .•.

The valueuf '1'Ii.e., fril:tion factor which is used in cqualion (ID.2A)] is given by

0.22 1

... ( I O.3·M )

/= 0.0032 + (R,t2l7

(ii) Va lu e of 1 ' fo r rough !liIJes. ror turbulent flow in rough pipes. the co-efficielll of frict ion is ,\ fu nction of rclmive roughness (kiD) and it is independent o f Reynolds number. This is because the val ue of laminar sub-layer for rough pipes is very s mall as compared to the heig ht of sutface roug hness. The average velocity for rough pipes is give n by (10.24) as

U

-

= 5.75 log lO (Rlk) + 4.75

",

8m

Substituting the val ue o f

Ij.

in the above equa tion, we get

U ~

= 5.75 Jog 1o (Rlk) + 4.75

U ,, /12 which is simplified But Nik uT
10

the fontt as

e~perintental

I r;;-;

, 4/

I

~

, 4/

= 2.03

10g lO

(Rlk)

+ 1.68

result gave for rough pipe the following

=2 tog lO (Rlk) + 1.74

... ( I 0.35) rel~t ion

for '1 '

~s

... (10.36)

(e) Va lu e of 'f ' fo r co mm ercial pi lles. The \'~Iue of '/' for commercial pipes such as pipes made of metal. connc tc and wood is obtained from NikuT
I I

Ii

~ I

IL Turbulent Flow

453 1

pipes. Accordi ng to Colebrook. by subtract in g 2 log,o (Rlk) fro m both sides of eq uati o ns (1 0.33) and (10.36). th e value of -I' is o btain ed fo r cO l11l11 ercia l s moo th a nd ro ugh pipes as : 1. Smooth pi lles

~

, 4f

- 2 laglO (Rlk)

= 2 log 10 (R,.j4j) - 0.8 - 2 log,o (Rlk)

Fl]

= 2 10gIO ( R, R ik

... ( I 0.37)

- 0 .8

2. KOIIRh pi llt'll I

~

,4 f

- 2 10&'0 (Rlk) = 2 log 10 (Rlk) + 1.74 - 2 10gIO (Rlk )

= 1.74 . ... (1 0.38) Problem 10.7 For tile problem /O.6,jind tile co-effie;ell! offriC/ion and Ihe orange 11t>;glll ofrol,gllness projections. Solution. From th c sol mio n of pro bl em 10.6. we have R=0. 15111 u. = 0. 1458 lOIs iJ = 1.4533 m's r or co-e ffi cien t o f fric tio n. we know Ih at II,

=

ff .J112

0'

0. 1458 = 1.4533.JTi2

0'

JT12 = 0 .1458 '" 0. 1

1.4533 f= 2.0 x (. l)~ = .02. Aus. Hdg tll o f roughness project ion is o btained fro m equa tion ( 10.36) as

I

r:=-: ,4 f

~

2 loglO ( Rlk) + 1.74

Su bstitut ing the va lu es o r R andf. we get

I

.)4 x 0.2 0'

.15 ) logl o ( k

~

(0.")

2 loglo - - + 1.74 or 3 .5355 = 2 loglo k

(k.")

+ 1.74

3.~ 53~5~5~-~I~ .74C = 0.8977 == logl o 7.90 = -~ 2

0 .1 5 == 7.90 k k == 0.15 == 0.0 1898

7.90

In

== 18.98 111111. AilS.

Problem 10.8 IYater is flowitlg thmllgll a rOl jgll pipe of diometer 5()() mm atld Ie/lgtll 4(X){) m ot tile rate of 0.5 m 3Is. Fitld the powe r required 10 maillwin this flow . Take ti,e m'erage heig/II of mug/mess as k == 0.40 mm.

~ I

I~

~ I

IL 1454

Fluid Mechanics

Solution. Given: Dia. of rough pipe.

D", 500

:, Radius,

R"'2=O.25m

111111 '"

0.50

III

D

Length of pipe.

L '" 4000 III DiM: hargc. Q'" 0.5 111 3fs 3 Average height of roughness. k '" 0.40 II1Ill '" 0.4 X 10- III

First fi nd the value of ,",o-efficient of friction. Then cakulmc the head lost duc 10 friction ilnd then power required. ror a rough pipe. Ihe value of l' is given by the equation (10.36) as

'2

I":"":" '"

.;4 J

, ( .25 )

]oglo (Rlk) + 1.74",. Ing lC

= 2 1og lo (625.0) + 1.74 = 5.591

.4 xlO

J

+ 1.74

+ 1.74 = 7.331

I 1 '" 0.1364 or /= (0. 1364t14 , = .00465 ,,'4/" = 733

"'

Also the average velocity.

Head

lo~t

fJ

= Discharge =~=~ '" 2.546 If 0 2 It (.5)! Area 4 4

duc to friction. "1=

4. f . L . V' d x2g

'"

4 x .00465 x 4000 x 2.546!

O.Sx2x9.81 [": V=U =2.546.d=J)=Q.5J

=49 .1 6m

Power required.

p =

=

IV x 11/

w.Q.iI!

pxgxQxll,

IlXlO

IlXlO

11100

kW

1000 x 9.8 1 x 0.5 x 49.16 1000 "" 241.1 3 kW. Ans.

Problem 10.9 A JmOOlil pipe of diameler 400 mm and /ellglll 800 m carries ...aler al Ille mIl' of 0.04 m)/s. D elermine Ille head IOJ'I due 10/riCliOIl ....all shear Slres~·. ce"lre-/ille \'e!ocily (md Ihickness o/imlli/llir sub-layer. Take Ihe killematic l"iSfMily of \I'll/a as 0.018 Siokes. Solution. Givcn : Dia. of pipe. Radiu s. Length of pipe. DiSl:hargc. Kincmatic viSl:osity. Avcragc velocity.

D "" 400 mm "" 0.40 m

D R""2,,0.20m L"

800 m

Q" 0.04 mlls v " 0.018 stokcs " 0.018 l:m 2ls " 0.0 18 x 10.... m 21s

fJ "" ....Q..... '" Arl:a

0.04

"" 0.3183 m/s

It (0.4 )1

4 Rey nolds number.

I I

R, '" VxD" U xD '" 0.3J83 x 0.4 ,,7.073 x 104 V V .0 18 x 10-4

Ii

~ I

IL Turbulent Flow

455 1

The flow is turbul ent.

J'

The co-effi cie nt of fric tion

f

is obtained from equation ( 10 .30) as = .0791 =

0.0791

( H, ) '"

(7.073x 10 ' )'"

f".__L:c.~V_l

,4 ~ .

(,) Head lo~t du e to friction. "! "'-

'"

= .0791 '" .00485 16'0 ..,

4. f . L. fJ;:'

Dx2g

Dx 2g

4 x .00485 x 800 x (.3183)l 0.40x2x9.81

'" 0.20 m. An s.

( ii) Wall she ar stress (tol is give n by equation (10.5) as f

_ (I -

f· p· Vl _ 2

j. p. fP

[-o'

2

v= U I

,

= 0 .00485 x 1000 x (.3 I84 )l N/m = 0.245 Nlm',' AilS. 2.0

(iii) The ce ntre- Jin e ve locit y (II"",) for smooth pipe is g ive n by equatio n ( 10 .20) as in which j j : II ....

at y "' R II

....!!:O!. '" II .

whe re th e she ar ve locit y

II , =

II . R

5.75 Jog 1o -

v

J¥ = J~:

+ 5.55 jPut in equation ( 10 .20),

II '"

II".., at y '" HI

= JO.OOO245 = 0.01 56 Ill/s

Substitutin g the valu es o f u •• R and v in th e abo ve equation, we gel

~: 5. 7 5Jogl o 0.0 156xO,;0 + 5 .55 :24.1 73 0.0156

.0 18 xlO

II ....... : 24. 173 x .0 156 '" 0.377 mls. A DS. (i ,') Th e lh ickn ess o f laminar s utl·la yer (0') is give n by

0': 11.6xv '" 11.6x.0 18xlO-4 '" .001338 m '"

1.33H 111m. An s. .0156 Problem 10.10 A rOl jgh pipe of i/iaml!ler 400 mm wid leng/II /000 II! carries »"ater (If /he rUle of 0.4 ", J/s. The 1\"(11/ rOllghness is 0.012 mm. Determine the co-efficie'll of f riction. \\"(11/ s/JelJr ~·Iress. centre-line "d ocity and "e/ocity (It a distan ce of 150 mm f rom the pipe 11'(11/. Solut ion. Given: D ", 400mm ",0.4 m Dia. of rough pipe. II,

Radiu s.

2

2

u:n glh of pipe.

L", 1000 m

DiSl:hargc.

Q : 0.4 ml/s

Wall TOu ghness.

I I

R ", 0 = 0.4 = 0 .20 m

I: = 0.0 12 rnm '" 0 .0 12 x IO- J III

Ii

~ I

IL 1456

Fluid Mechanics

'f'

(i) T he va lue o f co-cfricicllt of fricti on

1.0 = 2 log

~

, 4[

1.0 ri"7 = 2 ,,4[

=2 4/ =

10

]ogl o ( IO~IO

for ro ugh pipe is given by the eq uat ion (10.36) as

(RII<;) + L74

0.20 1 ) + 1.74 .012x l0

(16666.67) + 1.74 = 10. 183

(_,_)2 '"

.00964

10.1&1

_ .00964 _ 00' [ _ - - - _ . _41. Ans. 4.0 (ii) Cen t,..., -li"" velocity (II ..... ) for rough pipe is give n by equation ( 10.21 ) in which u is made :=

"m.., at )' '" R and hence

"nu, ".

= 5.75 log lO (Rlk) + 8.5

...(il

where shear velocity.

and t o'" wall shear stress '" where V",

~[ PC'.V_' 'C.

2

Di sc harge Q Q "'7'''''''" --" --= 3 . 183 m /s . A ns . Area ~D2 ~ ( .4 )l

4

4 = .24 00 I x l 000

11.=

f:=Jlt~

=0.11

3.1831

x -- " 2.0

Illis

Substituti ng the value of II .. R, k in equation (I). we get II"",. '" 5.75 log lo (

0.1 1

0.2

.0 12 x 10

J) +

8.5 '" 32.77

II ...., = 32.77 x 0.11 = 3.60 m/s. A ns.

(i l') Velocity (u)

~t

a distance y= 150 111m

= 0.15m

The velocity (u) at any poinT for rough pipe is giv en by equation ( 10.21) as

- " = 5 ,75 Jog 1o (ylk) + 8.5

,,'

whae

I I

It." 0.11 m/s and y" 0.15 111. k = 0.0 12 X 10-J m

Ii

~ I

IL Turbulent Flow 0.15 -, - '" 5.75 10);10 ( 0.11 .01 2 x 10

J

457 1

) + &.5 '" 32.05

/I = 32.05 x 0.11'" 3.52 m /s . An s. Problem 10.11 A smooth pipe line of 100 mm diameter c(!rries 2.27 m3 per millule ai,,'mer (If 20°C wilh kinem(llic "iscosily of 0.()098 Ilokes. Calculate III I' friction Judor. m(l.limllIM refoei/)' (lJ we ll (l~'

x/wur SlrellS

(l/

IiiI' bo undary.

SOlution. G ive n : Dia. of pi pe.

D = lOO rnm =O. 1 111 R '" 0.05 In

Radiu s of pipe. Di ~ h arg".

Kine/lHll k viscos ity, v '" O.{l098 stokes'" 0.0098 e m'ls = 0.0098 x 10--1 m~ls . g .ive n ..vy Now average vI' IOC .lt y IS

u '" -Q-

" c,O~.O~3~78'C

Area U xD

Rey nolds nu mbe r is give n by. R,= -

v

,

~( O .l )'

=

0.0378x4 -_4.8 17 m/, If

x om

4.817xO. l .... =4.9 154 x lOs, 0.0098 x 10

The fl ow is lurbul clll and R, is mo re (hall lOS, Hence for smooth pipe. th e co·crfkicnl of fric ti on

'f ' is o bla incd from equ ation (10.33) as I

J4i == 2.0 logl o (R, J41) - 0.8 I fA"i

"'

,,4/

" - O.S == 2.0 logl o (4.9 154 x 10, x ..."4[)

== 2.0 Il og lO 4.9 154 x lOs + loglo ,f4i 1- 0.8

= 2.0 15.69 15 + loglo J4l J - 0.8" 2 x 5.6915 + 2 loglo " 11.3830 + 10g\ 0 (,f4i)1 I

,f4i -log 1o (4j)"

"'

- 0.8 =

,f4i - 0.8

11.383 + loglO (4j) - 0.8

11.383 - 0.8 " 10.583

...(i)

(i) FriCli01I[aClo r

Now. friction fac tor (J* ) = 4 x co·cfficic nt of friction " 4[ Substitu ti ng the va lu e of '4F in equa ti o n (i), we gel I

~

, ['

- log 10

r"

10.583

.•. (i i)

The above eq uati o n is so lved by hit and trial method. Lei = o. t, th e n L.H.S. o f equation (ii), bn'()mes as

r

L. H.S."

I I

I f'i<:O - log 1o 0. 1 ,,3. 16 - (- 1.0) ,, 4.16 ... 0.1

Ii

~ I

IL 1458 Let

Fluid Mechanics

r

== 0.01. then L.H.S. of equation (ii). becomes as

I = ",0.01

L.H.S.=

log1 oO.Ol '" [0 - (- 2)= 12

BUI (orexael solution. L.H.S. sliould be 10.583. Hence value ofr lies betwccn 0.1 and 0.01. Le!.t" '" 0.013 the n L. H.S. of equation (ii), becomes as

=I

L. H.S. '"

"O.Ot3

- 101: )0 0.013", 8.77 - (- 1.886) '" 8.77 + 1.886 == 10.656

which is approximately .-:qualto 10.583. Hence the va lue of is equal \00.013. Friction factor. = 0.0 13. An s. (ii) M" ximum "e/ueil), (u"",,) Now we kROW that = 4/

r

r r

f· '" 0.0 13 = 0.00325

Co-efficient of friction. ! =

4

4

Now the shear ve locity (If.) in krrns of co-eFfIcient of friction and average velocity is given by cqu31ion (IO.3IA) as

u.==u-

fi f7 =4.817x

r·

OO325 2 =4.817xO.0403=O.J94

For smooth pipe. the velocity al any point is g iven by equation ( 10.20)

,,=

Il.

[S.75 10g lO

II. : ) '

+ 5.55]

The velocity will be maximum ~t the ce ntre of the pipe, where y = R = 0.05 i.e.. radius of pipe. Hence the above equation becomes as

1

U"",-,='" [ 5.75 10g 1o -,.xR ,- + 555 = 0.194 [ 5.75 1og lO (iii ) Sllear

1

0. 194 x 0 .05 • + 5.55 0.0098 x 10

'" 0.194 [22.974 + 5.55[ (t o)

J'lres,~ allhe bOllllliary

".=

We k now that

= 5.528

m ls. Ans.

Pi or,,~= ~

t o = P ,, 2. = 1000 X 0.194 2 = 37.fi3 N/m ~ . An s. Problem 10.12 (p = J()()() kg/m

J

fly
v = I~ m2/s) wilh a head loss of

diamerer. Use f = 0.0031 +

I I

0.221

(R,)

0.2J7

eqlwlion fo r f.

/englh of pipe. Determine Ihe pipe /xLxV z pVD wltere /If = ~""~~ and R< = - . D X2g ~l

3 III ill

100

111

Ii

~ I

IL 459 1

Turbulent Flow Solul io n. Gi ven: Dis.: hargc.

Q = 300 lis ", 0.3 mlfs

Density. Kine mat ic viscosit y.

p = 1000 kg/m l v '" 1O-~ ml/s

~!cad

Leng th o f pipe.

I,, "" 3m L '" 100 m

Va lue of fri ctiun fa(;lor.

f '" 0.0032 +

loss.

Rey no lds numbe r.

pVD

0.221 n ",

(R.)""

V)( D

R, : - , - : -,-,-

vx D 6 : - - , - = VX D XIO 10-

Find: Diamcler of pi pe. Let D = Di ameter o f pipe I'!cad loss in ten us of friction fad or is g iven as

' !J: -,J'cX
Dx 2g

3=

LJ~X~I~OO~X""V~'

C:

D:><2 :><9 .81

3)(D)( 2 )( 9.8 1

100 V ~

/= No w

or / =

/', = 3, L = 100 Ill)

0.58860 V2

...( i)

Q = AxV

O.3 = ~ D2 xVor dXV ", 4x 0.3 = 0.382

,

4

v = 0 .382 D' A lso

"'

_

f-

... Ui)

'} O.!X)3_ +

022 1

• '"

(R, )-

0.588,6 D = 0.0032 + _ _O~,2~2~1",,,, VxDxlO '

V' I, , )Oll1

C

From equation (i) .f '"

"'

0.5886 D '" 0 0032 ,

( O~~2 r

.

+

0.22 1

(O~~2 x Dx

10

05~S: 0

IO~ )

0 '37 6

) -'

( ":

I I

and R, '" V x D x

. ..

0.382 ) D'

From CqU3UOU (U ), V = - -'-

Ii

~ I

IL 1460

Fluid Mechanics

0.5886 X, D~ '" 0.0032 + 0.382'

-;-_~O.~22~',",= (0382 x IO ~

t!J1

D U.!J1- -

4.0.33 rY '" 0.0032 + 0.0 105 x Do.m 4.033 D~ - 0.0105 DD.m - 0.0032 '" 0

0' 0'

... (iii)

T he 3OO\'C equat ion (iii) will be solved by hit and trial method .

(il Assume D " I m. then L. H.S. of equation (iii). hecomes as L. H. S. '" 4.033 x 13 _ 0.0105 x 1°. 137 - 0.0032

= 4.033 - 0 .0105 - 0.0032 = 4.0 193 By inc re asing the val ue of f) more than 1 111. the L.H.S. will go on in creasing. He nce decrease the val ue of O. ( ii) Assume D = 0.3 Ill. then L.H.S. of eq uati on (iii). becomes as L.H.S. '" 4. 033 x O.3~ - 0.0 105 x 0.]°·231 - 0.0032 = 0.0098 - 0.00789 - 0.0032 = - 0.00129 As this va lue is negative. th e val ue of D wi ll be s li gh tl y more th an 0.], (iii) Assume D '" 0.306 m. Ihen L. H.S. of equati on (iii). becomes as

L. H.S. = 4.033 x O.306~ _ 0.0 I 05 x 0.306 o.m - 0.0032

= 0.0108 - 0.00793 - 0.0032 = - 0.00033 This value of L. H.S. is approximatel y equa l to zero. AClllally the va lu e of f) wi ll he sligh tl y more than 0.306 m say 0 .308 m . Ail S. Problem 10.13 Waler is /lowing IllrOl18h a rough pipe of diameler 600 mill (jf file rme of 600 Ii/res/second. The 'mil roug/lne.ls is J 111111. Find Ille power /OSI for I kill leng/h of pipe. Solution. Gi\"en :

Dia. of pipe. Radius uf pipe. Discharge. Wall roughness.

f) '"

600 mm == 0.6 m

8= °·6 :0.3m

2 Q = 600 litrels = 0.6 m 3 /s k == 3 mm '" 3 x 10- 3 m '" 0.003 m

Length of pipe. L = 1 km '" 1000 1Il Fur rough pipes, the co-efficienl of fri<.:tiun in lenns of wall roughness, k is gi\"en by equalion (10.36)

as

,

( 0.3)

fA7 =2 10g lo (Rlk)+ 1.74=2Io8 1O - - +1 .74= 5.74 -.J4 f 0.003 0'

Fl '" -5.76 '- == 0. 174 2 or 4f == (0. 174 2)2 '" 0 .03035 1 4 fXLXV The head loss due 10 friction is given tly. hf = -"--;;-";"-D x2 8

I I

Ii

~ I

IL Turbulent Flow w here

v:

Q ", A

0.6

~ (O.62)

'" 2. 122

461 1

111/5

", ,,,

0.03035 x HlOO x 2. 122 ' 0.6x2:><9.81

: 11.6 m

T he pow er* 10SI is give n by, P '" pg x Q x h! '" 1000 x9.8 1 x 0.6 x 11.6

lOOO

~w '" fiS.27 kW. An s.

1000

HIGHLIGHTS I . If the Reynold number is less Ih:m 2000 in a pipe. the flow is laminar while if the Reynold number is more than 4000. the flow is turbule!!! in pipes. 2. Loss of pressure head in a laminar flow is proporlionallo the mean velocity of flow. while in case of turbulent now il is approximately proportional to the square of wlocity.

3. Ex['ression for head loss due to friction in pipes is gh"en by Darcy -Weisbach equation. Z h '" 4 x f x Lx V . where! '" co-eff,cicll t of friction J dx 21(

'" J x Lx V' . where! '" friction factor dx 2g

4. Co-<:fficicnl of friction is expressed in tenns of shear Slress as ..

3

'v

where V = mean velocity of now. p .. mass densi ty of nuid. 5. Shear Stress in turbulent now is sum of shear StreSs due to vis.cosity and shear Stress due to turbulence. i.t .• r '" t ,. -+ r,. where t, '" shear stress due to viscosity t,. shear stress due to turbulence dii (fii

., - +, -

d)' d}' 6. Turbu lcnt shcar Siress by Reynolds is given as

where

7.

t '"

P w'y'

u' and ,r. nuctuating component of velocity.

TIle expression for shear

stres.~ in turbulent now due tQ I>r:mdtl is t ~ pI" [d")l. where 1_ mixing length. ,Iy

8. The velocity distrib uti on in the turbulent now for pipes is given by the expression u ~ u_ -+ 2.5 u· log, (yfR) where 11_ .. is the centre-line velocity. }' ~ distance from the pipe wall, R .. radius of the pipe. and

· Pow~r

I I

11. '" shear velocity wbich is equal 10

.. pg x Qxhf wa1t ..

Iff.

1000

Ii

~ I

IL 1462 Fluid Mechanics 9 . Velocity defect is the difference between the ma~ jmurn vdocity (u_,l and local velocity (u) at any point and is gi"cn by (u ..... - u) _ 5.75 X w. logIn ( Nlv). 10 . The ooundary is k.nown as hydrooynamically smooth if t. Ihe average h<'ighl of Ihe irregularities project ing from thc surface of Ihe Doundary is small compared \0 Ihc thickness of Ihc laminar sub-layer (0') and

boundary is rough if k is large in comparison with Ihe thickness of Ihc sub-layer. Or if

,< , 6'

~

and if

,

0 .25. the boundary is smoolh : if 0' > 6.0, the boundary is ro ugh

lie Dctwcen 0.25 \0 6.0 , thc boundary is in transition.

II. Velocity distribution for lurbulent flow is

•

'.

_ 5.75 [og lO u.y + 5.55 for smooth pipes

"

~

5.75 10SI0 0'11::) where " " velocity at 3ny point in Ihe turbulent flow.

u.

~ shear vcb::ity nnd ~

J*.

\I

*

+ 8.5 for rough pipes

~inematic viS<.'()sity of fluid.

y .. disunce from pipe wall. and k .. roughness factor. 11. Veloc ity distribution in tenns of avernge velocity is

U

-

'.

w.N

.. 5.7510g tO- - + 1.75 for smooth pipes.

"

'" 5.7510)\' 0 NI t + 4.75 for rough pipes. Il. Difference uf local velocity and ave"'~l!e velocity for smooth and ro ugh pipes is u

-fJ

- - '" 5.75 10gt O(yIR) + 3.75.

'.

14. The co-efficient of friction is given by

f., •

"

R; ..... for lami nar flow • 0.07~~

for turbulent flow in smooth pipes for R,

(R,)

.05525 n' •.0008 + -----.... for R, S hE but

(H.)"'

~

'"


4 x 10


4000 by S

1O~

,

• 2 10l! to (Rlk) + 1.74 for rough pipes whcre R, . Reynolds number.

EXERCISE (A) THEORETICAL PROBLEMS 1. What do you understand by turbulem fluw ? What factor decides the tyPe of flow in pipes " 2. (tI) De.i,·c ~n exprcssion for the loss of head due to friction in pipes. ad. S 20(2) J . E ~plain the tcnn c'O-efficicnt of friclion. On what factors docs this co-cfticient depend ?

I I

Ii

~ I

IL Turbulent Flow

463 1

4 . Obtain an expression for the co-.,fficienl of friction in the lenns of shear ,tress. 5. What do )'OU mea" by Pmndll mixing u,ngth Theory? Find an expression for shear slress duc to l>randt!. 6. Dcriw un cxpre3>ion for Prundl!'. universal ,-"I<>city distribution for lurbulent now in pipes. Why this

vciocity distribution is called universal ? 7. What is a velocity derect ? Dcrive an expression for velocity defect in pipc •. g, How wou ld you distinguish between hydrodynamically smooth and rough boundarie s?

9. Obta in an expression for the "c!ocity distrib ution for turbulent flow in smooth pipes. III . Show that "eloc.ily distribution for turbu lent flow through rough pipc is gi "en by

-"

~

5.75

log l~

(,-Ik) + 8.5

". where u.", shear velocity.)," distance from pipe wall. k., roughness factor. II . Obtain an expression for wlodty distribution in temts of ""crage velocity for (
" -u - '" 5.75 10g lO 6lR) + 3,75,

'.

13. Obtain an

e~pression

for velocity distribution in turbulent now for (I) smooth pipes and (iiI roug h pipes. (Delhi Uni,'er.,ily. I)eceml>er. ZOO])

(B) NUMERICAL PROBLEMS J. A pipe-line carrying water lias average height o f irregUlarities projecting from the surface of the boundary of tile pipe as 0.20 rum. What type of the boundary is it ? The shear stress development is 1 7.848 N/m . Take value of kinematic viscosity for water as 0.01 stokes. IAn s. Boundary is in transition I

2. Detenninc the average height of the roughness for a roullh pipe of diameter 10,0 em when the velocity at a I"'int 4 em from wall is 40% more than the "elocity at a point I em from pipe wall. [Ans. 0_94 eml J . A smooth pipe of diameter 10 em and 1000 m long carries water at the mte 0[0.70 m1/minute, Calcu late the loss of head. wall _,hearing stress. centre line velocity. velocity and shear stress at 3 em from pipe wall. Al so calcu late the thickness of the laminar sub-layer. Take kinematic viscosity of water as 0.015 stokes and value of co-efficient of friction J as

f· 4.

S.

6.

7.

I I

.0791 --,,-4 . where R, '" Reynolds number.

(I<)

IAns. 20.05 m. 4,9 Nlml ; 1.774 mls ; 1.65 mls ; 19.62 N/m1 : 0.248 mml The velocities of water thro ugh a pipe of diameter 10 cm. are 4 mls and 3.5 mls at the centre of tMe pipe and 2 cm from the pipe centre respectiYely . Delennine the wall shearing stress in the pipe for turbulent now. [Ans. 15.66 kgflnt' l f or turbulent flow in a pipe of diameter 200 mm. find the discharge when the centre· line velocity is 30 m/s and veloc ity at a point 80 mm from the centre as measured by pilOt·tube is 2,0 m/s. [Ans. 64,9 litres/51 For problem 5. find the co-efficient of friction and the average height of roughness projections. IAns. 0.029. 25 ,2 mml Water is nowing through a rough pipe of diameter 40 em and length 3000 m at the rate of 0.4 ml , •. (:ind the power required to maintain this flow . T"kc the average height of roughness as K _ 0.3 mm. [Ans. 278.5 kNI

Ii

~ I

IL 1464

Fluid Mechanics

K. A ,mooth pipe of diameter 300 111m and length 600 m carrie. waler at rate of 0.04 ",2/s. OClcnninc thc

head losl due [0 friction. wall shear stress. cenlre-line ~elOCilY and thickness of laminar sub·luycr. Take the kinematic viscosity of water as 0,018 stokes. lAos. 0.588 m, 0.72 NiemI, 0.665 m/s. 0.779 mml 9. A rough pipe of diameter 300 mm and length 800 111 carries water at the rate of 0.4 mIls. The wall roughness is 0 ,015 mm. DClcnninc thc co-dflcicn! of friction. wall shear stress. centre line "elocil), and velocity at a distance of I ()() min fmm the pipe wall. IAn s. f .. .flO263. TO", 42. 08 N/cm'. w.... '" 6.457 mIs, U = 6.2 49 m/s] I ll . Dctcmlinc the d;,lance from the

~-en(re

velocity for turbulent now in pipes.

I I

of Ihe pipe. a( which the local

~elocity

is equal to Ihc :l"cragc

[Ans. 0.7772 RI

Ii

CHAptER

to

I 1. 1 INTRODUCTION

In chapters 9 and 10. la minar flow and turbulent flow have ~"n discussed. We have ..;;cen lilat when th" Reynolds number is less than 2000 for pipe flow. the flnw is known as laminar flow whereas when the Reyno lds number is more than 4000. the flow is known ;IS turbulent flow. In this chapter. 1he lUrbulent flow of fluids through pipes running full will be considered. If 1he pipes are panially full as in the case of sewer lines. the pressure inside the pipe is same and equ:1110 atmospheric pressure. Then the flow of fluid in the pipe is not under pressure. This case will be taken in the chapter of flow o f water through open channels. Here we will consider flow of fluids 1hrough pipes under pressure only.

to I 1.2

LOSS OF ENERGY IN PIPES

When a fluid is flowing through a pipe. the fluid expe riences some resistance due 10 which some of the energy of fl uid is losl. This loS1S of energy is dassified as : Energy LolSes

I

I

1. Major Energy Losses

2. Mioo< Energy Losses

I

I This is d~ to friction and it is cak:ulated by the f,"lowing formula .. ,

This is du.. to (s) Sudden expansion of pipe (b) Sudd
(s) Darcy·Weisbach Formula (b) Ch .. zy's Formula

(e) An

to 11.3

obslruc~oo

in pipe .

LOSS OF ENERGY (OR HEAD) DUE TO FRICTION

(a) Darcy· Wels bach formula. The luss of head (or energy) in pipes due IU fri<.:liun is calcul;ncd from Darcy-Weisbach equ~lion which has been derived in chapter 10 and is given by

,

Ie'."L".V,-

,,4 ". 11=, d x 2g

whe re

... ( 11.1}

II, : loss of head due to frict ion

'65

I I

Ii

~ I

IL 1466

Fluid Mechanics

f

== co-efficient of friction whic h is a fUllction of Reynolds number ~

16 for R, <

.

2000 (VISCOUS

R,

now)

0.079,or R, varymg . f w m 4000 == ----;!4

R,

10'

!O

L == length of pipe. V == mean velocity of flow.

d ==

diamc l~r

of pipe.

(b) Ch el ~" s I' UI-mulll for loss uf IIl."lI d du .. to fr ldkm in p ili...". Rdcr \0 chapler 10 ankle 10.3.1 in which expression for loss of head duc \0 friction in pipes is derived. Equation (iii) o f anide 10.3.1. is

f'

P

pg

A

Ilf == - x~XLXV2

whe re II ", loss of head duc to fril:tion. f A == ~rca of noss-scdiun uf pipe. and V = mean ve locity o f flow. A [ == Now the ratio of ~

P

... ( I 1.2)

P == welled perimeter of pipe. L = length of pipe.

Area of flow Perimeter (wetted )

1

is I:allcd hydraulil:

UlCilll

depth or hydmuli<.: radius and

is denoted hy m.

Hydraulic mean depth,

SubstilUting

In

A P

_

==

:0

A

~d 2 4

d

P

'"'

4

- = -- ~ -

In

l In . equntlOn . (II .2). we get or -P :0 A III

pg II xmx _I ", _xlnx --.L L

f'

L ... (11.3)

Let

rei = C, whe,"" C is a constant known

~~

per unit length of pipe. Substituting the valucsuf

ff

and

JEt

3.'i

Che zy's constant and!!.L '" i. where i is loss of head L

in equation (11.3), we get

..r;;;;

V=C ... ( I 1.4) Equation (11.4) is known as ChCl.y'S formula. Thus thc loss o f head due to friction in pipe from Chczy's formula can be obtained if thc veloc ity ofllow through pipe and also the value of C is ~nown. The value of III for pipe is always equal to d14.

I I

Ii

~ I

IL Flow Through Pipes

467 1

Problem 11.1 Find tile head la,I'/ due 10 fricliol! ill a pipe of diameler 300 mm alld lellg/II 50 III, Ihrough which Wafer is flowing a/ a re/ociry of 3 mls /Ising (i) Darc), [orlllll/o, (ii) Che;.y'sformula for .... hich C '" 60.

Take y jar water'" 0.01 sioke. Solution. Given: Dia. of pipe. d", 300 mill '" 0.30 m L", 50 In lA:ngth of pipe. V:3 rnls Veloc ity of flow. Clw zy's con,tan!. C= 60 Kinematic viM'oshy, 1'= 0.0 1 stoke = 0.01 cm!/s = 0.01 x lO--4m Z/s. (i) Durey ~'ormul" is given by equation (Il.l) as

where

II =

~4~.f'c" . L~.~V_'

r

d x2g

J' = co-efficient of friction is a function o f Reynolds number. R, = 3.0 x 0.30 =9 x lOS

R = Vxd

But R, is given by

,

.01 x 10"'"

v

Value of

.00256 x 50 x 3! "1=4 x03x2.0x9.8J '" .78211 rn. Ail S.

Head los!.

(il) C helY's Formul a. Using equation (1 1.4)

V= C where C '" 60.

d 4

III '" -

0.30 4

'" - - "

.r;;r

0.075

1ll

3", 60 ,).075 x I or i"

,

h,

hi

L

50

(3)' I 60

x .075 '" 0.0333

,= - = -

/,

Equating the two valu es of i. we ha ve 5~ '" .0333 II! '" 50 x .0333 '" 1.665 m . t\ns. Problem 11 .2 Find the dlallieter of a pipe of lengrh 2IXXJ m .... Ilen rhe rate off/ow of\\'arer rhrougll the pipe is 200 litreYs (md the hemf losr due to friction is 4 III. Take rile \'alue of C '" 50 ill Chezy's fo rmulae.

I I

Ii

~ I

IL 1468 Fluid Mech a nics Solution. Given: L : 2000 111 Length o f pipe. Di sc harge. Q == 200 litre/s == 0.2 m3/s Head lost due to frict ion. I,! '" 4 m Value of Chczy's constant. C = 50 lei the diamete r of pipe '" II Velocity of flow.

V'" Disdlarg c Area

H ydmulic mean depth.

m= -

d

4 4

/If

Loss of ltead pcr unit leng th . i = -

'" - - = .002 2000

L

Chczy's formu la is given by eq uation ( 11.4) as

v=

c,;;;;

Substituting the values o f V. m. i and C. we ge l 0.2 )(• 4 _ _.,cO 'ltd -

. bo t h SI·d CS, -d Squarmg

4

X .

~~ x .002 ur J~ x. 002 __ 4

4

.OO509 ~ •

002 _

.~259 or d

d

d= >lJO.0518 =( .0518) " ~ ==

,r '"

0.2, x 4 __ .00509 ,

nd- )(50

11-

4 x .0000259 '" 0.0518 .002

O.553111 == 553 111m. Ans.

Problem 11 .3 A crude oil of kinematic I'iscosiry 0.4 sto lce is flowing Iitmugil a pipe of diameter 300 mm at 111l' mil' of 300 lilres pc, sec. Find (he head 1051 due ro friction/or a {'!IIg/h of 50 In oflhe pipe.

Solution. Given: Kine m;l1ic viscosi ty,

v = 0.4 stoke = 0.4 em'/s =.4 x 10--1 It/Is

Dia. of pipe. Discharge. Lenglh of pipe.

d = 300 Illm = 0.30 nI Q = 300 lilrcs/s = 0.3 ur' /s

Velocity of fl ow.

V= -

L = 50

III

~O~.3~ = -;; = 4.24 m/s

Q

~ ( 0.3 )1

Area

4 .". Rey nolds numbe r.

R = Vxd = 4.24x OJO =3.18 x 10~ , v 0.4 X 10--1

As R, li es t>ctween 4000 and 100000.

th~

.079

v;tlue of f is g ive n by .079

f = - -,,-, = "' iN,) (118x 10' )

I I

= .00591

Ii

~ I

IL Flow Through Pipes He ad lost du e to fr ictio n.

"J ~

4. / .L.V'dx2g

'"

4691

4 x.0059 I x50x 4.24 ' '" 3.61 m . An s. 03 x 2 x9.81

Problem 11 .4 An oil of Sf!. gr. 0.7 is flowing through" pipe of diameter 300 m", a/ the mil' of 500 li/re:i/s. Find Ille head /OJ'l dill' /0 friction amI power req"ired 10 mail/luin Ihe flow for a lenglh of 1000 m. r ake v '" .29 stokes. Solution. G iven: Sp. gr. of oil. s= 0.7 Dia. of pi pe, d= 300 m in '" 0.3 III Di sc h il rg~. Q '" 500 lilresls = 0.5 1I1 )/s Le ngth of pipe. L = IOOOm

v= ~ = ~_

Veloci1y,

Area

1'1:

d' -

0.5 )( 4 = 7.073 m/s X 0.3 '

If

4

7.073 )(0.3 = 7.3 16)((10)4 0.29 X 10 -4

Rey no ld s numbe r.

. . , .079 0.79 Co "cffic lcnl o f fnctlO n. /= ---w- = ~ = .0048 R, (7.3 16 )(104)"

He ad lost d ue to fri cti on. IIj =

Power required

'"

4x

f x L X V· 4 x .0 048 x 1000 x 7.0 73'd x2g

pg .Q. ll j I ()()()

0.3 x 2 x9.81

'" 163.1 R In

kW

where p = density of oil = 0.7 x 1000 = 700 kg/ml Powe r requircd = 7()() x 9.8 1 x 0.5 x 163.18 = 560.28 kW . Ans. I ()()()

Problem 11 .S Ca/cu/(1Ie Ihe t/isc/wrge Ihrougl, II pipe of dhm'e/er 200 """ Irhen IIII.' difference of pressure helld be/wee" Ihe IIro ends of (l pipe 500 iii oporl is 4 /Ii of 1I·Il/er. Tilke llie 1·llli,e of "/. == 0.009 ill Ihe formula h == f

c4~,f'c'"L~,~V_' d x 2g

Solution. Givc n : Dia. of pi pe.

d=200 mm =0.20 m

Lc nglh of pipe.

t=500 m

Differe nce of pressure head.

hI '" 4 m of water f== JI09

Usi ng equation ( 11.1). we ha vc

0'

I I

Ilf '"

4.0 '"

4x f xLxV l d x 2g

4x.OO9x500x Vl or vl = 4.0'1'.0.2'1'. 2 '1'. 9.8 1 ",0.872 0.2 '1'.2 '1'.9.8 1 4.0 x.{)()9 x 500

Ii

~ I

IL 1470 Fluid Mechanics V'" J0.872 '" 0.9338 '" 0.934 Illis Dischar£c,

Q'" veloci ty x area 11

It

,

,

= 0 .934 x - d- = 0.934 x - (0.2)4

'" 0.0293

111 3{S

4

= 29.3 Jilres/s. ,\"s.

Problem 11.6 Water i~' flolt'i"g through II pipe 0/ diameter 200 mm lI"ilh Ii nducil), of J m/J'. Filii! the head lost due /0 [ric/ion [o r a /ength of 5 II! if the co ·«fficiell/ of f riction i~' gil''''' by f = 0.02

+

~O:j' ,

where R, ;$ ReYllolds number. Th e kinemaric I'iscosiry a/waler '" .0 1 .Hoke.

Solution. Given: = 0.20

Dia. of pipe. Ve locity,

d = 200

Length, Kincrn:J.Iic vL'iCosil y.

L =5 m v = 0.01 stoke = .01 x lQ---I m1fs

:. Reynolds number .

R = Vxd = 3)(0.20 :6x 10'

Value of

111111

III

V = 3mls

,

\'

f~

.0 1 x 10"'"

09

02 + -0-)=02+ R,

0.09

.09

(6XIO' )"

1

=.02+ - 54.1 3

'" .02 + .00166 '" 0.02166 .. Head lost due 10 frict io n. Ilr

4 x! x 1- x V~ =24".0~X~.~ 02cIC66 ::;;:Xc5~.O~X7"-3' dx2g

0.20x2.0x9.8 1

== 0.993 III of Wilie r . Ail S.

Problem 11 .7 A" oil of sp. gr. 0.9 and \·isw sity 0.06 poise is /loII'i"g IllrOllg/l a pipe of diameter 200 mm at Ihe rale of6O liIles/s. Fimf Ihe /,,'ad losl dlle to friction for {/ 500 m "'''Slir of pipe. Find Ihe po"·er required to ",aintoi" this/low. Solution. Gi ve n : Sp. gr. of o il == 0.9

Viscosity.

~=

0.06 poise = 0.06 Ns/m 2

10

Dia. of pipe. Disdarge .

Q = 60 litresls = 0.06 m 31s

Leng th. [knsity

L = 500m p = 0.9 x 1000 == 900 kg/m3

.". Reynolds number.

d = 200 mm == 0.2 m

pVd

R, = - - =900x

" whe re

I I

V xO.2 006 10

1.909 lOIs '" 1.9 1 nils

Ii

~ I

IL Flow Through Pipes

471 1

R '" 9(0)( 1.91 )(0. 2 )( 10 "" 57300 , 0.06 - As R, lies be tween 4CHXl and 1O~. the va lu e of en-effic ient of fricti on./ is given by f= 0.079 '"

R,o. !)

Head lost d ue to friction. "1 =

4x j xLxV l 4x.OOSl x SOOxL91 l '" "-C"~"-,",""~o"'''-dx2g 0.2x2x9.8 1

'" 9 .411 III of wa te r. A il s.

Power required

... 11.4

pg. Q. h!

'"

1000

'"

900 x 9 .81

x 0.06 x 9.48

1000

= 5.02 kW . Am _

MINOR ENERGY (HEAD) LOSSES

The loss of h ~ad or energy due to frictioll in a pip.: is !;nown as major loss while the loss of energy due to change of vdocil y of the following nuid in magnitude or directio n is c all ed minor loss of energy. The minor loss of energy (or head) includes th e following C3.'>eS: I . L oss of head d ue 10 sudden l-nlargcrncn t.

2. Loss of he ad due to sudden contraction . 3 . Loss of head at the clltrancc of a pipe. 4. Loss of head at the exit of a pipe. 5. Loss of head due to an obstruetiun in a pipe. 6. Loss of he~d duc to bend in th e pipe. 7. Loss of he ad in various pipe fittings. In (;asc of long pipe the above losses are small as compared with the loss of head due to fri(;t ion and hen(;e they are called minor losses and even may be neglected with out seriuus e rror. But in casc of a short pipe. these losses arc comparable with the loss of head due to friction. I 1.4. 1 lou of Head Due to Sudden Enlargement. Consider ~ liquid nowing through a pipe which has sudden enlargemenl a~ shown in Fig. 11.1. Consider two sec lions (1) -(1) and (2) -(2) before and after the en largemen l.

o

Sudden eniargemelll. PI : pressure intensi ly at seclion I- I. Vt : ve locity of flow at sectio n I- I. At : area of pipe at sect ion I- I. Fig. 11.1

I I

Ii

~ I

IL 1472 Fluid Mechanics P2' V 1 and A 2 '" correspond in g values at section 2·2. Due to sudden change of diameter of tlie pipe from VI 10 D2_tile liquid l10w in g from the sma ller pipe is not able to follow the abrupt cha nge of the boundary. Thus the now separates from the boundary and turbulcm eddies are fomK-Q as shown in Fig. II. J. The loss of hc'1I.1 (or energy) takes pla",e due to the fonn mion of these eddies. Let p' = pressure inlclI~ity of the liquid eddies on the area (Al - AI) Ii , = lo.'\S of IIcad due to sudde n ~nlargernenl Apply ing Bernoulli's equation at sections I-I and 2·2.

v,'

fil

VI!

fI,

p,l;

2g

pg 2 g -

- + - + I I = - ' + - '- + l, + loss of head due to sudden enlargement BUl

ZI

= l, as pipe is horizontal

.

b.. + ~=

{J,

,

+ V, +11

pg2gpg2g

0'

'

h, =U~ - ~;)+(~~ - ~~)

... ( i)

Consider the contml vo lume of liqllid he twecn sections 1-1 and 2 ·2. Then the force acting on the liquid in Ihe comrol volume in the dire"tion of flow is given by ,.-~ '" PIAl + p'(A 2 - AI) - P02 Bllt experi mentally it is fOllnd that p' '" PI

F, '" /lIAI + PI(A 1 - AI) - P?"t-2 '" /lIA, - /102 ... (1 i)

'" (PI - P,JA 1 Momentum of liquid/sec al sec tion 1-1 = mass x velocily

,

'" pAl VI X VI '" pAIV I Momentum of liquid!sec al seclion 2·2 '" pA,V2 x V 2 = pA 2V/ Change of mome ntum/sec", pA 2 Vl 2 _ pA l VI2 Bllt from continuity equation. we have

>

Change of momcnt ulnfscc '" PA2V~- - P x

A;Vj Ie

,

>

X

,

VI- '" pA,V1- - pA1VIV1

'" pA 2 [V/ - VI V,I ... (iii) Now ncl force acting on the control volume in the direction of now must be eq ual co Ihe rate of change of mom~ntum or change of momentum iX'r secoJld. Hence equaling (Ii) and (iii) (PI - p z)A j '" pA1[Vl' - VI Vl l

Dividing by g on bolh sides. we have PI - Pl.

pg

I I

1', '" ,V",'~~ --:V""VL' = 'V./_~CV""VLl. or -P I - -'--'g

pgpg

g

Ii

~ I

IL Flow Through Pipes Substituting the va lue of

(li -Ji2) pg

pg

473 1

in equation (0. we get

v,--

2 ' 2 V, V, +V,'"- V,' V,, - V,V, V," V; , ' + - , -'- = ' . . g

2g2g

28

V , V' 'V V = ' + ' - - 1;

2,

II ,

=

(ltj _ V, )'

2,

.

. .. ( I 1.5)

I 1.4 .2 Lou of Head due to Sudden Contraction . Consider a liquid flowing in II pipe which has a sudden COll trn clion in area as shown in Fig. 11.2. Co nsider two sect io ns I- I and 2-2 before and af!cr (;ontraction. As Ihe liquid flows from large pipe 10 small er pipe. Ihe area of flow goes on decreasing and beco mes minimum ,11 a sect ion C-C as shown in Fig. 11.2. This sect ion C-C is called Vcna-comracta. After section e·c, 11 sudden cnl~rgcmc n l of the area takes place. The loss of head duc 10 slldd~ n contraction is actually due to sudden ClllaT)! Cmcnl from Vcna -cnntracla to smaller pipe. Lei

Ac '" Area o f now at section C-C V,. '" Velocily of flow .11 se<: lion C-C

A! '" Area of flow al section 2-2 V 2 ", Velocity of flow at seclion 2-2 ~,~ lIe '" Los~ of head due 10 sudden contraction. Now iI,. '" actu al loss of he ad due to en largement from section C-C to section 2-2 and is given by eq uat ion (1 1.5) as

'"

(~. ;gV.)l ~; l ~

-Ir

. .(il

_ ._ '- '....,.-

Fig. 11.2

~

,

o Sudden con/radion.

From continuity equation. we have

Suh~lituling

the valu e of

V

....£..

V,

;n (I). we gel

... ( 11.6)

'"

k~l ,wherck'" l ~c -Ir

If lhe value of Ce is assumed 10 be cquallo 0.62. lhclI k

= [ _ 1_ , 0.62

I I

Ill ", 0.375

Ii

~ I

IL 1474 Fluid Mechanics Th eil /1< beco mes as

t V,'

V,"

2g

28

h< '" -"- '" 0.175 - "-

If the value of Co is not give n th en the head loss duc 10 co ntractio n is take n as

V}

: 0.5 -

2g

V' o r ll<",0.5 - 1 .

. .. (11.7)

2g

Find the Ion" of head ",hen (' pipe of dilllneta 200 m m is ~'udde"ly en/Mgt!tI to diameter of 400 mm . Tile raIl" of flow of '\"lIler {hroug" the pipe iJ' 250 lilres/s. Solution. Gi ve n : Dia. of small er pipe. VI'" 200 nun ", 0.20 III Problem 11.8

1t

2

1t

,-

'

A I'" - D I = - (.2) = 0 .031 4 1 111-

Area.

4

4

Dia. of large pipe.

D!", 400mm =0.4m

:. Area.

A 2 '"

~

DiSl: hargc.

X (0 .4)' = 0. 12564 m' 4 Q = 250 litrcsls '" 0 .25 1I1 )/s

Ve loci ty,

Q 0.25 YI=-: - - =7.96 m!s AI

.03 14 1

v,,,, iL ,,,

Ve locity.

u:>ss of head

(l

du ~

A,

0.25 = 1.99 mls .12564

10 e nlarge me nt is give n by equat ion (1 1.5) as

/', =

(VI - Vll'

=

(7.96 _ 1.99)'

= 1.816m o f wut cr. An s. 2g 28 Problem 11 .9 AI (I sudden etl/argemelll of II W
Solution. Give n : Dia. of small er pipe.

D , "" 240 nlln "" 0 .24

:. Area.

A, ""

Dia. of large pip".

D2 "" 480 mm "" 0.48 m

Area.

Al ""

Ri se o f hydrauli c gradi ent·. ;.1'.•

It

It

2

11\

4D, "" 4 (.24)

1

~ (0.48)2 4

(Zl + ~;) - ( ;~ + l,) =

Let th e rate o f now = Q Appl ying Berno ulli 's eq uati on 10 b01h se ctio ns. ;,1' .. small e r pi p.: sec ti on. and large pi pe sec ti o n. ~2

Ji + _'_ + l , = pg

2g

V"

Pl

pg

+ _,_ + I I + He ad loss du e 10 e nlarge me nt 2g

...( i)

• Pica"" refer An. 11.5, I.

I I

Ii

~ I

IL Flow Through Pipes

475 1

BUl head toss due 10 en largement,

... (ii)

Substituting this va lu e in (ii) , we get

Now s uhSlitulin ll th e va lu e of II, alld VI in equati on (il. PI

-

(4V2 )'

P2 V" 9V/' + - 2 + Z, + -pg 2g ' 2g

+ - - - +Z, == -

pg2g

I~~} _ ~~ _ 9~" =(~; + Zl) - (~~+ZI)

0'

Btu hydraulic gradient ri sc

==

16V,'

V,"

9 V,"

2g

28

2g

- - - - - . -. - - . - :

(~; + Zl ) - (;~ +ZI) I ~ ==

-

I

100

Vl ",

6V,1

I

28

[00

0< -

-. : -

=O. 1808=O. 181mfs .I~26x~'~9~.8!' 100

Disctlargc. It

2

= 4 D, X V,-=

"4J'[

1

'-

(.48) x. 181 =O.OJ275m Is

= 32.75 lilres/s. Ails. Problem 11 .10 HIe rafe offlow a/li'a1er IllrQuglt a horizontal pipe is 0.25 mJIJ. Ti,e diameter of IIII' pipe whic/I is 200 111111 is suddenly enlarged 10 400 111m. The pressure intensifY in Ille mwl/a pipe is 11.772 N/cIl1". Delallline : (il IOJJ of/lead dill' 10 SI4ddell enlargement. (ii) preSSllre ill/ensif)' (iii) !,ower 1051 dlle /0 elliargemelli.

Solution. Given: Disch~rge,

Dia. of small er pipe.

I I

ill

Ihe large pipe.

Q = 0.25 11I 31s D I = 200 111111 = 0.20 111

Ii

~ I

IL 1476 Fluid Mechanics A,'"

Area.

'4" {.2)-=O.OJI 41

m

,

Dia. of large pipe. D2 = 400 mm = 0.40 III

"

:. Area,

Al"' "4 (O.4)- '" 0.12566 111 ,-

Pressure in smaller pipe,

PI = 11.772 Nfc11l 1 = 11.772 X 104 Nlm"

Now ve loc ity.

VI'"

Veloc ity .

Q 0.25 V,=-= - - : 1.99 m/s • A2 .12566

Q

0.25

AI

.03141

- =- - '" 7.96

mfs

(i) Luss of head duc 10 sudden c llJargcmc nl,

(7.96 _ 1.99 )2 2x9.81

= 1.816 m. Ans .

(ii) Let the pressure in te ns ity in large pipe '" Pl" Then applying Bernoulli"s cq untio ll before and aft er the sudden c nlarg cme nl ,

""'

,

,

.!!J....+ V,P8

_ fll

2g - pg

+ V1 +/1 28

11.772

X

10"

lOOOx9.81

h PI ' or pg = P8 +

, V,

+"'2g 1.99 "

2x9.81

2x9.8 1

, V,

(G ive n horizontal pipe)

2~ -II,

- 1.816

= 12.0 + 3.229 - 0.2018 - 1.81 60 = 15.229 - 2.0178 = 13.2 1 m of water P2 '" 13.21 x pg '" 13.21 x [000 x 9.8 1 N/m l 4 '" 13.2 1 x 1000 x 9.81 X 10- Nlcm 2 = 12.96 Nfl,m l , Ans.

(iii) Power loS! due to suddell en largemem. p=

pg .Q.h, I()OO

x 0.25 x 1.81 6 _ 4 453 kW = 1000 x9.811000 - . . AilS.

Problem 11 . 11 A horizoilial pipe of diameter 500 mm i$ .lIIddl'lll), cOlltracled to a dilllneler of 250 mm. 1'111' pressure illlelisities ill 1/11' large allil smaller pipe is gil'ell as 13.734 Nkm< alld J /. 772 Nkm" respectil·ely. Find the 10.iJ of head due 10 con traction if C~ = 0.62. Also determille Ihe r ail' of flow ofwlllu. Solution. G iven: Dia. of la rge pipe. DI = 500 U1m = 0.5 11\

Area.

I I

AI =

~ (0.5)"

= 0.1963

m~

Ii

~ I

IL FlowThrough PipeS Dia. of s maller pipe.

°

:. Area.

A 1 ",

Pressure in large pipe.

PI'" 13.734 NIemi. '" 13.734 x 104 Nfm! PI = 11.772 NIemi. = 11.772 x 104 Nlml Cc = 0.62

2 ",

Pressure in smaller pipe,

Head 10M due to comraction

4771

250 mill '" 0.25 m

"4" (.25)- '" 0.04908 Ill',

V' [ = ....L.

2g

1 - 1.0]1 = _V,___ 2 [1 _ 1.0 ]" = 0.375 ---.L VI

Co

0.62

2g

28

From continuity equation, we ha "e A ,V ,:: AlV!

,

,

4 D1X Yl

V, [ V,

~ Dl

J' xV,-~0(0.'']' V, ~ V, .50-4

4 ' Apply ing Bernoulli's equation before

~nd

~1

VI

pg2g

pg2s

after contral:tion.

.fL+ -'- +z '" 1:2+ _,_ + z, + II,. 8m

'

(pipe is horizontal)

l, '" <':2

~+

VI! '" P2 + V} + II C pg2gpg2g

v,l

VI

2,

4

" 0=0.375 - '- and VI '"

Substituting the se

valu~s

13.734 x 10' + 9.81 x 1000 14.0 +

"'

in

th~

above equation. we gel

(V! 14)2 = 2g

11.772 x 10' + V} + 0.375 VI' lOOOx9.81 2g 2g

VI!

V,·

cc'''-o-= 12.0 + 1.375-' 16x2g 28 1 V,l

V,'

V,2

14- 12 = 1.375 -'- - - -'- '" 1.3125 - ' 2g 16 2g 2g 2.0 '" 1.3125 x -v':-' or V 2 ",

"'

2g

(i)

LO!iS

.I~2~82x~2[x;;9.~8Il 1.3 125

'" 5.467 mfs.

V,2 0.375 X (5.467)2 of Ilc ad duc to contraction.il e '" 0.375 - '- '" '" 0.571 m . 2g 2x9.8 1

(ii) Rate of flow of watcr.

An.~ .

Q '" A 2V 2 '" 0. 0490& x 5.467 '" 0.26&3 ml/s '" 26!U liUs. AilS.

Problem 11 .12 If in Ille problem II.ff. Ihe rare of flow of lI"arer is J()() lilres/s. alher (Imil remO;1Ijllg Ihe some. fi/ld the I"(llue of CQ·efficiell/ of call/rocliall. Ce.

I I

Ii

~ I

IL 1478 Fluid Mechanics Solution. Gi ven:

VI == 0.5 m. l) 2 " 0.25 m. PI" 13 .734 X 104 Nfml. 4 P2"" 11.772 X 10 N/m 2, Q '" 300 li tis", 0.3 1I1 3/s Also from rmbi cill 11. 1 I . VI

= VI. WhH~ VI = .Q. = -i0~."30,,-:: = 1.528 4 AI ~ ( O.5) 1

mfs

4 V 2 =4 X VI = 4 x 1.528= 6.112 mfs

From Bernoulli 's c!.juntiu ll. we

P

\C

1

h~ vc

P

VI

- ' + -'-=~ + -'- +" c' pg 28 pg 28

"' "'

,13~.7~3=4~,,,I~Oc· (1.528)2 11.772 x (10)' (6.1 12f -: + = + + /l c 9.8 1)(1000 2)(9.8 1 9.8 1x1 000 2x9.81 14.0 + 0.1 19 = 12.0 + 1.904 + he

[4.11 9= 13.904+11.. li e'" 14. 11 9 - 13.904:0.215

[I ]'

v' Bm fro m cqum ion ( 11.6). 11=-'----1 c

28 Co

Hence eq uali ng th e twO va lu es of he. we get

V ,' [I ]'

2~

Co - I

== 0.2 15 1

6 .11 2 [_I 2x9.81 Co

_1]" =0.2[.'1

]1 == =;~=--~" 0.215)(2.0)(9.8 1 :0. 11 29

I [ -C,. - 1

6.(12)(6.112

'rJ'=9 LO = 1.0 + 0.336 = 1.336 -LO - 1.0 == "v.! [2'1 = 0.336 or Co Co LO Cc = - - = 0.748. An s.

1.336

Problem 11 .13 A 150 mill Jiamerer pipe reduces ill diame/er abrllplly 10 100 mm diameter. If IIII' pipe carries 'Hiler m 30 lilres per second. caJculme IIII' preSSllre loss across IIII' cOil/rae/ion. Take the co·effieielll of con/rac/ion as 0.6.

I I

Ii

~ I

IL Flow Through Pipes

479 1

Solution. Gi ven: Dia. of large pipe.

v, '"

150

of liorgc pi pe,

A,'"

'4" (.15)

Dia. of smalle r pipe.

D1

100 111111 = LI Om

Area of smalle r pi pe.

A , '"

Ar~a

",

-

I11Ill '"

0. 15

III

=0.0 1767111"'

~ (. 1 0)~ = 0.007854 Illl 4

Q = 30 litrcsls = .03 m1/s

Dis.;hargc.

Co-cffi<:i c l11 o f contract ion .Cc '" 0.6

QO VI'" - = - .G3 - - '" 1.6 9 7 Ill is A, .0 1767

v, '" .Q. = -:;;.~03;;:-

and

•

Al

= 3.82 Ill/s

J)()7854

Apply ing Bernoulli's equation before and after co ntrac ti on,

,

1i+l+ Z _ pg

'- pg

2g

o.

.

fI, + V,"

Z, '"

2g

+2:1+

/1

... ( i) C

Z:!

and h c' the head loss due [0 contraction is gi ve n by eq uat ion (11.6) as

he

v' [= -;_8

1-I ]' =-,!'-""o 3.82 ' [-1 - - I ]' =0.33 2 x9.8 ! 0.6

Co

S ubMitulin g these va lues in eq uat io n (I). we gel

P,

1.697'

PI

3.82 ~

- + =- + + 0.33 pg 2 x9.8 1 pg 2 x9.8 1 ~ +O. 1467= P1 +.7438+.33 pg pg

li _ pg

p, '" .7438

+ .33 _ .1467 '" 0.927 1 m of wate r

pg

(PI - P2)

= pg x 0.9271'" lOoo x 9.81 x 0.927 1 Nfm 2 '" 0.909 x 10 Nfm 2 = 0.909 N/cm 2 4

Pressure loss across con trac tio n = PI - II! '" 0.909 Nfem 2. AilS.

~ I

I~

~ I

IL 1480

Fluid Mechanics

Problem 11 .14 In Fig. 11.3 below, 11'/11'11 a suddell cOlltraction is ,illlmdllced in a, horizontal pipe, line from 50 em ro 25 em. tile pressure c/umge.! from fO,500 /(g/l11- (103005 NinO fa 6900 kg/m(67689 Nlm!). Ca/cuiare Ihl' rale a/flow, Alsllme co·.jjicielll of cOlI/rac/ioll of jet ro be 0.65. FollolI'ing Ihis if Illere is a sudden enlargemellf from 25 em to 50 em alld if Ihe pressure (1/ Ihe 25 em seclio/l is 6900 kg/",l (67689 Nlm') wlul! is the pressure allhe 50 em elilarged sec/ion? Solu tion. Given:

Dia. of large pipe.

DI '" 50cm '" 0.5 III

Arc~.

AI = ~ (.5)'''' 0.1963

4 D, '" 25 em '" 0.25

Dia. of smaller pipe. Arca.

A 2 ""

Pressure in large pipe. Pressure in small er pipe.

'4" (.25)

m'

III

'" 0.04908

III

,

PI'" 10500 kgfm' ur 103005 Nlm' Pl '" 6900 kg/m' OJ 67689 N/m'

CD

' 0

-

pJE6900 kglm' D,= 25 em

D , ~50cm

p, zlOSOO kglm l

@

f-----' Fig. l1.J

Hc~d

los! duc to l'omraction is given by equation (I 1,6).

h~ =

l Vl [ _I _ 2g

Cc

Loll = 2gv/ (_'0.65_ _,)1 =0.2899 v/2g

... ( i)

From continui ty equation. we Ital'c

" ( 0.

°l'

5

0.25

x V, '" Y1 •

4

... (i i)

Apply ing Bcmoulli's equation at sections I·] and 2·2.

B",

I I

2 1 ", 21 (as pipe is horizontal)

Ii

~ I

IL Flow Through Pipes \C ~ ~ + _,_

pg

4811

1

V _ Pl + ---..L + II

2g-pg

2g

C

Substi tutin g th e val ues of PI_ fi 2. lic and VI _ we get

103005

+ (V,f4)2 =

IOOOx9.81

28

10.5 +

,.,;6~7~68~9c;-;-+ _V,_" + .2899 v/ 1000 )(9.81

2~

1.2899

V," V,' _0- _I , _0'"

28 3.6 '" 1.2274 x

"'

V1 ",

28

v'

v/ ,..c"-o-= 6.9 + 16 x 2 8

10.5 - 6.9 "" 1.2899

"'

28

16

V 1.2274 _,_'

2g

28

vo' _02,

I '" 7.586 m/s .I~3~6~'~2~'~9~8I 1.2274

(r) Rate of flow o f wa ter. Q'" A1V1 '" 0.04908)( 7.586 '" 0.3723 m J/s or 372.3 liUs. A ns . (ii) Applying Be rno ulli' s equati on to sections 3-3 and 4-4.

, V, _ P. v.;° + -"- + ZJ '" - + - + Z4 + head loss d ue to sudden enlar£crncnl (II,) pg 28 pg 2g Pl

-

Pl '" 6900 kg/Ill ' , or 67689 N/m 1

""'

VJ = V 1 ", 7.586 Ill/s V, 7.586 Y4 = V, '" = - - '" 1.8965 _0

4

4

Z}= Z4 And head Joss due 10 sudden enlargement is given by equa tion ( I 1.5) as

(VJ - VJ

II, '" '--'-:;-"-'- '"

28

S ub~tituling

(7.586 - 1.8965/ 2x9.8 \

= 1.65

III

the se va lues in Berno ulli' s equation. we ge l

~6,, 76:c89~

7.586 ' =

1000 x9.81 + 2 x 9.81

:;;;~I~"=~ + 1000 x9.81

1.8965' + 1.65 2 x 9.81

6.9 + 2.933 = -;:;;;!-I'o,"",,," -;+ 0.183 + 1.65 1000 x 9.81

I I

Ii

~ I

IL 1482 Fluid Mechanics -;c=PC''-c= "" 6.9 + 2.'0133 - 0.1 &3 - 1.65 '" 9.833 - 1.833 '" 8.00 1000 x 9.8 1

p~ '" 8 x 1000 x 9.8 1 '" 711480 N/m l. An s .

I 104. 3

Lon of Head at the Entrance of a Pipe.

This is the loss of energy which occurs

when a liquid enters a pipe which is connected to a large tank or reservoir. This Joss is simila r to Ihe loss of head duc (0 sudden contraction. This loss depends on Ihe fonn of entrance. For a sharp edge cntram:e. this loss is sligtnly more ttJan a rounded or be ll mouthed entrance. In practice the v,duc of

V'

loss of head at the cntram;c (or ililct) of a pipe wit h sharp cornered Cnlrance is taken", 05 - . where V", velocity of liquid in pipe. 2g This loss is denoted by h i

v'

/I j = 0.5 -

...( I 1.8)

2,

I 1.4.4

Loss of Head at the Exit of Pipe .

velocity uf liquiu

~t

This is the 105s of head (or energy ) duc \0 the

outlet o f the pipe which is uissipawd either in the fonn

of~

free jet (i f oullcl

ofth~

pipe is free) or it is lost in the tank or reservoir (if the outlet of the pipe is connected to the tank or

V'

reservoir). This loss is equal to - . where V is the velocity o f liquid 3t the outlel of pipe. This loss is 2g denoted

"0.

. ..(11. 9)

where V'" veloc ily al oUi let of pipe. I 1.4 .S loss of Head Due to an Obs truction in a Pipe. Whenever there is 3n obstruction in a pipe. tbe loss of enagy takes place due to reduction of tbc area of tbc cross-section of the pipe at the place where obstruction is present. There is a sudden enlargement of the are~ of flow beyond the obstruction due to which loss of heau takes place as shown in Fig. 11.3 (a) Consider a pipe of area of cross-section A hav ing an obstruction as shown in Fi g. 11.3 (a). u,t (I = Maximum area o f obstruction A = Area of pipe V= Velocity of liquid in pipe Thcn (A - a) = Area of flow o f liquid at section I-I. As the liqu id flows and passes through section I-I. a vcna-l·ontracta is formeu beyond $Cction I-I. after which tbe stream of liq uid widens again anu ve locity o f flow at section 2-2 becomes un iform and equal to velocity. V in the pipe. This situa tion is si milarto the flow of liquiu throu gh sudden enlargement. Let

®,

:

:

-~-

•

I

-

Fig . 11.3 (a)

- -

~-

.- - -

An obstruction in a pipt .

Vc '" Veloci ty of liqu id 3t I'cn a-contrac!a.

Then loss of heau due to obstruction = loss o f heau due 10 enlargement from l'ena·l"Unlracta to sect ion 2-2.

I I

Ii

~ I

IL Flow Through Pipes

(v, - V)'

.. .( i)

2, From co ntinuit y. we ha ve a e x V( '" A x V whe re a e " area o f cross-scctio n at vc na-contracta If Co '" l'O -c ffi cicnt o f contracti o n

...(i i)

c _ area al vena - CO nlracta

Th en

'"

( A - d)

c -

483 1

Q<

(A

a)

" c =Cc X( A-a)

Substitutin g lhi s value in (ii). we get C~ X(A - a)x

Su bstitutin g Ihi s value of

V~

Vr =A x V

in eq uat ion (i). we get

, ( AxV - v)'

Head loss due 10 ohstrl,.clion '"

(V.. -V )28

C« A - a)

'" "-'''''''7''---"2g

V ' ( A -,)1... 28 Ce (A - a)

( 11.1 0)

I 1.4.6 lo5S of Head due to Bend in Pipe. Wh e n there is an y bend in a pipe. the veloc ity of flow c hanges. du e 10 whic h Ih e se parat io n o f th e fl ow from th e bound ary and al so fo rmation of eddi es takes place. Thu s th e ene rgy is lost. Loss o f Itcad in pi pe due 10 be nd is ex pressed as kV ~

lib

= --

2,

where " b ::: loss of he ad due [0 he nd . V= ve loci!y of flow . k "" co·cffi cie nt of be nd The valu e of k dqlC nds o n (i) Ang le o f be nd.

I 1.4.7

(iil Radius of curva ture o f bend.

(iiI) Dia meter of pipe .

loss of Head in Various Pipe Fittings . Th e loss of he ad in th e vario us pipe fillin gs

such as va lves. couplin gs etc .. is expressed as

...( I 1. 1 I) where V = ve loci ty of tl ow. k = co-efficie nt o f pi pe fil li ng. Problem 11 .15 IViller is flowillg Ihrollgll a horizolllal pip/' of diometer 200 mm at a "e/ocity of 3 mls. A circuli" solid plute of diameter' 50 mm is placed ill IIII' pipe /0 obstruct IIII' flow. Filld Ihe loss of Ilead due 10 oDslrue/ioll ill Ihe flipe if Co = 0.62.

Solullon. G iven: Di a. of pi pe. Ve loci ty.

D = 200 mill = 0.20 m V= 3.0 m's

Arca o f pi pe.

A=

Dia. of obstru cti o n.

d= 150m lll =0. 15 m

Area of o bstruc tio n.

I I

~ 0' = ~

"

(0.2) ' = 0.03 141 m'

u= "4 (. 15) =0.0 1767 m

,

Ii

~ I

IL 1484

Fluid Mechanics

The head los! due [0 obstruction is give n by equati o n ( II 10) as

1.0

J

3x3 [ .03 141 2x9.81 0.62 [.03 141- .01767]

LOr

9

""':;-;;713.687 - 1.011 = 3 .3 11 m. An s. 2 x 9.81 Problem 11 .16 Determine the rale of flail' of Iruler Ihrougll (l pipe of diameter 20 em and length 50 III when olle end of Ihe pipe is connected 10 (I limk (Ind other end of Ihe pipe is open /0 111ft atmosphere. The pipe j~. horiZo,,'a/ {lIId Ihe height of wo ler in the lank is 4 m aboI''' Ihe centre of Ihe

pipe. Consider all minor losses and /(Ike f == .009 ill Ihe formllio h! " ./. f . L. V I d x2g Solution. Dia. of pipe.

d= 20 cm == 0.20 m

Lengt h of pipe,

L = 50

In ","fER SUflf"'"

of waler.

H=4 m

Co-efficient of friction.

/= .009

H ~igh l

Let the ve loc ity o f water in pipe

_ ~~

__'-4

" V mls.

Appl ying Bernoulli 's eq uation al the top of th e wate r su rface in the tank and at the o utl et of pipe, we have ITakin g point I on the top and point 2 at the oU llet of p ipe l·

Fig. 11.4

~,

E.!... + _,_ +<, '" ~ + -v''- + l, + all pg

2g

pg

2g

•

losses

Considerin g datum line pass in g through th e centre of pipe 0+ 0 + 4.0= 0+

v;

4.0= - '- + h;+ ",

2,

But the ve locity ill pipe = V,

:. V = V1

V'

4.0 = -

2,

V'

From equatio n (11.8). h; '" O.S -

2,

~ I

+11; +",

...(i)

and h, fro m eq uat ion ( 11.1) is g ive n as

I~

~ I

IL Flow Through Pipes

4851

Substituting these val ues. we have

2 4.0= V" + O.5V + 4 .!.L.V"

2g

=

2g

dx2g

~ [ 1.0+0.5 + 4 X.OO9X50 l= ~ 28

0.2

= 10.5 x

V' 2,

4 x 2 x 9.81

V=

[1.0 + 0.5 + 9.0]

2g

10.5

= 2.734 m/sec

Q = A x V = ~ X (0.2)2 x 2.734 = 0.08589 111 3,s

Rate of now,

4

= 85.89 lil res/s. AilS. Problem 11 .17 A iloriw,,/a/ p ipe /i",' 40 m /Oll8;S cOllllec/cd 10 II ...Mer /
L

= 40 m

lA:ngth of 1M pipe.

LI=25m

DiJ. of lSI pipe. lA:ngth of 2nd pipe.

L 2 =40-25= 15m

ill = 150 111111 = 0. 15 m

=

=

Dia. of 2nd pipe. d 2 300 mm 0.3 m Hcight of wate r. fI = 8 m Co-e flici ent of fric tio n. /= 0.01 Applying Bernoulli' s th corem 10 the free surface of water in the tank and oUllet of pipe as shown in Fig. I 1.5 and t<[king refe rence line passing through the centre of pipe.

'm j L__ _

"V~, __r---~::;,~-

L, " 25 m Fig . 11 .5

v'

0+0+8= p~ +-'- + O+all iosscs pg 2g

I I

Ii

~ I

IL 1486 Fluid Mechanics v,!

8.0:0+ 2~ +/J i +l!f, +11, +11"

0'

... (i)

~' whe re h j = loss of head al enlfance = 0.5 -'-

2,

/'f,

= head

InS! due \0 friction i ll piP<" I

4X[ X L,XV;l

= :::'-'f:~:'-'"­ d 1 xl,

(V, _ ¥1)1 2,

I"~ = loss head duc \0 sudden enlargement '" ,",-;;---"'-

hf = Head lost duc to friction in pipe 2 =

,

4xfxL xv,l 2_ dz X2g

BUI from continui ty equation. we have A ,V, '" A 2V 2

I, . = _0_' _"_,1 '" "0.,, ' "x ,, (4"V"J_' •

2,

2,

28

4 xO.O I x 25 X(4Vlr II" =

=

11, '"

0.15x2x8

4x.0 ] x25xI6 0.1 5

v/

x~

2,

V' 2,

= 106.67 -'-

(V, - V2)2

(4Vl - Vj

1 9V1

28

2,

2,

_ 4x.O l x l 5xYll _ 4x.0 I x [5 I' f -

'

Suh~liluling

0.3x2,

Ihe values of tllese

lo~scs

0.3

V}

x ~

2,

-20 - . x

V}

~

2g

in equation (i ), we gel

V,l 8V,' V," 9V," 1 8.0= -'- + - - +106.67 -'-+-'- + 2x V:

2,

v'

2,

= -'- [I 2,

-+-

8

2,

-+-

106.67

8.0x2xg '"

126.67

..

~ I

-+-

2,

2,

v',

9 + 2J = 126.67Tg

8.0x2x9.8 1 ", JL2391 = J 113 IIl/s

126.67

, , , Ralc of flow, Q == A, x V, == - (0.3) ' x I 113 " 0.01861 III Is" 18.61111,..,s/s, ADS. ,

4

I~

~ I

IL Flow Through Pipes

4871

Problem 11 .18 D etermine Ihe difference in Ille <'iel'lliioll ,~ bet ..... een the ..... lIIer sllrfaces in Ille 110'0 tallks which are cOlineCled by a horizontal pipe of diameter 300 mm (//11/ Jeng lll 400 m. The rate 0/11QI<' ofwmef through Ihe pipe is 300 li/res/s. Consider all/asses (lnd take Ihe ra/lle of! == .008.

Solution. Given : Dia. of pipe,

Length. Disc h.arge . Co-cffjcicnl of fric tion.

11= 300 l111n == 0.30 III L = 400111 Q = 300 li tis = 0.3 m 3,~

/= 0.008 v

Vc locily,

~ _ Q_ ~ -;;-~O.~3--:-

= 4.244 IlIfs

~ X {.3 }l

Are a

4

Le t the Iwo tank s im: conne<:tcd by a piP<' as show n in rig. 11 .6.

-0-1

!"' H, -

l: 400 m d- O.30m

H,

l

t

- v Fi g . 11.6

Let HI = he ig h1 of wmer in lSI lank above th e cent re o f pipe Hl = he igh1 of wme r ill 2nd lank :Itxwc th e ",,"He of pipe The n difference in elev ati ons belwee n wat er surfaces = H I - Hl Appl ying Be rno ulli 's equaliOIl to lh" fr"" surface of wa leT in the lw o tallks , we hav" H , '" Hz + losses ", f1 1 +

...( i)

h i + H f , + 110

v" 0.5 )( 4.244 " where I. i ", Loss of head at e ntran ce'" 0.5 ~ 28 2 x9.8 [

"':;c",,,,'-- '" 0.459 III

hJ, '" Loss of head due to frict ioll '"

4 x [ x L x Vl 4 x .00&x 400x 4.244 ' '" ""~';;';'::",";;';;'C''--dx2g 0.3x2 x 9.81

v" = 4.244 '

h" '" Loss o f head at out le t ", -

2g

2 x 9.81

'" 0.9 18

'" 39 . 16 III

III

Substitutin g lhese values in (i ). we get H , '" Hl + 0.459 + 39.[6 + 0 .9 18 '" Hz + 40.537

Problem 11 .19

H , - Ill '" Differe nce in e lev ati o ns '" 40 .537 rn. Ans. The Irk/ioll jtKlor [or /Ilrblliell/ flow l" rOllgh rOllgh pipes CUll be delermined by

Kamw,,· Pramlll elfllalin",

h

= 2 loglo (Rolk) + I .N

...llere ['" [ric/iOI' factor, Ro '" pipe radiuI, k = aI'erage rouglmess.

I I

Ii

~ I

IL 1488

Fluid Mechanics

Two resen'oirs Willi a ,I'ur/ace lel'eI difference 0[20 melre.1 are to be (onne(fed by f melre diam eter pipe 6 kill long. What will be Ille tli.leilarg" when a caSI iron pipe of rOllglllle$s k '" 0.3 mill is IUi'd ? Whm will be Ihe percel1lt1ge increase in rhe discharge iflhe caS I iron pipe is replaced by a Sleet pipe of roughness Ii. = 0. 1 mm ? Neg /eel aI/locH/losses. Solution. Given:

1,:20m <1= I.Om

Di ffere nce in le vel s.

Dia. of pipe.

:. Radius. Ro = 0.5

11\

= 500

111111

Length of pipe. L = 6 kill = 6 x [000 '" 6000 In Roughn ess of c ast iron pipe. Ii. = 0.3 mm Roughness of st~cI pipe, Ii. = 0.1 mm lst Case. Ca~llro" Pipe. First fi nd the v alue of friction factor usin g

,

"Jj=2 10 g lO (R J k)+ 1. 74

= 210g IO (500/0.3) + 1.74

f~

(_, _)1 8.1837

...(1)

= 8.1837

= 0. 0 149

Local losscs arc to be negl ected . Th.is means onl y head loss due to frict ion is 10 be considered. loss due to friction is

~lcad

20",! XLXV ' d x2g [Here f is the friction factor and not co-efficie nt o f friction .: Friction factor", 4 x co-efficient o f friction]

20 '"

v'"

.0149 x 6000 x V ' '" 4.556 LOx2x9.81

J4.~~6

.

v2

'" 2.095 rn ls 11

2

II

,

3

DlschHrge_ Q 1 = Vx Area = 2.()95 x - x If '" 2.()95 x - x 1-= 1.645 rn Is

4

4

2nd Case. Sleell'ipe. k",O.1 mm.R o = 500mm Substitutin g th ese va lues in equatio n (I). we get

, J1 = 2 loglo (50010. 1) + 1.74 '" 9. 1379 f'"

(9J~79

r'"

0. 0 11 9

f xLxV 1 .0119x6000xV 2 Hcad loss duc 10 fricti on. 20 '" L",~,,_ or 20 '" "C"c"'-;"'~"-_ " 3.639 Ifx2g 1.0x 2x9.81

I I

v2

Ii

~ I

IL Flow Through Pipes

:. Discharge,

v=

J3.639 20

'" 2.344 mls

01 =

VxAn:~

'" 2.344 x

pcn;cllwgc inncasc in the di scha rge ==

- 0I

Q

1

'4" X

X 100:=

Q,

I '" 1.841

( 1.84 1- 1.645 )

1.645

111

4891

,Is

x 100 '" 11 .9 1<,t,. An s.

Desig" IIII' dill"'eler of (l sleel pipe 10 ( .Try IIlIl/er II",';ng kinematic 1';KO~ily IO-l! ",lis willi II mean I"e/Ocily of I m/s. The /,eat! loss iJ" /0 be limited /05 m per I()() m /ength of pipe. Con sider the ",/,,;m"'"( sa",/ rollg/mess Iwig/II of pipe k, := 45 x fO • em. AISllme I/Wlille Darcy Problem 11.20 1' =

Weisbach friction jaClor oreT Iile whole rmlge of lurbulent jlOII" ellll be e.l'(Jressed
w/lere D = Diameter of pipe and R, = ReYliolds lIumber. Solution. Gi ven: y:= IO-~ 11111s

Kinematic viscosi ty. M~an velocity. Head loss.

V=lmfs

h[ =5 m in a knglh L= 100m k, := 45 X 10---1 em = 45 x 10...... m

Value of

Value of

f = 0.0055 [ I + ( 20 X tO

Usin g Darc y Weisbach equa ti on. hf =

i=

J

~ + IZ,~ )"l]

... ( i)

4X f XLXV 1

Dx2 g

h, xDx2g

, = 5xDx 2x9.81 :0.2452D 4xlOOxi l

4x LxV -

Now the Reynolds number is given hy.

= IXD =I0 6 D IO -~

Substitutin g the va lu es oFf. R, and k, in

cqu~t io l1

(i), we gel

l 10'D 0.2452D=0.0055 1+ ( 20X IO x 45 xD10-' + I O~ [

0.2452 0.9+ I - - 0 =[1+(0.0055

I I

0

0

J'' ]

J'' ] Ii

~ I

IL 1490

Fluid Mechanics

44.581J :

0'

[ (i19)'''] 1+

Of

44.58 1J - 1 '"

(-t-19)'"

(44.58 D - IY'= .!.:2. or D(44.58D-ll= 1.9 D

... ( i i)

Equation (ii) is so lved by h.i! and trial method. (il Let D = 0. 1 111. then LH .S. of equation (ii) oc(:omcs as L.H.S.=O.I (44.58xO. I - l »)=O.1 X 3.45S3 = 4. 135

This is mo re than the R. Il .S. ( ii) LeI D '" 0.08 m. tllen L. H.S . of equation (iJ)

bcCQIll CS

as

L. H.S. '" 0.08 (44.58 x 0.08 - 1)3", 0.08 (2.5664)3 '" l.J52 This is less than the R.H.S . He nce va lu e of D Ik s between 0.1 ~[l(l 0.08 (iii) LeI D "" 0 .085. th e n L.H.S. of equation (il) bcl'Omcs as L.II.S. == 0.085 (44.58 x 0.085 _ 1)3 == 1.S44

This val ue is sli ght ly less than R. H.S. He nce increase the va lue of 0 slightl y. (i,') LeI D '" 0.0854 m. then L. H.S. o f equati o n (iil hecorncs a~ L.H.S. '" 0 .0854 (44.58 x 0.0854 _ 1)3 '" 1.889 Thi s va lue is nea rl y equa l to R.H.S. :. Com~et va lu~ o r D = 0.0854 m . Ans. Problem 11 .21 A pipe line AR of diameter 300 mm and of lenglll -100 m ,'arries waler al Ilw HilI' of 50Iilres/s. Tlwflow lakes place from A 10 B wi,erI' poinl B iJ' 30 melres aho\"(! A. Fi",l lile press"re al A if Ihe pUJs" re al B is 19.61 Nkm !. Take f = .008. S o lution. G ive n: Dia. or pipe. d = 300 mm = 0.30 m Len gth or pipe.

L =400 m

Di sc h~rge.

Q = 50 litresls = 0.05 m 3fs

:. Velocity.

V=

--.SL = Area

0.05 =

0.05

~d 1

~ x (.3)1

4

4

= 0.7074 mls

/Is = 19.62 Nfcm 1 = 19.62 x 104 Nfm l

Pres,ure at n .

/= .008 Apply ing Bernoulli's eljuatio ns at points A and lJ and taking datum line passing throu gh A. we have

vi

PA V; P8 - + - + lA = - + pg

Bo'

2g

pg

VA

2g

+ ZII+ h!

= VB

[.: Dia. is same [

ZA=O.lf/ = JO

"od

I I

4 2X~f~X~L:;-::X ~V_' 11= -~ f dx2g

Ii

~ I

IL Flow Through Pipes [9.62x I0

4

491 1

+30+ 4 x.008x 400x.7014 '

1000:><9.8 1

0.3 x 2x9.8 1

:20+30+ 1.088:51.088 f!A '"

=

51.08& x 1000 x 9.81 5 1.088 x 1000 x 9.81

'" 50. 12 N/cm 1, An s.

... 11 .5

\

Fig. 11 .7

HYDRAULIC GRADIENT AND TOTAL ENERGY LINE

The concept of hydraulic gradien! line and total energy line is very us.cful in the siudy of flow of

fluids through pipes. They 3fC defined as: [1 .5 . 1

Hydr;Julic Gradient Line.

h is defined as the line which gives the sum uf pressure head

( ~ ) and datum head (2) of a flowing fluid in a pipe with respect to some reference line or it is the line which is obtained by joining the lOp of all venical ordinates. showing the pressure head (plw) of a flowing fluid in a pip
I 1.5 .2 Total Energy Line . II is defined as the line which gives the sum of pressure head. datum head and kinetic head o f a flowing fluid in a pipe with respect to some referen~ line. II is also dernwd as the tine which is obtained by joining the tops of aU I'ertical ordinates showing the sum ofpressurc head and kinetic head from the centre of the pipe. It is briefly written as TEL. (Total Energy Line). Problem 11.22

For tile problem 11./6. draw Ih;: ffydrm,/ic G",di;:,1/ Line (H.G.L.) and Total Ellergy Lille (T.EL).

Solution. Given:

L" 50 m. d" 200 mm" 0.2 m H"

4 m./==.OO9

Velocity, Y through pip.: is calculated in problem 11.16 and its value is y" 2.7J4 mfs Now. == Head Inst at entrance of pipe y 1 0.sx2.734 1 == 0.5 - + ==O.19m 2g 2x9.8 1

"j

,

1:e_L 4m

1

- - hJ;,j.

----~ 2 ___..g,

B.... - - - - --- - -

38

L:50 m

d:20=

Fig. 11.8

and h, == Head loss due to friction

I I

Ii

~ I

IL 1492 Fluid Mechanics == 4 X j XLXV ':: 4xO.009x50x (2.734 )' = 3.428 n1. dx2g O.2x2x9.81 (a) To tal E ne rgy Lin e (T .E.L.). Conside r three points. A, Band C on Ihe free s urface of wat er in th e tank. at the inle t of the pipe and at th e o utlet of the pi pe respecti ve ly as show n in r ig. 11 .8. Lei us find total e ne rgy at these points. taking the ce ntre o f pipe as rcfc~ncc lin e.

I. Total e nergy alA

II

=:

-

pg

V"

+-

2g

+ l = 0 + 0+ 4 .0

2 . Total e nergy a1 B :: Total e nergy at A - Il j p 3. Total e nergy at C = - ' pg

V

:=

=:

4111

4.0 - 0 .1 9 == 3.81 III

V1

l

+ 0:: 2.734' '" 0.38 m. 2 x 9.8 1 2g Hence t Ol ~1 ene rgy line will coincide wi lh free surface uf water in th e tank. At the inlet of the pipe. it w ill decrease by II; (= O. [9 111 ) from free s urface ami at outle t of pipe IO ta l ene rgy is 0 .38 m. He nce in Fig. [1 .8, (i) Puin! D re prese nts total e ne rgy at A (;1) Po int E. where DE " I" , represc nts total e nergy at inl et o f th e p ipe (iii) Po int F. where CF ", 0 .38 represe nts total e nergy at outl et o r pi pe. Jo in D to E and E to P. T he n DEF represc nts th e to tal cncrgy lin c. (b) Hydra ulic Gra di ent Line ( H.G.L.). H.G. L. g ives the sum o r (plw + z) with refercnce to the

+-'- + Z,. :: 0 + 2g

dalUm · lin e. Hence hydrauli c gradient line is

V'

by subtractin g -

o htai n ~d

2g

from to tal ene rgy lin e. At

Vl

V'

outlet of th e pipe. tot al e ll crgy " - . By subtracting fro m total e nergy at thi s point. we sha ll ge t 2g 2g poin t C, whic h li es on the centre lin e o f pipe. Fro m C, draw a lill e CO p arall c l to EF. The n CO represc llts the hydraulic gradi cnt line . Problem 11 .23

For Ihe problem 11. 17. draw Ihe II),dralllic gradiem am/lOwl ellerg), lille.

Solution. Refer to problem 11. 17. Gi ve n : L ",25m. d l ",0.1 5 m I

L2 ", 15m. d 2 : O. 3 m. /" .0 1.1I :8 m The ve locity V2 as

e al c u l at~d

in probl e m! 1.1 7 is V2 ", 1.1 13 mfs VI '" 4 V2 '" 4 x l. ln '" 4.452 m/s

vi

The vario us he ad losses arc iI; '" O.5 x " -"O".5cx,,40.4,5"2~' " 0 .50 m 2g 2 x9.8(

"Ji

I I

=

~ 4 f,-,;X~~'7 X ~V,,' d l x2g

=

4 x .OI X25x (4.452)' 3 = 6 .7 III 0.15x2x9.8 1

Ii

~ I

IL Flow Through Pipes

hi,

_

~ 4 ~'2f~'~L, ~_~X~V ~,'C '" 4 x .oJ x 15 x (1.1 13f -

V,1 Ll l3' II == - -=

o

0.3x2x 9.8 1

d"x2g

2g

2 x9.81

493 1

",D. 126m

== 0.06] III

Alw

T otal E nergy Line (I) Point A lies on free surface of water. (ii) Take All "" h; == 0.5 m. (iii) From R. draw a hori zo nta l line. Tak e BL

~qual

[Ollie leng th of pipe. i.e .. 1- 1" From L draw a

vt' rtical line downward.

(il') CUllhc li ne LC" h" == 6.73 m. (.,) Join the point lJ \0 C. From C. take a line CD verticall y do wnward equa l 10 11, := 0.568 m. (d) From D. draw DM hori zontal and fro m point F whi c h is lying on the cenlre o f Ihe pipe. d ra w a vCT1ic~ l linc

in the upward direc tion, meeling at M. From M. tak e a dislam::e ME = h" = 0.126.

10il1 DE.

The n lin e ABCDE rc prcsc llIs th e total e nergy li ne.

I

Fig . 11.9

Hydrpulic Gradicnt Lin e (H.G.L.)

V'

0) From IJ. take BG '" - '-

2g

= 1.0 rn.

(ii) Draw th e line GH parallel to the line BC (iii) From F. draw a line Fl parallel to the lin e ED . (il') Join the poi nt H and I. Then th e line GHIF represen ts the hydraul ic gradicntline ( H.G .L.). Pro blem 11 .24 For Problem 11./8. draw Ihe hydrrw/ic gradienl "",{Io l,,/ e"ergy /i"e. So lution . Refer to Problem II.IS.

Given:

d= :\00 mm "" 0.3 m

L = 400 m. Q "" JOO liucs/s "" 0.3 11I 3/s

f=

I I

.OOS

Ii

~ I

IL 1494

Fluid Mechanics II, '" 50

11 2 ",

BUIU 1 - 11 2 '" 40.53 7 50 - 40.537 '" 9.463 Ill. Ill.

In

(Calculated in Problem 11.18)

The calcu lated losses arc: (!)

"i'" 0.459 m

(iii) 11,,:0.918 (a)

= 39.16 m

(ii) ""

III

T.E.L.

(i) Point A is on [h e fr<)c surface of W3Icr in 1st tank.. From A. take All '" h i '" 0.459 m. (ii) Draw a horizont al line Bf', Take HF equal 10 the length o f pipe. From F. dmw a vertical line in

the downward di rection. CUI Fe", h/, '" 39.16 m.

lie. From C take CD '" 110 = 0.918 m. T he point D should coinci de with free surface of water in 2nd tank. Then line ABeD is the IOtal e nergy li ne.

(iii) Join

-.L h,=0 .459 m

A

l"'4wm d=O.3m

Fig. 11 .10 (b) H. G.L. From D, draw a li ne DE parJlIcI to line Be. The n DE is the H.G.L.

0,

v'

From IJ, take BE '" -

2,

'" 0.9 18

III

and from Edraw a line ED parallel

10

BC. The poim D should

coi llcide with free ~urface of wate r in the 2nd tan k. Th~n line ED represcnls the H.G.L. Problem 11 .25 Ti,e rale ofjlow ofwllter pllmped illlo a pipe ABC. which is 200 mlong. is 20 litres/s. The pipe i~· laid on an upward slope of I in 40. TIl/' length of the porlion AB i5 100 m Imd it~· di",neler il· /00 mill. while Ihe length of the portion BC is also 100 m but il~· diameter is 200 mm. The change of di",nt!/er (It B i5 l·udden. Th e jlOII' is laking pia,·" from A 10 C. wlwre Ihe pressure al A is 19.62 Nkm 2 and end C i5 connecled to (I /imk. Find Ihe pressu'e al C (Illd draw IIII! /lyil,,""ic gradient lind lo/al energy line. Take f = .008. Solu t ion. Given: Leng[h of pipe. ABC = 200 m Discharge. Slope of pipe.

Leng[h of pipe. Lenglh of pipe, Pressure a[ A. Co·efficicnt of friction.

Q = 20 litrcsls = 0.02 m 31s .

I

= 111140 = -

40

AB = 100 Ill. Dia. of pipe AB = 100 mm = 0.1 m

Dia. of pipe BC = 200 mm = 0.2 PI! = 19.62 Nfcm" = 19.62 x 10· Nlm"

BC = 100

111.

111

f = .008

Velocity of water in pipe AB. VI =

Disch~rge

Area of AS

0.02 = - - = 2.54 lilts

~ ( .l )2 4

I I

Ii

~ I

IL Flow Through Pipes Velocity of water in pipe

1Jc.

Q

V2 '"

Area o f

Be

495 1

'" 0.02 '" 0.63 111/s '::' (.2)2 4

Apply ing Bernoulli's cqumion \0 points A and C. V1

~+~+

pg

28

P z,\ '"

_C

pg

V!

+ _<_ + Zc + total 2g

loss from A to C

... (i)

TOlal loss from A to C", Loss duc to friction in pipe AB + loss of head duc to en largement at B + loss of head duc to friction in pipe Be. ... (ii) Now loss of head duc 10 friction in pipe AB. /1 , =

4 J1-V2

4 x.OO8 X 100 X (2 54)2

" d x2g

Loss of head due 10 fric tion in pipe II

/,

==

O.lx2x9.8 1

=10.52m

Be.

4 X.008 X100 X(0.63)1 '" 0.323 0.2 x2 x9.8 1

III

Loss of Itcad due to en largeme nt at B,

TOl allos.~froIllA

IOC

= h/l +h, + hI,

= 10.52+ .186+ .323= 11.029'= 11 .03111

SubstitUTing this v:1luc in (I). we gel

V' JI V1 P --=l..+.....i.. + z - ....!:. + _c_ + pg 28 '\- P8 28

l

+ 11 .03

. ,.( iii)

c

Taking datum line passing through. A. we have Z,\ == 0

<;- '"

Also

4~

x lotal length of pipe '"

4~

x 200:0 5 m

PA'" 19.62 X 104 N/m" VA'" V I = 2.54 mls. Vc = V1 = 0.63 mls

Substituting these va lues ill (iit). we get

19.62 x (10)" + (2.54)" + 0 =.&+ (0.63)' + 5.0 + 11.03 lOOOx9.81 2 x 9.81 pg 2x9.81 20 + 0.328:: .& + 0.D2 + 5.0 + 11.03

pg

20.328:: .& + 16.05

pg

I I

Ii

~ I

IL 1496

Fluid Mechanics

Pc == 20.328 _ 16.05 == 4.278 m

pg

Pc = 4.278 x I OOOx 9. 8 1 N/m 2

4.278 X 1000 x 9.8 1

= -'~"""'~";'''''""' Nlc(11" == 4. 196 N/c l1l ~. ADS. (0' H ydrllullc Gradient and Total E""rg," Un., p

, 1 ~''1052 , f" '~ ." "'G,. , " 1 ,

tt(

20 328

~

~

Lhf ·n3 , 'll ' c

....

---~

-

__ -

,

'",

Fig. 11.11

PIp.: AU . Assumin g tlie datum lin e pass in g th ro ugh A. th en IOlal e nergy 3l A

V;

p~ pg 2g == 20.328 III

=

Total e nergy at B

4 19.62 X 10 + (2.54), + 0" 20 + 0.328

l 000x9.81

'" Tot al e nergy at A !

A lso

~ + ~ +"

Vc 11g

=

(0.63)2

1 x9.RI

" ft

=:

2x9.8 1

20.328 - 10.52", 9.&08

III

== 0.02.

Tota l En"rgy l.ine . Draw
,

40 x 200 '" 5

111

fro m th e lin e AX. Now draw a vc ni ca l line A D cq ua lto IOta l ene rgy at A . i.e .. A D

= 20.)28 m. f'ro m poi nt D. draw a ho ri zont al line and fro m poi nt /J. a I'c n ica l line. meetin g at Q. From

Q. lake I'e ni cal di stan ce QE == h" = 10.52 1Il. Join DE. f'rorn E. tak e EF = h, = 0. 186 m. Fro m F, draw a hori zo nt al line and from C. a ve ni ca l line meet in g at R. From R ta ke RG = h" == 0.323 Ill. Join F to G. Then DEFG re prese nts th e lota l energy line. 1I~' d raul ic Gr.ldl enl Line. Draw the line 1M parallel to the line DE at a d istan ce in the dow nward

V'

direc tion eq ua l to 0.328 Ill . Also draw the line PN para llel to the line GF at a d islan ce of - '- = 0.02.

2,

Jo in po int M to N. Then lin e UlINP re prese nts the hydrau lic g rad ie nt lin e. Problem 11.26 A pipe lille. 300 111111 ill diameter (md 3200 III 10llg is IIsed to plllllp lip 50 kg per second of (Ill oil ...llOse dellsity is 950 kg/m J (md ... llOse kinematic viscosiry is 2.1 stokt'S. Tile centre of Ihe pipe line oll/u: IIpper elld is -10 m abore th(l1l Illal al rile lo·....er end. The disclwrge (II Ihe IIpper end is armospheric. Fi/ld Ihe pressllre at IIII' lower elld WId draw the Il)'dralllir gradient alld Ihe total energy line.

I I

Ii

~ I

IL Flow Through Pipes Solution. Given: Dia. of pipe.

497 1

d = 300 IlUll '" 0.3 III L=32oom

u,ngth of pip!:,

Mass.

M =50kgfs=p.Q

50 50 3 Q= - = '" 0.0526 m Is p 950

Discharge,

Density. p '" 950 kg/ml Kin ematic viscos it y. \' == 2.1 stokes" 2.1 cmlls = 2.1 x 10-4 m1fs Hcigtn of upper end

" 40

Pressure at upper end

" atmospheric = 0

In

vxd Discharge R, == - - . where V ==

Reynolds number.

"'7"'''' Area

v

R, '"

0.744 x 0.30 2.1 x 10

=I

~ (O.3)1

== 0.744 mrs

06'

~.S

f=~= _I6_ =O.QI5

:. Co..efficicnl of friction.

R, Head losl due \0 friction.

•

0.0526

11 == f

=

1062.s

~4~X~f~X~L",X~V_' d x 19

4 x 0.015 )( 3200 x (0.7 44 )1

0.3)(2x9.8 1

= 18.05 m ofoil

Apply ing the Bern oulli 's equation ;1( the lowe r and upper cnd of the pipe and tak in g datum line through the lower end. we have

p~ssing

But ZI '" O. Zl '" 40 m. VI = V1 as diameter is M ill e Pl=O.llf = 18.05m :. SubSlituting lhese values. we have

l!.!. '" 40 + 18.05" 58.05 pg

In

of oil

PI'" 58.05 x pg '" 58.05 x 950 x 9.8 1

I': p for o il '" 950)

H.G.L. and T.E.I ,. VI 2g

~ I

( .744)2

"

2x9 .81

'" 0.0282 m

I~

~ I

IL 1498 Fluid Mechanics

J!J... = 58.05 111 of oil pg

,

0.0282 m

r

fH: G./.-- , i, l;, (

/

58.05 m

T
Draw a horizontal line AX as shown in Fig. 11.12. From / / A. draw the CClllrc line of llie pipe in such
.. I 1.6

flOW THROUGH SYPHON

Syphon is a long ben! pipe which is llscd to transfer liquid from a reservoir al a higher elevation to another reservoir a1 a lowe r level wilen the (WO reservoirs arc separated by a hill or high level ground as show n in Fig. I [.1 J.

:;;:C;:;:;~ SUMMIT A - ...,,--- -- --

,

-----

Fig. 11.13

The point C which is at the highest of the syphon is called the summit. As the point C is above the free surface of the water in the tan~ A, the pressure 3t C will be less than atmospheric pressure. Theoretically. the pressure at C may be r~-duccd to - 10.3 III of water but in actual pra~1ice this pressure is unly - 7.6 III of water or 10.3 - 7.6 = 2.7 10 of water absolutc. If the prcssure at C be~'Qllles less tl1:ln 2.7 III of water absolute , the dissolved ai r and other gases would eomc out from water and ~"OlIcct at the summi t. The flow uf W3ler will be obstructed. Syphon is used in the fullowing cases: !. To carry water from one reservoir tu another reservoi r sepa rmed by a hill or ridge. 2. To t:lke out the liquid from ~ tank which is not having nny outlet. 3. To cmpty n channel nOl provided with nny outlet s luice.

zoo

P ro blem 11 .2 7 A J)"p ilon of diomelU mm co"nects 11"0 reserl"oirs Iwdng a diffuence in efem· lioll of 10 m. The lellglh of I/Ie sypholl is 500 m amllhe Siumnil is 3.0 111 abOl'e Ille Waler {ere/ ill Ille upper reserl"oir. The le"glh of Ihe pipe from upper resu}"oir 10 Ihe su mmil is 100 m. Delermi"e the di~'charge Illrough Ille $)"1'/10" alld "Iso press"re
Solution. Give n: Dia. of syphon. Difference in level uf twu reservoirs. Length of syphon,

I I

d", 200 mm '" 0.20 III H=201ll

L=500m

Ii

~ I

IL Flow Through Pipes Hciglll of s Ulllmit frolll upper reservo ir. Le ngth of sypho n UplO su mmit. Co-efficient of friction.

4991

h '" J.O III L1 ", 100m f == ,OOS

c

, Fig. 11 .14

If minor losse s arc neglected thcI11hc luss of head lakes place only duc 10 friction. Apply ing Bernoulli' s eq uati on to poi nt s A and B.

V1

J!.ii + ~ + ZA pg

2g

VI

'"

~ + ~ + z~ + Loss o f head duc pg 2g

to friction from A 10 8

0+ 0 + LA '" 0 + 0 + l /l+ IIJ I'; PA '" I'D '" atmospheric pressure. VA '" VB'" OJ

"'

1... -Z8= /IJ =

B"

l ,\ -18'"

20

4xjxLxV 1 dx2g

III

20= 4X.OO5X500XV 0.20)(2)(9.8 1

v=~

l

=2.548V1

20 =2.80mls 2.548

Q'" Veloci ty x Area

:. Disc harge ,

'" 2.80 x : (.2)1", 0.0879 mJ/s '" 87.9 Iilresls. An s. Press ure at S ummit. Applyi ng Bernoulli' s equ ation \0 poin ts A and C,

P V1 --
28

P

VI

V'

"'

, .

= - ' + -'- + l c + Loss of head due 10 fneltOH b<:twccn A and C pg 2g

O+O+O",&+ - + 3.0+hj, pg 2g ,

ITaking datum line passing through AI

p, + ;;;-~"~'= 4 00~5cX:cc'00 ""X+.(2=8"--)' O" _pg + 3.0 + - .,X~.,, 2x9.8 1 0.2x2x9.81

=

&

pg

I I

lVc =V= 2 .801

& + 0.399 + 3.0 + 4.00 == & + 7.399 pg

pg

== - 7.399 m or wute- r. AilS.

Ii

~ I

IL 1500

Fluid Mechanics

Problem 11 .28 A SypllOIl of diameter 200 mill C()IweCf$ two re~·er\'oi'.1 hal'ing II differene<' in eiel'arion of 15 m. Ti,e 10iailenglh oflhe syphon is 600 In oll
d"" 200 nUll == 15 III

Differe nce o f level in two rC!i.ervoirs

TOlat ]englli of pipe Hei glll of summit from uppe r

= 0.2

III

== 600 III == 4 III

rc~rvoir

.E.::.. = 2.8 III of water absolute

Press ure head al su mmit.

P8

& = 10.3

Atmospheric prcssure licad .

III

of water absolute

P8

,'" .004

Co-efficient of frictio n. f

A

, Fig. 11.15 (a)

Apply in g Bernoulli's eq uation 10 points A and C and taking the datum line passing throu gh. A , v~ Vl l!..i. + ..:..d... + ZA '" .&. + _c_ + z.- + Loss of head due to frict ion between A and C pg 2g pg 2g Substitutin g the va lu es o f pressures in terms of absolute, we have

v'

10.3+0 +0 =2.8+ -

2g

+4.0+ II ,

vo! h~

,

[.: V< '" velocity in pipe '" V]

"

VI

'" 10.3 - 2.8 - 4.0 - - = 3.5 - -

2g

Apply ing Bernoulli's equation to points A and 8 and taking datum line =

Bm

~ + vi pg

2g

pa ~sin g

th rough 8.

+ Zs + Loss of head due to friction from A to B

l!..i. '" pg

...(i)

2g

Ps "at mospheric pressure

pg

VA" O. VB=O.z,,= IS. zB"'O 0+0+ 15 =0+0+0+h,

I I

Ii

~ I

IL Flow Through Pipes 11 =150r

J

4 x.()(}4 X 600 X VI

0.2)(2)(9.& 1

501 1

4x /X LxV 1 =15 dx2g

=150rV=

15)(0.2)(2)(9.81

4 x.OO4 X 600

2.47 m/s

Substituting Ihis value of V in equation (il. we get

B.

It!,

2.471 =3.5- 2)(9.81 = 3.5 -0.311 = 3.189

II

"' -

/,

... (; i)

III

o4~'~f~',-,:!"~'~V,-'

... (iii)


where L] = inlet leg of syphon or length of syphon fronl upper reservo ir 10 the It

~uJ1\mit.

_ 4)(.004)(L, x (2.47)1 0.2)(2)<9.81 =0.0248)(L 1

I, -

Substituting this value in cquntioll (iil , 0.0248 LI = 3.189 3.189 L, '" .0 248 '" 128.511 Ill. ADs.

Problem 11 .29 A syphon of di"meter 200 mm COlineCIS 111'0 rcsen'o;,s whoJ'c water J'llff
Solution. Gil'en : d",200mm =0.20 m Dia. of sypllOn. Difference in lel' els of two reservoirs. fI " 40 m Total le ngth of pipe, I." 8000 m Heig ht of ridge summit from wa ter level in upper reservoir = 8 III Let th e depth of the pipe below Ihe summit of rid ge = x

III

Hei ght of syphon from water surface in Ihe upper reservoir " (8 - .r) Pressure head a( C,

III

!!..:...." 3.0 m of water absolute pg

. pressure head . ....!L I' Atmospheric pg

"

10.3 m of water

Co-erfkient of friction f = .006 Length of syphon from up per rese rvo ir to the s ummit. 1.( = 500 m

I I

Ii

~ I

IL 1502

Fluid Mechanics

Fig . 11.15 (b)

Apply ing Bernoulli's ~qumion to points A nnd B and taking d:llul11 line passing through B. we have p VI P VI .....d.. + .....d... + z~ '" ....!!.. +....!!.... + ll/ + head loss due 10 friction A to B pg 2g pg 28 0+ 0 +40 = 0+ 0+ 0+

4XjXLXV l

~"o::C,'~~

d x2g

40= 4xQ.OO6x8000x V

1

0.2x2x9.81

v: ,~40~x3o~.2gx'122x~9~.8i!:1

= 0.904 Ill/s

4 x .006 x 8000

Now applyi ng Bernoulli 's cquatiol110 points A and C and assuming datum line passing through A. we have

~ pg

Substituting

l!..!.. and i!L pg

pg

+ V; + z~ = .f!..::... + V} + 'c+ head 28

pg

2g

loss due to fric tion from

A lu

C

in (erms of absolute p .........wre

Vl 4xfx~ xV l 10.3 + 0 + 0 = 3.0 + - + (8 - x) + :..:":;~C:'-'-28 d x2g

•

•

10.3 = 3.0 + (0.904 ) + (8 _ .r) + 4 :dK16 x 500 x (0.904)2x9.8 1 0.2x2x9.81

0'

=3.0+0.041 + (8- .r)+2.499= 13.54-x x= 13.54-10.3=3.24m. An5. Discharge.

.. 1l.7

Q = Area x Velocity = ..::. x (.2) 2 X 0.904 = 0.0283 m J/s. Ans. 4

fLOW THROUGH PIPES IN SERIES OR FLOW THROUGH COMPOUND PIPES

Pipes in series nr compound pipes are defined a~ the pip es of different lengths and different diam· etas connected end to end (in scries) to form a pipe line as show n in Fig. 11.16.

I I

Ii

~ I

IL Flow Through Pipes Let,

Lt_~.

503 1

L)" length of pipes ],2 and 3 respectively

d l _ d!. d)" diameter o f pipes 1. 2. 3 respectively VI_ V2• V)" velocity of flow through pipes 1,2,3 f l'!!' f J " co-efficient of frictions for pipes 1.2.3 H = difference of water leve l in the two tan~S.

,

, F ig . 11.16

The discharge passing through e ach pipe is same. Q=AIVj"AzVt=AlVl The d ifFt're nce in liquid surface levels is equal 10

H=O.5 Vl

l

th~

sum of tile lOla] head loss in the pipes.

+ 4Ji~V/ +05V/ + 4f!~V}

2g

2g

d 1 x2g

d 2 x 2g

+ (V2 ,

vS + 4fl~VJ~ + v/ d' 2

.tg

J

x.tg

g

...( 11. 12)

If minor losses arc ncgLcclL-d. then above equation becomes as

,

,

' H= 4fl L, VI + 4f,L,V~ + 4fJ~V; d 1 x2g d 2 x2g d J x2g

... ( 11.13)

If the co.efficknt of friction is smne for all pipes II '" h '" fl = f. then equation ( 1 1.1 3) beco mes as

i.e..

H= 4 fL,V/ d,x2g == 4f 2g

+ 4 JL'lV1l + C4~JL, ,,,,V,c' d 2 X2g

d ) x 2g

[ ' , '1 L,V,· + L 2V1 + L)V, d, dl d)

.. _(11. 14)

Problem 11 .30 The difference in ...aler !J'urface lel'els in 1"'0 lallks ....hich are conneCled by Ihree pipes ill series of {engl/,s 300 m. 170 III alld 210 III alld of dlamelerI 300 mill. 200 ,,"" allli 400 111m respeclil'ely, i,f 12 III. Delermine Ihe rale of flo ... of \l"(/(cr if co-efficielll of frietiOIl are .005, .0052 IIl1d .1J048 respeeli"ely, cOllsiderillg : (i) millor losses also (iO IIcglee/illg millor 10S,Ie.!. Solution. Given: Difference o f water le"eI. lA:ngth of pipe I. lA:ngth of pipe 2.

~ I

H

= 12

In

L , == 300 m and dia .. tI, == 300 mm= 0.3 m L1 == 170 m and dia., 1f1 = 200 111111 :0 0.2 11\

I~

~ I

IL 1504

Fluid Mechanics

Len gth of pi pe 3.

L J '" 210 m and dia .. d.l '" 400 n1ln '" 0 .4 m

Also.

I, '" .005./, '" .0052 and!} '" .OM S

(i) Co ns id e ring Minor Losses. Lei VI' V, and V3 are th e veloc iti es in the l SI. 2nd and 3 rd pi pe

respecti vel y. From co ntinui ty. we have AI VI '" A1 V1 '" A JVJ

"d,'

,

()' X VI '" 2.25 VI

A,V, 4 d, 0.3 V, : - - . - - V , = - , V, '" Al d O 2 - d; 1 .

' 0

o

4

V == A, V, '"

and

Aj

J

0

d,l d;

=(03 )'

\I:

0.4

I

VI '" 0 .5625 VI

No w us in g eq uati on (II 12). we li avc

Substitut ing V1,md Y1•

O.5IC ' 4x ,005 x 300 X \1: 1 O,SX (2.25\11lt 12.0", - - ' - + ' + ---''0--'-'28 0.3 x 28 28

12.0 : VI' [0.5+20.0+ 2.53+ 89.505+2.&4 7 +3. 189 + 0.3 16[

"'

2,

"'2,

• - '- 111 8.8871 12 x 2 x 9.81

v,-R ale of now,

11&887

= 1.407 mls

Q = A rea x Ve loc it y = AI x VI It

,

It

,

1

'" "4 (dlt X VI '" "4 (.3)- x 1.407 = 0.09945 111 Is = 99.45 litrcs/s . Ans. (ii ) Neg lectin g !\linor Losses. Us in g

Cq U3Ii Oll

( 11. 13), we have

1

H = 4.fiI4 V1 + 4f.'-!. V} + 4fl ~Vll tf1 x2 g d 1 x 2g d J x2 g or

[ 2.0= ~ o [ 4x.OO5x300

2g

I I

0.3

+ 4x.OO52xI70x (2.25) , + 4x.()()48x2 10 x (.5625j"oJ 0.2

0.4

Ii

~ I

IL Flow Through Pipes

vl

= _,_ [20.0 + 89.505 + 3. 189] 2g

2x9.81x 120

11 2.694

V'

5051

X 112.694

'=' _ L

2g

== 1.445 m/s

Di.o;chargc. Q'" VI X AI '" 1.445 x ~ (.3)'" 0.1021 1113/S == 102. 1 lilrcsls. Ans. 4 Problem 11 .30 ( A). Tllret' pi{!1'S of.Joo mm. 200 111m mill 300 mm dialllelers Iw,'c (engll,s of 400 m. 200 III. ,md 300 III re~peclil'l.
, ,

AV = 4 dl dl1, VI = (00 4)' - X VI '" -

V1 ", - ' -'

d; 4 .

Al

d;

7r

0.2

VI'" 4V I

and

Now using equ ati on ( ] 1.13). we have H",

4JiI4 Vt' + 4fl ~Vl1 + 4 fJ L JV/ d l x2g

"'

16=

4xO.OO5x400xVI' 0.4x2x9.8 1

1

l '" 2 x9.81 V

16=

~' I

2x9.81

+

d 1 x2g

4 x 0.005 x 200 x (4Vd1

( 4 x 0.005 x 400 0.4

d l x2g

0.2x2x9.81

+

4 x 0.005 x

(20+320 + 63.14)=

200 x 16

0.2 V,' I

2x 9.81

+

4 xO.005x300 03x2x9.8 1

, x (1.77 V)" I

c4CXCO coOO """cX "3"OO ,,,-,X C3coO"c' .)

+-

OJ

x403.14

16 x2 x9.81 '" 0.882 mfs 403.1 4

I I

Ii

~ I

IL 1506

Fluid Mechanics

"

,

Q=A , X VI '" - (0.4) x 0.882 ", O. '108m Is. Ails. 4 ( ri) Disc ha rge through th e compound ]Iipe co ns idering minor losses also. Discharge.

Minor IOHes (lrt': 0.5 VI~

II : - -

(a) At inl et.

,

2,

(b) Between 1st pipe ;lml2nd pipe. due to contraction.

h: _ 05_V_,1 =,-O'.':..,;-(4-, V,C')L 2,

c

=

-

2g

'

°

'

5x 16xV, =8x V,2g 2g

(e) Between 2nd pipe and 3rd pipe. du e 10 sudden enl argeme nt.

/, '"

(V, -

'

vS

2,

(4 VI _ 1.77V )l

I = "'"--~'-".L

28

V1 \.\ ' = (2.23)1)( _ ,_ '" 4.973 - '2g

v1

(t.77Vd"

2g

28

(d) At lh~ o lltl el o f 3rd pipe. 11" = - '- =

The major losses arc

=

=

28

4Ji X 1'1 XV/

,~1 \1: ' = 1.77' x - '- = 3. 1329 -'2g

+ 4 fl x ~

d1 x2g

4 x 0.005 x 400 x 0.4x2x9.81

X V1'

d1 x2g

2,

+ 4 f J x~

XV)'

d J x2g

v/ + 4 x Q.()()5 x 200 x (4 VI) 1 + -,-4,-XCOC,OO:;;::5CX"300 :;:::OX,,(~,,-77CV,, ,-)'

'" 4{)3.14

0.2x2x9.81

x

0.3 )(2x9.81

v'I

2 x 9.81

:. Sum of minor losses and major losses

2

:

2

05 V/ +8x-+ VI ~' VI ., V12] 4.973-+3.13_9 - + 403.14[ 2g 2g 2g 2g 2g

v'

'" 4 19.746 - '-

2,

But tOlalloss IIll1st b<: equal 10 H (or 16 m) 4[9. 746)( -V''- : 2g :. Discharge.

I I

16

I = O.864m/s VI:,I~16~X~2~X~9~,8I 4[9.746

Q:A I V 1 : -It (0 .4) 1 x 0.864 = O. 10HS m J Is. An s. 4

Ii

~ I

IL Flow Through Pipes

5071

EQUIVALENT PIPE

.. 11 .8

This is d~rlncd as the pipe of uniform diameter having loss of head and discharge equa l 10 [he loss of head and discharge of a compound pire consisting of several pipes of different lengthS and diam Ckrs. The uniform diameter of the equi va lent pipe is called equivalent size of lh e pipe. The length of

cqu;valcm pipe is equal to sum of lengths of the compound pir<: consisting uf different pipes. u.l L [ '" le ngth of pipe 1 and ill'" diameter o f pipe I L 2 ", le ngth of pipe 2 and ill" diameter of pipe 2 L3 = knglh of pipe 3 and d 3 = diameter of pipe 3 II '" 100ai head loss

L = length of equ ival ent pipe

d", diameter oflhe equivalent pipe Then L=LI+L2+L3 Total head Joss in the compound pipe. neglecting minor losses

,

,

,

11= 4JiL,V[- + 4fl. ~V; + 4 /l~ V) d l x2g d l x2g dJ x2g

... (11. 14A)

II '"'Il'"'h==1

Assuming Discharge,

4Q

4Q

4Q

red,

red;

redj

VI'" - , . V 2 ", - , 311d V 3 ", - ,

Substituting these values in equmioll (11.14A), we have

... ( 11.1 5)

Head loss in the equivalent pipe. where V '"

Q

Q

"A'" ~d!

H",

4/.L.V d x2g

l

IT aking sallie

value

0

ff

as in compound pipe]

4Q '" rtd l

4

4Q

H= 4/.L. ( ;;if dx2g

J' ,", 4 x,16Ql/[~1 re - x2g

d~

... ( 11. 16)

Head loss in compound pipe and in equivalenl pip.: is same hellce equating equations (I 1.15) and (11. 16),wchavc

~ I

I~

~ I

IL 1508

Fluid Mechanics

... ( 11.1 7)

0<

Equati o n ( I I 17) is kn ow n as Dupuit' scq u3t ion. In thi s equat ion L ", L ] + L2 + Ll and dl' 112 and d J are kn ow n. He nce the equi val ent 5i].e of Ihe pipe. i.e .. value of II ca n be o bta in ed .

Pro blem 11 .31 Tilfee pipo of ":ngrl!.~ 800 m, 500 In ami 400 In allli of dialneter.! 500 min. 400 mm and 300 mill respeCfil'ely are connecled ill serres. TilesI' pipes are 10 be replaced by a single pipe of length 1700 m. Find Ihe diameter of Ihe single pipe. Solullon. Given : Len gth o f pipe I. LI '" 800!ll and dia.• III = 500 111m '" 0.5 III L" nglh o f pipc 2. ~ '" 500 III and o ia .. d~ '" 4 00 mm '" 0.4 In Le ngth o f pipe 3, L 3 ", 400 III and « ia .. d] '" 300 mm '" 0.3 In Le ngth of singic pi pe . L = 1700 m Le t the diame ter of eq ui va icnt sin g le pirc = d L 1., Appl ying c quuli on (I J 17), - , = - s d ii,

L,

L"

<11

ill

+---t + - ,

1700 800 500 400 -,- = - , + - , + - s = 25600 + 48828.1 25 + 164609 == 239037 d .5 .4 0.3 1700 .007 118 239037 d == (.007 188)°·2 == 0 .3 71 8 == 37 1.8 nlnI. A ns •

tiS==

.. 1 1.9

FLOW THROUGH PARALLEL PIPES

Consider a ma in pi pe whic h di vides inlo two or more bran ches as s how n in Fi g. 11.1 7 and aga in join toge ther dow nstream to form a s in g le pipe. the n Ihe bran ch pipes arc sa id 10 be conncclcd in paralle l. The di scharge Ih ro ugh th e ma in is increa sed by co nn ecting pipes in parallel. B R A N CH PtP E :2

B R A N C H PIPE 1

Fig. 11 .1 7

The rate of fl ow in Ih e main pipe is equ a l 10 th e s um of rale of flo w Ih roug h bra nch pipes. Hence from l"' ig . 11. 17. we have ...( 11.18) Q=Q I +Q 2 In th is. arTang emcl11. the loss of head for each branch pipe is sa me. :. Loss o f h" ad for bran c h pipe I = Loss of head for hrauc h pipe 2

I I

Ii

~ I

IL Flow Through Pipes

4f,L,V,l == 4 f2~V/ tI, x2g til x2g

I, ==/"

If

,hell

•

5091

... ( 1 1.19)

L,V,'

L"V}

... ( 11.20)

d, xlg d 1 x2g

Problem 11.32 A main pipe tfi\'ides illll) /WI) parallel pipes which again formJ QlIe pipe (/$ .fholi'll ill Fig. f 1.17. The lellglh alld diameter for the first parallel pipe are 2000 m and 1.0 m respecril·e!y ....hile l/ie lengll. and dim/leler of ]"'/ parallel pipe
Q'"

I, == 12 =/= .005 0, '" disctlarge in

pipe 1

Q1 == discharge;n pipe 2 Q "" Q, + Q2 == 3.0

... ( i)

From equ ation ( II 18). Using eq uation (1 1.19). we It avc

4j, ~V," '" 4j , L..,V,l d , x 2g

d l x 2g

4 x .005 x 2000 x Vll. 4 x .005 x 2000 x V, = 0.8x2x9.81 1.0 x 2 x9.81

" = -Vi.· or V, 1 ", -V,',

V,

-

0'

1.0

0.8

0.8

Now and Substituling the va lu e o f Q, and Ql in equa tion (I). we get 1( VI. It "4 xO.894 +"4 x.64 V1 '" 3.0 or 0.8785 V1 +O.5026 V1 =3.0

0'

I I

or V =

...lQ... = 2.17

Ill/S.

1.3811

Ii

~ I

IL 1510 Fluid Mechanics Substituting this va lue in equation (ii) , V, 2.1 7 VI = - -- = - -

.894

Hence

11:

0.894

= 2.427 m/s

2

It

0 1 ", - d, )( VI '" -

4

4

l

,

x I x 2 .427 = 1.906 nOs. Ans .

Q1 =Q-Q 1 '" 3.0-1.906= I.094m J /s. AilS. Problem 11 .33 A pipe line 0/0.6 III dillmeTer is /.5 km lOll/!.. To ;/lCT/'(Ise Ihe disc/"" ge. (Jnolher line of JIl" J'ome diameter iJ' in/rot/fleed pnralld 10 Ille first iii Ihe second half of Ihe /ength. Neglecting minor IOSJ'cs, find Ihe increase in tlischarg(' if 4[ = 0.04. The head lI/ inlet iJ' 300 mm.

Solution. Given: Dia. of pipe line. Length of pipe line.

Head a1 inlet. Head at outlet. :. Head loss.

D =O.6 rn L = 1.5 kill = 1.5 x 1000", 1500

III

4f =0.04or/:.0 1 II = 300 rmn = 0.3 m = atmospheric head = 0

h, ,,, 0.3 III

Lenglh of another parallel pipe. LI '"

lS~

= 750 m

Dill. of anolher parallel pillt'. d l = 0.6 10 Fig. I 1.18 shows the arrangement of pipe system.

L :750m.d :O.6m

' l m<---='''''''---='== T~-_cQi;;ii'i~-~' ~::-...;. 0, C O:O.6m ..........

l___

Q,

,

L,"'750m.d,:O.6m

Fig. 11.18 1s t Ca se. Discharge for a sing le pipe of length 1500 III and dia. '" 0.6 m. 2

This head lost due to friction in singic pipe is

4JLy· "r=-'<~Cdx2g

whereV" = veloc it y of flow for s ingle pipe

.,

4x.OlxI500xY·

"'

0.3 = V"=

0.6 x2g 0.3xO.6x2x9.81 =0.2426m/s 4x .01x 15oo

Discharge. Q. = V" x Area = 0.2426 x ~ (.6) 2 "" 0.06&5 m3/s ... ( i) 4 2nd Case. When an additional pipe o f length 750 111 and diamete r 0.6 m is connected in parallel with the last half length of th e pipe.

I I

Ii

~ I

IL Flow Through Pipes Let

511 1

Q 1 '" di scharge in I SI paral lel pi pe

Q2 '" d ischarge in 2nd pa rall el pipe Q '" Q, + Q! where Q= d isc harge in main pi pe when pipes J rc pa rall el. BUI as the ic ngth and dimnctcrs of eac h pa r,tlie ) pi pe is same

Q, =Q2'" Q!2 Consider the now thro ugh pipe ABC o r ABO Head loss tll ro ugh A BC", H ead lost th rough A n + head losl l hro ugh BC

...(ii)

BUl lic ad l ost du e to fri ct io n th rough ABC", 0.3 m gi ve n Head loss d ue to fric tion th ro ugh All '"

4 x/ x 750xV l

• w here V = ve locity o f fl ow t hrough A ll

0.6x 2 /9.81 Q 40

H ead loss d ue to f ric tion th rough. A lJ

= 4x .Olx 750

0.6 x 2 x9.81

X [~l2 =3 1.87 Q2 n:x .36

Head loss du e to frictio n tli rough. BC

= 4x / xL. XV,l d, x 2g

=

~; ~o;: ;~~ x[ xt(61'1 2

4x.oJx 750

. . V _ Distance [

•

I -

~(.6f

Q

1

4

16

'" 0.6 x 2 x 9.81 x 4xl1:'" x .36" Q2:7.969Q2 Substitut ing these va lues in cqu atioll (ii). we ge l 0.3 = 31.87 Q1 + 7.969 Q"

01 = 39.839 0 2

¥'153 39.839 "" 0 .0867 m,Is

Incrc aSoe in discharge", Q - Q. = 0.0867 - 0.0685 = 0.0 182 m 3 /s.

AD S.

Problem 11 .34 A I'umpili g pfali/ fvrces ,mler Ih rough a 600 mm di(/meler main. Ihe friClivn head being 27 III. III order IV retlllce Ihe po,..", COII~'umplion . il is proposed 10 lay {mother main of "ppropri· ale tli",ne/'" afOiIS Ihe "ide of Ille <'xiS/ins one. so Ilwl 1"'0 pipes may ,..ork ill parallel fo r Ihe enlire Jeng/II (/lid reduce 111<' frie/ioll 11<'(ld 10 9.6 m ollly. Filld Ihe diameter of Ille lIeli' maill if. ,..ilh 111<' e.(cepli011 of diameter, il i,~ simi/M 10 Ihe exiJlillg Ol1e in el'''')' respect. Solution. G ivc n : Dia. of si ng lc main pipe. d = 600 mm = 0.6 m hf = 27 m Frictio n head. Frictio n he ad for two parall el pipes "" 9.6 m 1s t Case. For a sin g lc mai n [Fig. 11. I 9 (a) ]

I I

Ii

~ I

IL 1512 Fluid Mechanics /1 '" 4 . f

I

2

. L. v or 27. 0 == c~4~X~f~X:;-::L~X~V.,-'

dxlg

0.6x2x9.8 1

f LV2 = 27.0 x O.6x2x9.8 1 ",3 17.844

4

5

=79.46 1. whcrcV= Q

A

Q'

n(i)

fL A' = 79.461 2nd Case. Two pipes arc in parallel [Fig. 11.19 (b) ] Loss of head in anyone pipe == 9.6 m :. For 1st pipe.

111,= ,

4 .!. L. V,"

-1

d, x2g

9 =.6

0 ---

n

-

- - -

n

n

-I

(a) Single main

-

----

-~'- - - - --- - -- - - - h,- "'9 .6m

, " - _. _ _ _

_Q_'I. ____ _ • __ • _ _ __ _ _ _ _ _ _ _ • __ _

(b) Two para llel pipes

Fig. 11.19

Om

.fl=~

L, =L . V, '"

A,

A

d,,,,d=O.6

4c·~fc'"L= Q,' ,c-:c Al '" 0.6 0.6)(2)(9.81

f

"' For the 2nd pipe,

Q,! - 9.6 XO.6x 2 x9.8 1_ '8 xX L -:-r - 282 .5_ A

4

I,f, -

... (ii)

4 jxL, XV,'

• • == 9.6,

dl x 2g

i,

4 / X LX Q =9.6 d,x2gx A;

/ xLxQ; '" 9.6 x 2 x 9.81 '" 47.088 ill X A~

Di vidi ng

I I

~qualion

4

... (iii)

(i) by equal io n (iiI, we gel

Ii

~ I

IL Flow Through Pipes Q2

79.461

Q I! '"

28.2528 '" 2.8 125

..£ ",./2.8125 Q,

513 1

'" 1.667

Q

Q 1 '" 1.667 '" .596 Q

QI+Q 2 :Q

Ql= Q- Q 1 = Q- .596 Q=O.404 Q Dividing eqU3lion (ii) by cqumion (iii),

ai' xd, xA; A l XQ;

= 28.2528 '" 0.6

47.088

Bw

.596 Q )' x di, =0.6 ( A04 Q .)6"

d/ '" 0.6

1

X

.36 x

(::~r

= 0.03537


Problem 11 .35 difference of

"'at'"

If wi

Ille <'.1"1.11;118 pipe, Jim/rile illerea,51! ill Ihe discharge. Take f

'"

.015 lind lIegleC/mitlQr 10$,ro.

Solution. Given: Dia. of pipe. Len glh of pipe,

d=20cm =O.20m

L=2000m Difference of waler levels. H = 20 In Co-cfflcicn! of frictio n. /=0.Ql5 1s t Ca se. When a si n g l ~ pipe connects the two reservoirs 11= 4./ . L .V dx2}:

1

4[. L [ Q ]' = dx2g : d1

32f·L.Q1 = rr. " xgxd '

I I

Ii

~ I

IL 1514 Fluid Mechanics

20 = "32""xc·Oc'c'o'c'cOOOccoxCiQ,-) '

" 30985.01 Q-,

I( ' X9.8 I X(0.2)

Q ==

~ o

309X5.07

,

,, 0.0254 m Is. An s.

2nd Case.

Lei Q 1 "diSl: hargc through pipe CD. Q1 == di sch arge through pipe DE. QJ = d isc harge through pi pe DF.

Le ngth of pipe CD.

LI == 800 111 and its dia .• d l " 0.20 111 L2 ", 1200 III and its dia., (/2 " 0.20 III L) = 1200 11\ and its dia .. d1 == 0.20 m.

Length of pipe DE.

Length of pipe DF.

Since the diam eters and leng th s o f th e pipes DE and DF arc equal. Hell ec Q2 will be equal 10 Ql' Also for parallcl pipes. we have Q 1 == Q! + QJ" Q1 + Q! = 2Q!

Q,=~ , 2 A

c

T

0,

;:-..;.

°

,

20m

2

•

800"" 20ciii ~ -----

'200",03

. 20c",

FL----'

Fig. 11.20 Apply in g Bernoulli' s equati on (0 points A and B and tak i ng th e fl ow throu gh CDE. we ha l'e

20 =

4/ .t.. v/ + 4/ .L,. V/ d t x2g

d 2 x2g

,

,

- 4x.015x800 , ( -4Q, - - )" + 4 x.015xl200 0.2x2 x 9 .81

Ifx.04

0.2x2x9.81

lQ, )' x( --/tx.04

'" 12394 01 1 + 46470/ '" 1704 1 Qi l 0

1 '"

~ 1 7~1

'" 0.0342 ml/s

Increase in di scharge'" Q I - Q '" 0.0342 - 0.0254 = .00811 mJ/s. AilS.

I I

Ii

~ I

IL Flow Through Pipes

515 1

Problem 11 .36 Two I'ipes /rare a /ellglh L each . aile of I/Ielll lias {/ diameter D. and Ille otller a diolllefer d. If Ihe pipes (Ir,. arwnged ill powllef. Ihe loss of head. ",lleli a IOral 'II/anllty of Willer Q flows II/rougll (llem is h. bU!. iflhe pipe.! are arrtmged ill series and the s(lme qlwlllif), Q flows throll8'1 loss of head ;.1 H. If II = 0 , find Ille ralio of H 2 assuming 1110' pipe co-efficient f Iws a COIUlalll \'al'le.

them.

Jill'

10

II, n ..glee/ing

.~econd(lfy

IOHeJ' {md

Solution. Given u:ngth of pipe I.

L, '" L and its dia. il, '" D Length of piP<' 2. L2 '" L and its dia .• d 2 '" d Total discharge '" Q ~1c'1(.l loss wlien pipes are arranged in parallel'" II Head loss when pipes arc arranged in se ries", H d

= 0 and f is conSlant

2 ht Case. Whell pipes are connected to parallel Q'" Q , + Q2 Loss of head in each pipe = II

... ( i)

For pipe AB, d, =0

co Fig.

4JL x [ nO' 4Q~)'

]1.2]

32fLQ,!

---,"'-";'--'-= " or n 10 ) xg '" " Dx2g ['or pipe AC.

32jf·Qi n ! d ) x8

= II

... (ii)

...(iii)

1

32jLQ , '" 32fLQi n 1 D )g

n l d )g

[':0=2d[

'" 2) '" 32

t '" ./Yi '" I I

5.657 or Q , '" 5.657 Q1

Ii

~ I

IL 1516

Fluid Mechanics

Substituting Ihe valu es of Q, in eq uation 0). we get Q '" 5.657 Q: + 0 1 '" 6.657 Q1

Ql : ~ :::O. 1 5Q

... (il·)

6.657

Q,=Q - Q1=Q - O. 15Q=0.85Q

From (i) 2nd Case.

Wh ~ n

... (v)

th e pipes arc ron ncclcd in se ri es. To tal

los_~

= Sum of he ad losses in lIie two pip.:s

1/ : 4 j .L.V,2 + 4 f ·L.V~! d,x2g u 1 x2g

WhcreV, :~= 4Q"Vl = ~= 4~ ~D l

rrD -

7Uf

!!.- dl

4

4

C.'

C. D

-

_ v,

0

-

0

Fig. 11.22

II =

11 =

"' From eq uation (ii).

4J. LX (-,:-;-ZY 4fL (~~r

---;;-cC2'-"+ -;=:-:"Dx2 g dxlg 32fl_Q ! S'

Dn "

32JLQ!

+ S ' x g d n·

xJ\

32jL -,,'?f"-- -" n 1 S x g - Q,!

and from equ ation (iii).

D

h ; " 2f ::: Q £ tr' d x g

SubstituTin g th ese valu es in equation ("i), we have

Q1

[Q -1 ,+ -Q,l]

1 Il ! II Q' H = Q x - , +Q X- , =-,11 +-,11 =" Q,Qi Q,Qi 0,

11

-

II

Q2

= - , +- , 0, Qi

But from equa tio ns (il') a nd (v). Ii -= h

Q2

Ol

0, = .85 Q and 0, = 0. 15 Q 1.384 + 44.444 = 4S.H2H. Ans.

Problem 11 .36 (A). Tlm·e piJll·s of the same /ength L, diameter D. WId friction faCiO r f Me connec/ed in p"",IIe1. Determine l/ie diwneter of Ille pipe of length L GIld friction filClor f ...lIich ...iII carry ti,e SlIIlIe discharge for the .tame Ilead loss. Use ti,e form ulllilf '" f x Lx V"'/2g D.

I I

Ii

~ I

IL 517 1

Flow Through Pipes

Solution. Given: Length of cacti pipe '" L Diamete r of c 31:h pipe == 0 Friction factor of cadI pipe =f 1 Head loss, Ilr '" fX L x V f2gD When th e th ree pipes arc connected in parallel. then head loss in each pi P<' will be same. Allu IOtal head loss will be equa l to Ihe head loss in each pipe. Let h ", Total he ad loss. f hi, '" Head toss in lSI pipe.

hi, '" Head loss in 2nd pipe. and IIf, '" Head loss ill3 rl1 pipe.

f xLx

Then Let

hI "" IIf, == Il f , == 1If, or II! '"

V~

'-'';,O',"D '-'--

...(il

Q1 '" Discharge through IS! pipe. Q2 '" Discharge Th rough 2nd pipe. Q3 = Discharge throug h 3rd pipe. and Q = Total discharge.

When th e three pipes arc con nected in parallel. Ihen Q = 0, + Q1 + QJ = 3 X Q, =)xA , xV t

:3)(

'4" 0 2 )( V (wlicrcA, = '4D " ', and V, = V)

... (ii)

For a si ng le pipe (or leng th L: friction facior j) wtJ ich will carry same discharge as lhe lh ree pipes in parallel LeI d = dia. of lhe sing le pipe ,. = ve loc ily through si ngle piP<' Then dischurge. Q = Area x Velocity

= (:

d')

...(iii)

x v

Equatillg the two values of discharge. given by equations (ii) and (iii). we get If

!

If

D' ,·

,

3x - D xV= -
But head loss for lhe sing le pipe of lengt h L, dia.
,

~f_'cL~'~"_'

... ( 1')

d x 2g

Equaling Ihe IWO valu es o f '"give n by equations (i) and (v). we gel

, f x L X V' = ~f~';--"L~'C'c"_ oc Dx2 g

dx2g If

-

D

, _V_'

,

= _"_

D

d

y.!

= -, V'

Substituting the va lu e o f "IV in eq uation (i,'). we gel

I I

Ii

~ I

IL 1518 Fluid Mechanics

(dJ'"

3 , -D',= dD

:!... '" 3;'5 "

0'

(d J'" (dJ' (
3° 0

D

x D

ur 3 = D '"

,, D

1.55

d" 1.55 D. Ans . He nce d ia. of singl e pipe sho uld be 1.55 ti mes tlie dia . o f the th ree pipes connected in pa rall c l. Pro blem ' 1,37 For a /O ll'n wU/«r supply, II main pipe /i" e ojdiomele r 0 ../ '" is required. As piPH more than 0.35 m drame/a are 1101 readily oWlilable. flVO parallel pipeJ of I/Ie same diameter were usedfar Waler supply. Iflhe IOral discharge ill Ihe parallel pipes is same as in the single moill pipe.Jind the (1;(IIlIeler of the parallel pipe. Assllml:' IIII.' co- efficienT of[rielion S(lme for (11/ pipes. So lution. Gi ve n : Dia. of sing le ma ill pip" line. d" 0 .4 m LeIth" length of single pipe lin e "L Co-cfticicnt of fric tio n =f

._ .

.

.

.

4JLVl d x 2g

4fLV '

Loss o f head duc to t nctlOn 111 sin g le P'PC " - - - • ~",,:o--

... ( i)

0.4 )(2)( g

where V:: Ve loc it y of n o w in the sin gle pi()C. In case o f parallel pilX'. as the diameters aud k llg ills o f th e two pipes are sam e. He!)ec di sc harge in each pi pe will he half the di scharge of s in g le main pipe. As di.<;ehargc in each paralie l pipe is same. hence ve loc ity will also be same . V. = Ve loc it y in eac h p~rallcl pipe Let d. = Dia. o f eac h paralle l pire The n loss of head du e 10 fri<:ti on in parallel pipes =

Equatillg Ihe two losses g ive n by equil ti o ns (i)

4 /. L.V l 0.4x 2g

v'

Cance ll ing 4 f L .

2,

0.4

~Ild

4/ X L XV,! J. x 2g

...(ii)

(ii). we ha ve

4/ xL x V.! d. x 2g

. -,v.! "' - ,' d.

V"

0.4

V.-

d.

... (iii)

From continuit y TOl al no w i n sin gle mai n :: SUIll of fl ow in two parallel pi pes Of Ve loc ity of ma in x Are a :: 2 X Veloc ity in each paralle l pi pe x Area II , 1( , V V x - (0.4)-:: 2 x V.)( - d-.or - ::

4

4

V.

2x!!.-d: 4

.~-;-

~ (O .4 ) l 4

u} 0. 16

Squarin g both sides. Comp ar ing equati ons (iii) and (il'). we get

I I

Ii

~ I

IL Flow Through Pipes

519 1

0.4 '" 4d; or ~ '" 0.4 x .0256 '" .00256 d. .0256 4 d. '" (J)0256)1/~ '" 0.303 III '" 30.3 cm. ADS. USC two pipes of 30.3 CIll diameter.

All old waIn supp!.,' diSlribUli01l pipe of 150 "'Ill di(alleler of (, cil)' is 10 be replaced by /"'0 p"ralld pipes of smfll/er eq""{ diameter having equal /""gllis "",/ idenlical friction jaclor I·u/aes. Find 011' tile new di",,,eler "-''1"ired. Problem 11 .38

Solution. Given: Dia. of old pipe. D '" 250 mm '" 0.25 rn Lei

= Di a. of each o f para lid pipes Q = Discharge in old pipe Q 1 = Discharge in first paralld pipe Q,- = Discharge in second paralic I pire d

f = Fri clion factor. When a single pipe is rcpla~'Cd by two parallcl pipes. the head loss will be same in the s in gle pipe and in eac h ufl he p3 rallcl pipes. Also the discharge in single pip e will be equal I() the total d ischarge ill twO p;lT
and Q" Q 1 + Q1 As the dia. of each parallel pipe is same and also length of each parallel pipe is equal. hence QI = Q1 or QI=Q1=Qf2 Now h, ,, Head luss ill single pipe =f X LXV ' Dx2g

JX LX[ =

It

wheref= Friction factor

Q X 0.25 '

J'

4

[.: V =

0.25 x 2 x 9.81 =

••• (ii)

f x tX (4Q )1 0.25 x 2 x 9.81 x( It x 0.25')"

A~a = : ~l ) ... (iii}

IlJi = Head loss in IS! parallel pipe "

=

JX Lx (~i II x2 g

(": Dia. of parallel pipe" II and VI is the velocity

2X~d'

I x LX [

r

in lSI parallel pipe)

dx2g

I I

Ii

~ I

IL 1520 Fluid Mechanics

=

! XLX C (4Q)' -d-,--:-,_,L9".8 ", X:-;(2"'X""CX-d~'~)"

[X Lx(4Ql

f X L X (4Q)! 2 '

0.25x2x9.81 x(n x 0.25

r

== d

x 2 ><9.8 1>«2x Itxd l

,

d x {2 Ittf~)2 '" 0.25 X ( It x 0.25 2)2 4 d X 4 x It '" 0.25 x 0.25

0'

"' "'

5

O.2 SS

0.25

d '" - - ord= - ,,-,

0.25

=-- =

0.1894m

-= 0.19m. An s.

4 (4 ) 1.31 95 Problem 11.39 A pipe of diam eter 0.4 In and of 1<'11gll1 2000 //I i,~ conllected 10 II reserl'Qir at 0111' end. '/'l,e oiller end oflile pip" is connected to a june/ioll from ..... hich IWO pipe.! of /ellglhs /000 1/1 alld diwneler 300 mill n Ul ill parallel. Tllese parallel pipes are connected /0 mlO/iler reurl'Oir. which is

IUll·ing lew/ a/waler /0

III

below Ille water lel'eI of Ihe (lbm"e re.~efl'ojr. Determine Iile lawl discharge

Iff'" 0.0 15. Neglecl millor losse.!. Solution. G i ven: Dia. of pi pe. d = 0.4 111 L ength of pipe.

L '" 2000 1lI

Dia. of parallcl pipes. d l '" d 2 '" 300 mm '" 0.30 111 Le ngth of parallcl pipes. LI '" ~ '" 1000 m Difference o f water leve l in two reservo ir. H = 10 m. ! = .015 Appl ying Bernoulli's eq uati on to po ints E and f -. Tak.ing fl ow throu gh AllC. 4 fL V1 -4,' 10= - + - f"XC~"'c:''-"'V'_ dx2g
'" 4 x.O 15 x 2000 x V 0.4)(2x9.81

2

+ :4C',-;;.O~,,,5"X:;-'"OOO ::;C;X;-:VL " 0.3)(2)(9.8 1 ...(i)

, ----.---

T

A

,

.om

-'-

("2000 0"04 m L -1 000-:n.
<0j"

~

----,,--------C

° 1

;>,,0.3

Fig. 11.23

From continuit y equat ion Discharge through All '" disc harge through Be + d isc harge throu gh llD

I I

Ii

~ I

IL Flow Through Pipes

521 1

"' 13m il, '" il2 and also the len gth s of pi pes Be and BD arc equ al and he nce discharge throug h Be and lJD wi ll be sallie. Tll is mea ns VI '" til also

(0.4/ x V", 2 X (O.3)!V, or .16V= O. HI V,

"'

V '" 0.16 V= 0.888 V I

0. 18

Substitutin g thi s va lu e o f VI in equ ati on (I). we get [0", 15.29 V! + ( I O. 1 9)(.888)~V2 = 15.29 V' + 8.035 V 1 = 23.325 VI

v=~ 21325 10 Discharge

= 0.654 I11ls

= Vx Area

= 0.654 x ~ d! = 0.6 54 x ~ (0.4)1 = .0822 ..h •. Am. 4 4 Problem 1 1.40 Two J'/w,p ended pipes of iliame/us 50 mm lim/ {OO mm respeClil'e/y. cach 0/ /eng/h 100 III are connected in p"ral/l'l betwN:1l IWO resen'oirs which lw,"e II difference of ie"ei of 10 m. If the co- efficienr of fricrion for each pipe iJ' (-If) 0.32, calcuh'lt' Ille rate of flow for eilcil pipe and (1/.10 Ihe dilllneler of II single pipe 100 //I long which would gil'e the same discharge. if i/ were subs/illll"d for rhe original 1\\'0 pipe.t. Solution . Gi l'~ n : d l '" 50 mm '" 0.051ll Dia. of 1st pipe. Le ngth of lst pipe. LI = 1{lO 1ll Dia. of 2nd pipe. d 2 '" 100 mm '" 0. 10 1ll L.cngth of 2nd pipe. L,. '" 100 111 Differe nce in level in reservoirs. H '" 10111 Co-e ffici ent of frict io n 4f'" 0.32

i

.om

L Fig. 11 .24

VI = ve locity of flow in pipe I. and V! = ve loci ty of flow in pipe 2.

I I

Ii

~ I

IL 1522

Fluid Mechanics

When the pipes arc connected in parallel. llie loss of head will be same in both the pipes. For tlie fir.>i pipe. loss of head is given as

-4fL"X"~~Xc"",'_ . 0.32 x 100 x V/

/f: -

dl x2g

10" 32.619

VI'"

VI

~ 32.61019

("; 4/" .32)

0.05x2x9.81

1

'" 0.5535 m/s

:. Rate of flow in 1st pire. Q, '" VI X AI"" 0.5536 x

~ (dill 4

'" .5536 x ~ (O.OS)! '" '(MJ]087 11I ]/s '" 1.087 litresls. A ilS. 4 For the 2nd pipe. loss of head is given by.

IO=H= 4f X~XV/ til. x 2g

'" 0.32xlOOxV} 0.lOx2x9.8 1

V2 ", " ' CIO=X~.I,OCX;;2~X;;-c9·c'c' .32 x 100

,

,

'" "4 (. 1)- x .783 "

'"

0.783 111/5

0.006 15

111

,Is'" 6. 15 lilres/s. Ans.

D" diameter of a single pipe which is substituted for the twO original pipes L" length ursinglc pipe '" ]00 V = velocity through pipe

III

The discharge ItJrough single pipe.

Q = Q 1 + Q1 = 1.087 + 6. 15 = 7.237 ]itrests = .007237 11I 3/s

v=

-.JL = Area

.007237 11: 0 "

4 x.007237 11:0 "

.009214 Illfs 0:

4

Loss of head through sing le pi pe is

O.32XIOOX(·()()D9~_ [4 )'

4 /X LXV 1

H"

0'

0'

"

0 x 2 x 9.8 [

10.0= 32x IOOx .0092 [4 ' '" .0001384 2 x9.81 x f) l D~ OS", .0001384 = .0000[384 f) '"

I I

-'-~DCxC20g-

10 (.00001384)"} = 0. 1067

III '"

106.7 mm. Ans.

Ii

~ I

IL Flow Through Pipes

5231

Pro blem 11.41 Two reJeFI'oir life COII/leeled by a pipe lille of diameter 600 mill and /englll 4000 m. 7'11,' di/fere/tce of water lel'e! in Ihe res(:'fmirs is 20 m. AI a diswnce of I (){)() III from the upper reserl'Oi r. a small pipe is connected /0 Ifle pip/' line. T ill! warer ntt1 be lak ell from {he small p ipe. Find Ihe discharge /0 II,e lower resen'oir. if (0 No wmer is {I' ken f rom the small pipe. and (ii) 100 /ilrl's/S of waler is /(I ke" f rom smol/ pipe. Take! '" .005 and "eg{ec/ mill or losses. Solution. Gi ven: Dia. of pi pe. d " 600 mm == 0.60 III Le ngth o f pi r<'. L = 400m Diff,,~n ce o f water le ve l. H == 20 rn .f= .005 (i) Nu w"t.,r Is taken from small pipe E

- =o.::~:i":---

t

A:-

"'m ~

C

l."4Qoo

1

OOOrn

~

rn.

<1'<0.6""

>

,

. ---

-

Fi g. ]1.25

4/ xLxV 1 4x .OO5 x 4000 x V " The head loss du e to fri cti on in pipe A B = -'--;cc-;;- - or 20 = 0.6)(2x9.8 1

d x2g

v= :. Disc harg<).

20xO.6x2x9.8 1 _ h .943 " 1.715rnls 4 x .005 x 4000

"

4 ( ii) 100 litrl'S o f wa ter is takc n from s mall pillC Le t Q I '" discharge through pipe AC

Ttwn for pnral kl pi""s

,

Q '" Ar<)a x V " - (0.6)" x 1.7 15" 0.485 rn Is. AilS.

Q1 '" di sc harge throug h pi p" CB Q I '" Q1 + 100 litresls '" Q, + 0. 1 mJls

-

,

Ql"(QI - O. I) m ls

...(1)

L I ", 1000 III Len gth of pi"" AC, Len glh of pipe Cll. L 1 '" 4000 - 1000 = 3000 III Appl ying Berno ulli' s equation 10 points E and F alld lak ing now th roug h ABC, we have

1

' "'-.11 + 4 jL,- Vl 20= -,,4,fL=, d l x2 g d 1 x2g where VI '" ve locity th roug h pi pe AC '"

dl

I I

'"

QI

~(0.6)'

... (ii)

=~ It

x .36

dia. of pipe A C = 0.6

Ii

~ I

IL 1524 Fluid Mechanics

-:- __ V2 ", velocity through pipe CB '" -'00,-, ..~(O.6) 1 4

4 Ql 1T x.36

d 1 == dia. of pipe CB '" 0.6

Substituting these values in equation (ii). we get

[4Q' )" + 4 x .005 x 3()()() x [4Q' - - ' -)'

"0- 4 x .005 x 1000 x - - - - O.6x2x9.81 1tx.36

1tx.36

20 = 21.25 QI2 + 63.75 Q/ Ql= Q1 - 0.1 or OJ '" Ql + 0.1

Bm from (i), Substituting the value of

or

0.6x2x9.8 1

0 , in

... (i ii)

equation (iii). we get

20", 21.25 (Q, + 0.1)' + 63.75 Q/ = 21.55 10/ +.01 + 0.2 OJI + 63.75 Q/ = 21.25 Q/ + 0.2125 + 4.250 Q 2 + 63.75 = 85 Q2" + 4.25 0 2 + .2125 19.7875'" 0

Q/

Q/

85 + 4. 25 0 2This is a quadrnlil: equaTion in Q 1

- 4.25± J4.2S l + 4 x 85 x 19.7875 QJ= 2x85

= "" 0.458 rn l ls

- 44.25 ± 82.13

=

82.1 3- 4.25

170 170 (Neg lecting negative TOOl)

Discharge 10 lower reservoir'" Q, = 0.458 rn J/s. Am.

to I 1. 10

flOW THROUGH BRANCHED PIPES

When th ree or more reservoi rs are connec ted by meanS of pipes. baving one or more junctions. tbe syste m is called a branclling pipt) system. Fig. 11.26 sbows tllrcc reservoirs at different levels connected to a single junctiOll. by means of pipes whicll arc called brallc bcd pipt)s. Tile lengtlls. diameters and co-efficient of friction of eacb pipes is given. It is required to find the disellarge alld direction of now in eacll pipe. Tile basic equations used for solving sucll problems arc: !. Conti nuity equatlol1 whicb means tbe il1f1ow of fluid at tile jUl1ctiol1 should be equal to tbe outflow of fluid. 2. Uerno lllll's .,quutJon. " nd 3. Darcy-Welsbach equ;.\lo l1 Also it is assumed tbat reservoirs are very large and tile water surface levels ill the rese rvoirs arc constant so that steady conditions exist in tile pipes. Also minor losses arc assumed very smull. The flow from reservoir A takes place to junction D. Tile flow from junction D is lowards reservo irs C. Now the now from jUllction D lowards reservoir B will wke place only when piezometric head al D ( WhiCh is equ al to

~; + Zn ) is mOre than the piezometric lIead at B (i.e .• ZIJ). Let us consi der that flow

is from D to reservoir B.

~ I

I~

~ I

IL Flow Through Pipes

r

525 1

---:-: -.-~ ~~

,

--=::

::::::::- , ~

L"

'.

~ '>-

a,. 't,

". " '" , "

,~

,

.-:.-

c

~

I 1 '.

'"

Fig. 11 .26

For flow from A to D from Bernoulli's equation

Z.t =Zo +ft+I,J,

pg

,

For now from D to B from BcnlOulli's equation Po ZB+ /If I ZO+ pg=

... (r i)

For now from D to C from Bernoulli's equation

Z,, + -Po : Zc + /'/

OK

...(iii)

•

From continuity equation, Discllarge through AD '" Disc harge through DR + Discharge through. DC

n,

It,

It,

4" til-V, '" '4 d 2- X V 2 + '4 d,·V.!

'"

d/V) '" iI/Vl + If/V,

There arc four unk nowns i.e., VI' V1 • Vj and

... ( iI')

.!!..IL and there arc four equations (i), (ii), (iii) and (11'). pg

H.:ncc unknown can be calculated. Problem 11.42 Three resen'uiTS A, Band C (lie connee/cd by" pipe S)'.~lem ,y/'oll'1I in ""ig . 11.27. Find ll1e discharge illlD or from lite rcurI'vir,l' Blind C if Ihe rafe of flow from rt'Jerl'oirs A i.~

60 Ii(res/s. Find (lie !leig/II of water lel'el ill lilt' ",serl'oir C. Take f '" .006 for al/ pipes. Solution. Given: lA:ngth uf pipe AD. L, '" 1200 III Dia. uf pipe AD. d , '" 30 elll '" 0.30 III Discharge through AD. 0 , '" 60 litres/s = 0.06 ml/s Height uf water level in A from reference line. Z... = 40 m For pipe DB. length L2 '" 600 m. dia .. d2 ", 20 cm = 0.20 m. ZI! '" 38.0 For pirc Dc' length L3 = 800 Ill. dia .. 113 = 30 Clll = 0.30 m

I I

Ii

~ I

IL 1526

Fluid Mechanics

Fig. 11 .27

Apply ing Bcmoulli's cqualio ll s to poinls E and D. Z... '" Zv +

where

Ii" '"

~ + h~, pg

l 4. f· ~ . Vi . where VI '" ~ '" 0.06 '" 0.M8 mlsec til x 2g Area IT (.3)2 4

"Ji '"

4 x.OO6 x 1200 x.848 2 = 3.518 m 0.3x2)(9.81

0 0 =Zo+ -flo + 3.518 Z... =Zv+ -p" + 3.5180r4. pg pg [

ZD

+.E.L] = 40.0 -

3.518 = 36.482

J1\

P8

~!cncc picwlllclric head at D = 36.482. But ZB'" 38 m. ]'[enee water Oow s from B w D. Apply ing Bernoulli's equmion to points lJ and D

Z8'"

(20+ :; )+hfl or 38= 36.482 + 1,1,

h!, =38-36.482= 1.518m

Om 1.:'i18 =

V2 =

Dischilrgc,

4 x .cXJ6 x 600 x V22

0.2x2)(9.8 1

1 = 0.643 m/s . ,I~1.~'~I8~'~O~.2~'~2~X~9~.8~ 4 x.OO6 x 600 1(

= 0.0202

I I

,

Q2 = V 2 X "4 (d 2t = 0.643 x 111

It

,

"2 x (.2)"

3

1s '" 20.2 lil resls. A ilS.

Ii

~ I

IL Flow Through Pipes

527 1

Appl ying Bcm o ul!i' s equati on 10 poillts D and C

-

I'D

ZD '~=Zc +hl, pg

"' Bu! from continuity

Q 1 + Q2

'"

QJ

Q3 = Q 1 + Q!" 0.06 + 0.0202 = 0.0802 m3 /s V =~ = 0.0802 = 1.I34m/s 3 /t , /t 4 (·3t "4(.09) ' 82 4 x.OIX> X !lOOX 1.I]4 ! 9 _, 6.4 = Zc + = Zc + 4.1 4 03x2x9.81

Zc " 36.482 - 4. [94 = 32.288 m . Ans. Problem 11 .43 Three THal'oirs, A. LJ and C are cOIll!ec led by a pipe .fys/em ,~ I/O\Vn in Pig . 11.2S. The leng/lls and dillmelers of pipes I. 2 alld 3 are 800 m. WOO III. 800 m. (md 300 mm. 200 mill all d 150 1II1n respeClil'eiy. Determ ine IIIe piezometric head m jUllc /ioll D. Take f " JXJ5.

Solution. Gi\'en : The length o f pipe I. LI = 800 III and its dia ., til = 300 mill = 0 .3 m The le ngth o f pire 2. L2 " 1000 In and it s dia .. d 2 " 200 mm " 0.2 111 The le ngth o f pipe 3. L} = 800 III and its dia .. til = I SO mill = O. IS 111 Hcig lll of rese rvoir, A from d
[

z. . -

"'

=

(Zo+ ~; )+ IIf'

f x ~ xV/ -_ 4 x.OO5x800xV (Zo + fpgll Jl-- IIf , _4x d x2g 0.3x2x9.&1

2 l

~

l

60 -

(Zo+ ~; ) = 2.7 1& V11

...( i)

Appl ying Bcrno ulli· s equali on to pnints D and B

( "oJ Z,, + ~

pg

= 28 + Il f = 40+ ,

= ..-v.,. .,,, ,

d 2 x2 g

c4cX~.OCOC;::X~IOC;COOCXCV~ !! = 40.0 + 5.09

(z/> +~;) -40.0 = 5.09 V/

I I

4f xL.,xV,~ -

0.2x2x9.81

V11

...(ii)

Ii

~ I

IL 1528 Fluid Mechanics

I·":··, l

'

'o;<>o~ @ ~,,~~

80m

T

':0>:' ...".,..

T'10m _____ ~~~~N.: ___ '.:a~~ ____ L -vr" a

C

30m

Fig. 11.28

Apply ing BcmooJ!i's equation to poillts D and C

[

'C·

ZD+POJ=4: +h pg

( Zv +

J,

OO ",,',,,,V,C' =30 + 4jxL.,xV/ :: 30+ -,4C ""5CX"8COO ilJ x2g O.l5x2x9.81

~; ) = 30.0 + 5.436 V/

... (iii)

Adding (il :md (ii), we hav e 60 - 40 = 2.711\ V12 + 5.09

V/ + 5.09 V/

20:: 2.718

or

VJ ... ( i v)

Adding (i) and (iii). we have 60 :: 2.71 R V I1 + 30.0 + 5.436 Vi

or

60 - 30:: 30 = 2.7 18 1'1 2 + 5.436 Also from cOll1inuily cqu~lion. we have Q,:: Ql + Q,l

"41t ill 2 X VI :: "41t

"' "'

2

2

iI! V! + 1

"4J!

D.3 V 1 '" O.2 V1 + 0.15

And from (v) ,

.. .{v)

222

d j V1 ur ill VI :: til V1 + tl3 V1 2

X

VJ or .09\', :: .04 V2 + .0225 V]

2.7 18 v/ 5.09

... ( vii)

30 - 2.7 18Vll 5.4 36

... (I'iii)

20

Now from (il'),

1

V/

Substituting the value of V" and VJ in (I'i), we gel

0.09 VI'" .04

20 - 2.7 18V/ + .0225 5.09

30

2.7 18 ~2

5.436

Squari ng bolh sides. we gel (0.09 V ) 2 '" (.04)2 X I

(20 -5.09 2.718 VI?] + (0.0225)2 X 30 - 2.718 V/ + 2 x.04 5.436 2~7~18~"CLll "v x .0225 X \1,~20,--~ 5.09

I I

Ii

~ I

IL Flow Through Pipes

529 1

.0081 VI! '" JlO628 - .000854 VI! + .00279 - .000253 VI! + .0018 .0081 VI! + .000854 VI! + .000253 V i l '" .00628 + .00279 + .0018", .01087 .009207 V12 ", .01087

or

or or

VI ==

' g08~7~ '" 1.0&6 mls J~.O~ .009207

Substituting this value o f VI in (1·;1) and (I'iii)

V = 2

,lo20,--~c2".7~'~8cXC~"-ll

=

5.09

30

2.718x 1.086

Jo2
= 1.816m/s

5.09

2

-222 _. InI S

5.436 Pi~zornClric

head 31

D = Zo + ~ = 30.0 + 5.436 x V/

pg

'" 30.0 + 5.436 X (2.22)2 '" 56.79 m. AilS. Pro blem 11 .44 A pipe line 60 em dil"neler bifurcales (II a Y-jU1IClion into Iwo branches 40 em (lnd JO em ill diameter. If tile rale of flow ill the mai" pipe is /.5 ",J/J' and mean "eloeily of flo .... in JO em iliameler pipe is 7.5 mis, determine tile rule of jlOK' ill Ihe"'O em diameter pipe.

Solution. Given: Dia. of main pip;.:. Dia. of branch pipe I. Dia. of brancb pipe 2.

D

= 60 em '" 0.6 10

DI = 40 cm = 0.4 m

D1 = 30 cm == 0.3 10

Velocity in brand piP<' 2. Vl = 7.5 m/s Rate of flow in main pipe. Q = 1.5 1I1 3fs

Fig. 11.29

Q I = Rate o f now in bwncb pipe J. Q1 = Rat~ of now in branch pipe 2. Q = Rate of flow in rnain pipe. Now rate of flow in main pipe is equal to the surn o f rate of flow in branch piP<'s. Let

Q=QI+Q 1 But Q2 = Area of branch pipe 2 x Velocity in branch pipe 2

I I

...( i)

Ii

~ I

IL 1530

Fluid Mechanics

,

,

",A,x V,: - D ,'x V,= • • 4 ' • Substituting the va lues Df

to I 1. 11

'4~ (0.3) ! x

3

7.5 "" 0.53 m Is

0

and Q! in equation (i) , we ge t 1.5 :Q 1 +0.53 0 1 = 1.5 - 0.53 = 0.97 Ill/s. AilS.

POWER TRANSM ISSION THROUGH PIPES

PlOwer is tran smill cd through pipes by flowing waler or mhcr liq uid s /lowing through Ilwm. The power Ira nsmillcu depends upon (i) the weight of liquid flowing through the pipe and (ii) the total head available at the end of the p ipe. Consider a pipe A B conncclcd to a tank as s hu wn in Fi g. 11.30. The ]lOwer ava ilable at the end B of th e pipe and t he condition for max i mullllransmi ssio n o f power w ill be obtained as lllc rl tioncd belDW :

....-.-.-.-.

Fig. 11.30

Pow('r trammi!!;on through pip('.

L = len gt h o f th e pipe. d == diamdcr of the pipe. H == to tal h~ad a vailahlc at the inle t o f pipe. V", ve loc it y of flow in pipe. II/ '" loss of head due to friction. and! '" co·efflcie nt of friction. The he ad available at th e out let of the pipe. if minor losses an: neglected '" Total he ad at inlet - loss o f head due to friction Let

J.: 1

", H - II =H- 4! X I. xV " / dx2g

h '" 4! XLXV "1 j dx2g

Weigh t of wa te r n owi ng th rough pipe per sec. II' '" pg x vo lum e of wmer pe r sec = pg x Area x Velocity

,

,

=pgx~d-xV

4 The powe r tr3nSmilled at the outlet of the piP<' = weight of water per see x head

~t

o Ull et

Olpg X!!..d 1 XV ) X ( H - 4!XLXV 1) 4 dx2g

Wa".~

Power tran smitted at ou tlet of the pipe ,

2 1'=...ELX !!.. d 2 XV ( H _ 4 fL V 1000 4 dx2 g

I I

),w

...{ II.2I )

Ii

~ I

IL Flow Through Pipes 531 1 Efficie ncy of power tran smission. Powe r available at ou tlet of the pi pe

" ~ ~==~~~C'C=;"C:"'-'Power suppli ed al lIie inle t of the pipe == Weight of wat~r per seC)( Head avai lable al oullet We ight of water pe r sec x Head at inle t

IVX ( H-h/ ) '" If - h! IVxH

If

...( 11.22)

I 1. 11 . 1 Condition for Maximum Trilnsminion of Power. The condition for maximum trnnsmission of power is obtained by differentiati ng cquatio ll ( 11.2 1) with respect \0 V an d cquJling the sain e to zero.

..!!.....(P)=O

Thus


or

.d! .V. . [...£LX!!..d~(IfV_ 4JLVlll~o 1000 4 d x2g

or

~X!!.-d!( H - 4X3X f XLXV11~o 1000

"'

d x2 g

4

4jL V l 11 - 3)( - - - =0 or fI - 3xll/ =O d x2 g ...( 1 1.23)

Equaling ( 11.23) is the cond ition for maximulIl tran smi ssio n or power. It states tlla! power transmi tted throug h a pipe is maximum wile n the loss of head due to friction is one-third of the total head at inlet. I 1. 11 . 2 Maximum Efficiency of Transmission of Power. sion lhrough piJ}t' is given by eq uation ( 11.22) as

Effilciency of power transmis-

H - hI fI

I] :

For maximum power transmission through piJ}t' the cond ition is given by eq uati on (11.23) as H

", : 3

Subst ilUl ing the va lue of hf in emciency. we get maximum "1]. 1]" ... :

H - H I3 H

I

2

=' - 3"="3 0r66.7%.

...( 1 1.24)

Problem 11.45 A pipe of diameter 300 ""Il aml/ength 3500 '" is u~·ed for tile trallsmissioll of power by waler. The Iota/ "eat! at Ihe illiel of the pip'· i~· 500 Ill. Fi",llhe m(lxim,,," po,,·er a!"ai/ah/e at lile ouliet of Ihe pipe. if lile ,",,/lle off .006. Solution. Given: Diame ter of the pipe. t! = 300 mm = 0.30 III

=

I I

Ii

~ I

IL 1532 Fluid Mechanics 3500

Len gth of th e pipe. Tot~1 h e ~d m inl et.

H =500 m

Co-efficient of frict ion.

/= .006

For ma xi mum power

L ""

Iran~rnission.

III

usin g eq uatio n (1 1.23)

H

500

3

3

h, ,,, - '" Now

1,= r

'"

166.7 III

4X!X L XV 1

dx2g

=

4x.OO6x3500XV l 0.3x2)(9.8 1

= 14.27

v2

Equating the two values o f ",. we get 166.7= 14.27 V1 0r

V= t66.7

14.27

=3.41 7 mfs

Q= VXArea

Discharge.

= 3.4 11

It

,

1t

>

J

x - (dt = 3.417 x - (.3t = 0.24 15 m Is 4 4

Head available alth e end of th e pipe

= H- h = j

M a~ imum

power ava ilable =

I/ - !!... '" 3

2 11 "" 2x500 '" 333.33 111

3

3

pg x Q x head a1 the end of pipe kW 1000

= 1000x9.8 1 x .241 5 x 333.33 kW = 689.7 kW . Ah S. 1000 Problem 11 .46 A pipe /i1le 0/ /enstll 2000 m is used for power /rtlllsm iiiJ'iOI1. If 110.3625 kW power is 10 be {"",smilIe" through ti,e pipc iii wilie" ,w,'er hlll'ing a pressure of 490.5 N/cm l al ;lllel is

flowing. Find Ille "iameler 0/ Ille pipe and e/ficie'lcy O/lfllrlsmil·siOIl If the pressure drop orer tile Jength a/pipe is 98.1 Nkm ". Take/: .0065. Solution. Given: Len gth o f pipe. Powe r lransmineo Pressure al in let. :. Pressure head at inl el. Pressure drop :. Loss of head.

L=

2000 m

: 11 0.3625 kW p = 490.5 Nfcm-, = 490.5 x 10 N/m"

"

II ='!!"": 490.5 x 10 ' : 500 m pg 1(00)(9.8 1 .

[.: P'" 1000J

= 98. 1 Nlcm 2 : 98. 1 )( 10· N/m 2 98.1 x 1O~ pg

1000)(9.8 1

: 100m

Co"efficien t of fric tion. / '" .0065 He
I I

Ii

~ I

IL Flow Through Pipes

Now power transll1iucd is g i ve n by. P "

110.3625 "

pgXQX ( H- hj

1000

533 1

)

kW

IOOOx9.8 1 xQx 400 1()()()

1103625 x lOOl = 0.02812 lOOOx9.8lx 400

8m

d ischarge. Q" Area x Velocity ""

~ If x V 4

~ d' x V: .028 12 4

V"" .28 12)( 4 " 0.0358 rrd ' d" 4/XLxV

-"-';:";C:-'-'

The IIc ad lost do~ to fricti on, "1"

Bm

...(i)

d x2 g

Ill "" 100 m J00

= 4 x f x L x V1

""

,4"X,·,llOc605~'c2cllOO~cX,V,--' ")(2x9.8 1

dx2g

2.65 X V ' = 2.65 x (.0358 )1 = .003396 d d" d~

v=

': Prom equat io n (i).

.0358

d' 100= JXJ3396

d' d=

[

.003396)'" [00

""

8 ' .1_77

III

=

127.7 mm . An s.

Efficiency of power tran sm i.<;sioll is give n by equali on (1 1.22). T]

~HC-i'h'L = 500 - 100 = 0.80 '" 80%. An s.

= -

H

500

Problem 11 .47 For Problem /1.46. find : (i) rhe diameter of the pipe correIponding 10 maximum efficiency of lrilll5l11issiOll . (ii) diameter of the pipe corresponding /0 90% efficiency of Irallwni.~,~ioll.

Solution. 0) Diamctcr of pipe co rrespondin g to max imum e fficie ncy. Le t th e dia. of pipe for 1]m., "" d

But fro m e qumi o n ( 11.24), 1]"",,:0 66.67% "" H - h

2

II

3

- - -j = - or

I I

~ 3

500 - Ir j 2 - ""

500

3

Ii

~ I

IL 1534

Fluid Mechanics

1500 - 1000 500 = - - '" 166.7111 3 3

2 3

1,/ ",SOO - SOOx -",

0'

The other data given from Problem 11.46. Power transmincd '" 110.3625 Length of piP<'. L = 2000 111 Co-crticicn t of friction. f = .0065 Powe r transmiucd is give n by Ihe relation. pg XQX ( H - h/)

p==-~~-",

1()()()

110.3625= lOOOX9.8IXQx (500-166.7)

"'

I ()()()

Q=

1103625 x 1000

l000x9.81 x (500

'" 0.03375 n/fs

166.7)

Q = area of pipe x veloc ity of fl ow

"0<

=

~ 4

Ir x V {where V = velocity o f no w }

, ,

0.03375= "4 d x V

V = 0.03375 ,X 4 = 0.04297 /tx d "

... ( i)

dl

4jLV l

Now Ihe head lost duc \0 fricti on. h/ = - - -

d x2g

"1=166.7 III O6 CC'c20()()()~C'CV~' 166.7= -.4C'C·~Ooe5 IIx2x9.8 1

= 2.65 V' = 2.65 X ( .04297 )' = .00489 d d (/1 ~ dS= .00489

(.: V =

.~2197)

= .()(X)(}2933

[66.7 <1= (.0CKl02933)'~ = 0. 1240

111

= 124

111111.

Ans.

(il) Lei th e diame ter o f pi pe. wilen e fficiency of tran smi ssio n is 90% = d 11=90%=0,9 But

BOI

I I

I]

is given by equation ( I 1.22) as,

I] "

" 0.9

H,,500 m

Ii

~ I

IL Flow Through Pipes

535 1

500 - h, 500 '" 0.9 or 500 - 500 x 0.9 = h, or 500 - 450 = hI /'1 = 500 - 450 = 50

The

olh~r

JIl

P = 110.3625. L '" 2000. f = .0065

given data is.

Using relation fo r power trans mi ssion. P =

pgXQX (H - h,)

'-'--'c;;;';;-~'-'

1000

'~1~'~Q~'_(~5~ OO~-~ 50,-,) 110.3625 = ~IOOO~_'~9~ 1000

0'

62,,5,-,7c:'OOO "",c"c =. 0'_S m-fs ' Q= :-CCICI~O=.3" [000 x9.8 1 x (500 - SO)

rr , Q= - d' xV

Bo<

4

.!:.

tf~ x

v =.025

or V : .025x4 = 0.03 183 nd 1

4

... ( i)

tfl

, , 4 fLV 1 Now the head 10SI duc to fnC\lon. hi = - - d x2g

0'

50= 4 x.OO6S x 2000 x II x 28 dS = .002685 = .00Cl0537 50 If = (.OOOO537)l ll = . 1399 III

... I 1. 12

= 140 mm.

AIlS .

FLOW THROUGH NOZZlES

Fig. 11.31 shows a nozz le fitted a1 th e e nd o f a long pipe. The 100al e nergy at [h e e nd o flhe pipe consists of pressure energy and kin etic ene rgy. By rUling the nozz le at the end of the pipe. 111e \olal energy is conve ned into kin etic energy. T hu s nozzles are use d. where higher ve lociti es of now arc required. Th e exam pl es are :

---" 'c{}f H

--rr-'

OIA=O ,

__ l __ -~ - -1+ --

BASE OF

~E

! PIPE

1• f ig. lUI

I I

L

NOZZLE

·1

Nozzle [imd to a pipe.

Ii

~ I

IL 1536

Fluid Mechanics

L In case of Pe lton turbin e. the nozz ic is fmed at the end of the pi pe (callcd penstock) to increase ve loc ity.

2. In case oflhe ex tinguishing nrc. a nozzle is fined at th e end oflhe hose pipe to increase velocity. Lei D", diameter of tile pipe, L = le ngth of the pipe.

A ", area of the pipe = ~ 0 2, 4 v= ve locit y of flow in pi pe. H", to tal he ad at the inlet o f Ihe pipe. d == diam eter of nozz le al ou l]e1,

\' = ve locit y of flow at oU ile! of noale. a'" area of Ihe noz7.le at n Ulle! '"

~

Ifl,

4 f = co·cfficic nt of friction for pipe. 2 Loss of head duc to frictio n in pipe. h, = 4fLV 28 X D

Head available al lh" e nd of th e pipe or al th e ba..,., of nozz le '" Head al in ld of pipe - head lost due to fricti on

=

H- hf = (H_4JLV2] 2g xD

Neglecting minor losses a nd also assumi ng losses in th e nozzle neglig ibl e. we have Tota l Itead at inkt of pipe == totalltead (energy) at tlte ou tk t of nozzle + losses 1,2

But total Itead at o utlet of noa l" == kinetic

h ~ad

vl

== -

2,

4JLV l

V

11= - + hf = - , - - 2g 2g 2gD

(

0: h = 4JLVl] ... U) I 2gD

From continuit y eq uat io n in the pi pe and o utl et of nou k. A V==m' a l'

V= -

A Substituting this va lu e in equati on (I). we ge t ,.'

4 fL 2gD

H= - , - - ,

2g

I'

=

(,n.)' -

A

2gH

(I

4fL D

fU "( 1< 4DA' .'] -

-

2g

... ( I 1.25)

" ] A'

+ -- , ~

Discharge th ro ugh nozz le = I' X v.

I I

Ii

~ I

IL 5371

Flow Through Pipes

I 1. 12. 1 nozzle"

Power Transmitted Through Nozzle. 1

The kin etic energy of Ihe jet a1th" outlet of

,

"2 mv-

Now mass o f liquid aT Ihe outlet of nuzzle per second" pm' Kinetic c llcrgy of the jd allhc oUlle t p.:r SIX.

= -I 2

P(lV

2

1

X ,. = - pm'

2

1

Power ill kW at the oUllet of nozzle = (K. E.lscc) x

3

I~

=

,

pm·

.'c]])=OO-

Effi ciency o f power transmission through nozz ie, 1

- pm'

Powe r at outlet of nozzele 11 = Power at Ihe in let of pipe =

,

. ~'~]])~(](]~

1'g, Q. H 1000

~ p(n •. \,l

~p(l\ •. \ , l

pg.Q.H

pg.av.H

: ~'c-;;-,,: ,

= 2gH " = [ 1 4 fL1 III +, ()

1

Q=al')

... ( 11.26)

Al

[ I 1. 12.2

1':

F,"m,q"",;O" (ll '25)'~:[

...

2g11

Condition for Maximum Power Transmitted Through Nozzle.

' 11

1

1+ _4 /_1. _" _. D A'

We know Ihal,

Ihe lo[al head at inlet of pipe = total head at the olllle! of Ihe noale + losses

tota l head at outlel of nozzle '"

i.e .. /1, '"

4

f L

V

Dx2g

1

~ and

"

j

'" Joss o f liq UId III pI pe

: _v 1 + C4~.f~.L,,'''V_' 2g D x2g

;; =(11 _4.~~~~Vl) I - pa v [ ( But power transmiucd through nozzle = - ' - - : -'-- X 1, 1=_'__ 2g H

I I

! 3 - pav

I - pm'

1000

1000

1000

-

4

.

f •L.

V'l]

lJ x 28

-

Ii

~ I

IL 1538 Fluid Mechanics

l

LV1

'" Plla)' [H _ 4/ 1000 D x2g

...( 1 1.27)

Now from cOnTinuity cqumion. AV '" (IV

v=~ A

Substituting tile value o f V in equation (11.27). we get 1

1

pgm' 4 fLa ' -vPower transmitted through nozzle" - [ II - -, lOOO D x 2g A

.

.

d(P)

The power (P) will be maXllllum, when - - " 0 ill'

~

"- [ pg",, ( II 0':' dv 1000 D x2g A -

"'

II

"0

"[pga ("'dv 1000

"'

P
1l=oorfl _ 3x

4 fL D x2g

xv!=o( . . V"-""l A

4 /L · '" II == licad loss in pi pe -V-

"'

D x2g

!

1

... { 11 .28)

"'

Equatioll (11.28) gives the condition for maximum power transrnillcd through nozzle. II states 1hat power transmitted through nozzle is maximum when tlie head lost due to friction in pipe is one-third the total head supplied at the inle t or pipe.

11. 12.l

Diameter of Nozzle for Maximum Transmission of Power Through Nozzle . For

/I rna.
3

Bo<

/,/

1 4fLV Dx2g

= 4! .L .V' D x2g 1 H 4!t V orH=3x - - 3 D x2g

---= -

Bu( H is also = (Olal head at outlet of nozzle + losses

I I

Ii

IS34

Fluid Mechanics

Equaling the two values o f H, we get

3x

4jLV 1 v 2 4fLV! 12jLY' = -+ or O x 2g 2g O x 2g D x2g

8jLV 2

or

O x2g

4jLY'

D x 2g

v2 2g

=-

=2g

... (i )

But from continuity. AV = av or V = ~. A

Substituting this value of V in equation (I), we get 8jL

0'

o

A

_

-x-:-z -

or

_8 jL _ x

-:(~",,-d_'-7)" = 1 or

o (~o' r

8jL x

0

d= ( -0' ) '" 8jL From equation (ii).

8jL

1

( Divide by ;: ) ... (ii)

d: = lord' = 0' 0 ~ ... ( 11.29)

A'

- o =-, 0

~o =rjL 0

... ( 11.30)

Equation ( 11 .30) gives the ratio of the area of the supply pipe to the area of the nozzle and hence from this equation. the diameter of the nozzle can be obtained. Problem 11 .48 A nozzie is jilted 01 the end of a pi~ of lenglh 300 m and of diameter 100 mm. For the ma;cimllm transmission of poKIer through the noWe, find the diameter a/nozzle. Take f = .009. Solution. Given : Length of pipe. L=300m D= I00mm=O. lm Diameter of pipe. Co-efficient of friction. /= .009 Let the diameter of nozzle =d For rnuimum trolnsmission of power. the diameter of nozzle is given by relation ( 11 .29) as

d = ( 0' 8jL

)1'.=(8 x.OO9 OJ! )1'. = 0.02608 m = 26.08 mm. ADS. x 300

Problem 11.49

The hl!ad of watl!r al thl! inlet of a pip#! Z(J(X) m long and 500 mm dial1ll!rl!r is 60 m. A nOWI! of diaml!tl!r 100 mm at iu outll!t is fitted to thl! pi~. Find (hI! l'l!locity ofwall!r at thl! ourll!t of rhl! nOWI! iff = .01 for thl! pipl!.

SOlution. Given : Head of water at inlet of pipe. H = 60 m

~ I

IL 1540

Fluid Mechanics

Length of pipe. Dia. of pipe. Dia. of n01:zle at oUllet.

L '" 2000

D '" 500

III

111111 '"

0.50

III

<1=100111111=0.1111

Co-cflkicm of frict ion. /= .01 The velocity at outkt of nozzk is given by eqUal ion ( 11.25) as 2x9.8 1 x6(}

2gfl

(I

AIL "' ) +- ' CT D

A

1-r:====C2C'~9~>~'~'~60~==~~, '" 30.61 m/s. Ans. X.I)" 1 + 4 x.O lx2000 , (O.I -0.5 0.5 x.5 Problem 11.50 Filld Ille mllXinwm power IWI/smilled by a jel of waler disc/!{jfgi,lg freely ow of noule jilTed 10 a pipe'" 300 III long ami /(}() mill diameter willi co·efficienl of [riujOlI as O.OJ. Ti,e (H'ailllhie head (lIllie "ou le is 90 m.

Solution. Given: Length of pipe.

L=300m

Dia. of pipe. D", HlO Illlll '" 0.1 III Co-efficient of frict ion. f= .01 Head available at nozzle. '" 90 III For maximum pow er transmission through the nozzle. the diameter at the outlel of nozzle is given by cqu~lIiun (11.29) as

,,_(D')IN_[ 8fL Area allh e nozz le.

/t,/t (I" -

4

(O.l)S 8x .0 I x300

]114".0254 m

,

,

d-" - (.0254,-" .0005067 m-.

A

The nozz le allhe oU IIet. discharges waler into almosphere and hence the 10lal head availabl e allhe nou le is converted inlO kinetic he ad. Head available at outlet " 1,1'2g ur 90" 1,212g

v" .)2 x9.8 1 x90 ,,42.02 mls DiSl:hargc through nozzle.

Q" a x v" .0005067 x 42.02" 0.02129 m Jls

HC'~'"dC';'oOc'e'ele"'"Oerc""""'='""'" Max imum power transrniued " jP"g",'"Q,-"", 1000 _ IOOOx9.8 IxO.02 129x90 _ 8796kW 1000 _ I . . Am.

Problem 11 .51 The rate of flow of ",(lfer throllgll (I pipe of length 2000 m and diallieter I III is 2 IIIJh. At the end of the pipe (I nozzle of olltIhle diallieter 300 mill is filled. Find 'he power trimsmilled

I I

Ii

~ I

IL Flow Through Pipes

Illrough Ihe Iioule

If Ihe

head of !I'll/a at inler of Ille pipe is 200

//I

541 1

(lnd co-efficient of frielioll for

pipe 1$ 0.0 1,

Solution. Given u:ngth uf pipe. Dia. of pipe, Discharge, Dia. of noule, H~ad a1 iold of pipe.

L =2000m D = 1m Q=2m 3/s

,,== 300 111m '" 0.3 111 H",

200

Co·dOcicn! of friction.

f= .01

Now area of pipe.

A

==.::. 4

rn

D 2 ='::' x 12 == O.7S54 m l

4

~ = 2546 m/s A 0.7854 Power lransmiltcd thro ugh nozz le is given by equation ( 11.27) as Velocity of water throu gh pipe. V", Q ==

p=

=

4jl.V' ]

pg .a, l'[ H- - 1000

Dx2 g

WOO x9.81 x 2.0 [ 4 x .01 x 2000 x (2.546)2] 200 (.: av= Q) ]()OO lx2x9.81

= 3405.43 kW . A n s .

.. 11. 13

WATER HAMMER IN PIPES

Consider a long pipe All as shown in Fig. 11.32 conncclcd at one end lO a lank cOlUaining water at a he ight of H from the CCnlrc of the pipe. At the other end of the pipe. a valve 10 rcgulalc thc n ow of water is providcd. When the valvc is complctcly open. the water is nowing wit h a veloci ty. V in the pipe. If now the vJlve is suddenly dosed. the momentum of th c flowing wJler will be destroycd ~nd ("()nsequcntly ~ wave of high pressure will be sc t up. This wave of high pressure will be translllillcd along the pipe with a velocity eq ual to the ve locity of sound wav e and may creale noise called knock ing. Also this wa,'e of high pressure has lhe effect of hamme ring action on the wa ll s of the pipe and hence il is also known as water hammer.

, FI g. 11 .32

VALVE

Waltr ba mmt!r.

Thc prcss urc ri se duc to water hammer depends upon: (i) the velocity of flo w of water in pipe. (ii) Ihe length of pipe. (iii) tim c taken 10 c lose the va lve. (il'l claslic properties of Ihc malerial of Ihc pipe. The following cases of water hammer in pipes will be considcred : I. Gradual closure of valve. 2. Sudden closure of valve :md I:onsidering pipe rigid, and

I I

Ii

~ I

IL 1542

Fluid Mechanics

3 . Sudden cl osure of va lve and considerin g pipe clastic. I 1. 1J. I Gradual Closure of Valve. Let the wate r is flowing through the pipe AB s hoWIl in Fig. 11.32, and the valve provided ill the end of the pipe is dosed gradually. Let A = area of cross· scClion of the piP<' AB, L = length o f pipe. V = ve locit y of flow of water through pipe, T = lime in second required \0 cl ose th e va lve. a nd p = intensity o f pressure wave produced. Mass of water in pipe All '" P x vo lum e of water = p x A x L The valve is closed g radu all y in tim e 'r seconds ami hellce the wate r is brought from init ial veloci ty V to ze ro velocit y in time sc~"oilds. :. Rem rdat ion of Waler

Change of ve locity

v- 0

V

Time

T

T

= Ma'is x

Retarding force

Retardation

= pAl.

x

!.

...( i)

T If P is the intens it y of pressure wave produced due \0 closure of the va lve. the force due to pressure wave. ... (ii) "'P x area of pipe '" p x A Equating the twO fo rces. givcn by equat ions (i) and (ii).

V

pALx - "'pxA T

pLV

...( 11.31)

/!"' - T

Head of pressure.

P pg

pLV pgx T

H ~ - ~--

~

pLV oc pxgxT

(i) The val ve c losure is said to be gradual if T> 21.

e

LV gT

H~ ­

... ( I 1.32)

.. .( 1 1.33)

where I " time in sec. C" veloc ity o f pressure wave 2L (ii) The valve c los ure is said to be s udden if T < -

e

.. .{ 11.34 )

where C", ve locit y of pressure wave.

11.13.2 Sudden Closure of Valve and Pipe is Rigid. Equation ( 11.31 ) gives the relation betw.:.:n iner.:ase of pressur.: duc to wa ter hammcr in pipe and th.: time r.:quircd to close th.: valve. If 1= O. tit.: increase in pressu re will be infinit e. Bul from cxpcri m.:nt s. il is ob""rved thaI thc increase in pressure due to wa ler hamlll~r is finite. eve ll for a very rapid closure of \·alve. Thu.~ eq uati on ( I 1.31) is valid on ly for (i) incompressible nuid s and (ii) whe n pipe is rig id. But when a WaVe of high pressure is crea ted. the liqui ds ge l co mpressed 10 some ext en t a nd also pipe material ge lS stretched . Por a sudden closure of valve [lhe valve of I is small and hence a wave o f high pressure is created ] the following lwo cases wi ll he conside red: (i) Sudden closure of va lve and p ipe is rigid, and (ii) Sudd~n closure of va lve and pipe is c lastic.

I I

Ii

~ I

IL Flow Through Pipes

543 1

Consider a pipe An in which wate r is flowing as s hown in Fig. 11.32. LeI the pi pe is rigid and va ll'c fitted at Ihe end IJ is dosed suddenl y. LeI A" Area of cross-sect ion of pipe AB. L", Length of pipe. V= Velocity of flow of water through pipe, P'" InlcIISi er of pressure wave produced. K = Bulk modulu s of water. When Ih e valve is closed suddenl y. Ihe kinetic energy of [he flow ing water is converted into strain ene rgy of wa ter if the effect of friction is neglected and pipe wa ll is assumed perfect ly rigid. I , Loss of kineti c energy'" x mass of wa ter in pipe x V'

'2

I , = - xpALxV'

2

Gain of strain ene rgy '"

(': mass" p x vo lume" p x A xL)

i

~ ( J; J x volume", I; x AL

Equati ng loss of kinetic energy 10 gain of stra in energy I , I 1/ - pALxV' = - xAL

2

2 K , I 1 2K p'= -pALxV x = pK v' 2

p=

AL

,

~PKV- =

JKp! P

... ( I 1.35)

V.,JKP=V

( .. ,JKlp =

=pVxC

cj ...(1116)

where C" vclocily* of pressure wave. I 1. 13. 3 Sudden Closure of Valve and Pipe is Elastic. Consider the pipe AB in wh ich water is flowing as shown in Fig. 11.32. LeI the thickness .r" o r th e pipc wall is s ma ll compared 10 the diameter 0 of the pipc and also let the pipe is clastic. Let E = Modulus of Elasticity of the pipe material. I

- " Pois;;on·s ratio for pipe matcrial. m p" Increase of pressure due 10 water hammer, I" Thkkness of the pipe wa ll. 0" Diameter of the pipe. When the valve is closed suddenly, a wave of high pressure of in knsity p will be produced in the water. Due to this high pres~urc p. circumferen tial and lo ngitudinal stresses in Ihe pipe wall wi ll be produced. li '" Lo ngillld inal stress in pipe Let Ie" Circu mferent ial stress in pipe, pO

The magnitude ofthesc stresses arc given aS li= -

pD

and / c " 41 21 Now from the knowl~dge of strength of material we know , strain cn~rgy slored in pipe material per unit volume • For derivation of velocity of pressure wave. please refer 10 chapler 15.

I I

Ii

~ I

IL 1544

Fluid Mechanics

=

_'2£! _ [t:1 + f}, _ 2fi X fc] m

2, pD , PO ]

, [( PD)' [PD)' '" 2£ 4r + 2r

-

Taking

m

41

21

m

(i.e.,

= 4

Poisson ratio '" ~)

:. Strain energy stored in pipe material pe r unit vo lume ,

1 [ p1 lJ

'" 2£

pl /) l

/

1)1

+"""'4T - 4/ " x4

16/ "

1

1 pl D l pl D " = 2£ x----:1T'" 8EI l

Total volume of pipe material", rt f) x I X L. TOl al strain energy stored in pipe material '" Strain energy per unil vulume x total volume

plnD'L P"01 = , x itO X IX L '" '-c~-' 8Er

8£1

c- n;~l '" Now loss of kmetlc energy of wa ter

Gain of strai n energy in water '"

= -I

2

1

mV

= -I

2

1

pAL x V

~ ( ~ ) x volume = +~

Area o f pip:: =

A)

C: '" = pAL)

x AI.

Then. loss of kinetic energy o f waler = Gain of strain energy in wale r + Suain energy stored in pipe 111<11crial.

..!...PAl. x 2 Divide by AL.

Vl= .2!. . (L] xAL+ p l AXDL K 2EI

"'..!... p! + p" O = p! [~ + .!::] 22 K2EI2KEI

pV"

,

p =

pv' ~+.!:: K

I I

EI

Of P=

/f.'iV' ~+.!:: '" K £1

Vx

JH ~+~ K

... ( 11 .37)

B

Ii

~ I

IL Flow Through Pipes

I 1. 13.4 LeI

5451

Time Tilken by Pressure WOlve to Trilvel from the Villve to the Tank i1nd from Tank to the Valve T", The required time taken by pressure walle

L = Length of the pipe C = Velocity of pressure wave 1'11<,," lotal distance = L + L = 21. :, Time,

2L

Distance

T:

=='7'=,--,----= : -C . Velocity of pressure walle

. ..( l l ,}8)

Problem 11 .52 Tile w(l/er isf/owing wilh (l "e/oCil)' of 1.5 ",Is ill (1 pipe oJ/eng/h 2500 m (wd of diameter 500 mm. AI/he end of Ihe pipe. " \'GI,''! i~' prorided. Find Iile rise i"pressure if the ,"u/,"e is closed ill 25 seCOllds. Take Ille ..a/"e olC = 1-160 mls. Solution. Gil'i.:n : V=l.5m/s Ve locity of water. L ", 2S00 III Length of pipe. D = 500 111111 = O.S III Diameter of pipe. T", 25 seconds Time to close the \'alve. Value of. C = 1460 m/s :p Let the rise in pressure The ratio .

2L = 2 x 2500 =3.42 C 1460

From equation ( 11.33). we have if T> 2L. the closure of valve is said to be gradual.

C

Here

T ", 25 sec and

'L

~

C

'" 3.42

T> 2L and hcnce valve is dosed gr:ldually. C

For gradually closure of va ll·c. the rise in press ure is given by equation (I [.31 ) as p

=

P~L

'" 1000 x 2500 x

~~

'" 150000 N/m

1

150000 N N '" 15.0 - ,' An s . [0' em 1 ern Problem 11 .53 If itl Problem 11.52. lit e mire is c1o ,~ed in 2 ICC. Jiml lite rise itl pre.l$ure behind the I'oll'e. Assume lite pipe 10 be rigid one (lnd /(Ike Bulk modulus of ",mer. i.e" K = 19.62 x 1(1 N/cm~. Solullon. Given: v = l.5mfs. L =2500 m f) = 500 mm '" 0.5 1ll T",2sec Time to close the valve. K '" 19.62 X [0' N/cml Bulk modu lu s of water. '" 19.62 x [0 4 )( [0 4 N/m ' "" 19.62)( [0 8 N/m' '"

Velocity of press ure wave is g iven by.

I I

Ii

~ I

IL 1546

Fluid Mechanics

c= fK=

19.62x 108 '" 1400 mfs 1000

Vp

The ratio.

(": p

C '"

1400

1000)

n

2x25oo

lL

=

' 7 < -

'" 3.57

C

.

:. From cq u:lli on ( I 1.34). ifT<. 2 L. va lve iscJoscd sudde nl y. For s udde n closure o f va lve . when C pipe is rigid. the risc in pressure is g iven by equatio n ( 11 .35) o r (I [.36) as P'" V.[KP '" 1.5

~19.62 x 109 x

("; P'"

WOO

[(00)

'" 210. 1 X [04 N/m 1 '" 210. 1 N/c m ~. Ans.

Problem 11 .54 Ifin Problem 11.52. Ille lhicknes,f ofrhe pipe is IV mm and Ihe mll'e is suddenly closed (1/ Ihe end of Ihe pipe. find IIII' rise in p re.fsure if Ille pipe is considered 10 be eias/le. Take E '" 19.62 X {OiO Nlm 1 for pipe lila/erial a llli K '" /9.62 X 104 Nkm! for water. Cu/cu/me Ihe circumferential srress and longiluriinal stress tll'I'e/oped ill IIII' pipe Will/. Solution. Given: V'" 1.5 m/s. L", 2500 m. D", 0.5 m r:lO mm : .OI m Thic kn ess of pipe. E: 19.62 x 10'0 N/m' Modu lus of elastici ty. Bulk modulu s, K : 19.62 x 10~ Nlcrn 1 = 19.62 x 10 8 N/rn' For sudden c losu re of the va lve for an c1aSlic pipe. the rise in pre:;sure is g ivc n by equat ion ( 11 37) as p = Vx

( ~:~) K

'" 1.5 x

= 1.5 x

Et

[ 1x lOR+ 19.62

1000

05)

19.62 X 1010 x .0 1

1000

(S.Cl9XIO 1O +2.54xlO 10 )

: 17 15510 Nlm 1 Circu mferential stress ife) is given by

'"

171.55 N/cm !. AilS.

'" pxD = 17 1.55 x 0.5 = 4286.9 Nlm l 2, 2 x .OI .. . . p xD 171.55xO.5 2 wngltudlnal stress tS given by,/, = - - = = 2 143.45 N/m • AilS. 4r 4 x.Ol

Problem 11 .55 A ,·,d..e is provided al rile end of a casr iron pipe of dialllerer 150 111111 m,d of rllid:lless 10 mm. Tile warer is [lowing rllrougll rile pipe, which is suddenly sropped by closing Ille \"a/l·t'. Pind rile maximum I"t'locify of Wafer. wilen rile rise of pre.~surc due 10 slIddeli clO.lllre of \"all"e is 196 .2 Nlcm J• Take K for wmcr as 19.62 x 104 Nkm ' allil E for casr irOiI pipe as 11. 772 x ul Nkml. Solution. Givcn : Diameter of pipe. Thic kn ess of pipe.

I I

D", 150 111111 '" 0. 15 m I '" 10 111111 '" .0 1 III

Ii

~ I

IL FlowThroughPipeS

Bulk modulu s.

p= 196.2 Nlc rn 2 : 196.2 x 104 Nfm ~ K = 19.62 x t o'! N/cm 2 '" \9.62 x 10 8 N/m 2

Modulus of elastic ity.

E" 11.772 X 106 Nlcm 2 " 11.772 x 10 10 N/m 2

Risc of pressure .

5471

For sudden c losure of va lve and when pipe is clastic. the pressure risc is given by equation (11.37) as

p

P" Vx

196.2 10 X

"'

4

",

Vx

=Vx

v=

(-'K- +,,) "

1000

Vx

I 0.15 ( 19.62xlO! + 1 1.772 X 10 10 x.OI

1

J:;::=:;::~I ~OOO ~;;::=:;::~ 5,(19 x 10 10 + 1.274 x 10 10 ~;:;;: w =VxI2S.27xI04 .Ir;:;;I~OOO 6.364 x to

196.2

X

10

4

125.27 x 1 0~

= I 566 mls

= 1.566 m/s. An s .

Ma!
... 11 . 14

'"

PIPE NETWORK

A pipe network: is an imc rcon nCl1cd system o f pipes funning several loops or cin: ui[s. T he pipe netwurk is s hown in Fig. 11.33. The examp les of suell networks of pipes arc the municipal wa ter distribution systems in cities and labormury suppl y syste m. In such syste m, it is requin-d to determine the distribution of flow through the various pipes of th e network. The follow in g are the necessary cond itions for any network of pipes: (,) The flow into eac h junction must be equa l to the flow o ut of th e junction. This is due to con tinui ty eq uati on. (ii) The a lgeb ra ic sum of head losses round each loop must be zero. This mea ns that in each loop. the loss of head due to flow in cloc kwise direction must be equa l to the loss of he ad due 10 flow in anticlockwisc direction. (iii) T he head loss in each pipe is expressed as hJ ==rQ". The value of r d~pends upon th~ leng th of pipe. diam eter of pipe and co-cfficient of fri ction o f pipe. The va lue of II for turbulent fl ow is 2. We know that.

II ==

J

==

~4~X~f~X~L:ccX~V_' : Dx2 g

4fLXQl DX 2g

==

4JLX ( QA' )'

X(: D l y

Dx2 g

==

4/LXQ '

DX2gX(:)l

XD4

4f xLxQ'

2gX(~r xD l

I I

Ii

• if.'

•

Row Through Pipes

5431

...( 11.39)

This head loss will be positive. when the pipe is a part of loop and the now in the pipe is clockwise. Generally. the pipe network problems arc: difficult to solve analytically. Hence the methods of successive approximations are used. ' Hardy Cross Method' is one such method which is common ly used.

I 1. 14. 1 Hardy Cross Method.

The procedure for Hardy Cross Method is as follows : I . In this method a trial diSiribution of disc harges is made arbitrarily but in such a way that

conti nuity equation is satisfi ed at each function (or node). 2. With the assumed values orQ, me head loss in each pipe is calculated according to equation ( 11 .39). 3. Now consider any loop (or circuits). The algebraic sum of head losses round each loop must be zero. Th is means that in each loop, the loss of head due to flow in clockwise direction must be equal 10 the loss of head due to fl ow in anliclockwise direction . 4 . Now calculate the nel head loss around each loop considering the head loss to be positive in clockwise fl ow Ilnd to be negative in anticlockwise flow . If the net head loss due to a.'isumed values of Q round the loop is zero, then the assumed values or Q in that loop is correct. But if the nel head loss due to assu med "alues of Q is not zero. then the assumed values of Q are corrected by introducing It correction 6Q for the flows. till thc circuit is balanced. The correction factor 6Q· is obtained by dQ- -

~> a;

... ( 11.40)

- L,rn Q; -I

For turbulent flow. lhe value of n = 2 and hence above correction factor becomes as

- L ,QJ

dQ= ~

... ( 11.41 )

.... 2'0.

5. If the value of 6Q comes out to be positive. then it should be added to the flows in the clock· wise direction C'; the flows in clockwise direction in a loops nt"e considered positive) and subtracted from the flows in the anticlockwise direction. 6. Some pipes may be common to two circuits (or two loops). then the two corrections are applied to these pi pes .

• L.e:t for any pipe Qo = assumw discharge and Q = correct discharge. then Q=Qo+dQ : . Head loSJ forthr pipe. h,- nr=r(Qo+tJ.Q)2. For complete circui!. the net head loss. IiI, = t (~) = II' (Qo + dQ)l = &(~ + 2Qo.1.Q + dQl) = !r (Q02 + 2Qo dQ) As dQ is small compared wilh Qo and hence dQl can be neilect~.

.. I ref '"' I rQ02 +!r x 2QoAQ For the conttl distributton. the net head loss for Il circuit should be uro (i.r .• I (rtf) '" 0) :. IrQJ +!r X 2QoAQ=O + 4Q II' x 2Qo " 0 (As AQ is same for one circuit. hence it can be taken 001 of the summationJ

f.h,-

or

IrQJ

AQ=

-}; , ot . ); 2,

a.

~ I

IL Flow Through Pipes

5491

7. Afte r tlie correctiuns have been applied to each pipe in a loop and \0 all loops. a second trial (;akulalion is made for all loops. The procedure is repeated Till t..Q becomes negligible.

~-~--I "

Fig.I1.33 Pipellei1J}Qrk. [Loops arc, ABCGFA- FEGf'. GEHG. GHDG

"III) GCDG]

Problem 11 .56 C('/cll/ale Ihe diIchllrge ill each 1';1'1' of IIII' Ildwork SilO"''' ill Fig. '/.3-1. Th e pipe network f Oll sis/S of Spires. The head loss "f ill a pipe is g;,'en by hI = ..Q2. The l'alw:s of r fOT "Mious pipes ami {,Iso IiiI' inflow or UifljIOWS at "odes are silo,.." ill Ihe figure.

,.,

.. •

,. ,

, '" ,. ,

,. , Fi g . 11.34

Solution. Given: Inflow at node A '" 90. outflow at B "" 30. at C '" 40 and at D '" 20. Values of rfor AB = 2. forBC = 1. for CD", 2. forAD = 4 and for liD = 1. Por the first trial. the discharges arc a.%umc(/ as sliown in Fig. 11.34 (a) so that continuity is satisfied at each node (i.e .. flow into a node == flow out of the node). rOT this distribution of diSl:hargc. the corrediuns IlQ for the loops ABD and BCD are calculated.

'~----~----~~ 90 60 8 30

Fig . 11.34(a)

I I

Ii

~ I

IL 1550

Fluid Mechanics

fi rs t Tri:1I LoopADB

Pipe

,

Q,

, , '" " '" '"

'
h," rQo'

Pil'"

AD

4 X 30' = 3600

08

_lxIO' ~ _ IOO

2xlxlO=20

C8

-2x60' . - 7200

2><2><60 .. 2-10

80

2

2x4x30=140

r." d

= - 3700.

AQ~

-I,o,; :l: 2r

o.o

DC

l:2rQo = .500,

..

-(-3700)

un(iciockwisc.

As toQ is positive for loop ADIl. hence it soould be lId
flow for second trial for loop ADn will be as follows: Pipe AD = 30 + 7.4 = 37.4 (flow is doc~ wi..,) Pipe All = 60 - 7.4 = 52.6 (flow is amic10dwisel Pipe nD .. 10 - 7.4 .2.6 (flow is an1icloc~wisc)

,

Loop OCB

Q.

, , "

2rQ.

hJ" rQo'

20

2x20" =800

2x2"20~80

20

_ l x21i-~ _ 400

2x I

1><10'., 100 t2 = .500

2><1><10 ., 20. !2rQo= 140

f(!;

x20~40

aQ .. - I rQi' -500 __ 3.57 1: 2r<4;. 140

. 7A

500

[n the loop AIlR. the head loss h, is Jl/:8~li\'e in pipes DR nnd All as the direction of dischargu in lh~
,

= - 3.6.

The head loss in pipe Be for loop OCR;s negat;ve ,", d irection pipe BC di.< d'a'g" nnlidockwise . As lIQ is negative for loop VCR. hence ;1 should be subtracted from the flow ;n {he dock"'i"" direchon a"d added 10 Ihe flow in lhe illllic1""~wise dirrelion. li enee correeled flow for second lrial for loop ocn will be as follows: Pi pe DC = 20 - 3.6 = 16.4

,,.

"

'"

'"

Pi pe Ile = 20 + 3.6 = 23.6 Pipe BO· .. 26 - 3.6. - I

N"I ~. 11lc pipe IlD is (."Qmm<>n 10 (wo loops (i.e .. loop ADn and loop (>e8). Hence Ihis pipe will gel Iwo corm:li'ms. Afler Ihe Iwo corm:li{ms. lhe r"sul!;.",( flow in pipe 80 is ncgali"e in I(JOp OC8. Ik nce Ihe di ....i:I;on of flnw will be anlidockwisc in pipe nD for loop DCn.

37.4

,. ,

The di
~ I

I~

~ I

IL Flow Through Pipes

Pire

AD DB

'"

,

,

,

w.,

ADB hr'" rQ/

Q,

,

37,4

.j

'''',

x 37.4' ~ 5595

2 x.t X 37..1

I'ip.e ~

299.2

DC

CB 2 52.6 _ 2 x 52_6' .. - 5533.5 2x2x52.6 .. 210A RD 2><1><1 .. 2

I X I ' .. 1

rrQ o' " 62.54.

",I.'IQ "

&'0, DCB Q, h,. 1001

,

,

L 2r
2rQ o

16.4 2" 16.4' ~ 537.9 2><2>< 16.4~65,6 ~ I x 23.6' - 5.56,9 2 x I x 2... 6_ 47.2 _ 1><1' .. _ 1 2><1><1 .. 2

, , ,

23.6

"irQo' .. - 2il.

UrQo" 511.6

- I, rQJ

551 1

-~> a;

62j~

,,- - -

,'.<.\Q" L/ 2rQ

-5 \ 1.6

o

nrQo'" II.t S

- (_20) .~

=....!£.... =0,174

=-0.122"'-0,1

11 4.8

::: 0.2 A. AQ is ""gati"~, hence;1 should be subtracted from lhe flow in (he doclwis<- dirtttion and added to the flow in the

As dQ is I"'siti\·e. heoce i1 should be added 10 1he flow in the clock",i ... direction and subtracted from

""tidockw;", di=tion

the flow in the antidockwise direction.

As the correction \l!.Q) i. small (i. ~ .. "'Q = - 0 . 1). this

As the correction ("Q ) is small (i.e .• t.Q = 0.2). this

rorrcclion is applied and further trial, are discontinued. lIeoce colTCCled I10w for loop ADB will be as follows: For pipe AD. 0 0 = 37,4 - 0. 1 = 37.3 (as flow is clockwise) For pipe /)B. Qo = I - 0.1 ~ 0,9 (as flow is clockwise)

""""'tion is applied and further trials"", discontinued. li enee COlTCCtoo flow for loop DCB ,,"'ill be as follows: For pipe DC. Qo = 16,4 + 0.2 = 16.6 (clockwi ... flow) For pipe eB. 0" = 23.6 - 0,2 = 23,4 (amiclockwise flow) For pipe 1m. QQ'" 09 - 0.2 .. 0.7 (antidockwise flow)

For pipe All. QQ'" 526 + 0.1 '"' 52 ,7 (as flow is anticlockwisc )

Th" final distribution o f dis.charges in each pipe is as follows: Di s.c harge in pipe AD = 37.3 from A to D AB = 52.7 from A to B DB = 0.7 fro m D to B DC = 16.6. from D 10 C BC = 23. 4 from B to C T he final discharge in each pi pe is sho wn in Fig. 11.34 (c)

ok-"'____",;.,___-..:;c, ,, 37.3

23.4

A~

90

____

~

____

52,7

~~_

B 30

Fig. 11 .34 (el

I I

Ii

~ I

IL 1552 Fluid Mechanics No te. The pipe DB is CommOn to two loop (i,t' .• loops ADB and loop DBC). Hence this pipe will get tWO

corrections. For loop AOB, lhe correction ""Q" - 0. I ami hence lhe rorTe<:ted flow in pipe DB is I - 0. I .. 0.9. Now again. the corrc~lion is applied to pipe DB when we consider loop DBe. For loop DBC. the ~'Orrcdion IlQ .. 0.2 but flow is antic1oc~wisc and hence the final correct 110'" in pipe DB will be 0.9 - 0.2 .. 0,7 ,

HIGHLIGHTS 1. The energy loss in pipe is classified as major energy loss and minor energy losses. Major energy loss is due \0 friction whi le minor energy losses are duc to sudden expansion of pipe. sudden C01\lr~Cl;on of pipe. bend in pipe and an obstruction in pipe.

_ _ . _ 4/ LV' 2. Energy loss duc to fflcllOn IS gtven by Darcy Formula.it _ - - - . l
C.

Ch~lis

V ~ c..r;;li Che~y's fonnula Constant

m _ Hydruali, mcan depth

V ", Velocity of flow

~~

(for pipe running full)

4

h

i_Loss of he'ld per unit length.. ;

hi '" L )( i. where i is oblai ned from Chezy's fonnula.

(V, _ V,j2

4 . Loss of head due to sudden expansion of pipe. he "

2,

-

where V, ,. Velocity in small pipe. VI = Velocity in large pipe. S. Loss of head due 10 sudden eo"tmetion of pipe.h, '"

[..!... - llYlC _II c

V'

where Cc _ co·efficient of contraction • 0.375 ~

... (For Co _ 0.62)

2,

~0.5 :i.

"

V'

6. Loss of head allhe ell lrJnCe of a pipe. hi .. 0.5 7. Loss of head atlhe exit of pipe. ho

V' 2,

" -.

"

" .(if value of Co is not gi ven) .

8. The linc representing Ihe sum of pressure head and datum head wilh respttl 10 some reference line is c~tlC<.l hydraulic gradient line (H.GL) while the linc representing the sum of pressure head. datum head and velocity head wilh resp<."t to SOme reference line is known as toml energy line (T. E.L). 9. Syphon is a long bent pipe used 10 transfer liquids fron' a reservoir at a higher level \0 another reservoir ~t a lower level. whcn the tWO reservoirs are separated by a high level ground. 10. The maxim um vacuum created allhe summit of syphon is only 7.4 m of waler. II. When pipes of different length. and different diameters are eonneeled end to end. pipes are called in series or compound pipes. The ratc of now through each pipe connected in series is same. 12. i\ single pipe of unifonn diameter. having same dischnrge and same loss of head as compound pipe consi,ting of .>cwml pipes of different lengthS und diameters. is known as equivalent pipe. The diamet~r of equivalent pi pe is called equivalent site of the pipe.

I I

Ii

1548

Fluid Mechanics

13. The equivalent sile of the pipe is

obtal~

from

L _I, +L,+.!:l. -.J" d - , d'J ,

7 d!

where L = equn'alent length of pipe = L, + ~ + ~ d l• d z• d J = art. diameters of pipes connected in series d = equivalent size of the pipes. 14. ~n the pipes are connected in parallel. the Joss of Mad in each pipe is same. The rate of now in mam pipe is equal 10 sum of the rate of now in each pipe. connected in parallel. 15. For solving problems for bnmchcd pipes, the three basic, equations I. ~.. continuity. Bemoulli"s and Darty's equations are used _

pgxQX(H - h, )

.6. Po~er transmitted in kW through pipe is given by P = ::::..-~=-= 1000 where Q = discharge through pipe :: area )( "elocity '"

~ JJ x Y 4

H • total head at Inlet of pipe hi = head lost due to friction

4/LV'

• - - - . when: L = Length of pipe

dx2g

pgXQ X(H - h, ) In SI. unJlS. po",er transmitted is ai,en by. Power'"

11. Efficiency of

po~er

transmISsion through pipes. 11 '"

kW.

1000 H-h, H

.

18. Condition for maximum transmission of power through Pipe. hl :ll'

H

3

and maximum efficiency

",66.61Q

19. The , 'clocIIY of water at the outlet of the noule is v =

2gH

4JL . ' [ 1+ o-)( • 2

where H "" head at the inlet of the pipe. L = length of the pipe. a = area of the nOlzle outlet. D • dmmetcr of the pipe. A = arra of the pipe.

po~er trunsmitted through nozzle. P= pg)( Q [H _ 4jLy 1(0) D)(2g 1

20. The

j

and the efficiency of po~er transmission through nonle. J'j '" ,..---'-....,.,

1+

11 . Condition for m.u.imum

4jL

)(

a~

o •1

po~er transmission through noule. hI = ~ .

j

~ I

IL 1554

Fluid Mechanics

22 . Diameter of !lou;]e for maximuill power transmission through nOlzle is. II ..

(;~ )".

where" z diameter of (he nozzle at oUlict, D .. diameter of Ihe pipe. L .. IcnJ!lh of (lie pipe., ~ co-efficicn! of friclion for pip<:. 23 . When a liquid js flowing through
high intensity is having Ihe effect of hammering action on (he walls of Ihe pipe. This phenomenon is known as waler hammer. 24. The inlcnsity of pressure rise due to watcr hammer is given by

P •

pLV T

'" when valve is closed gradually .

'" v.JKP

... when '-alve is closed suddenly and pipe is assumed rigi"

cv xJ<1J K

... when ,'al,·c is closcd suddenly and pipe is ciaslic.

£1

L = L<:ngth of pipe, T " Time required to close the '·al"c. D • Diameter of the pipe. I '" Thickness of the pipe wall. 25. If the time required to close the VJh'c: where

T>

3.!:

... the val"e closure is said 10 be

c

T< 2L C where

V,. Velocity of flow. K .. Bulk modu lus of water. E. Mod ulus of ciasticity for pipe material.

gr~dual.

... the valve closure is S:lid to be sudden

L .. length of pipe.

e m "clocilY of pressure wa,'e produced due to wala hammer

~ ~.

EXERCISE

(A) THEORETICAL PROBLEMS J . How will you detem,inc the loss of head due to friction in pipes by using (i) Darey Formula and (ii) Chczy's formula? 2. (a) What do you understand by the tenlls : Major eTlcrgy loss and minor energy losses in pipes? (b) What do you understand by tolal energy line. hydraulic gmdicnt line, pipes in scric~. pipes in p.~r.. ncl

and equivalent pipe

J.

?

(0 : (I) Sudden enlargement and (ii) Sudden contraction of a pipe. (b) Obtain expression for head loss in a sudden expansion in the pipe. List all lhe a!isumptions made in the derh·ation. 4. Define and explain tile tenTls: (I) Hydraulic gradient line and (iI) Total energy line.

I I

(a) Derive an expression for the loss of head due

Ii

~ I

IL Flow Through Pipes 5. Show that Ihe loss of head due to sudden 6 . What is a syphon? On

e~pansion

555 1

in pipe line is a fun ction of veloci ly head.

wh~l

principle it works? 7. What is a compound pipe? What will be loss ()f head when pipes are connected in series ? 8. E ~plajn the lcnns: (i) Pipes in pnrallcl Ui) Equivalent pipe and (iii) Equivalent size of the pipe .

'.I. Find an expression for Ihe pow"r transmission through pipes. Wh at is Ihc cond it ion for maximum lruns-

mission of power and corresponding efficiency of transmission? Ill. Prove Ihm the head loss duc 10 friction is equal to one -third of the lo\al head a\ inlet for max imum power transmission through pipes Or nouJes. II . Prove that Ihc velocity through nozzle is gj"en by \'

~

2gH

,i ,.--"-, D ,. 4fL

where a" Area of nozzle a1 outlet. A '" Area of the pipe. 12. Show that the diameter ofthc noule for maximum tran,mission of power is gi"en by d '"

(~:)"~

where /) Diameter of pipe. L Leng th of pipe. 13. Find an expression for the rat;o of the oUllet area of the nO'-1.1e 10 the area of the pipe for maximum transmission of power. 14. Explain thc phenomenon of Watcr Hammer. Obtain an expression for the risc of press ure when the flowing water in a pipe is bmught to rest by closing the valve gradually. IS. Show that the pressure risc due to suddcn closure of a "al\,c at the cnd of a pipe. through which water is E

flowing is given by I'

E

Z

V

' JH ~

+!2

K

E<

where V. Velocity of flow, 0 .. Diameter of pipe. E z Young's Modulus. K .. Bulk Modulus and I .. Thickness of pipe. 16. Three pipes of different diameters and di ffcrent lengths are connected in series to make a compound pipe. The ends of this compound pipe arc eonn«led with IWO tan ks whose difference of waler le"cl is H. If co--efficicnt of friction for these pipes is S:lme. then deri>'c the formula for the total head loss. ncgl«ting first the minor losses and then including them. 17, For the two c"ses of flow in a sudden contruction in a pipeline and flow in a sudden cxp.1nsion in a pipe linc. draw the flow p.1ltCm. piclomctric gr~dc line and total energy line. III, What do yo u mean by "l"
(8) NUMERICAL PROBLEMS 1. fi nd the head loss due to friction in a pipe of diameter 250 mm and length 60 tn. Ihrough which watcr is llowing at a "~Iocity of 3.0 mls using (i) Darcy formula and (ii) Chny"s I'onnuia for which C. 55. Take v for wafer" .01 sfokc. IAn s. (I) 1.182. (ii) 2.8561 2. Find fhe diameter of a pipe of lenllth 2500 m when the rute of flow ofwafcr through the pipe is 0.25 mJls and head loss duc to fricfion is 5 m. Take C z 50 in Chezy's fomlU la. I,\ os. 605 mml

I I

Ii

~ I

IL 1556

Fluid Mechanics

J . An oil of Kinematic ViSCQsily 0.5 ~loke is Oowing through a pipe of diameter 300 nlln at the rate of 320 1;lres per sc<;. Find thc head lost duc to friction for a length of 60 111 of thc pipe. [Ans. 5.14 m] 4 . Calculate the rale of flow of waler through a pip" of diameter 300 nnn, when the diff""'n,,e of pressure head belween the two ends of a pipe 400 m apart is 5 m of wat .. ,. Take the value off ~.009 in Ihe fonnula

h • 4 flyl f d x 2g

IAn •. 0101 m'/sl

5. The discharge through a pipe is 200 lilres/s. Find Ihc loss of head when Ihe pipe is suddenly enlarged from

150 mm to 300 mm diameter. [A ilS. 3.672 Inl 6. The ralc of flow of waler through a horizontal pipe is 0 .3 mJ/s. The diameter of thc pipe is suddenly enlarged from 250 mm to 500 mm. The pressure intensity in the smaller pipe is 13.734 N/Clnl. Determine: (.) loss of head due to sudden enlargement. (ii) pressure intensity in the large pipe and (iii) power lost due 10 enlargement. [Ans. (i) 1.07 m, (ii) 14.43 Nfcml. (iii) 3.15 kWI 7. A horizonl~1 pipe of cs. IAns.15.41ilre>lsl Ill. A horizontal pipe -line 50 m long is connected to a water tank at one end and diseharges freely into the atmosphere at the othcr cnd f or the first 30 m of its length from the tank. the pipe is 200 mm diameter and its diameter is sulsl II . !)etenninc the difference in the elevations between the waler .urface, in the two tanks which are connecled by a h"riwnlal pipe of diameter 400 Illm and lenglh 500 Ill. The rale of flow of waler through the pipe is 200 litres/s. Cons ider alllo'i"'s and takc Ihc value of!= .009. [Ans_ 11.79 ml 12. For the problems 9. 10 and II draw the hydraulic gradient lines (H.GL) and tO(a1 energy lines (TEL ) 13. A syphon of crvoir to Ihc .um",it i. SO ",. !)clenlline the discharge Ihrough Ihe syphon and also pre"ure at the ~Ulnl11i1. Neglect minor losses. The co-efficient of friction_! '" .005. [Ans. 41 .52 lim:>I•. - 7.281 m of walerl 14. A syphon of diamNer 200 mm connects two re<;ervoirs having a difference in elevalion a., 20 m. The lotal length of Ihe syphon i, ROO m and Ihc sUlmnil is 5 m above Ihc waler le,'cl in Ihe upper reservoir. If the scparation takes place at 2.8 m of water absolulC find thc maximum length of syphon from upper reservoir to Ihc summit. Ta~e f",.OO4 and almospheric pressure = 10.3 m of water. [,\ns. 87.52 IllI IS. Three pipes of lengths ROO Tn. 600 m and 300 III an
~ I

I~

~ I

IL Flow Through Pipes

557 1

16 . Two pipes of lengths 2500 In each and diameters 80 em and 60 em respectively. are connected in ""ralle!. The co -efficient of frktion for each pipe is 0.006. The total now is C
18 .

19 .

20.

21 .

22 . 23 .

24 .

25 .

26 . 27.

211 .

29.

~ I

In.

Dc\cnninc thc discharge thmugh thc pipe. If an additional pipe of dian\e!", 300 mm and

length 600 In is auachcd 10 Ihc last 600 m Icn~th of the cxisling pipe. find the increase in the discharge. Take / _ .02 and neglcct minor losses. IAns. 0.0742 m'ls. 0.0258 m'/sl Two sharp ended pipes of diameters 60 mm and 100 nun respectively. each of length 150 In arc CQnncrted in parallel betw",," two reservoi" which have a diff~r~nc~ of lev~1 of 15 m. If w-eff,ci~nt of friction for each pipe is 0.08. culculute the rate of flow for each pipe and also the diameter of a singlc pipe 150 m long Which would give the same discharge if it were subst it uted for Ihe original IWO pipe •. (Ans.0.0017 .. 00615, 110 mml Three reservoirs A. Band Care rormccled by a pipe syslem having length 700 m. 1200 m and 500 m and diameters 400 mm. 300 mm and 200 mm respectively . The water Ic"cis in reservoir A and B from a datum line arc 50 111 and 45 m respeeli,·ely. The level of waler in reservoir C is tlelow thc level of waler in rescH'oir 8. Find the dischar~e inlo or from the rescH'oirs lJ and C if the rale of flow from reservoir A is 150 lilres per '!Cc. Find Ihe height of waler lewl in the r6crvoir C. Take! ~ .005 for all pipes. [Ans . .005 mJls •. 095 m)ls. 24.16 ml A pipe of diameler 300 mm and kn~lh 3000 m is used for Ihe lTansmission of power by water. The lolal head al Ihe inlet of Ihe pipe is 400 m. Find the maximum power available al Ihe oUllel of Ihc pipe , Take !~.005. (An o.667.07l.:WI A pipe li ne of length 2100 m is used for Iransmining 103 kW. The pressure at the inlet of the pipe is 392.4 Nlcm l. If Ihe efficiency of transmi.sion is 80%. find the diameter of the pipe. Take f .. ,005. (Ans. l36mml A noule is filled at the end of a pipe of len ~lh 400 m and of diameter 150 mm . For the maximum lransmi,sion of power Ihrough Ihe noale. find Ihe diameter of Ihc noale . Take f ~ ,008. [Ans. 41.5 mml The head of waler at the inlet of a pipe of length 1500 m and of dialneter 400 mm is 50 m. A nOllle of diameter 80 nl111 at the outlet. is fitted to the pipe. Find the velocil y of water at the outlet of the n01.zle iff" .01 for the pipe. [Ans. 28.12 mlsl The mte of flow of water throu~h a pipe of length 1500 m and diameler I!OO mIll is 2 m)ls. At the end of the pipe a nozzle of outside diameler 200 nl1n is filled , Find the power tmnsmilled through Ihe nonle if the head of water at the inlet of the pipe is 180 m andf: .01 for pipe. (Ans. 2344.7I: WI The waler is flowing with ,) "eloci ty of 2 mls in a pipe of length 2000 m and of diameter 600 mm. At thc end of lhe pipe. a valve is provided, Find thc rise in pressure if thc val"e is closed in 20 s«onds, Take the value of C ~ 1420 mis, IA ns. 20 N/cml l If the \'al,·c in the problem 25 is closed in 1.5 sec. find the rise in pressure. Take bulk modulus of water = 19.62 x 10' Nk,n' and consider pipe as rigid one. (Ans.I86,75 Nlcm1 1 If in the prohlem 25. the thickne ... of the pipe i.' 10 mm and the valve is closed sudden ly. Find the ri'iC in pressure if th~ pipe is wnsidcred to be clastic. Take value of E ~ 19,62 X 106 Nkm l for pipe material and K ~ 19.62 X 10' Nlcm 2 for water. Calculate the cireumferentia l stress and longitudinal stress (An•. " = 221.47 Nkm1.f," 6644.1 Nicm './I" 3322 Nlem1 1 deve loped in the pipe wnll. The difference in water surface levels in two tanks. which are connected by two pipes in -"cries of lengths 600 m and 400 m and of diameters 30 em and 20 em respecti,'ely. is 15 III . Detennine Ihe mte of flow of water if Ih~ co-efficient of friction is 0.005 for both the pipes. Neglect mi nor losses. Water is flowing ,·erti.al1y downwards through a 10 Cm diameter pipe al the rate of 50 I.p.s. At a particular location the pipe suddenly enlarges 10 20 em diameler. A point P is located 50 cm above the section of enlar~emcnl and another point Q is located 50 em below it in Ihe enlarg~"
I~

•

Flow Through Pipes

5531

neglecting friction loss between P and Q but (:oRsidering the lou due to sudden enlargement. What will be the pressure al Q if the same discharge flows upwards assuming that the pressure P remains the same ? Consider the lou due 10 contraction with C~ _ 0.60 but neglect friction loss bet.....een P and Q. lAos. 21 .36 N/cm . 23.4 N/cm J (A.M.I .E.. Summer, 1985) 30. Two tanks are connected with the help of two pipes in series. The lengths of the: pipes are 1000 m and 800 m whereu the diameters are 400 mm and 200 rom respectively. The c~flicicnt of friction for both the pipe5 i. 0.008. 1be difference of water level in the two tanks il 15 m. Find the rJte of flow of water througb the pipes. considering aU losses. Also draw the total energy line and hydraulic gradient lines for the system. (Delhi University. May /998) (AM. 0.0464 ml /s1 [Hint. L, = 1000 m : Lz = 800 m :d1:: 400 mm = 0.4 m ;d1 = 200 mm =0.2 m,f= 0.008 : H - 15 m. Now H =h/ +hfl+hc+hJ2+ho.

,

.,

H _ O.SV,2 + 4/xL,X \.jl + O.5Vl + 4/ 'Xl:;zxVl 2g d)(2g 28 d z x2g or

1

1

IS = 0.5l'J + 4xO.OO8XI(XlO.xV. + 2x9.81

OAx2x9.81

o.svl 2x9.81

+::L 2g

+ 4 xO.OO8x800xvl +Y1.. O.2x2x9.SI

28

AI",

31 . A pipe of diameter 25 ern and length 2000 rn conneclS two reservoirs, having difference of water level 2!i m. Det~nnine the di5Charg~ throuSh the pipe. If an additional pipe of diamct~r 2S em and length 1000 m is auached lO the last 1000 m length of the existing pipe, find the increase in discharge. Tale! = O.OI!i. Neglect minor losscs. (IHlhi Univu.fity. lkcem~r 2(02) lAD&. (1149.62 I/s, (m 13.14I1sl