Solution Manual for Elementary Hydraulics 1st Edition by Cruise

CHAPTER 1 INTRODUCTION

3

1.1. A

certain control volume of porous media received a discharge of 0.1 m /s during a 3 period of 10.0 10 .0 minutes. The volume of o f outflow was measured to be 58.0 m through the same period of time. Determine the change in storage in the control volume using equation 1.1. Solution: Inflow

3

to the control volume = 0.1 × 10 × 60 = 60.0 m . Outflow from the 3 3 control volume = 58.0 m . Using equation 1.1, change in storage = 60.0-58.0 = 2.0 m .

1.2. In

problem 1.1, if the change in storage in the control volume is zero, what is the percentage of error encountered in measuring the outflow? 3

Solution:

In this case the outflow must be equal to the inflow = 60.0 m /s. % of error in measuring the outflow = {(60.0-58.0)/60.0} × 100 = 3.33%.

1.3. A

certain reach of an open channel 1.0 mile in length is excavated through permeable 3 3 soil. The rate of inflow to this reach is 100 ft /s and the rate of outflow is 95 ft /s. How much of the water is being lost through this reach every year?

Solution:

The difference between inflow and outflow is lost to the soil. Therefore, the volume of water lost through this reach every year = (100-95) × 60 × 60 × 24 × 365 = 3 157,680,000 ft /year. 1.4. A

cylindrical tank is supplied by volume flow rate of 80 liter/min. Two outlet pipes are connected to the base of the tank extracting water at the flow rate of 50 liter/min. Calculate the rate of change of water volume inside the tank per min. If the diameter of the tank is 60 cm, determine the change in the water level after one hour.

Solution:

Vin-Vout = Vstorage Rate of change of volume = 80 – (50 + 35) = -5 l/min. i.e. water volume will reduce at the rate of 5 l/min. 2 π d h 2 volume of tank = π r h = 4 Height of water in the tank is reduced in one hour by 4 = × 5000 cm 3 × 60 = 106.1 cm = 1.06 m. 2 π (60)

1.5. Water

is flowing from tap 8 mm in diameter and is collected in a 10 liter bottle. Determine the exit velocity of the water from the tap of diameter 8 mm, if the bottle is filled in: a) 30 sec. b) 5.0 min.

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Elementary Hydraulics Hydraulics

Solution:

2

2

-6

1.6. The

dimensionless parameter Reynolds number is defined as R n = V.d/ν V.d/ν, where V is the water velocity, d is the diameter of the pipe, and n is the kinematic viscosity of water -6 2 (1.005 × 10 m s). Determine R n for the two cases (a and b) of Problem 1.5. Solution:

a) R n =

b) R n =

Vd 6.63× 0.008 = = 52,776 V 1.005 ×10 −6 Vd 0.663 × 0.008 = = 5278 V 1.005 × 10 −6

The friction losses through a 10.0 km horizontal pipe of a certain diameter is given by the relation hf = 0.005 L Where hf is the head loss due to friction, and L is the length length of the pipe. The total energy head at the end of the pipe is 21.0 m. m. Estimate the total energy head at the pipe inlet. 1.7.

Solution:

1.8.

The total loss through the pipe is 3 hf = 0.005 × 10 × 10 = 50.0 m Energy head at pipe inlet = energy at pipe outlet + loss = 21.0 + 50.0 = 71.0 m

Prove the following ratios are dimensionless: Vy

ρ Vy

V

p

v

µ

gy

ρ V

2

gy

p

σ

V 2

γ y

γ y

2

Q y 2 gy

where V is the velocity, y is the flow depth, v is the kinematic velocity, ρ is the fluid density, µ is the dynamic viscosity, p is the pressure, γ is the specific weight, σ is surface tension, and Q is the discharge. Solution:

2

Cross section area of the tap = π/4d = π/4 × (8/1000 m) = 50.265 × 10 m Flow rate Q = volume/time. -3 3 a) Q = 10 litre/30 sec = 0.333 × 10 m /s -3 3 -6 2 Velocity = Q/area = (0.333 × 10 m /s) / (50.265 × 10 m ) = 6.63 m/s. 0.01 / (5 × 60) b) Velocity = Q/area = = 0.663 m/s 50.265 × (10) −6

Vy v

=

ρ Vy

[ LT −1 ][ L] [ L2T −1 ] =

µ

V gy

=

=1

[ ML−3 ][ LT −1 ][ L] −1

−1

[ ML T ] [ LT −1 ] −2

1/ 2

([ LT ][ L ])

= [ MLT ]o

= [ MLT ]o

2 Full file at https://testbanku.eu/Solution-Manua https://testbanku.eu/Solution-Manual-for-Elementaryl-for-Elementary-Hydraulics-1st-Edi Hydraulics-1st-Edition-by-Cruise tion-by-Cruise

Chapter 1. Introduction

p

=

2

ρ V

gy

=

2

V p

=

y γ σ γ y

2

y

−3

−1

[ ML ]([ LT ])

[ LT −2 ][ L ] −1

([ LT ])

2

−3

= [ M 1−1 L1− 2− ( −3) − 2T − 2− ( −2 ) ] = [ MLT ]o

2

= [ L1+1− 2T − 2 −( −2 ) ] = [ LT ]o

[ MLT −2 ][ L−2 ] −2

[ ML ][ LT

][ L ]

= [ M 1−1 L1− 2+3−1−1T − 2+ 2 ] = [ MLT ] o

[ MLT −2 ][ L−1 ]

=

−3

−2

[ ML ][ LT ]([ L ])

Q 2

[ MLT −2 ][ L−2 ]

= gy

2

[ L3T −1 ] 2

−2

[ L ]([ LT

][ L ])

= [ M 1−1 L1−1+3−1−2 T 2+ 2 ] = [ MLT ] o

3 − 2 − (1+1) / 2

1/ 2

= [ L

−1+ ( 2 / 2 )

T

] = [ LT ] o

1.9. A

commonly used equation for calculating the velocity of uniform flow in open channels, V, is 1 V = R 2 / 3 S 01 / 2 n Where V is the velocity of uniform flow, n is the Manning coefficient, R is the hydraulic radius, and So is the bed slope. Find the dimensions of the Manning coefficient to ensure that the equation is homogeneous. Solution:

V=

1

R 2 / 3 S 01 / 2

n -1 The dimensions of V and R are LT and L, respectively, and the bed slope, So, is a dimensionless number. Therefore, substituting in the above equation: -1

LT =

1 n

L2 / 3

,

1 n

= L1 / 3T −1

1.10. From

example 1.1 and problem 1.9, find the relationship between the Chezy and Manning coefficients. Check the homogeneity of this relationship. Solution:

From example 1.1 V = C R 1 / 2 S 01 / 2

and from problem 1.9, 1 2 / 3 1/ 2 V = R S 0 n Therefore,



C R 1 / 2 S 01 / 2 = C=

1 n

1 n

R 2 / 3 S 01 / 2 ,

R 1 / 6 1/2 -1

-1/3

The dimensions of C, n, and R are L T , L T, and L, respectively. respectively. For the the above 1/2 -1 equation, the dimensions of the left hand side are L T , and the dimensions of the right 1/3 -1 1/6 1/2 -1 hand side are L T T = L T . Hence, the homogeneity of the equation is satisfied. satisfied. 1.11. The

water power developed by a pump is given by the equation P = γQH, where P is the power, γ is the specific weight, Q is the discharge, and H is the water head raised by the pump. Check the condition of homogeneity of the equation. 2 -3

Solution:

-2 -2

3 -1

The dimensions of P, γ, Q, and H are ML T , ML T , L T and L, 2 -3 respectively. For the given equation the dimensions of the left hand side are ≡ ML T . -2 -2 3 -1 2 -3 The dimensions of the right hand side are (ML T ) (L T ) (L) = ML T . Therefore, the homogeneity of the dimensions of the given equation is satisfied. 1.12.

The Darcy equation for groundwater flow is written as ∂h

q = -K

∂ x -1

where q is the specific discharge (LT ), K is the hydraulic conductivity, and ∂h/∂ h/∂x is the gradient of the piezometric head. Find the dimensions of K. Solution:

The gradient of the piezometric piezometric head ∂h/∂ h/∂x is a dimensionless dimensionless number. The -1 dimensions of K are the same as those of the specific discharge, q ≡ LT .

1.13.

Determine the dimensions of the coefficients A and B in the following equation: 2 ∂ x ∂ x + A + Bx = 0 2 ∂t ∂t

where x is the length and t is the time. -2

Solution:

The dimensions of the first term of the given equation LT . Therefore, other -1 terms must have the same same dimensions. Thus the dimensions of A and B are thus T and -2 T , respectively. 1.14. Determine

whether the following equation is dimensionally homogenous or not. 2 -2 F = 16.83 µVd + 29.78 ρQ d Where F is force, ρ is the density, µ is dynamic viscosity, Q is the discharge, V is velocity and d is diameter. Solution:

Assuming that the constants have no dimension, following dimensional analysis is done: -2 -1 -1 -1 -3 3 -2 2 -2 [M.L.T ] = [ML T ].[LT ][L] + [ML ][L T ] .[L ]



-2

-3

6 -2

-2

= [MLT ] + [ML ][L T ][L ] -2

-2

= [MLT ] + [MLT ] -2

= [MLT ]. Therefore, the equation is dimensionally homogenous. 1.15. The

head loss for flow inside a pipe is often expressed as LV 2 h f = f d 2 g

where f is the friction coefficient, L L and d are length and diameter of the pipe, respectively, V is the average water velocity and g is the acceleration of gravity. Determine the units of friction coefficient f coefficient f . If the system system of units is changed from SI to BG, how this will affect the value of o f ‘ f f ’? ’? Solution: using

dimensional analysis. [L] [L2. T -2 ] [L] = [f] . . = [f] [L] [L] [LT -2 ] f is dimensionless, therefore, the value of f will not change by changing the system of units. 1.16. A

person weighs 150 lb at the earth’s surface. surface. Determine the person’s mass in slugs and kilograms. 2

Solution: Weight

(lb) = mass (slugs) x gravity (ft/s ) 150 = m × 32.2 The person’s mass = 150/32.2 = 4.658 slug = 4.658 × 14.594 = 67.98 kg ≅ 68.0 kg 3

1.17. The

discharge of water through an open channel is 10.0 m /s. What is the discharge discharge in liters/min, and gallons/min? gallons/min? What is the volume of water (in gallons) delivered by this channel per year? Solution:

3

Q = 10.0 m /s 3

5

= 10.0 × 10 × 60 = 6.0 × 10 lit/min 5

= 1.3216 × 10 gallons/min 5

Volume of water per year = 1.3216 × 10 × 60 × 24 × 365 10

= 6.946 × 10 gallons 1.18. A

certain object weighs 150.0 Newton at the earth’s surface. Determine the mass of the object (in kilograms) and its weight (in Newton) when located on a planet with an 2 acceleration of gravity equal to 4.0 m/s .



2

Solution:

150 (N) = m (kg) × 9.81 (m/s ). Therefore, m = 15.29 kg. The mass of the object is not affected by any variation variation in the acceleration due to gravity. The object will have the same mass irrespective of where it is placed. Weight of the object on the other planet = 15.29 × 4.0 = 61.16 N. 1.19. A

person having a total total mass of 80.0 kg rides an elevator. Determine the force (in (in 2 Newton) exerted on the floor of the elevator when it is accelerating upward at 5.0 m/s . 2

Solution:

W (N) = m (kg) × g(m/s ) W = 80.0 × (9.81 + 5.0) = 1184.8 N 2

1.20. In

problem 1.19, if the elevator is accelerating downwards at 5.0 m/s , determine the force (in Newton) exerted on the floor. 2

Solution: If

the elevator is accelerating downwards at 5.0 m/s , g will be equal to (9.812 5.0) = 4.81 m/s . Therefore, W = 80.0 × 4.81 = 384.8 N. 1.21. If

the force exerted on the floor of the elevator in problem 1.19 is zero, what will be the direction and acceleration of the elevator?

Solution: The

2

elevator must be accelerating downward at 9.81 m/s .

1.22. A

tank of oil has a mass of 30 slug. Determine its weight in pound and in Newton at the earth’s surface. 2

Solution:

W (lb) = mass (slug) x gravity (ft/s ) W = 30 × 32.2 = 966 lb = 966 × 4.4482 = 4296.96 N

1.23.

If the tank in problem 1.22 is placed on the surface of the moon where the gravitational attraction is one-sixth of that at the earth’s surface, what would be its weight in Newton? Solution:

1.24.

the weight on the moon will will be one-sixth of that at the earth’s surface W = 4296.96/6 = 716.16 N 2

How many hectares and acres in an area of 1 km ?

Solution: 1 1.25.

2

km = 100 hectare = 247.1 acre

Convert the following: 3 a) Discharge of 60 ft /min to lit/s. b) Force of 20 pounds to dynes. 2 2 c) Pressure intensity of 30 lb/in to N/cm . 3 3 d) Specific weight of 62.4 lb/ft to N/m . 2 2 e) Dynamic viscosity of 255 dyne.s/cm to lb.s/ft . f) Dynamic viscosity of 100 gm/cm.s to slug/ft.s.



Solution: 3

a) b) c) d) e)

3

Discharge of 60 ft /min = 60 (0.3048) × 60/1000 = 0.1019 lit/s. 6 Force of 20 pounds = 20 × 444,822 dynes = 8.896 × 10 dynes 2 2 2 Pressure intensity of 30 lb/in = 30 × 4.448/(2.54) = 20.68 N/cm . 3 3 Specific weight of 62.4 lb/ft = 62.4 × 4.448/(0.02832) = 9800 N/m . 2 2 Dynamic viscosity of 255 dyne.s/cm = 255 × (30.48) /(444,822) 2 = 0.5326 lb.s/ft . f) Dynamic viscosity of 100 gm/cm.s = 100 × 30.48/14,594 = 0.209 slug/ft.s.

1.26. Making

use of the conversion factors given in Appendix A, express the following 3 2 quantities in BG units: (a) 16.7 km, (b) 9.30 N/m , (c) 1.95 kg/m , (d) 0.045 N.m/s, (e) 6.35 mm/hr. Solution:

a) b) c) d) e)

16.7 km = 16.7/1.6093 miles = 10.377 miles. 3 2 3 9.30 N/m = 9.3 × 0.2248/(3.28) = 0.059 lb/ft . 2 2 2 1.95 kg/m = 1.95 × 0.06852/(3.28) = 0.0124 slug/ft . 0.045 N.m/s = 0.045 × 0.2248 × 3.28 = 0.0332 ft-lb/s. 6.35 mm/hr = 6.35/25.4 = 0.25 in/hr.

1.27.

Express the following quantities in SI units using the conversion factors in 2 Appendix A: (a) 10.25 in/min, (b) 4.20 slugs, (c) 4.0 lb, (d) 82.2 ft/s , and (e) 0.0315 2 lb.s/ft . Solution:

a) b) c) d) e)

10.25 in/min = 10.25 × 2.54 × 60/100 = 15.621 m/s. 4.20 slugs = 4.2 × 14.6 kg = 61.32 kg. 40.0 lb = 4 × 0.4536 = 1.814 kg. 2 2 82.2 ft/s = 82.2 × 0.3048 = 25/05 m/s . 2 2 2 0.0315 lb.s/ft = 0.0315 × 0.4536/(0.3048) = 0.1538 kg.s/m .

1.28. Verify

the conversion factors for (a) acceleration, (b) density, (c) specific weight and (d) dynamic and kinematic viscosity. Use the basic conversion relations, 1 ft = 0.3048 m, 1 lb = 4.4482 N, and 1 slug = 14.594 kg. Solution: 2

2

2

a) acceleration: 1 ft/s = 1 × (0.3048 m)/s = 0.3048 m/s . 3 3 3 b) Density: 1 slug/ft = (14.594 kg)/(0.3048 m) = 515.38 kg/m . 3 3 3 c) Specific weight: 1 lb/ft = (4.4482 N)/(0.3048 m) = 515.38 N/m . 2 2 2 d) Dynamic viscosity: 1 lb.s/ft = (4.4482 N.s)/(0.3048 m) = 47.88 N.s/m . 2 2 2 Kinematic viscosity: 1 ft /s = (0.3048 m) /s = 0.0929 m /s. 1.29. Water

flows from a large drainage pipe at a rate of 1350 gal/min. What is the flow rate in m /s and lit/min. 3



Solution: -3

3

3

1350 gal/min = (1350 × 3.7854 × 10 m )/(60 s) = 0.0852 m /s = (0.0852 × 1000 l)/(1/60 l)/(1/60 min) = 5,110 l/min l/min 4

1.30. The

viscosity of a certain fluid is 3.5 × 10 poise. Determine its viscosity in both SI and BG units. Solution: -4

-5

-5

3.5 × 10 poise = 3.5 × 10 kg/(m.s) = (3.5 × 10 )/47.88 = -7 7.31 × 10 slug/(ft.s).


Solution Manual for Elementary Hydraulics 1st Edition by Cruise

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