Fluid Mechanic Answer

CIVE1400: Fluid Mechanics

Examples: Answers


Examples: Answers

b)

Pressure and Manometers

p   gh

1.1

 103   9.81  0.4  3924 N / m2

What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m below the surface? water = 1000 kg/m3, and p atmosphere = 101kN/m2. 2 2 [117.72 kN/m , 218.72 kN/m ]

c) 

a) p gauge

  gh 81  12  1000  9.81  117720 N / m2 , ( Pa ) 117.7 kN / m 2 , ( kPa )  117

 7.9  10 3   0.4  3160 N / m2 d)

p   gh

 520  9.81  0.4  2040 N / m2

b) pabsolute



p gauge

1.4 A manometer connected to a pipe indicates a negative gauge pressure of 50mm of mercury. What is the absolute pressure in the pipe in Newtons per square metre is the atmospheric pressure is 1 bar? [93.3 kN/m2]

 patmospheric

720  101) N / m  (117 72 . kN / m 2 , ( kPa )  2187

2

, ( Pa )

 1bar  1  10 5 N / m 2 pabsolute  p gauge  patmospheric   gh  patmospheric

patmosphere

1.2 2 At what depth below the surface of oil, relative density 0.8, will produce a pressure of 120 kN/m ? What depth of water is this equivalent to? [15.3m, 12.2m]

 13.6  10 3  9.81  0.05  10 5 . kN / m2 , ( kPa )  9333

a)  water

p  g



120  10 3 800  9.81 81

patmosphere

. m of oil  1529

5

10

 1bar  1  105 N / m 2   gh

b) h

3   1000 kg / m

h

120  10

105 1000  9.81 81

. m of  1019

water

5

3

1000  9.81 81

N / m 2 , ( Pa )

1.5 What height would a water barometer need to be to measure atmospheric pressure? [>10m]

 0.8  1000 kg / m 3 p   gh h

  g

p   gh

h

. m of water  1223

10

(13.6  10 3 )  9.81

 0.75m of

mercury

1.3 2 What would the pressure in kN/m be if the equivalent head is measured as 400mm of (a) mercury =13.6 (b) water ( c) oil specific weight 7.9 kN/m3 (d) a liquid of density 520 kg/m3? 2 2 2 2 [53.4 kN/m , 3.92 kN/m , 3.16 kN/m , 2.04 kN/m ] a)  water

 13.6  1000 kg / m3 p   gh  13.6  10 3   9.81 0.4  53366 N / m2

Examples: Answers


1

Examples: Answers


2


Examples: Answers

1.6 An inclined manometer is required to measure an air pressure of 3mm of water to an accuracy of +/- 3%. The inclined arm is 8mm in diameter and the larger arm has a diameter of 24mm. The manometric fluid has density 740 kg/m 3 and the scale may be read to +/- 0.5mm. What is the angle required to ensure the desired accuracy may be achieved? [12 39’]

p

p 1

diamet diameter er d


Examples: Answers

1.7 Determine the resultant force due to the water acting on the 1m by 2m rectangular area AB shown in the diagram below. [43 560 N, 2.37m from O]

O

P

2

1.22m

1.0m 45° C

A diameter D

x

r e a d R e e l e a S c

2.0 m

2.0 m

z 2

D

B

Datum line z1

The magnitude of the resultant force on a submerged plane is: R = pressure at centroid

= z1 A1

 z 1 z 1



area of surface

 1000  9.81  1.22  1  1  2   43556 N / m2

p1  p2   man gh   man g z1  z 2  Volume moved from left to right



R   gz A

θ



 D

This acts at right angle to the surface through the centre of pressure.

z 2 sin

2

4

A2

 xA2 2



z 2 d sin 4 2

z 2

d

sin D 2

Sc

 x

 d

Ax



2nd moment of area about a line through O 1st moment of area about a line through O nd

4

d

D 2

Sc 

 d 2  p1  p2  man gx  sin  2  D   2  d  ater gh  man gx  sin  2  w D   2   0008 .  2 w ater gh   0.74  w ater  g x  sin  0024 .   h  0.74 x (sin  0.1111)

I GG Ax

 x

b

G

G d For a rectangle I GG

As the wall is vertical, Sc  D and x

The head being measured is 3% of 3mm = 0.003x0.03 = 0.00009m This 3% represents the smallest measurement possible on the manometer, 0.5mm = 0.0005m, giving

Sc

0.00009  0.74  0.0005 (sin   0 .1111)



1 23 121  2 1.22  1



bd 3 12

z,

 1.22  1

 2.37 m from O

sin   0132 .   7.6

I OO

By the parallel axis theorem (which will be given in an exam), I oo  I GG  Ax 2 , where I where I G GG G is the 2 moment of area about a line through the centroid and can be found in tables.

2

 x



2



[This is not the same as the answer given on the question sheet]

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3

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4


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1.8 Determine the resultant force due to the water acting on the 1.25m by 2.0m triangular area CD shown in the figure above. The apex of the triangle is at C. [43.5103 N, 2.821m from P]

Forces on submerged surfaces

G d

d/3

For a triangle I GG

Depth to centre of gravity is z

 10 . 2

2 3



Examples: Answers

2.1 Obtain an expression for the depth of the centre of pressure of a plane surface wholly submerged in a fluid and inclined at an angle to the free surface of the liquid. A horizontal circular pipe, 1.25m diameter, is closed by a butterfly disk which rotates about a horizontal axis through its centre. Determine the torque which would have to be applied to the disk spindle to keep the disk closed in a vertical position when there is a 3m head of fresh water above the axis. [1176 Nm]

b

G


The question asks what is the moment you have to apply to the spindle to keep the disc vertical i.e. to keep the valve shut?

bd 3

So you need to know the resultant force exerted on the disc by the water and the distance x of this force from the spindle.

36

cos 45  1.943m .

We know that the water in the pipe is under a pressure of 3m head of water (to the spindle)

R   gz A

 2.0  1.25 81  1.943    1000  9.81   2.0  2  23826 N / m Distance from P is x

2.375

748m  z / cos 45  2.74

F

I oo



I oo

 I GG  Ax

Sc 

Ax

x

Diagram of the forces on the disc valve, based on an imaginary water surface.

Ax I GG

3 h’

Distance from P to centre of pressure is Sc

h

h

2

 x 

1.25  2 3

2.74 361.25 25  7 48 

 3m , the depth to the centroid of the disc

h’ = depth to the centre of pressure (or line of action of the force)

748  2.748

Calculate the force:

m .  2829

F

  gh A  1.25  1000  9.81  3      2  kN .  36116

2

Calculate the line of action of the force, h’ . h' 

 nd

2nd moment of area about water surface 1st moment of area about water surface I oo Ah

By the parallel axis theorem 2 moment of area about O (in the surface) Ioo

where I G  IGG  Ah 2 where I GG G is the

r 4 /4. 2nd moment of area about a line through the centroid of the disc and I and I G GG G =  r

Examples: Answers


5

Examples: Answers


6


Examples: Answers h' 



I GG Ah

4( r 2 )3

DR  fd1  fd 2

r

12



fd 3

D 3 f   f d1  d 2

3

12  d1  d 2

2



Examples: Answers

If we take moments from the surface,

h

 r 4


0326 m  3  3.03

 d3 

 d 3

Taking the first beam, we can draw a pressure diagram for this, (ignoring what is below),

So the distance from the spindle to the line of action of the f orce is x  h' h

2H/3

0326  3  0.03 0326 m  3.03

And the moment required to keep the gate shut is

moment

H

F=58860

Fx  36.11 116  0.03 0326  1.176 kN m  Fx

2.2 A dock gate is to be reinforced with three horizontal beams. If the water acts on one side only, to a depth of 6m, find the positions of the beams measured from the water surface so that each will carry an equal load. Give the load per meter. [58 860 N/m, 2.31m, 4.22m, 5.47m]

We know that the resultant force, F

H 

First of all draw the pressure diagram, as below:

2 F  g





1 2

2 F

 gH 2 , so H 

2  58860 1000  9.81 81

 g

 3.46 m st

And the force acts at 2H/3, so this is the position of the 1 beam, d1

2h/3

h

position of 1st beam f

R

f

d2



2 H 3

 2.31m

Taking the second beam into consideration, we can draw the following pressure diagram, d3

f

f

The resultant force per unit length of gate is the ar ea of the pressure diagram. So the total resultant force is 1

0 .5  1000  9.81  6 2 R   gh 2 = 0. 2

h R   g   h  1 2

F=258860

 176580 N ( per m length)

Alternatively the resultant force is, R = Pressure at centroid force per m)

H 

 176580 N ( per m length)

3

2 F  g



2  (2  58860 ) 1000  9.81 81

 4.9 m

(2  58860)  3.27  58860  2 .31 31  58860  d 2 depth to second beam d 2

 4.22 m

For the third beam, from before we have,

 58860 N

12  d1  d 2  d 3 depth to third beam d 3

Examples: Answers

f

The reaction force acts at 2H/3, so H=3.27m. Taking moments from the surface,

The three beams should carry an equal load, so each beam carries the load f load f , where

R

d2

The reaction force is equal to the sum of the forces on each beam, so as before

 Area , (take width of gate as 1m to give

This is the resultant force exerted by the gate on the wa ter.

f 

d1=2.31

2H/3

H


7

Examples: Answers

 12  2.31  4.22  5.47 m


8


Examples: Answers

2.3 The profile of a masonry dam is an arc of a circle, the arc having a radius of 30m and subtending an angle of 60 at the centre of curvature which lies in the water surface. Determine (a) the load on the dam in N/m length, (b) the position of the line of action to this pressure. 6 [4.28  10 N/m length at depth 19.0m] Draw the dam to help picture the geometry,

R a


Examples: Answers

As this force act normal to the surface, it must act through the centre of radius of the dam wall. So the depth to the point where the force acts is, y = 30sin 39.31 =19m 2.4 The arch of a bridge over a stream is in the form of a semi-circle of radius 2m. the bridge width is 4m. Due to a flood the water level is now 1.25m above the crest of the arch. Calculate (a) the upward force on the underside of the arch, (b) the horizontal thrust on one half of the arch. [263.6 kN, 176.6 kN] The bridge and water level can be drawn as:

60°

1.25m

y

FR

h R

2m

Fh

a) The upward force on on the arch = weight weight of (imaginary) water above above the arch.

  g  volume of water 2   2  volume   (1.25  2)  4    4  26.867 m3 2   81  26.86 867  263.568 k Rv  1000  9.81 N Rv

Fv h  30 sin 60  25.98 m a  30 cos 60  15.0 m

b)

Calculate Fv = total weight of fluid above the curved surface (per m length) Fv

The horizontal force on half of the arch, is equal to the force on the projection of the curved surface onto a vertical plane.

  g (area of sector - area of triangle) 60   25.98 98  15   = 1000  9.81    30 2      360   2   2711375 kN / m .

1.25

Calculate Fh = force on projection of curved surface onto a vertical plane

Fh



1 2

2  gh

2.0

81  25.98  3310 .681 kN / m  0.5  1000  9.81 2

F h  press pressure ure at centro centroid id  area area

The resultant,

FR

81  2711.37 375  F  F  3310.6 81 4279.27 kN / m  4279 2 v

2 h

2

  g 1.25  1  2  4 . kN  17658

2

acting at the angle tan  

F v F h

.   3932

Examples: Answers

2.5 The face of a dam is vertical to a depth of 7.5m below the water surface then slopes at 30 to the vertical. If the depth of water is 17m what is the resultant force per metre acting on the whole face? [1563.29 kN]

.  0819 


9

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10


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Examples: Answers d1 d2

h2

d3

f 1 F f 2

f 3

h1 60

density of oil oil = 0.9water = 900 kg/m3. Force per unit length, F = area under the graph = sum of the three areas = f = f 1 + f 2 + f 3

x h2 = 17.0 m, so h1 = 17.0 - 7.5 = 9.5 .

Vertical force = weight of water above the surface, Fv

  gh2  x  0.5h1  x   9810  7.5  5.485  0.5  9 .5  5.485  kN / m .  659123



1 2



f 3



F



2  gh

f1  f 2

DF



 3  33109

2 f 3  165544 N



f 1d1  f 2 d 2

N

 f 3d 3

15 . 2 2  79461  (2  )  33109  (2  15 .) 3 2 3 347m ( from surface) D  2.347

165544 D  52974 

 0.5  1000  9.81  17 2  1417545 kN / m .

2

. m ( from base of wall)  1153

The resultant force is

FR

(1000  9.81  1.5)  1.5

To find the position of the resultant force F, we take moments from any point. We will take moments about the surface.

The horizontal force = force on the projection of the surface on to a vertical plane. Fh

f 1

(900  9.81  2 )  2

 3  52974 N 2 f 2  (900  9.81  2 )  1.5  3  79461 N

x = 9.5/tan 60 = 5.485 m.

1232  1417.54 545 2  Fv2  Fh2  659.12  156329 . kN / m

And acts at the angle tan  

F v F h

.  0465

.   2494



2.6 A tank with vertical sides is square in plan with 3m long sides. The tank contains oil of relative density 0.9 to a depth of 2.0m which is floating on water a depth of 1.5m. Calculate the force on the walls and the height of the centre of pressure from the bottom of the tank. [165.54 kN, 1.15m] Consider one wall of the tank. Draw the pressure diagram:

Examples: Answers


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12


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Examples: Answers

Application of the Bernoulli Equation

p B

p xxL p xxR p xxL

  k

u

2 g

 pB

  w gz B  p B  m gR p  w gzA 

 

w gR p  p A

p xxR m gR p



w gz A



w gR p

 pA

 p A  wg z A  zB   gR P  m w 14715  1000  9 .81 81  2 .5  9 .81 81 R p 13600  1000  . R p   0079 m

pB

3.2

B

A Venturimeter with an entrance diameter of 0.3m and a throat diameter of 0.2m is used to measure the volume of gas flowing through a pipe. The discharge coefficient of the meter is 0.96. Assuming the specific weight of the gas to be constant at 19.62 N/m3, calculate the volume flowing when the pressure difference between the entrance and the throat is measured as 0.06m on a water U-tube manometer. [0.816 m3/s]

Part i)

 0.15m d B  0.075m p B  p A  14715 N / m 2

d2 = 0.2m

Q  0.02 m 3 / s

d A



u 2 g

319 k  0.319

Rp

h f

u

2 A

Part ii)

A

dB = 0.2m

 z A 

2 A

1.5  2.5  1.045  0.065  0.065 k

w gz B

dA = 0.2m

 pA

1000 g

3.1 In a vertical pipe carrying water, pressure gauges are inserted at points A and B where the pipe diameters are 0.15m and 0.075m respectively. The point B is 2.5m below A and when the flow rate down the pipe is 0.02 cumecs, the pressure at B is 14715 N/m2 greater than that at A. v2 Assuming the losses in the pipe between A and B can be expressed as k where v is the velocity at A, 2 g find the value of k of k . If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres. [k = k = 0.319, 0.0794m]

2 B

d1 = 0.3m

kv 2 2 g

Taking the datum at B, the Bernoulli equation becomes: p A



u 2A

 z A 

 g 2 g z A

 2.5

p B



u B2

 z B  k

 g 2 g z B

Z2

u A2 2 g

Z1

Rp

h

0

By continuity: continuity: Q = uAAA = uBAB u A

 0.02 /  0.075 2   1.132 m / s

u B

 0.02 /  0.0375 2   4 .527 m / s

giving

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14


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What we know from the question:


Examples: Answers

3.3 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat ar e connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to be 2.5 H m/s, where H where H is is the manometer reading in metres of mercury. Determine the loss of head between inlet and throat of the Venturi when H when H is is 0.49m. (Relative density of mercury is 13.6). [0.23m of water]

. N / m 2  1962 C d  0.96 d 1  0.3m d 2  0.2m

 g g

Calculate Q.

u1

 Q / 0.0707



g

u2

 Q / 0.0314

For the manometer: p1

p1

gz  p 2





g z 2

g

 Rp

gR p

w

 p 2  19.62z 2  z 1   5 87.423

     (1)

For the Venturimeter

p1



u12

 g g 2 g

 z 1 

p1  p 2

p2



u 22

 g g 2 g

 z 2

Z2

 19.62 z 2  z 1   0.803u22

     (2 )

Z1

H

h

Combining (1) and (2) 0.803u 22

423  587 .42

047 m / s u 2 ideal  27.047 2

For the manometer:

 0.2  0.85m 3 / s  2   96  0.85  0.816 m 3 / s  0.96

047    Qideal  27.047 Q  Cd Qidea

p1   w gz1 p1  p2

 p2   w g  z2  H    m gH  w gz2  w gH  m gH  w gz 1

     (1)

For the Venturimeter p1



u12

 z 1 

w g 2 g

p1  p2



p2



u22

 w g 2 g  w u22 2

 z2  Losses

  w gz 2 

 wu12 2

 w gz1  L w g

     ( 2)

Combining (1) and (2) p1



u12

w g 2 g

 z 1 

L w g

p 2



u 22

 z 2  Losses

 w g 2 g

 Hg

m



w  

 w 2

u  u  2 2

2 1

     (3)

but at 1. From the question u1 u1 A1 175 . 

d 2 4 u2

Examples: Answers


15

Examples: Answers

 2.5 H  1.75m / s  u2 A2 2

 2 d   u2     10  937 m / s  10.937 CIVE1400: Fluid Mechanics

16


Examples: Answers


Examples: Answers

Substitute in (3)

Losses  L 

183 . 

0.49 49  9.81 81 13 13600  1000   1000 / 2  10 .93 937 2  1 .75 2 

2 g

 5.99 m / s

u3

9.81 81  1000

u

2 3

Q  u3 A3

233m  0.233 3.4 Water is discharging from a tank through a convergent-divergent mouthpiece. The exit from the tank is rounded so that losses there may be neglected and the minimum diameter is 0.05m. If the head in the tank above the centre-line of the mouthpiece is 1.83m. a) What is the discharge? b) What must be the diameter at the exit if the absolute pressure at the minimum area is to be 2.44m of water? c) What would the discharge be if the divergent part of the mouth piece were removed. (Assume atmospheric pressure is 10m of water). [0.0752m, 0.0266m 3/s, 0.0118m3/s]

0.02 02665  5.99 99  

p1  g

 p2

 z 1  u2



2

p 2



u2

 g 2 g 2 gz 1

Q  5.99

 5.99 m / s 0.05 2 4

0118 m 3 / s  0.0118

3.5 A closed tank has an orifice 0.025m diameter in one of its vertical sides. The tank contains oil to a depth of 0.61m above the centre of the orifice and the pressure in the air space above the oil is maintained at 13780 N/m2 above atmospheric. Determine the discharge from the orifice. (Coefficient of discharge of the orifice is 0.61, relative density of of oil is 0.9). [0.00195 m3/s]

h

2

4

0752m  0.0752

d3 If the mouth piece has been removed, p1

d 32

3 P = 13780 kN/m2

From the question: d2

 0.05m

minimum pressure  p1  g

p 2  g

0.66m

 2.44m

 10m 

oil

p 3  g

do = 0.025m

Apply Bernoulli: p1



2 1

u

g  2 g

 z 1 

p2



2 2

u

g  2 g

 z 2 

p3



2 3

u

 z 3

From the question

g  2 g

  0.9 

If we take the datum through the orifice:

z1

 1.83m

z2

 z3  0

u1

 negligible

 o

Between 1 and 2

C d 2

10  1.83  2.44  u2

u2

Between 1 and 3 p1

Examples: Answers

 900  0.61

Apply Bernoulli,

2 g

p1

. m/s  1357

 0.05 Q  u2 A2  1357 .     2 

 o  w



u12

 z 1 

 g 2 g

p2



u22

 z 2

 g 2 g

2

02665 m 3  0.02665

Take atmospheric pressure as 0,

/s

 p3 CIVE1400: Fluid Mechanics

17

Examples: Answers


18


Examples: Answers 13780  o g

 0.61  u2

u

CIVE1400: Fluid Mechanics p1

2 g



2 1

u

 g 2 g

 6.53 m / s

Q

Examples: Answers

2 2

p1

025  0.025 0.61 61  6.53 53     2 



p2



2 1

u



 h f 

 g  g 2 g

2

00195 m 3 / s  0.00195

25 

3.6 The discharge coefficient of a Venturimeter was found to be constant for rates of flow exceeding a certain value. Show that for this condition the loss of head due to friction in the convergent parts of the meter can be expressed as KQ as KQ2 m where K where K is is a constant and Q is the rate of flow in cumecs. Obtain the value of K if K if the inlet and throat diameter of the Venturimeter are 0.102m and 0.05m respectively and the discharge coefficient is 0.96. K =1060] [ K =1060]

3.7 A Venturimeter is to fitted in a horizontal pipe of 0.15m diameter to measure a flow of water which may be anything up to 240m 3/hour. The pressure head at the inlet for this flow is 18m above atmospheric and the pressure head at the throat must not be lower than 7m below atmospheric. Between the inlet and the throat there is an estimated frictional loss of 10% of the differ ence in pressure head between these points. Calculate the minimum allowable diameter for the throat. [0.063m]

3.77 2 2 g

 2.5  u2

p2



u

2 2

 g 2 g u

 h f

2 2

2 g u22 2 g

.  21346 m / s

Q  u 2 A2 0.06 0 667  21.34 346   d2

d 22 4

063m  0.063

3.8 A Venturimeter of throat diameter 0.076m is fitted in a 0.152m diameter vertical pipe in which liquid of relative density 0.8 flows downwards. Pressure gauges are fitted to the inlet and to the throat sections. The throat being 0.914m below the inlet. Taking the coefficient of the meter as 0.97 find the discharge a) when the pressure gauges read the same b)when the inlet gauge reads 15170 N/m2 higher than the throat gauge. [0.0192m3/s, 0.034m3/s]

d1 = 0.15m d1 = 0.152m

d2 From the question:

 0.15m u1  Q / A  3.77 m / s

3 Q  240 m / hr

p1

p2

d1

 g

 18m

 g

d1 = 0.076m

 0.667 m3 / s

  7m

Friction loss, from the question: h f

 0.1

 p  p  1

2

 g

From the question:

 0.152m  0.076m 3   800 kg / m C d  0.97

Apply Bernoulli:

d1

A1

d2

A2

 0.01814 m  0.00454 m

Apply Bernoulli:

Examples: Answers


19

Examples: Answers


20


Examples: Answers p1



2 1

u

 z 1 

 g 2 g a) p1

p2



u

2 2

u12

 z 1 

u22 2 g

4.1 A reservoir is circular in plan and the sides slope at an angle of tan -1(1/5) to the horizontal. When the reservoir is full the diameter of the water surface is 50m. Discharge from the reservoir takes place through a pipe of diameter 0.65m, the outlet being 4m below top water level. Determine the time for the

 z 2

water water level level to fall fall 2m assumi assuming ng the the disc dischar harge ge to to be 0.75a 2gH cumecs where a is the cross sectional area of the pipe in m 2 and H and H is is the head of water above the outlet in m. [1325 seconds]

By continuity: Q  u1 A1 u2 2

u1

2 g

.  0914  u1

 u1

Examples: Answers

Tank emptying

 z 2

 g 2 g

 p2 2 g


 u 2 A2

A1 A2

50m

 u1 4

16u12

r H

2 g 0.914  2  9.81



15

.  10934 m / s

x

Q  Cd A1 u1 96  0.01 01814  1.0934 Q  0.96

019 m 3 / s  0.01

1

b)

5 p1  p 2 p1

 p2

 g 15170  g

 

u

2 2

u

2 1

2 g

 15170 From the question:

914  0.914

43 2 Q 2  220.43

 55.11 2 

2 g

55.8577  Q 2  220.43 2

H = 4m

a = (0.65/2)2 = 0.33m2

Q  0.75a 2 gh .  10963

914  0.914

h

t the level in the reservoir falls  h, so In time  t the

 55.11 2 

Q t

Q  0.035 035 m / s 3

  A  h

t  

A Q

 h

Integrating give the total time for levels to fall f rom h1 to h2. T 

h2

A

1

Q

  h

dh

As the surface area changes with height, we must express A express A in terms of h. of h. 2

A =  r r But r varies r varies with h.

-1 It varies linearly from the surface at H at H = 4m, r = 25m, 25m , at a gradient of tan = 1/5.

r = x + 5h 25 = x + 5(4) x = 5

so

2

2

 (A5=+ 5h ) = ( 25 25 + + 25 25 h + 50 50 h )

Substituting in the integral equation gives

Examples: Answers


21

Examples: Answers


22

CIVE1400: Fluid Mechanics T 

h2

  h

Examples: Answers

 25h  50 h 2

25

10963 . h

1



25

h2

10963 .



1 h2

h1

h2

.  71641  h

1 h

h2



h 1/ 2

h

2

h



2h h

Q t

  A  h

t  

A

 h

Q

Integrating give the total time for levels to fall f rom h1 to h2.

dh

 h 3/ 2  2h 1/ 2

Examples: Answers

And we can write an equation for the discharge in terms of the surface height change:

dh

dh

h

1

 71.641 h

 2h


dh

T 

h2

A

1

Q

  h

dh

1

h2

  1.68 h

h2

2 4   716 . 412 h 1/ 2  h 53/ 2  h 3/ 2   71641 5 3  h

A h

1

       (1)

dh

1

From the question, h 1 = 4m

a) For the first 1m depth, A = 8 x 32 = 256, whatever the h.

h2 = 2m, so

So, for the first period of time:

2 4 2 4       4 53/ 2   4 3/ 2    2  2 1/ 2   2 53/ 2   2 3/ 2       5 3 5 3  828  2 .26 263  3 .77   71.6414  12.8  10.667   2 .82 6412 7.46 467  8.86 862    71.64  1333 sec

T  71.641 2  4 1/ 2

. T   168

h2



256

.   43008

4.2 A rectangular swimming pool is 1m deep at one end and increases uniformly in depth to 2.6m at the other end. The pool is 8m wide and 32m long and is emptied through an orifice of area 0.224m2, at the lowest point in the side of the deep end. Taking C d d for the orifice as 0.6, find, from first principles, a) the time for the depth to fall by 1m b) the time to empty the pool completely. [299 second, 662 seconds]

dh

h

h1

h1

 430.08  299 sec



h2



2.6  1.6



b) now we need to find out how long it will take to empty the rest. We need the area A area A,, in terms of h of h. A  8 L L

32.0m



32

h 1.6 A  160h So

1.0m 2.6m

T   168 .

h2



L .  2689 .  2689

The question tell us ao = 0.224m 2 , C d d = 0.6

p1



u12

 z 1 

p1 = p2 and u1 = 0. u 2



p2



u22

2

h

3 2

 h 

dh

3/ 2

1

16. 

3 . sec  36267

Apply Bernoulli from the tank surface to the vena contracta at the orifice:

 g 2 g

160h

h1

3/ 2

 h2 

3/ 2



  0 3/ 2 

Total time for emptying is,

 z 2

 g 2 g

T = 363 + 299 = 662 sec

2g h

We need Q in terms of the height h measured above the orifice. Q  Cd a o u 2

 Cd a o

 0.6  0.224  .  0595

Examples: Answers

2gh

2  9 .81 h

h


23

Examples: Answers


24


Examples: Answers

4.3 A vertical cylindrical tank 2m diameter has, at the bottom, a 0.05m diameter sharp edged orifice for which the discharge coefficient is 0.6. a) If water enters the tank at a constant rate of 0.0095 cumecs find the depth of water above the orifice when the level in the tank becomes stable. b) Find the time for the level to fall from 3m to 1m above the orifice when the inflow is turned off. c) If water now runs into the tank at 0.02 cumecs, the orifice remaining open, find the rate of rise in water level when the level has reached a depth of 1.7m above the orifice. [a) 3.314m, b) 881 seconds, c) 0.252m/min]

CIVE1400: Fluid Mechanics Q t

Examples: Answers

  A  h

t  

A Q

 h

Integrating between h1 and h2, to give the time to change surface level T  

h2

A

h1

Q



dh h2

.    6018 h

h

1/ 2

dh

1

. h  12036

3 Q = 0.0095 m /s

1/ 2 h2



h1

. h  h11/ 2   12036 1/ 2 2

h1 = 3 and h2 = 1 so T = 881 sec c) Qin changed to Qin = 0.02 m3 /s

h

From (1) we have Qout  0.00522 00522 h . The question asks for the rate of surface surface rise when h = 1.7m. i.e.

do = 0.005m

Q  Qin

u12

 z 1 

 g 2 g p1 = p2 and u1 = 0. u 2



p2

u22



Q  Au

 z 2

u

 g 2 g

2g h .

 Cd a o u2  Cd a o  0.05  0.6    2  0.00522

Q A



0.0132 0132

  2 2     4 

 0.0042 m / s  0.252 m / min

4.4 A horizontal boiler shell (i.e. a horizontal cylinder) 2m diameter and 10m long is half full of water. Find the time of emptying the shell through a short vertical pipe, diameter 0.08m, attached to the bottom of the shell. Take the coefficient of discharge to be 0.8. [1370 seconds]

With the datum the bottom of the cylinder, z 1 = h, z 2 = 0 We need Q in terms of the height h measured above the orifice. Qout

02  0.00 0068  0.01 0132 m3 / s  Qout  0.02

As Q = Area x Velocity, Velocity, the rate of rise in surface is

Apply Bernoulli from the water surface (1) to the orifice (2),



0068 m3 / s  0.00

The rate of increase in volume is:

From the question: Qin = 0.0095 m3 /s, d o=0.05m, C d d =0.6

p1

00522 1.7 Qout  0.00

2 gh

2

2  9.81 h

h

d = 2m

      (1)

32m

For the level in the tank to remain constant: inflow = out flow Qin = Qout 0.00 0095  0.00 00522 h h  3314 m .

do = 0.08 m

(b) Write the equation for the discharge in terms of the surface height change:

From the question W = 10m, D = 10m d o = 0.08m C d d = 0.8

Examples: Answers


25

Examples: Answers


26


Examples: Answers


Apply Bernoulli from the water surface (1) to the orifice (2),

p1



u12

 z 1 

p2

 g 2 g p1 = p2 and u1 = 0. u 2



u22



T  

 z 2

 g 2 g

2g h .

 0.08   0.8   

A Q

h2

2h  h 2

1

h

T  

h1

Q



2  h dh 2

h

4.5 Two cylinders standing upright contain liquid and are connected by a submerged orifice. The diameters of the cylinders are 1.75m and 1.0m and of the orifice, 0.08m. The difference in levels of the liquid is initially 1.35m. Find how long it will take for this difference to be reduced to 0.66m if the coefficient of discharge for the orifice is 0.605. (Work from first principles.) [30.7 seconds]

Integrating between h1 and h2, to give the time to change surface level A

dh

1

 h

h2

dh

 2  3/ 2 h .    2  h   11236 h  3  072.82 828  1  1369 .6 sec  749.07

2  9.81 h

  A  h

t  

h

1

Write the equation for the discharge in terms of the surface height change: Q t

2h  h 2

1

 1123.6  h

2

2

0178  0.0178

h2

h2

2 gh

dh

01078 . h

.   11236 h

We need Q in terms of the height h measured above the orifice.

 Cd ao u2  Cd a o

20 2 h  h

h1

.   11236 h

With the datum the bottom of the cylinder, z 1 = h, z 2 = 0

Qout

h2



Examples: Answers 2

d1 = 1.75m

d2 = 1.0m

dh

But we need A need A in terms of h of h

h = 1.35m

2.0m

1.0m a L h do = 0.108m

. 2

10L, so need L Surface area A area A = 10L, need L in terms of h of h

 L  12  a 2     2  a  (1  h) 12

do

A2

2

 1  . m2       0785 2

 0.08m,

ao

 0.08       0.00503m2 2

Cd  0.605

by continuity,

2

 A1h1  A2h2  Q t

2

2h  h 2 

     (1)

defining, h = h1 - h2

 h  h1   h2

2 A  20 2 h  h

Substituting this in (1) to eliminate  h2

Substitute this into the integral term,

Examples: Answers

.   175       2.4m2 2

 L   (1  h) 2    

L2

A1

2

2


27

Examples: Answers


28


Examples: Answers

 A1h1   h1

 A1

A2 h A1  A2



A2 (h1  h)  A2 h1  A2 h

CIVE1400: Fluid Mechanics Q t

A2 h

 Q t

T  

Q  Cd ao 2 gh



So A2 h A1  A2

t  

Q

 h

h2

A

h1

Q



60000 0678 .

dh h2



h1

1 h 3/ 2

58h  2  88495.58

 Cd a o

2 gh  t

A1 A2

 A  A C 1

A

Integrating between h1 and h2, to give the time to change surface level

      ( 2)

From the Bernoulli equation we can derive this expression for discharge through the submerged orifice:

 A1

  A  h

t  

A1  A2

Examples: Answers

2

d

1

ao 2 g

h

dh

1/ 2 h2



h1

From the question T = 3600 sec and h1 = 0.6m 3600  176991.15h21/ 2

 h

h2

 0.6 1/ 2 

.  05815 m

Total depth = 3.4 + 0.58 = 3.98m

Integrating T  

A1 A2

 A  A C a 1



d

o

o

2g

2 A1 A2

 A  A C a 1



2

h2

2

d



2g

1 h

h1



h2

dh



h1



 2  2.4  0.785 0.81 8124  1.1619  2.4  0.785  0 .605  0.00503 2  9 .81

 30.7 sec 4.6 A rectangular reservoir with vertical walls has a plan area of 60000m2. Discharge from the reservoir take place over a rectangular weir. The flow c haracteristics of the weir is Q = 0.678 H 3/2 cumecs where H where H is is the depth of water above the weir crest. The sill of the weir is 3.4m above the bottom of the reservoir. Starting with a depth of water of 4m in the reservoir and no inflow, what will be the depth of water after one hour? [3.98m]

From the question A question A = 60 000 m 2 , Q = 0.678 h

3/2

Write the equation for the discharge in terms of the surface height change:

Examples: Answers


29

Examples: Answers


30


Examples: Answers


Examples: Answers

5.2 Derive an expression for the discharge over a sharp crested rectangular weir. A sharp edged weir is to be constructed across a stream in which the normal flow is 200 litres/sec. If the maximum flow likely to occur in the stream is 5 times the normal flow then determine the length of we ir necessary to limit the rise in water level to 38.4cm above that for normal flow. C d . =0.61 d=0.61. [1.24m]

Notches and weirs

5.1 Deduce an expression for the discharge of water over a right-angled right-angled sharp edged V-notch, given that the coefficient of discharge is 0.61. A rectangular tank 16m by 6m has the same notch in one of its short vertical sides. Determine the time taken for the head, measured from the bottom bottom of the notch, to fall from 15cm to 7.5cm. [1399 seconds]

From your notes you can derive: From your notes you can derive:

Q

8 15



C d tan

2

2gH

Q  144 . H 5/ 2

Q

T  



h1

Q



16  6 144 .

/s,

1

=x h

3

/s,

2

= x +h 0.384

0.2  10 .

Integrating between h1 and h2, to give the time to change surface level A

3

So we have two situations:

 h

h2

Q1 = 0.2 m

where x where x is the height above the weir at normal flow.

  A  h

t  

Cd b 2 g h3/ 2

Q2 = 1.0 m

Write the equation for the discharge in terms of the surface height change: A

3

From the question:

For this weir the equation simplifies to

Q t

2

Q 5/ 2

3 2 3

Cd b 2 g x 3/ 2

 1.801bx 3/ 2

Cd b 2 g  x  0.384

3/ 2

 1.801b x  0.384  3/ 2

   (1)    (2 )

x From (1) we get an expression for b for b in terms of x

dh

. x b  0111

h2

1

h1

5/ 2





2

h

dh

2

h2

3

1

.  x  0384   x 

. h  3/ 2 h   6667

3/ 2

1.0  1.801  0.111

h1 = 0.15m, h2 = 0.075m T  44.440.075 3/ 2

3/ 2

Substituting this in (2) gives,

52 / 3

 0.15 3/ 2 

.  x  0384      x

. x  01996 m

 1399 sec

So the weir breadth is 011 . 101996 0199 . 6 b  0111

3/ 2

. m  124

Examples: Answers


31

Examples: Answers


32


Examples: Answers

5.6 Show that the quantity of water flowing across a triangular V-notch of angle 2 is 8 Q  Cd tan  2 g H 5/ 2 . Find the flow if the measured head above the bottom of the V is 38cm, when 15 = =45 and C d d=0.6. 0.6. If the flow is wanted within an accuracy of 2%, what are the limiting values of the head. 3 [0.126m /s, 0.377m, 0.383m]


Examples: Answers

Application of the Momentum Equation

6.1 The figure below shows a smooth curved vane attached to a rigid foundation. The jet of water, rectangular in section, 75mm wide and 25mm thick, strike the vane with a velocity of 25m/s. Calculate the vertical and horizontal components of the force exerted on the vane and indicate in which direction these components act. [Horizontal 233.4 N acting from right to left. Vertical 1324.6 N acting downwards]

Proof of the v-notch weir equation is in the notes. From the question: H = 0.38m  =  = 45 45

C d d = 0.6

45

The weir equation becomes:

25

. Q  1417 H

5/ 2

From the question:

5/ 2 14 . 17 038 .   141 3 m /s .  0126

075  0.02 025  1.875  10 3 m 2  0.07 u1  25 m / s 3 3 Q  1.875  10  25 m / s a1  a 2 , so u1  u 2 a1

3

Q+2% = 0.129 m /s 0.12 129  1.41 417 H

5/ 2

H

Calculate the total force using the momentum equation:

.  0383 m

FT x x

  Qu2 cos 25  u1 cos 45  1000  0.046925 25 cos 25  25 cos45 . N  23344

FT y y

  Qu2 sin 25  u1 sin 45

Q-2% = 0.124 m 3 /s 0.12 124  1.41 417 H 5/ 2 H

m .  0377

2 5 sin 25  25 sin 45   1000  0.0469 25 1324.6 N  1324 Body force and pressure force are 0. So force on vane: R x R y

Examples: Answers


35

Examples: Answers

. N   F t x x  23344   F t y  1324 1324.6 N


36


Examples: Answers

6.2 A 600mm diameter pipeline carries water under a head of 30m with a velocity of 3m/s. This water main is fitted with a horizontal bend which turns the axis of the pipeline through 75 (i.e. the internal angle at the bend is 105 ). Calculate the resultant force on the bend and its angle to the horizontal. [104.044 kN, 52 29’]

CIVE1400: Fluid Mechanics FRy

 FTy  FPy  FBy 457  80.376  0  82 .833 kN  2. 45

These forces act on the fluid The resultant force on the fluid is

u2

y

FR



  tan

x

Examples: Answers

FRx

104.44 k N  FRy  104

F 1  Ry 

   52  F Rx 



29 '

6.3 A horizontal jet of water 2103 mm2 cross-section and flowing at a velocity of 15 m/s hits a flat plate at 60 to the axis (of the jet) and to the horizontal. The jet is such that there is no side spread. If the plate is stationary, calculate a) the force exerted on the plate in the direction of the jet and b) the ratio between the quantity of fluid that is deflected upwards and that downwards. (Assume that there is no friction and therefore no shear force.) [338N, 3:1]

u1 θ

y u2

x From the question: 2

 0.6       0.283 m 2 d  0.6 m  2  u1  u 2  3 m / s Q  0.848 m 3 / s

h  30 m

a

u1

Calculate total force.

FTx

  Qu 2 x  u1x   FRx  FPx  FBx

θ

FTx  1000  0.848 3cos 75  3   1.886 k N

FTy

  Qu2 y  u1y   FRy  FPy  FBy

FTy

 1000  0.848 3 sin 75  0  2 .457 k N

u3

From the question

Calculate the pressure force p1 = p2 = p = h  h  g = 30 30  1000 1000  9.81 9.81 = 294.3 kN/m FTx

 p1 a1 cos 1  p 2 a 2 cos  2  294300  0.2831  cos 75 . kN  6173

FTy

 p1a1 sin 1  p2 a 2 sin  2  294300  0.2830  sin 75   80376 kN .

Examples: Answers

u = 15 m/s

Apply Bernoulli,

2

p1



u12

 z 1 

g  2 g

p 2



u 22

 z 2 

g  2 g

p 3



u 32

 z 3

g  2 g

Change in height is negligible so z 1 = z 2 = z 3 and pressure is always atmospheric p1= p2 = p3 =0. So u1= u2 = u3 =15 m/s By continuity Q1= Q2 + Q3 u1a1 = u2a2 + u3a3

There is no body force in the x or y directions. FRx

a2 = a3 =2x10-3 m2

 FTx  FPx  FBx  1.886  61.73  0  63.616 kN

so

1

= a2 + a3a

Put the axes normal to the plate, as we know that the resultant force is normal to the plate.


37

Examples: Answers


38


Examples: Answers


-3

Q1 = a1u = 2 2 10 10  15 15 = 0.03

Examples: Answers

F Tx Q( 0 - u1x ) Tx =  Q(

Q1 = (a2 + a3 ) u

F Tx 1000 0.11 0.11 ( 0 - 25 cos 30 ) = 2.39 kN Tx = 1000

Q2 = a2u

6.5 The outlet pipe from a pump is a bend of 45  rising in the vertical plane (i.e. and internal angle of 135). The bend is 150mm diameter at its inlet and 300mm diameter at its outlet. The pipe axis at the inlet is horizontal and at the outlet it is 1m higher. By neglecting friction, calculate the f orce and its direction if the inlet pressure is 100kN/m2 and the flow of water through the pipe is 0.3m3/s. The volume of the pipe is 0.075m3. [13.94kN at 67  40’ to the horizontal]

Q3 = (a1 - a2 )u Calculate total force.

FTx

  Qu 2 x  u1x   FRx  FPx  FBx

FTx

 1000  0.03 0  15 sin 60  390 N

y

Component in direction of jet = 390 sin 60 = 338 N

p2 u 2

As there is no force parallel to the plate F ty ty = 0

 u22 a2  u32a 3  a2  a 3  a1 cos  0 a1  a 2  a 3 a3  a1 cos  a1  a 3 FTy

4a 3 a3

 a1 



1 3

4 3

A2

x

u12 a1 cos  0 1m

p1 45°

u1

a2

A1

a2

Thus 3/4 of the jet goes up, 1/4 down 6.4 A 75mm diameter jet of water having a velocity of 25m/s strikes a flat plate, the normal of which is inclined at 30 to the jet. Find the force normal to the surface of the plate. [2.39kN]

1&2 Draw the control volume and the axis system

y u2

x

p1 = 100 kN/m 2,

Q = 0.3 m3/s

d1 = 0.15 m

d2 = 0.3 m

A1 = 0.177 m 2

A2 = 0.0707 m 2

 = 45

3 Calculate the total force in the x direction

FT x

u1

  Q u2 x  u1 x   Qu2 cos  u1 

θ

A1u1  A2u2 by continuity

Q

, so

u3

From the question,

d jet = 0.075m

u1=25m/ =25ms/

2

3

Q = 25  (0.075/2) (0.075/2) = 0.11 m /s

Force normal to plate is

Examples: Answers


39

Examples: Answers


40

CIVE1400: Fluid Mechanics 0.3

u1



u2



F T x x

2   0.15 / 4 

0.3 0.0707 0707

Examples: Answers

. m / s  1698

 4.24 m / s

c os45  16.98   1000  0.3 4.24 co . N  149368


Examples: Answers

F P x x

0177  225361.4 cos45  0 .07 0707  100000  0.01 34  9496 .37 37 kN  1770  11266.34

F P y y

0707  225361.4 sin 45  0.07 11266.37  11266

5 Calculate the body force The only body force is the force due to gravity. That is the weight acting in the y direction.

F B y

and in the y-direction FT y

   g  volume 81  0.07 075  1000  9.81 . N  1290156

  Qu2 y  u1 y   Qu2 sin   0  1000  0.34.24 sin 45 899.44 N  899

There are no body forces in the x direction, F B x

0

6 Calculate the resultant force

4 Calculate the pressure force.

FT x x

F P  pressure force at at 1 - pressure force at at 2

FT y y F P x x



p1 A1 cos 0  p2 A2 cos



F P y y



p1 A1 sin 0  p2 A2 sin 

  p2 A2 sin 

 F R x  F P x  F B x  F R y  F P y y  F B y

p1 A1  p2 A2 cos F R x

 FT x x  F P x x  F B x 37  4193.6  9496.37  5302 5302.7 N

F R y

 FT y y  F P y y  F B y 4 4  11266.37 37  735.75 75  899.44  1290156 . N

We know pressure at the inlet but not at the outlet. we can use Bernoulli to calculate this unknown pressure.

p1



u12

 g 2 g

 z 1 

p2



u22

 g 2 g

 z2  h f And the resultant force on the fluid is given by

where h f is the friction loss

FRy

In the question it says this can be ignored, h f =0

FResultant

The height of the pipe at the outlet is 1m above the inlet. Taking the inlet level as the datum: z1 = 0

z2 = 1m

φ

So the Bernoulli equation becomes: 100000 1000  9.81 81

2



1698 .

2  9.81

0 p2

Examples: Answers

p2 1000  9.81 81

.  2253614



4.24

2

2  9.81

FRx

 1.0

N / m2


41

Examples: Answers


42

CIVE1400: Fluid Mechanics F R



2 R x

F

F

2 R y

Examples: Answers

56  5302.7  12901.56 . kN  1395

Examples: Answers

6.7 A curved plate deflects a 75mm diameter jet through an angle of 45. For a velocity in the jet of 40m/s to the right, compute the components of the force developed against the curved plate. (Assume no friction). [R x=2070N, R y=5000N down]

2


2

u2

y And the direction of application is   tan

1

 F R y  1290156 .     tan 1  .    6766  5302 5302.7   F R x 

x



The force on the bend is the same magnitude but in the opposite direction

R   F R

u1

6.6 The force exerted by a 25mm diameter jet against a flat plate normal to the axis of the jet is 650N. What is the flow in m 3/s? [0.018 m3/s]

y

θ

u2

From the question:

  0.0752 / 4  4.42  10 3 m2 u1  40 m / s  Q  4.42  10 3  40  0.1767 m 3 / s , a1  a 2 so u1  u 2 a1

x u1

Calculate the total force using the momentum equation: FT x x

  Q u2 cos 45  u1  4 0 cos45  40   1000  0.176740 . N  207017

FT y y

  Q u2 sin 45  0

u2 From the question,

d jet = 0.025m

 1000  0.176740 sin 45  4998 N

F Tx Tx = 650 N

Body force and pressure force are 0.

Force normal to plate is

So force on vane:

F Tx Q( 0 - u1x ) Tx =  Q(

R x

650 = 1000 1000  Q ( 0 - u )

R y

Q = au = (  d d 2 /4)u 2

  Ft x x  2070 N   F t y y  4998 N

2

650 = -1000au = -1000Q /a 2

2

650 = -1000Q /(  0.025 0 .025 /4) Q = 0.018m3 /s

Examples: Answers


43

Examples: Answers


44


Examples: Answers

6.8 A 45 reducing bend, 0.6m diameter upstream, 0.3m diameter downstream, has water flowing through it at the rate of 0.45m3/s under a pressure of 1.45 bar. Neglecting any loss is head for friction, calculate the force exerted by the water on the bend, and its direction of application. [R=34400N to the right and down,  = 14] ρ2

y

CIVE1400: Fluid Mechanics F T x x

Examples: Answers

365 cos45  1.59   1000  0.456.36  1310 N

and in the y-direction

u2

FT y

A2

  Qu2 y  u1 y   Qu2 sin   0  1000  0.45 456 .36 365 sin 45  1800 N

x

4 Calculate the pressure force. ρ1 at 2 F P  pressure force at 1 - pressure force at

u1 A1

θ

F P x x



p1 A1 cos 0  p2 A2 cos



p1 A1  p 2 A2 cos

F P y y



p1 A1 sin 0  p2 A2 sin 

  p2 A2 sin 

We know pressure at the inlet but not at the outlet.

1&2 Draw the control volume and the axis system

we can use Bernoulli to calculate this unknown pressure. p1 = 1.45105 N/m2,

Q = 0.45 m3/s

d1 = 0.6 m

d2 = 0.3 m

A1 = 0.283 m2

A2 = 0.0707 m 2

 = 45

p1



u12

 g 2 g

 z 1 

p2



u22

 g 2 g

 z2  h f

where h f is the friction loss In the question it says this can be ignored, h f =0 =0

3 Calculate the total force

Assume the pipe to be horizontal

in the x direction

z1 = z2 So the Bernoulli equation becomes:

FT x

  Q u2 x  u1 x 

145000

 Qu2 cos  u1 

1000  9.81 81



159 . 2 2  9.81 p2

A1u1  A2u2 by continuity

u1



u2



0.45 2   0.6 / 4

0.45 0.0707 0707

Examples: Answers

Q



p2



2 6365 .

1000  9.81 2  9.81  126007 N / m 2

, so F P x x

283  126000 cos45  0 .07 0707  145000  0.28  41035  6300  34735 N

F P y y

0707  126000 sin 45  0.07

. m / s  159

.  6365 m / s

 6300 N


45

5 Calculate the body force Examples: Answers


46


Examples: Answers


Examples: Answers

The only body force is the force due to gravity.

Laminar pipe flow.

There are no body forces in the x or y directions,

7.1 The distribution of velocity, u, in metres/sec with radius r in metres in a smooth bore tube of 0.025 m bore follows the law, u = 2.5 - kr 2. Where k is a constant. The flow is laminar and the velocity at the pipe surface is zero. The fluid has a coefficient of viscosity of 0.00027 kg/m s. Determine (a) the rate of flow in m3/s (b) the shearing force between the fluid and the pipe wall per metre length of pipe. [6.14x10-4 m3/s, 8.49x10 -3 N]

 F B y  0

F B x

6 Calculate the resultant force

 F R x  F P x  F B x  F R y  F P y y  F B y

FT x x FT y y

The velocity at distance r from the centre is given in the question:

2

u = 2.5 - kr

 = 0.00027 kg/ms

Also we know: F R x

 FT x  F P x  F B x  1310  34735   33425 N

F R y

 FT y y  F P y y  F B y  1800  6300  8100 N

We can find k from the boundary conditions: when r = 0.0125, u = 0.0 (boundary of the pipe) 2 0.0 = 2.5 - k0.0125 k = 16000 u = 2.5 - 1600 r 2 a) Following along similar lines to the derivation seen in the lecture notes, we can calculate the flow  Q r : through a small annulus  r  Q  ur Aannulus

  ( r  r ) 2  r 2  2rr 2 Q   2.5  16000r 2  r  r

Aannulus

And the resultant force on the fluid is given by

FRy

2r = 0.025m

FResultant

0.0125 0125

Q  2

  2.5r  16000r  dr 3

0

0.0125 0125

 2.5r 2 16000 4  r   2   4  2 0 . m3 / s  614

φ

FRx F R



2 R x

F

F

2 R y

b)

The shear force is given by F =   (2 (2 r) r)

 334252  8100 2  34392 kN

From Newtons law of viscosity  du

And the direction of application is

 F R y  1    tan   F R x 

 8100  .  tan 1    1362   33425 

 

du dr

 2  16000r  32000 r

dr F  0.00027  32000  0.0125  (2    0 .0125 )



 8.48  10  3 N

The force on the bend is the same magnitude but in the opposite direction

R   F R

Examples: Answers


47

Examples: Answers


48


Examples: Answers

7.2 A liquid whose coefficient of viscosity is m flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean velocity u. Show that the pressure loss in a length of pipe is 32um/d2. Oil of viscosity 0.05 kg/ms flows through a pipe of diameter 0.1m with a velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m. [11 520 N/m2] See the proof in the lecture notes for


 Q  ur Aannulus

  (r  r ) 2  r 2  2 rr  p 1 2 2  R  r 2r r  Q  L 4 

r

 p

dr



 R 4

L 8



p  d 4

L128 

4

 d

,

 p    

du

 p 

dr

32  L 32  Lu d 2 32  u d 2

per unit length  = 0.05 kg/ms

d = 0.1m

d u

u = 0.6 m/s

L = 120.0m

dr  p r

32  0.05  120  0.6

  

L 2 du

 p

 p d 2

b) From the the question question r

3

L 128

u

L 2

Giving:

 p

2

uQ/ A

 p r

We are measuring from the pipe centre, so 

  R r  r  dr

To get pressure loss in terms of the velocity of the flow, use the mean velocity:

2  2r L  p A  p r

dy

R

So the discharge can be written

The fluid is in equilibrium, shearing forces equal the pressure forces.

du



L 2  0



Q

 

 p

Q

R

Newtons law of viscosity 

 R 2  r 2 

Aannulus

δr

 

1

L 4 

The flow in an annulus of thickness r

Consider a cylinder of fluid, length L, radius r, flowing steadily in the centre of a pipe

r

 p

ur 

Examples: Answers

 p 

01 .2

 11520 N / m2

L 2 

In an integral form this gives an expression for velocity, u

 p

1

L 2 

 r dr

The value of velocity at a point distance r from the centre ur  

 p

2

r

L 4 

 C

0, (the centre of the pipe), u = umax, at r = R (the pipe wall) u = 0; 0; At r = 0, C 

 p R 2 L 4 

At a point r from the pipe centre when the flow is laminar:

Examples: Answers


49

Examples: Answers


50


Examples: Answers

7.3 A plunger of 0.08m diameter and length 0.13m has four small holes of diameter 5/1600 m drilled through in the direction of its length. The plunger is a close fit inside a cylinder, containing oil, such that no oil is assumed to pass between the plunger and the cylinder. If the plunger is subjected to a vertical downward force of 45N (including its own weight) and it is assumed that the upward flow through the four small holes is laminar, determine the speed of the fall of the plunger. The coefficient of velocity of the oil is 0.2 kg/ms. [0.00064 m/s]

F = 45N Q


Examples: Answers

v piston = 0.00064 m/s 7.4 A vertical cylinder of 0.075 metres diameter is mounted concentrically in a drum of 0.076metres internal diameter. Oil fills the space between them to a depth of 0.2m. The rotque required to rotate the cylinder in the drum is 4Nm when the speed of rotation is 7.5 revs/sec. Assuming that the end effects are negligible, calculate the coefficient of viscosity of the oil. [0.638 kg/ms] r -1 = 0.076/2

From the question

d = 5/1600 m

r 2 = 0.075/2

Torque = 4Nm, L = 0.2m

The velocity of the edge of the cylinder is: ucyl = 7.5  2  2 r = 7.5 7.5 2 0.0375 0.0375 = 1.767 m/s udrum = 0.0 Torque needed to rotate cylinder

plunger

   surface ar area  L . N / m2   226354

T

0.13 m

4    2 r2

Distance between cylinder and drum = r 1 - r 2 = 0.038 - 0.0375 = 0.005m Using Newtons law of viscosity:

cylinder

 du

0.8m

  

du

dr 1.767  0

0.0005 0005 dr   2263.5   3534   0.64 kg / ms

( Ns / m 2 )

Flow through each tube given by Ha gen-Poiseuille equation Q

 p

4  d

L 128

There are 4 of these so total flow is Q4

 p

4

 d

L 128

  p

4 4 (  (5 / 1600)

0.13  128  0.2

601  10 10  p3.60

Force = pressure  area  area F

2   0.08  2  5 / 1600    45  p     4    2     2 

. p  9007206 N / m2 Q  3.24  10 6 m 3 / s Flow up through piston = flow displaced by moving piston Q = Av piston 3.2410-6 = 0.042v piston

Examples: Answers


51

Examples: Answers


52


Examples: Answers

Dimensional analysis

CIVE1400: Fluid Mechanics And the second group

8.1 A stationary sphere in water moving at a velocity of 1.6m/s experiences a drag of 4N. Another sphere of twice the diameter is placed in a wind tunnel. Find the velocity of the air and the drag which will give dynamically similar conditions. The ratio of kinematic viscosities of air and water is 13, and the density of air 1.28 kg/m3. [10.4m/s 0.865N]

M 0 L0 T 0

water 1.6m/s

uair

4N

Dair

 



13

1000 kg/m3

1.28 kg/m3

d

d

2d

The Reynolds number = Re 





0 = c2

0 = -a2 - 1 a2 = -1

2

 u 1d 1 0 



ud

So the physical situation is described by this function of nondimensional nondimensional numbers,

 D    1 ,  2     2 2 ,  0   u d ud 

  

For dynamic similarity these non-dimensional numbers numbers are the same for the both the sphere in wa ter and in the wind tunnel i.e.

ud 

1air  2 air

  1water   2 water

For 1

  u, d , ,D ,   0

 D   D       u 2 d 2  air   u 2 d 2  water

  1 , 2   0 1 2

 u d  D  u a d b  c  2

c1

2

Dair 1.28  10.4 2

2

0

M]

0

a1

  LT 1   Lb  ML3  1

c1

             ud air ud w ater

2

MLT

0 = c1 + 1

13

c1 = -1

uair  2d

L]

4 1000  1.6 2

 d 2

For 2

(note D is a force with dimensions MLT-2) 0

 (2 d ) 2



. Dair  0865 N

As each  group is dimensionless then considering the dimensions, for the first group, 1:

M LT

L2 T 1

-2 = a2 + b2

From Buckinghams  theorem we have m-n = 5 - 3 = 2 non-dimensional groups.

b1

3 c2

 ML 

0 = a2 + b2 - 3c2 + 2

Choose the three recurring (governing) variables; u, d, 

a1

2

b2 = -1

u Drag

 ud

a2

M]

air

Kinematic viscosity is dynamic viscosity over density =

2 :

  LT 1   Lb

L]

T]

Draw up the table of values you have for each variable: variable

Examples: Answers



 1.6  d

uair  10.4 m / s

0 = a1 + b1 - 3c1 + 1 -4 = a1 + b1

T]

0 = -a1 - 2 a1 = - 2 b1 = -2 1

 u 2 d 2  1 D 

D 2 2  u d

Examples: Answers


53

Examples: Answers


54


Examples: Answers

8.2 Explain briefly the use of the Reynolds number in the interpretation of tests on the flow of liquid in pipes. Water flows through a 2cm diameter pipe at 1.6m/s. Calculate the Reynolds number and find also the velocity required to give the same Reynolds number when the pipe is transporting air. Obtain the ratio of pressure drops in the same length of pipe for both cases. For the water the kinematic viscosity was 1.3110-6 m2/s and the density was 1000 kg/m3. For air those quantities were 15.110-6 m2/s and 1.19kg/m3. [24427, 18.4m/s, 0.157]

CIVE1400: Fluid Mechanics M]

0 = c1

L]

0 = a1 + b1 - 3c1 + 2 -2 = a1 + b1

T]

b1 = -1 1

water

air

u

1.6m/s

uair

p

pwater

pair

  

1000 kg/m3

1.19kg/m3

 

 



m s



1.28 kg/m3

0.02m

0.02m

1 1

 u d 

0

 



ud

And the second group

2 :

(note p is a pressure (force/area) with dimensions ML-1T-2)

m s

1000 kg/m3

d

0 = -a1 - 1 a1 = -1

Draw up the table of values you have for each variable: variable

Examples: Answers

0

0

M LT M]

0

a1

  LT 1   Lb  ML 3  1

c1

MT

 2 1

L

0 = c2 + 1

c2 = -1 Kinematic viscosity is dynamic viscosity over density = The Reynolds number number = Re 

 ud 



  

L]

0 = a2 + b2 - 3c2 - 1 -2 = a2 + b2

ud 

T]

0 = -a2 - 2

Reynolds number when carrying water: Re water 

ud 



1.6  0.02 1.31  10  6

a2 = - 2 b2 = 0

 24427 2

To calculate Reair we know, Re water



 Re air uair 0.02

24427 

 u 2  1 p p  u 2


15  10 6 . m / s uair  1844

  p  ,  0  ud  u 2 

 1 ,  2    

To obtain the ratio of pressure drops we must obtain an expression for the pressure drop in terms of governing variables. Choose the three recurring recurring (governing) variables; u, d, 

For dynamic similarity these non-dimensional non-dimensional numbers are the same for the both water and a ir in the pipe. 1air


 2 air

  u, d , ,  , p   0

We are interested in the relationship involving the pressure i.e. 2

  1 ,  2   0 1 2


 u a d b  c   u a d b  c p 1

1

2

2

1

2

As each  group is dimensionless then considering the dimensions, for the first group, 1: 0

0

M LT

0

a1

  LT 1   Lb  ML3 

Examples: Answers

1

c1

2

LT

1


55

Examples: Answers


56


Examples: Answers

 p   p       u 2  air   u 2  water pwater pair



1

Q

1000  1.6 2 1.19  18.44 44 2



1 0158 .

 Q

(note p is a pressure (force/area) with dimensions ML-1T-2) M 0 L0 T 0 M]

LT L



ML-1T-1

T]

0 = -a2 - 2 a2 = - 2 b2 = 4 2

We are told from the question that there are 5 variables involved in the problem: d, p, ,  and Q.

 Q2 d 4  1 p

 d  d 4 p  ,  0  Q   Q 2 

  1 ,  2    

From Buckinghams  theorem we have m-n = 5 - 3 = 2 non-dimensional groups.   Q, d ,  ,  , p   0

or

 1 ,  2   0

d 

 Q a d b  c   Q a d b  c p 1

1

2

2

Q 

1

a1

 d  1/2 p1 /2   d 2 p1 / 2         

The question wants us to show : Q  f 

  L3T 1   Lb  ML3  1

c1

1

1

ML T

Take the reciprocal of square root of 2:

0 = c1 + 1

1  2

1/ 2



 Q d 2 p 1/2

  2 a ,

Convert 1 by multiplying by this number

c1 = -1

1/ 2

0 = 3a1 + b1 - 3c1 - 1

1a

-2 = 3a1 + b1 T]

 d 4 p    1  2    Q 

2


L]

d4 p  Q 2

So the physical situation is described by this function of non-dimensional numbers, numbers,

Choose the three recurring recurring (governing) variables; Q, d, d, 

M]

MT  2 L1

0 = 3a2 + b2 - 3c2 - 1



0

c1

-2 = 3a2 + b2

i.e. Re is dimensionless.

0

1

0 = c2 + 1

L]

Re = ML-3 LT-1L(ML-1T-1)-1 = ML-3 LT-1 L M-1LT = 1

0

a1

  L3T 1   Lb  ML3 

c2 = -1

-1

d

2 :


327  6.327

ML-3

u

M LT

d  

2

The dimensions of these following variables are

2

Examples: Answers

1

 air uair

Show that Reynold number, ud/, is non-dimensional. If the discharge Q through an orifice is a function of the diameter d, the pressure difference p, the density , and the viscosity , show that Q = Cp 1/2d2/1/2 where C is some function of the non-dimensional group (d1/2d1/2/). Draw up the table of values you have for each variable:

1

1 1

d 



 w atat erer uw2 atat erer






0 = -a1 - 1

  1 2 a 

d   Q Q  d 2 p 1/ 2



 d  1/ 2 p 1/ 2

then we can say

a1 = -1 b1 = 1

Examples: Answers


57

Examples: Answers


58


Examples: Answers

 p  d d p , 1/2  1 /  1a ,  2 a        1/ 2

1/2

2

  0 

0

M]

 p 1/2  1/2 d  d 2 p1/ 2      1/2

 = 1.12kg/m3  = 1000kg/m3



3 c1

 L   ML b1



1

1

ML T

0 = a1 + b1 - 3c1 - 1 -2 = a1 + b1

T]

0 = -a1 - 1 a1 = -1 b1 = -1

 = 1610-6 kg/ms  = 810-4 kg/ms

1

 u 1d 1  1 

[11.55m/s, 0.057] Draw up the table of values you have for each variable: water

u

uwater

33m/s

F

Fwater

Fair

 

1000 kg/m3

1.12kg/m3





air

kgms

0.16m



0

M]

0

2 : a2

  LT 1   Lb

2

3 c2

 ML 

1

2

ML T

0 = c2 + 1

c2 = -1

kg/ms

L]

0 = a2 + b2 - 3c2 - 1 -3 = a2 + b2

T]

  

0 = -a2 - 2 a2 = - 2

ud



0

M LT

1m

Kinematic viscosity is dynamic viscosity over density =  ud






  ud

i.e. the (inverse of) Reynolds number

variable

The Reynolds number number = Re 

  LT

Examples: Answers

1 a1

0 = c1 + 1

L]

8.4 A cylinder 0.16m in diameter is to be mounted in a stream of water in order to estimate the force on a tall chimney of 1m diameter which is subject to wind of 33m/s. Calculate (A) the speed of the stream necessary to give dynamic similarity between the model and chimney, (b) the ratio of forces.

d

0

c1 = -1

Q   

Model:

0

M LT

or

Chimney:


1 /2

b2 = -1

 2

For dynamic similarity:

 u 2 d 1 1 F 

 Re air 1000uwater 0.16 1.12  33  1  8  10  4 16  10 6 uwater  1155 . m / s Re water

F 2

u d 


  F  ,   1 ,  2       ud  du 2 

To obtain the ratio of forces we must obtain an expression for the force in terms of governing variables.

0

Choose the three recurring recurring (governing) variables; u, d,  F,  For dynamic similarity these non-dimensional non-dimensional numbers are the same for the both water and a ir in the pipe.

From Buckinghams  theorem we have m-n = 5 - 3 = 2 non-dimensional groups.   u, d ,  ,  , F   0

1air

 1 ,  2   0

 2 air

1 2

To find the ratio of forces for the differ ent fluids use 2

 u a d b  c   u a d b  c F 1

1

2

2


1

2


Examples: Answers


59

Examples: Answers


60

CIVE1400: Fluid Mechanics  2 air

Examples: Answers

  2 water

M 0 L0 T 0

 F   F       u 2 d  air   u 2 d  water F water



M]

1000  11.55

2

 2 

2

  R  ,  1 ,  2       ur  ru 2 

R  ru 2

L]

2

1

2

1 a1







  ur     

f 

or

R   2

  ur   f     

1

Multiply the LHS by the square of the RHS: (i.e. 2(1/12) )

2

R

3 c1

 L   ML b1



1

 ru

1

ML T

2



2 2 2  u r

 2



R   2

So

0 = c1 + 1

R 

c1 = -1

 2

0 = a1 + b1 - 3c1 - 1

T]

     1     ur 

he question asks us to show R 

 u a r b  c   u a r b  c R

  LT

0

or

As each  group is dimensionless then considering the dimensions, for the first group, 1: M LT

R u 2 r 


  1 ,  2   0

M]

 u 2 r 1  1 R 

  u, r , , , R  0

0

0 = -a2 - 2

b2 = -1

  ur  f     


2



a2 = - 2

Choose the three recurring (governing) variables; u, r,  R, 

1



ML 1T 2

0 = a2 + b2 - 3c2 - 1

T]

Hence show that if at very low velocities the re sistance R is proportional to the velocity u, then R = k ru where k is a dimensionless constant. A fine granular material of specific gravity 2.5 is in uniform suspension in still water of depth 3.3m. Regarding the particles as spheres of diameter diameter 0.002cm find how long it will take for the water to clear. Take k=6 and =0.0013 kg/ms. [218mins 39.3sec]

0

3 c2

 ML 

-3 = a2 + b2

R 

0

2

0 = c2 + 1

L]

1 .  0057  0.16

8.5 If the resistance to motion, R, of a sphere through a fluid is a function of the density  and viscosity  of the fluid, and the radius r and velocity u of the sphere, show that R is given by

1

a2

  LT 1   Lb

c2 = -1

1.12  332

2

Examples: Answers

2 :


 F   F       u 2 d  air   u 2 d  water

F air


  ur   f     

The question tells us that R is proportional to u so the function f function f must must be a constant, k

-2 = a1 + b1

R 

0 = -a1 - 1



2

 ur

 k



R   kru

a1 = -1 b1 = -1 1

 u 1r 1 1  

The water will clear when the particle moving from the water surface reaches the bottom. At terminal velocity there is no acceleration - the force R = mg - upthrust.

  ur

From the question:

 = 2.5

i.e. the (inverse of) Reynolds number

Examples: Answers


61

Examples: Answers

so

 = 2500kg/m3

 = 0.0013 kg/ms

k = 6


62

CIVE1400: Fluid Mechanics r = 0.00001m mg 

4 3

3

00001  0.00

Examples: Answers depth = 3.3m

81  2500  1000   9.81

 6.16  10 11 kru  0.0013  6  0.00001u  6 .16  10

11

4

u  2.52  10 m / s

t 

Examples: Answers

3.3 2.52  10 4

 218 min 39.3 sec


63

Fluid Mechanic Answer

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