flow through pipes_4

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1

1st trial

LOOP ABC

Line r Q rQ2 2rQ

AB 6 15 -1350 180

CB 4 30 3600 240

CA 8 5 -200 80

∑ = 2050 500

Correction for loop ABC

∆Q = -4.1 ≈ - 4

LOOP ADC

Line r Q rQ2 2rQ

CD 6 5 -150 60

AD 7 10 700 140

CA 8 5 200 80

∑ = 750 280


∆Q = -2.67857 ≈ - 3

20

45 40

A D10

5

30

515

15

B C

20

45 40

A D7

9-3 = 6

26

819

15

B C

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2nd trial

LOOP ABC

Line r Q rQ2 2rQ

AB 6 19 -2166 228

CB 4 26 2704 208

CA 8 6 -288 96

∑ = 250 532


∆Q = -0.46992≈ - 0.5

LOOP ADC

Line r Q rQ2 2rQ

CD 6 8 -384 96

AD 7 7 343 98

CA 8 6 288 96

∑ = 247 290


∆Q = -0.85172≈ - 0.9

20

45 40

A D7

9-3 = 6

26

819

15

B C

20

45 40

A D6.1

6.5-0.9 = 5.6

25.5

8.919.5

15

B C8/19/201564

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2

3rd trial

LOOP ABC

Line r Q rQ2 2rQ

AB 6 19.5 -2281.5 234

CB 4 25.5 2601 204

CA 8 5.6 -250.88 89.6

∑ = 68.62 527.6


∆Q = -0.13006 ≈ - 0.13

LOOP ADC

Line r Q rQ2 2rQ

CD 6 8.9 -475.26 106.8

AD 7 6.1 260.47 85.4

CA 8 5.6 250.88 89.6

∑ = 36.09 281.8


∆Q = -0.12807 ≈ - 0.13

20

45 40

A D6.1

6.5-0.9 = 5.6

25.5

8.919.5

15

B C

20

45 40

A D5.97

5.73-0.13 = 5.6

25.37

9.0319.63

15

B C

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Network Analysis

� Find the flows in the loop given the inflows and outflows.

� The pipes are all 25 cm cast iron (ε=0.26 mm).

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

200 m

100 m

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Network Analysis

� Assign a flow to each pipe link

� Flow into each junction must equal flow out of the junction

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

0.320.00

0.10

0.04

arbitrary

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Network Analysis

� Calculate the head loss in each pipe

f=0.02 for Re>200000 hf =

8 fL

gD5π 2

Q 2

fh kQ Q=

339)25.0)(8.9(

)200)(02.0(8

251 =

=

πk

k1,k3=339k2,k4=169

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

1

4 2

3

hf1 = 34.7m

hf2

= 0.222m

hf3 = −3.39m

hf4

= −0.00m

hfii=1

4

∑ = 31.53m

Sign convention +CW

2

5

s

m

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4

Numeric Solver

� Set up a spreadsheet as shown below.

� the numbers in bold were entered, the other cells are calculations

� initially ∆Q is 0

� use “solver” to set the sum of the head loss to 0 by changing ∆Q

� the column Q0+ ∆Q contains the correct flows

∆Q 0.000

pipe f L D k Q0 Q0+∆Q hf

P1 0.02 200 0.25 339 0.32 0.320 34.69

P2 0.02 100 0.25 169 0.04 0.040 0.27

P3 0.02 200 0.25 339 -0.1 -0.100 -3.39

P4 0.02 100 0.25 169 0 0.000 0.00

31.575Sum Head Loss 8/19/201569

Solution to Loop Problem

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

0.218

0.102

0.202

0.062

1

4 2

3

Q0+ ∆Q

0.218−0.062−0.202−0.102

Better solution is software with a GUI showing the pipe network.8/19/201570

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5

Water Hammer in pipes

� When the water flowing in a long pipe is suddenly brought to rest by closing the valve or by any similar cause, there will be a sudden rise in pressure due to the momentum of moving water being destroyed. This phenomenon of sudden rise in pressure in the pipe is known as water hammer or hammer blow.

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� Gradual closure of the valve

� Rapid closure of the valve

� Other causes� Pump startup can cause the rapid collapse of a void space that exists

downstream from a starting pump.� Pump power failure can create a rapid change in flow, which causes a

pressure upsurge on the suction side and a pressure downsurge on the discharge side.

Water Hammer Analogy

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6

� Gradual closure of the valve� Axial force available to produce retardation

� Mass of liquid contained in the pipe

� Force bringing retardation = mass × acceleration/retardation =

ip a×

( )wa L

g×

( ) 0w V waLVaL

g t gt

−× =

ii

pwaLV LVp a

gt w gt× = ⇒ =

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Transmission of pressure wave along a pipe

due to instantaneous closure of valve

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� dqc = volume by which the liquid is compressed due to pressure pi

� dqe = additional volume provided by the stretching of the pipe walls under pressure pi

� dD = resulting increase in pipe diameter

� dt = time for pressure wave to traverse the pipe

� T = wall thickness of pipe

� E = Young’s modulus for the material of the pipe

� K = bulk modulus for the liquid

� ft = tensile hoop stress in pipe walls due to pressure pi

� 1/m = Poisson’s ratio

� Instantaneous closure of the valve2

4dQ D dL

π= × ×

11

2

tfdD

D E m

= −

Hoop strain

Hoop stress

2

it

p Df

T=

11

2 2

ip DdD

D TE m

∴ = −

21

12 2

ip DdDTE m

∴ = −

And 2

11

2 2 2

ie

p D DLdq

TE m

π = −

And

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( ) LdD

DLDdDDdqe244

22 πππ

=

−+= LD

K

pdq i

c

2

4

π=

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8

� But

e cdQ dq dq= +2

2 2 11

4 4 2 2 2

i ip p D DLD dL D L

k TE m

π π π × × = + −

1 11

2i

DdL p L

k TE m

= + −

Again dL Vdt=

1 11

2

i

Vdtp

DLk TE m

= + −

But

i

wLVdt

gp=

1 11

2

i

Vp

g D

w k TE m

∴ = + −

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� Velocity of the compression wave

01 1

12

L gV

dt Dwk TE m

= = + −

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9


� Critical time:

� Gradual closure of the valve (T > T0)

� Rapid closure of the valve (T ≤ T0)

i

wVLp

gt=

0

0

2LT

V=

1 11

2

i

Vp

g D

w K TE m

= + −

1i

V wKp V

gg

w K

= =

For rigid pipes

01 1

12

L gV

dt DwK TE m

= = + −

For rigid pipes 01

L g KV

dtwK

= = =ρ

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� As an approximation if Poisson’s ratio is not considered

� INERTIA HEAD,

� If the valve is rapidly closed partially such that the velocity of flow changes from V to V1, then

1i

Vp

g D

w K TE

= +

0ii

p VVVLh

w gdt g= = = Allievi Formula

00

ii i

p VVh p VV

w g

∆= = ⇒ = ρ∆

01

L gV

DdtwK TE

= = +

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10

Problems

� A valve is suddenly closed at the downstream end of a 0.90 m diameter pipeline carrying water in such a manner that the velocity is decreased from 4.0 m/s to 1.0 m/s instantaneously. Estimate the maximum pressure rise at the valve. Assume the pipe to be rigid and K for water = 2.20 × 103 MPa

0ip VV= ρ∆

9

0

2.20 101483.24 m/s

1000

KV

×= = =

ρ

6

0 1000 3 1483.24 4.45 10 Pa= 4.45 M Paip VV= ρ∆ = × × = ×

For rigid pipes,

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Problems

� Calculate the velocity of propagation of a pressure wave in a steel pipe (E = 2.07 × 105 MPa) of 2.5 m diameter carrying kerosene (sp.gr. = 0.80; K = 1.43 × 103 MPa). The pipe thickness is 2 cm. Neglect Poisson’s ratio.

0

9 11

1876 m/s

1 1 2.51000

1.43 10 0.02 2.07 10

L gV

DdtwK TE

= = = = + + × × ×

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11

Problems� A steel pipeline is 30 cm in diameter and has a wall thickness of 3 mm.

The pipe is 1000 m long and conveys a flow of 100 l/s of oil (sp. gr. = 0.82). The static head at the outlet is 160 m of oil. If the working stress of steel is 0.1 kN/mm2, find out whether the pipe will be able to withstand instantaneous closure of the valve. Also calculate the rise of pressure within the pipe if the valve is closed in 3 s. For oil: K = 1 ×109 Pa and for steel: E = 2.14 × 1011 Pa.

0

9 11

1928.38 m/s

1 1 0.30.82 1000

10 .003 2.14 10

gV

DwK TE

= = = + × + × ×

Velocity of the pressure wave,

Velocity of flow,

2

4

0.11.415 m/s

0.3V

π= =

×

CONSIDERING INSTANTANEOUS CLOSURE OF THE VALVE

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Solution contd…

0 0.82 1000 1.415 928.38 1077.2 kPaip VV= ρ∆ = × × × =

Static pressure = 160 × 0.82 × 1000 × 9.81 = 1287 kPa

Inertia pressure developed

Total pressure = 1077.2 + 1287 = 2364.2 kPa

Stress developed in the pipe, 52364.2 0.31.18 10 kPa

2 2 0.003

it

p Df

T

×= = = ×

×

But working stress = 0.1 × 105 kPa < ft

So, time of closure should be more than critical time, 0

0

2 2 10002.15 s

928.38

LT

V

×= = =

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12

Solution contd…

Pressure rise for valve closure in 3 s

1000 0.82 1.415 1000386.8 kPa

3i

wVLp

gt

× × ×= = =

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Practice Problems (Water Hammer)� A 20 cm steel pipe is 1500 m long and conveys 50 L/s of water

with a static head of 200 m at the downstream end of the pipe. If a valve at the downstream end is closed in 3 s, estimate the stress in the pipe wall at the valve. The pipe thickness is 6 mm. [K = 2.2 × 109 Pa for water; E = 2.11 × 1011 Pa for steel] [Ans. ft = 0.059 kN/mm2]

� Water flows at 10 fps in a 400 ft long steel pipe of 8-in diameter with 0.25-in thick walls. Calculate the critical time & maximum pressure rise theoretically caused by instantaneously closing the end valve

� a) completely� b) partially, reducing the velocity to 6 fps.� [Ans. V0 = 4110 ft/s; Complete closure, pi = 79,600 psf; Partial

closure, pi = 31,800 psf.]8/19/201586

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Practice Problems (Hardy Cross)

� Refer TEXT Book MODI and SETH, Illustrative example 11.17

� For the network shown below, the head loss is given by hf = rQ2. The values of r for each pipe, and the discharge into or out of various nodes are shown in the sketch. The discharges are in an arbitrary unit. Obtain the distribution of discharge in the network.

r=4

r=1

r=1

r=5

r=2

A

D

B

C

100 units

20 units

50 units

30 units

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42.1

32.7

20.6

17.3

57.9

A

D

B

C

100 units

20 units

50 units

30 units

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flow through pipes_4

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