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This analysis of water hammer allows for the compressibility of water and the expansion of the pipe

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Introduction

In the other submission on water hammer it was assumed that water is incompressible and whilst this assumption may be adequate in certain situations, it is clearly not true. It would be, for example impossible to control the depth of a submersible vessel unless the density of water increased with depth.

In this submission on Water Hammer allowance is made for the compressibility of water and for the fact that, in general pipes expand due to internal pressure

Pressure rise following Instantaneous Closure

$\text{The pressure rise at the valve}=\rho\; C\;v_0$

(5)

Where $\rho$ is the density and C the velocity of Sound for the fluid (Water)

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Pressure rise for Instantaneous Partial Reduction of Flow.

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$\text{The Pressure rise}\; \delta\,p = \rho\,C\,\delta \,v\;\text{where}\;\delta\,v\;\text{ is the change of velocity.}$

(10)

The increase in Pressure head $\delta\,h = \frac{\delta\,p}{w} = \frac{\delta\,p}{\rho} = \frac{C}{g}\:\delta\,v$

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Instantaneous Partial Closure of Valve

If the reduction in area of the valve is known rather than the reduction of flow or pipe velocity,then the pipe may be treated as a nozzle with a coefficient of discharge of $C_d$

From the Initial Conditions and from the change in Valve Area

$\text{The change of pressure head}\;\;\;\delta\,h = \frac{C}{g}\:\delta\,v$

(20)

For the proof of the above equation please click on the red button

The relationship between pressure rise, speed of sound and bulk modulus allowing for lateral expansion of the the pipe

Notes

$\text{Bulk modulus K}\; = \frac{\text{Increase of Pressure}}{\text{Volumetric strain}} = dp\;\div\frac{dv}{v}$
(39)

Assume a thin cylinder constrained longitudinally and subject to an internal

pressure rise p causing lateral expansion from diameter $d\; to\; d_1$

$C = \sqrt{\frac{1}{\rho\left(\frac{1}{K}\:+\:\frac{d}{t\,E} \right)}}$

(40)

And the equivalent Bulk Modulus $(K')$ allowing for Pipe Expansion is given by :-

$\frac{1}{K_1} = \frac{1}{K}\:+\:\frac{d}{t\,E}$

(41)

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Pipe with change of section

At the initial steady flow conditions it is assumed that friction; velocity head and contraction losses can be neglected, Therefore the pressure throughout both pipes is $h_0$

$\therefore\;\;\;Q_0 = A\,V_0\:=\:a\,v_0\:=\:Cd\,a_v_0\:\sqrt[]{2g\,h_0}$

(42)

Consider a sudden partial closure.

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The above diagram shows some instant when the pressure wave is between V & P

from equation (46 )

$\text{Flow through valve}\;=Q_1 = a\,v_1 = Cd\,a_{v1}\:\sqrt[]{2g\,h_1}$

(43)

and from equation ( 16 )

$\text{Also}\;\;\;h_1 - h_0 = \Delta\,h = \frac{C}{g}\,\Delta = C(v_0 - v_1)$

(44)

Equations ( 54 ) ( 55 ) and ( 56 ) can be solved for $\delta\,v\;\;\;\text{And hence}\:v_1\;\;\;\text{And}\;\;\; \delta\,h\:and\;hence\;h_1$

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Now consider an instant in time after the pressure wave has passed through the junction and a reflected wave has set off up the pipe from P to R.

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$\text{By continuity}\;\; A V_1=av_2$

(45)

Using equation ( 16 ) for the pressure wave in the large pipe.

$h_2 - h_0 = \frac{C}{g}\:(V_0 - V_1)$

(46)

similarly for the pressure wave in the small pipe

$h_1 - h_0 = \frac{C_*}{g}\;(v_1 - v_2)$

(47)

Note C_* is the velocity of sound in the small pipe

Adding equations ( 59 ) and(60 )

$h_1 - h_0 = \frac{C}{g}(V_0 - V_1) + \frac{C_*}{g}}\;(v_1 - v_2)$

(48)

Equate to equation ( 56 )

$\frac{C}{g}(V_0 - V_1) + \frac{C_*}{g}(v_1 - v_2) = \frac{C_*}{g}(v_0 - v_1)}$

(49)

substitute from ( ) and ( ) for V_0 & V_1

$\frac{C}{g}\left(\frac{a}{A}(v_0 - v_2) \right) = \frac{C_*}{g}(v_0 - v_1) - \frac{C_*}{g}(v_0 - v_2)$

(50)

Using the value for v_1 found from equations ( 14 ) ( 27 ) and( ) this equation can be solved for $v_2$ and hence $V_1$ and $h_2$

Notes

If the velocity of sound is the same for both pipes then $C = C_*$

$\frac{a}{A}(v_0 - v_2) = v_0 - 2v_1 + v_2$

(51)

$\therefore\;\;\;\left(1 + \frac{a}{A} \right) = 2v_1 - v_0\left(1 - \frac{a}{A} \right)$

(52)

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Putting A = infinity i.e.the pipe entering the reservoir

$v_2 = 2v_1 - v_0 = 2v_1 - 2v_0 + v_0$

(53)

$\therefore\;\;\;v_2 = v_0 - 2(v_0 - v_1) = v_0 - 2\,\Delta\,v$

(54)

Where $\delta\,v$ is the initial reduction of the pipe velocity due to a partial closure of the valve. i.e. When the pressure wave reaches the reservoir entrance the velocity is reduce still further by an amount equal to the initial velocity reduction at the valve.

Worked Examples

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Example 1

A pipe carrying water has a valve at the discharge end. The bulk modulus, K, is $2\times 10^5\;lb./in.^2$ a) If the initial velocity in a pipeline is $V_0$ show that an instantaneous closure of the valve will result in a pressure rise at the valve of $7470\;v_0lb/sq,ft.$

b) Given that the initial velocity is 12 ft/sec. and the pressure head in the pipe is 600 ft, find the increase in pressure head which results when the valve is instantaneously closed by 1/5 th.

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Example 2

Show that the pressure rise in a pipe-line caused by the sudden displacement of a valve is given by $dp=\frac{w\,C\;dv}{g}$ where $dv$ is the decrease in velocity of the water, $w=\text{The water density}$ and $C$ is the velocity of sound transmission in the pipe concerned.

A valve at the outlet end of a pipe-line 4000 ft. long through which water flows at $8\;ft./sec.$ is closed in 5 seconds in such a way that the pressure on the upstream side of the valve increases at a uniform rate. Assuming at $C=4800\;ft./sec.$ find the increase of pressure caused by the valve closure.

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Consider a pound of water at a pressure p which is moving with a velocity v. If K is the bulk modulus then:-

$\text{The kinetic energy}=\frac{v^2}{2g}$

(67)

$\text{And the Strain Energy}=\frac{p^2}{2Kw}$

(68)

If the velocity v of the water changes by dv and there is a corresponding increase in pressure of dp the, neglecting pipe elasticity, the change in Kinetic Energy must equal the change in Strain Energy.

$i.e.\;\;\;\;\;\frac{(v+dv)^2}{2g}- \frac{v^2}{2g}=\frac{(p+dp)^2}{2Kw}-\frac{p^2}{2Kw}$

(69)

$\frac{2vdv}{2g}+\frac{dv^2}{2g}=\frac{2pdp}{2Kw}+\frac{dp^2}{2Kw}$

(70)

But the rate of change of Kinetic Energy = the rate of change of Strain Energy

$\therefore \;\;\;\;\frac{2v}{2g}\times \frac{dv}{dt}=\frac{2p}{2Kw}\times \frac{dp}{dt}$

(71)

Combing the above two equations:-

$K=\frac{dp^2}{dv^2}\times \frac{g}{w}$

(72)

From Equation (40)

$C=\sqrt{\frac{K}{\rho}}=\sqrt{\frac{gK}{w}}$

(73)

$\therefore\;\;\;\;\;K=\frac{wC^2}{g}$

(74)

$\therefore \;\;\;\;dp=\frac{w\;C}{g}dv$

(75)

If the valve is closed gradually the water has a time to decelerate and hence:-

$h=-\frac{L}{g}\times \frac{dv}{dt}$

(76)

$\therefore \;\;\;\;\frac{dh}{dt}=-\frac{L}{g}\times \frac{d^2v}{dt^2}$

(77)

$But\;\;\;\;\frac{dh}{dt}\;\;\;\text{is constant}\;=\frac{h_1}{5}\;\;\;\;\text{Where }\;h_1=h_{max}$

(78)

From equations (77) And (78)

$\frac{h_1}{5}=-\frac{L}{g}\times \frac{d^2v}{dt^2}$

(79)

Integrating

$\frac{h_1\;t}{5}=-\frac{L}{g}\times \frac{dv}{dt}+A$

(80)

Integrating again

$\frac{h_1\;t^2}{10}=-\frac{L}{g}\times v+At+B$

(81)

$\text{When}\;\; t=0,\;\;\;v=8\;\;\;\text{And when}\;\;t=5,\;\;\;v=0$

(82)

Substituting into equation (81)

$B=\frac{8L}{g}\;\;\;\;\;\text{And}\;\;\;\;\frac{25}{10}h_1=5A+\frac{8L}{g}$

(83)

$\therefore \;\;\;\;A=\frac{1}{5}\left ( \frac{25}{10}h_1-\frac{8L}{g} \right )$

(84)

From equations (76) (79) and (84)

$\frac{h_1}{5}\;t_1=h+\frac{h_1}{2}-\frac{8}{5}\times \frac{L}{g}$

(85)

$\text{But}\;\;\;\;h=h_1\;\;\;\;\text{When}\;\;\;\;t=5$

(86)

$\therefore \;\;\;\;h_1=\frac{16}{5}\times \frac{L}{g}=\frac{16}{5}\times \frac{4000}{32.2}=398\;ft.$

(87)

Example 3

Develop a formula for the rise of pressure in a pipe through which water is flowing at a constant rate, due to a sudden closing of the valve, allowing both for the expansion of the pipe and the compressibility of the water.

A cast iron pipe of 6 in. bore and walls $\frac{5}{8}$ in. thick, contains water flowing uniformly. Calculate the maximum permissible flow in $ft.^3/sec.$ if a sudden stoppage is not to stress the pipe to more than $3\;tons/in^2$

$\text{K for Water}\;= 3\times 10^5\;lb./in.^2$

(88)

$\text{E for Cast Iron}\;=18\times 10^6\;lb./in^2$

(89)

(B.Sc. Part 2)

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Let:-

$p_1\;$ equal the rise in pressure due to the valve closure.

v be the initial velocity of flow before valve closure.

T be the thickness of the pipe wall.

f be the circumferential stress in the pipe under an internal pressure of $p_1$

y be the length of pipe needed to hold 1 lb. of water.

Then:-

$f=\frac{p_1D}{2T}$

(90)

$\text{And}\;\;\;\;y=\frac{4}{w\pi\;D^2}$

(91)

$\text{The strain energy in the pipe}\;=\frac{f^2}{2E}\times \text{volume of pipe steel}$

(92)

$=\frac{f^2}{2E}\times \pi DTy$

(93)

Substituting for f and y

$\text{The strain energy}\;=\frac{\pi^2 d^2}{4T^2}\times \frac{1}{2E}\times \pi DT\times \frac{4}{w\pi D^2}=\frac{\pi^2 D }{2TEw}$

(94)

$\text{The strain energy in the water}=\frac{{p_{1}}^{2}}{2Kw}$

(95)

$\text{The Kinetic energy of the water}=\frac{v^2}{2g}$

(96)

When the valve is shut the Kinetic Energy of the water is converted into the Strain Energy of the Water and the Strain Energy of the Pipe. i.e.

$\frac{v^2}{2g}=\frac{{p_{1}}^{2}}{2Kw}+\frac{{p_{1}}^{2}D}{2TEw}$

(97)

$\therefore \;\;\;\;p_1=v\;\;\sqrt{\frac{w}{g\left ( \frac{1}{k}+\frac{D}{TE} \right )}}$

(98)

When the water pressure in the pipe rises by $p_1$ the stress in the pipe, f, $=\frac{p_1D}{2t}$ and substituting in the values given in the question:-

$3=\frac{p_1\times 6}{2\times 0.625}$

(99)

$\therefore \;\;\;\;p_1=0.625\;ton/in.^2\;\;=0.625\times 2240\times 12^2=201,600\;lb./ft.^2$

(100)

Substituting this value for $p_1$ in Equation (98)

$201,600=\sqrt{\frac{62.4}{32.2\left ( \frac{1}{3\times 10^5\times 12^2}+\frac{6}{0.625\times 18\times 10^6\times 12^2} \right )}}$

(101)

From which:-

$v=23.7\;ft./sec.$

(102)

The maximum flow is given by cross sectional area of the pipe X the velocity of flow

$\therefore \;\;\;\;Q_{max}=\frac{\pi}{4}\times \left ( \frac{6}{12} \right )^2\times 23.7=4.65\;ft.^3/sec.$

(103)

Example 4

Prove that if a valve at the end of a pipe-line conveying water is suddenly but partially closed so as to change the speed in the pipe by $\Delta v\;ft./sec.$ the increase in the head $\Delta H\;ft.$ is given by $\Delta H=\frac{a}{g}\times \Delta v$ where a is the speed of sound in the pipe.

A valve at the end of a pipe-line of length $L = 2500\;ft.$ is closed in ten equal steps each of $\frac{2L}{a}$ where $a=4000\;ft./sec.$ The initial head at the valve, which discharges to atmosphere, is $400\;ft.$ and the initial speed in the pipe is $12\;ft./sec.$