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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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Although it is usual to consider water as incompressible and of a uniform density, this is clearly not true. It would, for example, be impossible to control the depth of a submersible vessel unless the density of water increased with depth.

Here we consider the impact of compressibility for the purpose of computing water hammer and the subsequent pipe expansion.

Pressure rise following Instantaneous Closure

The pressure rise at the valve $=\rho\; C\;v_0$ where $\rho$ is the density and $C$ the velocity of Sound for the fluid (Water)

Let the initial pipe velocity be $v_0$

At the instant that the valve is closed a pressure wave, moving at the speed of sound $C$ sets off up the pipe bringing the water to rest.

Consider the instant shown above, $t$ secs after valve closure. At this moment the length of the column brought to rest is $x$ .

$\therefore\;\;\;C = \frac{x}{t}$

(1)

By Newton'\b{s second law}, the pressure force equals the change of momentum per second.

i.e. $pa = \frac{wax}{g}\times \frac{v_0}{t}$

Pressure rise at valve $p = \frac{w}{g}\frac{x}{t}\;v_0;= \rho\,C\,v_0$ where $C = \frac{x}{T}$

Notes:

The pressure at the valve remains at $p$ above normal from the instant the valve is closed until the pressure wave is reflected from the open end of the pipe as a wave of normal pressure and velocity $(= - v_0)$ reaching the valve at time $\displaystyle\frac{2L}{C}$ .
A wave of reduced pressure, $p$ below normal will then set off up the pipe.
The time $\displaystyle\frac{2L}{C}$ is the period of the pipe.
The above proof still applies if the valve is closed in a time of closure $T\leq\;\displaystyle\frac{2L}{C}$ i.e. the valve is closed before the pressure wave returns to the valve.

Pressure rise for Instantaneous Partial Reduction of Flow.

The Pressure rise $\delta\,p = \rho\,C\,\delta \,v$ Where $\delta\;v$ is the change of velocity.
The increase in Pressure head $\delta\,h = \displaystyle\frac{\delta\,p}{w} = \displaystyle\frac{\delta\,p}{\rho} = \displaystyle\frac{C}{g}\delta\,v$

The initial velocity $v_0$ is reduced to $v_1$ by the sudden partial valve closure. $t$ seconds later the pressure wave will have traveled a distance $x$ up the pipe as before $C = \frac{x}{t}$

Applying Newton'\b{s second Law}:

$p\:a = \frac{w\,a\,x}{g}\;\left(\frac{v_0 - v_1}{t} \right)$

(2)

Pressure $p = \frac{w}{g}\:\frac{x}{t}\,(v_0 - v_1) = \rho\,C(v_0 - v_1)$

Which is commonly written as:

Pressure rise $\delta\,p = \rho\,C\delta\,v$ Where $\delta\;v$ is the change of velocity.

And:

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

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:

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

The initial flow through the valve $= Q_0 a_p v_0=C_d\:a_{v_0}\:\sqrt[]{2\,g\,h_0}$

After partial closure:

$Q_1 = a_p\;v_1 = C_d\;a_v_1\;\sqrt[]{2g(h_0 - \Delta\,h)}$

(3)

Thus: $\frac{Q_1}{Q_0} = \frac{v_1}{v_0} = \frac{a_v_1}{a_v_0}\sqrt[]{\frac{h_0 + \Delta\,h}{h_0}}$

But: $v_1 = v_0 - \Delta\,h$

$\therefore\;\;\;\frac{v_0 - \Delta\,v}{v_0} = \frac{a_v_1}{a_v_0}\sqrt[]{1 + \frac{\Delta\,h}{h_0}}$

(4)

But: $\Delta\,h = \frac{C}{g}\:\Delta\,v$

$\therefore\;\;\;1 - \frac{\Delta\,v}{v_0} = \frac{a_v_1}{a_v_0}\sqrt[]{1\;+\frac{C\;\Delta\,v}{g\:h_0}}$

(5)

For known initial conditions and change in the valve area this equation can be solved as a quadratic in $\delta\;v$

Hence: $\delta\,h = \frac{C}{g}\:\delta\,v$

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

Bulk modulus $K$ = Increase of Pressure/Volumetric strain $= dp\;\div\frac{dv}{v}$

Assume a thin cylinder constrained longitudinally and subject to an internal

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

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

(16)

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

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

(17)

Hoop stress/Hoop strain = Young's Modulus

Hoop strain

$=\frac{f_H}{E} = \frac{\pi(d_1\:-\:d)}{\pi\,d} = \frac{d_1 - d}{d}$

(6)

also: ${p\,d\,l} = {2\;X\;f_H\;X}\;t\,l$

or: $f_H = \frac{p\,d}{2\,t}$ where $t$ is the thickness of the pipe wall

$\therefore\;\;\;\frac{p\,d}{2\,t\,E} = \frac{d_1 - d}{d}$

(7)

Consider the situation in the pipe at a time $T$ sec.after an instantaneous valve closure. The pressure wave will have reached a point $P$ at a distance $x$ from the valve where $C = \displaystyle\frac{x}{T}$ The water in the length $x$ is now at rest and at a pressure $p$ above normal.

During time $T$ after closure of the valve a volume $v_0\, a \,T$ has passed point $P$ and this volume must be accommodated in addition to the initial volume $(a x)$ . This is possible due to the compression of the water and the expansion of the pipe.

$v_0\,a\,T = \frac{p}{k}\;X\;a\:x + \frac{\pi}{4}\left(d_1^2 - D^2 \right)x$

(8)

$= \frac{p}{K}\;X\;ax + \frac{\pi}{4}\:(d_1 + d)(d_1 - d)$

(9)

Assuming that $2d\approx(d_1+d)$ and substituting from equation 29

Then: $v_0\,a\,T = \frac{p}{k}\:a\,x + \frac{\pi}{4}\:2\,d\:\frac{p\:d^2}{2\,t\,E}$

$\therefore\;\;\;\;v_0\,T = p\,x\left(\frac{1}{K} + \frac{d}{t\,E} \right)$

(10)

Combining the above equation with $p = \rho\,v_0\,C$

From the above two equations we can :

Eliminate $C$ by putting $\;p = \rho\,v_0X\,\frac{v_0}{p\left(\displaystyle\frac{1}{k}\:+\:\displaystyle\frac{d}{t\,E} \right)}$

$\therefore\;\;\;p = v_0\;\sqrt{\frac{\rho}{\left(\displaystyle\frac{1}{K}\:+\:\displaystyle\frac{d}{t\,E} \right)}}$

(11)

And the increase in pressure head $= \frac{p}{w}$ or: $\frac{p}{\rho\,g}$

To eliminate $p$ :

$C = \frac{v_0}{\left(\displaystyle\frac{1}{K}\:+\:\displaystyle\frac{d}{t\,E} \right)\rho\:v_0\:C}$

(12)

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

(13)

By ignoring Pipe Expansion Equations (35 ) (38 ) (41) and (46 ) can be simplified so that:

$C = \frac{v_0 K}{p}$

(14)

$p = v_0\;\sqrt[]{K \rho}$ And The head rise $=\frac{v_0}{g}\sqrt{\frac{K}{\rho}}$

And: $C = \sqrt[]{\frac{K}{\rho}}$

Thus the equivalent bulk modulus allowing for pipe expansion is given by:

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

(15)

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 $h_0$ .

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

(18)

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

from equation (46)

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

and from equation (16)

also: $h_1 - h_0 = \Delta\,h = \frac{C}{g}\,\Delta = C(v_0 - v_1)$

Equations (54) (55) and (56) can be solved for $\delta\;v$ and hence $v_1$ and $\delta\;h$ and hence $h_1$ .

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 $R$ .

By continuity: $AV_1=av_2$

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

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

(19)

Similarly for the pressure wave in the small pipe:

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

(20)

Note: $C_*$ is the velocity of sound in the small pipe.

Adding equations ( 59 ) and(60 )

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

(21)

Equate to equation ( 56 )

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

(22)

substitute from ( ) and ( ) for V_0 & V_1

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

(23)

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 $C = C_*$

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

(24)

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

(25)

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

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

(26)

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

(27)

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.

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Example 1 [imperial]

Problem

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.

Workings

a) Using equation (39)