Pressure rise for instantaneous closure, elastic pipe

Pressure rise on sudden valve closure is found from the momentum (impulse) equation applied to the pressure wave, and the wave speed from the continuity (energy/volume) equation, including elasticity of the pipe.

Set-up

Pipe of diameter $D$, wall thickness $t$, Young's modulus $E$ of pipe, water density $\rho$, bulk modulus $K$, flow velocity $V$ before closure. The valve closes suddenly at the downstream end. A pressure wave moves upstream with celerity $c$, bringing the water to rest and raising pressure by $\Delta p$. Let the wave advance a distance $c\,\Delta t$ in time $\Delta t$.

 Reservoir  <---- wave front ----  Valve
 ======= (V) -> |  V=0, p+dp  | ====|
                  <-- c dt -->

1. Momentum (impulse) equation

Mass of water brought to rest in time $\Delta t$: $m=\rho A\,c\,\Delta t$. Change in momentum $=\rho A c\,\Delta t\,(0-V)$. The net force is $\Delta p\,A$ (acting against the flow).

$$-\Delta p\,A\,\Delta t=-\rho A c\,\Delta t\,V\;\Rightarrow\;\Delta p=\rho cV$$

2. Wave velocity (strain energy)

Kinetic energy lost $=\tfrac12 m V^2=\tfrac12\rho A c\Delta t\,V^2$.

Strain energy stored in water (volume $A c\Delta t$): $\dfrac{\Delta p^2}{2K}Ac\Delta t$.

Strain energy stored in pipe wall (hoop stress $\sigma=\Delta p D/2t$, strain energy per unit volume $\sigma^2/2E$, wall volume $\pi D t\,c\Delta t$):

$$\frac{1}{2E}\left(\frac{\Delta pD}{2t}\right)^2\pi D t\,c\Delta t=\frac{\Delta p^2}{2}\,\frac{D}{tE}\,A\,c\Delta t$$

Energy balance:

$$\frac12\rho V^2=\frac{\Delta p^2}{2K}+\frac{\Delta p^2}{2}\frac{D}{tE}$$

Substitute $\Delta p=\rho cV$:

$$\frac12\rho V^2=\frac{\rho^2c^2V^2}{2}\left(\frac1K+\frac{D}{tE}\right)\;\Rightarrow\;c=\sqrt{\frac{1}{\rho\left(\dfrac1K+\dfrac{D}{tE}\right)}}=\sqrt{\frac{K/\rho}{1+\dfrac{KD}{tE}}}$$

3. Pressure rise for elastic pipe

$$\boxed{\Delta p=\rho cV=V\sqrt{\frac{\rho K}{1+\dfrac{KD}{tE}}}}\qquad\text{(head rise }\Delta H=\frac{cV}{g}\text{)}$$

4. Rigid pipe

A rigid pipe does not stretch: $E\to\infty$, so $KD/tE\to0$ and

$$c=\sqrt{\frac{K}{\rho}},\qquad \Delta p=\rho cV=V\sqrt{K\rho}$$

This is the maximum possible pressure rise; it is larger than the value for an elastic pipe.