Water hammer causes and momentum equation

Water hammer phenomenon

When the velocity of flow in a closed conduit is changed rapidly, the moving water column is suddenly slowed down or accelerated. Its kinetic energy is converted into pressure (strain) energy, and a pressure wave travels along the pipe at the wave speed $c = \sqrt{\dfrac{K/\rho}{1 + KD/(Et)}}$, reflecting at the reservoir and at the valve. The resulting rapid rise and fall of pressure, with a hammering noise, is water hammer. For instantaneous closure, $\Delta p = \rho cV_0$.

Causes

1. Sudden closing or opening of valves or gates (the commonest cause).
2. Starting or stopping of pumps, or a power failure to the pump.
3. Sudden change of load on a turbine (governor action closing the guide vanes or nozzle).
4. Rupture of a pipe, or a sudden change in the demand at the end of a pipe.
5. Air pockets trapped in the pipe being suddenly expelled, or vapour cavities collapsing when a separated water column rejoins.
6. Slamming of a check (non-return) valve when the flow reverses after a pump trips.

Momentum equation for unsteady flow through a pipe

Consider a short element of fluid of length $dx$ in a pipe of cross-section $A$ and perimeter $P$, inclined at an angle $\theta$ to the horizontal, with $x$ measured along the pipe. Forces along the pipe:

• Pressure: $pA - \left(p + \dfrac{\partial p}{\partial x}dx\right)A = -\dfrac{\partial p}{\partial x}A\,dx$
• Weight component: $-\rho gA\,dx\sin\theta$
• Wall shear: $-\tau_0P\,dx$

Newton's second law, with mass $\rho A\,dx$ and acceleration $\dfrac{DV}{Dt} = \dfrac{\partial V}{\partial t} + V\dfrac{\partial V}{\partial x}$:

$$\rho A\,dx\left(\frac{\partial V}{\partial t} + V\frac{\partial V}{\partial x}\right) = -\frac{\partial p}{\partial x}A\,dx - \rho gA\sin\theta\,dx - \tau_0P\,dx$$

Divide by $\rho A\,dx$. With the piezometric head $H = \dfrac{p}{\gamma} + z$ ($\dfrac{\partial z}{\partial x} = \sin\theta$), the pressure and weight terms combine to $-g\dfrac{\partial H}{\partial x}$. The wall shear $\tau_0 = \dfrac{f\rho V|V|}{8}$ and $\dfrac{P}{A} = \dfrac{4}{D}$ give $\dfrac{\tau_0P}{\rho A} = \dfrac{fV|V|}{2D}$:

$$\boxed{\frac{\partial V}{\partial t} + V\frac{\partial V}{\partial x} + g\frac{\partial H}{\partial x} + \frac{fV|V|}{2D} = 0}$$

This is Euler's (momentum) equation for unsteady flow in a pipe. In water hammer problems the convective term $V\,\partial V/\partial x$ is small compared with $\partial V/\partial t$ and is usually dropped.

Use in water hammer. With $V\,\partial V/\partial x$ and friction neglected, the equation reduces to $\dfrac{\partial V}{\partial t} + g\dfrac{\partial H}{\partial x} = 0$. Together with the continuity equation $\dfrac{\partial H}{\partial t} + \dfrac{c^2}{g}\dfrac{\partial V}{\partial x} = 0$, it gives the wave solution $\Delta H = \mp\dfrac{c}{g}\Delta V$.