Water hammer, Euler and continuity equations

Water hammer phenomenon

When the velocity of water flowing in a pipe is changed quickly (a valve closed or opened suddenly, a pump or turbine started or stopped), the kinetic energy of the water is converted into pressure energy, and a high-intensity pressure wave travels through the pipe at the speed of sound in the water-pipe system ($c$). The wave is reflected at the reservoir and at the valve, and the pressure may rise or fall by many times the normal pressure, with a noise like hammering. This is water hammer. It can burst pipes (pressure rise) or collapse them and cause column separation (pressure fall).

For instantaneous closure, the pressure rise is (Joukowsky)

$$\Delta p = \rho c V_0,\qquad \Delta H = \frac{cV_0}{g},\qquad c = \sqrt{\frac{K/\rho}{1 + \dfrac{KD}{Et}}}$$

Euler (momentum) equation for unsteady pipe flow

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.

Continuity equation for unsteady flow (elastic water and pipe)

Consider a pipe element of length $dx$. Mass inflow minus outflow equals the rate of change of mass in the element:

$$\frac{\partial(\rho AV)}{\partial x} + \frac{\partial(\rho A)}{\partial t} = 0 \quad(\text{or, equivalently, } \frac{D(\rho A)}{Dt} + \rho A\frac{\partial V}{\partial x} = 0)$$

Hence $\dfrac{1}{\rho A}\dfrac{D(\rho A)}{Dt} + \dfrac{\partial V}{\partial x} = 0$. A change of pressure $dp$ changes both the density and the area:

• Water compressibility (bulk modulus $K$): $\dfrac{d\rho}{\rho} = \dfrac{dp}{K}$
• Pipe elasticity (thin wall, hoop stress $\dfrac{pD}{2t}$, strain $\dfrac{dD}{D} = \dfrac{D\,dp}{2tE}$): $\dfrac{dA}{A} = 2\dfrac{dD}{D} = \dfrac{D\,dp}{tE}$

So $\dfrac{d(\rho A)}{\rho A} = dp\left(\dfrac{1}{K} + \dfrac{D}{tE}\right) = \dfrac{dp}{\rho c^2}$, which defines the wave speed

$$c^2 = \frac{K/\rho}{1 + \dfrac{KD}{tE}}$$

Therefore $\dfrac{1}{\rho c^2}\dfrac{Dp}{Dt} + \dfrac{\partial V}{\partial x} = 0$. Using $p = \gamma(H - z)$, $\dfrac{Dp}{Dt} = \gamma\left(\dfrac{\partial H}{\partial t} + V\dfrac{\partial H}{\partial x} - V\sin\theta\right)$:

$$\boxed{\frac{\partial H}{\partial t} + V\frac{\partial H}{\partial x} - V\sin\theta + \frac{c^2}{g}\frac{\partial V}{\partial x} = 0}$$

The terms $V\,\partial H/\partial x$ and $V\sin\theta$ are small, so the usual form is $\dfrac{\partial H}{\partial t} + \dfrac{c^2}{g}\dfrac{\partial V}{\partial x} = 0$.

Together, the momentum and continuity equations govern water hammer. With friction neglected they combine into the wave equation $\dfrac{\partial^2H}{\partial t^2} = c^2\dfrac{\partial^2H}{\partial x^2}$, whose solutions are waves travelling at $\pm c$; for a sudden change of velocity $\Delta V$ they give $\Delta H = \mp\dfrac{c}{g}\Delta V$.