Water hammer effects and gradual closure

Water hammer and its effects

Water hammer is the sharp rise of pressure in a closed conduit when the velocity of the flowing liquid is suddenly reduced, for example by closing a valve at the end of a penstock. The moving water column is brought to rest, its kinetic energy changes to pressure energy, and a pressure wave travels up and down the pipe with wave speed $c$ until friction damps it.

Effects:

• Very high pressure; pipe may burst or joints may fail.
• Low (negative) pressure phase may collapse the pipe or cause cavitation.
• Hammering noise, vibration, damage to valves, anchors and turbine parts.
• Penstock must be made thicker; surge tanks or relief valves are needed.

Pressure for gradual closure

Closure is gradual when the closing time $T > \dfrac{2L}{c}$ (the reflected wave returns before closure ends).

Consider a pipe of length $L$, area $A$, initial velocity $V$, valve closed uniformly in time $T$.

 Reservoir                         Valve
 ~~~~~|=============================|X
      |<------------ L ------------>|
          water mass = rho A L

1. Mass of water in pipe $= \rho A L$.
2. Uniform retardation $= \dfrac{V - 0}{T} = \dfrac{V}{T}$.
3. Retarding force by Newton's second law:

$$F = \rho A L \cdot \frac{V}{T}$$

4. This force is provided by the pressure rise $p$ acting on area $A$: $F = pA$.
5. Equating:

$$\begin{aligned} pA &= \frac{\rho A L V}{T} \\ p &= \frac{\rho L V}{T}, \qquad H = \frac{p}{\rho g} = \frac{LV}{gT} \end{aligned}$$

Example: $L = 2500$ m, $V = 1.5$ m/s, $T = 25$ s: $p = 1000 \times 2500 \times 1.5/25 = 150$ kN/m².

If $T < 2L/c$, the closure is sudden and $p = \rho c V$ (rigid pipe, $c = \sqrt{K/\rho}$).