Rapid closure time and peak hammer pressure

Data. $D = 0.3$ m, $t = 8$ mm, $L = 1500$ m, $Q = 200$ l/s, $K = 2200$ MPa $= 2.2\times10^9$ Pa, $E = 80\times10^9$ Pa for cast iron.

Velocity and wave speed

$$V = \frac{Q}{\pi D^2/4} = \frac{0.2}{0.070686} = 2.829\ \text{m/s}$$ $$c = \frac{\sqrt{K/\rho}}{\sqrt{1 + \dfrac{KD}{Et}}} = \frac{\sqrt{2.2\times10^6}}{\sqrt{1 + \dfrac{(2.2\times10^9)(0.3)}{(80\times10^9)(0.008)}}} = \frac{1483.2}{\sqrt{1 + 1.0312}} = 1040.7\ \text{m/s}$$

(a) Maximum time of closure for rapid closure

Closure is rapid if $t_c \le \dfrac{2L}{c}$ (the time for the pressure wave to travel to the reservoir and back):

$$t_c \le \frac{2(1500)}{1040.7} = 2.883\ \text{s}$$

(b) Peak water hammer pressure (rapid closure)

$$\Delta p = \rho cV = 1000(1040.7)(2.829) = 2.945\ \text{MPa}\quad(\Delta H = \frac{cV}{g} = 300.2\ \text{m of water})$$

(c) Length of pipe subjected to peak pressure for $t_c = 2.0$ s

Since $2.0\ \text{s} < 2L/c = 2.88$ s, the closure is rapid, but the closing takes a finite time, so only part of the pipe sees the full peak pressure. At a point $d$ from the valve, the last pressure increment of the closure arrives at $t = t_c + d/c$, while the negative wave reflected from the reservoir arrives at $t = (2L - d)/c$. The full peak $\rho cV$ is reached only if the last increment arrives before the reflection:

$$t_c + \frac{d}{c} \le \frac{2L - d}{c} \;\Rightarrow\; d \le L - \frac{c\,t_c}{2}$$ $$L_{peak} = 1500 - \frac{1040.7(2.0)}{2} = 459.3\ \text{m}$$

This is the length of pipe measured from the valve towards the reservoir that experiences the peak pressure; the remaining $c\,t_c/2 = 1040.7$ m next to the reservoir does not.

Answer: (a) $t_c \le 2.88$ s; (b) $\Delta p = 2.94$ MPa; (c) about $459$ m of the pipe adjacent to the valve experiences the peak pressure.