Water hammer in 1.2 m steel pipe, wall thickness

Data. $D = 1.20$ m, $Q = 1.40$ m³/s, static head 300 m ($p_0 = \rho gH = 2.943$ MPa), $E = 210$ kN/mm² $= 210\times10^9$ Pa, safe stress $\sigma_s = 0.1$ kN/mm² $= 100$ MPa, $K = 2.10$ kN/mm² $= 2.1\times10^9$ Pa. Sudden closure, so the full Joukowsky rise applies.

$$V_0 = \frac{Q}{\pi D^2/4} = \frac{1.4}{1.1310} = 1.238\ \text{m/s}$$

Water hammer pressure

The wave speed depends on the wall thickness $t$ (unknown):

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

First estimate (rigid pipe, $t \to \infty$): $c = \sqrt{K/\rho} = 1449$ m/s, so $\Delta p \approx 1.79$ MPa. This is the upper limit.

Thickness

The pipe must carry the maximum pressure $p_0 + \Delta p$ at the safe hoop stress:

$$\sigma_s = \frac{(p_0 + \Delta p)D}{2t} \;\Rightarrow\; t = \frac{(p_0 + \Delta p)D}{2\sigma_s}$$

Since $\Delta p$ depends on $t$, iterate. Start with $t$ = 10 mm and repeat: each new $t$ gives a new $c$ and $\Delta p$. The substitution converges to

$$t = 26.6\ \text{mm},\qquad c = 1202.9\ \text{m/s},\qquad \Delta p = \rho cV_0 = 1.489\ \text{MPa}$$ $$p_0 + \Delta p = 2.943 + 1.489 = 4.432\ \text{MPa}, \qquad t = \frac{4.432\times10^6(1.2)}{2(100\times10^6)} = 26.6\ \text{mm}$$

To the nearest (larger) millimetre: $t = 27$ mm.

Check with $t = 27$ mm: $c = 1205.8$ m/s, $\Delta p = 1.493$ MPa ($152.1$ m of water), maximum stress $= \dfrac{(2.943 + 1.493)\times10^6(1.2)}{2(0.027)} = 98.6$ MPa $\le 100$ MPa.

Answer: Water hammer pressure rise $\approx 1.49$ MPa ($\approx 152$ m of water; $1.79$ MPa if the pipe were rigid); minimum wall thickness $\approx 27$ mm.