Diameter to thickness ratio for stress limit

Data. $L = 500$ m, $V_0 = 2$ m/s, static head 300 m, instantaneous valve closure, increase of circumferential stress $\Delta\sigma \le 20$ MPa, $K = 2$ GPa, $E = 200$ GPa. Let $x = D/t$.

Relations

Wave speed: $c = \dfrac{\sqrt{K/\rho}}{\sqrt{1 + \dfrac{K}{E}\dfrac{D}{t}}} = \dfrac{1414.2}{\sqrt{1 + 0.01x}}$ (since $K/E = 0.01$).

Pressure rise: $\Delta p = \rho cV_0 = 1000(2)c = 2000c$.

Hoop stress increase (thin-walled pipe): $\Delta\sigma = \dfrac{\Delta p\,D}{2t} = \dfrac{\Delta p\,x}{2}$.

Solve for $D/t$

$$\Delta\sigma = \frac{x}{2}\cdot\frac{2000(1414.2)}{\sqrt{1 + 0.01x}} = 20\times10^6 \;\Rightarrow\; \frac{x}{\sqrt{1 + 0.01x}} = \frac{40\times10^6}{2000(1414.2)} = 14.142$$

Square both sides: $x^2 = 200(1 + 0.01x) = 200 + 2x$, so

$$x^2 - 2x - 200 = 0 \;\Rightarrow\; x = 1 + \sqrt{201} = 15.18$$ $$\boxed{\frac{D}{t} \approx 15.2}$$

Maximum time for rapid closure

Wave speed with this thickness:

$$c = \frac{1414.2}{\sqrt{1 + 0.01(15.18)}} = 1317.7\ \text{m/s}$$

(then $\Delta p = 1000(1317.7)(2) = 2.64$ MPa, and $\Delta\sigma = 20.0$ MPa, as required; the maximum pressure is $2.943 + 2.64 = 5.58$ MPa.)

Closure is rapid if its time $t_c \le \dfrac{2L}{c}$:

$$t_c \le \frac{2(500)}{1317.7} = 0.759\ \text{s}$$

Answer: $D/t \approx 15.2$ (wall thickness is about 1/15 of the diameter); closure is rapid for $t_c \le 0.76$ s.