Surge tank: tunnel 5 m dia, 28 cumec

Given and tunnel length

$A_t = \frac{\pi}{4}(5)^2 = 19.635\ \text{m}^2$, $V = 28/19.635 = 1.426$ m/s, $A_s = 300\ \text{m}^2$, $h_f = 0.025 \times 200 = 5.0$ m.

$$L = \frac{h_f\,2gD}{fV^2} = \frac{5.0 \times 2(9.81)(5)}{0.015 \times 1.426^2} = 16{,}080\ \text{m}$$

Total tunnel length $\approx 16.1$ km.

Maximum upsurge and downsurge

Frictionless amplitude: $Z_0 = 1.426\sqrt{\dfrac{16080 \times 19.635}{9.81 \times 300}} = 14.77$ m.

With friction ($h_f = 5$ m), the upsurge for full load rejection (measured above reservoir level) is found from the equations of motion and continuity:

$$\frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right]$$

Solving by trial gives $Z_{max} = 11.64$ m above reservoir level. Integrating the reversed-flow equations step by step, the first downswing after the upsurge reaches $8.57$ m below reservoir level.

Factor of safety (Thoma stability)

Head-loss coefficient $\alpha = h_f/V^2 = 5/1.426^2 = 2.459\ \text{s}^2/\text{m}$. Net head $H - h_f = 195$ m.

$$A_{th} = \frac{L A_t}{2g\alpha (H-h_f)} = \frac{16080 \times 19.635}{2(9.81)(2.459)(195)} = 33.56\ \text{m}^2$$ $$\text{FOS} = \frac{A_s}{A_{th}} = \frac{300}{33.56} = 8.94$$

Answer: $L \approx 16{,}080$ m; upsurge $= 11.64$ m; downsurge $= 8.57$ m; factor of safety $= 8.94$ (the tank is very stable, since $A_s \gg A_{th}$).