Surge tank: 7.5 m³/s through 2.5 m tunnel

Method used

Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$.

• Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$.
• Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$.
• Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity:

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

• The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level).
• Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

Calculation

$A_t = \frac{\pi}{4}(2.5)^2 = 4.909\ \text{m}^2$, $A_s = \frac{\pi}{4}(7)^2 = 38.48\ \text{m}^2$, $L = 2500$ m, $f=0.018$.

• $V = 7.5/4.909 = 1.528$ m/s
• $h_f = \dfrac{0.018 \times 2500 \times 1.528^2}{2(9.81)(2.5)} = 2.14$ m
• $Z_0 = 1.528\sqrt{\dfrac{2500 \times 4.909}{9.81 \times 38.48}} = 8.71$ m

Time period of oscillation

$$T = 2\pi\sqrt{\frac{2500 \times 38.48}{9.81 \times 4.909}} = 280.8\ \text{s} \approx 4.7\ \text{min}$$

Height of surge tank

Solving the upsurge equation with friction: $Z_{max} = 7.35$ m above reservoir level. The first downsurge is $5.75$ m below reservoir level, so the water level swings through about $13.1$ m. Adding a freeboard of about 2 m at the top, the surge tank must be at least about $15$ m high above the minimum level.

Answer: $T = 280.8$ s; maximum upsurge $= 7.35$ m above reservoir level (tank height about 15 m including range and freeboard).