Surge numerical: 4 m tunnel, 4000 m, 20 m³/s

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

$A_t = \frac{\pi}{4}(4)^2 = 12.566\ \text{m}^2$, $A_s = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $L = 4000$ m, $f = 0.018$.

• $V = 20/12.566 = 1.592$ m/s
• $h_f = \dfrac{0.018 \times 4000 \times 1.592^2}{2(9.81)(4)} = 2.32$ m
• $Z_0 = 1.592\sqrt{\dfrac{4000 \times 12.566}{9.81 \times 78.54}} = 12.86$ m

Instantaneous closure of the turbine gates means complete rejection of the discharge, so the water rises in the tank. The penstock (3.2 m, 700 m) lies downstream of the tank, so it does not affect the tank oscillation.

Maximum height of surge tank

From the upsurge equation with friction: $Z_{max} = 11.36$ m above reservoir level (the frictionless value is 12.86 m).

The tank must therefore extend at least 11.4 m above the reservoir water level (plus a freeboard of about 1 to 2 m). The first downswing is 9.36 m below reservoir level.

Time period

$$T = 2\pi\sqrt{\frac{4000 \times 78.54}{9.81 \times 12.566}} = 317.2\ \text{s} \approx 5.3\ \text{min}$$

Answer: maximum upsurge height $= 11.36$ m above reservoir level; $T = 317.2$ s.