Surge numerical: 4.5 m tunnel, 2500 m, 25 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

$A_t = \frac{\pi}{4}(4.5)^2 = 15.904\ \text{m}^2$, $A_s = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $L = 2500$ m, $f = 0.02$.

• $V = 25/15.904 = 1.572$ m/s
• $h_f = \dfrac{0.02 \times 2500 \times 1.572^2}{2(9.81)(4.5)} = 1.40$ m
• $Z_0 = 1.572\sqrt{\dfrac{2500 \times 15.904}{9.81 \times 78.54}} = 11.29$ m

Instantaneous closure means full rejection, so the water rises in the tank. The penstock is downstream of the tank and does not change the tank oscillation.

Maximum height of surge tank

With friction, $Z_{max} = 10.38$ m above reservoir level (frictionless value 11.29 m). The tank should rise at least 10.4 m above the reservoir level, plus freeboard of about 1 to 2 m. The first downswing is 9.01 m below reservoir level.

Time period

$$T = 2\pi\sqrt{\frac{2500 \times 78.54}{9.81 \times 15.904}} = 222.9\ \text{s} \approx 3.7\ \text{min}$$

Answer: maximum upsurge height $= 10.38$ m above reservoir level; $T = 222.9$ s.