Surge numerical: 13 m tank, 1.8 km 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: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

Assumption

The discharge is not given. Assume a design discharge $Q = 30\ \text{m}^3/\text{s}$ (4 penstocks of 7.5 m³/s, about 5.65 m/s each). Surge heights scale almost in proportion to $V$, so for another $Q$ repeat the same steps.

Calculation

$A_t = \frac{\pi}{4}(3.95)^2 = 12.25\ \text{m}^2$, $A_s = \frac{\pi}{4}(13)^2 = 132.73\ \text{m}^2$, $L = 1800$ m, $f = 0.016$.

• $V = 30/12.25 = 2.448$ m/s
• $h_f = \dfrac{0.016 \times 1800 \times 2.448^2}{2(9.81)(3.95)} = 2.23$ m
• $Z_0 = 2.448\sqrt{\dfrac{1800 \times 12.25}{9.81 \times 132.73}} = 10.08$ m

The penstock friction (0.025) acts downstream of the tank, so it does not enter the tank oscillation.

Results (full load rejection)

• Maximum up-surge $= 8.65$ m above reservoir level.
• Maximum down-surge $= 6.90$ m below reservoir level.
• Time of oscillation

$$T = 2\pi\sqrt{\frac{1800 \times 132.73}{9.81 \times 12.25}} = 280.1\ \text{s} \approx 4.7\ \text{min}$$

Answer (for $Q = 30$ m³/s): up-surge 8.65 m; down-surge 6.90 m; $T = 280.1$ s. For other discharges the surges grow roughly in proportion to $Q$ (e.g. $Q = 20$ m³/s: 6.07 m up, 5.16 m down; $Q = 40$ m³/s: 10.94 m up, 8.27 m down).