Surge and water hammer: 60 m³/s, 30 m surge tank

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: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

Surge in the tunnel

$A_t = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $A_s = \frac{\pi}{4}(30)^2 = 706.86\ \text{m}^2$, $L = 7000$ m.

• $V = 60/78.54 = 0.764$ m/s
• $h_f = \dfrac{0.016 \times 7000 \times 0.764^2}{2(9.81)(10)} = 0.333$ m
• $Z_0 = 0.764\sqrt{\dfrac{7000 \times 78.54}{9.81 \times 706.86}} = 6.80$ m

(i) Maximum up-surge and down-surge

• Up-surge $= 6.58$ m above reservoir level.
• Down-surge (first downswing) $= 6.19$ m below reservoir level.

(ii) Water hammer pressure

Discharge per penstock $= 60/3 = 20$ m³/s; $A_p = 3.142\ \text{m}^2$; $v = 20/3.142 = 6.366$ m/s.

$$h = \frac{a v}{g} = \frac{1800 \times 6.366}{9.81} = 1168.1\ \text{m}$$

$p = 11.46\ \text{MPa}$ ($\approx 116.8\ \text{kg/cm}^2$), for instantaneous (rapid) closure.

(iii) Time of oscillation

$$T = 2\pi\sqrt{\frac{7000 \times 706.86}{9.81 \times 78.54}} = 503.5\ \text{s} \approx 8.4\ \text{min}$$

Answer: up-surge 6.58 m; down-surge 6.19 m; water hammer head 1168.1 m (11.46 MPa); $T = 503.5$ s.