Surge and water hammer: 25 m³/s, 15 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

$A_t = 50.27\ \text{m}^2$, $A_s = \frac{\pi}{4}(15)^2 = 176.71\ \text{m}^2$, $L = 4000$ m.

• $V = 25/50.27 = 0.497$ m/s
• $h_f = \dfrac{0.016 \times 4000 \times 0.497^2}{2(9.81)(8)} = 0.101$ m
• $Z_0 = 0.497\sqrt{\dfrac{4000 \times 50.27}{9.81 \times 176.71}} = 5.36$ m

(i) Maximum up-surge

$Z_{max} = 5.29$ m above reservoir level.

(ii) Maximum down-surge

First downswing $= 5.16$ m below reservoir level.

(iii) Water hammer pressure

Flow per penstock $= 25/2 = 12.5$ m³/s; $v = 12.5/3.142 = 3.979$ m/s. Closure time for full load rejection is taken as less than $2L/a = 2(500)/1600 = 0.625$ s (rapid closure):

$$h = \frac{a v}{g} = \frac{1600 \times 3.979}{9.81} = 649.0\ \text{m}$$

$p = \rho g h = 6.37\ \text{MPa}$ ($\approx 64.9\ \text{kg/cm}^2$).

(iv) Time of oscillation

$$T = 2\pi\sqrt{\frac{4000 \times 176.71}{9.81 \times 50.27}} = 237.9\ \text{s} \approx 4.0\ \text{min}$$

Answer: up-surge 5.29 m; down-surge 5.16 m; water hammer head 649.0 m (6.37 MPa); $T = 237.9$ s.