Surge and water hammer: 45 m³/s, 20 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

$A_t = \frac{\pi}{4}(8)^2 = 50.27\ \text{m}^2$, $A_s = \frac{\pi}{4}(20)^2 = 314.16\ \text{m}^2$, $L = 4000$ m.

• $V = 45/50.27 = 0.895$ m/s
• $h_f = \dfrac{0.016 \times 4000 \times 0.895^2}{2(9.81)(8)} = 0.327$ m
• $Z_0 = 0.895\sqrt{\dfrac{4000 \times 50.27}{9.81 \times 314.16}} = 7.23$ m

a) Maximum up-surge

$Z_{max} = 7.02$ m above reservoir level.

b) Maximum down-surge

First downswing $= 6.63$ m below reservoir level.

c) Water hammer pressure

Each penstock carries $45/3 = 15$ m³/s. $A_p = \frac{\pi}{4}(2)^2 = 3.142\ \text{m}^2$, so $v = 15/3.142 = 4.775$ m/s.

Full load rejection means the gates close suddenly (closure time $\le 2L/a = 2(500)/1400 = 0.71$ s), so the Joukowsky rise is

$$h = \frac{a v}{g} = \frac{1400 \times 4.775}{9.81} = 681.4\ \text{m}$$

$p = \rho g h = 6.69\ \text{MPa}$ ($\approx 68.1\ \text{kg/cm}^2$).

d) Time of oscillation

$$T = 2\pi\sqrt{\frac{4000 \times 314.16}{9.81 \times 50.27}} = 317.2\ \text{s} \approx 5.3\ \text{min}$$

(The pressure wave in the penstock itself has a period $4L_p/a = 4(500)/1400 = 1.43$ s.)

Answer: up-surge 7.02 m; down-surge 6.63 m; water hammer head 681.4 m (6.69 MPa); $T = 317.2$ s.