Surge tank area, up/downsurge, invert level

Given data and losses

$A_t = \frac{\pi}{4}(5)^2 = 19.635\ \text{m}^2$, $V_t = 60/19.635 = 3.056$ m/s. Shaft: $A_p = 12.566\ \text{m}^2$, $V_p = 4.775$ m/s.

• Tunnel loss: $h_{ft} = \dfrac{0.014 \times 4000 \times 3.056^2}{2(9.81)(5)} = 5.33$ m
• Shaft loss: $h_{fp} = \dfrac{0.012 \times 600 \times 4.775^2}{2(9.81)(4)} = 2.09$ m
• Net head $H_0 = 250 - 5.33 - 2.09 = 242.58$ m

i) Surge tank area (Thoma)

Head-loss coefficient of the tunnel, $\alpha = h_{ft}/V_t^2 = 5.33/3.056^2 = 0.5708\ \text{s}^2/\text{m}$.

$$A_{th} = \frac{L A_t}{2g\alpha H_0} = \frac{4000 \times 19.635}{2(9.81)(0.5708)(242.58)} = 28.9\ \text{m}^2$$

Equivalent diameter $= 6.07$ m. Adopting $A_s = A_{th} = 28.9\ \text{m}^2$ (a larger area, e.g. $1.5A_{th}$, gives more safety).

ii) Maximum upsurge and downsurge

$Z_0 = 3.056\sqrt{\dfrac{4000 \times 19.635}{9.81 \times 28.9}} = 50.85$ m. With friction ($h_{ft} = 5.33$ m), for full load rejection:

• Maximum upsurge $= 47.36$ m above reservoir level, i.e. level $1048 + 47.36 = 1095.36$ m.
• Maximum downsurge (first downswing) $= 41.90$ m below reservoir level, i.e. level $1048 - 41.90 = 1006.10$ m.

For comparison, with $A_s = 1.5A_{th} = 43.4\ \text{m}^2$: upsurge 38.05 m and downsurge 32.88 m.

iii) Invert level of the headrace tunnel at the tank

The tunnel crown must stay below the lowest water level in the tank to prevent air entering the tunnel. Assume a minimum water cover of 1.0 m over the crown.

• Lowest tank level $= 1006.10$ m
• Crown $= 1006.10 - 1.0 = 1005.10$ m
• Invert $= 1005.10 - 5.0 = 1000.10$ m

Answer: $A_{th} = 28.9\ \text{m}^2$; upsurge level $= 1095.36$ m; downsurge level $= 1006.10$ m; invert level $\approx 1000.1$ m a.s.l.