Pumped storage plant: power, turbine, setting, surge

a) Maximum power output

Discharge for 48 h continuous: $Q = \dfrac{15 \times 10^6}{48 \times 3600} = 86.81\ \text{m}^3/\text{s}$.

Gross head $= 500 - 200 = 300$ m.

• Tunnel: $A_t = \frac{\pi}{4}(8)^2 = 50.27\ \text{m}^2$, $V = 1.727$ m/s, $h_{ft} = \dfrac{0.028 \times 4000 \times 1.727^2}{2(9.81)(8)} = 2.13$ m
• Each penstock: $q = 86.81/4 = 21.70\ \text{m}^3/\text{s}$, $V_p = 21.70/3.142 = 6.908$ m/s, $h_{fp} = \dfrac{0.016 \times 500 \times 6.908^2}{2(9.81)(2)} = 9.73$ m
• Net head $H_n = 300 - 2.13 - 9.73 = 288.14$ m

Turbine power: $P_t = \eta_t\gamma Q H_n = 0.9 \times 9.81 \times 86.81 \times 288.14 = 220{,}835$ kW.

Generator output: $P_g = 0.9 \times 220{,}835 = 198{,}752$ kW.

Maximum power output $\approx 198.8$ MW (about 49.7 MW per unit).

b) Specific speed and type

Speed from 16 poles at 50 Hz: $N = \dfrac{120 \times 50}{16} = 375$ rpm.

Power per turbine $= 220{,}835/4 = 55{,}209$ kW.

$$N_s = \frac{N\sqrt{P}}{H_n^{5/4}} = \frac{375\sqrt{55{,}209}}{288.14^{1.25}} = 74.2$$

This is a low specific speed at high head, so the machine is a Francis-type (reversible pump-turbine).

c) Safe turbine setting

$$\sigma = 0.043\left(\frac{N_s}{100}\right)^2 = 0.043(0.742)^2 = 0.0237$$

Taking $H_{atm} = 10.3$ m (given) and $H_v = 0.24$ m (assumed, water at 20 °C):

$$H_s = H_{atm} - H_v - \sigma H_n = 10.3 - 0.24 - 0.0237 \times 288.14 = 3.2\ \text{m}$$

The limit is about 3.2 m above the downstream reservoir level. A pump-turbine is normally set well below tailwater (negative $H_s$) because pumping needs more submergence, so this value is only the upper limit.

d) Surge chamber ($A_s = \frac{\pi}{4}(6)^2 = 28.27\ \text{m}^2$)

The steady friction loss in the tunnel is $h_f = k V^2$ with $k = \dfrac{fL}{2gD}$; the tank level $Z$ is measured from the upper reservoir level. Integrating the equations of motion and continuity for the low-pressure tunnel ($L = 4000$ m):

• Four units: $Q_1 = 86.81\ \text{m}^3/\text{s}$, $V = 1.727$ m/s; three units: $Q = 65.10\ \text{m}^3/\text{s}$, $V = 1.295$ m/s.
• (i) Rejection of one unit (flow $86.81 \to 65.10\ \text{m}^3/\text{s}$): maximum upsurge $= 9.72$ m above reservoir level; the following downsurge $= 11.13$ m below reservoir level (frictionless amplitude $\Delta V\sqrt{LA_t/gA_s} = 11.6$ m).
• (ii) Sudden demand of one unit (flow $65.10 \to 86.81\ \text{m}^3/\text{s}$): maximum downsurge $= 13.01$ m below reservoir level (about 11.8 m below the initial steady level, which is 1.20 m below reservoir level).

Period of oscillation $T = 2\pi\sqrt{\dfrac{LA_s}{gA_t}} = 95.2$ s.

Answer: (a) 198.8 MW; (b) $N_s = 74.2$, Francis (pump-turbine); (c) $H_s \approx 3.2$ m above tailwater (upper limit); (d) upsurge 9.72 m, downsurge 11.13 m (rejection); downsurge 13.01 m (demand).