Installed capacity, energy and daily peaking reservoir (2072)

The generator efficiency is missing in the question; it is assumed to be 99%, the same as the transformer (all other data as given).

Data and approach

Gross head $= 2250-1650 = 600$ m. Only friction loss is considered. Overall efficiency $\eta = 0.95\times0.93\times0.99\times0.99 = 0.8659$ (hydraulic × turbine × generator × transformer). Energy is computed month by month (730 h per month), with the friction loss recalculated for the flow used.

Head loss at design flow $Q_d = 18$ m³/s (Darcy-Weisbach)

Tunnel: $A=\frac{\pi}{4}(3)^2 = 7.069$ m², $V = 2.546$ m/s

$$h_{f,t} = f\frac{L}{D}\frac{V^2}{2g} = 0.014\times\frac{4000}{3}\times\frac{2.546^2}{2\times9.81} = 6.17\ \text{m}$$

Penstock: $A=\frac{\pi}{4}(2.2)^2 = 3.801$ m², $V = 4.735$ m/s

$$h_{f,p} = 0.012\times\frac{1000}{2.2}\times\frac{4.735^2}{2\times9.81} = 6.23\ \text{m}$$

Net head at $Q_d$: $H_n = 600-6.17-6.23 = 587.60$ m. (Loss varies as $Q^2$ for smaller flows.)

(i) Installed capacity, energy and plant factor

Environmental release $=10\%\times$ minimum monthly flow $=0.1\times3.4=0.34$ m³/s, so the flow available for power is $Q-0.34$.

$$P = 9.81\,\eta\,H_n\,Q_d = 9.81\times0.8659\times587.60\times18 = 89,846\ \text{kW} = 89.85\ \text{MW}$$

• Total annual energy $= 452.92$ GWh
• Primary (firm) energy, from the minimum available flow $3.06$ m³/s (available in every month): $136.54$ GWh
• Secondary energy $=452.92-136.54 = 316.38$ GWh
• Plant factor $=\dfrac{452.92\times10^6}{89.85\times10^3\times8760} = 0.575$ (57.5%)

(ii) Reservoir for 4-hour daily peaking

In the driest month the available inflow is $Q_{in} = 3.06$ m³/s. The plant is to run for 4 hours a day at the design flow of 18 m³/s and store water in the other 20 hours. The daily inflow is 264,384 m³ and the peak use is $18\times4\times3600 = 259,200$ m³, which is slightly smaller, so 4-hour peaking at 18 m³/s is possible.

During the 4 peak hours the draw-down from the reservoir is

$$V = (Q_d - Q_{in})\times t = (18-3.06)\times4\times3600 = 215,136\ \text{m}^3 \approx 2.15\times10^5\ \text{m}^3$$

The same volume must be stored in the 20 off-peak hours ($3.06\times20\times3600 = 220,320$ m³ available), so it fills in time.

Answer: P = 89.8 MW; total energy = 452.9 GWh; primary energy = 136.5 GWh; secondary energy = 316.4 GWh; plant factor = 0.58; pondage volume ≈ 215,136 m³ (≈ 2.15 × 10⁵ m³).