RoR installed capacity without pondage and with peaking

(i) Pure RoR plant (no pondage)

Only the minimum flow can be used continuously:

$$P_1 = 9.81\,\eta\,Q\,H = 9.81\times0.85\times30\times70 = 17,511\ \text{kW} = 17.51\ \text{MW}$$

(ii) Peaking plant, 6 h peak, two units

Let $Q_u$ be the discharge of one unit at full load. Off-peak (18 h) only one unit runs, using $Q_u$. In peak hours (6 h) both units run, using $2Q_u$.

Water stored in the 18 off-peak hours (inflow 30 m³/s) after 5% loss:

$$V_s = 0.95\,(30-Q_u)\times18\times3600$$

During the 6 peak hours the extra flow required is $(2Q_u-30)$, supplied from storage:

$$(2Q_u-30)\times6\times3600 = 0.95\,(30-Q_u)\times18\times3600$$ $$12Q_u - 180 = 513 - 17.1\,Q_u \Rightarrow Q_u = \frac{693}{29.1} = 23.814\ \text{m}^3/\text{s}$$

Peak discharge $=2Q_u=47.629$ m³/s.

$$P_2 = 9.81\times0.85\times47.629\times70 = 27,801\ \text{kW} = 27.80\ \text{MW}$$

Check: off-peak, one unit uses 23.81 m³/s (less than 30), so the surplus goes to the pond. Water put into the pond $=(30-23.814)\times18\times3600 = 400,825$ m³ (the pondage capacity, about 4.01 × 10⁵ m³). After the 5% loss, $0.95\times400,825 = 380,784$ m³ is usable, which equals the peak draw $(47.629-30)\times6\times3600 = 380,784$ m³.

Answer: (i) installed capacity = 17.51 MW (pure RoR); (ii) installed capacity = 27.80 MW with 6-hour peaking (2 units of 13.90 MW each), a gain of 1.59 times.