RoR pondage: capacity and pondage volume

Maximum generating capacity

The minimum flow gives the average power available throughout the day:

$$P_{avg} = \gamma Q H \eta = 9.81\times25\times30\times0.60 = 4,414.5\ \text{kW}$$

With pondage the plant can follow the daily load, and the load factor is

$$\text{Load factor} = \frac{P_{avg}}{P_{max}} = 0.65 \Rightarrow P_{max} = \frac{4,414.5}{0.65} = 6,791.5\ \text{kW} \approx 6.79\ \text{MW}$$

Volume of pondage (20 h average load + 4 h peak load)

Peak load: $P_{max}=6,791.5$ kW for 4 h. The average over the day must equal $P_{avg}$, so the off-peak load for 20 h is

$$P_{off} = \frac{24\times4,414.5 - 4\times6,791.5}{20} = 3,939.1\ \text{kW}$$

Flows needed at peak and off-peak (with $9.81\times30\times0.6 = 176.58$ kW per m³/s):

$$Q_{peak} = \frac{6,791.5}{176.58} = 38.46\ \text{m}^3/\text{s},\qquad Q_{off} = \frac{3,939.1}{176.58} = 22.31\ \text{m}^3/\text{s}$$

The river supplies 25 m³/s, so the deficit of $(38.46-25)$ m³/s for 4 hours must come from the pond:

$$V = (38.46-25)\times4\times3600 = 193,846\ \text{m}^3 \approx 1.94\times10^5\ \text{m}^3$$

(During the 20 off-peak hours the pond fills by $(25-22.31)\times20\times3600 = 193,846$ m³, the same volume.)

Answer: generator capacity = 6.79 MW; pondage volume ≈ 193,846 m³ (≈ 1.94 × 10⁵ m³).