Peaking RoR: max power and wet/dry energy ratio

Data and assumptions

Net head $H=250$ m. Dry months (Nov-May, 7 months) use stored water for 4 h/day of peaking. Wet months (Jun-Oct, 5 months) the plant runs as RoR for 24 h. One month $=730$ h $=2.628\times10^6$ s. Overall efficiency is not given, so power is first found for $\eta=1$ and then for a typical $\eta=0.85$; the energy ratio does not depend on $\eta$.

Design flow $Q_{25}$

Arrange the 12 monthly flows in descending order: 1200, 1100, 800, 500, 350, 200, 200, 120, 105, 100, 80, 80. The flow equalled or exceeded 25% of the time (3 months out of 12) is the 3rd value:

$$Q_{25} = 800\ \text{m}^3/\text{s}$$

Is the stored water enough for peaking at $Q_{25}$?

• Dry-season inflow volume $=\sum Q\times 2.628\times10^6 = 885\times2.628\times10^6 = 2,326\times10^6$ m³
• Plus live storage $=1100\times10^6$ m³, so water available $=3,426\times10^6$ m³
• Peaking hours in 7 months $=7\times730\times\frac{4}{24} = 851.7$ h
• Water needed to run at $Q_{25}$ for these hours $=800\times851.7\times3600 = 2,453\times10^6$ m³ (less than available)

So the turbines can run at $Q_{25}=800$ m³/s during the 4 peak hours every day, and the plant capacity is controlled by the design flow (the water available would allow up to 1,117 m³/s).

Maximum power

$$P_{max} = 9.81\,\eta\,Q\,H = 9.81\times\eta\times800\times250 = 1,962\,\eta\ \text{MW}$$

• With $\eta = 1$: $P_{max} = 1,962$ MW
• With $\eta = 0.85$: $P_{max} = 1,668$ MW

Wet-season to dry-season energy ratio

Wet season flows used: $\min(Q,800)$ = 500, 800, 800, 800, 350, total $=3250$ m³/s-months.

$$E_{wet} \propto 3250\times730 = 2,372,500,\qquad E_{dry} \propto 800\times851.7 = 681,333$$ $$\frac{E_{wet}}{E_{dry}} = \frac{2,372,500}{681,333} = 3.48$$

Answer: maximum power = 1,962 MW for $\eta=1$ (1,668 MW for $\eta=0.85$); wet season energy / dry season energy = 3.48.