RoR and PRoR from hydrograph equation (Q40)

Monthly flows from $Q_t = 5.589t^2 - 51.275t + 139.94$

$H=150$ m, $\eta=0.85$: $9.81\,\eta\,H = 1250.77$ kW per m³/s.

Design discharge $Q_{40}$

Descending flows: 329.46, 252.18, 186.09, 131.17, 94.25, 87.44, 59.75, 54.88, 36.42, 33.49, 24.26, 23.29. Weibull position for 40%: $m=0.4\times13=5.2$ (between the 5th flow 94.25 and the 6th flow 87.44):

$$Q_{40} = 94.25 - 0.2\,(94.25-87.44) = 92.89\ \text{m}^3/\text{s}$$

a) RoR project

$$P = 1250.77\times92.89 = 116,185\ \text{kW} = 116.18\ \text{MW}$$

The minimum monthly flow is 23.29 m³/s (February), the flow available in all months, so

$$E_{firm} = 1250.77\times23.29\times8760 = 255,183,616\ \text{kWh} = 255.2\ \text{GWh}$$

(Total energy with flows limited to $Q_{40}$ is 715.8 GWh; the part above firm energy is secondary energy.)

b) Peaking run-of-river (6 h daily peaking)

Daily water volume at $Q_{40}$: $Q_{40}\times24\times3600$. Half of it (50%) is used during the 18 off-peak hours and the other half is stored (pond volume $=0.5\times92.89\times86400=4,012,865$ m³) and used in the 6 peak hours:

$$Q_{peak} = \frac{0.5\,Q_{40}\times24}{6} = 2\,Q_{40} = 185.78\ \text{m}^3/\text{s}$$ $$P_{PRoR} = 1250.77\times185.78 = 232,370\ \text{kW} = 232.37\ \text{MW}$$

Off-peak the plant uses $0.5\,Q_{40}\times24/18 = 61.93$ m³/s. The installed capacity is therefore twice that of the RoR plant.

Answer: (a) RoR: installed capacity = 116.2 MW, firm energy = 255.2 GWh; (b) PRoR: installed capacity = 232.4 MW.