RoR best installed capacity (NRs, 45 years)

The optimum capacity of a RoR plant is where the marginal benefit of the last unit of capacity equals its marginal cost (equivalently, where the net benefit is maximum). The fixed cost does not change with capacity, so it does not affect the optimum.

Data and constants

• Fixed cost = NRs 10,00,00,00,000 (= NRs 10,000 million); variable cost = NRs 1,50,000/kW (present worth); price = NRs 10/kWh
• Net head $H = 150$ m; turbine 95% and generator 97%, so overall efficiency $\eta = 0.95\times0.97 = 0.9215$; no O&M cost is given
• "Power factor" in the question is read as the plant factor.
• Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.10139$ for $i=10\%$, $n=45$ years
• Energy per month: $E_m = 9.81\,\eta\,H\,Q\,(730\ \text{h})$ (8760/12 h per month)
• Annual cost $=(\text{fixed}+\text{variable}\times P)\,CRF$; annual benefit = energy × price

Flow available for power

Available flows (m³/s) in descending order: 1200, 1000, 800, 500, 450, 200, 150, 100, 95, 90, 85, 80.

Marginal check

Each extra 1 kW of capacity above a flow level is used only in the months in which the flow is greater than that level. A kW used for $k$ months produces $0.73\,k$ MWh per year. The benefit per kW for each month it is used = 730 kWh × NRs 10 = NRs 7,300, and the annual cost of 1 kW = 1,50,000 × 0.10139 = NRs 15,208. So a kW must be used for at least 2.08 months, i.e. at least 3 months in a year, to pay for itself. The 3rd highest flow is 800 m³/s, so capacity beyond 800 m³/s (used for only 2 months) is not justified.

Trial design discharges

The net benefit is greatest at $Q_d = 800$ m³/s.

Results

• Installed capacity: $P = 9.81\times0.9215\times150\times800 = 1,084,790\ \text{kW} = 1,084.79$ MW
• Firm (primary) energy, from the minimum flow of 80 m³/s all year: $9.81\times0.9215\times150\times80\times8760 = 950,275,865\ \text{kWh} = 950.28$ GWh
• Total energy: $4,107.96$ GWh
• Secondary energy $= 4,107.96-950.28 = 3,157.69$ GWh
• Plant factor $= \dfrac{E}{P\times8760} = \dfrac{4,107.96\times10^6}{1,084,790\times8760} = 0.432$ (43.2%)

Answer: best installed capacity = 1,084.8 MW (design flow 800 m³/s); firm energy = 950.3 GWh; secondary energy = 3,157.7 GWh; total energy = 4,108.0 GWh; plant factor = 0.43.