RoR best installed capacity (USD 2795 million)

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 = USD 2795 × 10⁶; variable cost = USD 645/kW; O&M = 2.5% of the variable cost per year; price = USD 38/MWh = USD 0.038/kWh for primary and secondary energy
• Net head $H = 48$ m (constant); overall efficiency $\eta = 0.89$
• Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.11101$ for $i=11\%$, $n=45$ years
• Energy per month: $E_m = 9.81\,\eta\,H\,Q\,(730\ \text{h})$ (8760/12 h per month)
• Annual cost $=(\text{fixed}+\text{variable}\times P)\,CRF + \text{O\&M}$; annual benefit = energy × price

Flow available for power

Available flows (m³/s) in descending order: 93, 73, 68, 53, 48, 43, 38, 36, 33, 28, 23, 18.

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 $=0.73\times38 =$ USD 27.74 per year. The annual cost of 1 kW $= 645\times0.11101 + 0.025\times645 = 71.60 + 16.13 =$ USD 87.73, so a kW must be used for at least 3.16 months, i.e. at least 4 months. The 4th highest flow is 53 m³/s, so the optimum design flow is 53 m³/s (capacity up to 68 m³/s would be used for only 3 months: 3 × 27.74 = 83.22 < 87.73).

Trial design discharges

The values for 48, 53 and 68 m³/s look tied at this rounding; the marginal check above shows the net benefit is greatest at $Q_d = 53$ m³/s.

Results

• Installed capacity: $P = 9.81\times0.89\times48\times53 = 22,211\ \text{kW} = 22.21$ MW
• Firm (primary) energy, from the minimum flow of 18 m³/s all year: $9.81\times0.89\times48\times18\times8760 = 66,081,039\ \text{kWh} = 66.08$ GWh
• Total energy: $146.54$ GWh
• Secondary energy $= 146.54-66.08 = 80.46$ GWh
• Plant factor $= \dfrac{E}{P\times8760} = \dfrac{146.54\times10^6}{22,211\times8760} = 0.753$ (75.3%)

Note: with the fixed cost taken exactly as stated (USD 2795 million), the net benefit is negative for every capacity (the project does not pay for itself). The optimum capacity is the same because the fixed cost is independent of capacity; the fixed cost is possibly a misprint in the question.

Answer: best installed capacity = 22.2 MW (design flow 53 m³/s); firm energy = 66.1 GWh; secondary energy = 80.5 GWh; total energy = 146.5 GWh; plant factor = 0.75.