RoR best installed capacity (USD 1200 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

• Net head $H = 229-12 = 217$ m (gross head minus the average transition loss)
• Overall efficiency $\eta = 0.94\times0.92\times0.99 = 0.8562$ (turbine × gearbox × generator)
• Fixed cost US$ 1200 × 10⁶; variable cost US$ 500/kW (present worth, annualised with the CRF); price US$ 48/MWh = US$ 0.048/kWh
• 10 m³/s is left in the river, so the available flow $=Q-10$
• Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.12042$ for $i=12\%$, $n=50$ 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

After deducting the 10 m³/s release, the available flows (m³/s) in descending order are: 2532, 2004, 1440, 1001, 820, 583, 199, 186, 108, 105, 83, 79.

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 per month of use $=0.73\times48 =$ USD 35.04 per year; the annual cost of 1 kW $= 500\times0.12042 =$ USD 60.21, so a kW must be used for at least 1.72 months, i.e. at least 2 months. The 2nd highest flow is 2014 m³/s (2004 m³/s after the 10 m³/s release); capacity beyond that would be used for only 1 month.

Trial design discharges

The net benefit is greatest at $Q_d = 2004$ m³/s (available flow after the release; river flow 2014 m³/s).

Results

• Installed capacity: $P = 9.81\times0.8562\times217\times2004 = 3,652,392\ \text{kW} = 3,652.39$ MW
• Firm (primary) energy, from the minimum available flow of 79 m³/s all year: $9.81\times0.8562\times217\times79\times8760 = 1,261,277,982\ \text{kWh} = 1,261.28$ GWh
• Total energy: $11,457.94$ GWh
• Secondary energy $= 11,457.94-1,261.28 = 10,196.66$ GWh
• Plant factor $= \dfrac{E}{P\times8760} = \dfrac{11,457.94\times10^6}{3,652,392\times8760} = 0.358$ (35.8%)

Answer: best installed capacity = 3,652.4 MW (design flow 2004 m³/s); firm energy = 1,261.3 GWh; secondary energy = 10,196.7 GWh; total energy = 11,457.9 GWh; plant factor = 0.36.