Best installed capacity at 10% interest

The best (optimum) installed capacity is the capacity at which the net annual benefit (value of energy sold minus annual cost of the capacity) is maximum. Equivalently, the last increment of capacity added should earn at least as much as it costs.

Data and assumptions

• Each month is taken as $8760/12 = 730$ h.
• Energy price = 40 $/MWh = 0.04 $/kWh.
• Capital recovery factor (10%, 40 years):

$$\begin{aligned} CRF &= \frac{i(1+i)^n}{(1+i)^n-1} = \frac{0.10(1.10)^{40}}{(1.10)^{40}-1} = 0.10226 \end{aligned}$$

• Annual cost per kW of capacity = capital charge + O&M:

$$\begin{aligned} C_a &= 800\,(0.10226 + 0.025) = 101.81\ \text{\$/kW/yr} \end{aligned}$$

Power duration (descending order)

650, 560, 510, 400, 320, 270, 200, 140, 110, 95, 80, 65 kW. A capacity $P_c$ produces $\min(P_i, P_c)$ in each month.

Annual energy for capacity $P_c$: $E = 730 \sum \min(P_i, P_c)$ kWh.

Benefit–cost table

Benefit = $E \times 0.04$; Cost = $P_c \times 101.81$.

Incremental check: going from 320 to 400 kW adds 80 kW (cost $8,145) and 233,600 kWh (benefit $9,344): worth it. Going from 400 to 510 kW adds 110 kW (cost $11,199) but only 240,900 kWh (benefit $9,636): not worth it. A kW is worth installing only if it runs at least $101.81/(8760 \times 0.04) = 29\%$ of the year (about 3.5 months); the 400–510 kW band runs only 3 months.

Results

• Best installed capacity = 400 kW (maximum net benefit $43,373/yr).
• Firm energy (minimum power 65 kW available all year):

$$E_f = 65 \times 8760 = 569{,}400\ \text{kWh} = 0.569\ \text{GWh}$$

• Total energy at 400 kW = 2,102,400 kWh = 2.102 GWh.
• Secondary energy = 2,102,400 − 569,400 = 1,533,000 kWh = 1.533 GWh.
• Plant factor:

$$PF = \frac{2{,}102{,}400}{400 \times 8760} = 0.60 = 60\%$$

Answer: Installed capacity = 400 kW; firm energy = 0.569 GWh; secondary energy = 1.533 GWh; total energy = 2.102 GWh; plant factor = 60%.