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Best installed capacity at 12% interest

Short answer

Best installed capacity is 300 kW with net annual benefit of $29,408. Firm energy 438,000 kWh, total energy 1,569,500 kWh, secondary energy 1,131,500 kWh, plant factor 59.7%. Found by evaluating net benefit at each power duration level using capital recovery factor at 12% over 50 years.

Determine the best-installed capacity of a hydropower plant based on the given power duration data. Interest rate = 12%, Energy price = US36/MWh,Variablecost=US 36/MWh, Variable cost = US 750/KW, Economic life of the project = 50 years.

MonthJanFebMarAprMayJunJulAugSepOctNovDec
Power (KW)70059053530025518015010580706050

The best installed capacity is the one that gives the largest net annual benefit (revenue from energy minus annualised cost of capacity). It is found by trying each power level of the power duration data as the plant capacity.

Annual cost of capacity

No O&M cost is given, so only capital recovery is used.

CRF=0.12(1.12)50(1.12)50−1=0.12042Ca=750×0.12042=90.31 $/kW/yr\begin{aligned} CRF &= \frac{0.12(1.12)^{50}}{(1.12)^{50}-1} = 0.12042 \\ C_a &= 750 \times 0.12042 = 90.31\ \text{\$/kW/yr} \end{aligned}

Energy price = 36 $/MWh = 0.036 $/kWh. Each month = 730 h.

Power duration (descending order)

700, 590, 535, 300, 255, 180, 150, 105, 80, 70, 60, 50 kW.

For capacity PcP_c: E=730∑min⁡(Pi,Pc)E = 730\sum \min(P_i,P_c), Benefit =0.036E= 0.036E, Cost =90.31Pc= 90.31P_c.

Capacity (kW)Energy (kWh/yr)Benefit ($)Cost ($)Net ($)
50438,00015,7684,51611,252
105803,00028,9089,48319,425
1501,032,95037,18613,54723,639
1801,164,35041,91716,25625,660
2551,438,10051,77223,03028,742
3001,569,50056,50227,09429,408
5352,084,15075,02948,31726,712
5902,164,45077,92053,28424,636
7002,244,75080,81163,21917,592

Incremental check

  • 255 → 300 kW: +45 kW costs $4,064; extra energy 131,400 kWh (4 months) earns $4,730. Worth adding.
  • 300 → 535 kW: +235 kW costs $21,223; extra energy 514,650 kWh (3 months) earns $18,527. Not worth adding.

Break-even running time = 90.31/(8760×0.036)=28.6%90.31/(8760\times0.036) = 28.6\% of the year (3.4 months). The band above 300 kW runs only 3 months, so it does not pay.

Results at 300 kW

  • Firm energy =50×8760=438,000= 50 \times 8760 = 438{,}000 kWh
  • Total energy = 1,569,500 kWh; secondary energy = 1,131,500 kWh
  • Plant factor =1,569,500/(300×8760)=59.7%= 1{,}569{,}500/(300\times8760) = 59.7\%

Answer: Best installed capacity = 300 kW (net benefit ≈ $29,408 per year).

✓ Answer checked by Claude Opus 5.5 (high effort)

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