--- title: "Best installed capacity at 12% interest" subject: "Hydro Power" chapter: 3 key: "semester_6/hydro_power:best-installed-capacity-12pct" source: https://www.bctengineer.com/q/semester_6/hydro_power/best-installed-capacity-12pct site: BCT Engineering --- # Best installed capacity at 12% interest Hydro Power · Chapter 3: Planning of Hydropower Projects - Asked in: 2082 Kartik, 10 marks - Video: [Determine the best Installed capacity for Nepalese river || Hydropower Engineering](https://www.youtube.com/watch?v=KZzhkqzBaHA) — OPengineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/best-installed-capacity-12pct - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3 ## Question Determine the best-installed capacity of a hydropower plant based on the given power duration data. Interest rate = 12%, Energy price = US$ 36/MWh, Variable cost = US$ 750/KW, Economic life of the project = 50 years. | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Power (KW) | 700 | 590 | 535 | 300 | 255 | 180 | 150 | 105 | 80 | 70 | 60 | 50 | ## 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. ## Answer 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. $$ \begin{aligned} CRF &= \frac{0.12(1.12)^{50}}{(1.12)^{50}-1} = 0.12042 \\ C_a &= 750 \times 0.12042 = 90.31\ \text{\$/kW/yr} \end{aligned} $$ 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 $P_c$: $E = 730\sum \min(P_i,P_c)$, Benefit $= 0.036E$, Cost $= 90.31P_c$. | Capacity (kW) | Energy (kWh/yr) | Benefit (\$) | Cost (\$) | Net (\$) | |---|---|---|---|---| | 50 | 438,000 | 15,768 | 4,516 | 11,252 | | 105 | 803,000 | 28,908 | 9,483 | 19,425 | | 150 | 1,032,950 | 37,186 | 13,547 | 23,639 | | 180 | 1,164,350 | 41,917 | 16,256 | 25,660 | | 255 | 1,438,100 | 51,772 | 23,030 | 28,742 | | **300** | **1,569,500** | **56,502** | **27,094** | **29,408** | | 535 | 2,084,150 | 75,029 | 48,317 | 26,712 | | 590 | 2,164,450 | 77,920 | 53,284 | 24,636 | | 700 | 2,244,750 | 80,811 | 63,219 | 17,592 | #### 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 = $90.31/(8760\times0.036) = 28.6\%$ 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 $= 50 \times 8760 = 438{,}000$ kWh - Total energy = 1,569,500 kWh; secondary energy = 1,131,500 kWh - Plant factor $= 1{,}569{,}500/(300\times8760) = 59.7\%$ **Answer: Best installed capacity = 300 kW (net benefit ≈ \$29,408 per year).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Since no O&M is given, what is the annual cost per kW? A. Variable cost x CRF = 750 x 0.12042 = 90.31 $/kW/yr B. Variable cost = 750 $/kW/yr C. Variable cost / 50 D. Variable cost x 0.12 Answer: A. Only capital recovery with CRF (12%, 50 yr) is used. (Key point: Annual cost without O&M) 2. For capacity Pc, how is annual energy found from the monthly powers? A. Pc x 8760 B. Sum of all monthly powers x 730 C. 730 x sum of min(Pi, Pc) D. Maximum power x 730 Answer: C. Each month yields the smaller of the available power and the installed capacity, over 730 h. (Key point: Energy at a capacity) 3. Which capacity gives the maximum net benefit? A. 535 kW B. 300 kW C. 700 kW D. 180 kW Answer: B. 300 kW gives about $29,408/yr, higher than the other capacities tried. (Key point: Best capacity) 4. Why is capacity above 300 kW not worth adding? A. The interest rate is below 10% B. Firm energy would be lost C. Power in that band is available only about 3 months, below the 28.6% break-even D. Annual cost is zero Answer: C. Break-even is 90.31/(8760 x 0.036) = 28.6% of the year (3.4 months), but the band runs only 3 months. 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