--- title: "Best installed capacity at 10% interest" subject: "Hydro Power" chapter: 3 key: "semester_6/hydro_power:best-installed-capacity-10pct" source: https://www.bctengineer.com/q/semester_6/hydro_power/best-installed-capacity-10pct site: BCT Engineering --- # Best installed capacity at 10% interest Hydro Power · Chapter 3: Planning of Hydropower Projects - Asked in: 2082 Chaitra, 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-10pct - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3 ## Question The monthly power output (in kW) for a proposed hydropower plant is given below. The interest rate is 10%, energy price is US$ 40/MWh, variable cost is US$ 800/kW, annual operation and maintenance cost is 2.5% of variable cost, and the economic life of the project is 40 years. Determine the best-installed capacity, firm energy, secondary energy, total energy, and plant factor. | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Power (kW) | 80 | 65 | 95 | 200 | 320 | 400 | 510 | 650 | 560 | 270 | 140 | 110 | ## Short answer Best installed capacity is 400 kW, maximizing net annual benefit at $43,373. Firm energy is 0.569 GWh, secondary energy 1.533 GWh, total energy 2.102 GWh, plant factor 60%. Determined by comparing incremental benefit and cost across capacity levels using capital recovery factor and O&M. ## Answer 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 | Capacity (kW) | Energy (kWh/yr) | Benefit (\$) | Cost (\$) | Net (\$) | |---|---|---|---|---| | 65 | 569,400 | 22,776 | 6,618 | 16,159 | | 140 | 1,073,100 | 42,924 | 14,253 | 28,671 | | 200 | 1,379,700 | 55,188 | 20,362 | 34,827 | | 270 | 1,686,300 | 67,452 | 27,488 | 39,964 | | 320 | 1,868,800 | 74,752 | 32,578 | 42,174 | | **400** | **2,102,400** | **84,096** | **40,723** | **43,373** | | 510 | 2,343,300 | 93,732 | 51,922 | 41,810 | | 560 | 2,416,300 | 96,652 | 57,012 | 39,640 | | 650 | 2,482,000 | 99,280 | 66,175 | 33,105 | 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%.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What defines the best installed capacity? A. The maximum monthly power B. The capacity with maximum net annual benefit (energy value minus annual cost) C. The capacity giving 100% plant factor D. The minimum monthly power Answer: B. The last increment of capacity must earn at least as much as it costs, which is where net benefit peaks. (Key point: Optimum capacity criterion) 2. How is the annual cost per kW computed? A. Variable cost only B. Variable cost x O&M 2.5% only C. Variable cost x (CRF + 0.025) D. Variable cost / economic life Answer: C. Ca = 800 x (0.10226 + 0.025) = 101.81 $/kW/yr, capital recovery plus O&M. (Key point: Annual cost per kW) 3. Why is going from 400 to 510 kW not worth it? A. The extra 110 kW costs about $11,199 but earns only about $9,636 B. The plant would exceed the river flow all year C. Interest rate is too low D. Firm energy would fall Answer: A. The added band runs only 3 months, below the roughly 29% of the year needed to break even. (Key point: Incremental check) 4. What are firm energy and plant factor for 400 kW (minimum power 65 kW, total 2,102,400 kWh)? A. 0.569 GWh and 40% B. 2.102 GWh and 60% C. 0.569 GWh and 60% D. 1.533 GWh and 60% Answer: C. Firm = 65 x 8760 = 569,400 kWh; PF = 2,102,400/(400 x 8760) = 0.60. 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