--- title: "RoR best installed capacity (USD 2795 million)" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-best-capacity-usd-2795" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-best-capacity-usd-2795 site: BCT Engineering --- # RoR best installed capacity (USD 2795 million) Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2082 Baishakh, 9 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-best-capacity-usd-2795 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question The mean monthly flow of a river in a typical year is as follows: | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Flow (m³/s) | 38 | 33 | 28 | 23 | 18 | 36 | 73 | 93 | 68 | 53 | 48 | 43 | The effective head of the ROR plant is 48 m and can be assumed constant. The other data are as follows: fixed cost = USD 2795 × 10⁶; variable cost = US\$ 645/kW; annual O&M cost = 2.5% of variable cost; energy price = US\$ 38/MWh for primary and secondary energy; interest rate = 11%; economic life of the project = 45 years; overall efficiency = 89%. Determine the best installed capacity, firm energy, secondary energy, total energy, and plant factor. ## Short answer The optimum design flow is 53 m³/s, giving installed capacity 22.2 MW. Firm energy is 66.1 GWh, secondary energy 80.5 GWh, total energy 146.5 GWh, and the plant factor is 0.75. ## Answer 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 - Fixed cost = USD 2795 × 10⁶; variable cost = USD 645/kW; O&M = 2.5% of the variable cost per year; price = USD 38/MWh = USD 0.038/kWh for primary and secondary energy - Net head $H = 48$ m (constant); overall efficiency $\eta = 0.89$ - Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.11101$ for $i=11\%$, $n=45$ years - Energy per month: $E_m = 9.81\,\eta\,H\,Q\,(730\ \text{h})$ (8760/12 h per month) - Annual cost $=(\text{fixed}+\text{variable}\times P)\,CRF + \text{O\&M}$; annual benefit = energy × price #### Flow available for power Available flows (m³/s) in descending order: 93, 73, 68, 53, 48, 43, 38, 36, 33, 28, 23, 18. #### 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 for each month it is used $=0.73\times38 = $ USD 27.74 per year. The annual cost of 1 kW $= 645\times0.11101 + 0.025\times645 = 71.60 + 16.13 =$ USD 87.73, so a kW must be used for at least 3.16 months, i.e. at least 4 months. The 4th highest flow is 53 m³/s, so the optimum design flow is 53 m³/s (capacity up to 68 m³/s would be used for only 3 months: 3 × 27.74 = 83.22 < 87.73). #### Trial design discharges | $Q_d$ (m³/s) | P (MW) | Energy (GWh) | Benefit (USD M) | Cost (USD M) | Net (USD M) | |---|---|---|---|---|---| | 18 | 7.54 | 66.08 | 2.5 | 310.9 | -308.4 | | 23 | 9.64 | 82.91 | 3.2 | 311.1 | -308.0 | | 28 | 11.73 | 98.20 | 3.7 | 311.3 | -307.6 | | 33 | 13.83 | 111.97 | 4.3 | 311.5 | -307.2 | | 36 | 15.09 | 119.31 | 4.5 | 311.6 | -307.1 | | 38 | 15.93 | 123.60 | 4.7 | 311.7 | -307.0 | | 43 | 18.02 | 132.77 | 5.0 | 311.9 | -306.8 | | 48 | 20.12 | 140.42 | 5.3 | 312.0 | -306.7 | | 53 | 22.21 | 146.54 | 5.6 | 312.2 | -306.7 | | 68 | 28.50 | 160.31 | 6.1 | 312.8 | -306.7 | | 73 | 30.59 | 163.37 | 6.2 | 313.0 | -306.8 | | 93 | 38.97 | 169.49 | 6.4 | 313.7 | -307.3 | The values for 48, 53 and 68 m³/s look tied at this rounding; the marginal check above shows the net benefit is greatest at $Q_d = 53$ m³/s. #### Results - Installed capacity: $P = 9.81\times0.89\times48\times53 = 22,211\ \text{kW} = 22.21$ MW - Firm (primary) energy, from the minimum flow of 18 m³/s all year: $9.81\times0.89\times48\times18\times8760 = 66,081,039\ \text{kWh} = 66.08$ GWh - Total energy: $146.54$ GWh - Secondary energy $= 146.54-66.08 = 80.46$ GWh - Plant factor $= \dfrac{E}{P\times8760} = \dfrac{146.54\times10^6}{22,211\times8760} = 0.753$ (75.3%) Note: with the fixed cost taken exactly as stated (USD 2795 million), the net benefit is negative for every capacity (the project does not pay for itself). The optimum capacity is the same because the fixed cost is independent of capacity; the fixed cost is possibly a misprint in the question. **Answer: best installed capacity = 22.2 MW (design flow 53 m³/s); firm energy = 66.1 GWh; secondary energy = 80.5 GWh; total energy = 146.5 GWh; plant factor = 0.75.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the net benefit negative for all capacities but the optimum is still valid? A. The optimum is computed wrongly B. Fixed cost is independent of capacity, so it shifts all net values equally C. The efficiency is negative D. The price is zero Answer: B. The large fixed cost lowers every net value by the same amount, so the best capacity is unchanged. (Key point: Net benefit with fixed cost) 2. What is the annual cost of 1 kW of capacity here? A. 645 x 0.11101 only, ignoring O&M B. 645 x 0.025 C. 645 x 0.11101 + 2.5% of 645 = about USD 87.73 D. 38 x 0.73 Answer: C. CRF cost 71.6 plus O&M 16.1 gives about 87.7 USD/yr. (Key point: Annual cost per kW) 3. A kW earns USD 27.74 per month of use. How many months are needed? A. At least 4 months (3.16 rounded up) B. 1 month C. 12 months D. At least 8 months Answer: A. 87.73/27.74 = 3.16, so at least 4 months, making the 4th highest flow (53 m³/s) the design flow. (Key point: Months of use) 4. What is the installed capacity at design flow 53 m³/s? A. 38.97 MW B. 22.21 MW C. 15.93 MW D. 9.64 MW Answer: B. P = 9.81 x 0.89 x 48 x 53 = 22,211 kW. (Key point: Installed capacity) 5. How are firm and secondary energy obtained? A. Firm from maximum flow; secondary from minimum B. Firm from minimum flow 18 m³/s all year; secondary = total minus firm C. Both equal D. Firm from design flow Answer: B. Firm = 66.08 GWh; secondary = 146.54 - 66.08 = 80.46 GWh. 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