--- title: "RoR best installed capacity (NRs, 45 years)" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-best-capacity-nrs-10-percent" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-best-capacity-nrs-10-percent site: BCT Engineering --- # RoR best installed capacity (NRs, 45 years) Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2082 Bhadra, 10 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-nrs-10-percent - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question The mean monthly flow of a river at the headworks site of a ROR hydropower project is as follows: | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Q (m³/s) | 90 | 85 | 80 | 95 | 200 | 500 | 1,000 | 1,200 | 800 | 450 | 150 | 100 | The other data are as follows: net available head = 150 m; efficiency of turbine = 95%; efficiency of generator = 97%; fixed cost = NRs. 10,00,00,00,000; variable cost (present worth value) = NRs. 1,50,000/kW; energy price = NRs. 10 per unit; interest rate = 10%; economic life of the project = 45 years. Estimate the best installed capacity, firm energy, secondary energy, total energy, and power factor of the project. ## Short answer The best capacity is found by maximizing net annual benefit (energy revenue at NRs 10/kWh minus annualized cost using CRF=0.10139). Optimum design flow is 800 m³/s. Results: installed capacity = 1,084.8 MW; firm energy = 950.3 GWh; secondary energy = 3,157.7 GWh; total energy = 4,108.0 GWh; plant factor = 0.43. ## 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 = NRs 10,00,00,00,000 (= NRs 10,000 million); variable cost = NRs 1,50,000/kW (present worth); price = NRs 10/kWh - Net head $H = 150$ m; turbine 95% and generator 97%, so overall efficiency $\eta = 0.95\times0.97 = 0.9215$; no O&M cost is given - "Power factor" in the question is read as the **plant factor**. - Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.10139$ for $i=10\%$, $n=45$ years - Energy per month: $E_m = 9.81\,\eta\,H\,Q\,(730\ \text{h})$ (8760/12 h per month) - Annual cost $=(\text{fixed}+\text{variable}\times P)\,CRF$; annual benefit = energy × price #### Flow available for power Available flows (m³/s) in descending order: 1200, 1000, 800, 500, 450, 200, 150, 100, 95, 90, 85, 80. #### 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 = 730 kWh × NRs 10 = NRs 7,300, and the annual cost of 1 kW = 1,50,000 × 0.10139 = NRs 15,208. So a kW must be used for at least 2.08 months, i.e. at least 3 months in a year, to pay for itself. The 3rd highest flow is 800 m³/s, so capacity beyond 800 m³/s (used for only 2 months) is not justified. #### Trial design discharges | $Q_d$ (m³/s) | P (MW) | Energy (GWh) | Benefit (NRs M) | Cost (NRs M) | Net (NRs M) | |---|---|---|---|---|---| | 80 | 108.48 | 950.28 | 9,502.8 | 2,663.7 | 6,839.0 | | 85 | 115.26 | 1,004.72 | 10,047.2 | 2,766.8 | 7,280.3 | | 90 | 122.04 | 1,054.21 | 10,542.1 | 2,870.0 | 7,672.2 | | 95 | 128.82 | 1,098.76 | 10,987.6 | 2,973.1 | 8,014.5 | | 100 | 135.60 | 1,138.35 | 11,383.5 | 3,076.2 | 8,307.3 | | 150 | 203.40 | 1,484.81 | 14,848.1 | 4,107.3 | 10,740.7 | | 200 | 271.20 | 1,781.77 | 17,817.7 | 5,138.5 | 12,679.2 | | 450 | 610.19 | 3,019.11 | 30,191.1 | 10,294.1 | 19,896.9 | | 500 | 677.99 | 3,217.08 | 32,170.8 | 11,325.3 | 20,845.5 | | 800 | 1,084.79 | 4,107.96 | 41,079.6 | 17,512.1 | 23,567.5 | | 1000 | 1,355.99 | 4,503.91 | 45,039.1 | 21,636.6 | 23,402.5 | | 1200 | 1,627.18 | 4,701.89 | 47,018.9 | 25,761.2 | 21,257.7 | The net benefit is greatest at $Q_d = 800$ m³/s. #### Results - Installed capacity: $P = 9.81\times0.9215\times150\times800 = 1,084,790\ \text{kW} = 1,084.79$ MW - Firm (primary) energy, from the minimum flow of 80 m³/s all year: $9.81\times0.9215\times150\times80\times8760 = 950,275,865\ \text{kWh} = 950.28$ GWh - Total energy: $4,107.96$ GWh - Secondary energy $= 4,107.96-950.28 = 3,157.69$ GWh - Plant factor $= \dfrac{E}{P\times8760} = \dfrac{4,107.96\times10^6}{1,084,790\times8760} = 0.432$ (43.2%) **Answer: best installed capacity = 1,084.8 MW (design flow 800 m³/s); firm energy = 950.3 GWh; secondary energy = 3,157.7 GWh; total energy = 4,108.0 GWh; plant factor = 0.43.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What overall efficiency is used in the calculation? A. 0.95 B. 0.97 C. 0.9215 (turbine 0.95 x generator 0.97) D. 0.95 + 0.97 Answer: C. Efficiencies in series are multiplied: 0.95 x 0.97 = 0.9215. (Key point: Overall efficiency) 2. Why does the fixed cost not affect the optimum capacity? A. It is always zero B. It is the same for all capacities C. It is larger than the benefit D. It is paid by the government Answer: B. Fixed cost is independent of capacity, so it does not change the marginal comparison. (Key point: Role of fixed cost) 3. A kW earns NRs 7,300 per month of use. How is the minimum use determined? A. Months of use must be at least annual cost per kW divided by 7,300, about 3 months B. One month is always enough C. Twelve months are needed D. Use is not considered Answer: A. Annual cost of 1 kW is 1,50,000 x 0.10139 = about 15,200, so about 2.08 months, rounded up to 3. (Key point: Minimum months of use) 4. Which design flow gives the maximum net benefit? A. 1200 m³/s B. 1000 m³/s C. 500 m³/s D. 800 m³/s Answer: D. The net benefit table peaks at 800 m³/s (NRs 23,567.5 M). (Key point: Selected design flow) 5. How is firm energy found? A. Using the minimum flow of 80 m³/s all year B. Using mean flow C. Using the maximum flow D. Using Q_d for 2 months Answer: A. Firm energy uses the lowest flow (80) all year: 950.28 GWh. (Key point: Firm energy) 6. What formula gives plant factor? A. Installed capacity divided by energy B. Energy divided by (installed capacity x 8760 h) C. Firm energy divided by secondary energy D. Q_d divided by mean flow Answer: B. Plant factor = E/(P x 8760) = about 0.43 here. 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