--- title: "RoR best installed capacity (USD 1200 million)" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-best-capacity-usd-1200" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-best-capacity-usd-1200 site: BCT Engineering --- # RoR best installed capacity (USD 1200 million) Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2081 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-usd-1200 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question The mean monthly flows of a river in a typical year are as follows: | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sept | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Flow (m³/s) | 93 | 89 | 115 | 209 | 593 | 1011 | 2014 | 2542 | 1450 | 830 | 196 | 118 | The gross available head is 229 m, and the average head loss in transition is 12 m. The efficiencies of the turbines, gearboxes and generators are 94%, 92% and 99% respectively. 10 m³/s of water must be left in the river for downstream users. The other data are as follows: fixed cost = US\$ 1200 × 10⁶; variable cost (present worth value) = US\$ 500/kW; energy price = US\$ 48/MWh for primary and secondary energy; interest rate = 12%; economic life of the project = 50 years. Determine the best installed capacity, firm energy, secondary energy, total energy and plant factor of a run-of-river plant. ## Short answer With 10 m³/s left in the river, the optimum design flow is 2004 m³/s, yielding installed capacity 3652 MW. Firm energy is 1261 GWh, secondary energy 10197 GWh, total energy 11458 GWh, and plant factor 0.36. ## 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 - Net head $H = 229-12 = 217$ m (gross head minus the average transition loss) - Overall efficiency $\eta = 0.94\times0.92\times0.99 = 0.8562$ (turbine × gearbox × generator) - Fixed cost US\$ 1200 × 10⁶; variable cost US\$ 500/kW (present worth, annualised with the CRF); price US\$ 48/MWh = US\$ 0.048/kWh - 10 m³/s is left in the river, so the available flow $=Q-10$ - Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.12042$ for $i=12\%$, $n=50$ 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 After deducting the 10 m³/s release, the available flows (m³/s) in descending order are: 2532, 2004, 1440, 1001, 820, 583, 199, 186, 108, 105, 83, 79. #### 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 per month of use $=0.73\times48 = $ USD 35.04 per year; the annual cost of 1 kW $= 500\times0.12042 = $ USD 60.21, so a kW must be used for at least 1.72 months, i.e. at least 2 months. The 2nd highest flow is 2014 m³/s (2004 m³/s after the 10 m³/s release); capacity beyond that would be used for only 1 month. #### Trial design discharges | $Q_d$ (m³/s) | P (MW) | Energy (GWh) | Benefit (USD M) | Cost (USD M) | Net (USD M) | |---|---|---|---|---|---| | 79 | 143.98 | 1,261.28 | 60.5 | 153.2 | -92.6 | | 83 | 151.27 | 1,319.82 | 63.4 | 153.6 | -90.3 | | 105 | 191.37 | 1,612.52 | 77.4 | 156.0 | -78.6 | | 108 | 196.84 | 1,648.44 | 79.1 | 156.4 | -77.2 | | 186 | 338.99 | 2,478.65 | 119.0 | 164.9 | -45.9 | | 199 | 362.69 | 2,599.72 | 124.8 | 166.3 | -41.6 | | 583 | 1,062.55 | 5,665.11 | 271.9 | 208.5 | 63.5 | | 820 | 1,494.49 | 7,241.70 | 347.6 | 234.5 | 113.1 | | 1001 | 1,824.37 | 8,204.96 | 393.8 | 254.3 | 139.5 | | 1440 | 2,624.47 | 9,957.18 | 477.9 | 302.5 | 175.4 | | 2004 | 3,652.39 | 11,457.94 | 550.0 | 364.4 | 185.6 | | 2532 | 4,614.70 | 12,160.42 | 583.7 | 422.3 | 161.4 | The net benefit is greatest at $Q_d = 2004$ m³/s (available flow after the release; river flow 2014 m³/s). #### Results - Installed capacity: $P = 9.81\times0.8562\times217\times2004 = 3,652,392\ \text{kW} = 3,652.39$ MW - Firm (primary) energy, from the minimum available flow of 79 m³/s all year: $9.81\times0.8562\times217\times79\times8760 = 1,261,277,982\ \text{kWh} = 1,261.28$ GWh - Total energy: $11,457.94$ GWh - Secondary energy $= 11,457.94-1,261.28 = 10,196.66$ GWh - Plant factor $= \dfrac{E}{P\times8760} = \dfrac{11,457.94\times10^6}{3,652,392\times8760} = 0.358$ (35.8%) **Answer: best installed capacity = 3,652.4 MW (design flow 2004 m³/s); firm energy = 1,261.3 GWh; secondary energy = 10,196.7 GWh; total energy = 11,457.9 GWh; plant factor = 0.36.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the net head determined? A. 229 + 12 = 241 m B. 229 m C. 217 m D. 12 m Answer: C. Net head = gross head 229 minus loss 12 = 217 m. (Key point: Net head) 2. What is the overall efficiency? A. 0.94 x 0.92 x 0.99 = 0.8562 B. 0.94 C. 0.99 D. 0.94 + 0.92 + 0.99 Answer: A. Series efficiencies are multiplied. (Key point: Efficiency) 3. How is the 10 m³/s downstream release handled? A. Added to flow B. Ignored C. Deducted from each monthly flow to get flow available for power D. Subtracted only in the dry season Answer: C. Available flow = Q - 10, e.g. 2014 becomes 2004 m³/s. (Key point: Environmental release) 4. A kW earns 0.73 x 48 = USD 35.04 per month of use; the annual cost is USD 60.21. What minimum use results? A. 12 months B. 6 months C. At least 2 months (1.72 rounded up) D. None Answer: C. 60.21/35.04 = 1.72, so at least 2 months, so the second highest flow gives the design flow. (Key point: Minimum months of use) 5. Which design discharge gives maximum net benefit? A. 2532 m³/s B. 2004 m³/s C. 1440 m³/s D. 1001 m³/s Answer: B. Net benefit peaks at USD 185.6 M for 2004 m³/s, giving P = 3,652 MW. (Key point: Best capacity) 6. What is the firm energy based on? A. Maximum available flow B. Mean flow C. Design flow D. Minimum available flow 79 m³/s all year (1,261.28 GWh) Answer: D. Firm energy corresponds to the lowest flow available all year. 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