--- title: "RoR capacity, peaking capacity and benefit" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-three-sets-peaking-benefit" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-three-sets-peaking-benefit site: BCT Engineering --- # RoR capacity, peaking capacity and benefit Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2070 Ashad, 2+4+2 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-three-sets-peaking-benefit - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question The mean monthly flow of a typical Nepalese river is as follows: | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Q (m³/s) | 80 | 74 | 83 | 100 | 130 | 222 | 600 | 800 | 590 | 240 | 120 | 100 | i) Calculate the installed capacity of a plant based on the minimum flow of the river without pondage (if the plant is designed for a pure run-of-river plant) with net head of 200 m and overall efficiency of the plant 85%. ii) The plant has three sets of units (turbine and generator) such that one unit with full capacity is operated during off-peak hours. If the plant is designed as a peaking plant with 4 hour peaking (morning 2 hour and evening 2 hour), what will be the installed capacity of the plant? iii) What will be the increase in benefit from peaking if the peak hour energy rate is Rs 12/kWh and the off-peak energy rate is Rs 6/kWh during the minimum flow month? ## Short answer (i) 123.4 MW. (ii) 277.7 MW. (iii) Extra benefit ≈ Rs 3.70 million per day (≈ Rs 111 million per month). ## Answer #### i) Pure run-of-river plant Minimum flow is 74 m³/s (February). $$ P_1 = 9.81\times0.85\times74\times200 = 123,410\ \text{kW} = 123.41\ \text{MW} $$ #### ii) Peaking plant with three units, 4 h peaking One unit (discharge $Q_u$) runs at full load for the 20 off-peak hours; all three units ($3Q_u$) run for the 4 peak hours. Daily water balance on the minimum flow of 74 m³/s: $$ 20\,Q_u + 4\,(3Q_u) = 24\times74 \Rightarrow 32\,Q_u = 1776 \Rightarrow Q_u = 55.50\ \text{m}^3/\text{s} $$ Peak discharge $=3Q_u=166.50$ m³/s. $$ P_2 = 9.81\times0.85\times166.50\times200 = 277,672\ \text{kW} = 277.67\ \text{MW} $$ (Each unit is 92.56 MW. Check storage: off-peak stored $=(74-55.5)\times20 = 370$ m³/s·h; peak draw $=(166.5-74)\times4=370$ m³/s·h, equal.) #### iii) Increase in benefit during the minimum flow month Daily energy is the same in both cases ($24\times123.41 = 2,962$ MWh). - **RoR (constant output):** 4 peak hours at Rs 12 and 20 off-peak hours at Rs 6 per kWh: $$ B_1 = 123,410\times(4\times12+20\times6) = \text{Rs }20,732,846\text{ per day} $$ - **Peaking plant:** $B_2 = 92,557\times20\times6 + 277,672\times4\times12 = \text{Rs }24,435,140$ per day Increase $=3,702,294$ Rs/day $\approx$ Rs 3.70 million per day, or about Rs 111.1 million for a 30-day month. **Answer: (i) 123.4 MW; (ii) 277.7 MW; (iii) extra benefit ≈ Rs 3.70 million per day (≈ Rs 111 million per month).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which flow is used for the pure run-of-river installed capacity? A. Mean flow B. Minimum monthly flow, 74 m3/s in February C. 800 m3/s, the maximum D. Flow exceeded 40% Answer: B. P1 = 9.81 x 0.85 x 74 x 200 = 123.4 MW. (Key point: Pure RoR basis) 2. What equation gives the unit discharge Qu with 3 units and 4 h peaking? A. 20 Qu + 4 (3Qu) = 24 x 74 B. 3Qu = 74 C. 24 Qu = 74 D. 20 x 3Qu + 4 Qu = 24 x 74 Answer: A. One unit runs for 20 off-peak hours and three units for 4 peak hours, using the daily minimum-flow volume, so Qu = 55.5 m3/s. (Key point: Water balance for peaking) 3. What is the installed capacity of the peaking plant? A. 123.4 MW B. About 277.7 MW, for the peak discharge 3Qu = 166.5 m3/s C. 92.6 MW D. 55.5 MW Answer: B. P2 = 9.81 x 0.85 x 166.5 x 200 = 277.7 MW. (Key point: Peaking capacity) 4. Why does peaking give extra benefit when the daily energy is the same? A. More energy is generated B. Energy is shifted into peak hours that are paid Rs 12/kWh instead of Rs 6/kWh C. Efficiency increases D. The head increases Answer: B. Same daily energy, but a greater share is sold at the peak rate, giving about Rs 3.70 million more per day. 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