--- title: "Peaking RoR: max power and wet/dry energy ratio" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:peaking-ror-reservoir-250m-q25" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/peaking-ror-reservoir-250m-q25 site: BCT Engineering --- # Peaking RoR: max power and wet/dry energy ratio Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2078 Bhadra, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/peaking-ror-reservoir-250m-q25 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question A peaking ROR project in western Nepal with net head of 250 m has the following river flow data: | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | River flow (m³/s) | 100 | 80 | 80 | 105 | 200 | 500 | 1100 | 1200 | 800 | 350 | 200 | 120 | The storage capacity available for this project is 1100 million m³. This storage capacity is utilized for dry months (Nov-May) during which the plant is used as a peak load plant operating 4 hours a day. Considering design flow as Q₂₅, calculate the maximum power generation (in MW) and the ratio of wet season energy to dry season energy. ## Short answer Design flow Q25 is 800 m³/s. Maximum power is 1962 MW for 100% efficiency (1668 MW for 85% efficiency). Wet season to dry season energy ratio is 3.48. ## Answer #### Data and assumptions Net head $H=250$ m. Dry months (Nov-May, 7 months) use stored water for 4 h/day of peaking. Wet months (Jun-Oct, 5 months) the plant runs as RoR for 24 h. One month $=730$ h $=2.628\times10^6$ s. Overall efficiency is not given, so power is first found for $\eta=1$ and then for a typical $\eta=0.85$; the energy ratio does not depend on $\eta$. #### Design flow $Q_{25}$ Arrange the 12 monthly flows in descending order: 1200, 1100, 800, 500, 350, 200, 200, 120, 105, 100, 80, 80. The flow equalled or exceeded 25% of the time (3 months out of 12) is the 3rd value: $$ Q_{25} = 800\ \text{m}^3/\text{s} $$ #### Is the stored water enough for peaking at $Q_{25}$? - Dry-season inflow volume $=\sum Q\times 2.628\times10^6 = 885\times2.628\times10^6 = 2,326\times10^6$ m³ - Plus live storage $=1100\times10^6$ m³, so water available $=3,426\times10^6$ m³ - Peaking hours in 7 months $=7\times730\times\frac{4}{24} = 851.7$ h - Water needed to run at $Q_{25}$ for these hours $=800\times851.7\times3600 = 2,453\times10^6$ m³ (less than available) So the turbines can run at $Q_{25}=800$ m³/s during the 4 peak hours every day, and the plant capacity is controlled by the design flow (the water available would allow up to 1,117 m³/s). #### Maximum power $$ P_{max} = 9.81\,\eta\,Q\,H = 9.81\times\eta\times800\times250 = 1,962\,\eta\ \text{MW} $$ - With $\eta = 1$: $P_{max} = 1,962$ MW - With $\eta = 0.85$: $P_{max} = 1,668$ MW #### Wet-season to dry-season energy ratio Wet season flows used: $\min(Q,800)$ = 500, 800, 800, 800, 350, total $=3250$ m³/s-months. $$ E_{wet} \propto 3250\times730 = 2,372,500,\qquad E_{dry} \propto 800\times851.7 = 681,333 $$ $$ \frac{E_{wet}}{E_{dry}} = \frac{2,372,500}{681,333} = 3.48 $$ **Answer: maximum power = 1,962 MW for $\eta=1$ (1,668 MW for $\eta=0.85$); wet season energy / dry season energy = 3.48.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is Q25 found from 12 monthly flows? A. The average of the three largest flows B. The flow equalled or exceeded 25% of the time, i.e. the 3rd largest of 12 flows (800 m3/s) C. The 25th largest flow D. The flow equal to 25% of the mean Answer: B. 25% of 12 months is 3 months, so the third highest flow, 800 m3/s, is Q25. (Key point: Design flow Q25) 2. How does the plant operate in the dry months Nov-May? A. Run-of-river for 24 h B. As a peaking plant for 4 h a day using inflow plus stored water C. It is shut down D. At 50% load all day Answer: B. Stored water plus inflow is used for 4 h/day peaking in the dry months; in the wet months it runs 24 h as RoR. (Key point: Dry season operation) 3. Why is the water needed at Q25 for the peak hours compared with the water available? A. To check that inflow plus storage can supply Q25 during peaking hours, so the design flow controls capacity B. To find the efficiency C. To determine the head D. To compute the spill energy Answer: A. Needed 2,453 Mm3 is less than available 3,426 Mm3, so the turbines can really run at 800 m3/s. (Key point: Check storage sufficiency) 4. What is the maximum power for eta = 1? A. 9.81 x 800 x 250 = 1,962 MW B. 9.81 x 1200 x 250 = 2,943 MW C. 9.81 x 500 x 250 = 1,226 MW D. 9.81 x 800 x 250 x 24/4 = 11,772 MW Answer: A. P = 9.81 Q H with Q = 800 and H = 250 gives 1,962 MW. (Key point: Maximum power) 5. Why is the wet season total taken as 3250 m3/s-months? A. Wet flows are capped at 800: 500+800+800+800+350 B. It is the sum of all wet flows (5,750) C. It is the mean wet flow D. It is the flow of the dry season Answer: A. Flow above the design flow is spilled, so Q used = min(Q, 800) in Jun-Oct. 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