--- title: "Full regulation: reservoir, capacity, annual energy" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:storage-full-regulation-monthly-hydrograph" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/storage-full-regulation-monthly-hydrograph site: BCT Engineering --- # Full regulation: reservoir, capacity, annual energy Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2071 Chaitra, 3+2+3 marks - Video: [Storage Capacity of Reservoir Using Mass Curve](https://www.youtube.com/watch?v=SCYbpf6EvvA) — ENG-School - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/storage-full-regulation-monthly-hydrograph - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question A hydropower project is planned to develop in a Nepalese river having net head of 150 m, turbine efficiency of 90% and generator efficiency of 95% with the monthly hydrograph as shown below: | Months | Oct | Nov | Dec | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Q (m³/sec) | 100 | 80 | 60 | 50 | 40 | 30 | 40 | 50 | 70 | 110 | 150 | 120 | As an environmental flow, a minimum flow of 10% of each month is mandatory. If the storage project is designed with full regulation of the annual hydrograph, find out the capacity of the reservoir, installed capacity of the power plant, and annual energy generation. ## Short answer Reservoir capacity ≈ 437.6 Mm³, installed capacity ≈ 84.9 MW, annual energy ≈ 744 GWh. ## Answer #### Data $H=150$ m, $\eta=\eta_t\eta_g = 0.90\times0.95=0.855$. Environmental flow = 10% of each month's flow, so the flow available for power is $0.9\,Q$. One month $=2.628\times10^6$ s. Sum of flows $=900$ m³/s, mean $=75$ m³/s. Mean available flow $=0.9\times75 = 67.5$ m³/s. With full regulation the design discharge is this mean flow, so the plant runs at a constant $Q_d=67.5$ m³/s. #### Reservoir capacity (mass-curve method) The reservoir is full at the start of December, when the inflow drops below the draft. Cumulative of (0.9Q − 67.5): | Month | Q (m³/s) | 0.9Q | 0.9Q − mean | Cumulative (m³/s-month) | |---|---|---|---|---| | Dec | 60 | 54.0 | -13.5 | -13.5 | | Jan | 50 | 45.0 | -22.5 | -36.0 | | Feb | 40 | 36.0 | -31.5 | -67.5 | | Mar | 30 | 27.0 | -40.5 | -108.0 | | Apr | 40 | 36.0 | -31.5 | -139.5 | | May | 50 | 45.0 | -22.5 | -162.0 | | Jun | 70 | 63.0 | -4.5 | -166.5 | | Jul | 110 | 99.0 | +31.5 | -135.0 | | Aug | 150 | 135.0 | +67.5 | -67.5 | | Sep | 120 | 108.0 | +40.5 | -27.0 | | Oct | 100 | 90.0 | +22.5 | -4.5 | | Nov | 80 | 72.0 | +4.5 | +0.0 | The cumulative reaches its lowest value of $-166.5$ m³/s-months in June; from July the surplus refills the reservoir. $$ V = 166.5\times2.628\times10^6 = 437.6\times10^6\ \text{m}^3 \approx 438\ \text{Mm}^3 $$ #### Installed capacity $$ P = 9.81\,\eta\,H\,Q_d = 9.81\times0.855\times150\times67.5 = 84,924\ \text{kW} = 84.92\ \text{MW} $$ #### Annual energy $$ E = P\times8760 = 743.9\ \text{GWh} $$ (The head is taken as constant at 150 m; in practice the head varies with the reservoir level.) **Answer: reservoir capacity ≈ 437.6 Mm³; installed capacity ≈ 84.9 MW; annual energy ≈ 744 GWh.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which flow is available for power when 10% of each month's flow is mandatory release? A. Q B. 0.9 Q C. 0.1 Q D. Q - 10 Answer: B. 10% is released, leaving 0.9 Q. (Key point: Flow after environmental release) 2. What is the design discharge for full regulation of the annual hydrograph? A. Minimum flow B. Maximum flow C. Mean available flow, 0.9 x 75 = 67.5 m3/s, so the plant runs constantly D. Q40 Answer: C. Full regulation levels out the year, so the plant discharges the mean available flow all year. (Key point: Design discharge with full regulation) 3. Overall efficiency used is A. 0.90 + 0.95 B. 0.90 x 0.95 = 0.855 C. 0.95 only D. 0.90 only Answer: B. Turbine and generator efficiencies multiply. (Key point: Overall efficiency) 4. Which gives the installed capacity and annual energy? A. 9.81 x 0.855 x 150 x 67.5 = 84.9 MW and E = P x 8760 B. 9.81 x 150 x 100 MW C. P x 730 D. 9.81 x 0.855 x 150 x 120 Answer: A. P = 9.81 eta H Qd and annual energy = P x 8760, about 744 GWh. 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