--- title: "Minimum reservoir capacity and wasted water" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:reservoir-minimum-capacity-wasted-water-1000hp" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/reservoir-minimum-capacity-wasted-water-1000hp site: BCT Engineering --- # Minimum reservoir capacity and wasted water Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2080 Bhadra, 4+4 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/reservoir-minimum-capacity-wasted-water-1000hp - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question The water turbine at a hydel storage plant produces 1000 HP when working under a net head of 30 m and with an overall efficiency of 80%. The inflow in the reservoir during a year is given below. | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Q (Mm³) | 90 | 80 | 73 | 80 | 70 | 98 | 120 | 80 | 96 | 105 | 100 | 75 | Find (i) the minimum capacity required, (ii) the total quantity of water wasted. (Assume the reservoir is full at the beginning of the year.) ## Short answer For 1000 HP, minimum capacity = 0, wasted water = 967 Mm³. For 10,000 HP (probable intended), minimum capacity ≈ 29.9 Mm³, wasted water ≈ 76.4 Mm³. ## Answer #### Water needed by the turbine Take 1 HP = 0.7457 kW. Output $=1000\times0.7457 = 745.7$ kW, net head 30 m, overall efficiency 0.80: $$ Q = \frac{P}{9.81\,\eta\,H} = \frac{745.7}{9.81\times0.80\times30} = 3.167\ \text{m}^3/\text{s} $$ Monthly demand volume $D = 3.167\times2.628\times10^6 = 8.32\times10^6$ m³ $=8.32$ Mm³ (constant draft). #### With the data as given (1000 HP) The demand (8.32 Mm³/month) is far below the smallest monthly inflow (70 Mm³), so the inflow exceeds the demand in every month and the reservoir is never short of water. - (i) Minimum capacity required $=0$ Mm³ (no storage is needed for this load). - (ii) Water wasted $=\sum(\text{inflow}) - 12D = 1067 - 99.88 = 967.1$ Mm³ per year. #### If the turbine is 10,000 HP (probable intended load) The inflow data (mean 88.9 Mm³/month) match a turbine about ten times larger, so the working for 10,000 HP is also shown. $P=7,457$ kW, $Q=31.67$ m³/s, monthly demand $D=83.24$ Mm³. | Month | Inflow | Demand | Opening | Closing | Wasted | |---|---|---|---|---|---| | Jan | 90 | 83.24 | 29.94 | 29.94 | 6.76 | | Feb | 80 | 83.24 | 29.94 | 26.71 | 0.00 | | Mar | 73 | 83.24 | 26.71 | 16.47 | 0.00 | | Apr | 80 | 83.24 | 16.47 | 13.24 | 0.00 | | May | 70 | 83.24 | 13.24 | 0.00 | 0.00 | | Jun | 98 | 83.24 | 0.00 | 14.76 | 0.00 | | Jul | 120 | 83.24 | 14.76 | 29.94 | 21.59 | | Aug | 80 | 83.24 | 29.94 | 26.71 | 0.00 | | Sep | 96 | 83.24 | 26.71 | 29.94 | 9.53 | | Oct | 105 | 83.24 | 29.94 | 29.94 | 21.76 | | Nov | 100 | 83.24 | 29.94 | 29.94 | 16.76 | | Dec | 75 | 83.24 | 29.94 | 21.71 | 0.00 | The reservoir (full at the start of the year) falls to empty in May, so the **minimum capacity** is the deficit accumulated from February to May: **29.9 Mm³**. The inflow after June refills it, and the water that cannot be stored (spill) adds up to **76.4 Mm³** per year. (The same table applies if the load is 1000 HP but the inflows are really in units of 10⁵ m³; all volumes are then divided by 10, giving a capacity of about 2.99 Mm³ and wasted water of about 7.64 Mm³.) **Answer: for the 1000 HP load stated, minimum capacity = 0 and wasted water = 967 Mm³. For a 10,000 HP turbine, minimum capacity ≈ 29.9 Mm³ and wasted water ≈ 76.4 Mm³.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the turbine's water demand found from 1000 HP? A. Convert HP to kW (x 0.7457) then Q = P / (9.81 eta H) B. Q = HP / H C. Q = HP x eta D. Q = P x 9.81 Answer: A. Q = 745.7 / (9.81 x 0.8 x 30) = 3.167 m3/s, i.e. 8.32 Mm3 per month. (Key point: Flow demand from HP) 2. For the 1000 HP load as stated, why is the minimum capacity zero? A. The monthly demand (8.32 Mm3) is below even the smallest inflow (70 Mm3), so inflow always exceeds demand B. Reservoir is empty C. The head is too small D. Evaporation is high Answer: A. No month has a shortage, so no storage is needed. (Key point: Result for 1000 HP) 3. How is wasted water computed in the 1000 HP case? A. Total inflow minus total demand (1067 - 12 x 8.32 = 967 Mm3) B. Total inflow C. Demand minus inflow D. Maximum inflow Answer: A. All inflow not used by the turbine is wasted when no storage is needed. (Key point: Wasted water) 4. For the 10,000 HP case, how is the minimum capacity found? A. From the greatest accumulated deficit when demand exceeds inflow, starting full (29.9 Mm3) B. Largest monthly inflow C. Mean inflow D. Sum of all inflow Answer: A. The reservoir drains from February to May; the accumulated deficit of 29.9 Mm3 is the minimum capacity. (Key point: Minimum capacity for 10,000 HP) 5. When does water spill in the 10,000 HP simulation? A. When inflow minus demand would take the reservoir above its capacity B. Whenever inflow is positive C. Only in July D. Never Answer: A. A full reservoir cannot store the surplus, so the excess is wasted (76.4 Mm3 per year). 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