--- title: "Mass curve and critical period storage" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:mass-curve-critical-period-inflow-table" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/mass-curve-critical-period-inflow-table site: BCT Engineering --- # Mass curve and critical period storage Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2081 Baishakh, 12 marks - Video: [estimation of storage capacity of reservoir using mass curve](https://www.youtube.com/watch?v=YbjPdjo_dKY) — NavAeon (Civil Engineering studies) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/mass-curve-critical-period-inflow-table - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question The table below shows the inflow for one reservoir. | Month | Oct | Nov | Dec | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Inflow (m³/s) | 38 | 32 | 27 | 26 | 15 | 10 | 8 | 7 | 6 | 4 | 3 | 3 | i) By assuming the mean inflow as the draft, develop the mass curve of the reservoir. ii) Determine the storage capacity and length of the critical period. iii) Determine the storage capacity if the draft is only 80% of the mean inflow and compare it with the storage capacity of (ii). ## Short answer With mean inflow as uniform draft (14.92 m³/s), storage capacity is 63.42 m³/s-months (166.7×10⁶ m³) and the critical period is 7 months (March to September). Reducing draft to 80% of the mean (11.93 m³/s) lowers storage to 42.53 m³/s-months (111.8×10⁶ m³). ## Answer Units are m³/s-months; one m³/s-month $=2.628\times10^6$ m³. #### i) Mass curve Mean inflow $=\dfrac{\sum Q}{12} = \dfrac{179}{12} = 14.917$ m³/s, taken as the uniform draft. | Month | Inflow | Cum. inflow | Cum. draft (mean) | Cum. inflow − draft | |---|---|---|---|---| | Oct | 38 | 38 | 14.92 | +23.08 | | Nov | 32 | 70 | 29.83 | +40.17 | | Dec | 27 | 97 | 44.75 | +52.25 | | Jan | 26 | 123 | 59.67 | +63.33 | | Feb | 15 | 138 | 74.58 | +63.42 | | Mar | 10 | 148 | 89.50 | +58.50 | | Apr | 8 | 156 | 104.42 | +51.58 | | May | 7 | 163 | 119.33 | +43.67 | | Jun | 6 | 169 | 134.25 | +34.75 | | Jul | 4 | 173 | 149.17 | +23.83 | | Aug | 3 | 176 | 164.08 | +11.92 | | Sep | 3 | 179 | 179.00 | +0.00 | ```text Cum. volume 179 | ____* inflow mass curve | ___--- | _--* draft line (slope = mean) | __.-* .-' | _.-* _.--' | * _.--' +--------------------------- Oct Sep ``` The mass curve of inflow lies above the draft line from October to February (surplus) and below it from March to September (deficit, the reservoir is drawn down). #### ii) Storage capacity and critical period The largest vertical gap between the draft line (tangent to the mass curve at the end of February) and the mass curve is at the end of September. This is the sum of the deficits from March to September: $$ S = \sum(\bar Q - Q)_{Mar-Sep} = (4.917+6.917+7.917+8.917+10.917+11.917+11.917) = 63.42\ \text{m}^3/\text{s-months} $$ $$ S = 63.42\times2.628\times10^6 = 166.7\times10^6\ \text{m}^3 $$ **Critical period** $=7$ months (March to September), when the reservoir goes from full to empty. It refills by about February of the next year. #### iii) Draft = 80% of mean inflow $0.8\times14.917 = 11.933$ m³/s. Deficit months are those with inflow below this value (Mar to Sep: 10, 8, 7, 6, 4, 3, 3): $$ S_{80} = (1.933+3.933+4.933+5.933+7.933+8.933+8.933) = 42.53\ \text{m}^3/\text{s-months} = 111.8\times10^6\ \text{m}^3 $$ The critical period is still March to September (7 months). The storage required is reduced from 166.7 Mm³ to 111.8 Mm³, a saving of 33%, but the water supplied also falls by 20%. **Answer: storage for the mean-flow draft = 166.7 Mm³ (63.42 m³/s-months), critical period = 7 months; storage for 80% draft = 111.8 Mm³ (42.53 m³/s-months).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When the draft is taken as the mean inflow, how is the draft value obtained here? A. Sum of the 12 monthly inflows divided by 12 (179/12 = 14.92 m³/s) B. The minimum monthly inflow (3 m³/s) C. The maximum monthly inflow (38 m³/s) D. The median of the deficit months only Answer: A. The mean inflow is the total inflow divided by 12 months, giving 14.917 m³/s, which is applied as a uniform draft. (Key point: Draft equals mean inflow) 2. On the mass curve, in which months does the inflow curve lie below the draft line (reservoir being drawn down)? A. October to February B. March to September C. June to August only D. January to April Answer: B. From October to February inflow exceeds the mean (surplus); from March to September it is below the mean (deficit), so the reservoir is drawn down. (Key point: Reading the mass curve) 3. How is the required storage capacity found for the mean-flow draft? A. Add all monthly inflows over the year B. Take the largest vertical gap between the draft line tangent at the peak and the mass curve, i.e. the sum of the deficits Mar-Sep (63.42 m³/s-months) C. Multiply the mean inflow by 12 months D. Subtract the smallest inflow from the largest inflow Answer: B. Storage equals the maximum cumulative deficit below the draft line, which is 63.42 m³/s-months, or 166.7 Mm³ after multiplying by 2.628x10^6. (Key point: Storage from maximum deficit) 4. What happens when the draft is reduced to 80% of the mean inflow? A. Storage rises because the reservoir stays fuller B. Storage stays 166.7 Mm³ and the critical period doubles C. Storage falls to about 111.8 Mm³ (a 33% saving) but water supplied also falls by 20% D. No storage is needed at all Answer: C. With 11.933 m³/s draft, the deficits sum to 42.53 m³/s-months (111.8 Mm³); the critical period stays 7 months, but less water is supplied. 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