--- title: "Flow duration curve and power at mean flow" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-capacity-fdc-flow-table-79" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-capacity-fdc-flow-table-79 site: BCT Engineering --- # Flow duration curve and power at mean flow Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2079 Bhadra, 4+4 marks - Video: [HOW TO DRAW FLOW DURATION CURVE (FDC)||HYDROPOWER ENGINEERING||](https://www.youtube.com/watch?v=lH2CO83TuVo) — CIVIL TALKS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-capacity-fdc-flow-table-79 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question In a Nepali river the mean monthly flow in the year 2021 is given below. | Month | Discharge (m³/s) | Month | Discharge (m³/s) | |---|---|---|---| | Jan | 50 | July | 125 | | Feb | 40 | Aug | 150 | | March | 30 | Sept | 120 | | April | 25 | Oct | 100 | | May | 10 | Nov | 75 | | June | 75 | Dec | 70 | a) Draw the flow duration curve. b) Find the power available at the mean flow of water if the available head is 100 m at the site and the overall efficiency of the plant is 85%. ## Short answer The mean monthly flow is 72.5 m³/s. Power at mean flow with 100 m head and 85% overall efficiency is 60.45 MW. ## Answer #### a) Flow duration curve Arrange the 12 monthly flows in descending order and assign the exceedance probability by the Weibull formula $p = \dfrac{m}{N+1}\times100$ with $N = 12$. | Rank | Q (m³/s) | Exceedance (%) | |---|---|---| | 1 | 150 | 7.7 | | 2 | 125 | 15.4 | | 3 | 120 | 23.1 | | 4 | 100 | 30.8 | | 5 | 75 | 38.5 | | 6 | 75 | 46.2 | | 7 | 70 | 53.8 | | 8 | 50 | 61.5 | | 9 | 40 | 69.2 | | 10 | 30 | 76.9 | | 11 | 25 | 84.6 | | 12 | 10 | 92.3 | Plot Q (vertical) against percentage of time (horizontal) and join the points with a smooth curve: ```text Q (m³/s) 150 |* 125 | * 100 | * 75 | * * 50 | * 25 | * * 10 | * +-------------------------- 0 20 40 60 80 100 % time flow is exceeded ``` #### b) Power at the mean flow Mean flow $\bar Q = \dfrac{\sum Q}{12} = \dfrac{870}{12} = 72.5$ m³/s $$ P = \gamma Q H \eta = 9.81\times72.5\times100\times0.85 = 60,454.12\ \text{kW} = 60.45\ \text{MW} $$ The mean flow of 72.5 m³/s is exceeded about 46% of the time (six of the twelve months have flows above it). **Answer: mean flow = 72.5 m³/s; power at mean flow ≈ 60.45 MW** (the FDC is tabulated above). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the points of the flow duration curve obtained from the monthly flows? A. Flows are kept in calendar order and plotted against month B. Flows are sorted in descending order and the exceedance p = m/(N+1) x 100 is assigned to each rank m C. Flows are sorted in ascending order and plotted against m/N D. Flows are averaged in groups of three Answer: B. The FDC is made by ranking flows from highest to lowest and using the Weibull formula to get the percentage of time each flow is equalled or exceeded. (Key point: Building the FDC) 2. On the FDC, the point (150 m3/s, 7.7%) means that A. 150 m3/s occurs exactly 7.7% of the time B. 150 m3/s is the average flow in 7.7% of months C. 150 m3/s is equalled or exceeded about 7.7% of the time D. 7.7% of the flow is lost Answer: C. Exceedance probability is the percentage of time the flow is equalled or exceeded. (Key point: Meaning of exceedance) 3. What is the mean flow of the river? A. 72.5 m3/s, the sum of 870 divided by 12 months B. 75 m3/s, the median C. 150 m3/s, the maximum D. 87 m3/s, the sum divided by 10 Answer: A. Mean flow = 870/12 = 72.5 m3/s. (Key point: Mean flow) 4. Which expression gives the power at the mean flow with H = 100 m and efficiency 85%? A. P = 9.81 x 72.5 x 100 = 71.1 MW, efficiency ignored B. P = 9.81 x 72.5 x 100 x 0.85, about 60.45 MW C. P = 9.81 x 72.5 x 0.85 / 100 D. P = 1000 x 72.5 x 100 x 0.85 W Answer: B. Power is gamma x Q x H x efficiency = 9.81 x 72.5 x 100 x 0.85 = 60,454 kW. 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