--- title: "Flow duration curve; firm and secondary energy" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:fdc-firm-secondary-energy-69" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/fdc-firm-secondary-energy-69 site: BCT Engineering --- # Flow duration curve; firm and secondary energy Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2069 Chaitra, 8 marks - Video: [Power Duration Curve ,Firm Energy ,Secondary Energy Numerical || Hydropower Engineering](https://www.youtube.com/watch?v=OPfxdiDJc6A) — OPengineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/fdc-firm-secondary-energy-69 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question The stream flow record for a hydropower development site is given below. Draw a flow duration curve and determine firm and secondary energy if the available head is 60 m, design discharge capacity is 45 m³/s and overall efficiency is 82%. | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Q (m³/s) | 30 | 38 | 28 | 22 | 16 | 32 | 56 | 72 | 54 | 46 | 38 | 36 | ## Short answer With head 60 m, design discharge 45 m³/s, and 82% efficiency, firm energy is 67.6 GWh, secondary energy is 80.3 GWh, and total annual energy is about 148.0 GWh. ## Answer #### Flow duration curve Rank the monthly flows in descending order and take $p=\dfrac{m}{N+1}\times100$ with $N=12$. | Rank | Q (m³/s) | Exceedance (%) | Q used = min(Q, 45) | |---|---|---|---| | 1 | 72 | 7.7 | 45 | | 2 | 56 | 15.4 | 45 | | 3 | 54 | 23.1 | 45 | | 4 | 46 | 30.8 | 45 | | 5 | 38 | 38.5 | 38 | | 6 | 38 | 46.2 | 38 | | 7 | 36 | 53.8 | 36 | | 8 | 32 | 61.5 | 32 | | 9 | 30 | 69.2 | 30 | | 10 | 28 | 76.9 | 28 | | 11 | 22 | 84.6 | 22 | | 12 | 16 | 92.3 | 16 | ```text Q (m³/s) 72 |* 56 | * 45 |-+--*-*----- design flow 38 | * * 30 | * * 22 | * 16 | * +---------------------- 0 20 40 60 80 100 % ``` The FDC is flat at 45 m³/s for about 31% of the time (months with flow above the design flow). #### Energy Constants: $H=60$ m, $\eta=0.82$, $Q_d=45$ m³/s, 730 h per month. - Installed capacity $P=9.81\times0.82\times60\times45 = 21,719$ kW $=21.72$ MW - **Firm energy** is based on the flow available all the time (minimum flow 16 m³/s): $$ P_{firm} = 9.81\times0.82\times60\times16 = 7,722\ \text{kW},\quad E_{firm} = 7,722\times8760 = 67.65\ \text{GWh} $$ - **Total energy** with $Q_{used}=\min(Q,45)$: sum of used flows = 420 m³/s-months, so mean used flow = 35 m³/s $$ E_{total} = 9.81\times0.82\times60\times35\times8760 = 147.98\ \text{GWh} $$ - **Secondary energy** $=E_{total}-E_{firm} = 147.98-67.65 = 80.33$ GWh **Answer: firm energy ≈ 67.6 GWh; secondary energy ≈ 80.3 GWh; total ≈ 148.0 GWh per year.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With a design discharge of 45 m3/s, what flow is used for power in a month when the river carries 72 m3/s? A. 72 m3/s B. 45 m3/s, because the turbines cannot pass more than the design discharge C. 27 m3/s, the excess D. 58.5 m3/s, the average of the two Answer: B. Flow is limited to the design discharge, so Q used = min(Q, 45). (Key point: Used flow with design discharge) 2. On what flow is firm energy based? A. The mean flow of 35 m3/s B. The design flow of 45 m3/s C. The minimum flow of 16 m3/s, available all the time D. The flow exceeded 50% of the time Answer: C. Firm energy is what can be generated continuously from the flow available all the time, here the minimum flow of 16 m3/s. (Key point: Firm energy basis) 3. How is secondary energy obtained? A. Total energy (with Q used) minus firm energy B. Firm energy minus total energy C. Energy from flows above 45 m3/s D. Installed capacity x 8760 Answer: A. Secondary energy is the total energy less firm energy: 147.98 - 67.65 = 80.33 GWh. (Key point: Secondary energy) 4. Which calculation gives the installed capacity? A. 9.81 x 0.82 x 60 x 16 = 7.7 MW B. 9.81 x 0.82 x 60 x 35 = 16.9 MW C. 9.81 x 0.82 x 60 x 45, about 21.7 MW, using the design discharge D. 9.81 x 60 x 45 = 26.5 MW, ignoring efficiency Answer: C. Installed capacity is based on the design discharge of 45 m3/s with efficiency 0.82. 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