--- title: "Firm and secondary energy from average monthly flows" subject: "Hydro Power" chapter: 3 key: "semester_6/hydro_power:firm-secondary-energy-2075" source: https://www.bctengineer.com/q/semester_6/hydro_power/firm-secondary-energy-2075 site: BCT Engineering --- # Firm and secondary energy from average monthly flows Hydro Power · Chapter 3: Planning of Hydropower Projects - Asked in: 2075 Bhadra, 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_6/hydro_power/firm-secondary-energy-2075 - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3 ## Question Determine the firm and secondary energy for the following set of data of average monthly flows of river over a year. | Months | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Q (m³/s) | 150 | 250 | 300 | 425 | 500 | 100 | 1250 | 1100 | 1000 | 1300 | 1600 | 500 | ## Short answer With assumed head 100 m and efficiency 85%, firm energy is 730.5 GWh/yr from June minimum flow (100 m³/s). Secondary energy is 4428.4 GWh/yr; total 5158.8 GWh/yr. ## Answer **Firm energy** is produced by the minimum flow (available 100% of the time); **secondary energy** is the extra energy from flow above the minimum. #### Assumptions The head, efficiency and plant capacity are not given, so assume: - Net head $H = 100$ m, overall efficiency $\eta = 0.85$ - The plant can use all the river flow (no spilling) - Each month = 730 h Note: the June flow (100 m³/s) is taken as printed, so June is the minimum-flow month. #### Power per unit discharge $$ P = 9.81 \times Q \times 100 \times 0.85 = 833.85\,Q\ \text{kW} $$ #### Firm energy Minimum flow = 100 m³/s. $$ \begin{aligned} P_f &= 833.85 \times 100 = 83{,}385\ \text{kW} \\ E_f &= 83{,}385 \times 8760 = 730.45\ \text{GWh} \end{aligned} $$ #### Total energy | Month | Q (m³/s) | Power (MW) | |---|---|---| | Jan | 150 | 125.08 | | Feb | 250 | 208.46 | | Mar | 300 | 250.16 | | Apr | 425 | 354.39 | | May | 500 | 416.93 | | Jun | 100 | 83.39 | | Jul | 1250 | 1042.31 | | Aug | 1100 | 917.24 | | Sep | 1000 | 833.85 | | Oct | 1300 | 1084.01 | | Nov | 1600 | 1334.16 | | Dec | 500 | 416.93 | | **Sum** | **8475** | | $$ \begin{aligned} E_{total} &= 833.85 \times 8475 \times 730 = 5158.82\ \text{GWh} \\ E_s &= 5158.82 - 730.45 = 4428.37\ \text{GWh} \end{aligned} $$ In general form: $E_f = 9.81\,\eta H \times 100 \times 8760$ and $E_s = 9.81\,\eta H \times (8475 - 12\times100) \times 730$ kWh, so the answer scales directly with the assumed $\eta H$. **Answer (H = 100 m, η = 85%): firm energy ≈ 730.5 GWh/yr; secondary energy ≈ 4428.4 GWh/yr.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must the head, efficiency and capacity be assumed? A. The question does not give them B. They cancel out C. They do not affect energy D. Firm energy needs no head Answer: A. They are not given, so H = 100 m, eta = 0.85 and full use of flow are assumed. (Key point: Assumptions) 2. Which flow gives the firm energy as printed? A. 150 m^3/s in January B. 100 m^3/s in June C. 1600 m^3/s in November D. 500 m^3/s in May Answer: B. The smallest listed value is 100 m^3/s (June), so firm power is 83,385 kW. (Key point: Minimum flow) 3. How do the answers change with the assumed eta x H? A. They do not change B. They change with the square C. They scale directly in proportion to eta x H D. They scale inversely Answer: C. Energy is 9.81 x eta x H x flow-hours, so it is directly proportional to eta x H. 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