--- title: "Power duration curve and primary/secondary energy" subject: "Hydro Power" chapter: 3 key: "semester_6/hydro_power:power-duration-curve-energy" source: https://www.bctengineer.com/q/semester_6/hydro_power/power-duration-curve-energy site: BCT Engineering --- # Power duration curve and primary/secondary energy Hydro Power · Chapter 3: Planning of Hydropower Projects - Asked in: 2070 Magh, 10 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/power-duration-curve-energy - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3 ## Question From the following data of flow at a given site for an average year compute and draw a power duration curve. Assume an average net available head of 100 m and a combined turbine generator efficiency of 89%. Determine the primary and secondary energy available during a year if the plant capacity is fixed at power corresponding to the flow available for 45% of time. | Flow (m³/s) | 900 | 600 | 500 | 450 | 400 | 360 | 340 | 300 | 280 | 200 | 140 | 100 | |---|---|---|---|---|---|---|---|---|---|---|---|---| | % time | 1 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 93 | 100 | ## Short answer Installed capacity at 45% exceedance flow (380 m³/s) is 331.8 MW. Primary energy from minimum flow (100 m³/s) is 764.8 GWh/yr; secondary energy 1705.6 GWh/yr from area under the capped power duration curve. ## Answer A **power duration curve** is the flow duration curve with the flow axis converted to power, $P = 9.81\,Q H \eta$. The area under it gives the energy. #### Power for each flow $$ P = 9.81 \times Q \times 100 \times 0.89 = 873.09\,Q\ \text{kW} $$ | % time | Q (m³/s) | Power (MW) | |---|---|---| | 1 | 900 | 785.78 | | 10 | 600 | 523.85 | | 20 | 500 | 436.55 | | 30 | 450 | 392.89 | | 40 | 400 | 349.24 | | 50 | 360 | 314.31 | | 60 | 340 | 296.85 | | 70 | 300 | 261.93 | | 80 | 280 | 244.47 | | 90 | 200 | 174.62 | | 93 | 140 | 122.23 | | 100 | 100 | 87.31 | ```text P (MW) 786 |* | \ 524 | * 437 | * 393 | * 349 | * 332 |======*===================== <- installed capacity 331.8 MW (Q45) 314 | * 297 | * 262 | * secondary 244 | * 175 | * 122 | * 87 |=====================*===== <- firm power 87.3 MW | primary (firm) +------------------------------> 0 45 93 100 % time ``` #### Plant capacity (flow at 45% time) $$ \begin{aligned} Q_{45} &= 400 - \frac{45-40}{50-40}(400-360) = 380\ \text{m}^3/\text{s} \\ P_{inst} &= 873.09 \times 380 = 331{,}774\ \text{kW} \approx 331.8\ \text{MW} \end{aligned} $$ #### Primary energy Flow available 100% of the time = 100 m³/s. $$ \begin{aligned} P_f &= 873.09 \times 100 = 87{,}309\ \text{kW} \\ E_p &= 87{,}309 \times 8760 = 764.83\ \text{GWh} \end{aligned} $$ #### Total energy (area under capped curve, trapezoidal rule) Flow used is capped at 380 m³/s from 0% to 45%. | Interval (%) | Mean Q used | Δt (%) | Area | |---|---|---|---| | 0–45 | 380 | 45 | 17,100 | | 45–50 | 370 | 5 | 1,850 | | 50–60 | 350 | 10 | 3,500 | | 60–70 | 320 | 10 | 3,200 | | 70–80 | 290 | 10 | 2,900 | | 80–90 | 240 | 10 | 2,400 | | 90–93 | 170 | 3 | 510 | | 93–100 | 120 | 7 | 840 | | **Total** | | | **32,300** | Mean flow used $= 32{,}300/100 = 323$ m³/s. $$ \begin{aligned} E_{total} &= 873.09 \times 323 \times 8760 = 2470.39\ \text{GWh} \\ E_s &= 2470.39 - 764.83 = 1705.56\ \text{GWh} \end{aligned} $$ Plant factor $= 323/380 = 85\%$. **Answer: Installed capacity ≈ 331.8 MW; primary energy ≈ 764.8 GWh/yr; secondary energy ≈ 1705.6 GWh/yr.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a power duration curve obtained from the flow duration data? A. By plotting flow against time B. By converting each flow to power with P = 9.81 Q H eta C. By dividing flow by head D. By ranking power in ascending order Answer: B. Each flow is multiplied by 9.81 H eta to give power. (Key point: Power duration curve) 2. Which flow gives primary energy? A. 900 m^3/s B. 380 m^3/s C. 100 m^3/s available 100% of time D. 323 m^3/s Answer: C. Flow exceeded 100% of the time is 100 m^3/s, giving 764.8 GWh. (Key point: Primary energy) 3. How is total energy found? A. Area under the curve capped at 380 m^3/s, giving mean used flow 323 m^3/s B. Area under the uncapped curve C. Capacity x 8760 D. Primary energy x 2 Answer: A. The curve is capped at 380 m^3/s; mean flow 323 gives E = 873.09 x 323 x 8760. 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