--- title: "Primary/secondary energy and pondage volume" subject: "Hydro Power" chapter: 3 key: "semester_6/hydro_power:energy-and-pondage-volume" source: https://www.bctengineer.com/q/semester_6/hydro_power/energy-and-pondage-volume site: BCT Engineering --- # Primary/secondary energy and pondage volume Hydro Power · Chapter 3: Planning of Hydropower Projects - Asked in: 2075 Baisakh, 10 marks - Video: [Calculating Power, Primary & Secondary Energy | Power and Energy potential || Hydropower Engineering](https://www.youtube.com/watch?v=3v4OCvSD9fU) — OPengineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/energy-and-pondage-volume - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3 ## Question Determine the primary and secondary energy if the hydrograph as shown below is given net head = 200 m and efficiency = 80%. What is the volume of the pondage required if the firm capacity is increased by 150%. | Month | Jan | Feb | Mar | Apr | May | Jun | July | Aug | Sept | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Discharge (m³/s) | 100 | 120 | 140 | 300 | 320 | 1800 | 2000 | 2510 | 2000 | 900 | 500 | 300 | ## Short answer Primary energy is 1375 GWh/yr, secondary 11,217 GWh/yr. Power per unit flow: 1569.6 Q kW. For firm capacity increased by 150% (to 250 m³/s), pondage required is 1.025×10^9 m³. ## Answer **Primary energy** is produced by the minimum flow that is available all year; **secondary energy** is produced by the flow above this minimum. No design discharge is given, so the plant is assumed to use all the river flow (secondary energy is then the maximum possible). Each month is taken as 730 h ($2.628 \times 10^6$ s). #### Power per unit discharge $$ P = 9.81 \times Q \times 200 \times 0.80 = 1569.6\,Q\ \text{kW} $$ #### Primary energy Minimum flow = 100 m³/s (January). $$ \begin{aligned} P_f &= 1569.6 \times 100 = 156{,}960\ \text{kW} = 156.96\ \text{MW} \\ E_p &= 156{,}960 \times 8760 = 1374.97\ \text{GWh} \end{aligned} $$ #### Total and secondary energy | Month | Q (m³/s) | Power (MW) | Energy (GWh) | |---|---|---|---| | Jan | 100 | 156.96 | 114.58 | | Feb | 120 | 188.35 | 137.50 | | Mar | 140 | 219.74 | 160.41 | | Apr | 300 | 470.88 | 343.74 | | May | 320 | 502.27 | 366.66 | | Jun | 1800 | 2825.28 | 2062.45 | | Jul | 2000 | 3139.20 | 2291.62 | | Aug | 2510 | 3939.70 | 2875.98 | | Sep | 2000 | 3139.20 | 2291.62 | | Oct | 900 | 1412.64 | 1031.23 | | Nov | 500 | 784.80 | 572.90 | | Dec | 300 | 470.88 | 343.74 | | **Total** | **10,990** | | **12,592.43** | $$ E_s = 12{,}592.43 - 1374.97 = 11{,}217.46\ \text{GWh} $$ #### Pondage for firm capacity increased by 150% "Increased by 150%" is read as new firm capacity = $1 + 1.5 = 2.5$ times the original. $$ \begin{aligned} P_{f,new} &= 2.5 \times 156.96 = 392.4\ \text{MW} \\ Q_{f,new} &= 2.5 \times 100 = 250\ \text{m}^3/\text{s} \end{aligned} $$ Months with flow below 250 m³/s must be supplied from storage: | Month | Q | Deficit (250 − Q) | |---|---|---| | Jan | 100 | 150 | | Feb | 120 | 130 | | Mar | 140 | 110 | | **Total** | | **390 m³/s-month** | $$ \begin{aligned} V &= 390 \times 2.628\times10^6 \\ &= 1.025 \times 10^9\ \text{m}^3 \end{aligned} $$ This storage is easily refilled in the monsoon months, when flows are far above 250 m³/s. **Answer: Primary energy ≈ 1375.0 GWh/yr; secondary energy ≈ 11,217.5 GWh/yr; storage (pondage) required ≈ 1.025 × 10⁹ m³ (1025 million m³).** (If "increased by 150%" is taken as 1.5 times, i.e. firm flow 150 m³/s, the deficits are 50 + 30 + 10 = 90 m³/s-month and the storage is $90 \times 2.628\times10^6 = 2.37\times10^8$ m³.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Since no design discharge is given, what is assumed? A. The plant uses all the river flow, so secondary energy is the maximum B. Only the minimum flow is used C. The plant uses the mean flow only D. Flow is capped at 900 m^3/s Answer: A. The plant is assumed to use all river flow, giving the maximum possible secondary energy. (Key point: No design discharge assumption) 2. What gives the primary energy 1374.97 GWh? A. 1569.6 x 2510 x 8760 B. 1569.6 x 100 x 8760 kWh with minimum flow 100 m^3/s C. 1569.6 x 10,990 x 730 D. 1569.6 x 500 x 8760 Answer: B. Minimum flow 100 m^3/s gives 156,960 kW, times 8760 h. (Key point: Primary energy) 3. How is the required storage found? A. Sum of flows above 250 times 2.628x10^6 s B. Sum of deficits (250 - Q) for months below 250 m^3/s, times seconds per month C. Difference between max and min flow D. Firm energy divided by head Answer: B. Deficits 150 + 130 + 110 = 390 m^3/s-month times 2.628x10^6 s gives 1.025x10^9 m^3. 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