--- title: "Primary and secondary energy (Nepali calendar months)" subject: "Hydro Power" chapter: 3 key: "semester_6/hydro_power:energy-nepali-months-155m" source: https://www.bctengineer.com/q/semester_6/hydro_power/energy-nepali-months-155m site: BCT Engineering --- # Primary and secondary energy (Nepali calendar months) Hydro Power · Chapter 3: Planning of Hydropower Projects - Asked in: 2073 Bhadra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/energy-nepali-months-155m - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3 ## Question Following table shows the mean monthly flow of a typical river of Nepal. Estimate the primary and secondary energies available in the river throughout the year. Assume 10% of minimum flow of the series has to be released for environmental flow to the downstream from the dam. Take overall efficiency of the plant is 80% and effective head is 155 m. Note: The months are as per Nepalese calendar i.e 1 as Baishakh and so on. | Month | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Discharge (m³/s) | 1150 | 1140 | 1450 | 1830 | 1700 | 1550 | 1150 | 1000 | 850 | 400 | 480 | 960 | ## Short answer Primary energy is 3836.2 GWh/yr, secondary 7867.7 GWh/yr. Minimum usable flow after 40 m³/s env. release is 360 m³/s; firm power 437.9 MW. Total energy 11,703.9 GWh/yr using all usable flow. ## Answer **Primary energy** is the energy available throughout the year from the minimum usable flow; **secondary energy** is the extra energy from usable flows above that minimum. No design discharge is given, so the plant is assumed to use all the usable flow. Each month is taken as 730 h (Nepali months vary from 29 to 32 days, but the yearly total is the same). #### Environmental flow Minimum flow of the series = 400 m³/s (month 10, Magh). $$ Q_{env} = 0.10 \times 400 = 40\ \text{m}^3/\text{s} $$ Usable flow = Q − 40. #### Power per unit discharge $$ P = 9.81 \times Q \times 155 \times 0.80 = 1216.44\,Q\ \text{kW} $$ #### Primary energy Minimum usable flow = 400 − 40 = 360 m³/s. $$ \begin{aligned} P_f &= 1216.44 \times 360 = 437{,}918\ \text{kW} \approx 437.9\ \text{MW} \\ E_p &= 437{,}918 \times 8760 = 3836.17\ \text{GWh} \end{aligned} $$ #### Total energy | Month | Q | Usable Q | Power (MW) | |---|---|---|---| | 1 Baisakh | 1150 | 1110 | 1350.25 | | 2 Jestha | 1140 | 1100 | 1338.08 | | 3 Asar | 1450 | 1410 | 1715.18 | | 4 Shrawan | 1830 | 1790 | 2177.43 | | 5 Bhadra | 1700 | 1660 | 2019.29 | | 6 Ashwin | 1550 | 1510 | 1836.82 | | 7 Kartik | 1150 | 1110 | 1350.25 | | 8 Mangsir | 1000 | 960 | 1167.78 | | 9 Poush | 850 | 810 | 985.32 | | 10 Magh | 400 | 360 | 437.92 | | 11 Falgun | 480 | 440 | 535.23 | | 12 Chaitra | 960 | 920 | 1119.12 | | **Sum** | | **13,180** | | $$ \begin{aligned} E_{total} &= 1216.44 \times 13{,}180 \times 730 = 11{,}703.86\ \text{GWh} \\ E_s &= 11{,}703.86 - 3836.17 = 7867.69\ \text{GWh} \end{aligned} $$ **Answer: Primary energy ≈ 3836.2 GWh/yr; secondary energy ≈ 7867.7 GWh/yr (total ≈ 11,703.9 GWh/yr).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the minimum flow and the environmental release? A. 400 m^3/s and 40 m^3/s B. 850 m^3/s and 85 m^3/s C. 1830 m^3/s and 183 m^3/s D. 400 m^3/s and 400 m^3/s Answer: A. Minimum is 400 (Magh); 10% of it, 40 m^3/s, is released. (Key point: Minimum flow and environmental release) 2. What flow gives primary energy? A. 400 m^3/s B. 1100 m^3/s C. 40 m^3/s D. 360 m^3/s (minimum usable flow) Answer: D. Usable minimum is 400 - 40 = 360 m^3/s, giving 437.9 MW and 3836 GWh/yr. (Key point: Primary energy) 3. Does the Nepali calendar change the method? A. Yes, months must be 30 days each B. No, 730 h per month is used since the yearly total is the same C. Yes, the head changes D. Yes, flows must be reordered Answer: B. Month lengths vary but the total hours per year are the same, so 730 h per month is used. 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