--- title: "Primary and secondary energy at 45% exceedance" subject: "Hydro Power" chapter: 3 key: "semester_6/hydro_power:primary-secondary-energy-45pct" source: https://www.bctengineer.com/q/semester_6/hydro_power/primary-secondary-energy-45pct site: BCT Engineering --- # Primary and secondary energy at 45% exceedance Hydro Power · Chapter 3: Planning of Hydropower Projects - Asked in: 2080 Asoj, 8 marks - Video: [Primary Energy And Secondary Energy](https://www.youtube.com/watch?v=H_5a9WnsHvY) — BEST MECHANICAL ENGINEERING - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/primary-secondary-energy-45pct - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3 ## Question The mean monthly flow of a river in Nepal is given in the below table. Calculate the primary and secondary energy that can be generated from the river if the plant capacity is fixed at power corresponding to the flow available for 45% of the time. Consider the plant's overall efficiency as 85% and the Net head as 170 m. | Months | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Flow (m³/s) | 30 | 28 | 24 | 29 | 18 | 37 | 65 | 90 | 74 | 56 | 42 | 39 | ## Short answer Plant capacity is 55.9 MW based on flow at 45% exceedance (39.45 m³/s). Primary energy 223.5 GWh/yr from minimum flow 18 m³/s. Secondary energy 192.7 GWh/yr, total 416.3 GWh/yr. Computed using flow duration curve, power equation, and monthly energy summation with 730 hours per month. ## Answer **Primary (firm) energy** is the energy that can be produced all year from the minimum (dependable) flow. **Secondary energy** is the extra energy produced when flow is above the minimum, up to the plant capacity. #### Step 1: Flow duration (Weibull plotting position $P = \frac{m}{N+1}\times100$, N = 12) | Rank m | Q (m³/s) | % time exceeded | |---|---|---| | 1 | 90 | 7.69 | | 2 | 74 | 15.38 | | 3 | 65 | 23.08 | | 4 | 56 | 30.77 | | 5 | 42 | 38.46 | | 6 | 39 | 46.15 | | 7 | 37 | 53.85 | | 8 | 30 | 61.54 | | 9 | 29 | 69.23 | | 10 | 28 | 76.92 | | 11 | 24 | 84.62 | | 12 | 18 | 92.31 | #### Step 2: Flow at 45% time (linear interpolation) $$ Q_{45} = 42 - \frac{45-38.46}{46.15-38.46}(42-39) = 39.45\ \text{m}^3/\text{s} $$ #### Step 3: Plant capacity $P = \gamma Q H \eta = 9.81 \times Q \times 170 \times 0.85 = 1417.55\,Q$ kW $$ P_{inst} = 1417.55 \times 39.45 = 55{,}922\ \text{kW} \approx 55.9\ \text{MW} $$ #### Step 4: Primary energy (minimum flow 18 m³/s, May) $$ \begin{aligned} P_f &= 1417.55 \times 18 = 25{,}516\ \text{kW} \\ E_p &= 25{,}516 \times 8760 = 223.52\ \text{GWh} \end{aligned} $$ #### Step 5: Total energy (flow used = smaller of Q and 39.45; 730 h per month) | Month | Q | Q used | Power (kW) | |---|---|---|---| | Jan | 30 | 30 | 42,526 | | Feb | 28 | 28 | 39,691 | | Mar | 24 | 24 | 34,021 | | Apr | 29 | 29 | 41,109 | | May | 18 | 18 | 25,516 | | Jun | 37 | 37 | 52,449 | | Jul–Nov | ≥ 42 | 39.45 | 55,922 each | | Dec | 39 | 39 | 55,284 | $\sum Q_{used} = 402.25$ m³/s-months. $$ \begin{aligned} E_{total} &= 1417.55 \times 402.25 \times 730 = 416.25\ \text{GWh} \\ E_s &= 416.25 - 223.52 = 192.73\ \text{GWh} \end{aligned} $$ **Answer: Plant capacity ≈ 55.9 MW; primary energy ≈ 223.5 GWh/yr; secondary energy ≈ 192.7 GWh/yr (total ≈ 416.3 GWh/yr).** (If $P = m/N$ is used for the duration curve, $Q_{45} \approx 40.8$ m³/s and the secondary energy rises slightly; the method is the same.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does primary (firm) energy represent in this problem? A. Energy from the flow exceeded 45% of the time B. Energy from the minimum flow available throughout the year C. Energy from the mean annual flow D. Energy from flows above the plant capacity Answer: B. Primary energy is produced all year from the minimum (dependable) flow; secondary energy is the extra from higher flows up to capacity. (Key point: Primary vs secondary energy) 2. With N = 12 monthly flows, how is the percentage of time a flow is exceeded found? A. P = m/N x 100 B. P = m/(N+1) x 100, with flows ranked in descending order C. P = (N+1)/m x 100 D. P = m/(N+1) x 100, with flows ranked in ascending order Answer: B. Flows are ranked from largest (m = 1) down and P = m/(N+1) x 100 gives the exceedance percentage. (Key point: Flow duration with Weibull formula) 3. How is Q45 = 39.45 m^3/s obtained and used? A. It is the mean of all twelve flows and gives the firm power B. It is the minimum flow and gives the primary energy C. It is interpolated between 42 (38.46%) and 39 (46.15%), and the plant capacity is 9.81 x Q45 x H x eta D. It is the maximum flow and gives the installed capacity directly in m^3/s Answer: C. 45% lies between the 38.46% and 46.15% points, so linear interpolation gives 39.45 m^3/s, which sets the capacity. (Key point: Plant capacity from Q45) 4. When computing total energy, why are Jul to Nov flows taken as 39.45 m^3/s? A. The plant cannot use more than the flow corresponding to its capacity B. Those months are dry months C. The environmental flow is subtracted D. Efficiency drops at high flows Answer: A. Flow above the design flow Q45 is spilled, so used flow is the smaller of Q and 39.45. 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