--- title: "Installed capacity, energy and daily peaking reservoir (2072)" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-capacity-plant-factor-18cumec-peaking-72" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-capacity-plant-factor-18cumec-peaking-72 site: BCT Engineering --- # Installed capacity, energy and daily peaking reservoir (2072) Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2072 Chaitra, 5+2+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-capacity-plant-factor-18cumec-peaking-72 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question A hydropower plant is to be planned in a Nepalese river, where the mean monthly flows for a typical year are as follows: | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Q (m³/s) | 4.4 | 3.9 | 3.4 | 4.2 | 5.6 | 16.5 | 78.1 | 108.9 | 52.8 | 22.0 | 9.9 | 6.4 | Other data pertaining to the plant: design discharge = 18 m³/s; full supply level = 2250 masl; turbine center line = 1650 masl; diameter of 4.0 km long tunnel = 3.0 m, f = 0.014; diameter of 1.0 km long penstock = 2.2 m, f = 0.012; hydraulic efficiency = 95%; turbine efficiency = 93%; generator efficiency; transformer efficiency = 99%. Considering only frictional loss, a) Compute the installed capacity, primary and secondary energy to be produced from the power plant assuming that 10% of the minimum flow is to be released downstream. What is the plant factor? b) The developer is interested in developing a daily peaking reservoir for 4 hours. What will be the capacity of the reservoir to satisfy the daily peaking requirement? ## Short answer Installed capacity is 89.8 MW. Total annual energy is 452.9 GWh, with primary energy 136.5 GWh and secondary 316.4 GWh, giving a plant factor of 0.58. For 4-hour daily peaking, the required pondage volume is about 215,136 m³. ## Answer The generator efficiency is missing in the question; it is assumed to be 99%, the same as the transformer (all other data as given). #### Data and approach Gross head $= 2250-1650 = 600$ m. Only friction loss is considered. Overall efficiency $\eta = 0.95\times0.93\times0.99\times0.99 = 0.8659$ (hydraulic × turbine × generator × transformer). Energy is computed month by month (730 h per month), with the friction loss recalculated for the flow used. #### Head loss at design flow $Q_d = 18$ m³/s (Darcy-Weisbach) Tunnel: $A=\frac{\pi}{4}(3)^2 = 7.069$ m², $V = 2.546$ m/s $$ h_{f,t} = f\frac{L}{D}\frac{V^2}{2g} = 0.014\times\frac{4000}{3}\times\frac{2.546^2}{2\times9.81} = 6.17\ \text{m} $$ Penstock: $A=\frac{\pi}{4}(2.2)^2 = 3.801$ m², $V = 4.735$ m/s $$ h_{f,p} = 0.012\times\frac{1000}{2.2}\times\frac{4.735^2}{2\times9.81} = 6.23\ \text{m} $$ Net head at $Q_d$: $H_n = 600-6.17-6.23 = 587.60$ m. (Loss varies as $Q^2$ for smaller flows.) #### (i) Installed capacity, energy and plant factor Environmental release $=10\%\times$ minimum monthly flow $=0.1\times3.4=0.34$ m³/s, so the flow available for power is $Q-0.34$. $$ P = 9.81\,\eta\,H_n\,Q_d = 9.81\times0.8659\times587.60\times18 = 89,846\ \text{kW} = 89.85\ \text{MW} $$ | Month | Q (m³/s) | Q − 0.1Qmin | Q used | Net head (m) | Power (MW) | Energy (GWh) | |---|---|---|---|---|---|---| | Jan | 4.4 | 4.06 | 4.06 | 599.4 | 20.67 | 15.09 | | Feb | 3.9 | 3.56 | 3.56 | 599.5 | 18.13 | 13.23 | | Mar | 3.4 | 3.06 | 3.06 | 599.6 | 15.59 | 11.38 | | Apr | 4.2 | 3.86 | 3.86 | 599.4 | 19.65 | 14.35 | | May | 5.6 | 5.26 | 5.26 | 598.9 | 26.76 | 19.54 | | Jun | 16.5 | 16.16 | 16.16 | 590.0 | 80.99 | 59.12 | | Jul | 78.1 | 77.76 | 18.00 | 587.6 | 89.85 | 65.59 | | Aug | 108.9 | 108.56 | 18.00 | 587.6 | 89.85 | 65.59 | | Sep | 52.8 | 52.46 | 18.00 | 587.6 | 89.85 | 65.59 | | Oct | 22 | 21.66 | 18.00 | 587.6 | 89.85 | 65.59 | | Nov | 9.9 | 9.56 | 9.56 | 596.5 | 48.44 | 35.36 | | Dec | 6.4 | 6.06 | 6.06 | 598.6 | 30.81 | 22.49 | - Total annual energy $= 452.92$ GWh - **Primary (firm) energy**, from the minimum available flow $3.06$ m³/s (available in every month): $136.54$ GWh - **Secondary energy** $=452.92-136.54 = 316.38$ GWh - **Plant factor** $=\dfrac{452.92\times10^6}{89.85\times10^3\times8760} = 0.575$ (57.5%) #### (ii) Reservoir for 4-hour daily peaking In the driest month the available inflow is $Q_{in} = 3.06$ m³/s. The plant is to run for 4 hours a day at the design flow of 18 m³/s and store water in the other 20 hours. The daily inflow is 264,384 m³ and the peak use is $18\times4\times3600 = 259,200$ m³, which is slightly smaller, so 4-hour peaking at 18 m³/s is possible. During the 4 peak hours the draw-down from the reservoir is $$ V = (Q_d - Q_{in})\times t = (18-3.06)\times4\times3600 = 215,136\ \text{m}^3 \approx 2.15\times10^5\ \text{m}^3 $$ The same volume must be stored in the 20 off-peak hours ($3.06\times20\times3600 = 220,320$ m³ available), so it fills in time. **Answer: P = 89.8 MW; total energy = 452.9 GWh; primary energy = 136.5 GWh; secondary energy = 316.4 GWh; plant factor = 0.58; pondage volume ≈ 215,136 m³ (≈ 2.15 × 10⁵ m³).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The generator efficiency is missing in the question. What does the answer do? A. Ignores the generator efficiency completely B. Assumes it equal to the transformer efficiency (99%) C. Assumes 100% D. Assumes it equal to the turbine efficiency Answer: B. It is assumed 99%, same as the transformer, giving overall efficiency 0.8659. (Key point: Assumption for missing data) 2. What is the net head at Qd = 18 m3/s? A. 600 m B. About 587.6 m, after about 6.2 m loss in the tunnel and 6.2 m in the penstock C. About 575 m D. About 612 m Answer: B. Friction losses of 6.17 m and 6.23 m are subtracted from the 600 m gross head. (Key point: Net head at design flow) 3. In July (Q = 78.1 m3/s) how much flow is used for power? A. 77.76 m3/s B. 18 m3/s, the design discharge, with the rest spilled C. 9 m3/s D. 78.1 m3/s Answer: B. The turbines cannot pass more than the design discharge, so Q used = 18 m3/s. (Key point: Flow limited by design discharge) 4. Primary (firm) energy is calculated from A. The design flow of 18 m3/s B. The minimum available flow of 3.06 m3/s that is present in every month C. The annual mean flow D. The maximum monthly flow Answer: B. Primary energy is based on the flow available in every month, 3.06 m3/s after the release. (Key point: Primary energy) 5. Why is the pond volume (18 - 3.06) x 4 x 3600? A. The pond must supply the deficit between design flow and the dry-month inflow during 4 peak hours B. It stores the full design flow for 24 hours C. It equals the daily inflow D. It is the spill volume of July Answer: A. While the plant runs at 18 m3/s for 4 h, the river gives only 3.06 m3/s; the difference is drawn from storage and refilled in the other 20 hours. 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