--- title: "Installed capacity, energy and daily peaking reservoir (2079)" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-capacity-plant-factor-18cumec-peaking-79" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-capacity-plant-factor-18cumec-peaking-79 site: BCT Engineering --- # Installed capacity, energy and daily peaking reservoir (2079) Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2079 Baishakh, 2+2+2+2+4 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-79 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question A hydropower plant is planned to be designed in a Nepalese river, where 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 | 4.2 | 16.5 | 78.1 | 108.9 | 52.8 | 22.0 | 9.9 | 6.4 | Other data: design discharge = 18 m³/s; full supply level = 2250 masl; turbine centerline = 1650 masl; diameter of 4 km long tunnel = 3 m, f = 0.014; diameter of 1 km long penstock = 2.2 m, f = 0.012; hydraulic efficiency = 95%, turbine efficiency = 93%, generator efficiency = 99%, transformer efficiency = 99%. Considering only the frictional loss, (i) Compute the installed capacity, primary and secondary energy to be produced from the power plant assuming that 10% of the minimum monthly flow is to be released downstream. What is the plant factor? (ii) 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 447.7 GWh, with primary energy 136.5 GWh and secondary 311.2 GWh, giving a plant factor of 0.57. For 4-hour daily peaking, the required pondage volume is about 215,136 m³. ## Answer #### 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 | 4.2 | 3.86 | 3.86 | 599.4 | 19.65 | 14.35 | | 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 $= 447.73$ GWh - **Primary (firm) energy**, from the minimum available flow $3.06$ m³/s (available in every month): $136.54$ GWh - **Secondary energy** $=447.73-136.54 = 311.19$ GWh - **Plant factor** $=\dfrac{447.73\times10^6}{89.85\times10^3\times8760} = 0.569$ (56.9%) #### (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 = 447.7 GWh; primary energy = 136.5 GWh; secondary energy = 311.2 GWh; plant factor = 0.57; 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. How is the net head at design flow found when only friction loss is considered? A. Gross head 600 m minus tunnel and penstock friction losses (Darcy-Weisbach), giving about 587.6 m B. Gross head only, 600 m C. Gross head plus friction loss D. Half of the gross head Answer: A. Net head = 600 - 6.17 - 6.23 = 587.6 m at Qd = 18 m3/s. (Key point: Gross and net head) 2. How is overall efficiency combined from the component efficiencies? A. Sum divided by 4 B. Smallest of the four values C. Product: 0.95 x 0.93 x 0.99 x 0.99 = 0.866 D. Sum minus 3 Answer: C. Component efficiencies in series multiply. (Key point: Overall efficiency) 3. How does the environmental release change the flow available for power? A. Flow for power = Q - 0.1 x minimum monthly flow (0.34 m3/s) B. Flow for power = 0.9 Q in every month C. No change, the release comes from the spill D. Flow for power = Q - 10 m3/s Answer: A. Release is 10% of the minimum monthly flow, 0.1 x 3.4 = 0.34 m3/s, and is subtracted from each month. (Key point: Flow available for power) 4. How is the plant factor calculated? A. Installed capacity / annual energy B. Firm energy / total energy C. Annual energy / (installed capacity x 8760 h) D. Mean flow / design flow Answer: C. Plant factor is actual annual energy over the energy if run at full capacity all year, about 0.57. (Key point: Plant factor) 5. How is the reservoir volume for 4 hours of daily peaking found? A. (Qd - Q_in) x 4 x 3600 using the dry-month inflow, about 2.15 x 10^5 m3 B. Qd x 24 x 3600 C. Q_in x 20 x 3600 only, ignoring Qd D. Qd x 4 x 3600 without subtracting the inflow Answer: A. During the peak the plant draws design discharge while the river still supplies Q_in, so the pond supplies the difference. 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