--- title: "Firm power and plant load factors" subject: "Hydro Power" chapter: 3 key: "semester_6/hydro_power:firm-power-load-factors" source: https://www.bctengineer.com/q/semester_6/hydro_power/firm-power-load-factors site: BCT Engineering --- # Firm power and plant load factors Hydro Power · Chapter 3: Planning of Hydropower Projects - Asked in: 2077 Chaitra, 1+7 marks - Video: [Hydropower/ Load factor/ Capacity factor/ Utilization factor/ BE Civil/ 7th semester/ PU 2017, 2018](https://www.youtube.com/watch?v=emNMl8uLVRg) — Rupak Deo - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/firm-power-load-factors - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3 ## Question Define firm power and secondary power. The load on a hydropower plant varies from a minimum of 12,000 kW to a maximum of 35,000 kW. Two turbo-generators of capacities 22,000 kW each have been installed. Calculate: (i) total installed capacity of the plant, (ii) plant factor, (iii) maximum demand, (iv) load factor, and (v) utilization factor. ## Short answer Firm power is continuously available from minimum flow; secondary power is surplus above firm. Installed capacity: 44,000 kW, plant factor: 53.4%, maximum demand: 35,000 kW, load factor: 67.1%, utilization factor: 79.5%. ## Answer #### Firm and secondary power - **Firm (primary) power** is the power that is available continuously (100% of the time), even in the driest period. It is fixed by the minimum flow. - **Secondary (surplus) power** is the power above firm power that is available only part of the time, when flow is higher than the minimum. It is sold at a lower rate or used for non-essential loads. #### Data Minimum load = 12,000 kW; maximum load = 35,000 kW; two units of 22,000 kW each. The load is assumed to vary linearly (uniformly) between minimum and maximum, so the average load is the mean of the two. $$ P_{avg} = \frac{12{,}000 + 35{,}000}{2} = 23{,}500\ \text{kW} $$ #### (i) Total installed capacity $$ P_{inst} = 2 \times 22{,}000 = 44{,}000\ \text{kW} $$ #### (ii) Plant factor (capacity factor) $$ PF = \frac{\text{average load}}{\text{installed capacity}} = \frac{23{,}500}{44{,}000} = 0.534 = 53.4\% $$ #### (iii) Maximum demand Maximum demand = peak load = **35,000 kW**. #### (iv) Load factor $$ LF = \frac{\text{average load}}{\text{maximum demand}} = \frac{23{,}500}{35{,}000} = 0.671 = 67.1\% $$ #### (v) Utilization factor $$ UF = \frac{\text{maximum demand}}{\text{installed capacity}} = \frac{35{,}000}{44{,}000} = 0.795 = 79.5\% $$ Check: $PF = LF \times UF = 0.671 \times 0.795 = 0.534$. | Quantity | Value | |---|---| | Installed capacity | 44,000 kW | | Plant factor | 53.4% | | Maximum demand | 35,000 kW | | Load factor | 67.1% | | Utilization factor | 79.5% | **Answer: 44,000 kW; 53.4%; 35,000 kW; 67.1%; 79.5%.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is firm power? A. Power available only in monsoon B. Power available continuously, fixed by the minimum flow C. Power equal to the maximum demand D. Power sold at a lower rate Answer: B. Firm power is available 100% of the time in the driest period. (Key point: Firm power) 2. What is the plant factor? A. Maximum demand / installed capacity B. Average load / maximum demand C. Average load / installed capacity = 23,500 / 44,000 D. Installed capacity / average load Answer: C. Plant factor is 0.534, using average load of (12,000 + 35,000)/2. (Key point: Plant factor) 3. What gives the load factor 67.1%? A. Average load 23,500 kW / maximum demand 35,000 kW B. 35,000 / 44,000 C. 12,000 / 35,000 D. 23,500 / 44,000 Answer: A. Load factor = average load / maximum demand. (Key point: Load factor) 4. Which relation holds among the factors? A. PF = LF + UF B. PF = LF / UF C. PF = LF x UF D. PF = UF - LF Answer: C. 0.671 x 0.795 = 0.534, the plant factor. Utilization factor is maximum demand / installed capacity. (Key point: Relation of factors) ## Related in Hydro Power - [Best installed capacity at 12% interest](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Abest-installed-capacity-12pct) — Hydro Power, chapter 3 - [Annual firm and secondary energy (900 m³/s, Nov 650)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afirm-secondary-energy-2079) — Hydro Power, chapter 3, asked 2× - [Annual firm and secondary energy (900 m³/s, Nov 100)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afirm-secondary-energy-2078) — Hydro Power, chapter 3, asked 2× - [Best installed capacity at 10% interest](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Abest-installed-capacity-10pct) — Hydro Power, chapter 3 - [Firm and secondary energy from average monthly flows](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afirm-secondary-energy-2075) — Hydro Power, chapter 3 - [Power duration curve and primary/secondary energy](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apower-duration-curve-energy) — Hydro Power, chapter 3 ## Related in other subjects - [RoR best installed capacity (USD 1200 million)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-usd-1200) — Hydropower Engineering, chapter 3 - [Flow duration curve; firm and secondary energy](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afdc-firm-secondary-energy-69) — Hydropower Engineering, chapter 3 - [RoR best installed capacity (NRs, 45 years)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-nrs-10-percent) — Hydropower Engineering, chapter 3 - [Fault level: two 40 MVA generators, 2 Ω feeder](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-40mva-2-ohm-feeder) — Power Plant Equipment, chapter 4, asked 5×