---
title: "Load, capacity and utilization factors"
subject: "Hydropower Engineering"
chapter: 3
key: "semester_7/hydropower_engineering:load-capacity-utilization-factor-220gwh"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/load-capacity-utilization-factor-220gwh
site: BCT Engineering
---

# Load, capacity and utilization factors

Hydropower Engineering · Chapter 3: Power and Energy Potential study

- Asked in: 2072 Kartik, 2+2+2 marks
- Video: [How the Load Factor, Plant Factor, and Utilization Factor are calculated 📈🏭💡](https://www.youtube.com/watch?v=KLk-Eflqzl0) — Hydropower
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/load-capacity-utilization-factor-220gwh
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3

## Question

A hydropower plant receives design discharge of 25 m³/s from 150 m height. The annual output of the plant is 220 GWh. If the peak load demand is 30 MW, determine (i) annual load factor, (ii) capacity factor and (iii) utilization factor. Assume the overall efficiency of the plant equals 85% and neglect head loss in the penstock.

## Short answer

Installed capacity = 9.81×0.85×25×150 = 31.27 MW, average load = 220 GWh/8760 h = 25.11 MW. Annual load factor = 25.11/30 = 0.837; capacity factor = 220×10⁶/(31.27×8760) = 0.803; utilization factor = 30/31.27 = 0.959.

## Answer

#### Installed capacity

$$
P_{inst} = 9.81\,\eta\,Q\,H = 9.81\times0.85\times25\times150 = 31,269\ \text{kW} = 31.27\ \text{MW}
$$

#### Average load
Annual energy $=220\ \text{GWh} = 220\times10^6$ kWh.

$$
P_{avg} = \frac{220\times10^6}{8760} = 25,114\ \text{kW} = 25.114\ \text{MW}
$$

#### (i) Annual load factor

$$
\text{LF} = \frac{P_{avg}}{P_{peak}} = \frac{25.114}{30} = 0.837\ (\approx 83.7\%)
$$

#### (ii) Capacity (plant) factor

$$
\text{CF} = \frac{E}{P_{inst}\times8760} = \frac{220\times10^6}{31,269\times8760} = 0.803\ (\approx 80.3\%)
$$

#### (iii) Utilization factor

$$
\text{UF} = \frac{P_{peak}}{P_{inst}} = \frac{30}{31.27} = 0.959\ (\approx 95.9\%)
$$

Check: CF = LF × UF = 0.837 × 0.959 = 0.803.

**Answer: load factor = 0.837; capacity factor = 0.803; utilization factor = 0.959.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the installed capacity found from the given discharge and head?
   A. P = 9.81 x eta x Q x H = 9.81 x 0.85 x 25 x 150 = 31.27 MW
   B. P = Q x H = 3,750 kW
   C. P = 220 GWh / 8760 = 25.1 MW
   D. P = 9.81 x Q x H without efficiency = 36.8 MW
   Answer: A. Installed capacity is 9.81 eta Q H in kW, which gives 31,269 kW or 31.27 MW. (Key point: Installed capacity)

2. How is the annual load factor calculated?
   A. Average load divided by installed capacity
   B. Peak load divided by installed capacity
   C. Average load (energy/8760 = 25.11 MW) divided by peak load (30 MW) = 0.837
   D. Energy divided by installed capacity only
   Answer: C. Load factor is average load over peak load: 25.114/30 = 0.837. (Key point: Load factor definition)

3. Which expression gives the capacity (plant) factor?
   A. Peak load / installed capacity
   B. Annual energy / (installed capacity x 8760) = 0.803
   C. Average load / peak load
   D. Installed capacity / annual energy
   Answer: B. Capacity factor compares actual energy with the energy possible at installed capacity over the year: 220x10^6/(31,269x8760) = 0.803. (Key point: Capacity factor definition)

4. What is the utilization factor and how does it relate to the other two?
   A. Peak load / installed capacity = 0.959; CF = LF x UF
   B. Average load / installed capacity = 0.803; CF = LF + UF
   C. Installed capacity / peak load = 1.04; CF = LF / UF
   D. Peak load / average load = 1.19; CF = UF / LF
   Answer: A. UF = 30/31.27 = 0.959, and capacity factor equals load factor times utilization factor (0.837 x 0.959 = 0.803). (Key point: Utilization factor and relation)

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