---
title: "Wind power output formula"
subject: "Power Plant Design"
chapter: 1
key: "semester_8/power_plant_design:wind-power-formula"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/wind-power-formula
site: BCT Engineering
---

# Wind power output formula

Power Plant Design · Chapter 1: Energy Sources and Electric Power Generation

- Asked in: 2082 Shrawan, 4 marks; 2075 Bhadra, 4 marks (asked 2 times)
- Video: [Wind power - Part 2: Wind power equation derivation](https://www.youtube.com/watch?v=ZmRNUD1q97g) — LT Vault
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/power_plant_design/wind-power-formula
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-1

## Question

Derive the formula to estimate electric power output from a wind power system.

## Short answer

Electric power output is P_e = ½ ρ A v³ C_p η_g η_gen, where A = πr² is swept area. Power scales with cube of wind speed and square of blade radius; practical power coefficient C_p is 0.35–0.45.

## Answer

The power in wind is the rate of flow of kinetic energy of air passing through the area swept by the turbine rotor.

#### Derivation
Let $\rho$ = air density (kg/m³), $A$ = swept area (m²), $v$ = wind speed (m/s).

Kinetic energy of air of mass $m$:
$$
E = \frac{1}{2} m v^2
$$
Mass flow rate through area $A$:
$$
\dot{m} = \rho A v
$$
Power available in the wind (energy per second):
$$
P_w = \frac{1}{2}\dot{m}v^2 = \frac{1}{2}\rho A v^3
$$
A turbine cannot take all this power, because air must leave the rotor with some speed. The fraction extracted is the **power coefficient** $C_p$. By Betz's law, $C_p \le 16/27 = 0.593$.

Mechanical power on the rotor:
$$
P_m = \frac{1}{2}\rho A v^3 C_p
$$
Electrical output includes the gearbox and generator efficiencies:
$$
P_e = \frac{1}{2}\rho A v^3 C_p\, \eta_g\, \eta_{gen}
$$
For a horizontal-axis turbine with blade length (radius) $r$: $A = \pi r^2$.

#### Key points
- $P \propto v^3$: doubling wind speed gives 8 times the power.
- $P \propto r^2$: doubling blade length gives 4 times the power.
- $P \propto \rho$: output is less at high altitude (lower air density).
- Practical $C_p$ is about 0.35–0.45.

## Check yourself

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

1. What is the power available in the wind through swept area A?
   A. (1/2) rho A v^2
   B. (1/2) rho A v^3
   C. rho A v
   D. (1/2) rho v^3 only
   Answer: B. Mass flow = rho A v and power = (1/2) m_dot v^2, giving (1/2) rho A v^3. (Key point: Power in wind)

2. Why is the power coefficient Cp introduced?
   A. A turbine cannot take all wind power because air must leave with some speed
   B. To account for the generator only
   C. Because wind speed is constant
   D. To convert watts to kW
   Answer: A. Betz's law limits Cp to 16/27 = 0.593. (Key point: Power coefficient)

3. If the wind speed doubles, the power becomes
   A. 2 times
   B. 4 times
   C. 6 times
   D. 8 times
   Answer: D. P is proportional to v^3, so 2^3 = 8. (Key point: Effect of speed)

4. If blade length (radius) is doubled, the power becomes
   A. 4 times
   B. 2 times
   C. 8 times
   D. Unchanged
   Answer: A. A = pi r^2, so power is proportional to r^2: 2^2 = 4. (Key point: Effect of blade length)

## Related in Power Plant Design

- [Wind power pros/cons and output: 1.25 m blade](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-pros-cons-1-25m) — Power Plant Design, chapter 1, asked 2×
- [Wind power pros/cons and output: 1.5 m blade](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-pros-cons-1-5m) — Power Plant Design, chapter 1, asked 2×
- [Wind turbine merits and output: 5.5 m blade](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-merits-numerical-5-5m) — Power Plant Design, chapter 1
- [Wind turbine output: 12 m blade, 18 m/s](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-numerical-12m-18ms) — Power Plant Design, chapter 1, asked 3×
- [Wind turbine output: 6.5 m blade, 9 and 18 m/s](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-numerical-6-5m-9ms) — Power Plant Design, chapter 1, asked 3×
- [Wind turbine output: 6 m blade, 8 and 16 m/s](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-numerical-6m-8ms) — Power Plant Design, chapter 1, asked 3×

## Related in other subjects

- [Derive Newton-Raphson formula](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Anewton-raphson-derive) — Numerical Method, chapter 2, asked 2×
- [Wind turbine power, tip speed ratio and energy](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Awind-turbine-power-energy-numerical) — Energy Resources, chapter 4
- [Wind sources, turbines, parks, power control](https://www.bctengineer.com/question.md?key=semester_7%2Fenergy_environment_and_society%3Awind-sources-turbines-parks-control) — Energy Environment and Society, chapter 3
- [Types of electric wiring system compared](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Awiring-systems-comparison) — Industrial Power Distribution and Illumination, chapter 5