---
title: "Wind turbine output: 12 m blade, 18 m/s"
subject: "Power Plant Design"
chapter: 1
key: "semester_8/power_plant_design:wind-numerical-12m-18ms"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/wind-numerical-12m-18ms
site: BCT Engineering
---

# Wind turbine output: 12 m blade, 18 m/s

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

- Asked in: 2080 Chaitra, 4 marks
- Video: [Wind Energy | Maximum output power](https://www.youtube.com/watch?v=zXQQj_CIREo) — Sasthra
- 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-numerical-12m-18ms
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-1

## Question

Assume that we live in an area slightly above sea level that has an air density of 1.225 kg/m³ and we have installed a 45% efficient wind turbine which has a rotor blade radius of 12 m. Calculate the output power from the turbine at a wind speed of 18 m/s.

## Short answer

Swept area A = π(12)² = 452.4 m², power P = 0.5 × 1.225 × 452.4 × 18³ × 0.45 ≈ 727.2 kW.

## Answer

The output of a wind turbine is the power in the wind multiplied by the efficiency (power coefficient):
$$
P = \frac{1}{2}\rho A v^3 \eta
$$

#### Given
- $\rho = 1.225$ kg/m³
- $r = 12$ m
- $v = 18$ m/s
- $\eta = 0.45$

#### Step 1: Swept area
$$
A = \pi r^2 = \pi \times 12^2 = 452.39\ \text{m}^2
$$

#### Step 2: Power available in the wind
$$
\begin{aligned}
P_w &= \frac{1}{2}\rho A v^3 \\
&= 0.5 \times 1.225 \times 452.39 \times 18^3 \\
&= 0.5 \times 1.225 \times 452.39 \times 5832 \\
&= 1{,}615{,}980\ \text{W} \approx 1.616\ \text{MW}
\end{aligned}
$$

#### Step 3: Output power
$$
P = \eta P_w = 0.45 \times 1{,}615{,}980 = 727{,}191\ \text{W}
$$

**Answer:** Output power $\approx 727.2$ kW (about 0.727 MW).

Note: 45% is below the Betz limit of 59.3%, so it is a realistic value.

## Check yourself

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

1. Which formula gives the turbine output power?
   A. P = rho A v^2 eta
   B. P = 1/2 rho A v^2 eta
   C. P = 1/2 rho A v^3 eta
   D. P = rho A v^3 / eta
   Answer: C. Output = efficiency x 1/2 rho A v^3. (Key point: Formula for output)

2. What is the swept area for r = 12 m?
   A. 452.4 m^2
   B. 37.7 m^2
   C. 144 m^2
   D. 904.8 m^2
   Answer: A. A = pi x 12^2 = 452.39 m^2. (Key point: Swept area)

3. What is the power in the wind (before efficiency) at 18 m/s?
   A. About 0.727 MW
   B. About 1.616 MW
   C. About 3.6 MW
   D. About 0.5 MW
   Answer: B. 0.5 x 1.225 x 452.39 x 5832 = about 1.616 MW. (Key point: Power in the wind)

4. With 45% efficiency, the output is about
   A. 1.616 MW
   B. 727 W
   C. 727 kW
   D. 7.27 MW
   Answer: C. 0.45 x 1.616 MW = 0.727 MW = 727 kW. (Key point: Final output)

## Close variants set in other exams

- [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×

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