---
title: "Wind turbine output: 6.5 m blade, 9 and 18 m/s"
subject: "Power Plant Design"
chapter: 1
key: "semester_8/power_plant_design:wind-numerical-6-5m-9ms"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/wind-numerical-6-5m-9ms
site: BCT Engineering
---

# Wind turbine output: 6.5 m blade, 9 and 18 m/s

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

- Asked in: 2079 Shrawan, 4 marks
- 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-6-5m-9ms
- 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 35% efficient wind turbine which has a rotor blade radius of 6.5 m. Calculate the output power from the turbine at a wind speed of 9 m/s and again at double the velocity of 18 m/s.

## Short answer

A = π(6.5)² = 132.73 m². At 9 m/s: P = 0.5×1.225×132.73×9³×0.35 ≈ 20.74 kW. At 18 m/s: P = 0.5×1.225×132.73×18³×0.35 ≈ 165.95 kW (8× due to v³).

## Answer

Turbine output power:
$$
P = \frac{1}{2}\rho A v^3 \eta, \qquad A = \pi r^2
$$

#### Given
$\rho = 1.225$ kg/m³, $r = 6.5$ m, $\eta = 0.35$, $v_1 = 9$ m/s, $v_2 = 18$ m/s.

#### Swept area
$$
A = \pi \times 6.5^2 = 132.73\ \text{m}^2
$$

#### At $v_1 = 9$ m/s
$$
\begin{aligned}
P_{w1} &= 0.5 \times 1.225 \times 132.73 \times 9^3 \\
&= 0.5 \times 1.225 \times 132.73 \times 729 = 59{,}267\ \text{W} \\
P_1 &= 0.35 \times 59{,}267 = 20{,}743\ \text{W}
\end{aligned}
$$

#### At $v_2 = 18$ m/s
$$
\begin{aligned}
P_{w2} &= 0.5 \times 1.225 \times 132.73 \times 18^3 \\
&= 0.5 \times 1.225 \times 132.73 \times 5832 = 474{,}133\ \text{W} \\
P_2 &= 0.35 \times 474{,}133 = 165{,}947\ \text{W}
\end{aligned}
$$

#### Check
$$
\frac{P_2}{P_1} = \left(\frac{18}{9}\right)^3 = 8, \qquad 8 \times 20.743 = 165.95\ \text{kW}
$$

**Answer:** Output $\approx 20.74$ kW at 9 m/s and $\approx 165.95$ kW at 18 m/s. Doubling the wind speed gives 8 times the power because $P \propto v^3$.

## Check yourself

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

1. For r = 6.5 m, the swept area is about
   A. 20.4 m^2
   B. 42.25 m^2
   C. 132.7 m^2
   D. 265.5 m^2
   Answer: C. A = pi x 6.5^2 = 132.73 m^2. (Key point: Swept area)

2. The turbine output at 9 m/s (35% efficient) is about
   A. 20.7 kW
   B. 59.3 kW
   C. 166 kW
   D. 2.07 kW
   Answer: A. Wind power 59.27 kW x 0.35 = 20.74 kW. (Key point: Output at 9 m/s)

3. Why is the output at 18 m/s exactly 8 times that at 9 m/s?
   A. Power is proportional to v
   B. Power is proportional to v^2
   C. Power is proportional to v^3, and 2^3 = 8
   D. Efficiency doubles
   Answer: C. Same area, density and efficiency, so P ratio = (18/9)^3 = 8. (Key point: Cube law check)

4. The output at 18 m/s is about
   A. 41.5 kW
   B. 83 kW
   C. 474 kW
   D. 166 kW
   Answer: D. 8 x 20.74 = 165.9 kW (474 kW is the wind power before efficiency). (Key point: Output at 18 m/s)

## Close variants set in other exams

- [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×
- [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×

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