---
title: "Wind turbine power, tip speed ratio and energy"
subject: "Energy Resources"
chapter: 4
key: "semester_6/energy_resources:wind-turbine-power-energy-numerical"
source: https://www.bctengineer.com/q/semester_6/energy_resources/wind-turbine-power-energy-numerical
site: BCT Engineering
---

# Wind turbine power, tip speed ratio and energy

Energy Resources · Chapter 4: Wind energy

- Asked in: year not given, 8 marks
- Video: [Problem 1 on tip speed ratio (TSR) for wind Turbine/ Renewable energy engineering](https://www.youtube.com/watch?v=4_-syh1eb98) — Edutech Guru Engineering Learning-By Priyanka ma'am
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/wind-turbine-power-energy-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-4

## Question

A horizontal axis wind turbine has a rotor diameter of 40 m. At a wind speed of 12 m/s (air density 1.225 kg/m³) the power coefficient is 0.40, the mechanical-to-electrical efficiency is 0.90 and the rotor turns at 20 rpm. (a) Find the power in the wind, the mechanical power and the electrical power. (b) Find the tip speed ratio. (c) At a site the wind speed measured at 10 m is 5 m/s. Using the power law with exponent 0.14, find the wind speed at a hub height of 80 m and the wind power density there. (d) Taking the electrical power in (a) as the rated power and a capacity factor of 30 %, find the annual energy output.

## Short answer

Swept area 1256.6 m²; P_w = 0.5·1.225·1256.6·12³ = 1330 kW, P_m = 532 kW, P_e = 479 kW. Tip speed ratio λ = 3.49. At 80 m, v = 5(8)^0.14 = 6.69 m/s, power density = 183 W/m². Annual energy = 479×0.3×8760 = 1.26×10⁶ kWh.

## Answer

#### (a) Powers
Swept area $A = \dfrac{\pi D^2}{4} = \dfrac{\pi (40)^2}{4} = 1256.6\ \text{m}^2$.

$$
\begin{aligned}
P_w &= \tfrac{1}{2}\rho A v^3 = 0.5(1.225)(1256.6)(12)^3 \\
&= 1.330\times10^{6}\ \text{W} = 1330\ \text{kW} \\
P_m &= C_p P_w = 0.40 \times 1330 = 532\ \text{kW} \\
P_e &= 0.90 \times 532 = 479\ \text{kW}
\end{aligned}
$$

#### (b) Tip speed ratio
$$
\omega = \frac{2\pi N}{60} = \frac{2\pi(20)}{60} = 2.094\ \text{rad/s}
$$

$$
\lambda = \frac{\omega R}{v} = \frac{2.094 \times 20}{12} = 3.49
$$

#### (c) Wind shear
$$
\frac{v_h}{v_{10}} = \left(\frac{h}{10}\right)^{\alpha} \Rightarrow v_{80} = 5\,(8)^{0.14} = 5(1.338) = 6.69\ \text{m/s}
$$

Power density (power per unit swept area):

$$
\frac{P}{A} = \tfrac{1}{2}\rho v^3 = 0.5(1.225)(6.69)^3 = 183\ \text{W/m}^2
$$

#### (d) Annual energy
$$
E = P_{rated}\times CF \times 8760 = 479 \times 0.30 \times 8760 = 1.26\times10^{6}\ \text{kWh}
$$

**Answer:** $P_w = 1330$ kW, $P_m = 532$ kW, $P_e = 479$ kW; $\lambda = 3.49$; $v_{80} = 6.69$ m/s with 183 W/m²; annual energy $\approx 1.26$ GWh (1 258 000 kWh).

## Check yourself

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

1. For Pw = 1330 kW, Cp = 0.40, efficiency 0.90, the electrical power is...
   A. 532 kW
   B. 479 kW
   C. 1197 kW
   D. 1330 kW
   Answer: B. Pm = 0.40 x 1330 = 532 kW, Pe = 0.90 x 532 = 479 kW. (Key point: Power chain)

2. Tip speed ratio uses which radius and result?
   A. R = 20 m giving λ = 3.49
   B. D = 40 m giving λ = 6.98
   C. R = 20 m giving λ = 12
   D. R = 40 m giving λ = 1.75
   Answer: A. ω = 2.094 rad/s so ωR/v = 41.9/12 = 3.49. (Key point: Tip speed ratio)

3. Wind speed at 80 m from 5 m/s at 10 m with exponent 0.14 is...
   A. 5 m/s
   B. 40 m/s
   C. 13.4 m/s
   D. 6.69 m/s
   Answer: D. v = 5 x 8^0.14 = 5 x 1.338. (Key point: Wind shear)

4. Annual energy output is found by...
   A. Rated power x 8760 only
   B. Rated power x 0.30 x 24
   C. Rated power x capacity factor x 8760 h ≈ 1.26 x 10^6 kWh
   D. Rated power / 0.30 x 8760
   Answer: C. 479 x 0.30 x 8760 = 1.26 GWh. (Key point: Annual energy)

## Related in Energy Resources

- [Horizontal and vertical axis wind turbines](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Ahawt-vawt-types-characteristics) — Energy Resources, chapter 4
- [Derivation of Betz limit for wind turbine](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Abetz-limit-derivation-wind) — Energy Resources, chapter 4
- [Micro hydro net head, power and energy](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Amicro-hydro-power-output-numerical) — Energy Resources, chapter 5

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