---
title: "Wind power varies as cube of velocity"
subject: "Power Plant Design"
chapter: 1
key: "semester_8/power_plant_design:wind-power-cube-velocity"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/wind-power-cube-velocity
site: BCT Engineering
---

# Wind power varies as cube of velocity

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

- Asked in: 2079 Chaitra, 3+3 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-power-cube-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-1

## Question

Show that the power output of wind power plant is cube of the velocity of air. What are the other factors that may guide the possible output of wind power project?

## Short answer

Power in wind is P = 1/2 ρ A v³, so output is proportional to the cube of velocity. Other factors include swept area, air density, power coefficient (Betz limit), hub height, wind speed distribution, turbine cut-in/rated/cut-out speeds, mechanical/electrical losses, availability, wake effects, and grid constraints.

## Answer

#### Power varies as cube of wind velocity
Consider air of density $\rho$ moving at speed $v$ through the swept area $A$ of the rotor.

Mass of air passing per second:
$$
\dot{m} = \rho A v
$$
Kinetic energy carried per second (power):
$$
P = \frac{1}{2}\dot{m}v^2 = \frac{1}{2}(\rho A v)v^2 = \frac{1}{2}\rho A v^3
$$
The turbine extracts a fraction $C_p$ (power coefficient):
$$
P_{out} = \frac{1}{2}\rho A v^3 C_p \;\Rightarrow\; P_{out} \propto v^3
$$
So doubling wind speed gives $2^3 = 8$ times the power, and 10% higher speed gives $1.1^3 = 1.33$ times the power.

#### Other factors that guide the output of a wind project
1. **Swept area / blade length** – $A = \pi r^2$, so power $\propto r^2$.
2. **Air density** – lower at high altitude and high temperature; at about 3000 m density is roughly 25–30% less than at sea level.
3. **Power coefficient** – blade design, tip-speed ratio, pitch control; limited by Betz limit (0.593).
4. **Hub height** – wind speed increases with height above ground (wind shear).
5. **Wind speed distribution** – mean speed, variability and duration (Weibull distribution), turbulence.
6. **Cut-in, rated and cut-out speeds** of the turbine.
7. **Mechanical and electrical efficiency** – gearbox, generator, converter, transformer losses.
8. **Availability and wake losses** – downtime for maintenance; turbines in a wind farm shading each other.
9. **Grid availability and curtailment.**

## Check yourself

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

1. Which expression gives the mass of air passing through the rotor per second?
   A. rho A v
   B. rho A v^2
   C. 0.5 rho A v
   D. rho A / v
   Answer: A. Mass per second = density x swept area x speed, because a column of length v passes each second. (Key point: Mass flow rate of air)

2. Why does wind power depend on v^3?
   A. Mass flow is proportional to v and kinetic energy per kg is proportional to v^2
   B. Both mass flow and energy per kg are proportional to v^2
   C. Area of the rotor grows with v^3
   D. Density rises with v^3
   Answer: A. P = 1/2 (rho A v) v^2: one factor of v from mass flow and v^2 from kinetic energy. (Key point: Origin of the v^3 term)

3. If wind speed doubles, the power available changes by a factor of
   A. 2
   B. 4
   C. 6
   D. 8
   Answer: D. Power is proportional to v^3, so 2^3 = 8. (Key point: Effect of doubling speed)

4. How does the output vary with blade length (radius r)?
   A. Proportional to r
   B. Proportional to r^2
   C. Proportional to r^3
   D. Independent of r
   Answer: B. A = pi r^2, so power is proportional to r^2. (Key point: Other factors: swept area)

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