Question

If $\rho = $ air density; V = wind speed and A = swept frontal area of the machine (wind mill); then the amount of energy available in the wind is .

• A. $\frac{1}{2} \rho A V^2$
• B. $\rho A V$
• C. $\frac{1}{2} \rho A V^3$
• D. $\sqrt{\rho A V}$

Hint:

Just remember that "Power" for wind always involves the Cube ($V^3$). If you see $V^2$, that is just standard kinetic energy for a solid object; for a continuous flow like wind, you have to multiply by $V$ again to account for the speed at which the air is arriving.

Answer 1

The Correct Option is C

1. Basic Kinetic Energy Formula: The kinetic energy ($K.E.$) of any moving mass ($m$) is given by: $$K.E. = \frac{1}{2} m V^2$$

2. Calculating Wind Mass Flow: In the case of wind, we are interested in the power, which is the energy available per unit of time. The mass of air ($m$) passing through a swept area ($A$) in one second is the mass flow rate ($\dot{m}$): $$\dot{m} = \text{Density} \times \text{Volume Flow Rate}$$ $$\dot{m} = \rho \times (A \times V)$$

3. Deriving Wind Power (Energy available): Substituting the mass flow rate into the kinetic energy equation: $$\text{Power} = \frac{1}{2} (\dot{m}) V^2$$ $$\text{Power} = \frac{1}{2} (\rho A V) V^2$$ $$\text{Power} = \frac{1}{2} \rho A V^3$$