Question

A wind turbine on a wind farm turns in response to a force of high-speed air resistance, $\mathit{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{D}\mathit{\rho }\mathit{A}{\mathit{v}}^{\mathbf{2}}$. The power available is, $\mathit{P}\mathbf{=}\mathit{R}\mathit{v}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{D}\mathit{\rho }\mathit{\pi }{\mathit{r}}^{\mathbf{2}}{\mathit{v}}^{\mathbf{3}}$where v is the wind speed and we have assumed a circular face for the wind turbine of radius r. Take the drag coefficient as D = 1.00 and the density of air from the front endpaper. For a wind turbine having r = 1.50 m, calculate the power available with (a) v = 8.00 m/s and (b) v = 24.0 m/s. The power delivered to the generator is limited by the efficiency of the system, about 25%. For comparison, a large American home uses about 2 kW of electric power.

Short answer

(a) The power available with v = 8.00 m/s is $P_a=542.5W$.

(b) The power available with v = 24.00 m/s is$P_b=14.65KW$ .

Step-by-step solution

Introduction

The amount of energy transferred from one body to another in unit time interval is known as power. Its SI unit is Watts (W).

For a wind turbine, power is given by:

$\mathit{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{D}\mathit{\rho }\mathit{\pi }{\mathit{r}}^{\mathbf{2}}{\mathit{v}}^{\mathbf{3}}$

D = Drag coefficient

ρ = Density of air

r = Radius of the turbine

v =Speed of the object

P = Power of the turbine

Given Data

The power available is,$P=\frac{1}{2}D\rho \pi r^2{v}^3$

The drag coefficient is$D=1.00$

The radius of wind turbine$r=1.50m$

The velocity of the turbine is $v_a=8\text{\hspace{0.33em}m}/\text{s\hspace{0.33em}and\hspace{0.33em}}{v}_b=24\text{\hspace{0.33em}\hspace{0.33em}m}/\text{s}$

Calculation for part (a)

We use $1.2Kg/m^3$for the density of air, and calculate as shown below.

$\begin{array}{c}{P}_a=\frac{1}{2}Drp{r}^2{v}^3\\ =\frac{1}{2}\left(1\right)\left(1.2{\text{\hspace{0.33em}kg/m}}^{\text{3}}\right)\left(3.14\right){\left(1.50\text{m}\right)}^2{\left(8.00\text{m/s}\right)}^3\\ =2.17\times {10}^3\text{\hspace{0.33em}W.}\end{array}$

As the efficiency of the system is 25% the power available to the motor is calculated as shown below.

$\begin{array}{c}{P}_{available}=P_a\times \frac{25}{100}\\ =\left(2.17\times {10}^3\text{\hspace{0.33em}W}\right)\times \frac{25}{100}\\ =542.5\text{\hspace{0.33em}W.}\end{array}$

Calculation for part (b)

We solve part (b) by proportion:

$\begin{array}{c}\frac{P_b}{P_a}=\frac{v_b^3}{v_a^3}={\left(\frac{24\text{\hspace{0.33em}m/s}}{8\text{\hspace{0.33em}m/s}}\right)}^3=27\\ P_b=27P_a\\ =27\left(2.17\times {10}^3\text{\hspace{0.33em}W}\right)\\ =58.6\text{\hspace{0.33em}kW.}\end{array}$

The power available is calculated as

$\begin{array}{c}{P}_{available}=P_b\times 0.25\\ =\left(58.6\times {10}^3\text{\hspace{0.33em}W}\right)\times 0.25\\ =14.65\text{\hspace{0.33em}kW.}\end{array}$

The power available to the motor is 14.65 KW