Chapter 3, Problem 14P

To determine

The site that is better for the wind power generation.

Explanation of Solution

Given:

Velocity of the wind at first site $\left(V_1\right)$ is $7\text{m}/\text{s}$.

Velocity of the wind at second site $\left(V_2\right)$ is $10\text{m}/\text{s}$.

Time taken for the wind to blow at first site per year $\left(\Delta t_1\right)$ is $3000\text{hr}/\text{year}$.

Time taken for the wind to blow at second site per year $\left(\Delta t_2\right)$ is $1500\text{hr}/\text{year}$.

Density of the air $\left(\rho \right)$ is $1.25\text{kg}/{\text{m}}^3$.

Calculation:

Consider unit flow area for both sites.

  $A=1{\text{m}}^{\text{2}}$

Calculate the power generation potential of the wind at first site $\left({\dot{W}}_{\text{max,}\text{1}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{max,}\text{1}}={\dot{E}}_{\text{mech,}\text{1}}\\ ={\dot{m}}_1{e}_{\text{mech, 1}}\\ =\left(\rho V_{1}A\right)\left(\frac{V_1^2}{2}\right)\\ =\left[\left(1.25\text{kg}/{\text{m}}^3\right)\left(7\text{m}/\text{s}\right)\left(1{\text{m}}^{\text{2}}\right)\right]\left(\frac{{\left(7\text{m}/\text{s}\right)}^2}{2}\right)\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{max,}\text{1}}=\left[\left(1.25\text{kg}/{\text{m}}^3\right)\left(7\text{m}/\text{s}\right)\left(1{\text{m}}^{\text{2}}\right)\right]\left(\frac{{\left(7\text{m}/\text{s}\right)}^2}{2}\right)\left(\frac{\text{1}\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =0.2144\text{kW}\end{array}$

Calculate the power generation potential of the wind at second site $\left({\dot{W}}_{\text{max,}2}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{max,}2}={\dot{E}}_{\text{mech,}2}\\ ={\dot{m}}_2{e}_{\text{mech, 2}}\\ =\left(\rho V_{2}A\right)\left(\frac{V_2^2}{2}\right)\\ =\left[\left(1.25\text{kg}/{\text{m}}^3\right)\left(10\text{m}/\text{s}\right)\left(1{\text{m}}^{\text{2}}\right)\right]\left(\frac{{\left(10\text{m}/\text{s}\right)}^2}{2}\right)\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{max,}2}=\left[\left(1.25\text{kg}/{\text{m}}^3\right)\left(10\text{m}/\text{s}\right)\left(1{\text{m}}^{\text{2}}\right)\right]\left(\frac{{\left(10\text{m}/\text{s}\right)}^2}{2}\right)\left(\frac{\text{1}\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =0.625\text{kW}\end{array}$

Calculate the maximum electric power generation per year at first site $\left(E_{\text{max,}\text{1}}\right)$.

  $\begin{array}{c}{E}_{\text{max,}\text{1}}=\left({\dot{W}}_{\text{max,}\text{1}}\right)\left(\Delta t_1\right)\\ =\left(0.2144\text{kW}\right)\left(3000\text{hr}/\text{year}\right)\\ =643\text{kwh}/\text{year}\end{array}$

Calculate the maximum electric power generation per year at second site $\left(E_{\text{max,}\text{2}}\right)$.

  $\begin{array}{c}{E}_{\text{max,}2}=\left({\dot{W}}_{\text{max,}2}\right)\left(\Delta t_2\right)\\ =\left(0.625\text{kW}\right)\left(1500\text{hr}/\text{year}\right)\\ =938\text{kwh}/\text{year}\end{array}$

Therefore, the second site produces maximum electric power when compared to first site.

Thus, the best site for the wind power generation is $\underset{\_}{\text{second}\text{site}}$.