Chapter 2.8, Problem 14P

To determine

Choose the better site for wind power generation.

Answer to Problem 14P

The better site for wind power generation is $\text{second}$.

Explanation of Solution

Write the formula to calculate mechanical energy of the wind $\left(e_{\text{mech}}\right)$.

$\begin{array}{c}{e}_{\text{mech}}=ke\\ =\frac{V^2}{2}\end{array}$ (I)

Here, kinetic energy per unit mass of wind is $ke$ and velocity of wind is $V$.

Write the formula to calculate power generation potential of the wind $\left({\dot{W}}_{\max }\right)$.

$\begin{array}{c}{\dot{W}}_{\max }={\dot{E}}_{\text{mech}}\\ =\dot{m}{e}_{\text{mech}}\\ =\rho VA{e}_{\text{mech}}\end{array}$ (III)

Here, mechanical energy of the flowing fluid on a unit mass basis is ${\dot{E}}_{\text{mech}}$, mass flow rate of the wind is $\dot{m}$, area of the wind turbine is A and density of air is $\rho$.

Write the formula to calculate electric power generation per year $\left(E_{\max }\right)$.

$E_{\max }={\dot{W}}_{\max }\Delta t$ (III)

Here, total run time of the site is $\Delta t$.

Conclusion:

Consider unit area for the site I and site II respectively.

$A=1{\text{m}}^2$

Site I

Substitute $7\text{m}/\text{s}$ for $V$ in Equation (I).

$\begin{array}{c}{e}_{\text{mech}}=\frac{{\left(7\text{m}/\text{s}\right)}^2}{2}\\ =\frac{{\left(7\text{m}/\text{s}\right)}^2}{2}\left(\frac{\text{1}\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =0.0245\text{kJ}/\text{kg}\end{array}$

Substitute $1.25\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $7\text{m}/\text{s}$ for $V$, $1{\text{m}}^2$ for $A$ and $0.0245\text{kJ}/\text{kg}$ for $e_{\text{mech}}$ in Equation (II).

$\begin{array}{c}{\dot{W}}_{\max }=\left(1.25\text{kg}/{\text{m}}^{\text{3}}\right)\left(7\text{m}/\text{s}\right)\left(1{\text{m}}^2\right)\left(0.0245\text{kJ}/\text{kg}\right)\\ =0.2144\text{kJ}/\text{s}\\ =0.2144\text{kJ}/\text{s}\times \frac{\text{1}\text{kW}}{1\text{kJ}/\text{s}}\\ =0.2144\text{kW}\end{array}$

Substitute $0.2144\text{kW}$ for ${\dot{W}}_{\max }$ and $3000\text{h}/\text{yr}$ for $\Delta t$ in Equation (III).

$\begin{array}{c}{E}_{\max }=\left(0.2144\text{kW}\right)\left(3000\text{h}/\text{yr}\right)\\ =643\text{kWh}/\text{yr}\end{array}$

Site II

Substitute $10\text{m}/\text{s}$ for $V$ in Equation (I).

$\begin{array}{c}{e}_{\text{mech}}=\frac{{\left(10\text{m}/\text{s}\right)}^2}{2}\\ =\frac{{\left(10\text{m}/\text{s}\right)}^2}{2}\left(\frac{\text{1}\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =0.050\text{kJ}/\text{kg}\end{array}$

Substitute $1.25\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $10\text{m}/\text{s}$ for $V$, $1{\text{m}}^2$ for $A$ and $0.050\text{kJ}/\text{kg}$ for $e_{\text{mech}}$ in Equation (II).

$\begin{array}{c}{\dot{W}}_{\max }=\left(1.25\text{kg}/{\text{m}}^{\text{3}}\right)\left(10\text{m}/\text{s}\right)\left(1{\text{m}}^2\right)\left(0.050\text{kJ}/\text{kg}\right)\\ =0.625\text{kJ}/\text{s}\\ =0.625\text{kJ}/\text{s}\times \frac{\text{1}\text{kW}}{1\text{kJ}/\text{s}}\\ =0.625\text{kW}\end{array}$

Substitute $0.625\text{kW}$ for ${\dot{W}}_{\max }$ and $1500\text{h}/\text{yr}$ for $\Delta t$ in Equation (III).

$\begin{array}{c}{E}_{\max }=\left(0.625\text{kW}\right)\left(1500\text{h}/\text{yr}\right)\\ =938\text{kWh}/\text{yr}\end{array}$

Therefore, it is found that the site II generates more wind power when compared to site I.

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