Chapter 6, Problem 96P

(a)

To determine

The volume of air passes through the circular area swept out by the blades.

Answer to Problem 96P

The volume of air passes through the circular area swept out by the blades is $\underset{\_}{500{\text{m}}^3}$.

Explanation of Solution

Given that the length of the blade is $4.0\text{m}$, the speed of wind is $10\text{m/s}$, and the time is $1.0\text{s}$.

Write the expression for the volume of air swept out by the blade.

$V=Ad$ (I)

Here, $V$ is the volume, $A$ is the area swept out by the blade, and $d$ is the distance moved by the air.

Since the blade sweeps out circular area, the radius of the circle is equal to the length of the blade.

Write the expression for the area swept out by the blade.

$A=\pi L^2$ (II)

Here, $L$ is the length of the blade (radius of the circle)

Write the expression for the distance travelled by the air in the given time.

$d=v\Delta t$ (III)

Here, $v$ is the speed of air, and $\Delta t$ is the time interval.

Use equation (II) and (III) in (I).

$V=\pi L^{2}v\Delta t$ (IV)

Conclusion:

Substitute $4.0\text{m}$ for $L$, $10\text{m/s}$ for $v$, and $1.0\text{s}$ for $\Delta t$ in equation (IV) to find the volume $V$.

$\begin{array}{c}V=\pi {\left(4.0\text{m}\right)}^2\left(10\text{m/s}\right)\left(1.0\text{s}\right)\\ =502{\text{m}}^3\\ \approx 500{\text{m}}^3\end{array}$

Therefore, the volume of air passes through the circular area swept out by the blades is $\underset{\_}{500{\text{m}}^3}$.

(b)

To determine

The mass of $500{\text{m}}^3$ air.

Answer to Problem 96P

The mass of $500{\text{m}}^3$ air is $\underset{\_}{600\text{kg}}$.

Explanation of Solution

The density of air is $1.2{\text{kg/m}}^{\text{3}}$.

Write the expression for the mass of air in terms of its density and volume.

$m=V\rho$ (V)

Here, $m$ is the mass, and $\rho$ is the density.

Conclusion:

Substitute $500{\text{m}}^3$ for $V$, and $1.2{\text{kg/m}}^3$ for $\rho$ in equation (V) to find the mass of air $m$.

$\begin{array}{c}m=\left(500{\text{m}}^3\right)\left(1.2{\text{kg/m}}^3\right)\\ =600\text{kg}\end{array}$

Therefore, the mass of $500{\text{m}}^3$ air is $\underset{\_}{600\text{kg}}$.

(c)

To determine

The translational kinetic energy of the air.

Answer to Problem 96P

The translational kinetic energy of the air is $\underset{\_}{30\text{kJ}}$.

Explanation of Solution

Given that the mass of air is $600\text{kg}$, and the speed of air is $10\text{m/s}$.

Write the expression for the translational kinetic energy.

$K=\frac{1}{2}m{v}^2$ (VI)

Here, $K$ is the kinetic energy.

Conclusion:

Substitute $600\text{kg}$ for $m$, and $10\text{m/s}$ for $v$ in equation (VI) to find the kinetic energy $K$.

$\begin{array}{c}K=\frac{1}{2}\left(600\text{kg}\right){\left(10\text{m/s}\right)}^2\\ =30,000\text{J}\\ =30,000\text{J}\times \frac{1\text{kJ}}{1000\text{J}}\\ =30\text{kJ}\end{array}$

Therefore, the translational kinetic energy of the air is $\underset{\_}{30\text{kJ}}$.

(d)

To determine

The electric power output of the turbine.

Answer to Problem 96P

The electric power output of the turbine is $\underset{\_}{12\text{kW}}$.

Explanation of Solution

Given that the turbine can convert $40\%$ of the kinetic energy to the electrical energy. It is obtained that the kinetic energy of the air is $30\text{kJ}$.

Write the expression for the power output of the turbine (it is the energy converted per unit time).

$P=\frac{\Delta E}{\Delta t}$ (VII)

Here, $P$ is the power output, $\Delta E$ is the energy converted, and $\Delta t$ is the time interval.

Since the turbine converts $40\%$ of the kinetic energy to the electrical energy, $\Delta E=0.40K$.

Thus, the equation (VII) becomes,

$P=\frac{0.40K}{\Delta t}$ (VIII)

Conclusion:

Substitute $30\text{kJ}$ for $K$, and $1.0\text{s}$ for $\Delta t$ in equation (VIII) to find $P$

$\begin{array}{c}P=\frac{0.40\left(30\text{kJ}\right)}{1.0\text{s}}\\ =12\text{kW}\end{array}$

Therefore, the electric power output of the turbine is $\underset{\_}{12\text{kW}}$.

(e)

To determine

The power output of the turbine if the speed of the wind is decreases to half of its initial value and the conclusion about the electric power production by wind turbines.

Answer to Problem 96P

If the speed of the wind is decreases to half of its initial value, the power output of the turbine decrease to $\underset{\_}{1/8}$ of its previous value. It can be concluded that the power production of an individual wind turbine is inconsistent, since modest changes in the wind speed produce large changes in power output.

Explanation of Solution

Equation (VII) gives the power output of the wind turbine.

$P=\frac{\Delta E}{\Delta t}$

Assume that the entire kinetic energy of the wind is converted to electrical energy. Then the equation (VII) can be modified as,

$P=\frac{\frac{1}{2}m{v}^2}{\Delta t}$ (IX)

Use equation (V) in (IX).

$P=\frac{\frac{1}{2}V\rho v^2}{\Delta t}$ (X)

Use equation (IV) in (X).

$\begin{array}{c}P=\frac{\frac{1}{2}\left(\pi L^{2}v\Delta t\right)\rho v^2}{\Delta t}\\ P\propto v^3\end{array}$ (XI)

Conclusion:

From the expression (XI) it is clear that the power output of the turbine is proportional to the cube of the wind speed.

Let $P_v$ and $P_{v/2}$ be the power output of the turbine when wind speed is $v$ and $v/2$ respectively.

Take the ratio of $P_{v/2}$ and $P_v$.

$\begin{array}{c}\frac{P_{v/2}}{P_v}=\frac{{\left(v/2\right)}^3}{v^3}\\ ={\left(\frac{1}{2}\right)}^3\\ =\frac{1}{8}\end{array}$

This indicates that the power production of an individual wind turbine is inconsistent, since modest changes in the wind speed produce large changes in power output.

Therefore, If the speed of the wind is decreases to half of its initial value, the power output of the turbine decrease to $\underset{\_}{1/8}$ of its previous value. It can be concluded that the power production of an individual wind turbine is inconsistent, since modest changes in the wind speed produce large changes in power output.