Chapter 33, Problem 16SP

ELECTRIC GENERATORS

A 120-V generator is run by a windmill that has blades 2.0 m long. The wind, moving at 12 m/s, is slowed to 7.0 m/s after passing the windmill. The density of air is $1.29{\text{ kg/m}}^{\text{3}}$. If the system has no losses, what is the largest current the generator can produce? [Hint: How much energy does the wind lose per second?]

A 120-V generator is run by a windmill that has blades 2.0 m long. The wind, moving at 12 m/s, is slowed to 7.0 m/s after passing the windmill. The density of air is $1.29{\text{ kg/m}}^{\text{3}}$. If the system has no losses, what is the largest current the generator can produce? [Hint: How much energy does the wind lose per second?]

To determine

The largest current produced by the $120\text{V}$ generator, which is run by the windmill that has $2.0\text{ m}$ longblade. The wind, initially moving at $12\text{ m/s}$ is slowed to $\text{7 m/s}$ after passing the windmill. The density of air is $1.29{\text{kg/m}}^3$.

Answer to Problem 16SP

Solution:

$77\text{ A}$

Explanation of Solution

Given data:

The length of the blade is $2.0\text{ m}$.

Thevoltage of the generator is $120\text{V}$.

The density of air is $1.29{\text{kg/m}}^3$.

The initial speed of the wind is $12\text{ m/s}$.

The final speed of the wind is $\text{7 m/s}$.

Formula used:

The expression of mechanical power is expressed as,

$P=\frac{W}{t}$

Here, $W$ is the work-done and $t$ is the time.

The expression of electrical power is expressed as,

$P=VI$

Here, $V$ is the voltage and $I$ is the current.

The work-energy theorem is expressed as,

$W=\Delta K$

Here, $\Delta K$ is the change in kinetic energy.

The expression of kinetic energy is expressed as,

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

Here, $m$ is the mass and $v$ is the velocity.

The expression of density of the material is expressed as,

$\rho =\frac{m}{V}$

Here, $m$ is the mass of the materialand $V$ is the volume of the material.

The volume of cylindrical shape is expressed as,

$V=Al$

Here, $A$ is the cross-sectional area of cylindrical shape and $l$ is the length.

Explanation:

Consider the expression of wind power.

$P_w=\frac{W}{t}$

Consider the work energy theorem.

$W=\Delta K$

Substitute $\Delta K$ for $W$

$P_w=\frac{\Delta K}{t}$

Understand that the wind comes with speed $12\text{ m/s}$ and passed out from the blade of the windmill. In this process, the velocity reduces to $\text{7 m/s}$.

Therefore, the expression of wind power can be written as,

$\begin{array}{c}{P}_w=\frac{\frac{1}{2}m\left(v_f^2-v_i^2\right)}{t}\\ =\frac{1}{2}\frac{m}{t}\left(v_f^2-v_i^2\right)\end{array}$

Consider the expression of density of air.

${\rho }_{air}=\frac{m}{V}$

Rearrange the expression.

$m={\rho }_{air}V$

Substitute ${\rho }_{air}V$ for $m$

$P_w=\frac{1}{2}\frac{{\rho }_{air}V}{t}\left(v_f^2-v_i^2\right)$

Consider the expression of volume.

$V=Al$

Substitute $Al$ for $V$

$\begin{array}{c}{P}_w=\frac{1}{2}\frac{{\rho }_{air}\left(Al\right)}{t}\left(v_f^2-v_i^2\right)\\ =\frac{1}{2}\left({\rho }_{air}A\right)\left(\frac{l}{t}\right)\left(v_f^2-v_i^2\right)\end{array}$

Understand that the $\frac{l}{t}$ is the speed of wind that strikes on the blade of the windmill. Therefore, the expression of power can be written as,

$\begin{array}{c}{P}_w=\frac{1}{2}\left({\rho }_{air}A\right)\left(v_i\right)\left(v_f^2-v_i^2\right)\\ =\frac{1}{2}\left({\rho }_{air}\right)\left(\pi r^2\right)\left(v_i\right)\left(v_f^2-v_i^2\right)\end{array}$

Here, $A$ is the area of cross-section formed by the motion of the blades.

Substitute $2.0\text{ m}$ for $r$, $1.29{\text{kg/m}}^3$ for ${\rho }_{air}$, $12\text{ m/s}$ for $v_i$, and $\text{7 m/s}$ for $v_f$

$\begin{array}{c}{P}_w=\frac{1}{2}\left(1.29{\text{kg/m}}^3\right)\left(\pi {\left(2.0\text{ m}\right)}^2\right)\left(12\text{ m/s}\right)\left({\left(\text{7 m/s}\right)}^2-{\left(12\text{ m/s}\right)}^2\right)\\ =-9240.05\text{ W}\end{array}$

Negative sign indicates that the power is loss.

The magnitude of the power loss of the wind is,

$P_w=9240.05\text{ W}$

Consider the expression of electrical power.

$P_e=VI$

Understand that there is no loss, hence, this wind power is converted in to the electrical power completely. In this case, the current would be maximum.

$P_e=P_w$

Substitute $V{I}_{\max }$ for $P_e$

$V{I}_{\max }=P_w$

Substitute $9240.05$ for $P_w$ and $120\text{V}$ for $V$

$\left(120\text{V}\right)I_{\max }=9240.05$

Solve for $I_{\max }$.

$\begin{array}{c}{I}_{\max }=\frac{9240.05}{120\text{V}}\\ =77\text{ A}\end{array}$

Conclusion:

The maximum current that generator can be producedis $77\text{ A}$.