A wind-powered generator converts wind energy into electric energy. Assume that the generator conver

Q: Why do we use $\frac{dm}{dt}$ instead of just a fixed mass $m$?

Because wind is a continuous flow, not a single solid object. The turbine constantly intercepts new air. To find the power (energy per second), we must look at how much mass passes through the blades every second, which is the mass flow rate $\frac{dm}{dt}$.

Q: The book's solution uses $F = v \frac{dm}{dt}$. Where does this come from?

That approach calculates the hypothetical force required to completely stop the wind, using Newton's second law $F = \frac{dp}{dt} = v \frac{dm}{dt}$. The power is then $P = Fv$. While it gives the correct $v^3$ proportionality, using the kinetic energy rate $P = \frac{1}{2} \frac{dm}{dt} v^2$ is physically more accurate for representing the total energy available in the wind.

Q: Why is the power proportional to $v^3$ and not $v^2$ like kinetic energy?

This is the beautiful catch! The kinetic energy of a *fixed* mass is proportional to $v^2$. But for a wind turbine, as the wind blows faster, not only does each kilogram of air have more energy ($v^2$), but *more kilograms* of air hit the turbine every second (an extra $v$ factor). Multiplying these gives $v^2 \times v = v^3$.

Q: Does the density of the air matter?

Absolutely! The power equation is $P = \frac{1}{2} \rho A v^3$. Denser air (like cold air at sea level) carries more mass for the same volume, meaning more kinetic energy hits the blades. This is why wind turbines are often more efficient in winter.

Q: Can a turbine extract 100% of this wind energy?

No, and this is governed by Betz's Limit. If a turbine extracted 100% of the energy, the wind would completely stop right behind the blades, blocking any new wind from coming in! The theoretical maximum efficiency is about 59.3%.

Question

A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed $v$, the electrical power output will be proportional to

Options:

(A) $v$

(B) $v^2$

(C) $v^3$

(D) $v^4$

Sigma Percentile · Accepted Answer

### The Setup: Capturing the Wind

Imagine standing in front of a massive wind turbine. The blades sweep out a giant circular area, which we will call $A$.

The wind is blowing towards this turbine with a constant velocity $v$. Our goal is to figure out how much energy this wind carries, and more importantly, how fast that energy is being delivered to the turbine.

### The Flow of Mass

Wind is not a single solid object; it is a continuous fluid. To understand its energy, we need to look at the mass of air passing through the turbine's swept area every second. This is known as the mass flow rate.

Let the density of the air be $ho$. Consider a cylinder of air that passes through the area $A$ in a small time interval $dt$. The length of this cylinder is the distance the air travels, which is $dx$.

The volume of this cylinder is $V = A \cdot x$. Therefore, the mass of this air is $m = ho A x$.

Now, we can find the mass flow rate by taking the derivative with respect to time:

$$ \frac{dm}{dt} = \frac{d}{dt}(ho A x) $$

Since the density $ho$ and the area $A$ are constants, we can pull them out of the derivative:

$$ \frac{dm}{dt} = ho A \frac{dx}{dt} $$

And what is the rate of change of distance $\frac{dx}{dt}$? It is simply the velocity of the wind, $v$. Thus, our mass flow rate is:

$$ \frac{dm}{dt} = ho A v $$

### The Power Equation

Now comes the main point. The kinetic energy of any mass $m$ moving with velocity $v$ is given by:

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

Power $P$ is defined as the rate at which this kinetic energy is transferred. So, we take the time derivative of the kinetic energy:

$$ P = \frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2} m v^2 ight) $$

Because the wind velocity $v$ is constant, we can pull $\frac{1}{2} v^2$ out of the derivative, leaving us with the mass flow rate we just calculated:

$$ P = \frac{1}{2} v^2 \left( \frac{dm}{dt} ight) $$

Let's substitute our expression for the mass flow rate into this equation:

$$ P = \frac{1}{2} v^2 (ho A v) $$

Multiplying the terms together, we arrive at the beautiful master equation for the power available in the wind:

$$ P = \frac{1}{2} ho A v^3 $$

### The Final Verdict

The problem states that the generator converts a *fixed fraction* of this intercepted wind energy into electrical energy. Let's call this efficiency fraction $\eta$.

The electrical power output $P_{	ext{elec}}$ is then:

$$ P_{	ext{elec}} = \eta \left( \frac{1}{2} ho A v^3 ight) $$

Since $\eta$, $ho$, and $A$ are all constants, we can clearly see the relationship between the electrical power and the wind speed:

$$ P_{	ext{elec}} \propto v^3 $$

This cubic relationship is a fundamental principle of wind energy. It means that if the wind speed doubles, the power output doesn't just double—it increases by a factor of eight! This is why finding locations with high average wind speeds is absolutely critical for wind farms.

Select Answer:

Visualized Solution

The Wind Turbine Setup

Kinetic Energy of Air

Defining Power

Rate of Kinetic Energy

The Mass Flow Rate

Mass in terms of Density

Volume of the Air Cylinder

Calculating $\frac{d m}{d t}$

Velocity and Mass Flow

The Final Power Equation

Conclusion

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Setup

The Flow of Mass

The Power Equation

The Final Verdict

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