A Simple Way to Derive the Betz Limit

After I came up with the idea for my invention I wondered at what speed the paddles should be moving with respect to the outside wind speed to extract the maximum amount of work. It turns out, I was looking for one of the results of Betz Law.

According to Betz Law, no wind turbine of any design can capture more than 16/27 of the kinetic energy in the wind. And when the maximum power is extracted from the wind the speed of the outgoing wind is 1/3 of the speed of the incoming wind. In the case of my invention, that would be the speed of the paddle.

There are an enormous number of explanations and derivations of the Betz limit that seem to me more complicated than they need to be. This is partly because Betz limit is usually meant to determine how much energy a wind turbine can extract from the wind. Wind turbines are complicated devices with a variable number of blades of various shapes and orientation. Because the Betz Law doesn’t depend on the mechanism that extracts the energy, it can be helpful to use a simplified wind engine to derive it. This derivation uses concepts that are normally introduced in high school. The concepts include the quadratic equation, simple integrals and the definition of work as force times distance. In addition, because the parameters such as the air density and temperature that affect the wind’s kinetic energy factor out of the equations that are used to calculate the Betz Limit.

Ideal Wind Engine

Imagine a tube oriented parallel with the wind that has a thin disk inside it. As the wind enters the tube it hits the disk which causes a pressure on the disk that moves the disk down the tube doing work. One definition of work is force times displacement in the direction of the force. In our ideal wind engine, assuming the force is constant, the amount of work done by the disk is the force on the disk times the distance it moves. One might ask what the speed of the disk should be to extract the most work out of the wind. Obviously, if the disk is not moving the pressure from the wind would be the maximum but since the disk is not moving it doesn’t do any work. And if the disk is moving as fast as the wind there is no force on the disk and there is no work being done. Clearly, between those two extremes, there is an optimum speed at which the disk should move down the tube to maximize the amount of work done.

Figure 1

Most Efficient Disk Speed

Here are the calculations to determine the disk speed that will extract the maximum work out of the wind entering our imaginary tube.

Figure 1 illustrates our ideal wind engine system. The pressure of the wind on the surface of the disk is dependent on many factors such as the temperature of the air, the density of the air and other variables. For our purposes we will combine these variables into one and call it C. Let the wind speed be represented by the variable Ws and the disk speed by the variable Ds. We know the effective wind speed hitting the disk is Ws – Ds and that the force from the wind is proportional to the effective wind speed squared. So, the force on the disk is proportional to equation 1.

C (Ws - Ds)^{2}

Equation 1

Figure 2 is a plot of disk speed versus pressure on the disk. The units don't really matter but it is important to note that it is an exponential curve with a static pressure on the disk when its speed is 0 and a pressure of 0 when the speed of the disk is the same speed as the wind speed.

Figure 2

The work done is the force on an object, times the distance it moves. The distance the disk moves in time T is the speed of the disk Ds times the time it is moving T or DsT. Figure 3 is the plot of distance versus disk speed. We can choose an arbitrary time period such as one second. In electrical terms the work done by a force is expressed as Newton Meters per second, or Joules per second which is equal to one watt.

Figure 3

Keeping the horizontal axis the same, we can multiply the y axis values in figure 2 (force) by the y axis values in figure 3 (distance) to get the work done versus disk speed as shown in figure 4.

Figure 4

To determine the most efficient disk speed we need to find where the derivative of the disk speed with respect to the work done is zero.

Ws = Wind speed

Ds = Disk speed

T = time interval

W = Work

C = Constant

Work = C
(Ws - Ds)^{2} Ds T

Work = C T
(Ws^{2} 2 * WsDs + Ds^{2 }) Ds rearranging

Work = (Ds^{3}
2 WsDs^{2} + Ws^{2}) C T

The value of
the disk speed where the derivative of (Ds^{3} 2 WsDs^{2} +
Ws^{2}) over disk speed is zero is the optimum speed for getting the
maximum work out of this system.

Taking the
derivative, we have 3Ds^{2}-4WsDs+Ws^{2}

To solve the
general equation ax^{2}+bx+c = 0 we use the quadratic formula.

Plugging in our values into the quadratic equation we have:

So, when the disk is moving down the tube at 1/3 the wind speed outside the tube Ws, the maximum work is being done. The wind exiting the tube is also moving at 1/3 the speed of the incoming wind so this is in alignment with the results others have calculated in a slightly different way. [1]

Using the Most Efficient Disk Speed to Derive the Betz Ratio

The Betz ratio is the maximum possible work that can be extracted out of the wind entering an area divided by the total amount of work that can be extracted from the kinetic energy of the wind entering that area. To calculate the Betz ratio, we again only need two variables – the wind speed Ws and the disk speed Ds.

T = time interval

Ws = Wind Speed

Ds = Disk Speed

C times 1/2 Ws^{3 }= total kinetic energy in the wind. The constant C includes factors such as air density,
air temperature and air pressure but since both the top of the ratio and the bottom of the ratio
will be multiplied by that same value, we will leave it out.

As we have seen, the work done by the disk moving through the frictionless imaginary tube is DsT(Ws-Ds)^{2}.
We put this value on the top of the ratio. At the bottom of the ratio, we need the total amount of energy entering
the system that could do work. The total kinetic energy of the air entering the tube is proportional to the wind
speed cubed and is equal to 1/2Ws^{3}.

Equation 2

Going back to our ideal imaginary tube with a disk in it, we can think of the wind imparting its kinetic energy to the disk and then allowing it to do work until the disk’s kinetic energy is zero at the end of time T. This view only considers kinetic energy and nothing about wind as a fluid.

The kinetic energy imparted to the disk at the start is 1/2Ws^{3}.
To find the total work done to reduce the kinetic energy from 1/2Ws^{3} to zero
over time T we integrate the change in work from the start to end.
We know that the integral is the area under the function curve which is the line in Figure 5.
The area is 1/2 the total kinetic energy times the time interval T. So, we can put 1/2Ws^{3}1/2T at the
bottom of the ratio.

Figure 5

Work done by the disk divided by the total work possible.

Plugging the value we got for the most efficient disk speed, 1/3 Ws, into this ratio should give us the Betz ratio.

16/27 is the Betz ratio.