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1/3 Wind Speed Difference and the Betz Limit

When determining the efficiency of an engine it is often useful to imagine an ideal situation where friction is disregarded. For example, to generate mechanical work by means of thermal energy conversions the Stirling and Carnot cycles are used. In a similar way, it is helpful to consider an ideal wind engine when deriving optimum variable values. It can be shown in the case of a wind turbine, that the maximum work can be done when the speed of the wind after going through the turbine is 1/3 of the speed of the wind entering the turbine. This fact doesn't help much in designing the blades of a wind turbine. With an ideal wind engine, in this case a disk being blown down a tube, we can show that the 1/3 wind speed drop can derived using two variables - the wind speed and the speed of the disk.

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 pressure on the disk moving the disk down the tube. By definition, 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 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 again, there is no work being done. Clearly, 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.

The pressure of the wind on the surface of the disk is dependent on many variables such as the temperature of the air and the density of the air. For our purposes, we will combine these variables into one variable 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 as the wind speed.

Figure 2

Work done is the force on an object times the distance it moves and the distance our disk moves in time T is 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 expressed as Newton Meters per second, or Joules per second is equal to one watt.

A graph showing a line going up

Figure 3

If we multiply the y axis values in figure 2 (pressure) by the y values in figure 3 (distance) we get the work done. A plot of work versus disk speed is 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 (Ws2 2 * WsDs + Ds2 ) Ds rearranging

Work = (Ds3 2 WsDs2 + Ws2) C T

The value of the disk speed where the derivative of (Ds3 2 WsDs2 + Ws2) over disk speed is zero is the optimum speed for getting the maximum work out of this system.

Taking the derivative, we have 3Ds2-4WsDs+Ws2

To solve the general equation ax2+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 wind so this is in alignment with the results others have calculated in a slightly different way. When considering a wind turbine, the maximum efficiency occurs when the ratio of the downstream velocity of the wind is 1/3 the velocity of the upstream wind. [1] I find it easier to understand the optimum drop in wind speed when the maximum work is calculated using the simple equation of force times distance rather than the more complex calculations that go into how a turbine blade converts the kinetic energy of the wind into work.

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 kinetic energy of the wind entering that area.

T = time interval

Ws = Wind Speed

Ds = Disk Speed

1/2 Ws3 = total kinetic energy in the wind. This is multiplied by 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 value it can be ignored here.

As we have seen, the work done by the disk moving through the frictionless imaginary tube is DsT(Ws-Ds)2. The total kinetic energy of the air entering the tube is equal to 1/2Ws3. You will notice we have a time interval T on the top and bottom that will factor out.

The top of the ratio is the total work done integrated over the time interval T. The bottom of the ratio is the total possible amount of work that can be done which by definition is the same as the drop in kinetic energy. The kinetic energy imparted to the disk at the start is 1/2Ws3. To find the total work done to reduce the kinetic energy from 1/2Ws3 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. That area is 1/4 Ws3 times the time interval T.

Figure 5

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.



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