# Betz Limit

No wind turbine can capture more than 59.3% of the kinetic energy of incoming wind in an open channel. This is called the Betz limit named after the german scientist Alfred Betz who derived this limit in 1919.

## Betz Limit Explanation

Let’s understand why the wind turbine cant extracts 100% kinetic energy from the wind. If it does so, this means that the kinetic energy of wind after passing through the turbine must be zero and so is the velocity (V_{2} =0). The wind with zero speed would act as a blockage and no further incoming wind (V_{1}) could further pass through the turbine.

The air downwind of the turbine needs to be flowing to get out of the way of air that is entering the turbine.

## Betz law

Betz law only tells us about how much energy can be captured from the wind. This is not telling us how much electricity can be generated as this depends upon many engineering factors.

## Betz limit Derivation

Why there is a limit of only 59.3% (or 16/27), why can’t be any other number? So to understand this let’s derive the Betz limit.

We are considering an air stream that has an area and velocity (A_{1} and V_{1}) before and after passing the turbine (A_{2} and V_{2})

In general, we know that if air density remains constant then

**A _{1}** .

**V**

_{1}_{ }

**=**

**A**also

_{2}. V_{2}**A**and

_{2}>A_{1}**V**

_{2}<V_{1} Please note that if the airstream area before and after the turbine is the same then this would also lead to the same velocities (V_{2}=V_{1}) which means the turbine has not captured any kinetic energy. So, **A _{2}>A_{1}** and

**V**is a valid assumption.

_{2}<V_{1}But how much flow should be spread out or how much **V _{2}** should be slower compared to

**V**for maximum power extraction?

_{1}The power generated by the wind turbine of cross-section **S** can be expressed as

Here rho is the density of air and **V **is the velocity at the cross-section of the turbine itself.

To some reasonable approximation, Velocity **V** is the average of **V _{2} **and

**V**

_{1}To substitute this velocity into the power equation we get the following

With a bunch of algebra, we can express power equation as

let’s call the above expression as eq 1, we will need it later. The above eq looks complex but it is not, only V_{2 }here is a variable all others are constants.

To find the value of V_{2}, that maximizes power P, let’s take the derivative of the power equation and put equal to zero.

you can divide the V_{1 }to get

and with some algebra, we get the following expression

The above equation can be solved quadratically or by factorization and we will get the final expression as

**V _{2} / V_{1 } = 1/3**

So plug this result back into equation 1

So, the maximum power a turbine can capture is