## Windmills on my Mind

#### Wednesday, March 27, 2019 · 2 min read

Quick back-of-the-envelope analysis of a windmill, just for fun.

According to National Wind Watch a typical windmill is a GE 1.5-megawatt model which has 116-ft blades.

Let’s say the wind speed is ($ v $), which is typically 10-15 m/s on a good
day. Then in time ($ t $) the volume of air that passes across the windmill is
a cylinder of height ($ tv $) and area ($ \pi r^2$). Air’s density, ($ \rho $),
is around 1.225 kilograms per cubic meter. This gives a total mass of ($ M =
\rho tv\pi r^2 $). Taking ($ v $) to be a modest 10 m/s, this works out to 48
metric *tons* of air per *second*.

Now let’s apply conservation of energy on this mass of air. Initial energy is ($ Mv^2 / 2 $) and final energy is ($ Mw^2/2 $), and the difference is on the order of the energy output, which is a function of the power output and ($ t $) assuming reasonable efficiency in the turbine.

So, we have ($ \rho tv\pi r^2 v^2 / 2 - \rho tv\pi r^2 w^2 /2 = Pt $). Time
cancels, which is comforting since this is a continuous process. The equation
that’s left suggests that power output of a windmill is proportional to the
*cube* of the wind speed!

Another thing to think about is how much the wind slows down by. Solving for ($ w $) and taking ($ P $) to be 1.5 MW, we have ($ w = \sqrt{ \frac{2P/\rho \pi r^2 - v^3 }{-v}} $). For 15 m/s winds, this means wind slows down by around 10% because of the windmill.

Actually this wasn’t “just for fun,” it was to show off initial progress on my side-project. :-)

Oh, and an unresolved question: because air is coming into the windmill faster than it’s going out, the momentum flux across the plane of the windmill is net negative. Applying Gauss’ law, we would expect a buildup of air in the windmill, which obviously doesn’t happen. What’s wrong?