March 2008

## Blowin' in the wind: solution

In last issue's Outer space we developed a formula for the power output per unit time of a windmill:$$P \approx 0.38 \times (U/10ms^{-1})^3 \;\; kilowatts, $$ where $U$ is the speed of the approaching wind. We deduced that an average wind speed of $10ms^{-1}$ gives a power output approximately equal to 0.38. The question was why this calculation significantly underestimates the true output. \\ The answer lies in understanding that the average speed is not a good guide. If, for example, the wind speed was $20ms^{-1}$ for half of the time unit and 0 for the other half, then we would get an output approximately equal to $$\frac{1}{2}(0.38 \times 2^3) = 4 \times 0.38.$$ This is four times more than what we got in the original calculation.Read more about...

## Comments

## rotation speed of windmill blade

If forward wind speeds are 4, 6, 8, 10 m/sec What is the tip speed of a 40m blade at each input rate? (formula?)