Calculating Wind Turbine Radius for 571 W of Power

[physicsss]

Wind blows at an average speed of 5m/s 10% of the time. If the efficienty is 45%, what is the radius of the wind turbine that would be needed to generate 571 W of electricity. density of air = 1.25 kg/m^3

The formula is Power = η(1/2*pi*r^2*ρ*v^3), where η is the efficiency, ρ the density, and v the average velocity, r = radius.

Would the average speed be just 0.5 m/s ?

Thanks.

 

[mezarashi]

This is a bit ambiguous from my point of view because... alright, if wind blows at an "average speed" of 5m/s 10% of the time and at an "average speed" of 100m/s 90% of the time, then the answer would be kind of different wouldn't you agree? The term average would usually make me think that using 5m/s is justifiable. I really don't understand why there is another 10% when it says average.

Also to note, the power generated is not linear in v, so taking such an average would be a really rough estimate.

 

[physicsss]

Or better, use Power = 0.10* η(1/2*pi*r^2*ρ*v^3), and keep v at 5 m/s, any ideas?

 

[Pengwuino]

This is just a simple 'plug-in' equation. When it says that the wind blows at an average speed of 5m's 10% of the time... that's just an awefully useless piece of information. Technically... if its average speed is 5m/s, it can be spinning at exactly 5m/s only 0.0000001% of the time or it could be doing it 100% of the time. All you need is the average speed however, you don't even need to know how long the air is actually going at the average speed. So simply plug in everything and you get the answer. Do not multiply the average speed by .1 because that makes no sense.

 

[physicsss]

But I think you do need to account for that..here's the full problem:

Assume that the household consumes 5000 kWh/year of electricity. The house is on the coast where the wind blows at an average speed of 5 m/s 10% of the time (assume the wind speed is too low to produce electricity the rest of the time). If the efficiency of converting wind energy to electricity is 45%, what is the radius of the wind turbine that would be needed to generate the household’s electricity with wind? Assume the density of air is 1.25 kg/m3.

The formula is Power = η(1/2*pi*r^2*ρ*v^3), where η is the efficiency, ρ the density, and v the velocity, r = radius.

 

[Pengwuino]

It's a badly worded problem. Using that logic, you need to start working in divisions. For the lower 45%, there is no power but for the upper 55% (the 5m/s is the 50% mark and since its 10% of the time, we're talken about 5% on both sides), the power produced follows that equation. The problem with that is the following:

What if for that upper 45% above 5m/s, the wind only went at 10m/s?
What if for that upper 45% above 5m/s, the wind only went at 5.5m/s?

Both situations will allow the possibility of a 5m/s average airspeed 10% of the time but both will give you drastically different answers. The only way this question makes any sense is if the wind turbine will only work at the 5m/s mark and not any lower or any higher.

 

[physicsss]

But here's how this website I found does it:

Example calculation:

* Windmill efficiency = 42%
* average wind speed = 10 m/s (20 mph)
* Power = 0.0006 x 0.42 x 1000 = 250 Watts per square meter
* Electricity generated is then .25 KWH per sq. meter
* If wind blows 24 hours per day then annual electricity generated would be about 2200 KWH per sq. meter

* But, on average, the wind velocity is only this high about 10% of the time
* typical annual yield is therefore 200-250 KWH per sq. meter

http://zebu.uoregon.edu/2001/ph162/l11.html

I'm confused.

 

[mezarashi]

* If wind blows 24 hours per day then annual electricity generated would be about 2200 KWH per sq. meter

* But, on average, the wind velocity is only this high about 10% of the time

You would then have to assume that, 90% of the time then, there is NO wind at all. Or atleast "useless" amounts of wind. The question was worded with ambiguity in any case.

 

[Pengwuino]

There is a big difference between...

But, on average, the wind velocity is only this high about 10% of the time

and

The house is on the coast where the wind blows at an average speed of 5 m/s 10% of the time

The first means that at a sample taken over a few months or something, the wind will only reach 10m/s (and it is implied that it means that it is the maximum) 10% of the time. The second has no meaning because it is saying that the average wind speed is 5 m/s and it is 5 m/s 10% of the time. So basically, for the problem... yes you do actually multiply everything by 10% even though the instructions are rather ambiguous.

 

[physicsss]

OK, just to make sure, is this setup correct:

Power = 0.10* 0.45(1/2*pi*r^2*1.25*5^3) ?

And thank you so much for the help.

 

[Pengwuino]

Yes, that is the correct equation