Windmill power generation problem

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Homework Statement


A windmill is designed to operate at 20 rpm in a 15 mph wind and produce 300 kW of power. The blades are 1.75 ft in diameter. A model 1.75 ft in diameter is to be tested at 90 mph wind velocity. What rotor speed should be used, and what power should be expected.


Homework Equations


This question is in a chapter on "dimensional analysis" so I assume it wants me to use dimensional analysis to solve this. The problem is that I don't know how to begin.

Would I say that the rotor speed, w = f(rho, blade diameter, wind velocity)
and
power, P = f(rho, blade diameter, wind velocity, w) ?

I could then find the PI groups but I'm not sure how this helps me answer the question.
Could it be much more simple than that and have nothing to do with dimensional analysis?

Also, I'm aware of the equation: Power = 0.5*rho*swept area*Velocity^3 => for wind turbines



The Attempt at a Solution


Other than what I have above, I have been unable to come up with much.
Any help would be appreciated!
 
Last edited:

Answers and Replies

  • #2
Delphi51
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The power equation you wrote should do to find the power.
Check for typos in the question - the first 1.75 ft can't be right and the 90 mph seems unreasonable.

I don't know anything about the rotor rpm.
 
  • #3
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The power equation you wrote should do to find the power.
Check for typos in the question - the first 1.75 ft can't be right and the 90 mph seems unreasonable.

I don't know anything about the rotor rpm.

I thought 90 mph seemed high too, but I checked the problem again and those were the numbers given, unless the book has a typo.

Also if I use the power equation and solve for the radius of the blades I get r = 22.73 meters = 74.6 ft => does this seem too large?

For example: 300,000Watts = 0.5*PI*r^2 *(1.225kg/m^3)*(6.7056m/s)^3 =>solving for r, r = 22.73m

Would the Rotor Speed be equal to the angular velocity of the blade tip?
 
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  • #4
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Windmills can only be 58% efficient (betz limit). The rotor radius should be at least 29.8m.
I wonder if a diameter of 175 feet was meant, altough it isn't enough (26.57m radius) to
get the power if you include the betz limit to efficiency.

The speed of the blade-tips in m/s is independent of the radius, and proportional to the wind speed. (so you get the same angle of attack) The rpm of the small windmill at 90 mph
should be 600 times higher.

In reality the rotor speed of any practical windmill at 90mph wind speed (a cat 2 hurricane)
should be 0, so it won't fly apart.
 
  • #5
Delphi51
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I should have looked closer! That formula appears to be for the power of the wind going through the windmill. Of course the windmill will not capture all of that. Take a look at
http://www.ecolo.org/documents/documents_in_english/WindmillFormula.htm
where it shows the formula with an efficiency factor constant factor. I'm thinking this is what you would use. With the information given about the 300 kW windmill (which must have a diameter of 175 feet, not 1.75) you could find the constant. Then use the same formula again to find the expected power of the model windmill.

I see some sites are saying the speed of the blade tips is proportional to the wind speed. If you have something like that or can justify it with dimensional analysis, you can use it to answer the RPM question.
 
  • #6
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Ok, how does this look...
using the equation from http://www.ecolo.org/documents/docum...illFormula.htm [Broken]
(P = C*k*D^2 *V^3) where C=constant, k = efficiency = 0.58
and solving for the Constant, C, while assuming the blade diameter = 175ft(53.4m) and the power generated = 300kW, I get C = 0.603

If I use this same Constant for the model windmill at a wind velocity of 90mph = 40.234m/s I get:

Power = 0.603*0.58*(53.34)^2 *40.234^3 = 64.81 MW
=>this is assuming the windmill hasn't blown apart from those 90mph winds

I'm still a bit lost on how to get the rotor speed.

Thanks for your help!
 
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  • #7
Delphi51
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Looks good except I think the model was supposed to have a 1.75 ft diameter - that's great for the model, but the full sized one had to be bigger!

You could use rotor tip speed = k*windspeed to work out the second part.
 
  • #8
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oh ok, how bout this?

with D(model) = 1.75ft = 0.5334m
Power = 0.603*0.58*(.5334m)^2 *(6.7056)^3 = 30 W

rotor tip speed = k*6.7056 = 3.89m/s => if I then divide 3.89m/s by .5334m diameter I get (3.89/.5334) = 7.29s^-1 = 437 rpm

thanks again
 
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  • #9
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oh ok, how bout this?

with D(model) = 1.75ft = 0.5334m
Power = 0.603*0.58*(.5334m)^2 *(6.7056)^3 = 30 W

This is the power at 15 mph=6.7056 and not 90 mph

rotor tip speed = k*6.7056 = 3.89m/s => if I then divide 3.89m/s by .5334m diameter I get (3.89/.5334) = 7.29s^-1 = 437 rpm

thanks again

you should use the data about the large windmill to find the rotor tip speed, wich is
much larger than the wind speed and has nothing to do with k.
When you have it you should divide by the circumference = (pi * diameter) and multiply with 60 to get the rpm.
 
  • #10
Delphi51
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UFeng I am too lazy to work all that out and using my shortcuts I get different answers. For the power, I wrote P = k*A*v^3 for the model and divided by the same for the big one. The k's cancel and so do all the unit changes so I have that the power for the model is
P/300 = (1.75/175)^2*(90/15)^3 and get P = 6.48 kW

The rpm calc is complicated because you must convert RPM to tip velocity using something like v = rpm*2*pi*r/60. Maybe just say v = k*rpm*r and let the k cancel out. I agree with Willem2 that the model rpm is 600 times that of the big one, much higher than your answer.
 
  • #11
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oh ok. I plugged in the wrong wind velocity for the model. Thanks guys! I appreciate it! I think I got it now.
 

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