How is wind speed related to electricity generation in a wind turbine?

  • Thread starter Marcin H
  • Start date
In summary, the power output of a wind turbine is proportional to the cube of the wind speed, which means that small changes in wind speed can result in a significant increase in power output. This was not mentioned in the given information, but is an important concept to consider when solving the problem. The correct answer is option d, 162,000 kWh, and the reasoning behind this is that the input energy to the turbine is proportional to the wind speed cubed, while the output energy is more complicated and depends on factors such as turbine design.
  • #1
Marcin H
306
6

Homework Statement



19. A wind turbine generates 6000 kWh of electricity on a day when the wind speed was 3 meters per second for 24 hours. If the wind speed was 9 meters per second for 24 hours, how many kWh of electricity will be generated?a. 12,000 kWh b. 18,000 kWh c. 54,000 kWh d. 162,000 kWh

*This question is way below precalc, but I figured this is the best place to ask... Didn't want to bother the engineering thread... :/

Homework Equations


Basic math...

The Attempt at a Solution


Ok, I feel like a complete idiot for not understanding this, what appears to be, a very simple math problem... Why is this not as simple as doing a ratio...

6000/3 = x/9

==> x=18,000kWh.

But the answer is somehow 162,000 kWh according to the answer key. Why is that the answer and why am I so dumb?
 
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  • #2
The power output of a wind turbine is proportional to the cube of the wind speed. This relationship should have been provided to you in my opinion.
 
  • #3
I know nothing about such things but IF the generated power goes as the cube of the wind speed THEN the answer is 162,000
 
  • #4
phinds said:
I know nothing about such things but IF the generated power goes as the cube of the wind speed THEN the answer is 162,000

EDIT: HA ... according to Mr. Magoo, I nailed it :smile:
 
  • #5
You assume a linear dependence between wind speed and electrical power, but this is not the correct formula. You have to consider the mass of the air, velocity, radius, density and kinetic energy. Of course we can assume that radius and density are constant. But moved mass is not.
 
  • #6
fresh_42 said:
You assume a linear dependence between wind speed and electrical power, but this is not the correct formula. You have to consider the mass of the air, velocity, radius, density and kinetic energy. Of course we can assume that radius and density are constant. But moved mass is not.
So is Mr Magoo right or not?
 
  • #7
I do not want to write down the view steps needed in the homework section, but kinetic energy and volume is turned into power.
 
Last edited:
  • #8
fresh_42 said:
I do not want to write down the view steps needed in the homework section.
fair enough
 
  • #9
magoo said:
The power output of a wind turbine is proportional to the cube of the wind speed. This relationship should have been provided to you in my opinion.
Yeah, that was not given, but I did notice cubic eventually just didn't understand why... Good to know. Thanks!
 
  • #10
phinds said:
I know nothing about such things but IF the generated power goes as the cube of the wind speed THEN the answer is 162,000
Yup. Got it. Thanks!
 
  • #11
fresh_42 said:
You assume a linear dependence between wind speed and electrical power, but this is not the correct formula. You have to consider the mass of the air, velocity, radius, density and kinetic energy. Of course we can assume that radius and density are constant. But moved mass is not.
I will let my professor know.
 
  • #12
Marcin H said:
Yeah, that was not given, but I did notice cubic eventually just didn't understand why... Good to know. Thanks!
Did you understand why the proportion goes with velocity cubed and not linear?
 
  • #13
I think power generated goes as cube of wind speed because: (a) Kinetic energy of given mass of air is proportional to velocity square and (b) Amount of air that hits the turbine in given time is proportional to velocity. So overall the energy transferred to turbine is proportional to cube of velocity.
 
  • #14
shiv222 said:
I think power generated goes as cube of wind speed because: (a) Kinetic energy of given mass of air is proportional to velocity square and (b) Amount of air that hits the turbine in given time is proportional to velocity. So overall the energy transferred to turbine is proportional to cube of velocity.

That is an over-simplification of the real situation, as are most of the responses you have received. Real "power curves" are not really simple cubics; see, eg.,
http://www.wind-power-program.com/turbine_characteristics.htm .
Basically, the cubic formula gives the "input energy" to the turbine, but the output energy of the turbine is more complicated.
 
  • #15
Ray Vickson said:
Basically, the cubic formula gives the "input energy" to the turbine, but the output energy of the turbine is more complicated.
Yes, of course, however in the context, i.e. as multiple choice question and as to why cubic is the correct answer (here), the explanation is correct. I've read of something about roughly 60% efficiency. And in real life, density isn't constant either.
 
  • #16
fresh_42 said:
Yes, of course, however in the context, i.e. as multiple choice question and as to why cubic is the correct answer (here), the explanation is correct. I've read of something about roughly 60% efficiency. And in real life, density isn't constant either.

The link I included gives a typical power vs. wind-speed curve. It starts out at zero---and remains at zero up to a certain positive wind speed---then increases (perhaps cubically) for a while, then starts to level off again. So, it is "S-shaped", but may be cubic over a certain range, but of the form ##(s-a)^3## for some positive ##a##, not simply ##s^3##.

I agree, however, that in a multiple-choice question the OP has no choice but to go with the cubic.
 
  • #17
Ray Vickson said:
That is an over-simplification of the real situation, as are most of the responses you have received. Real "power curves" are not really simple cubics; see, eg.,
http://www.wind-power-program.com/turbine_characteristics.htm .
Basically, the cubic formula gives the "input energy" to the turbine, but the output energy of the turbine is more complicated.
In this type of objective question one looks at most important factors. For further, accurate answer, I think turbine design and lot of other information will be needed in question itself.
 

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4. How do I divide one number by another?

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