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Homework Help: Power proportional to speed

  1. Jul 13, 2016 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=8b0e10b6b264016654fe7227a3da5a31.png

    2. Relevant equations


    3. The attempt at a solution

    Power is given by F.v , but I am not sure how to calculate F in this case .Again 'v' is the speed of the point of application of force F . I also do not understand how force applied on blades is causing work .

    Any help is appreciated .

    Thanks
     

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  2. jcsd
  3. Jul 13, 2016 #2

    billy_joule

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    What's the equation for kinetic energy?
     
  4. Jul 13, 2016 #3
    I think you meant "expression" of KE . It is ½mv2 . Could you respond to my doubts in the OP ?
     
  5. Jul 13, 2016 #4

    billy_joule

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    Nope, I meant equation (ie contains an equality)
    EK = 1/2mv2


    Forces aren't really relevant here. Let's look at the definition of power:
    P = E/t
    We know that the winds energy is EK = 1/2mv2
    And that some fixed fraction of that is converted to power.
    A bit of algebra and we can see what relationship Power has to wind speed..
     
  6. Jul 14, 2016 #5
    @haruspex , @Doc Al Could you please share your ideas about doubts raised in OP . I can do this problem from energy concept . Interestingly the solution provided uses force approach .
     
  7. Jul 14, 2016 #6

    PeroK

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    Power is also Energy/second. Personally, I would think of the air as a series of particles impacting the generator. I can't see any reason to think about forces when you're given the facts about energy.
     
  8. Jul 14, 2016 #7

    haruspex

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    That seems strange. Doesn't that require making assumptions about how the energy is transferred from wind to turbine? Can you describe the solution in more detail?
     
  9. Jul 14, 2016 #8
    Please see the attached image .
     

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  10. Jul 14, 2016 #9

    haruspex

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    Maybe I'm missing something, but I see no compelling logic in the F=vdm/dt step. What does dm/dt mean here? What mass is changing? I could understand F=d(mv)/dt, i.e. the rate at which the air loses momentum, and from that you can certainly get the rest of the steps.
    As others have posted, this is a circuitous route, given the problem statement.
     
  11. Jul 14, 2016 #10
    Ok .

    An alternative solution is also provided . Please see the attached image . Do you think this one makes sense ?
     

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  12. Jul 14, 2016 #11

    haruspex

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    Yes, that's quite simple and persuasive.
     
  13. Jul 14, 2016 #12

    Doc Al

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    I agree.
     
  14. Jul 14, 2016 #13
    You're mixing a physical quantity with a unit of measure.
    You may have energy divided by time, or joules per second.
     
  15. Jul 14, 2016 #14
    I am not sure about the first line .Can we assume particles in wind to be uniformly distributed ?
     
  16. Jul 14, 2016 #15
    @Doc Al ,what do you think about the OP ?
     
  17. Jul 14, 2016 #16

    Nidum

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  18. Jul 14, 2016 #17

    Doc Al

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    I think that's a reasonable assumption.

    I think the problem is fine. The way I'd do it is to track a parcel of air that hits the fan. The volume will be x*A; the rate at which the air hits the fan will be v*A. (A is cross-sectional area.) That should allow you to see that the energy in the parcel is proportional to v2, and the power is proportional to v3.
     
  19. Jul 14, 2016 #18

    Doc Al

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  20. Jul 14, 2016 #19
    I meant what I had written in the "The attempt at a solution" :smile: .What do you have to say about issue raised in OP?
     
  21. Jul 14, 2016 #20

    Doc Al

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    Ah, sorry... o0)

    You can calculate the force by considering the rate of change of momentum of the air as it hits the fan. (That's the approach used in one of the solutions you posted.) The air pushes the blades, making them turn, thus doing work. (They are angled.) That approach should allow you to get the v3 proportionality.
     
  22. Jul 14, 2016 #21
    So , do you agree completely with the official solution posted in #8 ?
     
  23. Jul 14, 2016 #22

    Doc Al

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    It's OK. (Though I prefer the energy approach.)

    My version of that force solution would be along these lines:
    ##F = \frac{d(mv)}{dt} = \frac{d(\rho x A v)}{dt} = \rho A v^2 ##

    Again, I am thinking in terms of a voume of air moving at speed v.
     
  24. Jul 14, 2016 #23
    Thanks a lot Doc Al :smile:
     
  25. Jul 14, 2016 #24

    TSny

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    In the power formula ##P = \vec{F} \cdot \vec{u}##, the velocity ##\vec{u}## is the velocity of the point of application of the force. Wouldn't this be the velocity of a point of the blade? Of course, the speed of a point of the blade depends on the distance from the axis of rotation. So I guess you would need to use some average speed ##\overline{u}## for the points of the blade. Also, the force is probably different for different points of the blade. So, again you would need to use an average ##\overline{F}##. Even if you can argue that ##\overline{F}## is proportional to the square of the wind speed, ##v^2##, you would still need to argue that ##\overline{u}## is proportional to ##v## in order to get the power to be proportional to ##v^3##.

    The energy method avoids all this, but does rely on the assumption that the rate at which energy is transferred to the blades is proportional to the rate of flow of kinetic energy of the wind.
     
  26. Jul 14, 2016 #25
    Thanks :smile:

    Why is this an assumption ? Isn't this what is happening Or is it some simplified model ?
     
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