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Motor Efficiency

  1. Oct 28, 2012 #1
    Is this problem as simple as I think it is?

    1. The problem statement, all variables and given/known data
    A 532 Watt motor is 85% efficient. How much power cannot be converted to energy?


    2. Relevant equations



    3. The attempt at a solution
    [itex]85\%[/itex] of [itex]532W\,=\,452.2W[/itex]
    [itex]532\,-\,452.2\,=\,80W[/itex]
     
  2. jcsd
  3. Oct 28, 2012 #2

    rcgldr

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    Seems you have to assume that the 532 watts is the input power, and that the problem is asking for how much input power can't be converted into energy (assuming energy dissapated as heat isn't included as "energy"). If this is true, then the problem is that simple.
     
  4. Oct 28, 2012 #3
    Thanks, mate.
     
  5. Oct 28, 2012 #4

    rcgldr

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    The problem should have asked how much power can't be converted into mechanical energy.
     
  6. Oct 28, 2012 #5

    CWatters

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    I would have assumed the motor was intended to output 532W because that's what normally matters when you are specifying a motor.

    Efficiency = output power/Input power = 80%

    Therefore

    Input power = 532 / 0.8 = 665W

    Power that can't be converted into mechanical energy = 665-532 = 133W.
     
  7. Oct 28, 2012 #6
    So then the input power minus the output power is power that can't be converted to energy?
     
  8. Oct 28, 2012 #7
    Just had a moment of understanding. If you had to put 665W of into it, but it only returned 532W, then those Watts can't be turned into energy! Also, just because something is 80% effective does not mean you can just take 80% of the energy and subtract it from the total. Totally get it!
     
  9. Oct 28, 2012 #8

    CWatters

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    Yes but that's a slightly bad choice of words. Those 133W still exist as energy because energy is conserved. It's just that they aren't in the form of energy you want from a motor.

    You put in 665W of electrical power. The motor delivers 532W in the form of mechanical power (eg torgue * angular velocity) and 133 W comes out in the form of heat.

    On the types of motor you are probably familiar with most of the 133W lost as heat is due to the resistance of the copper windings.

    I'm not quite sure about that last line of yours but the important thing to remember is that the equation is normally:

    Efficiency in % = 100 * Power out / Power in

    where "Power out" only includes the energy that comes out in the wanted form.
     
  10. Oct 28, 2012 #9

    rcgldr

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    I wasn't sure if the original problem statement or the student was supposed to be aware that the normal practice is to rate a motor based on it's power output, not power input, so I went by the OP's initial answer.

    Also the stated efficiency is 85%, not 80%, so assuming 532 watts is the rated motor output, the input power would be 626 watts.

    The other issue is the awkwardly worded question "how much power cannot be converted into energy", as opposed to how much is the loss in power input versus mechanical power output.
     
    Last edited: Oct 28, 2012
  11. Oct 28, 2012 #10
    85% of the power is being converted to mechanical energy.
    The other 15% is going to become heat.
    I think it's a tricky question; all of the input power will be converted to energy, but 15%
    will be wasted as heat. Definition of efficiency of the motor.
     
  12. Oct 28, 2012 #11

    CWatters

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    Well spotted. My bad.
     
  13. Oct 28, 2012 #12
    Ah. I missed that too. Thanks.
     
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