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DC motor - power calculations

  1. Dec 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A permanent magnet DC motor has an armature resistance of 0.4 Ohms. When a voltage of 110 V is applied to the motor it reaches a steady-state speed of rotation of 126 rad/s and draws 39 A.
    (a) Find the power loss in the armature
    (b) the power input to the motor

    I really don't know what i'm doing. I just used the equations I already know. What is the significance of steady state speed being reached? What do i do with the 126 rad/s?

    2. Relevant equations
    P=IV (power = current x voltage)
    P=(I^2)(R) (power = current squared x resistance)

    3. The attempt at a solution
    (a) P = (39^2) x (0.4) = 608.4 W
    (b) P = 39 x 110 = 4290 W
     
  2. jcsd
  3. Dec 21, 2015 #2

    Hesch

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    Gold Member

    The armature current will be steady, thus the self induction of the armature can be ignored.
    The motor constant, Km, can be calculated.
    Back EMF = 110V - 39A * 0.4Ω = 94.4V →
    Km = 94.4V / 126 rad/s = 0.749 Vs.
    You are just not asked about that, but maybe the the missing questions could be:

    c) Find the steady state torque at 100 rad/s
    d) Find the steady state current at 100 rad/s
     
  4. Dec 22, 2015 #3
    Hi, thanks for your help!
    But why would I find Torque and current at 100 rad/s and not 126 rad/s?
    Continuing from your calculations: I find Torque = (0.749 Vs)x(39A) = 29.2 Nm. So using P=Tw; Power = 29.2 x 126 rad/s = 3681 W.
    Power input to motor = (39 x 110) = 4290 W. So power loss is (4290 - 3681) = 609 W

    Is this correct?
     
  5. Dec 23, 2015 #4

    Hesch

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    Gold Member

    That's just an example, how the exercise could be continued.
    Yes, that's correct.
     
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