DC Motor

Discussion in 'Homework Help' started by keviiii, Nov 14, 2008.

  1. keviiii

    Thread Starter New Member

    Oct 17, 2008
    8
    0
    hi to all,

    im doing a question on DC motors and i'm having some conceptual doubts.

    this is a seperately excited DC motor that operates at 1070rpm. a load will be attached to it such that the the new speed of the motor is 950rpm. i have found the new current that the motor will draw, which is 3.274A and also the new back emf that the motor will generate, which is 10.88V. now, i am going to find the rotational loss by the motor, i have used proportion method the find, using (950/1070)*(6.175W) = 5.482W. 6.175W is the rotational loss when the motor is operating without any load. this value of 5.482W is correct.

    however, i wondering why is it that this value of rotational loss is different from when i use (3.274A)*(10.88V) to find the rotational loss. this value would be 35.62W. notice the huge diff from the proportional method. why is it so? ):

    thanks.
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    (3.274A)*(10.88V) is the electrical power converted to mechanical power. In a frictionless motor (theoretically) its the output mechanical power. However, a in a real motor the mechanical power is a bit less than this electrical power due to friction losses.
     
  3. keviiii

    Thread Starter New Member

    Oct 17, 2008
    8
    0
    oh i see. can i also say that the difference is due the frictional losses?

    so, theoretically, 35.62W IS the rotational loss, just that friction accounts for the difference?

    i can also say that using the proportional method to find the rotational loss will take the frictional losses out of consideration?

    thank you so much.
     
  4. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Yes, the real mechanical is the the V*I of the armature winding minus the rotational losses.
     
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