Dc motor question

Discussion in 'Homework Help' started by skiddman, Oct 25, 2013.

  1. skiddman

    Thread Starter New Member

    Nov 7, 2012
    26
    0
    A 250 volt dc motor taken an armature current of 28A when running at 1000rpm the armature resistance is 0.25Ω. Determine the no load speed if the armature current is 4A at no load

    I need some help on this question When the armature current is at 0A I can solve the problem im not quite sure what to do when the no load speed of the armature current is 4A

    So far I have
    EC=250V-28Ax.25Ω=243V
    Kg=243V/1000rpm=.243

    not sure what to do next
    help on the next step would be much appreciated
     
    Last edited: Oct 26, 2013
  2. donpetru

    Active Member

    Nov 14, 2008
    186
    25
    You should tell us what kind of DC motor is about: with series-wound or shunt-wound ?
    You will need to use the following equations:
    n1 = (U - R*I0) / k*Fi;
    n2 = (U - R*I1) / k*Fi;
    where: n1 = 1000rpm; n2 is unknown; R = 0.25 Ohm, I0 = 28A, I1 = 4A and U = 250Vdc.

    If you take the ratio of the two equations above gives:
    n1 / n2 = (U - R*I0) / (U - R*I1)
    n2 = n1 * ((U - R*I1)/(U - R*I0)).
     
  3. skiddman

    Thread Starter New Member

    Nov 7, 2012
    26
    0
    it is series wound and what is k and Fi?
     
  4. donpetru

    Active Member

    Nov 14, 2008
    186
    25
    Product k * Fi is constant DC motor. Is a constant that can be found in the technical specifications of the engine, where Fi is the flux in the motor winding excitation.
    DC motor speed can be changed by varying the excitation winding flux. That would be one method.
     
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