DC shunt motor

Discussion in 'Homework Help' started by gsan, Apr 2, 2008.

  1. gsan

    Thread Starter Member

    Mar 4, 2008
    12
    0
    A dc shunt motor (50hp, 250V) is connected to a 230V supply and delivers power to a load drawing an armature current of 200A and running at a speed of 1200rpm, given Ra=0.2ohm and the rotational losses = 500W.

    Find the value of load torque.

    I got 2 different answer with two method. But actually which one is correct?

    Method 1
    Total mechanical power developed
    = Output power + Rotational losses Power
    = (50x746) + 500
    = 37.8kW

    Torque = 9.55x38.8k/ 1200 = 300.83Nm

    Method 2
    Eb = Vt - IRa = 230 - 200(0.2) = 190V

    Power developed by motor = Eb x Ia = 190x200=38kW
    Torque = 9.55x38k / 1200 = 302.42Nm
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    The half percent difference in the two estimates is less than real-world variation.
     
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