DC Shunt Motor

Discussion in 'Homework Help' started by Jess_88, Oct 19, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    Hey guys :)

    I keep getting a different answer to a question. Can anyone explain where I'm going wrong?

    220V DC shunt motor has an armature resistance of 0.2Ω and field resistance of 110Ω. At no-load, the motor runs at 1000rmp, and it draws a line current of 7A. At full-load, the input to the motor is 11kW. Consider the air gap flux remains fixed at its value at no load; that is armature reaction

    a) fined Ea, speed and speed regulation at full load

    b) Developed torque at full load

    c) Starting torque if the starting armature current is limited to 150% of full load current.

    Ea = 210.4V, N2 = 960.7, %SR = 41%
    Pd = 100.4N.m, Tst = 150.6N.m

    I can't get the first part...:(. Can anyone help me get started?
    My attempt
    Ra = 0.2, Rf = 110
    V = 220

    If = V/Rf = 220/110 = 2A
    Ia = IL -If = 7A - 2A = 5A
    Ea = V - Ia*Ra = 220 - 5*0.2 = 219
    ...wrong :(

    thanks guys
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    At no load.

    With 7A draw the armature current is 7A minus the field current [2A]

    Ia = 7 - 220/110=7-2=5A

    So the Ra voltage drop is Ra*Ia=0.2*5=1V

    The motor internal emf at 1000rpm is 220-1=219V

    The emf per rpm=219/1000=0.219 V/rpm

    At full load

    The armature current = 11kW/220V-field current=50-2=48A.

    The Ra drop is therefore 48*0.2=9.6V

    With no change in armature reaction the full load motor emf is therefore 220V-9.6V or 210.4V

    At 0.219V per rpm this equates to a full load speed 210.4/0.219=960.7 rpm.

    Hope that helps