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DC Motor

  1. Sep 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello I have a bit of a problem solving this
    A 20 kW, 500 V d.c. shunt wound motor draws a current of 45 A when running at full load with a speed of 600 rev min–1. On no load, the current drawn from the supply is 5 A. If the armature resistance is 0.3 Ω and the shunt field resistance is 220 Ω, calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full load value, assuming that the flux per pole does not change.


    2. Relevant equations
    Armature current: Ia=20000/500=40A
    Field current= 500/220=2.3A

    3. The attempt at a solution
    Now first i dont understand where is the missing 2.7A or is the field current=5A?
    If the torque falls half of its value the armature current falls to half as well Ia=20A
    Speed:
    On full load E=V-Ia x Ra=500- 40 x 0.3=488V
    When Ia falls to half= E=500-20 x 0.3=494V
    if the back e.m.f. increases the speed increases: 494/488=1.012 new speed= 600 x 1.012=607 rev/min
    Efficiency:
    V x Ia=Ia^2 x Ra + E x Ia
    500 x 20=20^2 x 0.3 + 494 x 20=10000W

    Can anyone tell me if I am right so far
    Thank you
     
  2. jcsd
  3. Sep 12, 2016 #2

    cnh1995

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    Homework Helper

    That's not correct.
    5kW is the mechanical power output of the motor and not electrical power input.
    That's 2.2728 A(≈2.3 A).
    So the armarure current would be simply supply current minus field current.
     
  4. Sep 12, 2016 #3
    Thank you for your answer!

    So the armature current Ia=45-2.3=42.7
    Then the power=V x Ia=500 x 42.7=21350W
    And that 1350W is a fixed lost?
    Ia on half torque = 42.7/2=21.35A
    Speed:
    On full load
    E=V-ia x Ra=500-42.7 x 0.3=487V
    Half torque
    E=500-21.35 x 0.3=494V
    The speed increases if the back e.m.f. grows (Ia decreases) 494/487=1.014
    n2=n1 x 1.014=600 x 1.014=608 rev/min
     
  5. Sep 13, 2016 #4

    cnh1995

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    Looks good!
     
  6. Sep 13, 2016 #5
    Thank you!
    And the efficiency :
    Total power used: V x Ia = 500 x 21.35= 10675W
    Losses:
    Armature loss: Ia^2 x Ra =21.35^2 x 0.3=136.7W
    Field loss: If^2 x Rf= 2.3^2 x 220=1163.8W
    Brush losses: 2 x Ia= 2 x 21.35=42.7W
    Fixed loss: 1350W
    All losses:136.7+1163.8+42.7+1350=2693.2W
    Is that looks right?
     
  7. Sep 13, 2016 #6

    cnh1995

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    I didn't qoute this part in my previous post. You should use the no load data to find the fixed loss.
    You need not take brush drop into account if it is not mentioned in the problem.
     
  8. Sep 13, 2016 #7
    Ok
    So the current drawn on no load is 5A the power consumed is 500 x 5=2500W
    Field loss is 2.3^2 x 220=1163.8~ 1164W
    Then the fixed loss: 2500-1164=1336W
    Efficiency at half torque :
    10675/10675+1336+1164+137=0.8
    Efficiency =80%
     
  9. Sep 13, 2016 #8

    cnh1995

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    To be accurate, you should also subtract armature copper loss from this. Usually, because of low armature resistance, armature copper loss is neglected on no load but here, since nothing is mentioned as such, you should calculate it.
     
  10. Sep 13, 2016 #9
    Ok thank you very much for the help!!!!
     
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