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Separately excited DC motor

  1. Dec 8, 2016 #1
    1. The problem statement, all variables and given/known data
    A 180V separately excited DC motor has an armature resistance of 1.2ohm. When the note runs at 1200rpm,t he armature current is 10A.
    1) Assuming that motor torque is constant at all speeds,d determine the mechanical torque and power developed in the motor.
    2)Find the speed of motor when the armature voltage is increased to 200V and magnetic flux is kept unchanged.

    2. Relevant equations
    P=VI
    P=I²R
    Mechanical power=torque•angular speed
    ω=V/KΦ



    3. The attempt at a solution
    I had tried the question few times,and not sure if I was doing it correctly.Here's my solution:

    1) power loss=I²R=120W
    Power developed in the motor=VI-I²R=1800-120=1680W
    Mechanical power=ω•torque
    ω=(2π•1200rpm)/60=125.7rad/S
    Since electrical power=mechanical power, 1680/125.7=13.7Nm

    2)ω=V/KΦ
    Since magnetic flux is kept unchanged, therefore KΦ is constant, KΦ=V/ω

    V1/ω1=V2/ω2
    180/125.7=200/X
    X=838rad/s=8002rpm

     
    Last edited: Dec 8, 2016
  2. jcsd
  3. Dec 8, 2016 #2

    cnh1995

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    Homework Helper

    That's more than 7 times the original speed. Check this calculation.
     
  4. Dec 8, 2016 #3
    I recalculated the whole thing,and found out that I did some calculating error. The speed I found was 1334rpm. Does it make sense?
     
  5. Dec 8, 2016 #4

    cnh1995

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    Homework Helper

    Yes.
     
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