DC Motor - Flux Per Pole

Discussion in 'Homework Help' started by Zaraphrax, Nov 1, 2009.

  1. Zaraphrax

    Thread Starter Active Member

    Mar 21, 2009
    47
    3
    Hi there,

    I'm studying for finals and can't quite get this question out. :(


    For a DC machine:

    Average air-gap flux density - B = 389 mT
    Effective length of rotor slots - L=0.2 m
    Rotor diamater D=0.2m
    Number of poles P=4
    number of parallel paths in armature winding - alpha=4
    Required rotational back EMF E = 220V
    Rotational speed N = 1500 r/min

    Calculate the flux per pole and the number of turns on the rotor winding.

    I'd be able to do it if I knew how many slots there were (because one could then find the number of conductors).

    Becuase in that case you could just rearrange

    E = (pz/alpha) * phi * N

    to

    phi = (alpha*E)/(pzN)

    However since I can't find Z (number of conductors), I'm lost.

    Any help is muchly appreciated. Thanks :cool:
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Would this be correct?

    Ec=BLv

    where
    B=0.389,
    L=0.2,
    v=n\piD/60=1500x\pix0.2/60=15.708 m/s

    Ec=0.389x0.2x15.708=1.222 volts

    E=ZxEc/a

    Z=axE/Ec=4x220/1.222=720 turns
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Clearly I forgot the number of poles (4) - so I guess Z=720/4=180 ...????
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Sorry about the confusion between number of conductors Z and turns ...?
     
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