DC Series Motor

Discussion in 'Homework Help' started by TrevorP, Dec 5, 2009.

  1. TrevorP

    Thread Starter Active Member

    Dec 8, 2006
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    I'm doing practice questions for an exam on electric machines and I'm doing a question on DC series motors. Instead of just putting the question here to get solved I'd really just like to know the correct way about using the equivalent circuit model.

    In DC shunt motors we could assume that field flux remained constant as long as the field coil current remained constant (this makes sense). Then we could use E = kphiω and T = kphi * armature current to evaluate the circuit.

    However, when the field coil is in series the flux is going to change at different loads/speeds as the induced motor voltage increases, thus changing the field current.

    What kind of information am I allowed to assume and when is basically my question?

    I apologize if this makes no sense...because it definitely doesn't to me right now.
     
  2. TrevorP

    Thread Starter Active Member

    Dec 8, 2006
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    Actually you know what here's the question:

    A 120 V, ¼ hp, dc series-connected motor similar to the ones used for lab experiments in this course has a full load speed of 1500 rpm. It has armature and field resistances of 9ohms and 7ohms respectively. If the motor efficiency is 51.8% determine the following

    a) What is the full-load torque?
    b) What is the full load line current?
    c) Assume friction losses are proportional to speed and show that theory predicts a no-load speed around 5000 rpm.

    I can get a) and b) no problem. The problem occurs when trying to figure out the new speed in c). I am able to say Power lost an no load = some constant A * speed, and then find the speed. But when I go to no load I feel as though I'm mising something...I somehow need to find the new lost power...but can't figure it out.
     
  3. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    Good choice.

    What did you find for FLA?

    Because your speed and Eg are propotional, you should be able to cross multiple your full load speed at it's Eg, to that of terminal.

    Your mechanical losses will mount, but I2R losses fall.

    I would simply take your calculated speed above and apply the overall efficiency, and you should get very close.
     
  4. TrevorP

    Thread Starter Active Member

    Dec 8, 2006
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    a) FL Torque = 1.187Nm
    b) FLA = 3A

    I think I may have figured out a way to do it...here's my method:

    We know: P_loss_noload / W_no_load = Td_noload = Ploss_full/W_full. So we can solve for developed torque...which is to say the frictional torque remains constant. Then I'm able to rearrange to get an equation for current at no load = sqrt(Td_noload * I_fullload / KPhi_fullload).

    Then I can use P_loss_noload = (120)(I_noload) - (16ohms)(I_noload)^2. Then use W_noload = P_noload_loss * W_fullload / P_fullload_loss. I got an answer of 5670...which seems a little high.

    I had assumed that KPhi is a linear function of current. I had mounting mechanical losses and reduced I^2R losses...so hmm,
     
  5. TrevorP

    Thread Starter Active Member

    Dec 8, 2006
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    0
    OK I just did your suggested method and got 4826RPM...which is a damn good number. Could you possibly just elaborate on the method? I think my problem was that I have a changing efficiency...where I should be keeping it constant.
     
  6. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    I have applied a 'view point' to the basic calculations.

    Eg = K(flux)(speed), indicating a fixed constant, while flux is current squared. However, this is largely summed in your given overall efficiencies.

    At your given efficiency, you found your Eg at speed. To move to another operating point, the Eg/speed formula is performed at the efficiency of the machine. You could probably tweak that up or down with the difference in I2R losses as a ratio to mechanical losses, but If I'm not mistaken the indication was they offset, contributing to an overall fixed effeciency.
     
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