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DC shunt motor flux control

  1. Sep 8, 2015 #1

    cnh1995

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    Suppose a motor is run at 1500 rpm and field rheostat is at minimum position (say 10Ω). Then the field rheostat resistance is increased to 100Ω and the speed obtained is 1700 rpm. Now the motor is shut down and started again. Will it now run at 1700 rpm?
    My logic: No it won't, because:
    the field current is less⇒starting torque is less⇒load torque is same⇒torque difference is less⇒acceleration is less⇒steady state speed is less.

    Please correct me if I'm wrong.(I recently had an argument with my friend with whom I'm going to give a presentation on this method.He disagrees and my professor backed him.)
    I don't understand how it will run at higher speed when started at lower field current. By that logic, at no field current, the motor should run at infinite speed. Am I missing something?
     
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  3. Sep 8, 2015 #2

    jim hardy

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    What is nature of the source for armature current?
    The motor will simply draw more armature current to make whatever torque is necessary.

    Long thread on that here:
    https://www.physicsforums.com/threa...c-motor-increase-when-flux-is-reduced.804006/


    It'll try. Remember this picture?
    pot.com%2F-_pGliNArb3Q%2FUPwvsdT_RHI%2FAAAAAAAAO1s%2FXhTXTlDj64U%2Fs640%2Fimage_1358694250175816.jpg
     
  4. Sep 8, 2015 #3

    cnh1995

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    So it won't run at 1700 rpm and will draw more current to produce same torque right?
     
  5. Sep 8, 2015 #4

    jim hardy

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    Or will it just accelerate more gently? Looks to me like you've already established (by measurement) the rpm for that particular applied voltage, field current and torque.


    does your observation agree with the math ?

    Go back to your basic motor equations

    Counter EMF = KΦ X RPM
    so RPM = Counter EMF / KΦ
    RPM = (Vapplied - Iarmature X Rarmature ) / KΦ

    Torque = 7.04 K X Φ X Iarmature (Torque in foot pounds)
    Iarmature = Torque / 7.04KΦ

    Substitute that Iarmature into RPM formula

    RPM = (Vapplied - (Torque/7.04KΦ) X Rarmature) / KΦ

    RPM = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2


    first term is unloaded speed
    second term is how much slower it runs because of load torque.

    field rheostat sets KΦ,
    load's speed-torque curve sets torque...

    Play with the algebra ? What's RPM/Torque ?
     
  6. Sep 8, 2015 #5

    cnh1995

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    Whatever values I mentioned are not experimentally obtained. I just took some random values, just to reduce wording.

    Will this train of thought work? For both the cases below, starting armature current will be same.
    For case 1(rheostat at minimum):

    Starting field current is higher ⇒Starting torque is higher⇒acceleration will be higher⇒steady state speed is higher⇒Steady state armature current is smaller.
    For case 2(rheostat value increased):
    Starting field current is smaller⇒Starting torque is smaller⇒acceleration will be smaller⇒steady state speed is smaller⇒steady state armature current is higher.
    To produce the same load torque in both the cases,field flux*armature current product,i.e. Φ*Ia product should be same.
    In case 1, Φ is more, Ia is less and in case 2, Ia is more,Φ is less, keeping the product constant.
    So no way the motor should run at higher speed when started at smaller field excitation.

    Is this logic okay? This can be proven by algebra.
     
  7. Sep 8, 2015 #6

    jim hardy

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    go back to those two equations, they won't let you down.
    CounterEmf = KΦRPM
    Torque = 7.04 KΦIarmature

    Acceleration has nothing to do with final speed, only how quickly you reach it.

    show that algebra ?

    my algebra says
    RPM = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2

    Halving KΦ doubles the first term and quadruples the second.
    RPMcase1 = Vapplied / KΦ - (Torque X Rarmature) / 7.04K2Φ2
    RPMcase2 = 2Vapplied / KΦ - 4(Torque X Rarmature) / 7.04K2Φ2

    RPMcase2 - RPMcase1 = Vapplied / KΦ - 3(Torque X Rarmature) / 7.04K2Φ2

    That's the speed increase from halving excitation.
    So long as Torque X Rarmature/7.04KΦ is less than 1/3 Vapplied, it'll be positive.
    Speed increase will be zero when Vapplied / KΦ - 3(Torque X Rarmature) / 7.04K2Φ2 = 0
    Vapplied / KΦ = 3(Torque X Rarmature) / 7.04K2Φ2
    Rarmature = Vapplied X 7.04KΦ/(3 X Torque)
    .
    Assume some per-unit values
    Torque = 1
    Vapplied = 1
    KΦ= 1

    Rarmature = 1 X 7.04 X 1 / (3 X 1) = 234 % ?
    Real motors have small Rarmature on purpose, a few percent


    better check my algebra, i'm challenged.
    Show me yours?
     
  8. Sep 9, 2015 #7

    jim hardy

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    Actually i should have treated KΦ as a single term
    and written

    [Q RPM = Vapplied / KΦ - (Torque X Rarmature) / 7.04(KΦ)2
     
  9. Sep 9, 2015 #8

    cnh1995

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    Well you are right as always! I made three major mistakes while thinking, which now make me want to kill myself...
    1. I assumed back emf to be constant in both the cases, which actually isn't since the field is reduced. This was the reason I was saying speed won't increase..
    2. Wrong understanding of starting torque.
    3. I totally overlooked
    P=2*pi*NT
    If the field is reduced and load torque is same, that must draw more power from the source. So with T constant, N must increase with reduction in field.

    Thats how my "train of thoughts" got derailed..

    Thanks a lot for putting up with my nonsense..My apologies..!
     
  10. Sep 9, 2015 #9

    jim hardy

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    Don't apologize at all - you made me straighten out my thinking !

    Thanks for working through my ineloquent presentation.

    It is heartwarming to see "The Light Come On"

    I distinctly remember the morning Professor Grimm derived those two formulas for us boys , my immediate thought was "How Delightfully Intuitive ! " . Both fall out naturally from QVcrossB and right hand rule.
    Complications from armature reaction you add in later.

    Thanks for the feedback ! helps an old guy feel useful.

    old jim
     
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