High torque cage induction motor

Discussion in 'Homework Help' started by Jess_88, Sep 26, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    Hey guys :)
    I'm a bit confused with this question.

    Given the high torque cage induction motor name plate data
    [​IMG]

    and the full equivalent circuits parameters
    [​IMG]

    The friction and windage loss is 0.910 kW and may be considered constant
    over speed range 1500 rev/min to 1000 rev/min.

    It is required to control the speed of the motor over a limited range by voltage control.

    I need to determine for a motor speed of 1355 rev/min:
    a) the required stator line voltage;
    b) the motor efficiency at 1355 rev/min;
    c) comment on the suitability of running the motor in this reduced voltage mode.

    I'm finding it difficult to fined much information on this kind of question. For part 'a', isn't the stator line voltage = (√3)(Phase voltage)?
    Is phase voltage = 415V

    I can't see how how the motor speed relates to the voltage:confused: Can anyone show me the formulas required in this?

    I also can't fined anything on determining the efficiency at a given speed.:(

    Thanks a lot guys
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Which equivalent circuit model are you expected to use? What have you been taught as the working schematic form? There are a couple of options. The model is usually derived on a per-phase basis.

    If you know the speed you can find the slip 's'.

    The per phase rotor power is related to the rotor equivalent resistance which is normally given as R2/s.

    With no mechanical load, the rotor power is simply equivalent to windage and friction losses - which must be per phase. So you can find the effective rotor current [per phase]. From the model you can then derive the stator input line-to-neutral voltage. This line to neutral value can then be converted to a line-to-line value.
     
    Jess_88 likes this.
  3. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    I'm think this is the equivalent circuit model
    [​IMG]

    so would the line current be I1?
    Sorry, I'm still a bit lost
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    OK - it's important to know which model you have.

    It's possible you may have been taught that the following relationships hold. For a poly-phase induction machine.

    Total Air Gap Power=p*I2^2*R2/s

    Total Rotor Copper Loss=p*I2^2*R2

    Total Shaft power=Total Air Gap Power- Total Rotor Copper Loss

    Hence Total Shaft Power=p*I2^2*(1-s)/s

    'p' is the number of phases & 's' is the slip.

    In this case p=3.

    You are asked to find the voltage that would run the machine at 1355 rpm - presumably with no mechanical load.

    The only mechanical load then that the motor 'sees' is 910W - due to windage and friction loss.

    This would be the total shaft power

    So

    910=3*I2^2*0.293(1-s)/s

    The slip will be s=1-1355/1500=0.0967

    This will therefore allow you to find the unknown I2.

    Once you have I2 it's relatively simple circuit theory to find I1 - which is indeed the AC line current. Hence you will also be able to find the input phase voltage V_\phi. You will have the voltage and current as complex phasors so you should then be able to deduce the input power factor knowing the phase displacement between the current I1 & voltage V_\phi.

    Efficiency can also be deduced by calculating the input power and doing the efficiency calculation with the shaft power as the result - in this case simply the windage & friction losses as the notional shaft mechanical output power.
     
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  5. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ah ok.
    so I get I2 = 9.08A
    Then I would use I1 = 64A -I2 - IM right?
    I was thinking I would determine IM from the Air gap Power loss. But I see you have
    Total Air Gap Power=p*I2^2*R2/s. I would have thought you would use IM
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Yes - You must include the magnetizing current to find the current I1.

    How did you arrive at 9.08A for I2?

    If the mechanical output is 910W

    then from the aforementioned formula (my post #4)

    I2=√(910/(3*0.293*0.9345))=10.525A
     
  7. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ah, yeah I made a mistake in my math.
    How would I go about determining Im
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    E1=I1*(R2/s+jX2)

    Im=E1/Zm

    Where Zm=Rc||jXc
     
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