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Using the Locked Rotor Test to find parameters of DC motor?

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A locked rotor test is performed on a permanent magnet DC machine with the following results:
    Terminal Voltage 5 V DC
    Input Current 10 A DC
    Locked rotor torque. 20 Nm
    a) Making suitable approximations deduce the effective rotor resistance in ohms?
    b) Making suitable approximations deduce the torque constant kφ in Nm/A?
    c) If a light running test was now performed with a terminal voltage of 50V DC what approximate shaft speed in rpm would you expect?
    I did some calculations, however I am not sure if what I am doing is correct.
    2. Relevant equations

    Short circuit current at normal voltage
    c7c71b1051eb6ba3e37499a84c69aeee.png is the short circuit current at voltage 205bab72f2efb131a5cec9117644b8ce.png
    8820d417d95175d75158b62f70402674.png is the short circuit current at normal voltage 5206560a306a2e085a437fd258eb57ce.png
    ef32b1d342278d7c54baedadcf1fc569.png

    Short circuit power factor
    7c53fabc7f5af3cdc4c53c3706b23b81.png is the total input power on short circuit
    b35b840654fba65b0b276f3108246d91.png is the line voltage on short circuit
    395526544fd3848a2c6d1ed85310c540.png is the line current on short circuit
    69b72934530949d8c5ec16ba95a94a09.png is the short circuit power factor
    [PLAIN]http://upload.wikimedia.org/math/8/4/3/843762fe06455b51d90ae61f2ee1c042.png[URL='http://en.wikipedia.org/wiki/Blocked_rotor_test#cite_note-12'][12][/URL] [Broken]
    Leakage reactance
    a6258b227b3059f9f4cabce4e53644a6.png is the short circuit impedance as referred to stator
    3775fc72c558a374b0dd81d437ffa16c.png is the leakage reactance per phase as referred to stator
    a89b5229f9bce23c87ccc5760b8afec5.png

    8630bdd537dcfae88502258873f537ff.png is the total copper loss
    de699019f02ab5912838f36ea086b153.png is the core loss

    827363811517f78e90b3ed7ed53bd0dd.png
    20d81dcd679c4f24dd9be640ba6cd0c6.png

    3ae31053f7dfaa8b00d3d54cccefbee6.png

    c2b34b95f006ab9ec85f56aa77a67a9f.png

    3. The attempt at a solution

    a) V/I= R => 0.5 ohms
    b)Torque = k*Ia => Torque/Ia = k = 20 N*m/10 = k = 2
    c) w= k*v => 50v*2= 100 rpm

    I am not too sure if these are correct, it would be great if someone can help! Thank you :)
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 8, 2015 #2
    It looks like you've copied in some equations from a Wikipedia page, but note that it deals with an induction machine, not a PM DC machine.

    Your answers look correct to me. You could perhaps qualify them a bit by using your motor model, which I guess, in steady state, is given by:
    $$
    V = RI_a + k\omega
    $$
    What is it you're not sure about?

    Edit:
    I missed the (c) part, which you have the wrong answer for. You should consider what the back-EMF should be for it to balance the terminal voltage of 50 V.
     
    Last edited: Feb 8, 2015
  4. Feb 8, 2015 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    Equations that involve power factor and leakage reactance per phase are for AC motors.

    Your answer to (b) wil be correct when you give it units.
    I think you have insufficient information to be able to answer part (c).
     
  5. Feb 8, 2015 #4
    I assume 'light running test' is equivalent to 'no-load test', so it'll accelerate until the terminal voltage is balanced by its back-EMF, but otherwise, I'd agree.
     
  6. Feb 8, 2015 #5
    Thanks guys! I was confused because I searched through my textbook and it said that the no-load test was only for the AC machines and didn't mention anything about the DC. Hence, I used those equations I found on Wikipedia. I assumed that for a locked rotor test, the w would be 0. so V=RIa and that would make a=0.5 ohms and then for C I have no idea how to approach it. Even with a back-emf (how would I even calculate that? haha) Thank you!
     
  7. Feb 8, 2015 #6
    The motor will accelerate until its terminal voltage equals the back-EMF it's generating, because at that point there isn't any voltage to drive current into the armature, i.e. there's no torque to continue acceleration.

    If the back-EMF is equal to ##k\omega##, what does that tell you the angular velocity ##\omega## is in steady state with a terminal voltage of 50 V?
     
  8. Feb 8, 2015 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    Your approach would work, but it needs a different k. Had you preserved the units in your calculations, you would see that your calculation cannot possibly produce the answer, e.g.,
    V/k ➜ volts per newton metre, and this looks nothing like radians/sec or RPM. You'd need a different k, one having units of volts per radian/sec.
     
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