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Speed regulation of a dc motor

  1. Apr 24, 2015 #1
    I am having confusion about the no load and full load speeds of a dc motor. What exactly does 'loading' stand for? How do we change the loading of a dc motor ?What is the relationship between loading and the speed of the motor?
    Which situations does no load and full load refer to?
     
  2. jcsd
  3. Apr 24, 2015 #2

    cnh1995

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    No load means no mechanical load on the rotor.Just the rotor is rotating alone. Loading means adding something on the rotor (via shaft) like an engine,machine tool etc. DC shunt motor is a speed regulating motor. Its no load and on load speeds are ideally same. Practically, there is a difference but it is negligible. Back emf regulates the speed. Its a beautiful mechanism..
     
  4. Apr 24, 2015 #3

    Merlin3189

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    I was puzzled at first by cnh1995 comments, as, depending on load, DC shunt motor will run at any speed from stall to free run. But now I notice he is talking about the "on load" speed being nearly the free running speed. So I thought this graph might be of interest to Rounak.
    ScottMotorCharacteristic.gif Presumably the "on load" operating point is normally at the peak efficiency (though I'd have thought that in some circumstances one might wish to operate at higher power, at least intermittently.)

    So although the no load and "on load" speeds are very close, this is due to the choice made about the operating point - if you choose to operate (for reasons of efficiency) at small load very close to the free running speed, then obviously there is no difference! But generally the speed is inversely proportional to load.
     
  5. Apr 24, 2015 #4

    cnh1995

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    I haven't worked much with dc shunt motor practically .Maybe you could point me in the right direction. Here's what I have understood about dc shunt motor. Suppose the motor is free running with armature current 2A and suddenly mechanical load is applied. Speed will reduce due to opposing load torque, reducing back emf. Now armature current will increase to say 6A and thus torque will increase, increasing the speed and back emf.Meanwhile the current will start decreasing with increase in speed. At a point where running torque of motor becomes equal to the load torque, motor will stabilize at that speed. The new armature current will be say 3A(greater than free running current) which will cancel the load torque. So yes, speed reduces with the load. Is my understanding right so far?
     
  6. Apr 24, 2015 #5

    Merlin3189

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    That seems right to me.

    My only quibble would be about the motor slowing then speeding up. If a new constant load is applied, then there will be a new point of equilibrium and the motor will slow to the speed of that equilibrium point, with lower back emf, higher current and higher torque. I see no reason why the motor should slow below that new equilibrium speed and then speed up again. As the motor starts to slow, the decrease in back emf is immediate and so then is the increase in current and torque to counter the load.

    If there is a temporary underspeed, I think this must be due to the inertia of some real loads. If, instead of applying a friction brake to the motor as might be done in a dynamometer test, you changed gear so that a mass had to accelerate, then you are not applying a constant load and the strange speed load graph shown for example here at Electrical4U Elec4U-shunt-motor.png is really showing the variation in load as this inertial load accelerates to the new equilibrium speed.
     
  7. Apr 24, 2015 #6

    cnh1995

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    That's the graph I have in my textbook.Yes,you are right.That inertia point didn't occur to me. Thanks..
     
  8. Apr 24, 2015 #7

    cnh1995

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    I have a big confusion (maybe stupid) about series motor too..When it is started, high starting torque is generated and motor starts rotating. Due to rotation,back emf will be produced and it will reduce the current. As Ia=If, both will be reduced. This will reduce the field flux. So,back emf will get reduced. This will again increase the current and flux. This increase-decrease in current will carry on forever..Where am I getting it wrong?? How is the steady state attained??
     
  9. Apr 24, 2015 #8

    Merlin3189

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    You are not getting it wrong! This is a problem with lightly loaded series wound motors.' http://en.wikipedia.org/wiki/DC_motor#Series_connection

    I'm not an expert on this, but my understanding is that as long as there is an adequate load, the motor cannot overspeed because as the speed increases, the torque of the motor reduces. For any load there will be an equilibrium speed where, if the speed increases slightly, the emf increases, reduces the current and torque, so the motor slows down. Even though the reduced field would allow a higher speed for the same emf, the motor cannot accelerate the load to reach this point.
    At any given operating point the motor is not unstable because the feedback, described in the runaway scenario, is not positive: the torque is still decreasing with speed. The danger comes when the load drops too low (typically a break in the transmission decouples the load) and the equilibrium point is now at too high a speed for the motor to tolerate.
    The torque speed characteristic is non linear. At low speeds the torque drops rapidly with increasing speed leading to stable operating points, but at higher speeds it is less stable because torque falls slowly with large increases in speed.
     
  10. Apr 24, 2015 #9

    cnh1995

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    That helped...Thanks a lot..:smile:
     
  11. Apr 24, 2015 #10

    Hesch

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    For a conventional shunt-motor:

    https://www.google.com/search?q=shu...tpub.com%2F14177%2Fcss%2F14177_52.htm;336;259

    I have found (after some calculations) that the stationary change in speed due to a stationary change in torque can be expressed:

    Δω = ΔT*R / Km2, ω=rad/s, T=Nm, Km=motor-constant [ Nm/A ], R= resistance in the rotor-turns.

    Also that the motor may have an undershoot/overshoot when changing the torque, if R is small compared to self-induction (L) in the rotor.
     
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