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Thread: Motor Torque and HP vs Current

  1. #1
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    Motor Torque and HP vs Current

    The motor I am referencing in this thread is the same motor I have referenced in another thread regarding the thermal capacity of the motor.

    This motor information is: 2300HP, 1140rpm, 4.16kV, 312FLA

    We have recently been running this motor at around 265A as mentioned in my previous thread. I had someone state to me today that weather we run this motor at 265A or at its full load of 312A we are outputing the same amount of HP on the motor shaft. I am no sure if I believe this statement and my understanding is as follows:

    Ouput HP is a function of HP= Torque x Speed. For this mill grinding motor the speed of the motor is kept constant at 60Hz and 1140rpm.

    The torque on the motor is then dependent on how loaded the motor is. The current that goes along with this toruqe is the current that would align with any given torque on a motor speed vs torque curve and speed vs current curve. So for example if we are running at a given speed, there would be a given torque and current value for that speed. The given torque x the motor speed would give the motor HP output.

    So if we are running with a given load on this motor at 265A then this current value equates to a given toruqe value on the motor curve. At this torque value we can multiply it by the speed to get the motor output hp. Now lets say we start loading the mill more. As we load the mill more the torque requirement on the motor increases. As the torque requirement increases we move left on the motor motor speed vs torque curve. At the new required torque value we will have an increased current value since we also move to the left on the motor speed vs current curve. So with a higher current value we have a higher torque value. Since torque increased and speed is kept constant I would expect the motor output hp to also increase, and prove that as current increases then torque and therefore HP increases.

    The only part that I'm unsure about is how slip effects my explanatin above. As we are moving to the left on the motor torque curve we are also increasing the slip with is slowing the motor. So as we increase torque and slow the motor speed is this essentially a wash with the motor output HP staying the same? Or is the slip speed and speed decrease small compared to the torque and current increase and therefore the overall output HP is increased.

    Can anyone help me out with grasping this concept?

  2. #2
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    Quote Originally Posted by philly View Post
    The motor I am referencing in this thread is the same motor I have referenced in another thread regarding the thermal capacity of the motor.

    This motor information is: 2300HP, 1140rpm, 4.16kV, 312FLA

    We have recently been running this motor at around 265A as mentioned in my previous thread. I had someone state to me today that weather we run this motor at 265A or at its full load of 312A we are outputing the same amount of HP on the motor shaft. I am no sure if I believe this statement and my understanding is as follows:

    Ouput HP is a function of HP= Torque x Speed. For this mill grinding motor the speed of the motor is kept constant at 60Hz and 1140rpm.
    You are right here, at 312 A the Output HP have to increase, unless the speed had decreased somehow. But this is not what you are saying. You say the Speed stays constant. Right?

    Quote Originally Posted by philly View Post
    The torque on the motor is then dependent on how loaded the motor is. The current that goes along with this toruqe is the current that would align with any given torque on a motor speed vs torque curve and speed vs current curve. So for example if we are running at a given speed, there would be a given torque and current value for that speed. The given torque x the motor speed would give the motor HP output.

    So if we are running with a given load on this motor at 265A then this current value equates to a given toruqe value on the motor curve. At this torque value we can multiply it by the speed to get the motor output hp. Now lets say we start loading the mill more. As we load the mill more the torque requirement on the motor increases. As the torque requirement increases we move left on the motor motor speed vs torque curve. At the new required torque value we will have an increased current value since we also move to the left on the motor speed vs current curve. So with a higher current value we have a higher torque value. Since torque increased and speed is kept constant I would expect the motor output hp to also increase, and prove that as current increases then torque and therefore HP increases.

    The only part that I'm unsure about is how slip effects my explanatin above. As we are moving to the left on the motor torque curve we are also increasing the slip with is slowing the motor. So as we increase torque and slow the motor speed is this essentially a wash with the motor output HP staying the same? Or is the slip speed and speed decrease small compared to the torque and current increase and therefore the overall output HP is increased.

    Can anyone help me out with grasping this concept?
    The slip, also takes part in this issue. But since the slip is relation between the frecuence of the system (60Hz), and the speed of the machine, and you have stated that they both stay the same. Then theres no reason for the Output Hp to stay similar after an increase of the current.

    From my notes, I have that for induction motors, this the Output Power (p.u):

    P={3*Vth^2*Rr*[(1-s)/s]}/{[(Rth+Rr/s)^2]+Xth^2} (1)

    I = Vth/[(Rth + Rr/s)^2 + Xth^2] (2)


    where,
    P = output power
    Vth = Thevenin Voltage from the rotor perspective.
    Rr = rotor resistance
    s = slip
    Rth = thevenin resistence from the rotor perspective.
    Xth = thevenin inductance from the rotor perspective.

    So, if you change the current value, automatically you change the thevenin value Vth (see (2)). And then the P value change as well due to that change.

    Also if you change the slip value, by changing the speed for example, the P value will change as well.
    Last edited by Mayimbe; 10-20-09 at 12:03 PM. Reason: adding stuff
    "I am the son and heir... Of nothing in particular"

  3. #3
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    Induction machine model


    ANd the simplified thevenin model

    "I am the son and heir... Of nothing in particular"

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    Quote Originally Posted by philly View Post
    The only part that I'm unsure about is how slip effects my explanatin above. As we are moving to the left on the motor torque curve we are also increasing the slip with is slowing the motor. So as we increase torque and slow the motor speed is this essentially a wash with the motor output HP staying the same? Or is the slip speed and speed decrease small compared to the torque and current increase and therefore the overall output HP is increased.

    Can anyone help me out with grasping this concept?
    I think your explanation is sound and you are right not to believe that someone. Increase the mechanical load and the current will increase.
    Assuming you stay within the rated operating envelope.

    Slip is generally quite low for most induction motors.
    Yours is rather higher than most, but that is probably because it was built to suit the application. It probably has deliberately high rotor resistance to improve starting characteristics and this results in higher slip.
    That said, the slip is still just 5% from no load to full load torque. It isn't negligible, but still a second order effect in the speed times torque calculation.

  5. #5
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    An induction motor is a torque matching machine. At synchronous speed (1200 rpm for your motor) it will produce zero torque. As the shaft speed slows down below 1200 rpm the interaction of the magnetic fields in the motor produces torque. The slower the speed, the higher the torque (within a small range). The shaft and rotor slow down until the torque produced by the rotor equals the torque required by the load plus the windage losses in the motor.

    At 1140 rpm your motor produces the rated torque to achieve rated output horsepower.
    HP= T (ft-lb) x rpm/5240. Torque is about 10,600 foot pounds for 2300 HP. At rated voltage the motor will draw rated current and deliver 2300 HP.

    Since your measured current is less, HP (or KW, kW=0.746 x HP) the load is drawing less horsepower and exerting less torque on the shaft. The motor is spinning at some speed faster than 1140 rpm, or in other words the slip is less. Zero slip = 1200 rpm.

    If you have a speed versus torque curve, you should be able to see this and the speed and horsepower the motor is putting out. Measure the KW to the motor and you will be able to estimate the horsepower out very closely.

    The statement that the motor always draws rated load HP is not correct. The motor only delivers the horsepower needed by the load by matching the torque at the running speed. If your amps are less than rated, so is the output HP or KW.
    Last edited by rcwilson; 10-20-09 at 01:27 PM. Reason: Corrected missing sentence.
    Bob Wilson

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    Quote Originally Posted by Mayimbe View Post
    Induction machine model
    ANd the simplified thevenin model
    I use the Steinmetz model for motors.
    It has served me well over many years on VSD systems.
    When we bid for a project, we routinely have to provide guaranteed performance figures for the drive and motor. That puts us between a rock and a hard place. If we quote a low but safe figure on efficiency, we don't get the job. If we make it higher and win the job but fail to deliver (yes, it has to be demonstrated) we face financial penalties that would turn the job into a loss making exercise.
    Getting it wrong isn't an option.

  7. #7
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    Thanks for all the responses guys. They were very helpful and help clear up the picture a great deal. I am going to try to use the motor speed vs torque/current curves to create a visual (will try to post)

    Quote Originally Posted by Mayimbe View Post
    You are right here, at 312 A the Output HP have to increase, unless the speed had decreased somehow. But this is not what you are saying. You say the Speed stays constant. Right?
    Yes the motor is operated off a starter so is always operated at 60Hz. The only speed decrease is the slip as I mentioned which it sounds like is very small in magnitude compared to torque increase

    Quote Originally Posted by Besoeker View Post
    I think your explanation is sound and you are right not to believe that someone. Increase the mechanical load and the current will increase.
    Assuming you stay within the rated operating envelope.

    Slip is generally quite low for most induction motors.
    Yours is rather higher than most, but that is probably because it was built to suit the application. It probably has deliberately high rotor resistance to improve starting characteristics and this results in higher slip.
    That said, the slip is still just 5% from no load to full load torque. It isn't negligible, but still a second order effect in the speed times torque calculation.
    O.k. I think I understand what you are saying here. From no load to full load the speed difference (slip) will only be about .05% P.U. of the motor speed. However over this range the torque will change from 0 P.U. to 1 P.U.. This torque magnitude (compared in P.U) change is much larger than the speed change P.U. value and therfore causes an increase in the HP equation. The slip is not neglegible and does cause a small decrease in speed, but this decrease is much much smaller than the torque increase. Do I have this right?

  8. #8
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    Quote Originally Posted by rcwilson View Post

    Since your measured current is less, HP (or KW, kW=0.746 x HP) the load is drawing less horsepower and exerting less torque on the shaft. The motor is spinning at some speed faster than 1140 rpm, or in other words the slip is less. Zero slip = 1200 rpm.

    If you have a speed versus torque curve, you should be able to see this and the speed and horsepower the motor is putting out. Measure the KW to the motor and you will be able to estimate the horsepower out very closely.
    .
    After I posted my O.P. I thought that I could simply use the KW = V * A * p.f.*1.73, and show that different current values corrospond to different kW outputs. I would of course also need account for the motor efficency at a given speed. So if I could estimate the % loading I could then use the efficency and p.f. at that % loading and calculate the power on the ouput shaft and electrical input power since power is linear over the % load range

  9. #9
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    O.K to take it a step further I'm trying to figure out how the V/Hz ratio compares to the current and torque.

    So we said earlier that the motor current was directly related to the motor torque which was determined by the required load torque. We also know that with a constant V/Hz ratio in the motor the flux density stays the same and it is this flux density that produces the torque.

    So if the V/Hz ratio stays the same in my motor will the torque produced in the motor always be the same? Or will the V/Hz ratio just allow the rated motor torque to be produced, and the actual torqe is that which is required by the load as discussed?

    How does current react to changes in this V/Hz ratio? For instance when a motor is driven over base speed and the voltage is maxed out then the V/Hz ratio is diminished? Will current increase when the V/Hz ratio decreases?

  10. #10
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    Quote Originally Posted by philly View Post
    O.k. I think I understand what you are saying here. From no load to full load the speed difference (slip) will only be about .05 P.U. of the motor speed. However over this range the torque will change from 0 P.U. to 1 P.U.. This torque magnitude (compared in P.U) change is much larger than the speed change P.U. value and therfore causes an increase in the HP equation. The slip is not neglegible and does cause a small decrease in speed, but this decrease is much much smaller than the torque increase. Do I have this right?
    Yes, that sums it up.

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