Motor Torque and HP vs Current

[philly]

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?

 

[Mayimbe]

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?

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.

 

[Mayimbe]

Induction machine model

ANd the simplified thevenin model

 

[Besoeker]

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.

 

[rcwilson]

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.

 

[Besoeker]

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.

 

[philly]

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)

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

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?

 

[philly]

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

 

[philly]

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?

 

[Besoeker]

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.

 

[Besoeker]

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.

There isn't a direct relationship between current and torque. For example, the motor will take typically 30% of full load current at no load. This is mainly the excitation current that produces the rotating field.
The output torque is a function of stator flux and rotor current.

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?

Constant V/f allows the motor to produce rated torque at different frequencies.

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?

Reduce the V for the same f and more current will be required for the same mechanical load.

 

[philly]

There isn't a direct relationship between current and torque. For example, the motor will take typically 30% of full load current at no load. This is mainly the excitation current that produces the rotating field.
The output torque is a function of stator flux and rotor current.

What equation is typically used to represent torque as a function of both the V/Hz ratio and current?

The only that I have seen is

T = (3V^2th * R2/S) / wsync [(Rth + R2/s)^2 + (Xth + X2)^2]

where th= the thevinin impedances of the stator.

Simplified for low slip this equation is:

T = 3sV^2th / wsyncR2

From this I guess you can seperate things and see that Torque is a function of both the V/Hz and V/R2 which is esentially the rotor current. So if the V/Hz is reduced then the rotor current will increase to try to maintain a constant torque output to the load?

Reduce the V for the same f and more current will be required for the same mechanical load.

So when we are above base speed of the motor the V/Hz will decrease and therefore the torque output will decrease. The motor current as a result will increase to try to keep providing the required torque. The motor will keep increasing until in can no longer produce enough toruqe and the motor will stall? Even though the V/Hz is decreasing and the current is increasing in this range, is the motor still producing the same output torque?

 

[Besoeker]

What equation is typically used to represent torque as a function of both the V/Hz ratio and current?

Rated torque can be calculated from nameplate speed and power.
To a fairly good approximation, available torque is a function of the square of the applied voltage.
This applies for rated V/frequency at any value of frequency.

So when we are above base speed of the motor the V/Hz will decrease and therefore the torque output will decrease.

Yes, the available torque will decrease.

The motor current as a result will increase to try to keep providing the required torque.

Generally true, I suppose. But there are applications where required torque falls off with increasing speed.

 

[philly]

Rated torque can be calculated from nameplate speed and power.
To a fairly good approximation, available torque is a function of the square of the applied voltage.
This applies for rated V/frequency at any value of frequency.

So at any given frequency if the V/Hz ratio is less than designed then the motor will draw more current to try to keep delivering the avaliable torque. For instance at a given speed lets just say that the motor torque speed curve had the motor outputting a certain torque. If for some reason the V/Hz ratio at that speed was diminished, then the current the motor would draw would be different than the current shown on the motor curve for the torque at that particular speed?

 

[philly]

On other question I had with this motor involved the fact that it is started with a pony motor (alhtough a little off topic)

I was always under the impression that this pony motor was to get the load and motor spinning to help reduce the starting current on the motor. Does this work by getting the load spinning to a speed on the load/torque curve that would corrospond to a diminished current value?

For instance if you start the 2300hp with load at rest, the load torque corrsponds to a very large current on the motor speed/current curve. But if you start the load with the pony motor and wait until it is spinning up close to full speed at a point on the motor speed/current curve where the current starts to diminish, then will the starting current be reduced to whatever value corrosponds to that speed on the speed/current curve.

 

[rcwilson]

A large but very short inrush current will occur to build up the magnetic field in the rotor.

Look at your Thevanin equivalent circuit. With no excitation on the motor, the slip = 1 and the current is only limited only by the resistance and inductance.

The motor will be spared the long high current needed to get the load up to speed.

 

[Besoeker]

So at any given frequency if the V/Hz ratio is less than designed then the motor will draw more current to try to keep delivering the avaliable torque?

Yes, assuming the load torque doesn't change, that's correct.

 

[Besoeker]

On other question I had with this motor involved the fact that it is started with a pony motor (alhtough a little off topic)

I was always under the impression that this pony motor was to get the load and motor spinning to help reduce the starting current on the motor. Does this work by getting the load spinning to a speed on the load/torque curve that would corrospond to a diminished current value?

For instance if you start the 2300hp with load at rest, the load torque corrsponds to a very large current on the motor speed/current curve. But if you start the load with the pony motor and wait until it is spinning up close to full speed at a point on the motor speed/current curve where the current starts to diminish, then will the starting current be reduced to whatever value corrosponds to that speed on the speed/current curve.

Yes. For large values of slip you get very high current. If you can get the motor up to near operating speed with low slip, that will reduce starting current. As rcwillson has said, there will still be a magenitising transient.

 

[philly]

Yes. For large values of slip you get very high current. If you can get the motor up to near operating speed with low slip, that will reduce starting current. As rcwillson has said, there will still be a magenitising transient.

I went and looked at all the information for this pony motor/main motor combination. The pony motor is a small 60hp motor with an rpm of 1760rmp, however it goes through a gearbox before it couples to the main 2300hp motor. The output speed of this gearbox is only 25.14rpm. This means that the main motor and therefore load is only turning at a maximum of 25rpm right before the main motor is energized.

This 25rpm is only about 2% of the rated motor & load speed. Looking at the motor torque/current curve this 2% is looks like there will barely be any current difference when starting the motor. Yes the motor will already be turning, but it will still be energized with the load at a very low speed and will still have a high current associated with the current at this speed?

So it appears from looking at this arrangement that this pony motor is really not reducing the starting current magnitude. It also appears that it is not reducing its duration by much either since the speed before its engages is only 2% of main motor speed? Do you think then that there may be other reasons for this pony motor aside from current reduction?

 

[Besoeker]

Do you think then that there may be other reasons for this pony motor aside from current reduction?

Well, given that it won't make any appreciable difference to either starting current or run up time I'd take it that the pony motor is there for another function.

One of the applications I have come across is in paper making and particularly tissue machines.
The main drying cylinder is a big beast, runs very hot, and it needs to be kept turning to avoid distortion during cooling and shut down.
Just turning. And that's where the pony motor and drive is used.
For that application, it gets referred to as the Sunday drive.