Peak Torque of 3 phase motor when overspeeded with VSD

[vpxxx]

What is the formula for calculating the peak torque available when a standard induction motor is overspeeded with a standard pwm inverter drive with 150% overcurrent capacity. For example, for a 4KW, 4 pole, 50Hz motor with a DOL pull up torque of 3 X FLT overspeeded to 80Hz.

 

[redtrumpet]

Motor will operate in constant horsepower mode above rated speed, so torque will decline with increasing speed above rated speed.

HP=(Torque x rpm)/5252, HP is fixed above base rpm.

By overspeeding 33% (from 60 Hz to 80 Hz), your steady-state torque decreases to 75% of rated torque at base speed.

The VFD will drive the motor from the opposite end of the speed-torque curve as opposed to DOL starting, so your 3 x FLT pull-up torque is irrelevant in VFD operation. Think of a DOL start as starting from 100% slip, and a VFD starting from 0% slip. This is why the VFD can limit inrush current to rated current or less.

Your 150% peak current capacity will roughly correspond to 150% peak torque. Assuming the same relationship holds true in the overspeed condition, peak torque will be 150% of 75% of rated torque, or approximately 112% of rated torque at base speed.

I suggest you familiarize yourself with the basic mechanical power and torque relationships, the motor speed-torque curve, and elementary VFD application theory.

 

[redtrumpet]

Small clarification to my previous post:

P = torque x rpm/5252 where P is in horsepower and torque is in foot-pounds. Not sure about SI, but I think power (W) = torque (N) x angular velocity (rad/s).

Also, motor operates in constant power mode above rated speed because the VFD is flux-limited and cannot maintain the base V/Hz ratio in the overspeed condition.

 

[redtrumpet]

I'm reading these posts too fast. Overspeeding from 50 Hz (not 60 Hz as I thought) to 80 Hz is a 60% overspeed, so continuous torque will reduce to 62.5% of rated. Peak torque will only be in the neighborhood of 90-95% of rated.

 

[jOmega]

[red]RedTrumpet[/red],

I don't want to nit pick but technically, your statement about operating in constant HP above base speed only applies to dc machines; where the power is a direct product of the volts and amps.

In the case of the ac machine, when you go above base speed by holding the voltage constant and increasing the frequency, there is a corresponding change in the power factor. And as we all know, power in an AC machine is the product of the volts, amps and power factor (neglecting efficiency...as it has the same relevance in either case...ac or dc)......

So, to be technically correct, the HP drops off when an ac machine is operated in field weakening (reduced V/Hz) and instead of constant HP.... actually operates at constant kVA. The drop-off of HP is attributed to a decrease in in PF.

(Borrowing from JBartos)
Hint : what happens to the impedance of the machine in field weakening ?)

rated torque reduces by 1/N and BREAKDOWN torque decrease by (1/N)².
where: N = Motor RPM
or you can use the frequency ratio of

Hz @ base speed / Hz at field weakend speed

Hz at base speed will be either 60 for NA and Brazil, or 50 for most of the rest of the world.

Multiply (Hz @ base speed / Hz @ field weakend speed) x motor rated torque at base speed... to obtain the available rated torque at the field weakened speed.

Square the ratio and multiply it times the motor rated breakdown torque to obtain the available breakdown torque of the motor at the field weakend speed.

Above base speed, the rated machine torque drops off at a 1/N slope. The breakdown torque, while starting at a higher value, drops off at a (1/N)² slope.... which is a much steeper slope.... if the frequency is run up high enough, there will be a point at which the breakdown torque curve crosses over the rated torque curve... at that point the load torque must be reduced by the squared curve.... for to prevent problems, the load torque should be reduced when the breakdown torque curve is within 20 of the rated torque curve in the field weakend region. This will prevent the machine from going over the hump and pulling-out on transient load disturbances..

(this is much easier to see when it is graphed) .

HTH

 

[electricpete]

good comments by all. I think in this case j&[ignore]omega[/ignore]; is correct that Breakdown Torque is proportional to (volts/hz)^2.... If we clamp V at rated and increase f then &[ignore]tau[/ignore];breakdown goes with (1/f)^2

 

[jbartos]

Suggestion: Visit

http://www.controldepot.net/vfd1.pdf

for Figure 20 VFD Voltage to Frequency Ratio plot

 

[vpxxx]

Thank you all for your positive contributions.
The point I was trying to fully understand out of this is, as confirmed Peak torque of motor when overspeeded by drive could be calculated from
Peak torque= rated torque x breakdown torque x X% safety margin x (rated speed squared/ new increased speed sqaured)

where X% is the safety margin of system to act as buffer between run torque and peak capacity.

In above breakdown torque is only given for DOL motor rating ( 3xFLT in above example) which in the case of DOL if allowed to develop by the control gear would result in a current much higher than rated current.

With the VSD limiting the current to typically 150%, what value should be used for breakdown torque in above formula.

 

[electricpete]

I would assume that during start that frequency and voltage both start low and gradually increase. Throughout the whole start the volts/hz ratio should be normal until you exceed rated speed/rated voltage. In that case there is no reduction in breakdown torque until you exceed rated speed and voltage.

Maybe someone else can comment.

 

[redtrumpet]

It has been my experience that a motor operated on VFD will exhibit a substantially constant HP characteristic to about 150% overspeed, which is roughly the range vpxxx mentioned in his post. Above that speed the horsepower will decrease as stated by jomega.

Also, a motor on VFD will drive the VFD into current limit long before it reaches the breakdown torque point unless the VFD is grossly oversized.

 

[redtrumpet]

vpxxx - you cannot use the formula given by jomega with a VFD. The breakdown torque point will never be reached. The VFD operates the motor only in the linear portion of the speed-torque curve, from 0% to perhaps 5% slip depending on the motor. In this range torque will be nearly proportional to current. Your "breakdown" torque with a VFD will be roughly 150% at 150% current. However, this is not the same as the breakdown torque on constant voltage/constant frequency ie. DOL operation, which occurs at high slip and draws a high current.

I am attempting to give you an engineering approximation for your problem. The AC squirrel cage motor will operate, for all intents and purposes, in a constant power mode to around 150% of base speed - 75 Hz for your 50 Hz motor. If you are going to operate above that range, then yes, you will have to consider the rapid decrease in power beyond that point. Below that range, the simple relationships I gave in my original post should suffice.

 

[redtrumpet]

"Multiply (Hz @ base speed / Hz @ field weakend speed) x motor rated torque at base speed... to obtain the available rated torque at the field weakened speed."

My apologies - the above formula from jomega can be used. It yields the same answer for rated torque that I gave in my original post - 50 Hz/80 Hz x 100% rated torque = 62.5% torque available. On this it appears jomega and I agree.

"Square the ratio and multiply it times the motor rated breakdown torque to obtain the available breakdown torque of the motor at the field weakend speed."

This relationship cannot be used with the VFD for the reasons stated in my last post.


You cannot use the breakdown torque relationship, of given by jomega, however, for the reasons I stated in my last post.

 

[electricpete]

Isn't it 2 separate issues:
1 - max steady state torque above rated speed ~ 1/f
2 - breakdown torque above rated speed ^ 1/f^2

It is agreed that we cannot operate continuously near breakdown torque, but the breakdwon torque is still of interest in determining how the motor will initially accelerate the load, and also how the motor will respond to a momentary increase in load torque after reaching steady state, isn't it?

 

[electricpete]

To further my previous comments, aren't we closer to breakdown torque when we operate at rated power at 150% speed than we are when we operate at rated power at 100%.

After all, in increasing from 100%-speed/rated-power to 150%-speed/rated-power, the breakdwon torque has fallen by (2/3)^2~44% while the operating torque has fallen only 2/3~ 67%

 

[jOmega]

Need to set some things straight.

First to RedTrumpet's 2nd post of Dec. 12:

[blue]Also, a motor on VFD will drive the VFD into current limit long before it reaches the breakdown torque point unless the VFD is grossly oversized.[/blue]

This would be the case, RedTrumpet, IF, and only if the torque load was at rated at base (rated) speed, and the torque load is not reduced as the speed is increased.

However, in applications that are designed to operate into the field range of the motor, the torque load is not allowed to exceed the available torque from the motor.

With regard to your 3rd post of Dec. 12, RedTrumpet, where you challenged the formula that I presented, I offer the following in rebuttal.

I have just spent the morning running a motor on a dynamometer from a VFD source and can tell you that as long as the load torque is less than or equal to the 1/n derated value of torque, current stays in-bounds right around 100% or less. And if the field range is extended to where the breakdown torque approaches the running torque, it must further be reduced as a squared function, eg. 1/(n^2).

I know you disagree with this, but Dr.s Novotny & Lipo at UW believe this enough to have published it in several different papers and in course material for specific classes on the subject. You can also find it in some text books on the subject as well.

... Note: Reference pages 2.33 - 2.37 from the following publication....
IEEE Industryt Applications Society Milwaukee Section
Tutorial Seminar
Variable Frequency Induction Machine Drives
April 12, 1986
presented by Drs. T.A. Lipo and D.W. Novotny
Department of Electrical and Computer Engineering,
University of Wisconsin, Madison

ElectricPete also concurs and I am appreciative of his comments. After all, the objective is truth, accuracy, clarity, etc. of the principles involved.

It is unfortunate that this forum doesn't have a means of displaying graphic content or even attachments such as PDF files. Not having such facility is a handicap in discussions such as this. The graphs and curves make this easier to see and explain.

One last comment, and this comes from the motor manufacturer side: If operation in the field weakened range is to be prolonged, (more that a few minutes), it is advisable to check with the motor manufacturer for the suitability of the bearings to operate at the elevated speed.

.... and we'll save for another day, the way in which you can get rated constant torque operation to 2x base speed from a machine.

 

[redtrumpet]

So a pissing match it shall be. I have never disputed your theory. I poorly worded my response on Dec. 13 in saying that the breakdown torque relationship didn't apply with a VFD. The VFD can't change the basic motor parameters, so the motor will operate the same on VFD as off it. That includes the (1/n^2) reduction in breakdown torque, as well as the (1/n) reduction in running torque.

What changes, however, is at what point on the speed-torque curve the motor is operated. Sizing a VFD to provide enough current to reach the motor's breakdown torque point (at base frequency) is usually prohibitively expensive. All VFDs that I am familiar with are designed to operate the motor in the linear portion of the speed-torque curve, where current is roughly proportional to torque. When applied in this manner, the artificial torque limit imposed by the VFD is approximately 150% of rated torque at 150% of rated current. The VFD can't provide enough current to slip the motor to the point of breakdown torque, where 300% or more of rated current can be drawn.

As vpxxx asked on Dec. 12, "With the VSD limiting the current to typically 150%, what value should be used for breakdown torque in above formula?" It was clear that at the range in which he was operating the VFD, the motor breakdown torque point could not be reached without current limiting by the VFD. That is why I suggested the (1/n^2) reduction shouldn't be considered - the VFD under the most commonly encountered overspeed conditions couldn't provide the current to reach that point.

I don't dispute your test lab results. I am guessing, however, that you had to overspeed past 150% for the breakdown torque to be the limiting factor. I always said that past 150% overspeed the reduction in motor power (more correctly torque) should be considered.

Perhaps chalk this up to a theoretical vs. engineering application viewpoint. In a lab with a large VFD on a small motor - breakdown torque will limit every time. In the real world with a matched VFD and motor - VFD current will limit until a substantial overspeed condition is reached.

Or maybe I'm just full of it and got lucky with all those VFDs I successfully applied in the past.

 

[jbartos]

Suggestion: Visit

http://www.saftronics.com/pdf_fils/library/AN_VFD_GEN018.pdf

etc. for more info including graphics

 

[jOmega]

Jbartos,
I looked at what Saftronics presents in your reference. Their depiction is incorrect. Constant HP and linear torque derate are applicable to DC motors (Saftronics also makes DC drives) but it is not correct for AC motor applications.
As I stated before, in holding the voltage fixed when going above base frequency effectively reduces the V/Hz. As a consequence, the impedance of the motor is changing which results in a change in power factor. For constant HP to be obtained, E x I x CosØ must be constant. But CosØ decreases above base speed and so, the power decreases.
Operation above base speed is constant kVA.

Also, the portrayal of torque droppiing off linearly is not correct for an AC motor. The torque drops off as 1/n and that's not a linear function.

 

[jbartos]

Suggestion to the previous posting: What about

http://www.controldepot.net/vfd1.pdf

for Figure 20 VFD Voltage to Frequency Ratio plot