VFD in constant torque mode but without a constant load?

[P123]

Hello,

So attached I have a photo of a tentative motor (0.37 kW, 4 pole, 3 phase, rated speed = 1705 RPM, rated torque = 2.07 Nm). I also have my load. I aim to use a VFD, as my load Is over a range of speeds. Using a V/Hz ratio of 7.67 I have worked out that my rated motor torque will stick at 2.07 Nm regardless of the frequency (or synchronous speed) that I set the motor.

Under this logic my motor will not be able to provide enough rated torque at the lower speeds, however the peak torque is still above the load (sticking at 5.5 Nm for different frequencies). Is this OK to do?

The grey line is the motor curve at 60 Hz. The yellow curve is the load. I also know that things can get a bit funny below a 20 Hz frequency...

The only other option I can see is to pick a larger motor with a rated torque of around 5 Nm (for max load). Even then, my motor will have a much larger torque than is required at most frequencies and I am afraid I will end up accelerating the load.

Kind regards

Patrick

 

[LionelHutz]

How exactly did you produce that grey line?

Using a VFD, the minimum requirement is to pick a motor with a rated torque higher then the load requires throughout the speed range you'll be running. To run that load from 0rpm to 1450rpm, you'd need a motor that has a rated torque around 5 Nm.

Don't worry about the motor accelerating because the load is too low. The motor will run at the speed which corresponds to the VFD output frequency.

 

[P123]

Thank you LionelHutz!

The motor is from WEG, they publish the motor curve. I found a motor that was rated above 5 Nm. It's a 1.1 kW motor (4 pole) with a rated speed = 1735 RPM and a rated torque = 6.4 Nm.

Your help has been great, but the main trouble I am now having is not the motor accelerating. The main trouble is the motor curve is higher than the load curve, should it not match it? If the motor torque is greater than the load torque, surely the motor will accelerate the load? Or am I thinking in the wrong way?

I was also thinking if I wanted to make the motor curve match the load I could lower the voltage at each frequency (ie change the V/Hz ratio). Thank you once again!

Kind regards

Patrick

 

[LionelHutz]

The motor speed is tied to the frequency. Run it at 30Hz and it will run at 1/2 speed.

If that initial grey curve was published data when operating the motor on a VFD then that motor should work. If it was some other data like a full-voltage starting curve then you looked at the wrong data.

Take a look at the motor efficiency vs load data. I bet you'll find the efficiency remains high at 1/2 load meaning you're not causing undue losses by running the motor at less than full load.

You could lower the V/Hz ratio a little and it would reduce the losses a little. Be careful you don't go too low and lower the torque capability of the motor too much causing operational issues.

 

[ScottyUK]

Don't equate available torque with delivered torque. The motor has a torque capability which varies throughout the speed range as you have observed. The torque demanded by the load at any given speed is a function of the load, not of the motor. The motor simply delivers what the load demands, until the demand exceeds the motor's capability at which point the motor usually stalls. The motor tries to run at synchronous speed (1800 rpm in the case of the 4-pole motor in your example) but it always runs slightly below synchronous speed in order to produce torque: this is fundamental to how an induction motor works. At synchronous speed an induction develops zero torque.

 

[BrianE22]

I think the OP wants to know what happens if you are in torque control and the load changes. I'm guessing that if the load decreases then the motor would accelerate up to some speed alarm limit.

 

[jraef]

Patrick Gallagher,
Please understand that many of us denizens of this forum are "drives people" for whom terminologies have specific meanings in our industry. So using the term "constant torque mode" for many of us is a very specific operating profile, which may or may not mean the same thing to you. If you can more accurately describe what your MACHINE is intended to do, it may help clear this up before this goes off onto an irrelevant tangent.

If you meant to say it is a "Constant Torque Drive", that is more of a marketing term having to do with how the drive is rated for particular types of load profiles. "Torque Control Mode" is a way of using Vector Control capable drives to maintain a highly accurate control of shaft torque rather than speed (velocity control). A. Laddie example is a winder application when speed will vary on a roller as product is added or removed, but precisely maintaining a constant shaft torque is required to keep from breaking or distorting the product being wound.

By you using the term "constant torque mode" in your title you have used a sort of "mixed metaphors" of terms that, to us, mean different things.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[P123]

Hello!

Thank you for your help. "Torque control mode" is what I am referring to.

Without going into too much detail: A mass is being hauled by a cable hooked up to a "main winch". This is not what I am designing however. My design is for another winch, which provides a constant back tension via a cable as the "main winch" hauls the mass. The mass changes velocity as it crosses the desired distance. It should be noted that my winch is really there to keep the mass under control and not to move it.

My job is to create a VFD solution that is able to provide a constant tension (varied torque due to an increased radius from adding rope around the drum) against a range of velocities. This is where I got the load curve from.

Using the V/Hz ratio I was able to work out what my chosen motor's output torque is across this speed range. The load curve is almost like a target that I want to achieve. Since I am creating tension in a large mass, if my VFD is set to a constant 6 Nm then that will be reflected in the cable's tension. If I set the motor torque to 10 for example, the cable tension would increase. This is unlike a common winch application where the load is set by the mass it is hauling (as in the case of the "main winch").

Is it possible I have gone about this in a "speed control mode" way rather than a "torque control mode" way? For torque control is a V/Hz ratio even used?

Thank you guys for your help so much! I'm a student intern so this is really helpful.

Kind regards

Patrick

 

[BrianE22]

Wouldn't it be easier to control "force" instead of "torque"? I'm guessing VFDs have an option to control an external input (volts from the force transducer).

 

[LionelHutz]

That's an interesting application.

Yes, V/hz is still used, not directly as a setting in the VFD but rather the V/Hz applied to the motor has to remain relatively constant to be able to speed control the motor.

If the VFD has a torque limit parameter then you can use it. Say you set the VFD to operate at 1 Hz and 6 N-m. The VFD will attempt to run the motor at 1 Hz but if the pull on the cable exceeds 6 N-m (at the motor shaft) then the VFD will allow the motor speed to increase since it limits the motor torque to 6 N-m.

 

[jraef]

In a Flux Vector Control capable VFD, one aspect that makes them better (and more complex) than less expensive drives is the fact that in the drive control architecture there are multiple "control regulator" loops. In a simple V/Hz drive there is just a velocity regulator loop. In a "Sensorless Vector Control" drive, there is a velocity regulator that is augmented with torque optimization, but that is not always a separate regulation loop, it's a "best guess" (for lack of a better term). In a "Flux Vector" drive (aka (Field Oriented Control" or FOC) there is a fully separate torque regulator loop along with a velocity regulator loop, as well as the ability to fully separate the flux producing and torque producing current vectors so that the drive's ability to maximize the output torque at any given instant without also over fluxing the core is enhanced to the maximum capability of the motor. So when you use a FOC capable drive to accomplish torque control mode, which IS what you want to do, it's a bit of a misnomer to think of it in "V/Hz" terms. The concept of FOC maintaining a constant torque output on the motor shaft results in the "V/Hz ratio" being manipulated by the drive mP at a sub-cycle level, as in WITHIN each sine wave cycle. You therefore can't really equate the torque output to a specific V/Hz value on a continuous basis, it will be whatever it needs to be to provide that torque at the instant it needs it based on the load requirement.

But back to your original issue, and one mentioned very early in this discussion. You MUST chose a motor based on its CONTINUOUS torque output capability suitable for the highest torque required at whatever speed you are running at. The FOC based VFD can put out 200-220% of that for VERY brief periods for the purpose of accelerating or re-accelerating the load after a step-change, but more than a few seconds every few minutes is going to over heat your motor. So what you need to do is determine your maximum torque requirements as a continuous load and pick your motor from there. The drive will allow it to handle most transients in the load.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[P123]

Thank you for the response once again guys.

The reason I ask about the V/Hz ratio is the following: As you decrease the Hz (and consequently voltage) the power the motor can supply decreases as per

P=SQRT(3)*I*V*eff*PF.

If P decreases as RPM decreases, this is how you obtain the constant torque (P=T*RPM).

If there were no requirement for the power to be limited at lower speeds (ie not running in V/Hz mode), would the available torque increase as RPM decreases? According to my understanding, the reason a V/Hz ratio exists in the first place is to keep the current the motor draws constant as RPM decreases.

Any light shed on this is appreciated.

P

 

[LionelHutz]

If you keep increasing the V/Hz ratio then the magnetic circuit in the motor will saturate at which point any further current is not achieving anything useful except overheating the motor. Also, the motor can only handle rated current for long periods of time without overheating it and causing it to fail.

 

[waross]

V/Hz ratios;
If the voltage fed to a motor is too high the magnetic circuit will saturate and the back EMF will not be able to track the increased voltage to aid in limiting the current.
Described another way;
If the voltage is too high then the current increase will be disproportional and the motor will overheat.
The inductive reactance is related to frequency. At lower frequencies the maximum safe voltage without saturation will be lower.
We use the V/Hz ratio to determine the safe voltage at any frequency.
Before VFD use became widespread, we mostly encountered a need for V/Hz calculations when motor supply frequencies were changed between 50Hz and 60 Hz.
My first use of V/Hz ratios was many years ago when one project involved running machines with 50 Hz rated motors on 60 Hz and another project where machines with 60 Hz motors were exported to a 50 Hz country.



Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

It's that saturation that is addressed by the FOC aspect of a Vector drive, allowing the sub-cycle control of flux current to avoid saturation as much as possible, thus maximizing the amount of current available to produce torque. But regardless if being able to do that, torque is still directly related to current and much of the heating in a motor is based on I²R losses in the copper, so because increasing torque means increasing current, your continuous torque is still bound by the thermal capability of the motor.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington