Slip and Speed error at low speeds in a VFD 2

[rockman7892]

I had posted a previous thread where as a result of some of the responses I had questions regarding why VFD open loop control had a larger slip and speed error when operating at low speeds. I thought I would start a seperate thread to discuss this issue.

I understand how the magnetizing current (flux current) and rotor current and field interact in the induction motor to produce a slip and therefore the motor torque. I also understand the motor equivelent circuit. I have been trying to think however what casues the slip error calcuated by the VFD to increase with decreasing speed. I cannot seem to come up with a reasonable explanation? Can someone help explain why slip error increases with decreasing speed, and why below 4Hz or so the drive can not acurately control the speed or torque of the motor in open loop control?

At first I thought it was because the angle between the magnetizing current and the rotor current would be small, and the drive couldn't acurately distinguish between the two, but then realize that this angle is only small for low load values. Even at low speeds we can have high torque requirement and therfore have these two vectors at a pretty large angle.

 

[LionelHutz]

It seems that say a 5rpm absolute speed error is a lot larger percentage error running at 50rpm than running at 1750rpm.

 

[rockman7892]

LionelHutz

That makes sense that as speed decreases the error is a larger percent of the speed.

So is the slip and speed error constant throught the speed range, and it is this constant error that becomes a larger percent as speed decreases?

Why does this slip error exist, is it due to the way the drive estimates the actual slip and speed? How is this done?

 

[DickDV]

rockman, now you've got it. The slip is constant for a given shaft torque over the speed range so, at low speeds, it is a larger percent of the speed than at high speeds.

For example, if the motor needs 50rpm slip to magnetize its rotor to full torque, and the drive in one way or another manages to compensate for half the slip, then that is 25rpm.

At 2500rpm, 25rpm slip or error really, is 1%. At 250rpm, it is 10%, and at 25rpm it is 100%.

Now, if the drive is being told to run at 15rpm, and the motor needs more than that to just magnetize its rotor, I thinks it's fairly easy to see why the drive simply can't properly control the shaft speed anymore. In effect, the speed signal is completely lost in the error!

Finally, as I said in your other post on this subject, you need to get educated on how an induction motor works. The slip speed is an essential part of making the motor produce torque in the shaft. It has nothing to do with the drive. The motor slips the same way across the line on sinewave power.

 

[LionelHutz]

I'll expand a little on DickDV's response. He's given the motor 50rpm to magnetize the rotor. This is the full-load slip, for example the synchronous 1800rpm minus the rated speed of 1750rpm.

The drive will do calculations to compensate for most of this slip. So, if you told the drive to run the motor at 1800rpm, the drive could compensate so the real motor speed is within 25rpm of the desired setpoint. Or, in other words it would run the motor at a speed somewhere between 1775rpm and 1825rpm. The drive would actually output slightly more than 60Hz to do this.

Now, if you told the VFD to run at 15rpm then the VFD would run the motor at a speed somewhere between 40rpm and -10rpm (-10rpm would only occur if the load was able to drive the motor backwards).

 

[cswilson]

Rockman, you say this is an open-loop VFD. If so, it is not paying attention to what the motor is actually doing. Instead, based on some "presets", it puts out its best guess of voltage and frequency to achieve a commanded motor velocity.

So if the presets are optimized for 2 Hz slip at nominal load, if you want to run a 4-pole motor at 1740 rpm (58 Hz), it would output 60 Hz at the motor's rated voltage. If you wanted to run this motor at 840 rpm (28 Hz),it would output 30 Hz at half the rated voltage.

This proportionality between voltage and frequency (which keeps flux constant) leads these drives to be called V/Hz drives. And the proportionality works well over most of the range of the drive. But at low speeds it breaks down, for various reasons, and many drives have a "voltage boost" algorithm to compensate at these speeds. I have not actually been involved in the design of these algorithms, so I am not expert on them, but I know they are not exact (and they are probably of varying effectiveness in different drives).

The case for closed-loop control is different. VFDs attempting to sense what the motor is doing without a shaft sensor have trouble at low speeds because the key "signal" they are looking for (phase back EMFs) get drowned out by the noise at low speeds. You said in the OP that this was an open-loop VFD, though.

Curt Wilson
Delta Tau Data Systems

 

[LionelHutz]

I think it's a terminology thing.

V/Hz is nothing more than following a V/Hz curve from 0/0 to rated/rated. It might be a straight line or it might be some fixed curve. Using a curve can help with certain limitations of the V/Hz control method.

I believe what you are calling open loop is really more commonly called flux vector or sensorless vector. Having no direct shaft feedback sensor, you're calling it open loop. However, the VFD does have an internal feedback loop using calculations based on measuring the motor current and voltage.

There are also encoder feedback based systems which use an shaft mounted encoder for direct shaft speed and position feedback.

 

[cswilson]

Lionel -- Please re-read my post carefully. I tried to be very clear about what open-loop and closed-loop schemes do. I even emphasized that open-loop drives are "not paying attention to what the motor is actually doing".

What you call the "straight line", I called proportional. What you called "some fixed curve" is usually implemented as what I called a "voltage boost" at low speeds.

My last paragraph talked about "closed loop" drives and the difficulty they have in handling low speeds without a shaft sensor. I separated this out to emphasize that this is a different issue.

Curt Wilson
Delta Tau Data Systems

 

[BigInch]

Somebody needs to get their terminology right.

I didn't actually realize the slip to torque ratio was constant at any speed. A bit counterintuitive at first, but totally logical once somebody rings your bell. Thanks for that.

At those low rpms, arn't the VFD efficiencies pretty terrible anyway, not to mention the motor.

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[rockman7892]

Thanks for all the help guys its starting to become clearer.

To clear things up, what I am referring to as "Open Loop" is a sensorless vector control on the drive. If that is not the correct terminology I apoligize, but for this thread I am referring to sensorless vector control.

I understand the concept of slip in a motor and that the slip is what essentially creates the torque in the motor. For DickDV's example, the slip of 50rpm is for full torque and corrsponds to full current, and for lesser values of torque this slip value will decrease down to a small value of slip but never zero (synchronous speed motor wont run). I also understand that the torque produced in the rotor is the sum of the magneitzing current phasor and the rotor current phasor. The summation of these two phasors leads to Bnet and is the torque producing phasor in the motor. The difference between the magnetizing phasor and the rotor current phasor is esentially the slip.

I assume the 50rpm you guys are referring to as magnetizing the rotor above is esentially the difference between these two phasors, and as stated the difference between the full torque speed and synchronous speed.

Now let me see if I understand what has been said above about the speed error in a VFD:

With the motor running at full toruqe the slip is going to be 50rpm and therefore the motor is going to have an actual speed of 1750rpm regardless of weather the motor is across the line or running from a drive. Running across the line you are pretty much stuck with the slip speed of 1750rpm and there is really nothing you can do to maintain an actual speed of 1800rpm if desired. However with a drive you can try to make up for the slip speed and run close to to 1800rpm by having the VFD account for some of this slip speed. To do this the VFD takes its calculations and makes its best guess at what the slip speed is, and obviously it wont be perfect in an open loop and there will be some error or difference as has been stated above. So as the drive realizes there is a slip it tries to account for this by outputting an additional amount of speed which for this example is 25Hz. Again the drive can not get the slip amount correct so even after compensating there will still be an error between the actual commanded speed and the actual slip speed the drive is running at (25Hz in this case). Hopefully I'm on the right track so far

The drive will continue to compensate for this slip throughout the entire speed range of the drive. However as discussed above, this 25Hz will become a higher percentage of error as the drive runs at lower speeds because 25Hz will become a bigger portion of the overall speed as speed is decreased. Eventually at lower speeds the drive will get lost and the amount of error and my not be able to control the speed at all.

So assuming I am on the right track here are the questions I now have.

1) Does the slip compensation, which in this case is 25Hz stay constant throughout the entire speed range of the drive? In other words at full speed the drive ouputs and additional 25Hz to account for slip, however at lower speeds does it output this same 25Hz or is it constantly changing its slip compensation?

2) Does the amount of load torque on the motor effect the speed error? In other words at full load we are saying that the slip is 50rpm and the drive accounts for 25rpm of this and this 25rpm dictates the error throught the speed range of the drive (maybe depending on answer to #1) However with a smaller load on the torque this slip will change so I'm assuming the compensation will also change? If this is true will lower load torques increase or decrease speed error for a given speed.

3) As stated because of the error at small speeds the drive has trouble accurately controlling the speed and it was mentioned that below 5Hz or so it may not be able to control the speed at all. If it cannot cotrol the speed at all and its lost in the error what will happen? Will the drive shut down?

4) Is the fact that the drive gets lost in the speed error the reason why drives have trouble picking up speed when using the flying start feature at low rpm's?

Thanks again for sticking with me, and I hope I'm on the right track.

 

[cswilson]

rockman -- The key thing to understand is that, for low values of slip frequency, the torque generated by an induction motor is basically proportional to the slip. The torque/speed curve for an induction motor near synchronous speed is almost a straight line.

The slip frequency is the electrical frequency that the rotor "sees", and this is independent of the mechanical speed of the rotor. The transformer coupling into the rotor is proportional to this electrical frequency, and for low frequencies, this means that the resulting rotor current and magnetic field are proportional to slip, as is torque. (At higher slip frequencies, the inductance of the rotor reduces the current, so the proportionality breaks down.)

In true open-loop control, the VFD outputs a frequency (with roughly proportional voltage), and the motor "finds" a speed with a slip where the generated torque matches the load torque, in the same way as operating straight off the line. If the load torque increases, the motor decelerates until a matching torque is generated.

Vector control is closed-loop, and the chain of causality is reversed. A desired torque is computed, usually to maintain a desired velocity. Then, based on the measured/estimated rotor velocity, it adds the slip frequency necessary to generate the desired torque, resulting in the net electrical frequency.

With a shaft sensor, determining the rotor velocity is straightforward. However, in "sensorless vector" (what the technical papers more properly call "shaft-sensorless vector", the rotor velocity must be backed out from electrical measurements. The key thing that must be calculated is the phase back EMF, which is the voltage left over after the IR and L*di/dt drops are subtracted.

This voltage is proportional to the speed and varies sinusoidally with the rotor angle. The problem is that at low speeds, it is almost non-existent, and small errors from noise etc. can completely screw up the calculations, compromising the quality of the control. Usually the drive keeps trying to control, but does not do it very well.

Curt Wilson
Delta Tau Data Systems

 

[LionelHutz]

Something to realize is that some times, the error correction depends more on the quality of the the particular sensorless vector implimentation than anything else. It's much harder to impliment the mathematical model (which is inherently analog) in a digital processor than most people realize.

1,2. The slip compensation is constantly changing as the load and speed changes. I'm sure you meant 25rpm and not 25Hz. In reality, a good VFD should compensate the slip to a much smaller error than 25rpm.

3. Depends on the drive. It is quite frankly a bad implimentation if the drive gets so lost at low speeds it has no idea of the motor speed.

4. Flying start is another feature that is implimented many ways and it just depends on how well it's implimented.

Gunner was involved here a while ago recommending a VFD for sensorless vector hoist operation. I believe it was an Invertek VFD. The drive would hold the hoist stopped at 0 speed, showing that the sensorless vector algorithm in that particular drive worked very well proving that a lot depends on the motor model in use and how well it's implimented in the VFD.

 

[DickDV]

Lionel, I'm not sure why you are discussing slip compensation as it applies to a sensorless vector drive.

As far as I am aware, slip compensation only applies to V/Hz drives. And, being able to compensate for about half of motor slip is about right. Increasing slip compensation above that level usually leads to speed instability and oscillation.

In sensorless vector drives, the speed is calculated by resolving the vector triangle for torque producing amps and then and then comparing it with the full load slip speed. The resulting slip speed is then subtracted from the synchronous speed to get estimated shaft speed.

If the motor model is reasonably good and the calculation update rate is fast, the resulting estimated shaft speed should be very accurate, certainly better than could ever be acheived by slip compensation alone.

In my experience, sensorless vector in the commodity version is good down to about 1/4 of motor slip and Direct Torque Control is good down to at least 1/10 of motor slip. Below that, you need an encoder and flux vector control.

 

[LionelHutz]

When the VFD calculates the actual shaft rpm then isn't it compensating for the slip?? You can tell a good sensorless vector VFD to run at a certain rpm and it will attempt to turn the motor at that rpm. Another terminology issue I guess...

 

[cswilson]

Slip compensation is embedded within vector control algorithms, sensorless or not. (Vector control is more than just slip compensation, though.) Basically, they measure/estimate rotor velocity, then add a slip frequency proportional to the torque required to hold the commanded velocity in order to establish the stator current vector.

 

[DickDV]

Slip compensation in the form I was referring to occurs only in V/Hz drives. The process is to measure motor lead amps and add a little extra Hz to the motor frequency reference to compensate. It never involves a motor model or computation of vectors. A parameter in the drive software sets the level of compensation with too much compensation resulting in speed instability and hunting.

It is true that motor slip is being compensated for in sensorless and flux vector drives but the process is completely different.

Hopefully, this will resolve any terminology issues that might have crept into this discussion.