AC drive question 1

[guilio2010]

I have never used an AC drives before but I kind of understand the basics of it. I guess my question is more on the operation of the drive.

We have a SVX9000 AF drive. I am going to take a trip out to see the unit and get more info, but curious as to how I can use this to my benefit. Currently it's setup for 1 phase but we want to set this up to run on 3 phase 480V. The concern I have is startup amps of the motor, but it seems that the af drive, from my understanding, has the ability to lower that current draw at startup by adjusting the frequency. Is this correct? If so, what are things I need to look at to determine what would be best setup for my application? I want to make sure I get the right size genset in place and calculate the anticipated startup current using this drive. When looking at the motor, it seems that the required startup current exceeds the SVX9000 ratings so if the above can be achieved, I think this would work. Any help appreciated. If you need more info, I can provide what I have now.

Thanks,
TWK

 

[Skogsgurra]

In general, the VFD will be fine with a suitably sized and loaded motor. If VFD rated Power is equal to or greater than motor rated Power, there will be no problems with starting current. A VFD actually makes it possible to have a better starting torque than a DOL start.

That is the simplistic, but true, answer. Tons of literature available if you want to dig deper.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[eeprom]

The drive does not supply inrush current. I assume you are referring to inrush current as the required current to start the motor. When started across the line, the motor is subjected to full line voltage while the motor rotor is stationary. Therefore the motor circuit is going to draw locked rotor current, plus whatever magnetizing inrush is required. When a VFD starts a motor, the voltage to the motor is mostly proportional to frequency. The voltage to the motor is approximately V = f*Vmax/60. So at 5 hertz, the motor will see approximately 1/12 of the full voltage.

EE

 

[jraef]

The generalized rule is that if you NEED to accelerate a load with no more that FLC of the motor, ONLY a VFD is capable of it, because you can stretch out the accelerating time indefinitely; 2 minute, 2 hours, 2 days, it's all the same to the VFD.

When starting Across-The-Line, the starting current is high because of not only what was mentioned, but the fact that the power factor is extremely low. So when starting current reaches 600% ATL, at first, 590% of that current is Reactive (VARs), 10% is active, meaning working to create torque. So torque is low, but current is high in the beginning. That percentage shifts as the motor accelerates and at roughly 80% speed, most of the current is working to make torque, less is going to VARs and heat. With a VFD, it is capable of making the motor deliver 100% rated torque from the outset, so right away, most of the current is being used to accelerate the motor, the VARs are controlled immediately. The VFD can then accelerate the motor with less current by trading time for current.

My concern for your statement was the mention of single phase. What's that about? Are you now feeding the VFD with single phase 480V? If so, was the VFD properly sized to do that? The VFD would need to be 2x the motor size at a minimum. If that's the case, and it is working, what's the reason for changing to 3 phase? My point on that is that if the VFD is already working fine now with a single phase input, going to 3 phase will get you nothing.


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

 

[guilio2010]

eeprom: Yes I am referring to inrush on the motor. From readings, the inrush requirements is from the initial magnetism needed? Is this related to the amount of force needed in order to spin? From reading below, the VARs is the inrush?

Jraef: the single phase is the incoming from utility we have at location. The installed SVX has a single phase kit that is installed to operate in this situation. The motor is connected 3 phase to the output of the SVX but input is single. With that being said, I have to verify the model number as the manual has a 230 and 480 model of the SVX on the same frame but I think it is the 480 model but they have the program setup to 230max on the motor. The motor is operating on 230 as well but can be run on a higher voltage per the data plate. Why I am looking to change is we are relocating the equipment to our other facility temporarily and there is no utility yet so to cut down on wire size, was looking to do the 480. Maybe I should K.I.S.S. and just get a generator to match what we have, but there is a part of me that wants a challenge and learn something new.

I guess my next question is there an equation to use that incorporates the V/Hz to figure the inrush current or is it a rule of thumb 150% FLA on the motor?
Also, the ATL (across the line), is that equal to DOL mentioned above?

 

[LionelHutz]

In very simplistic terms, you can think of a VFD as always operating the motor at full-speed. Whatever frequency you're running at, the torque as a percentage of motor rate torque can be used equivalently to motor load.

So, say you have motor data like the following;
0% load = 80A
25% load = 110A
50% load = 140A
75% load = 170A
100% load = 200A
125% load = 230A

So, say you want 125% torque or the load requires 125% torque when accelerating - the VFD would have to source 230A.
So, say you want 50% torque or the load requires 50% torque when accelerating - the VFD would have to source 1400A.

The load may require the torque because of frictional drag in the load that must be overcome.
You may want to supply a certain torque because you want to accelerate the load at a certain acceleration rate.

 

[itsmoked]

You want to edit that LionelHutz?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Skogsgurra]

-0, done!

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[guilio2010]

LionelHutz: So lets say I take a load draw while starting the motor before we disconnect and move the equipment. I get a current reading of 180A. If my FLA is 200A, I can take the V/Hz of 7.67, which from my understanding would allow the 100% Torque which is my FLA, and decrease this to we'll say 7.4 (just for this purpose) as this would limit the amount of torque during ramp up since I do not need 100%?

I guess I'm failing to see where if I apply 7.67V@1Hz, 76.7V@10Hz, etc... that this would provide me my 100% Torque all the way to 60Hz. The one equation I found that makes sense with reading is that the constant V/Hz keeps the flux density constant?

 

[itsmoked]

I guess I'm failing to see where if I apply 7.67V@1Hz, 76.7V@10Hz, etc... that this would provide me my 100% Torque all the way to 60Hz.

Because the V/Hz is kept correct as you surmised you will have the full torque available at all the intermediate speeds. Do keep in mind that HP is torque X speed so you certainly won't have full HP all the way up.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[guilio2010]

So I went back to some basics in some old book and looked more into this as an equivalent circuit of an indication motor to understand why this happens. I think its the air gap flux is what I want to hold constant, assuming if I apply 480V/60Hz, the frequency I am applying will affect the total impedance of the circuit, but I want to limit the voltage drop across the air gap? so I need to adjust the voltage to accommodate this?

If so, lets say I have a starting torque requirement of 100%. Then I can anticipate that the current would be FLA and due to following the V/Hz (480/60 for example), I will always maintain the same voltage drop as long as I increase Voltage and Frequency together since my impedance will increase. Does my acceleration of the motor happen quicker?

Again, thanks everyone. This definitely isn't my strongest part.

 

[cswilson]

giuilio:

You are picking this up quickly! Yes, you are really trying to keep the air gap flux constant.

Another way of looking at it may help in your understanding. Let's say you have a 4-pole motor for 60 Hz with a rated speed of 1740 rpm. If you look at a standard torque/speed plot for this motor, you will see torque that is low at low rpms, rising to a peak, then falling roughly linearly to zero at the synchronous speed of 1800 rpm.

Now replace the rpms on the horizontal axis with the rotor slip frequency. It will be 0 Hz at the right end, 2 Hz at the rated speed, 5-10Hz at the peak torque point, and 60 Hz at the left end.

Electrically, this slip frequency, not actual rpms, is the important variable. If you get rated torque at 1740 rpm and 60 Hz input (2 Hz slip), you will also get rated torque at stop from a 2 Hz input.

That's the frequency part. For the magnitude, just realize that the motor is producing a back EMF voltage proportional to its speed, and any current-producing (and airgap-flux-producing) torque must be from voltage greater than that. (The back EMF is modeled as a speed-dependent resistance in the equivalent circuit, which is fine as long as you realize that it is not a real physical resistance.) This is responsible for the V/Hz relationship.

 

[itsmoked]

Great lesson CSW.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[LionelHutz]

I think you have it figured out, but you generally don't adjust the V/Hz to match the required torque.

Loads have their own torque vs speed curve or requirement. It's the amount of torque the load requires just to maintain a certain speed. Any extra torque you apply over this level is used to accelerate the load, always no matter what speed the motor is operating at. Obviously, more acceleration torque will accelerate the load faster.

Now, think of it the opposite way. To accelerate the load at a certain rate requires a certain amount of acceleration torque. So, using the ramp time to set the acceleration rate will determine how much torque the motor has to produce.

Once again, think of the motor as always running at full speed. If you accelerate slowly then the torque required is lower so the motor runs with less slip which means a lower torque and lower current.

The motor doesn't control the torque put into the load - the motor just produces the torque the load requires.