Calculating Amp requirements when sizing a VSD. 4

[pirates]

So I have a reciprocating refrigeration compressor that is attached to a 250kW motor. For various reasons it has been decided that the compressor should no longer be run at 1480rpm and that it should be stepped down to 1000rpm. For various reasons I ended up with a drive which I believe is too small. Whenever I have asked various people if there still is a workable solution the usual answer has been “try to determine what the motor amps are for long term operation and what they might be for short term upsets etc. compare this with the drive manufacturers claims” Here is where I came stuck not really knowing how to work this out. Now I think I have the answer and really need clarification as to if my workings (methodology) are correct.

From the compressor manufacturer I obtained Absorbed power figure(Pe=164.4kW) at the RPM (1000) that we want to run this compressor. Using these figures, the electrical data from the motor ( 250kW 400VAC 50hz 440A 1480rpm cos0.86 efficiency0.95 or 95%)
and the formula Amps = I deduced the following: guessing that we would need a frequency of around 34Hz to attain 1000rpm (IS THIS CORRECT) and keeping the same Volt/Freq ratio as 400:50 I came up with a volt/freq ratio of 272:34 substituting these figures I calculated out that amps needed were 427 amps.

IS THIS CORRECT??

If this is the correct answer then I now know that the drive is way too small. The other problem is the short term upsets. The drive manufacturer has overload claims of 139% for 60 seconds but this is at 400volts. So how do I bring this down to the 272volts that the drive should be operating at?

I would really appreciate any help with this problem.

 

[DickDV]

pirates, I would calculate the lower speed amps a little different.

The load at 1480rpm is assumed to be 250kw. Since kw is proportional to both speed and torque, at constant torque the kw would drop 1000/1480 X 250 = 169kw. Since this is essentially the same as the 164kw stated, one can conclude that the load torque doesn't change much by slowing down to 1000rpm.

As to current, since the load torque is about the same at the two speeds, you can expect that the motor lead amps will also be about the same. That means your drive needs to have an output current rating of at least 440amps.

You may have another option if the drive is too small. If the motor can be belt or chain connected to the compressor, you could increase the pulley ratio so the motor stays at full speed while the compressor slows to 1000rpm. This will reduce the torque on the motor and the amps will fall about in proportion since we are near full load. If you aren't looking at further changing the speed, then the VFD might not be needed at all.

 

[Skogsgurra]

Hi,

I wouldn't complicate things more than necessary.

You have the new absorbed power 164.4 kW. The new speed is 1000 RPM. That figures quite nicely since 250 kW times speed ratio 1000/1480 = 169 kW. Which indicates that power is quite linearly dependent on speed - and that is typical for reciprocating volume deplacement devices.

What can your 250 kW motor deliver at that new speed? The answer is that it can deliver about the same power as you need. With constant torque, which is what your compressor seems to have, it can deliver 250 kW times speed ratio. And that is exactly what your load demands. No problems at all, I would say.

The speed is reduced from 100 percent to 68 percent and that usually does not mean that your cooling gets reduced so that you need to reduce the output power of the motor.

Sorry, I cant see your problem. What do you think that it is?

Gunnar Englund

http://www.gke.org

 

[Skogsgurra]

Ok, so it's the drive electronics you are worried about? How big is it? I mean, if it would have worked with 1480 RPM, then it surely will work with 1000 RPM? No?

Gunnar Englund

http://www.gke.org

 

[Marke]

Yes, I agree,
Your torque requirement is essentially the same, so the current requirement is also the same. You need to select your drive for the full load current rating of the motor.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[pirates]

I must thank DickDV , skogsgurra and Marke for their input However this still leaves me in the dark.

I have a reciprocating refrigeration compressor which in normal operation at 1480rpm would absorb about 258kW calculating that out using Amps =
This equates to 455amps. Which is borderline for motor on both the kW at 250 and the amps at 440. Again this is at a worse case scenario so it would work quite well.

However the powers that be have decided that for mechanical reasons they now want the compressor to run at 1000rpm. So they hired a “knowledgeable contractor” to install a two speed motor. He convinced them that the cheapest alternative would be an AC drive and that is what he installed. However he never commissioned the drive and they asked me to sort it out. This is where I asked the question thread237-117167

So after much deliberation I finally got to figure out that the drive is too small. So we have a very sophisticated PLC and Scada system which we can set-up to make sure that the compressor stays within it’s absorbed power limits by rather loading or unloading the compressor under certain limits and keep it running at the required 1000rpm.

So I can use the compressor manufacturers data compilation programme to tell me the absorbed power under certain operating conditions. I then need to know how to convert this into the AMPS so that this can be compared to the data that is available from the AC Drive manufacturer.

I cannot change the drive I need to work with the equipment that we have.

Please help!

 

[Skogsgurra]

No. You don't have to work with what you have.

If it isn't possible, then don't.

But, there are many nuances of black. You may be able to run the compressor with limited output as you say. But, is it a good solution? Will it work also when air demand gets higher?

I would question the whole design process. Why on Earth have they decided to run the compressor at lower than design speed? Vibrations? Resonance? Or something else mechanic? If that is so - and if you were shorted out in the process of selecting drive and motor - then I think that you have to tell these folks that they either fix the mechanical problem or get a new drive. The supplier will probably understand the problem and offer you a larger one at some extra cost - he might even be willing to take the small one back without any discussion.

Have you tried that?

Gunnar Englund

http://www.gke.org

 

[Marke]

Hello pirates

OK, so you have an under sized drive and you want to know what maximum power you can get out of it on you compressor application at 1000RPM?
As was discussed above, the current at 1000RPM and 164 KW would be in the order of around 440A.

I think that you can assume an aproximately linear relationship between current and power down to less than half of the rated torque. For example if the drive has a rating of 220A, then I would look for around 82KW maximum load at 1000RPM.
Depending on the motor, you will have a magnetising current in the order of 20%-30% of the rated current. Provided that the motor current is signifantly higher than the magnetising current, I believe that a linear relationship is not inappropriate as a means of estimating the power draw.
The current rating of the drive will remain the same at 34Hz as at 50Hz.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[DickDV]

How about telling us what the drive output rating (current and KVA) is and we can probably help you estimate how far you must derate the system.

 

[pirates]

Thank You to all who have submitted comments etc. I appreciate it. To DickDV I have found out the information that you said I should submit (kva). This comment put me on a path and now I think I can work it out. Can you tell me if I am on the right path or not?

Motor = (250kW 400VAC 50hz 440A 1480rpm cos0.86 efficiency0.95 or 95%)
VSD = 200kva
Compressor = 185.6kW at 1000 rpm

Using the formula rpm = (120 x Hz)/No.of poles I conclude the frequency to be 33Hz.
Using the volt/freq ratio I deduce 1Hz = 8volts
Therefore 33Hz = 264volts
I substitute this into the formula Amps = (KW*1000) / (1.73*E*pf*eff) and come out with 496amps.

Using this figure and the formula for kVA = (I*E*1.73) / 1000 I get 226 Kva.

So therefore the drive is way too small.??

Am I on the right path? if so then I will be able to put in some further variables to calculate out what the derating should be.

Many thanks

 

[jraef]

You are on the right path. You drive is tooo small for this application. Most likely, the contractor bought the drive assuming it was a "Variable Torque" application, where the power used would decrease by the cube of the speed, so the drive was essentially derated for that. If the 185.6kW came from your compressor mfgr is correct, that is more linear, in fact even a little higher than linear (which would have been 165.6kW). So your application is definately what would be called Constant Torque, and the drive should have been sized based on what you have just done.

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[aolalde]

I agree with Jraef your driver is short in power and current capacity. Since your load is constant torque, the current is around constant while the Volts/Hertz ratio is kept constant. Note that the motor efficiency (EFF) and power factor (PF) are not necessarily constants all the range of speeds, but should be close to the nominal at 250 kW, when V/Hz is constant.

It is important to know what the VSD current capacity is.

To de-rate the system you have to modify the compressor setup to reduce load. Lets say the output pressure reduces until the torque required drops to match the VSD current capacity. The motor current will follow the torque on its shaft as it is demanded by the compressor.