Variable Speed Drive

[aida2011]

Hi,

I am a mechanical engineer with limited knowledge in electrical engineering.

Please someone help me, can we measure power usage (in kW) for machines using VSD by using this formula?

P = 1.732 x Volt x Ampere x Pf

The ampere (A) is obtained from the VSD's panel view.

Thanks

Miss Aida Hanani
Malaysia

 

[ScottyUK]

Can you also read the voltage and PF from the VSD display? You need all three parameters, measured at the same point in the circuit. Measurements on VFD outputs are not simple and without the correct equipment and measurement technique you will get erroneous results.

Most VFDs can provide either a display of true power output, or an analogue output signal proportional to true power. What VFD are you using?

 

[waross]

The basic formula is Volts x Amps x Efficiency x Power Factor = mechanical Watts output.
Line fed motors experience displacement Power Factor. (Absent harmonic distortion on the incoming supply.)
VFD fed motors may also suffer from distortion Power Factor. The rated PF of the motor will not include the losses of distortion Power Factor.
I agree with ScottyUK.

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

 

[jraef]

A Variable SPEED Drive could technically be anything, such as an AC VFD, a DC drive, even a mechanical vari-drive. Having a display implies it is electronic, but still could be AC or DC. This only matters in terms of WHAT power you are referring to. If you are in North America where the output shaft power would be described in HP, your reference "as in kW" would then imply the electrical power CONSUMED by the complete circuit. Just be aware however that in most places in the rest of the world, kW is also used to describe the output shaft power. That's where the TYPE of VSD becomes relevant.

So assuming you meant CONSUMED electrical power in kW, then you must add the losses in the drive itself, and understand the complexities of how Power Factor is different in an electronic drive system and that the drive losses as a percentage of the total losses increase as speed decreases. That's why, as ScottyUK said, you want to get that more complete picture directly from the VSD if it is available. Most good quality drives have all of the necessary capabilities to do that inherent to their ability to function correctly in the first place, but doing it externally requires attaining that capability to handle those complexities, which means a very expensive meter.

If you have a cheap drive what does not provide true power metering, then the best you can do is an educated guess using assumptions. You will not care about the actual power factor and efficiency at the motor, those are already accounted for in your drive current reading. So you can relatively safely assume that the drive input power factor is .90 and you can assume the drive throughput efficiency is 90%. Both values will probably be better at full speed, but not likely much worse.


"Will work for (the memory of) salami"

 

[aida2011]

I am sorry for having to ask this simple question to you all fine engineers.

I have done field data collection for power consumed in kW for several machines, of which some of them are running on VSD. I asked a senior electrical engineer to validate the calculation. However, he said calculation for power from machines with VSD is not as simple as using the formula. He asked me to revise my calculation, giving me no direction whatsoever.

I want to know whether the claim by the electrical engineer valid or not, and if yes, what formula should I use to calculate power consumed by machines using VSD.

Thanks for your concern, everyone.

Best regards

Miss Aida Hanani
Plant Engineer

 

[jraef]

Well, technically the FORMULA is still valid, although your formula is for calculating MECHANICAL power, not ABSORBED (consumed) power. To get that, you must also include efficiency.

Consumed 3 phase kW = 1.732 x A x V x PF x Eff.

But aside from that it is the data collection that becomes problematic, not the formula itself. Assuming (because you are asking) that you do not have a proper kW meter or cannot read it from the VSD* directly, then how are you going to know what the power factor and efficiency is for your formula? That is where the problems come in and there is no simplistic formula to help you determined that.

* Again, just to be pedantic, VSD means nothing specific here...

"Will work for (the memory of) salami"

 

[Skogsgurra]

Jeff, you need to be careful with your overunity ambitions!

"the division key on my ancient desk calculator is stuck!" doesn't mean that a bad efficiency reduces absorbed power. You still have to divide by efficiency for that. You know that better than most guys - but seem to have a problem with it IRL

And, Aida, what Jeff means by saying that VSD doesn't mean anything specific is that the use of VSD for "Variable Speed Drive" (which is common in general usage) perhaps should be replaced by VFD for "Variable Frequency Drive" so that it is clear what technology is being used to make the speed variable. It is a small detail, but it may be important in order to avoid mistakes.

Gunnar Englund

http://www.gke.org

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

 

[jraef]

Ok, I need to step away from the PC for a few days, maybe take a remedial math class...

"Will work for (the memory of) salami"