VFD Efficiency Measurement Problems

[rcw retired EE]

Our client's VFD specification for some 125 HP, 600V cooling tower fan drives requires a factory load test to demonstrate the VFD full load efficiency is 90% or better. (Motor not included.) The vendor's fabricator has set up a test system with a 125HP AC motor driving a DC motor as a variable load. True RMS multi-function meters on the input and output of the VFD drive measure power in and out.

But the KW in and out readings don't make sense. (KW greater than KVA, KW in less than KW out.) I am guessing harmonics are the problem. I have had them check and recheck the CT and voltage connections.

Now, the vendor is saying it is impossible to measure KW in and out on a VFD in this fashion.

Is there a recommended method of measuring full load efficiency of a variable frequency drive? Will true RMS meters work on the high harmonic content waveforms?

 

[davidbeach]

You'd have to connect the motor to a dynamometer so that you could measure the power out of the motor and then measure the power into the VFD. Unless you can accept the metering of the VFD for the power in the output circuit, you will probably need some very sophisticated metering to know exactly what is happening between the VFD and the motor.

 

[Skogsgurra]

Yes, the dynamometer works best. But I have also used a two-wattmeter setup with classic watt-meters (electromechanical) with 5 A secondary CTs. They seem to be able to handle very bad wave-forms.

The Yokogawa meters are used for real high accuracy.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[itsmoked]

Indeed! As davidbeach sez, and a thorough understanding of the motor and the dyno's own efficiencies.

Personally I think this endeavor is whack. You will not get any meaningful results EVER if you don not spend a small fortune on special equipment and someone who understands it well.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[ePOWEReng]

What type of PTs and CTs are you using for measurement? What type of meters are you using? If I understood your question correctly you don't care about the motor efficiency for this measurement, only the efficiency of the VFD? If this is the case a dynamometer is not useful.

 

[rcw retired EE]

Thanks for the input. The assembly shop just contacted another VFD supplier (major multi-national company) and found they only do watts loss measurements to prove efficiency. Their engineer said we would never get good measurements using true RMS meters.


BTW, the test set up was designed primarily for extended burn-in load tests and thermal cycling. The efficiency test was supposed to be a simple meter reading during the burn-in.

 

[Skogsgurra]

Watt loss measurement is even more difficult. How do they do it?

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[CJCPE]

To measure watts lost, you build a well insulated box and put the VFD box inside it. You duct air in and out of the box and measure the air flow and temperature rise. This is a 125 Hp drive. Say the output is 100kW at full load. Say you measure the losses at 7000W +/-15% or 5950 to 8050 watts. With that, you calculate the efficiency to be between 92.5% and 94.4%. What kind of direct measurement setup would you need to do better than that?

 

[rcw retired EE]

Gunnar - I'm not certain how a watts loss measurement is done. Put the equipment in an insulated enlcosure and measure the internal temperature rise or heat output? The factory engineer implied it was not an easy test. I would think so.

I believe we will settle for manufacturer's watts loss data and efficiency curves. My vendor's fabricator has spent too much money trying to prove the efficiency using expensive meters.

 

[itsmoked]

rcwilson; You do it just as described by CJCPE.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Skogsgurra]

Is building a well isolated box with a fairly well controlled air flow and measuring air flow plus taking two accurate temperatures readings any easier than measuring input and output power?

It all depends on what you already have, I think.

I have done fairly accurate comparative efficiency measurements for fan installations by first running a DOL fan and checking air pressure delta and motor speed. I have then put the inverter in the system and adjusted it to same pressure delta and speed. The increase in input power, which is easy to measure with moderately expensive watt meters, represents losses in inverter (plus PWM losses in motor).

The same technique can be used for other load types.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricuwe]

A value of only 90% does not look very demanding to me for such kind of drive. I would expect a value in the range of 94% to 96% (Assuming Diode Frontend, DC-voltage-link, IGBT-inverter)

What value do you expect from loss calculation and thermal design ? If this result is in that range as indicated above it may be sufficient to do some temperature measurements on main components and compare it to design data to prove that you have achieved at least 90%.

 

[rcw retired EE]

Thanks to all for the advice and education. I had just assumed that a high performance KW or KWH meter would measure power in and out. The drives are assembled in MCC lineups of six drives with an active harmonic filter for the bank of six, and individual input reactors and output reactors and filters and individual bypass contactors. (All needed to meet the client's equipment specification.)

My test with the meters was going to prove the complete assembly’s efficiency, two units at a time.

As suggested, we will use manufacturer's data to prove >90%. If that is insufficient for our client, I can run the fans in bypass mode and measure power input at the 600V substation, then run the same fans on the VFD's at 60Hz. The difference would be the VFD's losses plus any losses due to harmonics.

Thanks again for the help.