Calc Power for 3 phase power supply

[ENG2009]

I am trying to calculate the efficiency of a 3-phase 480V power supply that outputs 10.5 volts DC at 840 amps. The purpose is to see how much we can save in electricity by using a newer more efficient power supply. We are not charged for power factor by our electric utility. I've got a Fluke quality analyzer logging the inputs, which are:

Volts1 = 480.9
Amps1 = 17.29
Volts2 = 473.5
Amps2 = 17.4
Volts3 = 477.0
Amps3 = 14.6
Power Factor = .72

I want see how efficient the P/S is by comparing the input watts to the output watts. I wasn't sure if I needed to take into account power factor. I calculated:
Input KW = ((V1*A1) + (V2*A2) + (V3*A3))/1000 = 23.56KW
Output KW = (10.5 VDC x 840 Amps) / 1000 = 8.82 KW

The efficiency would then be 8.82/23.56 = 37%. This means that 63% or 14.7KW would have to be turned into heat, sound and vibration by the power supply. That sounds like way too much wasted energy. This power supply is about the size of a full server rack and it gets warm, of course, but not that hot. Am I missing something?

 

[jghrist]

You need the power factor to get from 23.56 kVA apparent power to the kW real power.

 

[itsmoked]

You need a proper power meter. The power factor is going to be distorted.

I suggest you rent one or buy a watthour meter then divide by the time.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[ENG2009]

I'm using a Fluke 435 Power Quality Analyzer; I think this should record all the data I need. Why will the power factor be distorted?

 

[itsmoked]

Because this isn't like a leading or lagging capacitive or inductive load with a uniform lead/lag. It will be a rectifier load that pulls power from the source in narrow gulps only at the points of maximum voltage. This will also change with loading.

Doesn't that Fluke have a POWER setting? Do not just multiply the voltage with the current as you will get garbage. Internally that meter will multiply the the instantaneous voltage with the instantaneous current and supply the actual watts.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[ENG2009]

Fluke softrare does have a power section. Below are links to three screenshots: Active Power, Reactive Power, Apparent Power

http://files.engineering.com/getfil...9e0-8427-0e4f195cb034&file=Reactive_Power.jpg

http://files.engineering.com/getfil...-494d-b46d-f383b340c76d&file=Active_Power.jpg

http://files.engineering.com/getfil...0ea-a36f-55365120a40a&file=Apparent_Power.jpg

 

[jraef]

Your meter is more than capable of measuring Power all by itself, you don't need to go through those gyrations. but read this app note, it discusses some of the potential pitfalls of working with VFDs. This meter will READ the displacement power factor but no mention of distortion power factor. It does imply that you can determine it my subtracting the displacement value from the overall PF value, but it doesn't say that it compensates for them automatically when calculating kW.

http://support.fluke.com/find-sales/Download/Asset/3433274_6112_ENG_A_W.PDF

"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[ENG2009]

Thanks for finding this article. I don't know if our power supplies behave the same way as a VFD or UPS, I'm guessing they probably do. I'll check with the guy who builds them. Do you know if I'd be able to use the Fluke Meter to verify if the P/S is actually pulling power from the narrow gaps at maximum voltage.

 

[rbulsara]

ENG2009:
Even as you heed other folk'd advice here, there is basic error in you formulas.

The KVA power of 23.56kVA is incorrect. For your method of adding power in each phase, you need to use per phase (phase to neutral) voltage, which is 277V or (Line to line voltage)/sqrt of 3. 480V is the line to line voltage.

or in other words you input kVA will be 23.56/1.732 =13.6kVA.

If your pf of 0.72 is correct then the input kW will be 13.6*0.72=9.8 kW. Now recalculate our efficiency!

 

[rbulsara]

Also here is a relevant thread, posted a few days ago:
thread238-246755

 

[elecedge]

The Fluke quality analyzers that I am familiar with do not measure DC current so I am not clear exactly what you are measuring. Could I hazard a guess that you are not measuring the DC output but rather assuming that the maximum output of a 1000A is being drawn when the input power is 23.56 kW? The input power will vary as the output power changes, so it seems to me that you need to measure both at the same time to get accurate results. The Fluke Norma product line is I believe designed to do just that.

 

[ENG2009]

With the Fluke I'm measuring the 480 VAC input. The DC output is constant and the power supply has DC Volts and DC Amps gauges built in. So the DC output wattage is easy to calculate.

--Scott

 

[rbulsara]

The first error is in the calcs. If there is measurement error it will be on top of that. See previous posts.

Rafiq Bulsara

http://www.srengineersct.com