Actual UPS efficiencies don't match factory claims...?

[Anonymous user]

I'm having trouble understanding the Factory/Manufacturers' claims on UPS efficiency.  All that I have seen claim min-90% efficiency of their UPS systems running in full dual-conversion mode.  However, what I am seeing in Actual efficiency (Amps out divided by Amps in) is in the low to mid 80% range.  Anybody have any clue why?  Or, how does APC actually "measure" their efficiency that allows them to claim ~95% efficiency starting at ~45% load.

For example, here locally we are being provided clean power by a bank of 5 APC MGE EPS 8000 625kva, 480v UPS.  They are connected to a common buss and are in an N+1 setup.

The average loading of the individual UPSs is 40-45%.  Based on an APC document I found online, these units should be 90% efficient starting at 20% load, with a peack efficiency of ~95% from 40-70% load.

However, what we are Actually seeing in production is Amps Out divided by Amps In = ~80%. 

I've seen elsewhere where KW Out/In is normally used for efficiencly calculations...but I can't see how that would make any significant difference?

Another, older, Liebert 750kva UPS we have is currently only loaded about 20-25%, and its measured Amps In/Out efficiency is ~85%...which I would expect with such a light loading.

A year old Eaton 550kva (2x 275kva) we recently looked at showed a similar 85% efficiency at ~35% load.

I guess my real question is this:

How do manufacturers measure "UPS Efficiency" in order to get mid-90% ratings?

[voidstar_apc]

Hi RT8,

I'm a little surprised you didn't measure 117% efficiency that way. If you figure the power factor of loads is 0.85 or so, and the UPS input is power factor corrected, then input_amps = output_amps*0.85 => output/input = 1/0.85 = 117%.

So yeah, there's a difference between W and A 🙂

I think of it like this: power is current * voltage, and that's sort of like you need both workers and available tasks to get work done. Both workers and available tasks go up and down with time. If they coincide, lots of work gets done. If you have workers when you don't have tasks and vice versa, no work gets done. So just measuring your average amount of workers and your average amount of tasks available can be misleading. Watts is like measuring the average amount workers who have tasks, while amps is like measuring the average amount of workers.

Even measuring volts or amps alone can get a bit tricky. Cheaper equipment assumes the standard ebb and flow of AC electricity. True-RMS meters make no such assumptions.

[BillP]

Schneider Electric measures UPS efficiency according to the procedures specified by IEC 62040-3 Ed. 2 and/or ENERGY STAR for UPS version 1.0.  Both standards are similar in that they measure at 100%, 75%, 50% and 25% balanced resistive (PF=1.0) load, with nominal voltage and frequency input, at a temperature near 25ºC, with a fully charged or disconnected battery (to represent steady state conditions). The IEC standard defines efficiency as the ratio of power out to power in. ENERGY STAR calculates efficiency as the ratio of energy out over energy in. Transformerless UPSs and newer server power supplies have power factor corrected inputs which typically achieve power factors of 0.95 or greater when operating at > 25% load.

Some of the UPSs you mentioned may have input power factors substantially below 0.95, resulting in additional current flow without additional power flow. Your load may not have have a high power factor either. Measuring power accurately requires a true RMS power meter capable of measuring voltage and current simultaneously

In summary, you can't determine efficiency if you only know voltages and currents. You need to know input power (or energy) and output power (or energy).

[Anonymous user]

Angela N.

Thanks!

The vast majority of the load on the 5x APC string is predominantly computers, most of which have a PF in the high .9s.

I have no idea about the PF of the 3 UPSs I mentioned.

Ok, so now I need to collect input & output KW?  or KVA?  Which one is considered "power"?

[voidstar_apc · Accepted solution]

You want W (or kW) for your efficiency calculation, not kVA. Watts are real power.