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Power Factor in determining the battery size

Power Factor in determining the battery size

Power Factor in determining the battery size

(OP)
Hi Guys,

Need some advice on a design stage of 240VAC UPS system.
I have 4 random loads of 19.8kW for the 1st 30mins, 10.85kw for the 1st hour 4.9kW for the next 240mins & 4kW for the final 480mins or 8hrs.

From the formula of Kilowatt = V x I x .8 pf x 0.91 inv.eff

Current I = kilowatt / V x pf x eff

using the IEEE 485 formula i have to convert first the kilowatt load into ampere.
My problem arises when the vendor submitted its sizing calculation, the pf was not taken into consideration or the load (in kilowatt) was converted to current using only the final discharge voltage of the battery bank and the efficiency of the inverter. Am I missing something here?
I believe the pf should also be taken into consideration.
please advise. Many thanks.

RE: Power Factor in determining the battery size

As you know, a complex load consumes active power and reactive power.

The active power (the kW) drains the battery.

The reactive power (the kvar) is exchanged between load and UPS sixty (or fifty) times per second. Half period reactive power is delivered to the load and the next half period, it is given back to the UPS.

So, the UPS vendor knows his business. That is why PF is not in the calculation.

You can also apply a simple energy view on this: If no energy is consumed by the reactive part of the load, then no energy is taken from the battery. No energy taken - no discharge.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Power Factor in determining the battery size

(OP)
Thanks skogsgurra, but the problem is the min discharge voltage of the battery bank which they say is 205.2VDC. but when I check their calculations for the 1st period (19.8kw x 110% spare = 21.77kw or 116.58)their conversion is 118.6A with a Kt factor of 7.75 giving 919.11amp-hrs. reversing the processing we can get 201.7 Vmin discharge, with an %eff of 0.91 the conversion does not jibe. but they are always saying that it is software generated.
I also agree with you they know their bizniz but they couldn't justify what they have in their calculations.
Thanks anyway will have to grill them more.

RE: Power Factor in determining the battery size

I'm not sure that's true skogs.

You can't take anything from the battery without going thru chemistry which is lossy.

It is common experience that on smaller UPSs that if your PCs have PF corrected supplies you get nearly double the run time off a UPS.  I have seen this many times.  Not sure why bigger UPSs wouldn't exhibit the same result.

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

RE: Power Factor in determining the battery size

Yes Smoked, that's right. There are always losses. But they are a "secondary effect". And there surely is a huge difference between a smooth sine-shaped mains current and the four to six times RMS current peaks that you feed into a PC PSU. And that creates lots and lots of losses.

But that isn't reactive power - only a bad power factor as defined by P/S and not by cos(phi1). The former is called distortion power factor. Distortion power factor has no vars so there are no vars to give back to the DC link either. That makes an UPS that supplies simple PC PSUs discharge its batteries a lot faster than one would expect looking at RMS current alone.

An induction motor never has that kind of distorted waveform, only a (slightly) higher current consumption due to displacement. I would not compare the influence of two so different loads.

The OP didn't say what kind of PF he is talking about. If it is distortion PF - then he has all reasons to question the UPS supplier's calculation. I didn't think of that possibility.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

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