Brian Teitelbaum wrote:
boB,

Not only does that make sense, it's the best explanation of the effect of power 
factor on an off-grid system that I've read.

I would than assume that this means that Drake's refrigerator is actually consuming more 
power from his battery system than is just measured in AC watts at the fridge. Some 
energy is lost as heat in the inverter system while the fridge is running, at least more 
than would be if the fridge had a higher power factor. I would also assume that measuring 
the DC amperage in the battery cables, and subtracting the "normal" inverter 
losses, would give a more accurate reading of the Fridge's true running watts.

There are a number of plug-in products on the market that are supposed to 
improve power factor, apparently through added capacitance. How would they work 
here, or would they?
Great question and I'm not sure I have a good answer.

You would have to modify the refrigerator itself to make it have a higher power factor, and I would hope that the appliance manufacturer has designed it to work as good as it can. Are those high efficiency refrigerators power factor corrected though, since Europe requires power factor close to 1.0 and the good ones
are also used there ?

I've seen many discussions about those power factor correction units for home use, and I think the consensus is that they really don't help much, if at all....
...Mainly because the homeowner only pays for real power.

I think that the higher the inverter efficiency, the better it will work with lower PF loads, because it has to keep its AC source impedance low when powering that refrigerator (or whatever), because it has to circulate and "slosh" around those reactive currents.

Lower resistance =equals= higher efficiency, =equals= better low power factor performance, is one way to look at that.

Adding extra and opposite reactance for power factor correction is something that utilities do at large reactive consumers such as steel mills. It makes sense for that type of power usage. BTW, if the primary power factor is low because of big inductive loads, they add capacitors, and if capacitive, they add inductors to pull the power factor back towards 1.0. At least that's the "passive" method that I'm familiar with for low Displacement Power Factor, (DPF).

But what happens if you have all that passive PF correction in place, and all of a sudden the load reduces or goes resistive ? Then you're left with a nice low power factor load again. (I think they take that into account and make it average out).

Then, there's active, electronic PFC. I probably won't be around to see, but I bet that high temperature
superconductors will play a big roll in reducing all losses.

boB

Brian Teitelbaum
AEE Solar



-----Original Message-----
From: [email protected] 
[mailto:[email protected]] On Behalf Of boB Gudgel
Sent: Wednesday, August 19, 2009 9:28 PM
To: RE-wrenches
Subject: Re: [RE-wrenches] power factor

Drake Chamberlin wrote:
One question about power factor:

We know it is the ratio of the real power to the apparent power, and
that most utility meters only register real power (watts).  Reactive
power doesn't dissipate as heat like power loss through a resistor.

The question is:  What is the quantitative effect that reactive power
has on a battery bank that is feeding loads through an inverter?  If
the power factor is 56% (like a refrigerator that I want to replace)
does that mean that only 56% of the power being drawn from the
batteries is being used to power the refrigerator?.  If so, where does
the rest of the power go?

If you just take your meter and multiply Volts X Amps, that's VA, or
apparent power.   If it's all real, then it's like the resistor and gets
counted by
the utility meter.  Some turns into heat (resistor) and some gets some
real work done.

The rest, is VA but is Reactive and is  VA Reactive or VARs.

So, the reactive current that comes out of the batteries, into the
inverter out the AC output, comes BACK into the battery.  (this is for
an inverter
that has the ripple reflecting back into the battery)

So, your batter kind of discharges and then recharges 120 times per
second for a 60 Hz system.
So, where does the power go ?   The real power gets used of course and
most of the rest, the VARs gets put back into
the battery.   HOWEVER, since the transformer and wire and FETs and
transormer and stuff have resistance, some of those VARS get turned
into heat by that inefficiency, or I-squared-R losses.  That's one
reason why low power factor isn't so good.

Does that kind of make sense ?

boB

Thanks,

Drake Chamberlin
Athens Electric
OH License 44810
CO License 3773
NABCEP TM  Certified PV Installer
Office - 740-448-7328
Mobile - 740-856-9648

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