[RE-wrenches] Power Factor (was Efficient step up, down transformers)

[Joel Davidson]

Power factor less than 1.0 can shorten the life of and damage equipment and 
reduce PV production. We installed a 200 kW PV system on a facility with a 
lot of 3-phase motors and monitored the PV and the grid. We measured around 
0.7 PF being delivered by the Los Angeles Dept of Water & Power. The data 
was used as evidence by the PV client to negotiate compensation for lost 
power and get better quality power at that site and other facilities. Now I 
always monitor utility power on commercial systems for trouble-shooting.

Joel Davidson

----- Original Message ----- 
From: "Exeltech" <exeltech at yahoo.com>
To: "RE-wrenches" <re-wrenches at lists.re-wrenches.org>
Sent: Tuesday, July 26, 2011 12:13 PM
Subject: Re: [RE-wrenches] Efficient step up, down transformers


Hello Ray,

Wow.  I could do an another thesis on the topic of
power factor alone...

There's no good way to fully explain this in depth
here .. so all we can do is scratch the surface.


First, a quick review.

Power factor less than 1.0 causes current flow in a load
that's out of phase with the applied voltage.  The actual
power delivered to the load (and thus consumed by the
source, disregarding efficiency for now), would be:
Vac * Iac * COS(theta) .. where theta is the phase angle
differential between the voltage and current (with the
values for V and I in rms).

Though the real power consumed in a reactive circuit
is going to be less than the real power consumed in
an equivalent resistive circuit (by virtue of the
phase angle), the amperage in the reactive circuit
is still whatever is flowing.

Now, to answer your question...

All inverters will operate at a lower efficiency
when connected to a load with lousy power factor
than if connected to a purely resistive load
under similar operating conditions (VA vs. watts).

The actual efficiency reduction will depend also
on whether the low power factor is due to reactive
components (e.g. inductors or capacitors), or
due to a non-linear load, such as a switching power
supply in a computer, microwave oven, etc.

Nonlinear loads present their own issues that I'll
skip for now.  Let's presume reactive...


That said...

The majority of the losses in inverters under
these conditions would be resistive.


Here's why:
Inverters made with power transformers have large
amounts of copper wire, and consequently, wire
resistance.  When connected to a load with a poor
power factor, transformer-based inverters will not
perform as efficiently as they do when connected
to a purely resistive load.

With a very low power-factor load, you could
theoretically be providing the max current from
an inverter at much lower real power output than
it's rated.  This additional loss, when taken
into account at an AC power with low real wattage
output, will make the inverter's efficiency
worse than with a resistive load.  Expect the
efficiency hit to be on the order of 2-5% for
transformer-based inverters (again, depending on
inverter's max wattage rating as well, as this
will impact the wire size used in its construction,
and thus the total wire resistance).

Transformerless inverters don't have this large
mass of wire (thus much lower internal resistance).
Subsequently, they aren't nearly as affected by
low-power-factor loads - but even here, I'd expect
around 0.5-1% efficiency reduction, all other
conditions being equal.


Power factor is deceptively simple .. but is one of
the more complex aspects of power systems and
power equipment design.


Hope this helps.


Dan Lepinski
Sr. Engineer
Exeltech


P.S. Thanks for being a long-time Exeltech customer.
Much appreciated.



--- On Mon, 7/25/11, R Ray Walters <ray at solarray.com> wrote:
> From: R Ray Walters <ray at solarray.com>
> Subject: Re: [RE-wrenches] Efficient step up, down transformers
> To: "RE-wrenches" <re-wrenches at lists.re-wrenches.org>
> Date: Monday, July 25, 2011, 4:10 PM
> Hi Dan;
>
> I agree that a 2000 VA inverter isn't equal to a 2000 watt
> unit.
>
> The focus on VA here though was not about looking at the
> rating of the inverter, but instead noting that a load with
> a high reactive component might draw more energy than a
> purely resistive load in a battery based inverter system. AC
> watts might be equal, but my understanding (perhaps false?)
> is that the inverter efficiency decreases with PF.
>
> SInce you're in the inverter manufacturing biz, perhaps you
> could better explain this to the rest of us.
>
> My fundamental question is: how does power factor effect
> the efficiency of an inverter?
>
> A better understanding of this issue would not only help us
> with the current issue of transformers vs. big wire,
> but also help us better size battery based systems to all
> types of AC loads.
>
> Thanks in advance for your time,
>
> R. Walters
> ray at solarray.com
> Solar Engineer
> (and long time Exeltech user)
>
>
>
> On Jul 25, 2011, at 1:41 PM, Exeltech wrote:
>
> > Focusing on VA ratings would not be advisable.
> >
> > Here's why.
> >
> > Watts are real power. This would be actual DC
> > volts times actual DC amps. Measuring DC volts
> > and DC amps as you're presently doing is a good
> > approach. It yields a valid result.
> >
> > On the other hand, volt-amps ("VA") are nothing
> > more than real AC watts at some unspecified power
> > factor. Without knowing the associated power
> > factor, a VA value is meaningless.
> >
> >
> > For example:
> > Let's say I'm a customer, and I tell you I need
> > a 1250 VA inverter.
> >
> > You offer to sell me a 625 watt inverter. Will
> > it work?
> >
> > Answer: Only if it's a true sine-wave inverter
> > and my load power factor is 0.5 or worse (<0.5).
> > I'm ignoring the possibility the inverter may not
> > be designed to work with a load power factor that
> > low .. but that's another topic.
> >
> >
> > Example #2: a 1 kW-rated sine-wave inverter
> > operating with a resistive load that has a PF of
> > 1.0 is truly generating 1,000 watts.
> >
> > At a load PF of 0.8, it becomes a 1250 VA
> > inverter. At a PF of 0.5, it's magically a 2000 VA
> > inverter.
> >
> > Unfortunately, "VA" is used by marketing
> > departments to> inflate the actual output wattage
> > and make an inverter (or other device) seem more
> > capable than it really is.
> >
> > In the end, the inverter won't generate more than
> > 1,000 watts, but incorporating a power factor of
> > less than 1.0 into the spec makes it seem to the
> > untrained eye like they're getting more bang for
> > the buck.
> >
> > Real watts = real watts. Period.
> >
> > ---
> >
> > Lars mentioned at the onset of this thread he
> > measured a no-load current of 15.6 amps flowing
> > into the transformer.
> >
> > It's very possible there could be that many amps
> > flowing through the transformer primary to build,
> > then reverse the magnetic field in the core.
> > This would mean, however, the current is 90
> > degrees out of phase with the voltage, hence
> > no real power being consumed.
> >
> > The majority loss at that point would be the
> > resistive> loss in the copper, which would be
> > on the order of a few watts - not kilowatts.
> >
> > If he's in fact measuring real power somehow, and
> > there's no load on the transformer secondary, he's
> > got a defective transformer.
> >
> >
> > Dan
> >
> >
> >
> > --- On Mon, 7/25/11, R Ray Walters <ray at solarray.com>
> > wrote:
> >
> > From: R Ray Walters <ray at solarray.com>
> > Subject: Re: [RE-wrenches] Efficient step up, down
> > transformers
> > To: "RE-wrenches" <re-wrenches at lists.re-wrenches.org>
> > Date: Monday, July 25, 2011, 12:20 PM
> >
> > I thought (based a previous thread) that we needed to
> > be concerned with VA output of inverters, not wattage,
> > at least on the transformer based machines like the SW.
> > I realize this wouldn't be an actual load if on grid.
> > In these situations, I just measure DC amps into the
> > inverter to determine the actual consumption,
> > irregardless of power factor, AC efficiency, etc.
> >
> > R. Waltersray at solarray.com
> > Solar Engineer
> >
> >
> >
> >
> >
> > On Jul 25, 2011, at 9:29 AM, Exeltech wrote:
> > Measuring current without the phase relationship
> > between the current and voltage isn't indicative
> > of the actual power consumed. Likely what you
> > were> seeing is the eddy current in the transformer
> > core.
> >
> > If so, it's 90 degrees out of phase with the voltage,
> > and the real wattage consumed with no load connected
> > is minimal.
> >
> > Dan
> > Sr. Engineer
> > Exeltech



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