[RE-wrenches] Efficient step up, down transformers

[R Ray Walters]

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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