SW inverter to battery cable calculation [RE-wrenches]
Joel Davidson
joeldavidson at earthlink.net
Wed Dec 3 20:00:57 PST 2003
Hi John,
You know this string started when a dear friend asked me for a simple
explanation for the fuse, breaker and wire loss calculations for the cable
between the SW inverter and the battery. I guess some PV questions do not
have simple answers. Onward through the fog.
Best regards,
Joel Davidson
John Berdner wrote:
> Wrenches:
>
> I spoke to John Wiles today regarding the example in his book.
> The following is his quick response:
>
> J. WIles: See the PV/NEC briefing sheets on the SW2024 that are part of
> my standard presentation.
> CALCULATED DC input currents are about 214 amps at 4 kW, 22 volts and
> .85.
> We measured a SW4024 under these conditions (4 kW and 22 volts) and got
> an average current of 254 amps (so much for 85% efficiency) and an RMS
> value of 311 amps.
> I suspect that the Heimenann 250 amp breaker would eventually trip on
> these currents depending on how it responds, as a magnetic hydraulic
> unit, to the ripple currents. For code purposes, it would have to be
> part of a listed assembly to use the 250 amp rating without an 80%
> derating.
> I have heard that the 250 amp breraker would trip (after a time) at
> Trace when the inverters were being burned in with 4 kW loads.
> Fuses as resistive devices would respond to the ripple. Conductor
> ampacity based on heating would also be based on the RMS value.
>
> >>>
>
> J. Berdner: After receiving the above email, I called John Wiles and
> we discussed inverter current measurement methods. This is presently a
> topic of some debate in UL 1741 Standards development circles. J. Wiles
> said he used a high speed Data Acquisition System (DAS) and then post
> processed the raw current measurement data to mathematically calculate
> the Average and RMS values of the dc current. Obviously this is an
> accurate, albeit costly and complex, way to get the real values of the
> inverter's dc input current. In the real world this is not a viable
> option so how the heck do you measure it becomes a valid question.
> As it turns out, you can't use any old RMS DVM. Most have >> either <<
> dc coupling or ac coupling but not both. The Fluke 180 Series (and
> possibly others) have a special position for RMS measurement of signals
> with combined ac and dc components. With one of these babies all you
> need is a current shunt, a big battery at it lowest voltage, and enough
> resistive load to drive the inverter to full power. Then you can
> measure the current and find numbers that are probably consistent with
> those reported by B. Brooks and J. Wiles. Some of you may be asking why
> all of this matters at all and why inverter manufacturers don't just
> give you the RMS value for the combined dc plus ac ripple current. I
> suspect it is not made more available because, if it were, the Code says
> you can't round the ampacity up to the next breaker size of 250 Amps.
> Instead you would need a 275 Amp breaker and 250 MCM cable as opposed to
> a 250 Amp breaker and 4/0 cable used today. Then 2" conduit becomes a
> problem and the existing wire bending space is an issue, and...
>
> Fear not!, You are not alone in your quest for truth. Many of the
> folks in the Standards development community are asking too.
> I would bet, with long odds, that this information will be a
> requirement for the manuals in the next version of UL 1741.
>
> Best Regards,
>
> John Berdner
>
> >>> allan at positiveenergysolar.com 11/28/2003 12:47:33 PM >>>
> John,
> OK, that helps a little. Wiles addresses it to a degree in Appendix F.
> Using
> the 4024 example, the calculated 267A is less than the 311A RMS value
> of the
> current waveform. Simply extrapolating: 311/267 X 184A = 213.4A RMS,
> which
> is 29.4A additional because of AC ripple, close to the 28A of the
> original
> 5548 calculation.
>
> But it leaves us with a circular (il)logic: Wiles' last paragraph reads
> "The
> systems designer should contact the inverter manufacturer in cases
> where it
> is expected that the inverter may operate at loads approaching the
> full
> power rating of the inverter. The inverter manufacturer should provide
> an
> appropriate value for the dc input current under the expected load
> conditions". Yet this whole discussion began because Joel "asked
> someone at
> Xantrex technical service for the fuse, breaker and wire loss
> calculations
> for the cable between the SW inverter
> and the battery, but the person I talked to did not know the
> calculations."
>
> It amazes me how one company's reputation can change so drastically and
> so
> quickly.
>
> Allan at Positive Energy, a Certified Dealer
>
> --
> [Non-text portions of this message have been removed]
>
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