SW inverter to battery cable calculation [RE-wrenches]

Fri Nov 28 14:31:21 PST 2003

Our technical support people are generally well versed in cable sizing and
rating requirements for our products. We are following up with Joel directly
to see who he talked to and why he did not get satisfactory follow up.

Regards,

Mark Edmunds
Xantrex Technology Inc.

-----Original Message-----
From: Allan Sindelar [mailto:allan at positiveenergysolar.com]
Sent: Friday, November 28, 2003 12:48 PM
To: RE-wrenches at topica.com
Subject: Re: SW inverter to battery cable calculation [RE-wrenches]

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

----- Original Message ----- 
From: "John Berdner" <jberdner at sma-america.com>
> I have discussed this issue many times with J. Wiles and the guys at UL
> and the rational is as follows:
> Normally we take the maximum ac output Watts / inverter efficiency at
> full power to get the maximum dc input Watts.
> Then we divide the maximum dc input Watts by the lowest battery voltage
> to get the maximum battery current.
> This current can theoretically be a continuous current so we need to
> multiply this number by 1.25 to get the NEC required ampacity for the
> battery cables.
> The problem with this approach is that it does not accurately reflect
> actual in field measurement of the current flowing in the battery
> cables.
> When you make RMS measurements of the current flowing in the battery
> cables you usually see a higher number than you would get from the
> straight dc calculation.
> The difference is due to the substantial 120 Hz ac ripple you see on
> top of the dc current when you measure the vast majority of inverter
> chargers.
> I suspect the 28 Amps John Wiles is talking about in his book is the
> difference between the calculated dc value and RMS value you get when
> you measure it.
> This will be addressed in the next release of UL 1741 which says you
> have to size the battery cables and over current devices for worst case,
> combined dc plus ac ripple, i.e. RMS value, of the current.
> I believe this makes sense because the NEC ampacity tables are based on
> RMS current values for calculating cable heating.
>
> Best Regards,
>
> John Berdner
>
>
>
> >>> allan at positiveenergysolar.com 11/28/2003 10:30:01 AM >>>
> Joel,
> The upper example has been the one that I have most often seen. It's
> used by
> Wiles on page E-16 of his PV/NEC book. If you extrapolate these numbers
> to
> the SW5548, I think you will get 5500 watts divided by 0.85 inverter
> efficiency divided by 44 volts = 147.05 x 1.25 NEC = 184 amps battery
> cable
> rating.
> I have no idea where you got the part about adding "28 amps RMS AC
> current"
> to the DC calculation.
> Allan at PosE
>
>
> ----- Original Message ----- 
> From: "Joel Davidson" < joeldavidson at earthlink.net >
>
>
> > Last week I 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.
> >
> > One system designer uses this calculation for the SW4024:
> > 4000 watts divided by 22 volts divided by 0.85 inverter efficiency =
> > 213.9 (for the 250 amp breaker) and an additional 1.25 NEC = 267
> amps
> > for the 300 amp fuse.
> >
> > Another system designer uses this calculation for the SW5548:
> > 5500 watts divided by 0.8 inverter efficiency divided by 40 volts =
> 172
> > amps DC plus 28 amps RMS AC current = 200 x 1.25 NEC = 250 amps
> battery
> > cable rating.
> >
> > Does anyone know the correct efficiency multipliers for the SW
> inverters
> > and the calculation to use?
>
>
>
>
>
>
> --
> [Non-text portions of this message have been removed]
>
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