Calculating string amperages [RE-wrenches]

Ray Walters walters at taosnet.com
Fri Feb 8 11:11:11 PST 2008 

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Well we do have some discrepancies here. William was saying we could  
go to 3 parallel Solar World strings, John Berdner here says only  
one, and I'm still saying 2 is good.
John I have 2 issues with your excellent analysis.
1) 690.8 B) (1) exception: "Circuits containing an assembly, together  
with its overcurrent device(s), that is listed for continuous  
operation at 100% of its rating shall be permitted to be utilized at  
100% of its rating."
This means that with the right breaker, we don't have to use the 1.56  
multiplier for array current, only 1.25 which is for edge of cloud  
effect, etc. I have seen Solar world modules exceed their nameplate  
rating, so this multiplier is appropriate.
2 x 5.1 A x 1.25 = 12.75 Amps on a 15A breaker that is rated for 100%  
duty is acceptable. (most Outback and Midnite solar offerings are  
100% rated BTW)

2) All this argument is based on the possibility of back feed from  
the grid or batteries, but my understanding is that both grid tied  
inverters, and charge controllers prevent back feed. So I don't think  
this is applicable to most work that we are doing. (I am aware of  
relay type controllers APT... that could allow backfeed possibly?) I  
suppose with large enough arrays, the backfeed would be from adjacent  
subarrays?

So unless I'm missing something (please help me out here) I think I'm  
going to keep paralleling the SW165s in lower voltage off grid  
systems. I like to set up 4 modules per breaker, with the modules  
wired 2 in series, 2 parallels. If I'm running series strings over 48  
v nom., I'll take another look at not paralleling.

Ray Walters

On Feb 7, 2008, at 3:10 PM, John Berdner wrote:

>
> William/Wrenches:
>
> The "other" they are talking about in 690.8 is the same as the "n-1"
> parallel strings in the fuses white paper I did.
>
> Consider a system with n strings in parallel feeding a single over
> current device.
> If you were to create a fault in one of the strings the faulted string
> does not contribute back flow current to itself.
> The fault current comes from the other strings that are connected in
> parallel, i.e. n-1, and from current flowing backwards through the  
> fuse.
>
> For the n-1, or other parallel connected, strings we have to add a
> safety factor of 1.25 for high irradiance. We do not have to consider
> the second safety factor of 1.25 for continuous duty.  (Why the  
> fault is
> not considered as possibly continuous still alludes me.)
> The resulting fault current is as described in 690.8: Ifault =
> ((n-1)*Isc*1.25 )+the rating of the over current fuse.
> Since the Code requires the over current fuse to be sized for a  
> minimum
> of n*Isc*1.56 this leads to a minimum wire ampacity / UL series fuse
> rating of >= ((n-1)*1sc*1.25)+(n*Isc*1.56).
>
> Using the example from another recent post with two parallel  
> SolarWorld
> modules with an Isc of 5.1 Amps and a UL series fuse rating of 15  
> Amps:
>
> 1) The fault current from the "other parallel connected strings" is
> 1*5.1*1.25 = 6.375 Amps.
> 2) The minimum over current fuse size / wire ampacity required by Code
> for 2 strings would be 2*5.1*1.56 =  15.91 Amps. Unfortunately the
> standard fuse sizes are 15 Amps (too small) and 20 Amps (next size
> larger) so the Code would require a 20 Amp fuse.
> 3) The total available fault current would then be 6.375 Amps for the
> modules plus another 20 Amps for the fuse or 26.375 Amps.
>
> Since the UL series fuse size for the module is 15 Amps there is no
> Code / UL compliant way to connect 2 parallel strings of this  
> module to
> a single fuse in an application where it is possible for current to  
> back
> feed the fault through the fuse.
>
> The minimum fault current of 26.375 Amps also applies to the wire
> ampacity of the module interconnecting wires per 690.8.  Since most
> modules have AWG 12 leads this would not pass 690.8 regardless of  
> the UL
> series fuse rating.  With a possible fault current of 26.375 Amps  
> AWG 10
> is marginal and likely would not fly after ambient temperature  
> derating
> even with 90C wire.  Modules with AWG 8 leads ?
>
> I hope this helped to add clarity rather than muddy the waters.
>
> Best Regards,
>
> John Berdner
>
>
>>>> wrmiller at charter.net 2/5/2008 01:23:54 PM >>>
>
> Jay:
>
> Begin citation:
>
> 690.8 (D) Sizing of Module Interconnection Conductors. Where a single
> overcurrent device is used to protect a set of two or more
> parallel-connected module circuits, the ampacity of each of the module
>
> interconnection conductors shall not be less than the sum of the  
> rating
> of
> the single fuse plus 125 percent of the short-circuit current from the
>
> other parallel-connected modules.
>
> End citation
>
> Note above where it says "other."
>
> What is not totally clear to me is that in an installation without a
> combiner, i.e. connecting leads through a Square DU HU361 directly to
> the
> inverter, are we meeting the "single overcurrent device" test of the
> above
> section? It seems to me that the internal current limitation
> characteristic of the modules makes the overcurrent protection
> inapplicable.
>
> This code interpretation is explained pretty thoroughly and
> convincingly by
> John Berdner in the SMA white paper. If you have lost the link, I have
> it
> on our web site:
>
> http://mpandc.com/resources/resources.html
>
> This paper has been universally accepted in the industry and has  
> passed
>
> muster by the local building inspector code wonk guys. I used to  
> submit
> a
> copy with permit applications to forestall questions.
>
> William Miller
>
>
> At 08:05 AM 2/5/2008, you wrote:
>
>> Hi William,
>>
>> Where is this subtraction of 1 string located in the code?
>>
>> thanks,
>>
>> jay
>> peltz power
>> On Feb 4, 2008, at 10:18 PM, William Miller wrote:
>
>
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R. Walters
Solarray.com
NABCEP # 04170442	




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