Nick,
Sorry for the delayed reply. I've been traveling and am just catching
up on e-mails.
Your method is absolutely correct. We're working on an appnote and a
calculator to provide you guys with some quick tools.
David also brings up a favorite technique of mine. When you center
feed the branch, the reduction of voltage rise is significant. Cutting
the # of inverters in half cuts the voltage drop to 1/4. It also
reduces the power loss in the branch wiring to 1/4 of the end-fed value.
A couple of examples (which use the M190 and the current 14awg wiring)
will demonstrate. This is at full power for a 240vac split-phase
circuit, and the current is 0.8Aac. These calcs also include contact
resistance of the connectors:
If you put 15 inverters on one branch circuit and they are end-fed, the
last inverter will see a voltage rise of 4.2 volts (1.75%). If you put
15 inverters (the max. allowed on one dedicated OCPD), split 7 and 8
per side, the last inverter on the side with 8 inverters will see a
1.3 volt (0.54%) rise. Acceptable in most cases.
If you must end-feed, a branch of 15 inverters will see a rise of 4.2
volts (1.75%) and
the end, but a branch of 10 inverters will only see a rise of 1.9 volts
(0.8%) at the end. Again, acceptable in most cases.
Ultimately, by center feeding a branch of 14 inverters, the end
inverters will see less than a 1 volt (0.42%) rise. If this is not
acceptable, there are other problems.
Of course, you must also add the voltage drop for the feeder to the PV
sub panel, and any wiring from the sub panel to the first pigtail in a
branch to arrive at the voltage at the PCC.
Please note that we are not accounting for terminations in the sub,
breaker losses, sub panel losses, intermediate connections, etc. These
can add a few tenths of a volt.
We are going to 12AWG wire in the upcoming Trunk and Drop wiring
system, which some of you have already previewed. This will allow a
20A OCPD, but will be limited by all of the same Physics that apply to
the existing wiring system. We just can't seem to get around that
Physics thing.
I hope this helps!
See ya!
Marv
707 763-4784 x7016
David Brearley wrote:
Maybe Marv can help out here. I’ve talked
with tech support at Enphase and am under the impression that you can
center tap their branch circuits to reduce voltage drop. This technique
was used on a project that we profiled in SolarPro:
http://solarprofessional.com/article/?file=SP3_2_pg92_Projects_1&search=
David Brearley, Senior Technical Editor
SolarPro magazine
On 6/11/10 2:50 PM, "Nick Soleil" <nicksoleilso...@yahoo.com>
wrote:
Did you
notice that in my calculation I divided the total amperage by 2 to get
the average ampacity across the string. This seems to be the best way
to deal with that. Otherwise the loss would have been over 2.8%
Nick Soleil
Project Manager
Advanced Alternative Energy Solutions, LLC
PO Box 657
Petaluma, CA 94953
Cell: 707-321-2937
Office: 707-789-9537
Fax: 707-769-9037
From: "ma...@berkeleysolar.com" <ma...@berkeleysolar.com>
To: RE-wrenches <re-wrenches@lists.re-wrenches.org>
Sent: Fri, June 11, 2010 7:02:00 AM
Subject: Re: [RE-wrenches] AC wiring loss on micro-inverters
Nick,
Good to see other wrenches looking at this issue. However, remember
that
in the interconnect cable between the last and the second to last
Enphase,
the current is small. If you really want to see how the loss is adding
up
you need a more detailed analysis that averages the voltage drop for the
entire system of interconnect cables.
Mark Frye
Berkeley solar Electric Systems
> Hello wrenches and Marv:
> How many Enphase inverters do you put on a single parallel
string?
> The Enphase manual allows for up to 15, but I have found that a
single
> string of 15 can have very high voltage rise on the AC wiring. Up
to
> 1.5% loss just on the Enphase cables, and that does not take into
> account resistance at the plug connectors or any home-run wiring.
> 15 inverters x 190 watts = 2850 watts
> 2850 watts / 240 Volts = 11.875 Amps (however at the beginning
of the
> string it is 0 Amps)
> Average Amps on the Enphase cables is 11.875 / 2 = 5.9375 Amps
> length of contacts + internal wiring = ~6'
> Total Enphase cabling length (6' x 15 inverters) = ~90'
> 5.9375 Amps @ 240 V running 90' on #14 AWG CU = 1.4% loss
> It is pretty hard to design for 1.5% voltage drop if you have to
size for
> 0% loss on your wiring! I hope Enphase will increase the size of
the wire
> in their cables soon. Have you experienced issues with
micro-inverters
> shutting off due to high AC voltages?
>
> Nick Soleil
> Project Manager
> Advanced Alternative Energy Solutions, LLC
> PO Box 657
> Petaluma, CA 94953
> Cell: 707-321-2937
> Office: 707-789-9537
> Fax: 707-769-9037
>
>
>
>
> ________________________________
> From: i2p <i...@aol.com>
> To: RE-wrenches <re-wrenches@lists.re-wrenches.org>
> Sent: Thu, June 10, 2010 4:50:47 PM
> Subject: Re: [RE-wrenches] Real world PV production
>
>
>
> On Jun 10, 2010, at 7:57:53 AM, "Kelly Keilwitz, Whidbey Sun &
Wind"
> <ke...@whidbeysunwind.com>
wrote:
>
> PV Watts VASTLY underestimate PV production
>> in our area.
> PV Watts underestimates in Central Ca too.
>
> Don
>
>
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