AC V-Drop, AC Current, and Wire Sizing Questions
arcturusk1
Solar Expert Posts: 26 ✭
Please forgive me if these questions are dumb as hell, but I'm reviewing and double-checking a design I did half a year ago and numbers are swimming around my head as I second-guess myself here.
I'm designing a 66.24kW solar array for a flat-roof-mounted setup using 6x Fronius IG Plus 12.0-3's @ 277V. It's a commercial building so it's 3-phase. There will be 288 modules @ 230W with 48 modules per inverter. Yes, I know it's a lot of inverters but the Fronius CL line *just* came out (or at least was visible to click and see on their USA page) and we submitted drawings and docs last year for grant approval. We can't make a change like that now.
The 277V is, I guess, actually 480V but the 277V is L-N while the 480V is L-L.
I'm going to aggregate the output of those six inverters in a load center, after which I'll run it through an AC disconnect and meter like usual before feeding it into the targeted building panel.
Fronius' site says these inverters put out 14.4A... per phase. For some reason, I had had a much larger wire size coming out of the inverters and coming out of the load center. I think this is because back then I simply used W=V*I with 12000W and 277V. That gives a current of 43.3A. So, as dumb as this is to ask: Which is it? This is important because I have to size my wires appropriately.
Preemptive-edit: Lightbulb moment, perhaps? I think I just realized that of course I have to size the wires for 14.4A. There will be one wire per phase coming out of the damn inverter (well five if you count the neutral and equipment ground), so that's three conductors that are carrying the current. Duh.
Ok so continuing along that thinking, now I need to concern myself with voltage drop. You see, the location of the inverters and load center are pretty far away from where the utility wants me to put the AC disconnect and meter. The two-way run length is going to be about 350' if not 400' so I have to seriously consider v-drop. With 14.4*6=86.4A, and 86.4*1.25=108A, and 90-deg-C insulation THWN-2, I should use 3 AWG from just an ampacity standpoint.
I used this calculator as well as this calculator and I'm getting some conflicting v-drop percentages. Givens: 277V AC 3-phase 4-wire, EMT conduit, 90-deg-C insulation, THWN-2 wire, 175' one-way distance, 86.A "load", copper conductor, 3 AWG. One says 2.5% and the other says 1.3%. Going from 3 AWG to 2 AWG gets me to 1% for the latter but not the former. In fact it takes 1 AWG to even get me under 2% on the former. I even tried to run some calcs by hand from this site as well as a few other places and I'm getting pretty close but still slightly differing values.
I'd like to keep cost down and make the install a bit easier for the electrician, but I really want to make sure I'm not losing power or compromising safety because I skimped on wire. And yes, it sucks that I have to run the darn wire that far and then back again, but the location of the inverters and utility requirement that the disc and meter be at person-level off the ground (reasonable, to be fair) means I'll have somewhat lengthy runs no matter which way I go.
As of right now, I had the wire upsized a bit to 2 AWG but I'm not sure if I should stop there. Thoughts and opinions? Am I missing some obvious math or something? Thanks guys!
PS - I also have some fun wire design coming up very shortly as I try to get ~100kW of DC power through wires across a couple-hundred foot soccer field! :roll:
I'm designing a 66.24kW solar array for a flat-roof-mounted setup using 6x Fronius IG Plus 12.0-3's @ 277V. It's a commercial building so it's 3-phase. There will be 288 modules @ 230W with 48 modules per inverter. Yes, I know it's a lot of inverters but the Fronius CL line *just* came out (or at least was visible to click and see on their USA page) and we submitted drawings and docs last year for grant approval. We can't make a change like that now.
The 277V is, I guess, actually 480V but the 277V is L-N while the 480V is L-L.
I'm going to aggregate the output of those six inverters in a load center, after which I'll run it through an AC disconnect and meter like usual before feeding it into the targeted building panel.
Fronius' site says these inverters put out 14.4A... per phase. For some reason, I had had a much larger wire size coming out of the inverters and coming out of the load center. I think this is because back then I simply used W=V*I with 12000W and 277V. That gives a current of 43.3A. So, as dumb as this is to ask: Which is it? This is important because I have to size my wires appropriately.
Preemptive-edit: Lightbulb moment, perhaps? I think I just realized that of course I have to size the wires for 14.4A. There will be one wire per phase coming out of the damn inverter (well five if you count the neutral and equipment ground), so that's three conductors that are carrying the current. Duh.
Ok so continuing along that thinking, now I need to concern myself with voltage drop. You see, the location of the inverters and load center are pretty far away from where the utility wants me to put the AC disconnect and meter. The two-way run length is going to be about 350' if not 400' so I have to seriously consider v-drop. With 14.4*6=86.4A, and 86.4*1.25=108A, and 90-deg-C insulation THWN-2, I should use 3 AWG from just an ampacity standpoint.
I used this calculator as well as this calculator and I'm getting some conflicting v-drop percentages. Givens: 277V AC 3-phase 4-wire, EMT conduit, 90-deg-C insulation, THWN-2 wire, 175' one-way distance, 86.A "load", copper conductor, 3 AWG. One says 2.5% and the other says 1.3%. Going from 3 AWG to 2 AWG gets me to 1% for the latter but not the former. In fact it takes 1 AWG to even get me under 2% on the former. I even tried to run some calcs by hand from this site as well as a few other places and I'm getting pretty close but still slightly differing values.
I'd like to keep cost down and make the install a bit easier for the electrician, but I really want to make sure I'm not losing power or compromising safety because I skimped on wire. And yes, it sucks that I have to run the darn wire that far and then back again, but the location of the inverters and utility requirement that the disc and meter be at person-level off the ground (reasonable, to be fair) means I'll have somewhat lengthy runs no matter which way I go.
As of right now, I had the wire upsized a bit to 2 AWG but I'm not sure if I should stop there. Thoughts and opinions? Am I missing some obvious math or something? Thanks guys!
PS - I also have some fun wire design coming up very shortly as I try to get ~100kW of DC power through wires across a couple-hundred foot soccer field! :roll:
Comments
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Re: AC V-Drop, AC Current, and Wire Sizing Questions
try the calculator i helped develop in my signature line.
given 277v at 90 degrees c with 86a and 350ft overall wire length per leg.
i get 2.73% with #3 and 2.17% with #2. #1 is 1.72% and you can be sure this is fairly accurate for the specs i listed, but i'm not 100% up on 3 phase system wiring. -
Re: AC V-Drop, AC Current, and Wire Sizing Questions
Hi Niel. Thanks for the reply and link to your v-drop calculator! I'm think I'm using it properly but I do have a few questions (exact spreadsheet wording in bold italics):
1) The array voltage input is apparently linked to panel (volts). Which is it? Do I input the panel Voc as array voltage or do I input my string voltage (Voc times quantity of panels in string)?
2) If my system doesn't involve batteries, it seems that I can just enter 0 for battery bank voltage without messing up the final calc. Is this a correct assumption?
3) I take it that panel (amps) is requesting the Isc of the panel without the 1.56 ampacity modifier applied?
4) The layout of the spreadsheet throws me off a bit:
4a) In the string section we're supposed to enter AWG, feet, and panel (amps). Now, a string of panels has to terminate at a J-box and then get run to a combiner. There is a changeover at that J-box. We're going from the wire type and size that are built into the panel to usually THWN-2 wire. I should technically include the intra-string length in addition to the THWN-2 wire length, but my intra-string wire gauge is naturally going to be smaller than the THWN-2 wire gauge due to length to combiner box. Basically, does the feet measurement include intra-string wire length?
4b) There is an entry section for string, then a dividing line, then an array to combiner box title. The calculations on that line do little more than duplicate the string values, except for the increased current due to parallel strings. Also, the resistance plummets. I'm sure this line is supposed to be the values of the system post-combiner-box but the radically lower resistance confuses me.
4c) I take it that combiner box to breaker means combiner box to inverter DC input?
Is there any way I would be able to access the guts of the spreadsheet or see the formulas which you used? Part of the reason is that we have yet to install a battery-based system and probably won't. I do not need any of the calculations that are touched in any way by the battery part of the system and would like to modify/streamline the spreadsheet. If you'd like we can take this to PMs to avoid unnecessary forum posting as I would love to learn how you guys made this. -
Re: AC V-Drop, AC Current, and Wire Sizing Questions
i gave the data and formulas while another did the spreadsheet and he laid it out in such a manner as to possibly confuse, but it really isn't all that complicated.
array voltage can change all pv voltage statises. play with it and you will see it do just that. go with the nominal panel volts as pvs usually don't stay high in their voltage when loaded down or used with mppt controllers. today it is difficult to determine nominal operating voltage as there are many tweeners that have been on the market for gt use. these should still be rounded downward to follow nominal battery voltages that they could match up to using a pwm type controller from the vmp of the pv(s) without nec multiplications.
examples:
17.6vmp is good for 12v nominal pv voltage
36vmp is good for 24v nominal pv voltage
26vmp i'd round down to 12v nominal pv voltage, but it wouldn't hurt that to only drop it down to say 18v as that pv would suffice quite well for such a nominal pv voltage through a pwm cc if there were 18v nominal battery banks and pwm controllers for such an application. i'm sure i'm going to far here and confusing you some.
as far as if you have no voltage, then you have no power. even without batteries you will be working with a voltage and this does not matter if ac or dc. this calculator relies on the voltage only to determine the percentages the voltage drop will have as the voltage drop will remain more constant given the same current over the same resistance.
use the imp or max power current and not the shorting current. you don't use the wire shorted so this doesn't enter any factors except for fusing.
you have found the whoops in the layout as each string may need to be calculated individually as they can have different wire gauges and lengths resulting in a different vdrop. it does get complicated because after the combiner it is involving all currents being inputted to the wire after the combiner. each section can be individually figured as i would rather do and just add up the vdrops and % losses later.
wire length includes inter pv connections as the current needs to pass through that resistance too.
it does mean combiner box to breaker (or fuse) as the wires need protected. this should be located at either the batteries or the gt inverter depending on system type used.
being i did not design the spreadsheet layout if you feel you can improve upon the calculator then i know the password to unlock it. i can license it to you under the conditions i can give my input to you on it plus approve of the improvement(s) and it can be beta tested here for feedback from the members. also, there is to be no monetary gain from this calculator in any way, shape, or form as anyone should be able to use it. how does that sound to you?
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