Three phase voltage drop

ggunn · July 2011

OK, here's the situation: I've got a boatload of SMA inverters (30) with their outputs gathered into a 990A (24 SB7000's @ 34A + 6 SB6000's @ 29A = 990A), 480VAC, three phase transmission line that has to go a considerable distance (405') to an interconnect point. My acceptable AC voltage drop is 1% (4.80V). How do I calculate the maximum resistance I can tolerate? I know it will require multiple conductors per phase and of course the resistance depends on the length of the conductor, but all I am looking for is the basic calculation for the resistance; I can get to the rest of it. I know there's a square root of three in there somewhere...

Is it Vd = (I)(Rmax)/(sqrt3)

Rmax = (sqrt3)(Vd)/(I) = (1.732)(4.8V)/(990A) = 0.0084 ohms ?

niel · July 2011

Re: Three phase voltage drop

ggunn,
you can use the calculator in my signature line as ohm's law applies to ac as well as dc. it will calculate any voltage you like for any distance that you like for any gauge of wiring and show both the voltage dropped and the percentage. i used to calculate voltage drops by hand for people on the forum until another member who removed himself offered to put my formulas and data into a spreadsheet that you can download onto your own pc with no need to even be online once it's downloaded.

BB. · July 2011

Re: Three phase voltage drop

I don't think you need sqrt(3) in there... Just the current per lead (total RMS current, not in-phase current) and the maximum voltage drop you will support.

Assume 480 VAC with 10% high voltage and 8% maximum operation voltage at the Inverter (to avoid trip):

[( 480 * 1.10 ) - ( 480 * 1.08 )] / 990 Amps = 0.0097 ohms

Of course, you will have a fair amount of voltage drop across the breakers and the feed back to the distribution transformer.

What will the utility "do for you"--Will they adjust transformer/drop/rise voltages for you to get better nominal voltage (closer to 480-500 instead of 532 volts)? Have you measured "typical line voltage" over the day/season there?

-Bill

RCinFLA · July 2011

Re: Three phase voltage drop

You need the sqrt(3) if the inverters are wired as a delta configuration.

The numbers you quoted are confusing.

Twenty four 7000 watt inverter could produce 168 kW peak
Six 6000 watt inverters could produce 36 kW peak.

If they are all putting out max power that would be 204 kW.

For a 480 volt three phase delta configuration that would be 245 amps r.m.s. per wire at full peak power output of 204 kW.
If using a 'Y' 277 vac inverter configuration the current would be (204kW / 3) /277 vac = 245 amps r.m.s. also.

I am not aware of a SB7000 that outputs 480 vac. Are you using a transformers?

ggunn · July 2011

Re: Three phase voltage drop

RCinFLA wrote: »

You need the sqrt(3) if the inverters are wired as a delta configuration.

The numbers you quoted are confusing.

Twenty four 7000 watt inverter could produce 168 kW peak
Six 6000 watt inverters could produce 36 kW peak.

If they are all putting out max power that would be 204 kW.

For a 480 volt three phase delta configuration that would be 245 amps r.m.s. per wire at full peak power output of 204 kW.
If using a 'Y' 277 vac inverter configuration the current would be (204kW / 3) /277 vac = 245 amps r.m.s. also.

I am not aware of a SB7000 that outputs 480 vac. Are you using a transformers?

No transformers. Inverters will be configured to 277VAC and connected to 480 delta/277 wye from phase to neutral in three balanced groups. I agree that current per phase is 245A. Figuring another way, each phase gets 8 SB7000's and 2 SB6000's operating at 277VAC. ((eight)(7000W)+(2)(6000W))/277V=245.5A. I was using nameplate maximum continuous output current, adding them, and dividing by sqrt3, but that was incorrect because output voltage was not considered, and the divide should have been by 3 instead of sqrt3. Thanks for your help.

OK, next question. Maximum voltage drop is 1%. Line voltage is 480V phase to phase, 277V phase to ground/neutral. I have my phase current. Is the value of Vd 4.80V or 2.77V? It seems to me that if I am considering phases independently it should be 2.77V.

It seems to me now that the maximum resistance in each of the three phase conductors should be:

Rmax = Vd/Iphase = 2.77V/245.5A = 0.01128 ohms.

ggunn · July 2011

Re: Three phase voltage drop

niel wrote: »

ggunn,
you can use the calculator in my signature line as ohm's law applies to ac as well as dc. it will calculate any voltage you like for any distance that you like for any gauge of wiring and show both the voltage dropped and the percentage. i used to calculate voltage drops by hand for people on the forum until another member who removed himself offered to put my formulas and data into a spreadsheet that you can download onto your own pc with no need to even be online once it's downloaded.

Thanks, but I need to understand this stuff instead of plugging numbers into a calculator.

BB. · July 2011

Re: Three phase voltage drop

I would agree with your post #5 if you are doing 277 volt phase to neutral inverter connections.

Is there anyway to run the DC from your arrays and install the SMA inverters closer to the utility drop?

-Bill

RCinFLA · July 2011

Re: Three phase voltage drop

Just FYI on neutral.

If all three phases have same power output from inverters then neutral current will be zero.

If one or two of the phases are at full power and other(s) at zero output then neutral will be 245.5 amps r.m.s. .

So worse case neutral current is 245.5 amps also. You might want to make sure the installation can handle that neutral current and the power company does not have a restriction on maximum neutral current relative to three phase lines.

If you want less then 1% voltage drop for 405 feet run then that would be close with 500 MCM copper cable (0.025 ohms/1000' hot).

That's 405' x 4 lines = 1620 feet. At about $10.50/ft for 500 MCM that is $17,000.

BB. · July 2011

Re: Three phase voltage drop

I believe some GT inverters have the ability to auto-shutdown if they lose 1 phase (and I think shutdown on phase loss may be requirement for many utilities).

-Bill

ggunn · July 2011

Re: Three phase voltage drop

BB. wrote: »

I would agree with your post #5 if you are doing 277 volt phase to neutral inverter connections.

Is there anyway to run the DC from your arrays and install the SMA inverters closer to the utility drop?

-Bill

Yes, we have done that, and the ~400' is what we have been able to reduce it to. We are looking into a dual transformer solution as well, but it remains that I must thoroughly understand these calculations because it will be a recurring issue for me. I appreciate the link to the calculator, but here is far too much money and responsibility on the line for me to blindly plug numbers into a calculation routine and accept the results without understanding them.

ggunn · July 2011

Re: Three phase voltage drop

RCinFLA wrote: »

Just FYI on neutral.

If all three phases have same power output from inverters then neutral current will be zero.

If one or two of the phases are at full power and other(s) at zero output then neutral will be 245.5 amps r.m.s. .

So worse case neutral current is 245.5 amps also. You might want to make sure the installation can handle that neutral current and the power company does not have a restriction on maximum neutral current relative to three phase lines.

If you want less then 1% voltage drop for 405 feet run then that would be close with 500 MCM copper cable (0.025 ohms/1000' hot).

That's 405' x 4 lines = 1620 feet. At about $10.50/ft for 500 MCM that is $17,000.

Thanks, and yes, I understand all that. Neutral must be the same size as phase conductors. The utility will have approval/modification/veto power on the project after reviewing all drawings; they are very much involved. And yes, I also understand that it will be a lot of copper no matter how we slice it.

solarix · July 2011

Re: Three phase voltage drop

May I ask why you are using a bunch of small inverters instead of a 250kW central inverter?

ggunn · July 2011

Re: Three phase voltage drop

solarix wrote: »

May I ask why you are using a bunch of small inverters instead of a 250kW central inverter?

Of course; that's the same question I asked when I got assigned to the project. It's because it's for a school and they want all the inverters on the roof. The roof won't support a central inverter that large.

BB. · July 2011

Re: Three phase voltage drop

In most (all?) cities--they have been installing cell sites on top of buildings for years--And they probably way as much (or more) than a large central inverter would.

Did they get an estimate to build a structure to support the Central inverter where needed?

-Bill

ggunn · July 2011

Re: Three phase voltage drop

BB. wrote: »

In most (all?) cities--they have been installing cell sites on top of buildings for years--And they probably way as much (or more) than a large central inverter would.

Did they get an estimate to build a structure to support the Central inverter where needed?

-Bill

I have no idea, but this decision was made long before I came aboard. My assigned task is to flesh out the design as proposed and not to open cans of worms that have been sealed for months. For the next one I will get my licks in early. ;^)

ggunn · July 2011

Re: Three phase voltage drop

OK, I am officially tearing out what is left of my hair. I am apparently doing something wrong but I can't figure out what it is. Again, I need to understand the fundamentals behind what I am doing, so I don't want to just plug and chug with a calculation script.

Here's the scenario: I have a bunch of Sunny Boys configured at 277V to be connected phase to neutral on a 277/480V three phase interconnect. For the sake of this exercise let the interconnect distance be 400' and the conductors be 700kcmil copper with a resistance of 0.0184ohms/1000', so R= (0.4)(0.0184) = .00736 ohms. What is my voltage drop? I do it two different ways and get two different answers.

Method 1: I have 42 SB7000s and 3 SB6000s. Each phase gets 14 7000's and 1 6000, which is 104kW. 104kW @ 277V is 375.5A per phase.

VD = IR = (375.5A)(0.00736 ohms) = 2.76V.

(2.76/277V)(100%) = 1% voltage drop.

This is how I proposed doing it earlier in this thread which I thought most of you agreed with and which seems logical to me. However, when I looked up three phase voltage drop, I found this page:
http://www.electrician-electricalcontractor.com/calculatevoltagedrop.html

This method calculates voltage drop on the whole three phase system, using the formula (Method 2):

VD = (2LRI/1000)(sqrt3/2)

The power for the total system is (42)(7000) + (3)(6000) = 312kW

312000W/480V = 650A.

substituting quantities:

VD = ((2)(400)(0.0184)(650)/1000)(sqrt3/2) = 8.286V

(8.286V/480V)(100%) = 1.726% voltage drop.

Note that this is within roundoff error of sqrt3 times the result I got using Method 1, which I don't think is a coincidence. The published formula is for loads rather than sources, but that shouldn't make any difference. Apparently I am off by a sqrt3 somewhere in my reasoning; can any of you guys point out where it is?

niel · July 2011

Re: Three phase voltage drop

ggunn,
i agree with method 2, but they multiply the run distance times 2 in the 1st part of the formula meaning a run of x will have a total length of 2x and i'm not sure you aren't confusing yourself here with the length of the wire. by calcs you listed it is the total wire length or 2x the run = that 400ft. if that's the case then drop the 2x in the 2nd formula and just run it from there as you already have the total wire length listed. the .866 multiplier i believe is there because while one leg may see your full voltage at +277 in an instant of time while the power being fed from the other section into that leg will be - voltage and something just off in time by 120 degree will roughly be -37v. it gets confusing as while one inverter has positive voltage going through a leg another inverter will be sending a portion of the sine waveform and will have a negative value subtracting from the 1st inverter.

maybe i am presupposing that you are confused here to a point, but for the benefit of others here too i will let you picture 2 12v batteries wired in series and there is a wire at the - and + of the string (of course), but there's also a wire to where they both interconnect in series. place a 12v pv to the 1st battery - and the 1st battery plus and another identical pv connected to the 2nd battery's - and + leads. straight forward here as you will get the rating of the current for the pvs and will either be + or - in polarity in any of those wires. now combine the middle wires from the pvs to the batteries and you will see 0 current because one pv is sending + current and the other is sending - current of equal value. although the 3 phase system isn't exactly like this, i just wanted to point out how power from 2 or more places can influence voltage and current seen on a wire. it's there and yet it's not, but 3 phase isn't able to totally make it seem to disappear like the batteries example and the net effect would be the .866 multiplier.

long story short is that i believe your answer of 1.762% if divided by 2 would be correct and would be .881%, if i miss my guess at an error on your part for the wire length, and that method 2 is valid. it stands to reason though that if it passes the 1st one that it'll pass the 2nd one as the 1st one would be higher as no negative voltage or currents are shown or accounted for in the 1st one.

am i making sense to you on this and do you now think you've got this down?

ggunn · July 2011

Re: Three phase voltage drop

niel wrote: »

ggunn,
i agree with method 2, but they multiply the run distance times 2 in the 1st part of the formula meaning a run of x will have a total length of 2x and i'm not sure you aren't confusing yourself here with the length of the wire. by calcs you listed it is the total wire length or 2x the run = that 400ft. if that's the case then drop the 2x in the 2nd formula and just run it from there as you already have the total wire length listed. the .866 multiplier i believe is there because while one leg may see your full voltage at +277 in an instant of time while the power being fed from the other section into that leg will be - voltage and something just off in time by 120 degree will roughly be -37v. it gets confusing as while one inverter has positive voltage going through a leg another inverter will be sending a portion of the sine waveform and will have a negative value subtracting from the 1st inverter.

maybe i am presupposing that you are confused here to a point, but for the benefit of others here too i will let you picture 2 12v batteries wired in series and there is a wire at the - and + of the string (of course), but there's also a wire to where they both interconnect in series. place a 12v pv to the 1st battery - and the 1st battery plus and another identical pv connected to the 2nd battery's - and + leads. straight forward here as you will get the rating of the current for the pvs and will either be + or - in polarity in any of those wires. now combine the middle wires from the pvs to the batteries and you will see 0 current because one pv is sending + current and the other is sending - current of equal value. although the 3 phase system isn't exactly like this, i just wanted to point out how power from 2 or more places can influence voltage and current seen on a wire. it's there and yet it's not, but 3 phase isn't able to totally make it seem to disappear like the batteries example and the net effect would be the .866 multiplier.

long story short is that i believe your answer of 1.762% if divided by 2 would be correct and would be .881%, if i miss my guess at an error on your part for the wire length, and that method 2 is valid. it stands to reason though that if it passes the 1st one that it'll pass the 2nd one as the 1st one would be higher as no negative voltage or currents are shown or accounted for in the 1st one.

am i making sense to you on this and do you now think you've got this down?

Thanks for taking the time to answer me, and as to whether you are making sense to me, yes and no(:D), but reading what you wrote sparked a thought process that I think yielded the solution. I was indeed discounting the return path, which was the root of my error. I thought it was correct to do so, since there is no current in the neutral in a balanced system and hence no voltage drop. That is incorrect, since the other two legs provide the return path and they must be considered, just as in a DC system where both the + and - conductors contribute to VD. But just like everything else in three phase power, the phase angle must be considered.

What it boils down to, I now believe, and I'm sticking with it until someone comes up with a better explanation, is while the effective length of a two conductor DC or single phase AC run is the distance multiplied by two, in a three phase system the effective length of the conductor for a single phase is the distance multiplied by sqrt3. When I apply that to what I was doing before (Method 1), it gives me a voltage drop of 4.78V on the 277V phase voltage, which is 1.73%, the same as the result from Method 2.

Thanks again for taking the time to respond. Now I have to go fix something that would have been a costly error...

niel · July 2011

Re: Three phase voltage drop

glad i helped even indirectly. good luck with it.

ggunn · July 2011

Re: Three phase voltage drop

niel wrote: »

glad i helped even indirectly. good luck with it.

Actually, you helped directly. The essence of my error, as you correctly pointed out, was that I was not considering the return path for the current. I knew I was doing that, but I erroneously thought that it was justified.

I had to fire off some emails confessing to the error, which wasn't a good feeling, but it's much better to catch it now than after the system is installed.

Life is a never ending educational process.

solarquestion2011 · July 2011

Re: Three phase voltage drop

I believe that in order to hook those SMAs up in a 277 configuration they need to be connected so that the system will shut down in the event that an array goes down and the phases become unbalanced. See NEC 690.63. Normally on a small system this isn't a problem, but for this number of inverters it could be a problem. On the systems I've seen with single phase SMAs feeding 277 the Inverters are daisy chained in balanced sets of 3 with connections at the communications boards to prevent loading of a phase.

If you are worried about the Inverter falling out of AC tracking due to high utility voltage and wiring losses. You can call SMA and with Utility Approval they will give you an access code to modify the AC Voltage tracking range on the inverters. You will need special permission from SMA to change these values.
I believe you also need to be a licensed electrical contractor as well.

Here is the link for the application from SMA
http://forums.sma-america.com/ubbthreads.php?ubb=showflat&Number=2199#Post2199

If there is still time I would suggest recommending a large central inverter for the roof. If you are using a ballasted racking system then the roof should have the load requirements to support a central inverter if you spread the load out sufficiently. It should be easy for the structural engineer on the job to design a steel rack or roof curb to distribute the load. The average weight for a central inverter is ~2200-2500 pounds. Even if the roof requires some bracing in this area it should be alot less than that large wire and all those SMAs. I know this can be done as I just had an engineer design a roof mounted inverter rack for two 100kW Satcon PGs.

ggunn · July 2011

Re: Three phase voltage drop

solarquestion2011 wrote: »

I believe that in order to hook those SMAs up in a 277 configuration they need to be connected so that the system will shut down in the event that an array goes down and the phases become unbalanced. See NEC 690.63. Normally on a small system this isn't a problem, but for this number of inverters it could be a problem. On the systems I've seen with single phase SMAs feeding 277 the Inverters are daisy chained in balanced sets of 3 with connections at the communications boards to prevent loading of a phase.

Thanks for the tip; I will look into that.

solarquestion2011 wrote: »

If you are worried about the Inverter falling out of AC tracking due to high utility voltage and wiring losses. You can call SMA and with Utility Approval they will give you an access code to modify the AC Voltage tracking range on the inverters. You will need special permission from SMA to change these values.
I believe you also need to be a licensed electrical contractor as well.

We have that covered, but the 1% voltage drop on the AC side is a requirement that was handed down to me from above.

solarquestion2011 wrote: »

If there is still time I would suggest recommending a large central inverter for the roof.

There isn't. Believe me, I lobbied for that; it would have made my life so much easier. I am dealing with bureaucratic inertia, though, and once the decision was made (long before I joined the project), there was no changing it.

Three phase voltage drop

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