voltage drop clarification

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ocoee

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Can someone please verify that I am doing this correctly. I am connecting 21 solar inverters to a 3 phase 4 wire 277/480 VAC panel. These inverters have a single pole 277 V output. 3 phases 7 inverters per phase 500 feet back to main service.

Assuming max/output "load" of 177 amps per phase
Here is the voltage drop formula that I am using

VD= 1.73 . 12.9 . I . L/CM

my inputs are
1. 1.73
2. 12.9 "approx" K value for copper
3. 177 amps max output I
4 500 feet distance back to service
5 CM = 300,000 300mcm

My results are this VD= 6.58 V or 2.3% vd

The engineer I am working with on this project comes out with a wire size that is 600 MCM to achieve the same VD and his methods are cryptic at least to me.

Please someone tell me that I have no idea what I am doing or confirm this is an acceptable result

thanks
 
I am not sure how the engineer got the results he/she got. I would have thought a 3/0 copper conductor would work but I am not an engineer. We do have engineers on this site however today being xmas you may not get a response til tomorrow. I would post his calculation and see if someone can decipher it.
 
I don'took handy but came up with a wire size of almost 170,000cm at 5%VD
 
I guess what I am asking is "Is my calculation wrong based on my inputs"

Also what is the difference between a 277/480v formula and a 277/480 4 wire formula I see "on line " calculators offer both
 
ocoee said:
Can someone please verify that I am doing this correctly. My results are this VD= 6.58 V or 2.3% vd

thanks
Click to expand...

I'm currently quite full of pumpkin pie so hope that doesn't have an impact on my reasoning ability but your calcs look correct as I got around 6.5 volts with the usual caveats on power factor etc.

Where you need to be careful is since you have the sqrt(3) in your formula, you're calculating a voltage drop against the line voltage (480V). In other words, the ~ 6.5 V is correct but that is about 1.3% of the line voltage.

I would still agree to ask the engineer what the rationale is behind his/her calculations.
 
My program results were 1.4% or 3.8 volts L/N for 300 kcm. Results for
3/0 were 2.5% and 6.8 volts L/N.

Merry Christmas
 
Last edited:
The engineer you are working with may be correct.

More information is needed, it depends on the inverter and if the inverter can compensate for the voltage rise above utility (not likely if you are using the smaller inverters). The inverters need to stay within the allowable voltage range for the service and the 'voltage drop' causes the inverters to operate at a higher voltage than the utility. Most inverter manufacturers recommend a maximum voltage drop in the wiring of 1.5%.

You also need to consider wire temperature as high temperature increases voltage drop.

You do not want your inverters to start dropping out in the summer due to a high voltage error.
 
To answer the above, we are using SMA 7000 us. They will automatically shut down if the grid goes out of range in either direction. nothing we can do about voltages that are too high. At this point my concern is the voltage drop, this we can address.

So.... Big picture We want the best possible production over time. The main variable being the wire size from the AC panel to utility interconnection, 500 feet.

The inverters are single phase outputs @277V
They are combined ( 21 of them) on to 3 phases of a 277/480 4 wire sub panel. Each inverter has a max output of 25.2 amps and there are 7 per phase. There is 177 amps per phase for the purposes of our calculation.

At the point where they are combined in my panel this becomes a 3 phase circuit. No?

Do I need to concern myself with the voltage drop from line to neutral?
or just line to line? being that the source is a single phase output as stated above.

We would like to limit VD to less than 2% for this run.
To be clear my question is about the best way to figure voltage drop without overdoing it .

Also I have done several systems this size and have others review my calcs and never a issue raised. I interest is to be better at what I do. ( not puffing chest out) If I can learn another method to figure VD for three phase please TELL ME.

Regard Geoff
 
ocoee said:
If I can learn another method to figure VD for three phase please TELL ME.

Regard Geoff
Click to expand...

There are a number of methods that are perfectly valid for determining voltage drop. Be a little wary with on-line calculators until you test them and make sure you understand the behind-the-scenes assumptions as they can turn out erroneous results at a furious pace.

Many people like to use an Excel formula which is also fine.

Your method above is fine assuming your line resistance dominates the inductive reactance and the power factor is reasonably good. Bob (above) got 1.4% vs. my 1.3% which for practical purposes are equivalent. In a balanced system the line-neutral drop is related to the line-line drop by sqrt(3), you just need to remember which one you are using when you calculate the voltage drop.

In your case, it appears you calculated the line-line drop of ~6.5 volts and divided by the line-neutral voltage to get the % which is incorrect.
 
ocoee said:
If I can learn another method to figure VD for three phase please TELL ME.
Click to expand...
Well how about using the method of Table 9? That way you don't have to wonder about some unknown internet calculator where you don't know what they are doing.

ice
I really got to go to work

Merry Christmas
 
ocoee said:
...

At the point where they are combined in my panel this becomes a 3 phase circuit. No?
Click to expand...
Essentially yes.

Do I need to concern myself with the voltage drop from line to neutral?
or just line to line? being that the source is a single phase output as stated above.
Click to expand...
Just 3? line-to-line. It is unlikely your "feeders" will ever see a full line-to-neutral load.
 
Consider NEC Art 690.64(B)(2) and Alternating-current resistance

Consider NEC Art 690.64(B)(2) and Alternating-current resistance

Consider NEC Art 690.64(B)(2) and Alternating-current resistance

If your PV output is 177A at 480V 3-phase, you will need a circuit ampacity of at least 221A (80% rule). Fuse would be 225A for the main disconnect.

Each inverter will be protected by a 40A breaker or fuse, total of 280A per phase.

Now consider 2008 NEC Art 690.64(B)(2) that requires an ampacity based on the sum of the overcurrent devices feeding the conductor be less than 120% of the conductor rating. This is (280 + 225)/1.2 or 421A.

Most likely the terminals in the AC Combiner and the OCPD at the utility end are not rated above 75?C, so the 75?C column in Table 310.16 will be needed. The minimum wire size if the ambient temperature does not exceed 30?C/86?F is 600 MCM CU or 900 MCM AL.

Because it will be possible to turn on only 7 inverters, all on the same phase, you need a full size neutral. With all inverters operating there will be very little current on the neutral, so all of the allowable voltage rise will be on the phase conductors.

SMA recommends a maximum 1.5% voltage drop. This is to include drops in inverter OCPD devices, inverter to AC Combiner, Main disconnect (with fuses) and the conductor resistance from AC Combiner to Disconnect (your 500'). Most PV installers aim for 1% for the cable. This is 2.77V in your case.

Power factor should be close to 1.0 at full power, so is not a concern.

V= I x R Therefore the allowable resistance over 500' is 0.0156 Ohm or 0.0313 ohm/1000'. Chapter 9, Table 9 shows 750 MCM (or 2 X 350 MCM) CU is required for voltage drop if in PVC conduit. I am not sure of the alternating current resistance if you use pole line for the 500'.

These need to be adjusted for ambient temperature, and for most areas will require larger cables.

Consider aluminum cable based on cost.
 
BillK-AZ said:
Consider NEC Art 690.64(B)(2) and Alternating-current resistance

If your PV output is 177A at 480V 3-phase, you will need a circuit ampacity of at least 221A (80% rule). Fuse would be 225A for the main disconnect.
Click to expand...
Uhhh... that'd be the 125% "rule". There is no 80% "rule", though it is 125% inverted. ;)

Each inverter will be protected by a 40A breaker or fuse, total of 280A per phase.
Click to expand...
35A is a standard ocp rating.

25.2A ? 125% = 31.5A (i.e. a 1P35A backfed breaker) minimum

25.2A ? 125% ? 7 inverters per phase = 220.5A (i.e. a 3P225A MCB) minimum​
Now consider 2008 NEC Art 690.64(B)(2) that requires an ampacity based on the sum of the overcurrent devices feeding the conductor be less than 120% of the conductor rating. This is (280 + 225)/1.2 or 421A.
Click to expand...
The premise is correct if, and I emphasize IF, the panel is a distribution panel... i.e it supplies power to loads via multiple branch and/or feeder circuits... though needs corrected...
((7 ? 35A) + 225) ? 120% = 392A minimum panel rating (i.e. a 400A-rated panel at a minimum)​

However, I'm getting the impression, and we'll need verification from Geoff (ococee), that the AC panel is being used solely as an inverter output combiner.

If this is the case, the second sentence of 690.64(B) states, "Where distribution equipment, including switchboards and panelboards, is fed simultaneously by a primary source(s) of electricity and one or more utility-interactive inverters, and where this distribution equipment is capable of supplying multiple branch circuits or feeders, or both, the interconnecting provisions for the utility-interactive inverter(s) shall comply with (B)(1) through (B)(7)."

As such, 690.64(B)(1) through (7) do not apply, and the panel bus need only be rated 220.5A or better... so 225A panel rating minimum and a connecting feeder of 4/0 Cu or 300MCM Al, 75?C or better conductors, before accounting for Vd.
 
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