# 3 phase voltage drop

ggunn
Posts:

**1,972**Solar Expert ✭
Here's the scenario:

I have 6 inverters totaling 112kW, whose 480VAC 3 phase 4 wire outputs are gathered in an AC combiner. The combiner will feed a wye connected stepdown transformer (480Y to 208Y) 205 feet away. What size single copper conductors do I need to run between the combiner and transformer to keep the voltage drop in that segment below 0.5%. Yes, I know that's very low, but I am under some unusual design constraints.

Please show your work (teach a man to fish) and don't just refer me to a calculator; I need to understand how this works.

EDIT: I have gone on line and found three calculators for three phase voltage drop. Interestingly enough, when fed the same input parameters their answers are significantly different.

Aside from that, I am somewhat confused as to what value to use for the current. The three inverters total 112kW at 480V, and their rated output current totals 135A. 112kW divided by 480V is 233.3A, which is 135A times sqrt3. When I am using a calculator, which value for current should I use, 233.3A or 135A?

To confuse me even more, one site I found said that for a delta connection line current equals phase current times sqrt3, but for a wye connection, line current equals phase current. I had thought, possibly erroneously, that a balanced wye connection was basically the same as a delta connection since there would be no current on the neutral.

(Also posted on Holt. Gentlemen, start your engines. )

I have 6 inverters totaling 112kW, whose 480VAC 3 phase 4 wire outputs are gathered in an AC combiner. The combiner will feed a wye connected stepdown transformer (480Y to 208Y) 205 feet away. What size single copper conductors do I need to run between the combiner and transformer to keep the voltage drop in that segment below 0.5%. Yes, I know that's very low, but I am under some unusual design constraints.

Please show your work (teach a man to fish) and don't just refer me to a calculator; I need to understand how this works.

EDIT: I have gone on line and found three calculators for three phase voltage drop. Interestingly enough, when fed the same input parameters their answers are significantly different.

Aside from that, I am somewhat confused as to what value to use for the current. The three inverters total 112kW at 480V, and their rated output current totals 135A. 112kW divided by 480V is 233.3A, which is 135A times sqrt3. When I am using a calculator, which value for current should I use, 233.3A or 135A?

To confuse me even more, one site I found said that for a delta connection line current equals phase current times sqrt3, but for a wye connection, line current equals phase current. I had thought, possibly erroneously, that a balanced wye connection was basically the same as a delta connection since there would be no current on the neutral.

(Also posted on Holt. Gentlemen, start your engines. )

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## Comments

3,112Solar Expert ✭✭✭✭We can do better than Holt on this one, he said optimistically. :-)

1. Make sure you distinguish between a circuit supplied by a delta or wye transformer and a load which is wired as delta or wye. If you do not keep this in mind, the confusion will be hard to recover from.

2. In a balanced wye source, there will be no current in the neutral. But you can either connect loads or sources from phase to phase or from phase to neutral.

In a balanced delta source, there will be no neutral. So your loads must connect from phase to phase only. (or you can derive a neutral, given a guaranteed balanced condition, but we will ignore that....)

3. The voltage drop is easily calculated as IxR once you know I. (No help there, I know....)

4. Your inverters are using three phases plus EGC or three phases plus neutral? If nominally wye, it is the latter. Each inverter is delivering a balanced output over all three phases or driving just one phase?

5. Assume for the purposes of this calculation that the 480 volt output of the inverters really is wye. That means that the phase to neutral voltage on each leg will be 480 volts.

To get the per-phase current, divide the output power by 3 and then calculate I from P=IV.

The voltage drop which concerns you will be in each of the phase wires only, as long as the source and load remain balanced. A calculator set up for a two wire, single phase circuit will automatically figure a voltage drop in both phase and neutral, giving you the wrong answer.

6. You can now get voltage drop in each single wire after finding the resistance per foot of each wire size. Note that getting the voltage drop you want will probably give you oversize conductors compared to the NEC minimum. I assume that if the wires will be very long you will also be calculating for using Aluminum as an alternative.

7. For a balanced delta, the load current is specified as the current from one phase wire to another, and the power to current relationship is based on that voltage. This is really where the wye versus delta differences all derive from, and the distinction between the transformer type and the load/source connected to it becomes critical.

In a 480 volt delta, the phase to (neutral) voltage will not be 480 when the phase to phase voltage is 480 (totally independent of whether the neutral exists or is carrying current. :-) ) That is where the square root of three comes in, when you draw the triangle vector diagrams with a 120 degree angle at the apex. The circuit is the same when considered as a black box, but the nominal voltage of the circuit is measured at a different place.

In a 480 volt wye, you have both the 480 phase to neutral voltage and the 480 time root 3 voltage from phase to phase available for use.

Did that do it, or are the fish not biting yet? :-)

3,112Solar Expert ✭✭✭✭Part two (caution flag for a few laps there....)

When you connect two delta transformers together (by "lines" of course), the loads will still be connected from phase to phase and therefore from line to line. But each line is supplying current to loads which run in to two different "directions" to the other phase conductors.

The end result of this, after vectors wreak their havoc, is that the current in each line will indeed be root three times the current in each of the two loads which are connected to it phase-to-phase.

In the case of two wyes connected to each other, the points of the triangle are still connected by "lines" but the nominal load current is specified as line to neutral, and only one load is connected to each phase/line wire. Hence the line current and the phase/load current are equal.

As far as the power/current/voltage calculation on the inverters goes, that is more troublesome. I suspect that the maximum current spec which you are looking at may be applicable to an unbalanced load condition, and if the load is balanced you can deliver the higher amperage. Or else the specification is mixing wye and delta nomenclature and the factor of root 3 is coming in because the voltage they are using is the phase to phase rather than the phase to neutral voltage. It all comes down to how the inverters are actually wired internally and how the spec voltages and currents are measured.

3,112Solar Expert ✭✭✭✭Part 3, have the checkered flag ready.....

The mysterious factor of root 3 in the power calculation has finally showed its origins. It is all in semantics.

Take a look at this site for reference: http://eece.ksu.edu/~starret/581/3phase.html

Note that in their diagram, the single letter subscripts denot line to neutral and the two letter subscripts denote line-to-line and that they show a source which is physically wye and a load which is physically delta.

For a balanced system, with power factor = 1 just to simplify the notation:

Total power = 3 (single phase power) = 3 (Va*Ia) which is identical to 3 (Vab*Iab) where Vab= root3 * Va and Ia = Iab / root3.

But if you describe the circuit by specifying the line current (Ia) and the phase-to-phase voltage (Vab), then the product of the two is root3 * Va*Ia. So to correct that, you have to divide by root3. But 3 = root3*root3, so 3 (single phase power) works out to be 3/root3 * Vab * Ia = root3*Vab*Ia. The mysterious root three comes from the fact that the conventional system of referring to 3 phase power mixes line to line voltage with line current. :-(

I am still playing around with all of the permutations, but I think this explains the results of your calculations. The line current (the only thing you can measure directly and the only one which counts for voltage drop) will be 135 amps. The 480 volts is in fact the phase-to-phase voltage, as expected from the designation 480/277. You got 233 amps because you forgot to divide by 3 and also did not multiply by root3. The actual line-to-line current in a delta-connected source or load would be 135/root3.

Fortunately for you, .5% of 277 is a respectable 1.4 volts. and .5% of 480 (2.4 volts) is even better. Choose which one you want to meet:-) But that voltage drop from 135 amps will still be tough if the cable is too long.

And he limps across the finish line running on fumes, but in first place.......

Seriously, I think I have successfully integrated all of the elements of your original question better than the host of responders at Holt. :-)

1,972Solar Expert ✭Three phases plus neutral. Each inverter (five REFUsol 020k's and one 012k) drives all three phases and has a neutral connection.

???? Unless everything I know is wrong (which is possible) phase to phase voltage is 480V and phase to neutral is 277V.

3,112Solar Expert ✭✭✭✭Right, as realized by my third post, and this is the explanation for the root3 in the power calculation. The nominal current is the line current, while the nominal voltage is the phase-to-phase voltage (a mismatch when trying to calculate power.)

1,972Solar Expert ✭OK, I still have much to learn about three phase power, but I am closing in on the solution to the problem at hand. I have established that the spec quoted maximum inverter current is the current in the individual conductors, and given that I can apply it and the values in Table 9 to calculate voltage drop. As to percent voltage drop, I compare Vd to 277V, right?

3,112Solar Expert ✭✭✭✭I would say that you have the option of comparing it to either 277 or to 480 depending on the results that you want to get. :-)

But if you want to estimate the effect of the wire voltage drop on the power delivered (to load or to POCO?), the easiest one to justify would be referencing to 277. Otherwise you would have to include a portion of the voltage drop on each line wire (good old root3 again) to get the effect on a 480 volt load connected phase to phase. And the end result of the calculation would be exactly the same. :-)

A reasonable, but harder to apply specification would be the percentage of output power lost in the lines, which would be totally unambiguous if that is in fact what you need to control. But based on the limited info in your initial post, there may be a specific reason for specifying the voltage drop instead.

In short, yes. Percentage of 277, the line (aka line to neutral) voltage. If this is grid tie, I think you are justified in using a power factor of 1, but that would require more thought for complete confidence.

3,112Solar Expert ✭✭✭✭I see that the guys at Holt are doing their best to confuse the issue. I will attempt to set you straight in a way that will allow you to tell them to go fish. :-)

The issue of power in a three phase circuit is not mystical, but can be confusing because of the terminology used. I will try to start with first principles and derive the correct result and then you will be in a position to re-derive it rather than remembering by rote the paradoxical appearing formula.

1. Your goal is to look at the power which is transmitted through the three lines. Anything else is of theoretical interest, but not relevant from the engineering side. I will consider only the balanced case, since it makes the math easier.

2. You can only measure the current in each of the lines. You have no access to the internals of either the source or the load to determine whether it is physically wye or delta. And in any case you cannot get inside it to measure any other current.

Accepted?3. You do have the luxury of measuring either the line to neutral voltage, Va=Vb=Vc, or the line-to-line voltage, Vab=Vbc=Vca. In the case of what everybody calls a 480 volt three phase circuit, Vab is 480 while Va = 277.

So far, so good?4. The power actually transmitted through each wire (setting PF=1 for convenience) is exactly the voltage times the current. And the power transmitted by all three wires is therefore

Power = VaIa + VbIb + VcIc = 3 VaIa since this is a balanced circuit. There is no need to consider the return path for those currents, since it does not enter into the power calculation.

I am sure you are still with me at this point, but hold on because the next step is the key.5. Notice that this is described as a 480 volt circuit, which corresponds to Vab. But the number we need to use in the power formula is just Va. Fortunately we know that Vab = root3 Va. So Va = Vab/root3. That means that Power = 3*(Vab/root3)*Ia. But that is just root3*Vab*Ia.

We multiply by root3 only because we actually multiplied by 3 and then divided by root3 to get the correct voltage.6. So rather than accept by rote the assertion that to get the power in the entire three phase circuit you must for some reason multiply by root3 instead of the more logical value of 3, I prefer as a teacher (which I have been) to let you see where the formula actually came from in small steps. If you have followed this so far, you can now tell the folks at Holt to go fish. :-)

If it starts to slip away, just return to the fact that, just as in a single phase 240 volt situation, a 20 amp circuit carries 20 amps on both line wires, not 10 amps on each, so a 20 amp three phase circuit has a current of 20 amps in each of the three lines, not 6.67 amps in each.

7. My work here is done.....

Who was that masked physicist anyway?

3,112Solar Expert ✭✭✭✭The entire discussion of transformer ratings and how that affects the current or the voltage drop or anything else is ONLY relevant in that there will be some idle current flowing through the windings of the transformer to create the magnetic field in the core. Although that current has a power factor of 0 (purely inductive) and will not affect the power transmission calculation at all, it will have to be counted to determine the voltage drop in the wires. So Ieffective = root(Iresistive**2 + Ireactive**2). If you were able to connect the inverter output directly to the grid, the grid would present a power factor of 1 and the voltage drop you would see in the wires would result exactly from Iresistive. But once you put one or more transformers into the circuit, the power factor seen on the line connecting the inverter to the first transformer will be slightly below 1 and the current will be larger. How large depends on how oversized the transformer is. :-)

Since you can determine the actual inductive reactance of the unloaded transformer, you can both calculate this current and, if you choose, add a fixed capacitor to each of the three lines at the transformer end to compensate exactly (at 60 Hz) for that inductive reactance. That will reduce the measured voltage drop and also the power dissipated as heat in the wires, if it becomes critical to do so.

1,972Solar Expert ✭I had an epiphany last night. The real issue is power loss in the conductors, and since the current from the inverters is a virtual constant, a 0.5% voltage drop approximately equates to a 0.5% power loss. The power loss is a simpler calculation and the voltage doesn't matter. Figuring on 0.25% power loss in the conductors between the combiner and the T'former (I will allocate the other 0.25% to the inverter to combiner runs):

D = 205'

I per conductor = 135A

Max AC power = 112kW

What resistance will limit loss to a combined 0.25% in the three CCC's when the system is running at full power?

(.0025)(112000)/3 = 93.3W loss per conductor

R = P/I^2 = (93W)/(135A)^2 = 0.00512 ohms/205' = 0.025 ohms/1000'

From Table 9 for uncoated copper in steel conduit, it will take 600kcmil or parallel runs of 300kcmil. Ouch.

I will use the same method for calculating the wire size between the inverters and combiner. Obviously, with these loss constraints, ampacity will not be the driver.

The transformer I'll need is new ground for me. Any suggestions? It needs to step down from 480Y/277 to 208Y/120 and handle 112kW. The inverters' PF is within 5% of 1.0. The inverters require a neutral, so I believe that that means I'll need a wye to wye transformer.

3,112Solar Expert ✭✭✭✭Ouch indeed. Quick question: Is the voltage drop spec only to be applied to the combiner to transformer leg or is it intended to be an overall voltage drop over the whole path from each inverter through the combiner to the transformer?

It looks like you have some long runs from some of the inverters to the combiner, and so you will be getting some voltage drop there, even though the current is lower and the wire size will be more manageable.

If one reason for limiting the voltage drop is the requirements of the anti-islanding circuit, you will need to consider the sum of the two voltage drops, although the resistance in the combiner leg will affect any interactions among the separate inverters!

As to the transformer, you need to use a wye on the inverter size, because of the neutral requirement. But you have a choice of delta or wye on the output side. Which you use there will depend on what the utility side wants to see and such concerns as phase shift, etc. Certainly wye-to-wye is the simplest to understand, since you can consider it to be three separate transformers combined into one package. (You can also build what you want using three separate transformers, properly rated for 1/3 of the total power!)