3PH VOLTAGE DROP CALCULATION

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molotov27

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Hello everyone

I'm having some problems with a 3 phase voltage drop calculation. I've looked in several places and my hand calculations never have the "right" answe when compared to an on-line calculator.

I have a 120/208V, 3ph, wye system 200 ft. length with two paralell feeders serving 700 AMPS total. Each feeder has 500MCM conductors and a 1/0 ground conductor and carries 350 AMPS. What is the correct way to do the voltage drop calculation?
 
Re: 3PH VOLTAGE DROP CALCULATION

Voltage Drop = 1.732 x K x D x I / cm

1.732 = 3-phase from source to load and back
K = 12.9 for CU and 21.2 for AL
D = total length of conductors
I = total load in amps
cm = circular mils of the conductors used
 
Re: 3PH VOLTAGE DROP CALCULATION

I found a reference that for conductors 2/0 and larger, there is a "Q" factor that needs to be put in. Any comments about it?
 
Re: 3PH VOLTAGE DROP CALCULATION

By the way, the reference was by Mike Holt himself....I guess that's where I'm getting lost.
 
Re: 3PH VOLTAGE DROP CALCULATION

One more thought....what about the equations of Article 215.2.(A).4?

According to that Article the equations are:
1PH VD=(2 x L x R x I)/1000
3PH VD=[(2 x L x R x I)(0.866)]/1000
L=one way length
R=conductor resistance from Table 8, Chapter9
I=load current

If you try both methods, the one suggested by bphgravity, which I have seen in other references and this one, you get different answers. Any thoughts?
 
Re: 3PH VOLTAGE DROP CALCULATION

Originally posted by molotov27: Anyone can shed any light on which is the correct way to go with this??
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I?ll give it a try.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Using the calculator from the Mike Holt Home Page ?Free Stuff,? I get a result of 3.128 volts. That equates to a 1.5% voltage drop in a 208 volt system. The spreadsheet uses the formula VD = 1.732 x K x L x I / CM</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Using the formula that is described in the NEC Handbook, Article 215.2(A)(4), I get the same result of 3.128 volts. The Handbook uses the formula VD = 2 x L x R x I x 0.866 / 1000</font>
<font size="2" face="Verdana, Helvetica, sans-serif">

If you try both methods, the one suggested by bphgravity, which I have seen in other references and this one, you get different answers.
Click to expand...
What I did differently is that I used the DC Resistance value from Table 8 (i.e., 0.0258 ohms/1000 feet), instead of the ?AC Effective Impedance at 0.85% power factor? value from Table 9 (i.e., 0.043 ohms/1000 feet).

Please note that the explanation in the cited Handbook article explicitly states that it is using the Table 8 information. Note also that the explanation following Table 9 in Handbook says,
Voltage drop calculations using the dc-resistance formula are not always accurate for ac circuits, especially for those with a less-than-unity power factor or those that use conductors larger than 2 AWG.
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The reason that ?larger than 2 AWG? comes onto the picture is that the ?self-inductance? of a conductor is closely related to its cross-sectional area. For smaller conductors, the inductance is so small that it can be neglected.

Finally, note that there is no code requirement limiting voltage drop to a maximum value. It is a design issue. My first suggestion is to use the 2 x K x L x I / CM (or for three phase, 1.732 x K x L x I / CM), or use the Mike Holt Free Calculator (which uses this same formula). My second suggestion is that you accept the result, and look no further into the issue, if one of the following are true:
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Your result is well below the 3% / 5% values suggested by the NEC.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Your conductor run is shorter than 100 feet.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Your conductor size is smaller than # 2.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">If you don?t meet any of these three, then you might take the next step of using the other formula (2 x L x R x I / 1000, with this formula multiplied by 0.866 for three phase), and that you use the Effective Z value from Table 9.
 
Re: 3PH VOLTAGE DROP CALCULATION

Whichever method you use, (k/CM, DC Resistance or Effective Z), how do you handle the specific situation posted?

With two 120/208Y parallel feeders at 350 Amps load each wired with 500 MCM Legs & 1/0 Neutral, how do you account for the different sizes of the phase legs and the neutral in the VD calculation?
 
Re: 3PH VOLTAGE DROP CALCULATION

Originally posted by kenjsil: . . . how do you account for the different sizes of the phase legs and the neutral in the VD calculation?
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That wasn?t the question. The 1/0 is a ?ground,? or more properly an ?Equipment Grounding Conductor,? and not a ?neutral,? or more properly a ?Grounded Conductor.?

But to answer your question, the simplest way I can think of doing a calculation with the phase conductors being a different size than neutral is to use the spreadsheet on the Mike Holt Free Stuff page. You have to enter data twice, and add the results of the two calculations. The first set of data has the full current, the size of the phase conductors, and one half the total distance (i.e., enter the one-way length). Look at the number to the left of ?volts drop,? not the percentage number shown to the right of ?volts drop,? and write that number down. The second set of data has the full current, the size of the neutral conductor, and (again) one half the total distance. Again, write down the number to the left of ?volts drop.? Add the two numbers, and you have the total volts dropped by the circuit. Divide by the nominal voltage of the system, and you will get the percentage of voltage drop.

You can use any of the other methods in a similar way. Do two calculations, using the phase conductor size for a one-way length, using the neutral conductor size for a one-way length, add the two voltage drop values, and then calculating the percentage voltage drop.
 
Re: 3PH VOLTAGE DROP CALCULATION

That wasn?t the question. The 1/0 is a ?ground,? or more properly an ?Equipment Grounding Conductor,? and not a ?neutral,? or more properly a ?Grounded Conductor.?
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Charlie B,
I apologize. I didn't read the posting carefully and thought the 1/0 was the neutral instead of the EGC. (I shouldn't post at 2 am ... :roll: )

Thanks for your response in any case. If we were talking about a 1/0 neutral, I understand calculating the VD seperately with a 1-way length for the 500 MCM and 1/0 conductors. The current in each of the phase legs is 0.866*I, where I is the 3-phase load. But what current would you use for calculating the neutral VD?

Thanks for your time,
Kenneth
 
Re: 3PH VOLTAGE DROP CALCULATION

Originally posted by kenjsil: But what current would you use for calculating the neutral VD?
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Zero amps! :eek:

If you are looking at a three phase load (such as a motor), or if you are looking at the total load on the panel, then you use the 3-phase version of the spreadsheet described above (or equivalently, use the formulas discussed above). But in this case, the conductors are all the same size. The neutral does not come into the picture.

On the other hand, if you are looking at a single phase load, and if the phase conductors and neutral are not the same size, you can use the process I describe above. In this case, the current going out to the load (along the phase conductor) has to be the same as the current returning from the load (along the neutral).
 
Re: 3PH VOLTAGE DROP CALCULATION

Zero amps! :)

But it occurs to me (if I can take this 1 more question): Suppose the 3-phase load isn't balanced. For example, if you're calculating for a 3-Phase 4-wire panel with an assortment of 3-phase and single-phase loads, you can expect some imbalance.

For that matter, if the loads are significantly non-linear then you have to consider the neutral current-carrying. And there's the possibility of additive triplen harmonics which I've read can cause significant neutral current.

I know it doesn't figure in the formulas I was taught to use, but doesn't the imbalanced current returning on the neutral have an effect on the VD? :confused:
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Re: 3PH VOLTAGE DROP CALCULATION

Originally posted by kenjsil:. . . but doesn't the imbalanced current returning on the neutral have an effect on the VD?
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Yes.

However, let us start with a balanced, fully-loaded system with no harmonics. You have the max current the system is designed to handle, and you therefore have the max voltage drop. All of the current is passing through the phase conductors, because the system is balanced, and no voltage drop along the neutral.

Now let us keep the same total load, but rearrange the load so that it is significantly imbalanced. There will be current, and therefore voltage drop, in the neutral. However, there will be lower currents, and therefore lower voltage drops, in one or more of the phase conductors. As a result, the total voltage drop (i.e., counting all four conductors) will be no higher than it was in the balanced condition.

What does harmonics do to the voltage drop? Now you would be into Fourier Analysis, and I no longer remember that branch of mathematics. I can say that the total current will be higher. Also, there will be current in the neutral, even if the loads are balanced (you heard correctly about the triplens harmonics). So I am willing to surmise that the overall voltage drop will be higher. Someone else will have to take up the story from here.
 
Re: 3PH VOLTAGE DROP CALCULATION

hey, thanks for all the information. Just to clarify, the 1/0 is indeed the grounding conductor. I'mm running a full sized neutral conductor (500 MCM).

Does anyone have any comments about the "Q" factor refered on the link that I posted. Apparently the "Q" factor should be somehow taken into account with the K (12.9 CU)....that's definitely where I'm getting lost.
 
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