# Voltage Drop (Three phase)

1. Jul 16, 2008

### lukas86

I was just looking for some insight on this problem. It is not a homework problem, my boss just asked me to figure it out.

Now, there is 120V supply, with 600 feet of cable (copper) running to a load, and 600 ft neutral coming back to the supply. Ignoring the resistance of the load, he asked what was the voltage at the load due to the voltage drop in the cables with a current of 12A.

Now, he wanted to know the values for #12 wire, and #8 wire. Below is what I came to...

VD = 1.732*k*Q*I*D/CM (equation from... h t t p : / / ecmweb.com/mag/electric_code_calculations_17/)

VD: Voltage drop
k: Direct current constant (12.9 for copper)
Q: Alternating current adjustment factor (skin effect), I thought it doesn't apply here...
I: Current (I think, it said "The load in amperes at 100%, not 125% for motors or continuous loads." I just put the 12A here)
D: Distance in ft
CM: Cable in Circ-Mils

#12 Wire
VD = 1.732*k*Q*I*D/CM
= 1.732*12.9 * 12A * 600ft / 6529.8CM
= 24.636V

Now, that is what I thought the voltage drop in the 600ft cable to the load was. Now I am confused because at first I thought that since there is also 600ft of cable back to the source, the voltage across the load would be 70.727V seeing as 120V - 24.636V (cable to load) - 24.636V (cable back to source) = 70.727V across the load.

Then I thought again, well... in 3 phase, if perfectly balanced, there is no current in the neutral lead back to the source. So the voltage across the load would be around 95V.
-----------------------------------------------------------------------------------------

For #8 Wire
VD = 1.732*k*Q*I*D/CM
= 1.732*12.9 * 12A * 600ft / 16509CM
= 9.744V

Now then the same situation as for the #12 wire, if I am doing any of this correctly... should be something like 120V - 9.744V *2 = 100.511V across the load or... 120V - 9.744V = 110.255V.

Any help or requests for further understanding so I could respond to get more help would be greatly appreciated. Thanks in advance.

2. Jul 16, 2008

### lukas86

Code (Text):
AWG       Dia-mils  TPI       Dia-mm    Circ-mils Ohms/Kft  Ft/Ohm    Ft/Lb     Ohms/Lb   Lb/Kft    *Amps     MaxAmps

8         128.49    7.7828    3.2636     16509    0.6282    1591.8    20.011    0.0126    49.973    22.012    33.018
12         80.807    12.375    2.0525    6529.8    1.5883    629.61    50.593    0.0804    19.765    8.7064    13.060

3. Jul 17, 2008

### stewartcs

That equation should work just fine.

In a balanced three-phase system the current in the neutral is zero. If it is unbalanced, the current in the neutral will be the phasor sum of the phases. Hence, the total cable drop for any phase would be the algebraic sum of the voltage drop in the hot wire plus the voltage drop in the neutral wire (keeping in mind this is phasor addition).

In a single-phase system you have to include the neutral wire resistance since there is current flowing in it and thus drops some voltage due to its resistance.

CS

4. Jul 17, 2008

### lukas86

Alright thanks.

If I took a different approach and took the Ohms/(1000feet) value of the cable, and multiplied that by the length (600 ft) then multiply that by the current... would that be another method?

= .6282 ohms/Kft * 600ft
= .37692 ohms

V = IR
= 12A * .37692 ohms
= 4.52304 V....

So that 4.52 V would be the drop over 600 feet of #8 wire. Is this another way, it gives less voltage drop though. It's just that I am not all that clear on these two approaches.

5. Jul 18, 2008

### stewartcs

That would give you the single phase voltage drop in a three-phase system. Since you have a balanced 3-phase system, the total drop would be multiplied by $$\sqrt{3}$$ to account for the phase difference (remember I said you have to perform phasor addition). That is just the equivalent for a balanced 3-phase system.

The approach I gave you the first time and the equation you posted are identical except that the approach I gave you will allow you to calculate an unbalanced load's drop also. Whereas the equation is for a balanced load.

Good luck.

CS

6. Jul 18, 2008

### stewartcs

BTW, I forgot to mention that those equations neglect the reactance in the cable. However, they are of course exact for DC.

If you want the to include the reactance, the approximate voltage drop for AC then use the general equation for line-to-neutral:

$$V = IR \cos{\phi} + IX \sin{\phi}$$

Then multiple by 2 for single-phase or $$\sqrt{3}$$ for three-phase to get the line-to-line drop.

Hope that helps.

CS

EDIT: If you need the exact voltage drop formula, let me know (it's longer to type!).

Last edited: Jul 18, 2008
7. Jul 18, 2008

### lukas86

Thanks again, in the 2nd part of that eqn. what is the X?

8. Jul 18, 2008

### stewartcs

The line reactance for one conductor (in Ohms).

CS

9. Jul 22, 2008

### lukas86

This relates to this problem at the mill. There were readings taken from around the mill, on the PDCs and the breakers/equipment relating to each PDC. Now what I was looking for by posting this PDF attachment was anything odd from the readings.

The only thing I really noticed was that in some cases, the equipment had higher readings than the PDC-MAIN voltages. Now the transformer feeds the PDCs, then the PDCs feed the power to the equipment, so I thought the PDCs should have the highest voltages.

Just looking for thoughts though. Thanks.