# Voltage Drop (Three phase)

by lukas86
Tags: phase, voltage
 P: 70 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
P: 2,284
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

 P: 70 Voltage Drop (Three phase) 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.
P: 2,284
 Quote by 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.
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
 Sci Advisor P: 2,284 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!).
 P: 70 Thanks again, in the 2nd part of that eqn. what is the X?
P: 2,284
 Quote by lukas86 Thanks again, in the 2nd part of that eqn. what is the X?
The line reactance for one conductor (in Ohms).

CS
P: 70
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.
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