Parallel Conductors - Voltage Drop

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Re: Parallel Conductors - Voltage Drop

Ok - Let me be sure that I understand. If I am running 150', 480V, 3 phase, 370A - and using parallel #3/0's, I calculate the voltage drop on the 3/0 with 2*(.079)*(150/1000)*370*sqrt(3)=15.92V so because I am using parallels the voltage drop is really 7.96V?
 
Re: Parallel Conductors - Voltage Drop

Caculate the VD using one conductor and divide by 2. I think you are multiplying by 2 to start with in your caculation
"2*(.079)*(150/1000)*370*sqrt(3)=15.92V "
Drop the 2 in the caculation.

[ February 07, 2005, 12:44 PM: Message edited by: bob ]
 
Re: Parallel Conductors - Voltage Drop

Ok. I was getting confused between two formulas for voltage drop. One was

VD=(2*K*L*I)/Cm
where K = resistance in ohms of one circular mil of conductor
Cm=Cross section area of conductor (circular mils)
I=current
L=one way length

and
VD=table value *circular length(1000) * circuit load
where table value = chapter 9, table 9 effective z at .85pf for uncoated copper wires

i guess i was using a portion of both calculations. oops. when do you use the table 8 properties vs. the table 9 properties??

and does circular length mean twice the one-way length of the circuit?
 
Re: Parallel Conductors - Voltage Drop

The values in table 8 are DC values, and may NOT be accurate for AC circuits (especially for those with less than unity power factor, or those with conductors larger than 2 AWG).

So use table 9 and your second equation. I think the "circular length" may be a misprint. It seems like "circuit length" is a better description. (In the handbook, they use "circular length" one time, and "circuit length" three times). Using the one-way distance seems to give the right answers.

Steve

Editied because I left out an important "not".

[ February 08, 2005, 12:19 PM: Message edited by: steve66 ]
 
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