voltage drop formula

[cvirgil467]

I tuypically use teh folowing VD formula for three phase feeders:

cm = kx1.72xIxD/vd

where,
cm = circular mils of conductor
k= resitvity of copper (12.9)
I = current
D = distance (one way)
VD = allowable voltage drop

for long runs with multile sets this formaul does not seems to work. For example, a 2000a feeder, 200 ft long, 480v with desired VD of 2%.

plugging in the values,

cm = 12.9 x 1.72 x 2000a X 200ft/9.6v = 929,875 cm

diving by 500,000cm i get 1.8 sets of 500kcm condcutors. Not right.

Anyone know what I'm doing wrong.

 

[haskindm]

First of all you can't install 1.8 sets of conductors, so the answer would have to be 2. You should use 1.732 for the sq. rt of 3. I don't see anything else that could be a problem unless there is some part of the question you are missing.

 

[winnie]

What do you mean by wrong? As best I can tell, you are using the correct equation, and getting the correct results.

For a 2000A, 480V three phase feeder, to meet the requirement of a 2% voltage drop at 2000A, you only need 2 500kcmil conductors per phase.

Perhaps you mean 'wrong' because when you look at the conductor ampacity requirements, you find that you need far more than 2 500kcmil conductors per phase. Remember that conductor ampacity and voltage drop are separate issues. A short run of undersized conductor will have low voltage drop, but will still overheat because of the excessive current flow. The above simply tells you that once you meet your ampacity requirements, your voltage drop requirements will be more than sufficiently met.

-Jon

 

[ramsy]

There is another formula for voltage drop based on NEC Tbl.9, which considers alternating-current variables not possible with conventional DC formulas.

Per NEC Tbl.9
Z = (R*PF+XL*SIN(ACOS(PF)))/1000*L

This formula considers changes in load "Power Factors" and, raceway "Reactance values" which alter the numbers.

The largest discrepancy, is between circuit balance vs unbalanced results. The only case-studies I found proving how to show this correctly were proprietary.

When solving for missing variables, such as Length, etc., we still need to work with K, but K doesn't precisely equal 12.9 for cu or 21.15 for aluminum. K = (R*CM/1000)

Ignoring errors in K, if either ferrous conduit, or PF < 1 exists in your installation, the differences between formulas exceed 2% voltage drop for a 200ft, 2000A, 480vac, balanced load, with #500's x2 in parrallel.

If this feeds a motor load, not corrected for 85% power factor, your looking at 3.14% voltage drop, add a ferrous raceway and its 3.6% VD. Additionally remove one phase from this 3-phase feed, 1-phase motors, it becomes unbalanced @ 4.16% VD.

So, conventional wisdom may be far removed from reproduceable results, based on case studies, even if the most-convienent applications are in proprietary form..

 

[tryinghard]

CM = 1.732x(KxQ)xIxD/VD. I'm not sure how exact your trying to get but the "Q" is for eddy currents and can be used for 2/0 and larger it can be calculated by dividing Table 9/by Table 8.