I have a situation where we have a 125hp 480v 3 phase motor being installed 600 ft from distribution block and power feed at a fascility of ours. My question is in calculating the voltage drop. I used the ugly's "alternate voltage drop calculations" and a .8 power factor (which is what the load is). Why does the power factor have anything to do with the vd, but is not factored in sizing wire in shorter run applications? We ended up choosing 4/0for the job just to be safe. Thanks.
Voltage Drop Calc's w/power factor
- Thread starter Cody K
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Sorry for the delay... I got sidetracked. Managed to dig out my Ugly's Revised 2002 Edition. It's alternate method is seemingly a vast departure from most methods I've come across. I haven't had a chance to evaluate it to any great length... but it appears that the table contains values from a partial calculation of vd. I'm somewhat perplexed when I get to the "Multiply by 1/(#/phase)" part. In the two examples given, both are 3? yet the example calculation shows values multiplied by 1/1... what gives???Cody K said:
For those that don't have an Ugly's on hand, the same method of calculation is on the Web here in pdf format, documernt pg 199 or pdf pg 8 of 10. Ugly's only published the "How to figure volt loss (drop):" and "How to select size of wire:" sections, and only the steel conduit section of copper conductors table on the next page.
To answer the question, sizing of wire has a two part solution: ampacity vs. voltage drop. In shorter runs the ampacity required dictates the size of the wire. All AC circuits have a reactance. However, in many cases the effect it has on voltage drop is negligible. Hence it is not included in the most basic of voltage drop calculations. Nevertheless, on longer runs it becomes a more substantial factor on voltage drop, as the magnitude of its effect is directly proportional to circuit length. As to why reactance adjustment isn't used more often, I can't say for sure. I'm of the impression it has something to do with having used the basic voltage drop calculation for decades and the "haven't had a problem" syndrome follows.
[edited a couple typo's and for easier readability]
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Smart:
I think the part about # per phase refers to parallel conductors. Say for example, you had (2) sets of 600KCM for a 800A feeder.
Cody:
"Why does power factor affect voltage drop". My best answer is because the voltages and currents are sine waves. Since they are sine waves, they are goverened by the trigonometric functions. And a power factor of other than 1.0 means the voltages are out of phase.
It is pretty easy to do calculations with sine waves that are in phase. Basically, 120V+120V =240V when they are in phase. But not so with sine waves that are out of phase. The amount that the waves are out of phase affects the results. 120V+120V that is out of phase might equal 208V. Or it might equal 0. Thats why the power factor affects the result.
I think the part about # per phase refers to parallel conductors. Say for example, you had (2) sets of 600KCM for a 800A feeder.
Cody:
"Why does power factor affect voltage drop". My best answer is because the voltages and currents are sine waves. Since they are sine waves, they are goverened by the trigonometric functions. And a power factor of other than 1.0 means the voltages are out of phase.
It is pretty easy to do calculations with sine waves that are in phase. Basically, 120V+120V =240V when they are in phase. But not so with sine waves that are out of phase. The amount that the waves are out of phase affects the results. 120V+120V that is out of phase might equal 208V. Or it might equal 0. Thats why the power factor affects the result.
kingpb
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Cody K said:
To get some info on this, look at the bottom of NEC Chapter 9, Table 9, Note 2.
LarryFine
Master Electrician Electric Contractor Richmond VA
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kingpb said:
2. Effective Z is defined as R cos(θ) + X sin(θ), where θ is the power factor angle of the circuit. Multiplying current by effective impedance gives a good approximation for line-to-neutral voltage drop. Effective impedance values shown in this table are valid only at 0.85 power factor. For another circuit power factor (PF), effective impedance (Ze) can be calculated from R and XL values given in this table as follows: Ze = R ? PF + XL sin[arccos(PF)].
Thanks for your help
Do all of you worry about power factor when you do day to day vd calculations? Or just doing the simple vd=(12.9*1.73*D*A)/cmil is good enough.
P.S. Without my book I dont know if that equation is right, but you get the idea.
Thanks again guys [and gals]
Cody
Do all of you worry about power factor when you do day to day vd calculations? Or just doing the simple vd=(12.9*1.73*D*A)/cmil is good enough.
P.S. Without my book I dont know if that equation is right, but you get the idea.
Thanks again guys [and gals]
Cody
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