Voltage Drop vs. Power Factor

[steve66]

I have a square D motor data calculator (slide rule type). It lists 3 phase voltage drop for each wire size for both 95% power factor, and 80% power factor. The numbers are given in "Line-Line voltage Drop Per Ampere Per 100 Ft. in Magnetic Conduit and copper wire".

The odd thing is how voltage drop varies with Power factor and wire size. For large wires (2/0 or bigger), Voltage drop is smaller for 95% power factor. This is what I would expect.

For small wires (#1 and smaller), the voltage drop is actually greater for 95% PF than for 80% PF. Does anyone have any insight why a better PF causes a greater Voltage drop on small wires?

Steve

 

[charlie b]

Re: Voltage Drop vs. Power Factor

It does seem counter-intuitive, doesn?t it? Without seeing your ?slide rule,? I can?t give you a useful answer. But I have one question. Can you tell if the 80% versus 95% power factor is intended to mean the power factor of the motor alone, or the power factor of the complete circuit (including the branch circuit conductors)?

The only reason I bring that up is that, as the power factor of the motor is increased towards 1.0, the overall power factor of the circuit becomes more and more driven by the inductance of the branch circuit conductors. I note that the impedance of a conductor gets larger, as the diameter of the wire gets smaller (until you deal with wires so small that they are constructed with strands). A #1 AWG has a higher impedance than does a #2/0. That fact might have something to do with your question.

But the unit of measure that you are citing, ?voltage drop per ampere per 100 feet of length,? reduces to a measure of the impedance of the wire (in units of ?ohms per 100 feet?). From Ohm?s Law and from the formulas for the resistance and impedance of a wire, if you divide voltage by amps and by length, you are left with the impedance per unit of length. But that value of impedance does not change with the power factor of the load. So I am left with no way of resolving your quandary.

 

[bob]

Re: Voltage Drop vs. Power Factor

Steve
The formula VD = IR cos Ǿ + IX Sin Ǿ is used to get
a close estimate of VD. VD is L/N, PF is the load PF, R & X are for 1 conductor, 1 way. As shown below, with the PF at 0.95 the R Cos Ǿ is much greater than the X Sin Ǿ. Looking at PF at 0.80 the Sin Ǿ factor has doubled an becomes a factor in the calculation since the R is now multiplied by 0.80. As you said 2/0 seems to be the cross over point in this calculation.

Pf - 0.95
----- x -------- sin ---- r --- cos ---- vd
14 .058 ------ 0.31 -- 3.1 -- 0.95 -- 2.96
12 .054 ------- 0.31 -- 2.0 -- 0.95 -- 1.91
10 .05 ------- 0.31 -- 1.20 -- 0.95 -- 1.15
8 .052 ------- 0.31 -- 0.78 -- 0.95 -- 0.75
2/0 .043 ------- 0.31 -- 0.10 -- 0.95 -- 0.10


PF = 0.80
---- x --------- sin ---- r ---- cos ---vd
14 .058 ------- 0.6 --- 3.10 -- 0.8 -- 2.51
12 .054 -------- 0.6 --- 2.00 -- 0.8 -- 1.63
10 .050 -------- 0.6 --- 1.20 -- 0.8 -- 0.99
8 .052 -------- 0.6 --- 0.78 -- 0.8 -- 0.65
2/0.043 ------- 0.6 --- 0.10 -- 0.8 -- 0.11

I can not get this table to line up

[ November 12, 2004, 01:55 PM: Message edited by: bob ]

 

[jim dungar]

Re: Voltage Drop vs. Power Factor

Charlie,

Some more information on the "slide rule". The formulas are for copper wire in magnetic conduit. This means they are actually using the impedance of the system not just that of the load, but the "look up" data is for the load power factor only.

More detailed tables showing power factor, non-magnetic duct, aluminum wire, and voltage system conversions for can be found in the "Ferraz Shawmut Book of Electrical Information".

 

[steve66]

Re: Voltage Drop vs. Power Factor

Charlie; I see what you are saying about units of impedence not changing with PF. I think Jim has some insight into why the impedence of the entire circuit changes with PF (but it hasn't completely clicked with me yet).


Posted by Bob:

I can not get this table to line up

Thats OK. Thanks for trying, and I see what you are saying. But I guess I was hoping for more of a simple insight. That may be too much to ask for something with such a complex equation.

Steve