Voltage drop formulas Help

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brother

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I was looking through one of my reference books, "electrcians guide to conduit bending" and their formula for Voltage drop is VD=(21.6 X I X D)/Cm (for copper). I was taught it was VD= (12.9 X I X D)/Cm. this is for single phase.

For 3 phase, I was taught 1.73 X I X D/Cm in this book they got VD= (18.7 X I X D)/Cm.

Where do they get this 21.6 for copper and 18.7 for 3 phase in this formula?
 
The formulas I have always been taught/used are:

Single Phase: Vd=(2K*I*D)/CM
Three Phase: Vd=(1.73K*I*D)/CM

K= temperature constant

12.9 for CU at 75 degrees
21.2 for AL at 75 degrees
See Chapter 9 Table 8 notes for different values of K for different temps
 
There are also DC resistance voltage drop formulas:

Single Phase
60 Degree Wire VD = DC Resistance x I x 2 x L / 1000 x 1.05
75 Degree Wire VD = DC Resistance x I x 2 x L / 1000
90 Degree Wire VD = DC Resistance x I x 2 x L x 1.05 / 1000

Three Phase
60 Degree Wire VD = DC Resistance x I x 1.73 x L / 1000 x 1.05
75 Degree Wire VD = DC Resistance x I x 1.73 x L / 1000
90 Degree Wire VD = DC Resistance x I x 1.73 x L x 1.05 / 1000

DC Resistance = Ohm per 1000 feet from chapter 9 table 8
I = Current in Amps
L = Length of wire in feet (one way)


I mostly use the standard
VD = 2KID / CM (Single Phase)
VD = 1.73KID/CM (Three Phase)
K = 12.9 for copper 21.2 for aluminum
 
mxstar211 said:
The formulas I have always been taught/used are:

Single Phase: Vd=(2K*I*D)/CM
Three Phase: Vd=(1.73K*I*D)/CM

K= temperature constant

12.9 for CU at 75 degrees
21.2 for AL at 75 degrees
See Chapter 9 Table 8 notes for different values of K for different temps
Click to expand...

Oops, It was too late for me to edit. Thats actually what I was taught on the single phase, I left out the '2' in my formula, and yes the 12.9 was the constant for K (resistance for copper at 75 degrees)
 
brother said:
Oops, It was too late for me to edit. Thats actually what I was taught on the single phase, I left out the '2' in my formula, and yes the 12.9 was the constant for K (resistance for copper at 75 degrees)
Click to expand...

Not trying to sharpshoot, but you also left out the K value in your 3 phase equation.

Maybe they used a different K value and multiplied it by 2 or 1.73. I don't have a NEC nearby to see if I can come up with 18.7 or 21.6.
 
mxstar211 said:
Not trying to sharpshoot, but you also left out the K value in your 3 phase equation.

Maybe they used a different K value and multiplied it by 2 or 1.73. I don't have a NEC nearby to see if I can come up with 18.7 or 21.6.
Click to expand...

Ok, Let me try this again, basically the formula mxstar211 has (see below) , is the what I was taught. I guess I got to do more 'preview' in my posting. :( I just dont see how this book got the different formula.


Single Phase: Vd=(2K*I*D)/CM
Three Phase: Vd=(1.73K*I*D)/CM

K= temperature constant

12.9 for CU at 75 degrees
21.2 for AL at 75 degrees
 
I think I have the book you are referencing as well. I will have to try and dig it out. Don't feel bad, I am the worst at previewing my posts. :)
 
I think the K value is only good up to 1/0. For 2/0 and large there is Q factor is applied due to skin affect. Maybe they used the Q-factor?
 
VD = 1.732 x K x Q x I x D/CM

?Q? = Alternating Current Adjustment Factor: You must adjust alternating current circuits No. 2/0 and larger for the effects of self-induction (skin effect). You can find the ?Q? adjustment factor by dividing alternating current resistance as listed in Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.

I found this on EC&M.
 
mxstar211 said:
VD = 1.732 x K x Q x I x D/CM

?Q? = Alternating Current Adjustment Factor: You must adjust alternating current circuits No. 2/0 and larger for the effects of self-induction (skin effect). You can find the ?Q? adjustment factor by dividing alternating current resistance as listed in Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.

I found this on EC&M.
Click to expand...
ok, just never heard of a 'Q' factor, I googled it and i suppose that means 'Quality of power' factor??

I thought maybe it might be a something off of Star Trek: Next Generation, have me wondering what does Q have to do with this?? is he up to his tricks again??
 
I am not positive, but I think the skin affect is where the conductor has more current at the outside of the conductor as opposed to the core of the conductor. Which would increase the heat, so the Q-factor is born into the equation for larger conductors. I am by no means an engineer, but that is my interpretation. Hopefully a more intelligent member can come and correct/expand on my statement. :)
 
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