is this right?

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jwest

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I read for voltage drop use 1 phase copper wire 2 x k x i x d divided by cir mills so .....2 times 12.9 ohms times amps times distance divided by cir mills is this correct for an ac circut is it always 12.9 for copper and 21.2 for alum.i looked at chapter 9 table 8 it says for dc is this ok for ac circuts also[i noticed different ohms in the ac area] ---thanks for the help also for 3 phase using copper would it be 1.732 x k x i x d divided by cir mills again 12.9 ohms for copper and 21.2 for alum in the -k- spot

[ March 13, 2005, 07:55 PM: Message edited by: jwest ]
 
Re: is this right?

Yes. You're correct on both points.

And the values for K you posted are typically used but they are generalizations.

Edit: What Table 8 means by direct current resistance is that inductive and capacitive reactances are not included in the resistance values. Usually those can be ignored. The resistance values still apply in ac circuits.

I don't think I'm doing very well explaining it. If I'm making it worse for you I'll try again. :D

[ March 13, 2005, 08:28 PM: Message edited by: physis ]
 
Re: is this right?

im confused so 12.9 [copper] and 21.2 [alum]is always used as k [ohms] no matter what the wire size#12 #14 1/0 #2 etc...thanks
 
Re: is this right?

No, 12.9 and 21.2 are averages based on the direct current resistance of copper and alluminum at 75? C. The only time these values should be used is when the conductor size is not known. I have seen some calculations use as low as 10.5 for the value of K.

To find true K, use the following formula:

K = R x cm / 1000

Ex. #12 solid. 1.93 x 6530 /1000 = 12.6029 Ohms.
(These vaules are using Table 8 of the NEC)
 
Re: is this right?

Yes, if you're using the formula you posted.

But be aware that that formula is a little inaccurate due to the sloppy K value used.
 
Re: is this right?

That's even better.

Bryan, you're an instructor arent you?

Why is this formula even used?
 
Re: is this right?

thanks .would 3 phase be 1.732 x r x cm /1000 for k so in table 9 chapter 8 if i had solid #12 bare wire it is 1.93 ohms and if it was coated it would be 2.01 ohms for k so for every time i need to check for v.d. refer to table 9 chapter 8 for k thanks again im studing to get my lience
 
Re: is this right?

Yes, replace 2 with 1.732 for 3 phase.

Vd = 1.732 x D x I x (R x cmil / 1000) / cmil

[ March 13, 2005, 09:25 PM: Message edited by: physis ]
 
Re: is this right?

Bryan, you've got the R for 14 ga. But you're math's good. :)

Edit: Sorry Bryan, I got dislexia from 3.19 of 14 ga.

I'm having a bad math day.

[ March 13, 2005, 09:37 PM: Message edited by: physis ]
 
Re: is this right?

in the mike holt book pg 102 elect.formulas k=1000 cir mills that is 1000 ft long at75degree c " the constant k value is 12.9ohms for copper and 21.2 ohms for alum" thats where i got my info who is right? this test is gonna be fun all differnt answers to the same question anybody got any ideas ?
 
Re: is this right?

The actual answer to voltage drop really comes from a rediculously simple Ohms law form.

V=IR

That's:

Voltage drop = Current times Resistance.

I don't have any of Mikes books. What I have as a definition for K is the resistance of 1 foot of material 1 circular mil in diameter, at a given temperature.

I think the way most people in the real world calculate votage drop is to get the resistance of the total wire run from table 8 (just divide the resistance per thousand feet by a thousand and multiply that by however many feet there will be) and multiply that by the expected current.

Personally I think this K business is over complicating things and throwing inaccuracy in on top of it.

Edit: Trouble is, this formula seems to be an industry standard that is forced on students with all it's lackings and this is how most appretentices are trained to calculate it and are tested on it. I think it's utterly stupid.

[ March 13, 2005, 10:42 PM: Message edited by: physis ]
 
Re: is this right?

Crossman, it is resistivity, and it's expressed in ohm's.

The whole object of using resistivity doesn't even work because the resistivity itself of a conductor changes as the dimension of the conductor changes from one circular mil. It's meant as some kind of one size fits all that doesn't fit any. :mad:

Edit: Ok, I feel better now.

Edit: And you still have to go to table 8 for cmil, so why not just get the stupid resistance value you need? It's right across from it.

I feel better again.

[ March 13, 2005, 10:56 PM: Message edited by: physis ]
 
Re: is this right?

Vd = 2 x I x D x Rf

Rf = Resistance per foot (From table 8, ohms/kFT devided by 1000)
D = distance
I = Current
Vd = Voltage drop

Forget K.
 
Re: is this right?

in the formula:

R = (K x L)/cma

where R = resistance of conductor in Ohms
L = length of conductor in feet
cma = cicular mil area of conductor

then the unit for K must be (cma x ohms)/foot

Yes this is the ohm value of a conductor with cma of 1 cmil and length of 1 foot, but the specified parameters of 1 foot long and 1 cma must be expressed in the unit of K, otherwise the math falls apart. The units have to work out too, not just the numbers.
 
Re: is this right?

I don't understand where you say the math falls apart?

1 mil foot of copper = 12.9 ohms approx.

1000 mil of copper at 1000 feet = 12.9 ohms aprox.
 
Re: is this right?

What I am pointing out is inconsequential as far as computing the numbers, but to get Ohms on the left side of the equation to equal a constant K times length divided by circular mil area, the unit of K has to cancel out the feet and the circular mil area, otherwise you don't end up with ohms on the left side of the equation.

The beauty of it is, in this case, the units work out and the numbers work out even though one may not be aware of the real units of K.
 
Re: is this right?

Calling for Rattus, Ed, and Al!

One thing I don't catch... the books say that K is the resistance per milfoot of a material.

This implies a unit of: ohms /(cmils x foot)

But the unit has to be : (ohms x cmils) / foot

Any mathematical insight on this?
 
Re: is this right?

Well, this is wrong.

This implies a unit of: ohms /(cmils x foot)
Click to expand...
But what's the question, it might just be me but I'm not sure what the point you're trying to make is.
 
Re: is this right?

I see. Translating the English directly into math.

Well I guess it doesn't work.

It's the resistance of 1 milfoot.

Edit: It works if you keep adding 1 foot lengths in parallel. And this is bothering me because resistance in parallel is the reciprical of the sum of the recipricals.

Edit: never mind, the resistances are equal.

[ March 14, 2005, 02:12 PM: Message edited by: physis ]
 
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