Voltage drop Question

[sparks27]

I am studying mike holt electricians math, I came across this question and cant find how the resistance for two #14 conductors comes out to 12.90 ohms

The question is stated: whats the maximum distance that two #14 can be run if they carry 16A and the allowable voltage drop is 10V?
I checked chapter 9 Tbl. 8 but cant find how he arrived at this number Help...

Dist= (4110 x 10V) / (2 x 12.90 ohms x 16A)

A=50ft
B=75ft
C=100ft
D=150ft

Thanks, Sparks

 

[don_resqcapt19]

The 12.9 is ohms per circular mil area per foot. It is commonly shown as "K" in voltage drop formulas and often called the "K factor". The 12.9 is based on a conductor temperature of 75°C and would be less for lower temperatures.

The 12.9 is the commonly used value and is a worst case as the operating temperature of most conductors is limited to 75°C by the temperature requirements of the terminations.

I believe the equivalent value for aluminum conductors is 21.2.

K = specific resistivity (Ω - circular mils/foot)

 

[sparks27]

Voltage drop

Voltage drop

The 12.9 is ohms per circular mil area per foot. It is commonly shown as "K" in voltage drop formulas and often called the "K factor". The 12.9 is based on a conductor temperature of 75°C and would be less for lower temperatures.

The 12.9 is the commonly used value and is a worst case as the operating temperature of most conductors is limited to 75°C by the temperature requirements of the terminations.

I believe the equivalent value for aluminum conductors is 21.2.

K = specific resistivity (Ω - circular mils/foot)

Don,

Are those values found in chapter 9 table 8 ?
I found that #14 has 3.19 ohms per 1000 ft. at 75c coated copper

Thanks, sparks

 

[don_resqcapt19]

Don,

Are those values found in chapter 9 table 8 ?
I found that #14 has 3.19 ohms per 1000 ft. at 75c coated copper

Thanks, sparks

They are not in the NEC as far as I know.

Take the resistance in the table and multiply it by the circular mil area and divide that result by 1000 to get the K factor.

Note the 3.19 ohms per 1000 feet is for "coated copper", it is 3.07 for uncoated copper and that would give you a K factor of 12.6

 

[WaxVolt]

Strange enough I am studying as well and tonight was vd night. considering this is evidence of pure capitulation, or maybe resonance would be more fitting, that it was actually the first topic that i saw in this forum. I figure i might as well post this here.

I noticed on figure 7.27 in mikes book that he has the cmils for 3awg as 83690. i may be mistaken but in tbl 9 it says its 52620. this seems like the fastest way to let them know.:bye:

 

[Dennis Alwon]

Strange enough I am studying as well and tonight was vd night. considering this is evidence of pure capitulation, or maybe resonance would be more fitting, that it was actually the first topic that i saw in this forum. I figure i might as well post this here.

I noticed on figure 7.27 in mikes book that he has the cmils for 3awg as 83690. i may be mistaken but in tbl 9 it says its 52620. this seems like the fastest way to let them know.:bye:

Correct 83690 is the cir mil for #1 awg. Unfortunately Mike doesn't post here much so you need to contact them at the website. Call them at 888-632-2633 and they can make the correction