VD for length

FranklinMade

Member
Location
Florida
Occupation
Electrician Apprentice
What would be the formula to find length of a circuit for VD?

VD x Cmil / 2 x K x I?????
Is this the correct way to calculate this?
 

FranklinMade

Member
Location
Florida
Occupation
Electrician Apprentice
L = Vd * CM/ 2KI so you have it for single phase!
Thanks….. i asked because i did a simple equation using #10 with 16a and it came out to like 90…. I was thinking that doesn't seem right.
So basically… how far could i run a 16a circuit using #10…. ????
 

HEYDOG

Senior Member
Thanks….. i asked because i did a simple equation using #10 with 16a and it came out to like 90…. I was thinking that doesn't seem right.
So basically… how far could i run a 16a circuit using #10…. ????
What’s the voltage? 120 volts….then that would be correct 90 feet ….240 volt would be about double that 180 feet!
 

FranklinMade

Member
Location
Florida
Occupation
Electrician Apprentice
What’s the voltage? 120 volts….then that would be correct 90 feet ….240 volt would be about double that 180 feet!
I used 120v and came up with the 90.something yeah. Just didn’t seem right. Seems like i have seen a lighting HR on a 20a circuit run further than 90’ before 🤷‍♂️
 

Tulsa Electrician

Senior Member
Location
Tulsa
Occupation
Electrician
My question for you is, K=?
Resistance/ ohms per…..
For exact K @75c you can find it by using resistance of a circular mill foot of conductor.
Most just use 12.9 for cu conductor. Which is exact K for a 500 mcm cu conductor.

You need two items table 8 ohms per foot and the area circular mills. Then divide by 1000.
For a #10 cu, 7 strand
1.24*10380/1000= 12.8712 for K
Using 12.9 is ok for K.
Now you know how to find K and what K is.

Then you can adjust K for 60c and 90c using 1.05

Example: 60c
(1.24*10380)/(1000*1.05)=12.2582

Example: 90c
1.24*10380*1.05/1000= 13.51476

For most applications 12.9 is fine.
Have fun playing with the numbers.
 

FranklinMade

Member
Location
Florida
Occupation
Electrician Apprentice
For exact K @75c you can find it by using resistance of a circular mill foot of conductor.
Most just use 12.9 for cu conductor. Which is exact K for a 500 mcm cu conductor.

You need two items table 8 ohms per foot and the area circular mills. Then divide by 1000.
For a #10 cu, 7 strand
1.24*10380/1000= 12.8712 for K
Using 12.9 is ok for K.
Now you know how to find K and what K is.

Then you can adjust K for 60c and 90c using 1.05

Example: 60c
(1.24*10380)/(1000*1.05)=12.2582

Example: 90c
1.24*10380*1.05/1000= 13.51476

For most applications 12.9 is fine.
Have fun playing with the numbers.
Thank you. Why is the 1.05 in different parts of the equation?
 

Tulsa Electrician

Senior Member
Location
Tulsa
Occupation
Electrician
Thank you. Why is the 1.05 in different parts of the equation?
This an effect on the conductors resistance at different temperature. I used 15 degree as it relates to 60,75 and 90 c.

Table 8 is based on 75c. For testing this change may effect an ansawer.

You get 1.05 from the coefficient for the conductor material in this case copper. Which is .00323

So let's go back to table 8 and go to the notes. Note 2 gives you this information on conductor coefficient @75c
What it also gives is a weird formula on how to correct for temperature difference Resistance.

What I provided was item to put in your book for testing. Here is a short cut for 1.05.

You can take the 15c change and the coefficient of cu and get close to 1.05.
CU
(15*.000323)+1=1.04845 round up to 1.05
AL
(15*.00330)+1=1.0495 round up to 1.05

I can walk you through the weird equation if you want. I just made a spreadsheet which had the formula account temperature chang in VD.

Once you understand what the formula is asking it's pretty simple.
R2 is the new resistance.

R1 is the table 8@75c.

The 1+ is what I used in the short cut.

a is the coefficient shown for either CU or ALat 75c or temperature coefficient. Which I also used in the short cut.

T2 is the new temperature.

-75 is the 75c.

Another good one is in 310.15 (3)(B)(2)
 
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