Nipple fill 60%

Therealcrt

Senior Member
Location
Kansas City
Occupation
Electrician
Just doing a for fun equation.

3” EMT nipple (under 24”)
Conduit fill is 60%
#12 THHN conductors

60% fill of 3” emt over 2 wires is 5.307in squared

#12 THHN area for 1 is 0.005in squared

So how many conductors can I fit without exceeding the 60% fill
 
60% fill of 3” emt over 2 wires is 5.307in squared
5.307 in^2 is 60% of the total area of 3" EMT, so you have the right number, but the way you worded it is confusing. Because normally in this context "over 2 wires" means a condition where you get to use 40% fill, whereas you have a condition where you get to use 60% fill.

Typo… #12 is 0.0133in squared sorry
So now all you have to do is divide. : - )

Cheers, Wayne
 
Table C1 lists #12 for 266 conductors in 3" EMT which would be 40 % fill. 60% fill would be 50% more or 133 more conductors. So 266+133=399.
That's so funny.
I remember my father teaching me as an lad.
6/4=1.5 (.60/.40)
I was so confused when he said it's 50% more.
266*1.5= 399
I swore it was only 20% more.
266*1.20 = He laughed and now I get to as well, I sure miss him.

Thank you for that.
 
That's so funny.
I remember my father teaching me as an lad.
6/4=1.5 (.60/.40)
I was so confused when he said it's 50% more.
266*1.5= 399
I swore it was only 20% more.
266*1.20 = He laughed and now I get to as well, I sure miss him.

Thank you for that.
You're welcome. I'm a path of least resistant guy and I loathe the conductor property tables. Looking up the 40% fill in C1 and adding 50% just seems easier.
 
There is a blindspot in the 60% fill rule, for 2 conductors. Since it is impossible to pack 2 identical circles in the same circle to more than 50% fill by area, you wouldn't be even able to fit them with a 60% fill. By the logic behind the 60% fill rule, it really should be a 41% fill for 2 conductors. The typical fill limits represent about 75% diametral fill, and the 60% fill rule represents about 90% diametral fill.
 
There is a blindspot in the 60% fill rule, for 2 conductors. Since it is impossible to pack 2 identical circles in the same circle to more than 50% fill by area, you wouldn't be even able to fit them with a 60% fill. By the logic behind the 60% fill rule, it really should be a 41% fill for 2 conductors. The typical fill limits represent about 75% diametral fill, and the 60% fill rule represents about 90% diametral fill.
Assuming both conductors are identical diameter. Though if both carrying same current they sort of should be if they are same conductor type.
 
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