application check

[copper chopper]

2 heads are better than one so can you tell me if this application is illegal or not..

a 6x6x4 box ( I dont know the cubic in of it yet) with 2- 3/4 inch emt entering it on the same side and 2 - 12/7 mc cables that fit in a connector which goes into a 1/2 inch k.o.



is it 6 x .75 =4.5
then 4.5 + .75 = 5.25
then 5.25+ .5 = 5.75
then 5.75 = .5 = 6.25

answer is 6.25 making this illegal or am I missing something???

 

[ptonsparky]

Explain what you are trying to do.

12/7?

 

[cadpoint]

12/7 is really 12/6 with ground ? Yes no maybe ....

 

[kwired]

Explain what you are trying to do.

12/7?

Sounds to me like he may be trying to size a box according to 314.28.

That section is only used if there are conductors 4AWG and larger.

He did not say what he has for conductors in the 3/4 inch raceways but is pretty good chance they are smaller than 4 AWG.

If all conductors he needs to use 314.16 to determine what minimum size box is acceptable.

He needs to use 2.25 cubic inches for each 12 AWG conductor.

6x6x4 box is 144 cu in. If all conductors are 12 AWG then there is a max fill of 64 conductors.

 

[cadpoint]

....
If all conductors he needs to use 314.16 to determine what minimum size box is acceptable.

He needs to use 2.25 cubic inches for each 12 AWG conductor.

6x6x4 box is 144 cu in. If all conductors are 12 AWG then there is a max fill of 64 conductors.

61 CCC, they have to account for all grounded condutors by adding one ground (2.25). and you didn't use 2 in your math either there joints.

 

[cadpoint]

6x6x4 =144 sq in

from your words
14 (from 2 sets-7 from one side) +[ Z (total number of free wires the other side] x 2.25 x 2 + 2.25* (for all grounds) ((use 2 because your make a joint))

or total number CCC of free wires that make a joint (assuming all #12's) x 2.25 x 2 + 2.25*

or
28 wires)assume( x 2.25 x 2 + 2.25 = 128.25 sq in

if it's 12/6 with ground then it'll be another total.

Start at Article 314 use Table 314.16(B)

 

[kwired]

61 CCC, they have to account for all grounded condutors by adding one ground (2.25). and you didn't use 2 in your math either there joints.

I knew somebody would bring up the grounds and other conditions. All I wanted to do is point out that OP is likely using wrong article to determine what size of box is required.

I could understand 63 conductors plus grounds, how did you arrive at 61?

 

[cadpoint]

I knew somebody would bring up the grounds and other conditions. All I wanted to do is point out that OP is likely using wrong article to determine what size of box is required.

I could understand 63 conductors plus grounds, how did you arrive at 61?

Well I worked it all up and then bit on your total... my thoughts are in...

 

[infinity]

6x6x4 box is 144 cu in. If all conductors are 12 AWG then there is a max fill of 64 conductors.

Sounds pretty simple to me. :happyyes:

 

[copper chopper]

I see now, there all 12 awg conductors each mc cable has 3 hots-3 neuturals- and 1 ground so that would be a total of 12 and 1 for all grounds

13 x 2.25=29.25

box is 144 cubic inches

so box size is fine, however the 3/4 emt having 6 hots and 6 neuterals in it is illegal unless you derate them to 50 percent.

 

[kwired]

I see now, there all 12 awg conductors each mc cable has 3 hots-3 neuturals- and 1 ground so that would be a total of 12 and 1 for all grounds

13 x 2.25=29.25

box is 144 cubic inches

so box size is fine, however the 3/4 emt having 6 hots and 6 neuterals in it is illegal unless you derate them to 50 percent.

Separate issue but likely yes. If you need 20 amp conductors yes. If you need 15 amp conductors you are OK as long as you have 15 amp overcurrent device.