Pull Box Sizing

J

Jacob Gerstner

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Location
Michigan
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Journeyman
Hello, I could use some help sizing this pull
box. I will attach a photo with what I need to accomplish. Thanks for any feedback.
 

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So this question is covered by 314.28(A)(2). But in order to apply the text there, we need to know what a "row" is.

For example, are the 12 left-side entries 1 row of 12, or 12 rows of 1? Likewise, are the back entries 2 rows of 6, or 6 rows of 2? Or if the two apparent rows are staggered, does that count as 12 rows of 1?

Once we have that sorted out, applying the rules there should be pretty straightforward. But reading the section fresh, I am surprised there is no clarification as to what a "row" is.

Cheers, Wayne
 
wwhitney said:
So this question is covered by 314.28(A)(2). But in order to apply the text there, we need to know what a "row" is.

For example, are the 12 left-side entries 1 row of 12, or 12 rows of 1? Likewise, are the back entries 2 rows of 6, or 6 rows of 2? Or if the two apparent rows are staggered, does that count as 12 rows of 1?

Once we have that sorted out, applying the rules there should be pretty straightforward. But reading the section fresh, I am surprised there is no clarification as to what a "row" is.

Cheers, Wayne
Click to expand...
Thanks for taking the time to help. Left side will be 1 row of 12. Back wall will be two rows of 6, somewhat staggered as I have a to work with the meter pack on the other side of the wall.

Does that make sense? Thank you again!

-Jacob
 
Jacob Gerstner said:
Any help?
Click to expand...
I'll take a stab at it, although I've not gone through 314.28(A)(2) before.

First, 314.28(A)(2) second paragraph says that the distance between raceways containing the same conductor shall be at least 6 time the nominal conduit size. So if you have an unspliced conductor going from say the top entry on the left side to the top left exit on the back, you'd need 12" between those two entries (presumably center to center). But if you can route your conductors so that the top on the left goes to the far right, and the bottom on the left goes to the top left, etc, you probably can comply with that requirement without having to make the box wider to have the clear space between the two groups of opening. Or if you are splicing, this is moot. Anyway, I will not include any further allowance for that requirement; if one is necessary, you'll have to determine that yourself and add it on.

For the height of the box, there are no entries on the top or bottom. So the height is not regulated by 314.28(A)(2). It would be determined by needing to fit the row of 12 2" entries into the left side. I don't know the diameter of a 2" locknut, but a quick google search turned up 2-7/8" or 2.9". So if 3" center to center is enough clearance, you'd need at least a 3*12" = 36" tall box.

As to the depth of the box (perpendicular to the page of the drawing), the exception covers that. Since your largest conductor is 2/0, Table 312.6(A) says that 3-1/2" is the minimum depth. So a 4" deep box would work. I have no hands-on experience, so can't say whether it would be useful to use a deeper box like 6" deep.

Then the width of the box is governed by the first paragraph of 314.28(A)(2) and the entries on the left. If you had (1) 2" entry, the minimum width would be 6*2" = 12". Now if you treat the 12 entries as 1 row of 12, then you need to add to that 11*2" = 22", for a minimum width of 34". Whereas if you treat the 12 entries as 12 rows of 1 entry, the answer would still be 12". [I guess a 1-1/2" locknut is a bit under 2-1/2" diameter, so for the the row of 6 1-1/2" entries along the back to fit, you'd need at least 15". And the first paragraph could force you to go a bit bigger.] Since I'm not clear on what a "row" is and what the point of the extra allowance is, I can't advise which of those two answers is correct. But certainly using the larger of the two is conservative.

So rounding up to a multiple of 6" on width and height, I get either a 18" w x 36" h x 4" d, or 36" w x 36" h x 4" d, depending on the interpretation of "row". Although you need to double check my locknut diameter figures.

Cheers, Wayne
 
wwhitney said:
So if you have an unspliced conductor going from say the top entry on the left side to the top left exit on the back, you'd need 12" between those two entries (presumably center to center). But if you can route your conductors so that the top on the left goes to the far right, and the bottom on the left goes to the top left, etc, you probably can comply with that requirement without having to make the box wider to have the clear space between the two groups of opening. Or if you are splicing, this is moot.
Click to expand...
And this is what really matters not the actual size of the box. IMO this needs to be rewritten because it's too complicated and confusing without providing any real world benefits. You do this dumb calculation to size the box but that calculation doesn't take into account the distance between the raceways (that's a separate component) which is all that really matters as far as pulling in the conductors.
 
As noted above, adhering to the 6" between conduits containing the same conductors might be a bit problematic so you will likely have conductors crossing over each other. For that reason I would likely use a 6" deep box.
 
Min wire bending radius for 2/0 is 3.5 in.

I think you would need at least a 8" deep considering the knockout location, and the room the conductors themselves take to keep the kinks out.
I would want a 12" or deeper if I could find one reasonably priced.
 
infinity said:
Do you think that the exception applies to this pull box?
Click to expand...
My understanding of the drawing is that the back of the box has the two rows of six, and the front of the box is a removable cover. So yes, the exception would apply to the distance between the back of the box and the front cover.

Cheers, Wayne
 
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