Pipe Fill

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topgone said:
#18 MTW - O.D. = 0.113" = 2.9464 mm
Wire area = (Pi X dia^2)/ 4 = 6.818 mm2
2" PVC, Sch. 80-I.D. = 1.939" = 49.2506 mm
Inside area of 2" PVC = 1905.079 mm2
40% of conduit area = 0.4 X 1905.079 = 762.03 mm2
no. of wires = 762.03/6.818 = 111.76; say 112 wires
Click to expand...
where did you find the area of #18 MTW?

Also the area of a circle is PI x radius squared

not PI x diameter squared
 
steelersman said:
where did you find the area of #18 MTW?

Also the area of a circle is PI x radius squared

not PI x diameter squared
Click to expand...
Or:
Code: 
Area of circle =  Pi X radius^2
               = Pi X (diameter/2)^2

                       diameter^2        diameter^2
Area of circle =  Pi X ---------- = Pi X ----------
                            2^2               4
 
topgone said:
#18 MTW - O.D. = 0.113" = 2.9464 mm
Wire area = (Pi X dia^2)/ 4 = 6.818 mm2
2" PVC, Sch. 80-I.D. = 1.939" = 49.2506 mm
Inside area of 2" PVC = 1905.079 mm2
40% of conduit area = 0.4 X 1905.079 = 762.03 mm2
no. of wires = 762.03/6.818 = 111.76; say 112 wires
Click to expand...
Ooops!

Redoing my math:

O.D. = 0.113 inches
O.D. = 0.113 X 25.4 mm/in. = 2.87 mm ( not 2.9464 mm, sorry)

# 18 MTW area = 3.1216 x (2.87^2)/4 = 6.47 mm2

40% of conduit area (see above) = 762.03 mm2

Finally, no. of wires inside conduit = 762.03/6.47 = 117.77; say 117 wires
Duh! A little bit rusty with calcs! This is what we get relying on tables too much.
 
topgone said:
Ooops!

Redoing my math:

O.D. = 0.113 inches
O.D. = 0.113 X 25.4 mm/in. = 2.87 mm ( not 2.9464 mm, sorry)

# 18 MTW area = 3.1216 x (2.87^2)/4 = 6.47 mm2

40% of conduit area (see above) = 762.03 mm2

Finally, no. of wires inside conduit = 762.03/6.47 = 117.77; say 117 wires
Duh! A little bit rusty with calcs! This is what we get relying on tables too much.
Click to expand...
I notice you are using 3.1216 for PI. Don't you think it should be 3.14159265 ?
 
nakulak said:
1.16 is the inside radius(oops) on 2" ?
Click to expand...
we don't really need to know the inside radius of any conduit to calculate conduit fill. The table tells you what 40% fill is. So after calculating the area of a certain conductor (usually you don't need to figure out the area as they are usually listed in the book) you just divide this into the 40% fill number that is given.
 
Thanks guys for the great info...and sorry Gus, I thought this would be a fun one. The conductors are for the fault feed back (24v) to 144 rectifiers. I was planning to install 96 per 2"
 
PVC Data might help

PVC Data might help

steelersman said:
we don't really need to know the inside radius of any conduit to calculate conduit fill. The table tells you what 40% fill is. So after calculating the area of a certain conductor (usually you don't need to figure out the area as they are usually listed in the book) you just divide this into the 40% fill number that is given.
Click to expand...

Please attached. Hope this helps.
FYI
 
Last edited:
topgone said:
Please attached. Hope this helps.
FYI
Click to expand...
As I said the internal diameter of this conduit is listed in table 4 pg. 674 of the 2008 NEC. Not so sure if you have one of these. Doesn't seem that you do. Also this isn't even necessary in this case as the same table already tells you how much 40% fill is. So the only thinkg you need to know is the area of the conductor (which in this case was not given in the same codebook for some reason which was what was creating all of the confusion). Hope this helps. :)
 
Excerpts from page 674 0f nec 2008

Excerpts from page 674 0f nec 2008

steelersman said:
As I said the internal diameter of this conduit is listed in table 4 pg. 674 of the 2008 NEC. Not so sure if you have one of these. Doesn't seem that you do. Also this isn't even necessary in this case as the same table already tells you how much 40% fill is. So the only thinkg you need to know is the area of the conductor (which in this case was not given in the same codebook for some reason which was what was creating all of the confusion). Hope this helps. :)
Click to expand...

I can't attach the page you referred to because of prohibitions by NEC re reproduction but attached is what I saw in that page 674 (actually p. 678 on my e-copy)(2" PVC, sch. 80).

As you can see, I.D. = 1.913 inches (different from what I used in previous computation, 1.939 in.).
We talk the same but different ways of doing things; you used tables, I tried to compute for the sake of understanding.
 
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