Help with multiconductor conduit fill

[brother]

I want to make sure Im doing this right in my calculation, there are 2 cables , each cable has 19 pairs. The o.d. on one cable is .603 inches. this is NOT square inches so I had to convert. I came up with using 1 1/2" emt conduit NEC 2008 TABLE 4 chapter nine.

Area= 3.14 * R(squared)
R= D/2

D= .603

R= .603/2=0.3015

Area= 3.14 X .3015 X.3015=.285

2X .285=.570 1 1/2 EMT conduit size needed. (0.631 sqr inches for emt)

since there are 2 cables and that means I have 2 conductors and have to use the 2 wires 31% fill.

Did I do this right?? I dont deal with multi cables much, just regular wire!

 

[brother]

Im sure some of the senior mike holt guys can help me out here!!

 

[480sparky]

T4 of Ch 9 give you the ID of 1?" EMT as 1.610", 2-conductor 31% fill as 0.631""

Note 5 of Ch 9 says: For conductors not included in Chapter 9, such as multiconductor cables, the actual dimensions shall be used.

Single conductor, 0.603"OD.

0.603? = 0.367. (Produces a square of 0.367)

0.367? * 0.7854 = 0.2856 in?. (Converts to cmil)

0.2856 in? * 2 = 0.5712 in?.

 

[Volta]

...
2X .285=.570 1 1/2 EMT conduit size needed. (0.631 sqr inches for emt)

since there are 2 cables and that means I have 2 conductors and have to use the 2 wires 31% fill.

Did I do this right?? I dont deal with multi cables much, just regular wire!

Yes, that is what I get. Min 1-1/2 EMT, or any other circular raceway, except PVC sched 80 would be 2" (and PVC type EB, too).