Question from journeymans exam book

[Bobhook149]

What size cable try width is required for six #500 kcmil THW conductors, four #750 kcmil THW conductors, and six #1000 kcmil THW conductors?

# 500 kcmil .7901x6=4.7406
# 750 kcmil 1.1652x4=4.6608
# 1000 kcmil 1.4784x6=8.8704

Now I add all these up and get 18.279

In the back of the book it shows this answer: 8.232x1.1 ( so they are multiplying the largest conductor by some sort of multiplier?) Then....

then they get (9.0552)**+4.7406+4.6608=18.4566

can someone explain this too me?

Also how to find fill for wireways.

Thanks
Bob

 

[Dennis Alwon]

I am not sure if their calculation is correct but look at Table 392.10(A) Col. #2 which reference art. 392.10(A)(3).

The multiplier 1.1 is in the table-- look at the notes also

 

[Dennis Alwon]

I got different numbers. Here is what I did and I have no idea if it is correct. Art.392.10(A)(3) says the sum of the conductors smaller than 1000kcm cannot be smaller than the calculation in the Table Col.2

The sum of the smaller conductors is 4.7406+4.6608= 9.4014.

Your example shows 8232 * 1.1. I am not sure where that comes from since the 1000kcm conductors is 8.8704.

So I used 1.1 * 8.8704 = 9.75744
9.4014+ 9.75744= 19.15584. If we look at the table col. 2 we see that an 18 in tray would work since 19.5 ( end column) is greater than 19.15584.