Sched 80 Fill Calc.

1793 · Apr 7, 2005

I would like to know the best way to calculate the fill for Schedule 80 for a service with different size conductors.

4/0 Al at 11/16"=.6875
.6875/2=.34375
Area R2*PI
Area .34375*.34375=.1181*3.14=3710349

2/0 Al at 9/16"=.5625
.5625/2=.28125
.28125*.28125=.0791015*3.14=.2483787

Now, how do I use these two answers?

Thanks in advance.

Norb

edit for calculations

[ April 07, 2005, 03:51 PM: Message edited by: 1793 ]

charlie b · Apr 7, 2005

Re: Sched 80 Fill Calc.

To start with, I do not know your information source for saying a 4/0 Al is 11/16 and a 2/0 Al is 9/16. If you measure actual conductors and get those diameter values, the rest of your calculation is ok. But I would have gone to Table 8 and read the area directly from the chart. Its answers are 0.219 for the 4/0 and 0.137 for the 2/0.

Your next steps are:

Multiply the area of the 4/0 by the number of 4/0 conductors you will have in the conduit.

Multiply the area of the 2/0 by the number of 2/0 conductors you will have in the conduit.

Add these two answers.

Divide by 0.40.

This tells you the minimum size for the internal area of a conduit that would be filled to no more than 40% of its area by this set of conductors. Call this your ?final result.?

Finally, look at Table 4 for the portion that addresses schedule 80 PVC. Find the smallest conduit that has an area at least as high as your ?final result.?

1793 · Apr 7, 2005

Re: Sched 80 Fill Calc.

Charlie:

Thank you for the walk through. It slipped my mind to go straight to Table 8. For some reason I had in my mind that I had to know the wire insulation type, THHN, TW etc.. I actually measured the overall conductor(insulation).

Thanks for setting me straight.

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[ April 07, 2005, 05:45 PM: Message edited by: 1793 ]

Sched 80 Fill Calc.

1793

Senior Member

charlie b

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1793

Senior Member