Samantha Koch, Anixter Wire and Cable Applications Engineer, walks us through how to calculate conduit fill.

How to Calculate Conduit Fill Video Transcript


Hi everyone. My name is Samantha Koch and I work as an Application Engineer in the Wire & Cable department here at Anixter.


Today, I’ll be talking about conduits, particularly how to calculate a conduit fill. So, today, we have with us two different cables. We have a VFD (variable frequency drive) and a tray cable. We are also going to assume we have a type-2 EMT (electrical metallic tubing) conduit.


What we are going to do is to see if these two cables can fit in that conduit. There are three different things to take into account when calculating a conduit fill:
1. The number of conductors
2. Total cross-sectional area
3. National Electrical Code (NEC) 2011 standar
ds



The most important variable when selecting your conduit is the number of conductors being installed. We need to take into account the number of conductors being installed now, as well as allow some wiggle room for future installations.


The diagram shows conductors/cables lying in a conduit in sets of one, two, three and four. NEC 2011 Chapter 9, table 1, shows that the maximum conduit allowance.


As you can see two conductors is the smallest maximum fill. Looking back to the diagram, this is due to the two conductors or cables forming an oval, rather than a circle.


According to the NEC, Chapter 9, note 9, a multiconductor or flexible cord containing two or more conductors shall be treated as one solid conductor for the calculation of conduit fill.


So looking at our VFD, we notice we actually have four conductors; however, the system treats it as one conductor. This is the same for our tray cable. Therefore, we have two conductors total for this calculation.


Moving on, we need to calculated the total cross-sectional area. This is done by using the formula:

A = piD2 or A = 0.79D2
        4


The VFD had an added diameter of 0.65 inches and the tray cable has a diameter of 0.44 inches. Inserting these into the equation we find that the cable tray has a total cross-sectional area of 0.15 and the VFD has a total cross-sectional area of 0.33 inches.

Cable 1 – VFD
A = 0.79D^2
A = 0.79(0.65)^2
A= 0.33^2

Cable 2 - CT

A = 0.79D^2
A = 0.79(0.44)^2
A= 0.156^2

Adding these together, we find that the total cross-sectional area is 0.48.
A = 0.33 + 0.156
A = 0.48 in^2


Now we move on to our NEC standards. Looking in the NEC 2011, Chapter 9, table 4, we see that there are various conduits and conduit allowances. You have to select you conduit material, which in our case is EMT(electrical metallic tubing). Looking at the EMT table, we look at our trade size, which is two. Then we look at our maximum fill allowance, which is 31 percent for using two conductors. The NEC requires less than 1.04; the number we calculated was 0.48, which is less than 1.04; therefore, it’s appropriate for this application. 


However, if your number is larger than in the NEC 2011, then you have to select a different material or size of conduit. When calculating conduit fill, not all cables being installed are completely circular; sometimes, they are more elliptical. When this is the case, you should use the eclipse as the outer diameter, and then use the equation we calculated before.  


Thank you very much for tuning in for learning how to calculate conduit fill. I hope you enjoyed your time. See you next time.

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