80 percent of MCA of AC for calculation ??

[mark henderson]

I am being told I can use 80 percent of the MCA while I am calculating the service of a residence. Is there an article that supports this?

Thanks
Mark

 

[jumper]

No.

The MCA is the sum of all the continuous and non-continuous loads of all the parts of a unit.

If the AC is larger than the heating load, tis simpler to just use the MCA and be done with IMO.

 

[mark henderson]

I have been through 220.40 through 220.84.just don't see any thing to support what they are telling me.

 

[jumper]

I have been through 220.40 through 220.84.just don't see any thing to support what they are telling me.

That is 'cause it ain't there.

Yoda says "Seek what does not exist and you will not find."

 

[Smart $]

IMO, if you are comparing A/C to Heating, the MCA tips the scale if watts of each are equal. I've never drawn any conclusions on how to equalize the two for a true comparison. If heating was totally a motor load, I'd multiply by 125% and then compare... rather than take 80% of the A/C. But that's just sheer speculation.

220.14(C) says branch circuit load is to be determined per 440.6 for A/C... and 440.6(A) Exception No. 1 says "Where so marked, the branch-circuit selection current shall be used instead of the rated-load current to determine the rating or ampacity of the disconnecting means, the branch-circuit conductors, the controller, and the branch-circuit short-circuit and ground-fault protection."