Continuous load

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madpenguin

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I'm a little confused here and know you guys can help me out... :wink:

I've been told several times that continuous load circuits need to be at 80% of the OCPD. That means:

15A = 12A
20A = 16A
30A = 24A

But the NEC tells me not to calculate by OCPD but by 125% of the actual load. Which is it because they don't compliment one another.

I'm going to install 4 - 5' baseboard heaters in my attic @ 240v. I'm using 10/2 on a 30A DP breaker.

250w per linear foot = 5000w draw
5000w/240v = 20.83A
20.83A x 1.25 = 26.038A

24A != 26.038A

I think I must be missing something here. The first method tells me I'm over. The second method tells me I have room to spare..... :confused:

Thanks!
 
Right. I was just struggling with that. Had it been 240w per linear foot then it would equal out.

However, I still seem to be struggling. I've been told not to go over 24A on a 30A OCPD continuous load circuit. I'm at 26A.... Oi. Don't know why this is confusing me.
 
madpenguin said:
Right. I was just struggling with that. Had it been 240w per linear foot then it would equal out.

However, I still seem to be struggling. I've been told not to go over 24A on a 30A OCPD continuous load circuit. I'm at 26A.... Oi. Don't know why this is confusing me.
Click to expand...

80% 0f 30 is 24 amps. If your circuit is a continuous circuit then it cannot be over 24 amps. You may however go up to 30 amps if it is not continuous.

Same holds true here--- 24 amps *125%= 30 amps. They are one and the same. One is 125% of the load the other is 80% of the ocpd.
 
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madpenguin said:
:confused::confused::confused::confused::confused::confused::confused::confused:

I didn't think I could get any more confused but I was wrong.

Sorry for being so dense. I must need more coffee or something.
Click to expand...

My mistake it should say over 24 amps. I will correct it. I think I was thinking it cannot be 26 amps but I didn't write that. My apologies.
 
madpenguin said:
Right. I was just struggling with that. Had it been 240w per linear foot then it would equal out.

However, I still seem to be struggling. I've been told not to go over 24A on a 30A OCPD continuous load circuit. I'm at 26A.... Oi. Don't know why this is confusing me.
Click to expand...

You don't have a 26A load.
You have a 20.83A actual load that you multliplied times 1.25 (125%) because it's continuous.
This gave you 26A (minimum circuit ampacity) which is less than the 30A breaker, so you're OK.
You could actually use a 26A breaker if there was one available.

If you had a actual load of 24A, you would multiply it by 1.25 (125%) which would give you 30A, hence a 30A circuit would be required.

You're over thinking it .:smile:

steve
 
I've been told several times that continuous load circuits need to be at 80% of the O

I've been told several times that continuous load circuits need to be at 80% of the O

Some one mentioned this in the past.

IE.. load to wire is 80%.. wire to load is 125%

OCP had best protect both.
 
hillbilly said:
You don't have a 26A load.
You have a 20.83A actual load that you multliplied times 1.25 (125%) because it's continuous.
This gave you 26A (minimum circuit ampacity) which is less than the 30A breaker, so you're OK.
You could actually use a 26A breaker if there was one available.

If you had a actual load of 24A, you would multiply it by 1.25 (125%) which would give you 30A, hence a 30A circuit would be required.

You're over thinking it .:smile:

steve
Click to expand...

Ah..... There it is. Thank you. Your right, I always over think things. Really frustrates my instructor (and myself).

80% method = actual load

DOH!

Thanks guys. Either way you slice it, I still have 3.9A to spare.
 
Very good Steve-- I couldn't figure out why the op was having a problem but I see what he was asking. This is why the forum is a great place. You almost always have more then one set of eyes helping out.
 
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