Protection of #10 Conductors

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vpower1989

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I am running into a situation where a piece of equipment came in with an FLA listed on the nameplate that is higher than we were anticipating. We were told to provide a 30A circuit with a L21-30R. #10 conductors have been installed. The equipment shows up and we have an FLA of 25.14A listed on the nameplate. The electricians are asking if they need to pull a new 40A circuit. So I have been going down the rabbit hole of splitting hairs to see if we can use the #10s.

  • So nameplate FLA is 25.14A, we treat it as a continuous load, 25.14 * 1.25 = 31.4A required rating for the conductors
  • We likely have 75C terminations, #10 is rated for up to 35A.
  • No temperature adjustments required.
  • Conduit Derating – 90C cable (THHN) – 40A * derating factor ( .8) = 32A.

So in my mind the ampacity of conductors are fine for the load. I may have to ditch the RCPT and go to a disconnect. One question I have is on the "next size up" rule. Could I protect the conductors described above with a 40A breaker?
 
Does this piece of equipment meet 240.4(G)? Does the equipment have a cut sheet requiring an MOCP of 40A?
 
Rock86 said:
Does this piece of equipment meet 240.4(G)? Does the equipment have a cut sheet requiring an MOCP of 40A?
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I have been told that the equipment has a set of 30A fuses inside of it. I do not know if it has an MOCP on the nameplate, I have asked that question. This piece of equipment falls under 670.
 
Why are you treating it as a continuous load ? Does it actually pull the full power for 3 hours ??
 
augie47 said:
Why are you treating it as a continuous load ? Does it actually pull the full power for 3 hours ??
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The tool itself is a little bit of a black box from my end. It is a manufacturing facility so they will be operating it for 3 consecutive hours all the time. Whether or not it pulls nameplate current is unknown. It is common practice on this site to size for continuous on these manufacturing tools.
 
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texie said:
I would call attention to 670.3 and 670.4
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Right. I've read those sections. There is some commentary in the handbook at the end of 670.4 that more or less states that the FLA is all you really need to know. Now one thing that I've always questioned is whether or not the nameplate FLA already accounts for any continuous aspects of the load. I have kind of fallen into line with how this company does things which is simply take the FLA, multiply by 125%, size your circuit accordingly.
 
zbang said:
Still not a CCC. Absent conditions like very high harmonics, the neutral can only carry the unbalance current of the three other leads, that's why it's not counted (as contributing heat to the wiring).
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Yeah I guess this probably falls under 310.15B(5)(a). I guess I'm not a 100% certain on the harmonics, or whether there are other non-linear loads.
 
Your concern is admirable but perhaps unwarranted,. If the machine has 30 amp fuses I would be comfortable supplying it with a 30 amp circuit and barring further information consider not consider the neutral as a current carrying conductor.
 
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