Response from RFI concerning current carrying conductors. Am I crazy

[jaggedben]

I understand why a 20 amp wire can't be used because of derating of course, but why can't a 20 amp breaker be used for each heat trace? Maybe I'm missing something

Nobody said a 20A breaker can't be used. But not with 12awg wire derated that much. If the circuits can use 15A breakers then the wire is fine. If the circuits require 20A breakers then they need larger wire or an arrangement with less of a derating factor.

 

[Tainted]

Nobody said a 20A breaker can't be used. But not with 12awg wire derated that much. If the circuits can use 15A breakers then the wire is fine. If the circuits require 20A breakers then they need larger wire or an arrangement with less of a derating factor.

Lets use Dsg319' example there are 18 CCC for heat traces at 15 amps per heat trace

that means 15x125% = 18.75A wire

to derate the wire it's 50% meaning you need 37.5A wires

that means you need a 20 amp breaker and a minimum 37.5A wire for each heat trace correct?

 

[Dsg319]

Lets use Dsg319' example there are 18 CCC for heat traces at 15 amps per heat trace

that means 15x125% = 18.75A wire

to derate the wire it's 50% meaning you need 37.5A wires

that means you need a 20 amp breaker and a minimum 37.5A wire for each heat trace correct?

I’ve never derated that way. I take the wire at its 90degree ampacity if applicable for example #12@ 30amp

30x.50=15

 

[Tainted]

I’ve never derated that way. I take the wire at its 90degree ampacity if applicable for example #12@ 30amp

30x.50=15

In my example, I use 90 degree column at the end of the calculation. In my example on post 22 you need a #10 wire

 

[Dsg319]

In my example, I use 90 degree column at the end of the calculation. In my example on post 22 you need a #10 wire

Same result. I guess that’s the same as the continuous load with 125% percent added Vs only loading the breaker to 80%

 

[jaggedben]

Lets use Dsg319' example there are 18 CCC for heat traces at 15 amps per heat trace

that means 15x125% = 18.75A wire

to derate the wire it's 50% meaning you need 37.5A wires

that means you need a 20 amp breaker and a minimum 37.5A wire for each heat trace correct?

There has been some lack of clarity whether the load requires at minimum 37.5A wire (before derating) on a 20A breaker or 30A (before derating) wire on a 15A breaker. But everyone agrees that 30A (before derating) wire on a 20A breaker is neither of those, and not protected.

In my example, I use 90 degree column at the end of the calculation. In my example on post 22 you need a #10 wire

If the load is continuous, then we all agree that's correct.

 

[James Williams]

I understand why a 20 amp wire can't be used because of derating of course, but why can't a 20 amp breaker be used for each heat trace? Maybe I'm missing something

If you run that number of #12s with a 90°C rating for the conditions in a single conduit, the adjusted ampacity is 15A for each wire.
The wire then needs to be protected at 15A, hence the 15A breaker.
With that number of wires of #10 with a 90°C , the adjusted ampacity is 20A. So you would have use #10 to be allowed to use a 20A breaker.

 

[Tainted]

If you run that number of #12s with a 90°C rating for the conditions in a single conduit, the adjusted ampacity is 15A for each wire.
The wire then needs to be protected at 15A, hence the 15A breaker.
With that number of wires of #10 with a 90°C , the adjusted ampacity is 20A. So you would have use #10 to be allowed to use a 20A breaker.

I understand why #10 wire is needed. What I am saying is that a 20 amp breaker is needed per heat trace regardless of how many wires are in a conduit because the continuous load for heat trace is 15 amps. 15x125% = 18.75amps

 

[infinity]

I understand why #10 wire is needed. What I am saying is that a 20 amp breaker is needed per heat trace regardless of how many wires are in a conduit because the continuous load for heat trace is 15 amps. 15x125% = 18.75amps

Two different requirements but you're correct if the load is 15 amp continuous you'll need a 20 amp OCPD.

 

[Tainted]

Two different requirements but you're correct if the load is 15 amp continuous you'll need a 20 amp OCPD.

To clarify: Reason I brought up continuous is because OP states that it's 15 amp continuous.

Generally, I don't really see how something like heat trace would be a continuous load, would it? Maybe if you're in Antarctica lol