I'm trying to calculate the maximum allowable footage allowed for 5 watt heat trace fed with a 20 amp cb at 277v. I can't figure out how the power adjustment factor needs to be applied to this because the manufacturers spec sheet only shows the maximum length at 240 volts but gives a paf of 1.16 if supplied by 277v.
There are two (closely related) components in the maximum length, IMHO:
1. The maximum length fed from one end based on voltage drop and still getting the rated output at the far end, and
2. The maximum length based on the total current pulled, in terms of the connections at one end and the longitudinal conductors inside the heat trace.
The third factor is just how much current it will draw on a 20a circuit. (I assume the manufacturer specifies a 20A max OCPD?)
Basically though, you will get the 5 watt per foot output at a lower average current when you feed the stuff from 277V. And with a higher temperature in the active element. If 1 and 2 are not factors, then you could go 1.16 times as far. (That is, put 1.16 times as much length on the fixed current breaker.)
But the one thing that I think you are assured of is that the maximum heat output of the trace will be (5 x 1.2) watts per foot when driven from the higher voltage. That means that if you have a high enough thermal load on it, it could end up using 1.2 times as much current for the same length. That would force you to cut the length by 1.2. Fortunately, it is unlikely that the design will be requiring maximum output from the tape. Unfortunately you cannot count on that without some engineering. :happysad:
