Re: I HAVE ANOTHER QUESTION
First, your use of the phrase ?allowable ampacity? does not make sense here. Are you asking for the size of conductor that you need to install, as a minimum, in order to be able to handle these four loads, presuming that there are no other loads? Or are you asking what the actual current will be from these four loads? Or are you asking how much current you need to take into account, when counting these loads as part of an overall service calculation? What, in short, is the purpose for which you are performing this calculation?
Let me assume that the lighting loads are 120 volts, and that the three lighting loads are installed in a balanced way over the three phases of the supply. You have clarified that the heater is 3-phase, so I infer it is rated at 208 volts. You start any such calculation by adding the power, not the current. That part is easy: 15 + 15 + 15 + 30 = 75 KW. I believe we can treat these lighting and heating as purely resistive loads, and so state that the total load is 75KVA.
The current in any one phase would be 75 KVA divided by 208 volts and divided again by the square root of three. The common shortcut is to divide 75 KVA by 360, and get 208 amps (you can ignore the coincidence that it matches the voltage number). To figure the smallest conductor needed to serve these loads, multiply by 125% (since both lighting and heating have been declared to be continuous). You need a minimum of 260 amps worth of conductor. A 300 MCM THHN Copper would do the trick. As to service load calculations, you go back to the 75 KVA, and multiply by the 125%. These four loads add 93,750 to the service calculation. Please note, however, that the lights might have been taken into account in another part of the calculation, in the form of VA per square foot, so take care to avoid duplication.
Does this answer your question?
