My method is to take the tabulated current times 480 (then times 1.732), for two reasons. First, that is the voltage at which the system will operate. Secondly, it gives me a slightly higher (i.e., more conservative) value for the motor load in KVA.
Well, both of you can't be correct about arriving at a more conservative (i.e., "better to big than too small") value for current by using 480v vs. 460v, unless you're each doing the math differently.I would say using 460V is not a bad idea since it takes into account possible voltage drop (depending on how far the motor loads are located) and also would result in a larger current (more conservative).
From the wording of the question, I think that the mission is to calculate load for the purpose of sizing branch circuits, feeders, and services. If that is the case, then "amps" is an input value, obtained from a table, rather than a result.It seems that Charlie is using V and I to derive KVA, while Skeshesh is using V and KVA to derive A.
This is a debatable issue. Part of the problem comes from the fact that we are supposed to take the "running current" values from table 430.250 (instead of from the nameplate), and the fact that that table shows the voltage as 460. OK, we all know that the power system is rated for 480, and that (for voltage drop considerations, perhaps) the motors are rated at 460. My method is to take the tabulated current times 480 (then times 1.732), for two reasons. First, that is the voltage at which the system will operate. Secondly, it gives me a slightly higher (i.e., more conservative) value for the motor load in KVA. I figure that it is better to provide a little more capacity than to provide too little capacity.
That said, do I think that the other engineer is doing the math incorrectly? I do not. It is a valid method as well.
Which calculation are you mentioning?Appreciate the responses. Do any of you have any thoughts on standardizing a voltage so there isn't confusion.
Also, wouldn't an engineer utilize the higher of the two voltages that would increase service size (thus creating a system that was inherently safer) due to the their responsibility to "...hold paramount the safety, health, and welfare of the public."
Thanks again.
The voltage drop itself is a load (I^2*R), so indeed, it is appropriate to use the supply voltage, not the utilization voltage.
Example: a 12AWG single phase circuit will produce ~200VA @ 16A, 250'