Re: panel calcs
Originally posted by gnu_mike: How does this balance, since dividing by 360 will give you 1/3 of the amps you installed?
I don?t understand what you mean by asking ?How does this balance?? Was your example of an 8 amp load intended to be just one load out of the many loads on the panel? Or are you saying put one load on the panel, and only that one load?
You are right in noticing that multiplying by 120 volts and then dividing by 360 will have the same result as simply dividing by three. That is fine. You are essentially taking the average current that that one load contributes to the three phases that are available. It is no different than having three shopping carts into which you can put your groceries. If you put nine apples into one cart, the average number of apples per cart is one third of nine, or three apples.
The problem gets tricky on a three-phase system when you consider that some loads can be single-phase 208 volts, and others can be three phase 208 volts. To make sure you are adding ?apples to apples,? you need to convert each load to its equivalent power (in units VA). Then, and only then, can you add load up (i.e., to get a total load). Before this point in the calculation, you would most likely make a math error, if you tried to add amps to amps, and use the result to get an average amps. The final step is to take the total load (in VA) and divide by the voltage (208) and again by the square root of three, to get the current per phase. If the loads are indeed balanced, then the result will be the current that is present on each phase. If the loads are not balanced, then the result will be an average current, and a calculation of the actual current on any given phase becomes a much harder math problem.
