- Thread starter rnnpila
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Code requires largest motor load be included in the summing at 125%. Whether you do it separately or just add the 25% separately is user's choice. The main issue is that the plans examiner or inspector be able to realize which way you did it...Regarding to the the total computation of load of line current.

Do I need to separate the highest motor rating current when I'm applying the demand factor or it must be combined on the demand factor in order to size correctly the breaker of the panel. thanks guys.

Starting current is not part of the service or feeder calculation. The reason for the extra 25% of the largest motor is not to provide for starting current. It is to cover the possibility of continuous and full loading of all the motors. Since it is only a possibility that all motors will be run continuously, and at full load, and more likely they will not, the extra 25% of the largest should cover the demand if it should ever be at its maximum. Perhaps not as stringent as 125% for all continuous loads, but historical evidence has shown the requirement, as is, to suffice in practically all scenarios.

I do not know what you mean using the word "formula" on this topic. There is no [magic? ] formula needed to determine the largest motor... its simply the one that'll use the greatest number of VA.Thank you so much for the information.. Can I ask the formula you using for this topic? I just need some reference and comparisons. Thank you.

The method (not formula) for such determination is prescribed in Article 220, with reference to other articles. If you're asking for a one line equation, it simply is not going to come from me. An entire Annex ("D") is devoted to providing examples of an Article 220 service or feeder calculation, which is where it all starts.:grin: im asking for the formula you using for finding the total line current of a panel in order to size the main breaker panel and to give the correct wire size to be used, of course there's no formula for the finding the highest motor rating, i know that.. hehe! thanx.

There are too many variables to the method to give you a simple "formula". I, and likely others can help much more if you were more specific...

...other than the previous aspect, the only thing I can fathom you asking is how to do so when there are single phase and three phase load combinations. Let me, us know if this is what you are after...???

I see you are wise beyond your earsok, i will just look up and read the article.

For single phase line to neutral it is rather obvious the VA load is assigned entirely to its connected line. Three phase loads (balanced) you assign one third of the load VA to each line. For single phase, line-to-line, you assign one-half the load VA to each connected line.You are right, Im also concern on the method you used if the load is a combination of three phase and single phase.

This method slightly flawed, but if the system is reasonably balanced out, it suffices... besides, engineers have been doing it this way without a problem since before I was born. No sense in breaking with tradition

Is the flaw you are referring to what is called the oregon fudge factor and due to the fact that there is a 120deg phase shift between currents in a 3 phase system?This method slightly flawed, but if the system is reasonably balanced out, it suffices... besides, engineers have been doing it this way without a problem since before I was born. No sense in breaking with tradition

If I recall correctly, yes. But I've never been one to recognize "o.f.f." as anything meaningful. It is just an peculiarity of the calculations.Is the flaw you are referring to what is called the oregon fudge factor and due to the fact that there is a 120deg phase shift between currents in a 3 phase system?

2/√3 = 1.1547 = sec(30?)

√3/2 = 0.8660 = cos(30?)

30? is the phase difference between Line-to-line and line-to-neutral voltages(wye, of course).√3/2 = 0.8660 = cos(30?)

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