jim dungar said:
There is no math that you can see.
The breaker needs to be able to hold the inrush of the motor during starting. Typically a motor has about 8-10 times inrush (the actual range can be found on the motor by referring to its starting code letter). So the 100A breaker needs to hold 520-650A instantaneously. According to their Digest this breaker will not trip instantly until the current is over 900A which is good for this motor. Now we need to look at the breaker trip curve versus the overload trip element. A standard NEMA overload realy is 6X current for 20 sec (there are otherpossibilities). The trip curve for this FA breaker shows it will hold 390A for about 8 secs, so it looks like there could be a selectivity conflict between the overlod relay and this breaker. When I plot a standard 65A overload relay with this FA breaker the actual conflict occurs at 318A or 4.9 times FLA. The end result is that except for the selectivity at true locked rotor conditions this 100A breaker should be useable with a 50HP 480V motor.
That's of course the right way to do it, although most of us don't bother. When you have a motor starter with an OL relay
AND a thermal-magnetic (inverse time) CB, the CB need only be the short circuit protection device; the OL relay becomes the long time over current protection. So the thermal trip elements of the CB become essentially irrelevant, other than having the potential for nuisance tripping as mentioned above. But most thermal-mag breakers don't come with adjustable (or widely adjustable) magnetic trips, so getting one that will hold in on startup means having to use a larger breaker. That's why starter manufacturers use mag-only breakers; they can get the higher trip settings without having to buy a bigger breaker. But we in the field can't use mag-only breakers; we have to use thermal-mag. So what most of us do is to select a thermal-mag breaker based on the magnetic trips only, while still fitting within the 400% (300% if over 100A) allowed in exception 2 of 430.52.C.