Okay, we'll assume that determining the equivalent horsepower ratings would apply to motor circuit switches as outlined in NEC 430.109(A)(1). These motor circuit switches are horsepower rated and therefore the 115% rule of NEC 430.110(C)(2) would not apply because NEC 430.110(C)(1) is titled Horsepower Rating.
I don't follow. The horse power rated motor circuit switch in 430.109(A)(1) will also have a Current Rating (see 430.110(A) exception.) The switch must meet the requirements for BOTH the ampere rating AND the horsepower rating - See the last sentence in 430.110(C): "The ampere AND horsepower ratings of the combined load shall be determined
as follows."
Both (C)(1) "Horsepower Rating" and (C)(2) "Ampere Rating" follow 430.110(C). Both must be applied for Combination Loads.
But the problem still remains that two different horsepower values would be calculated and that it's a given that the locked-rotor HP is always gonna be higher. Why include the rule of calculating the equivalent HP using FLC values?
Yes, the equivalent horsepower value will always be higher for the equivalent locked-rotor current than for the equivalent full-load current. So you would only use the
equivalent locked-rotor current when determining the required
Horsepower Rating of the disconnecting means in 430.110(C)(1). You would then use the
equivalent full-load current as determined by 430.110(C)(1) to determine the required
Ampere Rating of the disconnecting means as directed in 430.110(C)(2).
In the example I gave above, the calculations showed that a 30A, 240V, 7.5HP rated switch met the Ampere Rating requirements, but not the Horsepower Rating Requirements, requiring a larger Horsepower Rated switch. Here is an example where the opposite would be true; the switch would meet the Horsepower Rating requirements, but not the Ampere Rating requirements, requiring a larger Ampere Rating on the switch:
You have a 60A, 240V, 15HP rated disconnect switch. To this you would like to connect one (1) 5HP, 230V motor, one (1) 3HP, 230V motor, two (2) 0.5HP, 230V motors, and one 10kW, 240V, 3ph heater.
Looking at the Horsepower Rating first. The equivalent locked rotor current is 92A + 64A + 20A + 20A + 24.1A = 220.1A. From Table 430.251(B), you see that 220.1A falls above 10HP and below 15HP, so 15HP is the minimum required horsepower rating of the switch. The chosen switch has a 15HP rating, so the switch meets the Horsepower Rating requirement.
Then looking at the Ampere Rating. The equivalent full-load current is 15.2A + 9.6A + 2.2A + 2.2A + 24.1A = 53.3A. 430.110(C)(2) requires the switch to have an ampere rating of not less than 115% of the equivalent full-load current. 53.3A * 115% = 61.3A. You can see that this exceeds the Ampere Rating of the chosen switch. In this case, a larger switch (100A) would be required because the Ampere Rating requirement has not been met.
