I'm having some confusion reconciling 430.110 and fusing.
I'm currently designing an industrial panel being fed by 460/3/60. My loading is thirteen, 2 HP motors. According to my calculations and 430.24, my panel FLA is 45.70. (1.25 * 3.4) + (12 * 3.4) + (0.65 A for control power supplies) = 45.70
I always figure on at least an extra 25% for future expansion, so I'm sizing my conductors to 57.13 amps, which means #6 for the current-carrying conductors and #10 for the ground.
The issue comes when I'm sizing my disconnect. I'm using 430.110. So my first step is 115% sum of all currents at full load. Since these motors are all on Variable Frequency Drives, I remember reading somewhere I should use the drive input rating in that case, which is 4 amps a piece (1.15 * 3.4 is 3.91 anyway). That works out to be 52.52 amps. The thing is, though, according to 430.110 I also have to account for the locked-rotor current, and after re-reading it in the code I don't see any exceptions for inverter-driven motors. So my total locked-rotor current (using Table 430.251(B)) works out to 325.52. Going by that table, I need at least a 50 HP rated disconnect. By amp load, a 60 A disconnect will suffice, but by horsepower rating, I need a 100 A disconnect since the manufacturer I use (Allen-Bradley) only rates their 60 A disconnects at 30 HP and the 100A disconnects at 60 HP.
I don't know the relevant codes, but if I recall correctly, to calculate the correct size fuse for the feeder circuit, I need 150% of the largest motor and 100% of the rest of the loads. So 5.1 + (12 * 3.4) + 0.52 which is 46.42. Next fuse size up is 50A. Problem is, I don't think I can physically install a 50A fuse in a 100A Allen-Bradley fused disconnect. Minimum fuse size there is 70A because it's a 100A class disconnect. Unless there is some kind of adapter I can use.
So what do I do in this situation?
I'm currently designing an industrial panel being fed by 460/3/60. My loading is thirteen, 2 HP motors. According to my calculations and 430.24, my panel FLA is 45.70. (1.25 * 3.4) + (12 * 3.4) + (0.65 A for control power supplies) = 45.70
I always figure on at least an extra 25% for future expansion, so I'm sizing my conductors to 57.13 amps, which means #6 for the current-carrying conductors and #10 for the ground.
The issue comes when I'm sizing my disconnect. I'm using 430.110. So my first step is 115% sum of all currents at full load. Since these motors are all on Variable Frequency Drives, I remember reading somewhere I should use the drive input rating in that case, which is 4 amps a piece (1.15 * 3.4 is 3.91 anyway). That works out to be 52.52 amps. The thing is, though, according to 430.110 I also have to account for the locked-rotor current, and after re-reading it in the code I don't see any exceptions for inverter-driven motors. So my total locked-rotor current (using Table 430.251(B)) works out to 325.52. Going by that table, I need at least a 50 HP rated disconnect. By amp load, a 60 A disconnect will suffice, but by horsepower rating, I need a 100 A disconnect since the manufacturer I use (Allen-Bradley) only rates their 60 A disconnects at 30 HP and the 100A disconnects at 60 HP.
I don't know the relevant codes, but if I recall correctly, to calculate the correct size fuse for the feeder circuit, I need 150% of the largest motor and 100% of the rest of the loads. So 5.1 + (12 * 3.4) + 0.52 which is 46.42. Next fuse size up is 50A. Problem is, I don't think I can physically install a 50A fuse in a 100A Allen-Bradley fused disconnect. Minimum fuse size there is 70A because it's a 100A class disconnect. Unless there is some kind of adapter I can use.
So what do I do in this situation?