Re: Motor Circuit Protection Question
The motor plate is marked as 3.0 HP (2.2 KW), Code J. It is a single-phase, 240 VAC motor. The instruction manual for the saw states that the feed should be "not less than # 12 wire and protected by a 20 Amp timelag fuse." My NEC book is a 1984 edition, so I wouldn't be surprised if some of this had changed. According to table 430-148, the Full-Load Current in Amperes for a 230V single-phase motor is 17 Amperes. According to the instruction manual, the motor Branch-Circuit Short-Circuit and Ground-Fault Protection should be provided by a 20 Amp timelag fuse. When I go to Table 430-152, I find that the "Nontime Delay Fuse" value for a Code Letter J (full-voltage, resistor or reactor starting) is 300% of full load. In this case, it would be 17 Amps x 3 = 51 Amps. However, the "Instantaneous Trip Breaker" value is 700%, which would be 119 Amps. The intent is to have my electrical contractor install a sub-panel to feed a new woodworking shop that I am having built in an unfinished basement. So the motor feed will be supplied from that breaker panel. So, which do I specify, 50 Amps or 110 Amps? If he has to use a 110 Amp instantaneous trip breaker in the distribution sub-panel, then the motor branch-circuit and the fused disconnect switch have to be rated for 110 Amps. The fuses he will install in that disconnect should be 20 Amp timelag fuses, and the motor circuit conductor and the plug will be sized at 20 Amp. If I only need to have a 50 Amp breaker supplying the motor branch-circuit, then the rest of the motor branch-circuit will be sized accordingly. I'd hate to pay for a 110 Amp circuit if I only really need a 50 Amp circuit.