Motor Branch Circuit and Overload Protection
A 50 HP, 460 V, 3-phase, 60 Hz induction motor has a nameplate full-load current of 62 A, service factor 1.15, and temperature rise 40°C. The motor is supplied by a feeder with 4 AWG THHN copper conductors. The motor is controlled by a magnetic starter with a thermal overload relay. The motor drives a conveyor with a duty cycle of 30 minutes on, 30 minutes off. The ambient temperature in the room is 40°C. The motor is located 100 ft from the starter. The starter has a short-circuit protective device (SCPD) that is a 150 A circuit breaker. The motor is not hermetic.
1. Determine the minimum size branch-circuit conductors and the maximum rating of the overload relay.
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Branch-circuit conductors for a single motor must have an ampacity of at least 125% of the motor full-load current (FLC) per NEC 430.22. The FLC from Table 430.250 for a 50 HP, 460 V motor is 65 A (not the nameplate 62 A; use table value). So minimum conductor ampacity = 1.25 × 65 = 81.25 A. Using Table 310.15(B)(16) for 75°C THHN copper: 4 AWG = 85 A, so 4 AWG is sufficient. However, ambient temperature correction: 40°C ambient requires correction factor of 0.91 for 75°C conductors (Table 310.15(B)(2)(a)). So ampacity of 4 AWG at 40°C = 85 × 0.91 = 77.35 A, which is less than 81.25 A. Therefore, 4 AWG is not sufficient; need 3 AWG (100 A × 0.91 = 91 A). So minimum conductor is 3 AWG Cu. Overload relay: must be sized per NEC 430.32. For a motor with service factor 1.15, the overload trip setting can be up to 125% of nameplate FLC (62 A) = 77.5 A. But the overload relay must be selected based on the motor nameplate current, not table FLC. So maximum overload setting = 1.25 × 62 = 77.5 A. However, the relay must also be capable of carrying the motor current. So answer: conductors 3 AWG Cu, overload relay set at not more than 77.5 A.
2. Is the 150 A circuit breaker acceptable for short-circuit ground-fault protection? If not, what is the maximum allowable?
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For a standard induction motor, the maximum SCPD rating is based on NEC 430.52. For an inverse-time circuit breaker, the maximum is 250% of the motor FLC (Table 430.52). Using Table FLC of 65 A, 250% × 65 = 162.5 A. The next standard size is 175 A (240.6). However, the 150 A breaker is less than 162.5 A, so it is acceptable. But we must also consider that the breaker must be capable of carrying the starting current. The 150 A breaker is within the 250% limit, so it is acceptable. Therefore, the 150 A breaker is acceptable. But note: if the breaker was larger than 162.5 A, it would not be acceptable unless listed for motor protection. So answer: yes, 150 A is acceptable because it is less than the maximum 162.5 A (next standard 175 A).