countrytoad
Member
I thought we multiply by 1.25 to get the FLA of a CD (continuous duty) motor, but that doesn't appear to be the case. When do we multiply by 1.25? And why don't we here?
So I calculated: 65 Amps x 1.25 x 1.15 = 93.4 Amps
(65 Amps from NEC Table 430.250) (1.15 from NEC 430-32a1)
The correct solution is: 65 Amps x 1.15 = 74.75 Amps =(to maximum 5 Amp increment)=> 70 Amps
The original question #510: "A 50-hp, 460-V, 3-phase induction motor is served by a single feeder conduit... and it is rated for continuous duty... Starter running overload protection is available in 5-A increments... the initial maximum sized protection device (amperes) that can be applied is most nearly"
Thanks!
So I calculated: 65 Amps x 1.25 x 1.15 = 93.4 Amps
(65 Amps from NEC Table 430.250) (1.15 from NEC 430-32a1)
The correct solution is: 65 Amps x 1.15 = 74.75 Amps =(to maximum 5 Amp increment)=> 70 Amps
The original question #510: "A 50-hp, 460-V, 3-phase induction motor is served by a single feeder conduit... and it is rated for continuous duty... Starter running overload protection is available in 5-A increments... the initial maximum sized protection device (amperes) that can be applied is most nearly"
Thanks!