bcm
Member
- Location
- Atlanta, Georgia
- Occupation
- Engineer
I haven't done this in a long time, so I'm rusty and want to make sure my thinking on this is correct. I have a unit with the following specs:
Voltage: min 187 max 253
Compressor: RLA 14.3 LRA 58.3
Outdoor Fan Motor: FLA 0.8 NEC HP 0.125 KW out 0.09
Min ckt amps 18.7
Fuse/HACR Brkr Amps 30
So, min circuit ampacity is RLA x 125% + other loads. That = (14.3 x 1.25) + 0.8 = 38.3 amps. (I assume that the motor FLA is the "other loads" I needed to add for this calculation.)
Given the calculated 38.3 amps, I need wiring sized at #8 awg. Correct?
Now, unless I'm mistaken, the disconnecting means (switch that is next to the unit outside), is required to be rated at 115% of the RLA or branch circuit selection current. So, 14.3 X 1.15 = 16.45 amps. Branch circuit selection is a 50A circuit (that #8 I calculated earlier), so I need to provide a 50A disconnect by the unit since the branch circuit selection is greater. Is that correct?
Thanks so much, folks!
Voltage: min 187 max 253
Compressor: RLA 14.3 LRA 58.3
Outdoor Fan Motor: FLA 0.8 NEC HP 0.125 KW out 0.09
Min ckt amps 18.7
Fuse/HACR Brkr Amps 30
So, min circuit ampacity is RLA x 125% + other loads. That = (14.3 x 1.25) + 0.8 = 38.3 amps. (I assume that the motor FLA is the "other loads" I needed to add for this calculation.)
Given the calculated 38.3 amps, I need wiring sized at #8 awg. Correct?
Now, unless I'm mistaken, the disconnecting means (switch that is next to the unit outside), is required to be rated at 115% of the RLA or branch circuit selection current. So, 14.3 X 1.15 = 16.45 amps. Branch circuit selection is a 50A circuit (that #8 I calculated earlier), so I need to provide a 50A disconnect by the unit since the branch circuit selection is greater. Is that correct?
Thanks so much, folks!