I have a cutsheet of an air handler (ERU) by Trane. The cutsheet says:
- Circuit #1: Supply motor fan (each x2): FLA=3.8A, MCA=4.75A, MOCP=15A
- Circuit #2: Exhaust fan motor(s): FLA=6.7A, MCA=8.38A, MOCP=15A
It is rated for 460V, 3phase.
How can I estimate the total power withdrawn by the whole AHU, so that I can size my wires and MOCP correctly? I am not sure if I am using the following formula correctly:
MOCP = (2.25 x FLA of largest motor load) + (other motor loads) + (other resistive loads)
= 2.25*6.7 + 3.82*2 = 22.175 --> 25A breaker
Total load (kVA) withdrawn: 22.175*480*1.732 = 18.435 kVA
Please correct me if I'm wrong. Thank you!
- Circuit #1: Supply motor fan (each x2): FLA=3.8A, MCA=4.75A, MOCP=15A
- Circuit #2: Exhaust fan motor(s): FLA=6.7A, MCA=8.38A, MOCP=15A
It is rated for 460V, 3phase.
How can I estimate the total power withdrawn by the whole AHU, so that I can size my wires and MOCP correctly? I am not sure if I am using the following formula correctly:
MOCP = (2.25 x FLA of largest motor load) + (other motor loads) + (other resistive loads)
= 2.25*6.7 + 3.82*2 = 22.175 --> 25A breaker
Total load (kVA) withdrawn: 22.175*480*1.732 = 18.435 kVA
Please correct me if I'm wrong. Thank you!