Calculating Total Load of ERU with 3 Motors

[elecshop]

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How can I estimate the total power withdrawn by the whole AHU, so that I can size my wires and MOCP correctly?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Total load (kVA) withdrawn: 22.175*480*1.732 = 18.435 kVA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!

 

[LarryFine]

- Circuit #1: MCA=4.75A, MOCP=15A
- Circuit #2: MCA=8.38A, MOCP=15A

That's all you need. Two circuits, with conductors rated for 4.75a and 8.38a each, and on 15a breakers.

 

[suemarkp]

Calculating the power has nothing to do with calculating the max size breaker. To me, power would be the sum of the FLA's * 480*1.732. It appears you have 3 motors, 3.8 + 3.8 + 6.7 = 14.3A. 14.3 * 480 * 1.732 = 11.9 KVA

If you are tring to make one branch circuit to feed both circuit 1 and 2, I'm not sure that is possible. I would think a 20A circuit would hold and would have sufficient ampacity for the load (1.25*largest motor + the sum of all the other motors). However, if the manufacturer is putting a max breaker limit of 15A on each circuit then I don't think you may use a larger circuit.

 

[JoeStillman]

You could run a 25A circuit up to the ERU and feed 2 x 30/3 fusible switches with 15A fuses. That would pass inspection as long as the nameplate says Fuse or HACR breaker.