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View Full Version : Electrical Engineers save me :) Going from motor kw to FLA?

OVOleg
06-10-2011, 02:13 PM
1. Known Data
Fan Motor Output: 0.172 kW
Propeller Fan

Compressor Motor Output: 2.9kW
DC Inverter-driven Scroll

Voltage: 208/230, single phase

What I need to find is the FLA for each fan motor(2), and the FLA of the compressor motor.

Total MCA = 32A
2. Relevant equations

V=IR?
I'm not really sure, no other information is given.

3. The attempt at a solution

I'm pretty sure it goes something like this but I can't find any resources that would help me calculate this. I've taken electric circuits in physics and in electrical engineering but I do not recall.

P=V*I
P=V*I*eff*PF

Going to assume the power factor = 0.8
I have no clue as to the efficiency, and I'm going to guess that its incorporated into the power factor?

For the Fan,
0.172*10^3=230*FLA*.8

FLA = 172/(230*0.8)
FLA = 0.934783 Amps.

For the compressor I'm going to make generally the same assumption, that the power factor is 0.8 and that I have no clue as to the efficiency.

FLA = 2900/(230*0.8)
FLA = 15.76087 Amps.

So since the MCA is set to 32 Amps, I'm going to at least double check,

(Compressor FLA)*1.5+(Fan FLA) = MCA
^^Pretty sure this is the equation

(15.76087)*1.5+(0.934783) = 24.57 MCA

Obviously not 32. Anyone???

OVOleg
06-10-2011, 02:23 PM
Electrical Motor - Power

W3-phase = (E I PF 1.732) / 1,000 (6)

where

W3-phase = electrical power 3-phase motor (kW)

PF = power factor electrical motor

In theory I should be able to use this and and use 1.0 instead of 1.73 due to single phase?

RichardL
06-10-2011, 06:25 PM
First off, do not mistake a motors nameplate data with anything other than a motors "Rating".
All motors are required by NEMA to state "Correctly" a motors "Rating" for a given Load(HP)/Applied Voltage/Current & RPM. The FLA you refer to is in actuallity the "Rated-Load-Amps" for a stated condition of performance. As the load goes down, RPM increases, Input power decreases, and power factor goes down as well. On the other hand, as the load increases, the power factor goes up towards unity(1), current goes up and RPM goes down. Inverter driven motors on the other hand tend to operate at a much higher power factor as the applied voltage is much more constant while maintaining a given V/F ratio even in the vectorless feedback mode of operation.
The MCA must take into account the worst case scenerio in the power required to operate the motor(s) "Without" the use of any Inverter drive.