Hello! I am powering several RTU's for a commercial project (design) and my mechanical schedule I am working off of shows a 2 h.p. motor with 18.5 MCA and a 25A MOCP.. they are all 460v/3-phase. So, when I do my load calculation and use the h.p. and convert it to watts it is much lower than converting the MCA to watts. I am confused. Can you help? Thanks!
The discrepancy is due to efficiency and power factor of the motor, as well as safety factors that may apply (MCA vs FLA).
HP is the mechanical output power rating. If you convert this via the factor of 746 Watts = 1 HP, you are calculating the mechanical Watts. Fun fact about the history of the Watt and horsepower units. James Watt invented the horsepower unit, and the Watt was named in honor of him after his death.
For voltage and current, we relate these to electrical input Watts. Some of the electrical Watts are burned up as heat, and not all of it makes it to mechanical Watts. Since a motor isn't a pure resistive load, Volts*Amps doesn't necessarily equal Watts. There is a power factor, due to volts and amps being delayed relative to one another. Volts*Amps = Volt-Amperes, rather than Watts. Watts = VA*power factor.
MCA = minimum circuit amperes. This means the minimum ampacity of the terminations and the wiring, that supplies the motor. This may be larger than the actual amps that the motor is expected to see at full load. In any case, it tells you how much copper or aluminum is needed to safely carry the power from the source to the motor, per the NEC. Assuming circuit length is small enough for voltage drop not to matter.
FLA = full load amps. This is the datasheet value for how much current you get at full load from factory testing.
FLC = full load current. This is the NEC value for what full load current a motor of a given horsepower consumes in the general sense for all motors of that same horsepower. Sort of an upper limit of the FLA.
To help you remember: FLA = "full load actual", FLC = "full load code"