Motor power factor amperage

[DW98]

For a 225W exhaust fan motor, nameplate is 3.3A. 225W/115V = 1.95A. If you use .6 pf, 1.95/.6 = 3.26A. Is this how 3.3 is arrived at? What would a volt/amp meter read?

 

[Jraef]

For a 225W exhaust fan motor, nameplate is 3.3A. 225W/115V = 1.95A. If you use .6 pf, 1.95/.6 = 3.26A. Is this how 3.3 is arrived at? What would a volt/amp meter read?

The nameplate data is reflective of the WORST CASE scenario when the motor is running normally at full LOAD, meaning the load connected to is is USING 225 Watts of power.

What your meter will read will be the Watts that are ACTUALLY used by the fan to move the air, assuming you have a Wattmeter. It will not be the same. If you read the Amps, the reading will be reflective of ALL of the amps going into the motor, both the active (load based) and reactive (power factor based). That is generally difficult to get meaningful information from without other information.

What is your purpose in wanting to know the amps? Let's start there. People often incorrectly equate amps with power directly, it's just not that simple. Amps are just a COMPONENT of power in an equation with 4 variables:

1. Amps
2. Volts
3. Power Factor
4. Efficiency

You did not mention Efficiency of your motor by the way.

 

[gar]

130315-1732 EDT

DW98:

Jraef gave you a good answer. You need to expand your question to clarify what you really want to know.

Power Factor is defined as
PF = Real Power / (Volt*Amperes)

I will add that V and A need to be the RMS values. This definition of PF applies to any arbitrary wave shapes for V and A. Quite obviously A will be dependent on V based on the load.

Thus, a capacitor input DC power supply will have a poor power factor even though you might classify the current pulse near the peak of the voltage waveform to be in phase with the voltage. But there is really no phase angle to use in this non-linear case as there is between two sine waves of identical frequency and with a constant phase difference.

An unloaded single phase motor, meaning running with one oscillating magnetic field, has a particularly bad power factor, like 0.25 . This power factor improves with load. If you have a capacitor run single phase motor, actually a two phase motor, then the power factor is much better.

.

 

[Jraef]

130315-1732 EDT

DW98:

Jraef gave you a good answer. You need to expand your question to clarify what you really want to know.

Power Factor is defined as
PF = Real Power / (Volt*Amperes)

I will add that V and A need to be the RMS values. This definition of PF applies to any arbitrary wave shapes for V and A. Quite obviously A will be dependent on V based on the load.

Thus, a capacitor input DC power supply will have a poor power factor even though you might classify the current pulse near the peak of the voltage waveform to be in phase with the voltage. But there is really no phase angle to use in this non-linear case as there is between two sine waves of identical frequency and with a constant phase difference.

An unloaded single phase motor, meaning running with one oscillating magnetic field, has a particularly bad power factor, like 0.25 . This power factor improves with load. If you have a capacitor run single phase motor, actually a two phase motor, then the power factor is much better.

.

And thus is the tip of the iceberg, hence my asking whay the information is needed. I find that often times people ask these questions not realizing that the little piece of information they see in front of them on the VOM is but a minor exposed part of a deeper...

...oh the heck with it, it's Friday afternoon and I'm needlessly waxing off into metaphores again ... :slaphead:

What he said.

 

[Phil Corso]

Jraef...

Some metaphor :? The meter reads electrical power the motor needs to rotate and then converts most of it to mechanical power the driven machine requires, i.e, output power plus the motor's losses:slaphead:

Regards, Phil Coro