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why so much variation in amperage
- Thread starter cg 2005
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- Dec 30, 2011
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- 214
Any Far Eastern wonders in that list? As noted, you can play games with claimed horsepower ratings, especially for intermittent use. Efficiency shouldn't vary much, and power factor isn't important at full load.
- Joined
- Dec 30, 2011
- Messages
- 214
Not much experience with those, but I'd guess you could go by the continuous full-load current first, size second, and "horsepower" rating last.
There is no easy answer. it boils down to quality, efficiency, motor NEMA type (NEMA A,B,C or D), enclosure (open motors will draw more current than a totally enclosed motor due to fan losses), number of poles in the motor (RPM), Age, Single Phase vs 3 Phase. Motor slip (difference between synchronous speed and actual full load speed) As I said no straight forward answer.
What is important is to know the FLA rating of the motor. if for no other reason to size the fuse/ circuit breaker feeding the motor. The smart way to size a VFD is to size by AMP rating. Not all 5 HP drives are created equal.
Jim
What is important is to know the FLA rating of the motor. if for no other reason to size the fuse/ circuit breaker feeding the motor. The smart way to size a VFD is to size by AMP rating. Not all 5 HP drives are created equal.
Jim
- Joined
- Nov 13, 2012
- Messages
- 165
As mentioned below, your equation does not include the power factor (PF).
watts = volts * amps only works for DC or for purely resistive AC loads. For AC power the actual equation is watts = volts * amps * PF
A motor is a induction load. An inductor resists a change in current, so the current lags behind the voltage curve.
If you do all the calculus, the power factor (PF) ends up being the cosine of this "current lag" angle. So PF is always 1.0 or less.
It is not an efficiency since the motor does not actually use the additional current - it just bounces it back into the line.
Induction motors (like you have), usually run power factors in the range of 0.6. Sometimes the PF is even lower depending on the motor design. Straight induction motors have different PFs than capacitor run motors.
So, you have to multiply your answers by 0.6.
When you do this, the 3/4 and 1 HP fall pretty much in line.
The two 1/3 HP motors appear to have really low power factors which I can't explain.
Side note: Power factor is a big deal for manufacturing sites. The power company monitors the PF on large consumers and charges a penalty when the PF drops below a certain level. Many large plants with lots of large motors will often have a large capacitor bank to pull the PF back toward unity. The current being "bounced back" down the line adds to the resistive heating loss over miles of transmission line - which adds to the load at the generating station (but not at the consumer).
watts = volts * amps only works for DC or for purely resistive AC loads. For AC power the actual equation is watts = volts * amps * PF
A motor is a induction load. An inductor resists a change in current, so the current lags behind the voltage curve.
If you do all the calculus, the power factor (PF) ends up being the cosine of this "current lag" angle. So PF is always 1.0 or less.
It is not an efficiency since the motor does not actually use the additional current - it just bounces it back into the line.
Induction motors (like you have), usually run power factors in the range of 0.6. Sometimes the PF is even lower depending on the motor design. Straight induction motors have different PFs than capacitor run motors.
So, you have to multiply your answers by 0.6.
When you do this, the 3/4 and 1 HP fall pretty much in line.
The two 1/3 HP motors appear to have really low power factors which I can't explain.
Side note: Power factor is a big deal for manufacturing sites. The power company monitors the PF on large consumers and charges a penalty when the PF drops below a certain level. Many large plants with lots of large motors will often have a large capacitor bank to pull the PF back toward unity. The current being "bounced back" down the line adds to the resistive heating loss over miles of transmission line - which adds to the load at the generating station (but not at the consumer).