# why so much variation in amperage



## cg 2005 (Jan 22, 2015)

I have four ac electric motors and none of them are close to the w/v=a formula, even allowing for 81% efficiency.

examples from the motor data plates, all at 115v


1/3hp @ 5a
1/3hp @ 9a
3/4hp @ 10a
1hp @ 12a

using the power formula and adding for 81% efficiency they should be, respectively


2.68a
2.68a
6.05a
8.05a

why such variation from the theoretical and difference between the two 1/3 hp motors?


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## mzayd3 (Jan 22, 2015)

Are they under load?  A motor will draw considerably less current when just idling.


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## cg 2005 (Jan 22, 2015)

No, these are the specifications as indicated on the motor I.D. plate.

I am aware of the idle amperage being lower since not all the power is in use at idle. I believe the spec voltage is full load, correct?


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## mzayd3 (Jan 22, 2015)

yes, but that can vary +- 10% or so.  if you have high voltage, your current will be lower also.


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## den-den (Jan 22, 2015)

Your formula does not take power factor into consideration.  Explaining power factor is not simple.  In a DC circuit, volts X amps = watts; in an AC circuit the current reverses direction 120 times per second.  The peak amperage and the peak voltage do not occur at the same time, which complicates the calculation quite a bit.

An analogy is a clock pendulum; mass swings back and forth but the actual work done can not be calculated with just the pendulum mass and distance traveled.


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## countryguy (Jan 22, 2015)

Here is some really technical info on the PF ....  Complex indeed.  There is also a calc link on the page but I did not try it yet 

http://www.rapidtables.com/electric/Power_Factor.htm
Good post and neat theory!    Thanks all. 
CG. 



den-den said:


> Your formula does not take power factor into consideration.  Explaining power factor is not simple.  In a DC circuit, volts X amps = watts; in an AC circuit the current reverses direction 120 times per second.  The peak amperage and the peak voltage do not occur at the same time, which complicates the calculation quite a bit.
> 
> An analogy is a clock pendulum; mass swings back and forth but the actual work done can not be calculated with just the pendulum mass and distance traveled.


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## cg 2005 (Jan 23, 2015)

It appears that the power factor, PF, you reference is primarily a measure of efficiency since it is a ratio between power applied by the motor to power available in the circuit. In other words power output to power input.

Getting back to my original post. The specs as stated would then suggest that the 1/3 HP motor rated at 5A is more efficient than the 1/3 HP motor rated at 9A. The 9A motor is more than two times larger than the 5A motor in all dimensions. I suspect the 9A motor has much greater torque rating. I am going to research that a bit.

All of this was brought to light because I just bought a new Enco 5x6 band saw, model 505-6840, to replace my ten year old HF 4 1/2x 6.

The Enco has a 1/3 HP, 9A motor that is larger than the HF 3/4 HP, 10A motor.


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## den-den (Jan 23, 2015)

Power factor and efficiency are not the same thing.  Another issue here is the motor ratings.  "Horsepower" is well defined but people have been playing games with HP ratings since before electric motors were invented.


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## Skarven (Jan 24, 2015)

Here is a link to a table over current versus efficiency:
http://www.engineeringtoolbox.com/electrical-motor-hp-amps-d_1455.html

I also think there are different ways of specifying the power of an electric motor depending on how it is used.
You have 'Maximum continuous', 'intermittent with stops' and intermittent with idle'
http://www.engineeringtoolbox.com/iec-duty-cucles-d_739.html

Kai


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## cathead (Jan 24, 2015)

*on efficiency...*

Personally I would have an interest in the efficiency of a refrigerator or
anything that runs continuously.  On a band saw with intermittant use I
wouldn't give it a second thought.


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## JimDawson (Jan 24, 2015)

cg 2005 said:


> I have four ac electric motors and none of them are close to the w/v=a formula, even allowing for 81% efficiency.
> 
> examples from the motor data plates, all at 115v
> 
> ...




I'm guessing that the 9a motor is a slow speed motor,  maybe in the 900 RPM range?


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## cg 2005 (Jan 24, 2015)

JimDawson said:


> I'm guessing that the 9a motor is a slow speed motor,  maybe in the 900 RPM range?




Nope, 1720 rpm.


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## Alan Douglas (Jan 24, 2015)

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.


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## cg 2005 (Jan 24, 2015)

Alan Douglas said:


> Any Far Eastern wonders in that list?




All of them.


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## Alan Douglas (Jan 24, 2015)

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.


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## Jim Nunn (Jan 24, 2015)

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


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## twstoerzinger (Jan 26, 2015)

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).


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