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What is power factor in an induction motor?
Power Factor
What is power factor?
ELI the ICEman is a saying taught in electrical engineering schools, or at least used to be, and means that E, voltage, in an L, inductor, leads the I, current, and also that I, current, in a C, capacitor, leads E, voltage. This leading and lagging causes currents to be out of phase with the supply voltage, causing a phase shift described as power factor (PF). If both are 100% in phase, PF=1. If they are fully out of phase, there will be at 90 electrical degrees from each other, and PF=0. Any angle between is calculable as the cosine of the actual current to the voltage.
One more school comment then on to motors. If one has a perfect inductor with no resistance or capacitance, when voltage is initially applied across it, the current will not want to flow thru it so it is delayed until eventually it does. The result is the current is, as ELI above states, 90 degrees behind the voltage. Cosine of 90 degrees is 0, so power factor is 0. In reality all things have resistance and such so seldom is PF=0 in real life.
OK, our common induction motor. We put rated voltage at 60hz in USA into it and it runs at its nameplate base speed. Of course it draws current. It actually draws two separate currents at the same time: a current in phase with the voltage & this makes torque, & a current 90 degrees out of phase that makes the magnetic field inside the motor. At base speed, and below on a vfd, the magnetic field is a constant strength. So the motor draws a constant current at this 90 degree phase shift to make it and to make itself a motor.
But the torque producing in phase current varies depending on the load; no load and a 10hp or smaller motor is typically only drawing 1/2 amp to make torque enough to overcome bearing friction and windage to rotate. Load it up and this torque producing current increases LINEARLY with torque until the motor nameplate rated current is reached at rated load (torque). All this time, that magnetic field producing 90 degree shifted current is flowing at one constant value......
Taking a 10 hp motor as an example, 230v, 3ph, typically around 27 amps full load rating. That is 30#-ft torque (HP=N*T/5252). But at no load, it probably draws about .5 amps of in-phase current to make enough torque to overcome friction and rotate. The magnetic field is usually around 1/3 to 1/2 the nameplate amps, so lets assume 10 amps here. To see what a clamp on ammeter will read we have to vector add these 2 currents to get the sum. Luckily it is easy to do with Pythagorean theorem: C^2=A^2+B^2 or C=[A^2+B^2]^.5 where C is the resultant vector sum - what the ammeter reads - while A & B are the two currents 90 degrees apart.
So our clamp on ammeter will read [10*10+.5*.5]^2 or 10 amps on our 10hp motor with no load.
Likewise it will read 27 amps when fully loaded.. But wait, what part of the 27 amps is making torque? pythagorean to the rescue again: 27^2=10^2+x^2, so x=25 amps! Sounds like magic that the 10 amps there is STILL there too but only 'looks' like 2 amps until vector summed to the torque producing current.
Now that we have no load and full load current values, and know that the torque producing in-phase current increases linearly with load from .5 to 25 amps, and the magnetic field producing current is a constant at 10 amps, WE CAN DEFINE POWER FACTOR FOR THIS MOTOR.
Remember, power factor is the angle of the total current (vector sum of these two currents) that your ammeter will read at any given load. It is defined as the cosine of the angle of this resultant current from the voltage. Cosine is defined as adjacent/hypotenuse sides of a right angle triangle, which is what we have, since the magnetizing current is the opposite side, the torque producing current is the adjacent side, and the vector sum the ammeter reads is the hypotenuse.
So we can see the power factor of our 10hp motor is about .5/10 or .05 no load, and 25/27 or .93 fully loaded.
There is a lot of neat stuff we can do with this information once we understand PF, like understand why power factor capacitors work, understand why a 10 hp motor only draws about .5a*230v*1.73= 200 watts to spin unloaded, so our vfd with single phase input will only require us to pay the power company this 200/230v=.87amps to run it, even though our clammp on ammeter is reading 10 amps! Or how we can make that 10 amps not load go down to .5 amps with power factor caps, or how we can quickly see the magnetizing portion of our current, or how we can tell exactly what torque our motor is putting out at any given time by measuring the current, or, or....lots more neat stuff.
ELI the ICEman is a saying taught in electrical engineering schools, or at least used to be, and means that E, voltage, in an L, inductor, leads the I, current, and also that I, current, in a C, capacitor, leads E, voltage. This leading and lagging causes currents to be out of phase with the supply voltage, causing a phase shift described as power factor (PF). If both are 100% in phase, PF=1. If they are fully out of phase, there will be at 90 electrical degrees from each other, and PF=0. Any angle between is calculable as the cosine of the actual current to the voltage.
One more school comment then on to motors. If one has a perfect inductor with no resistance or capacitance, when voltage is initially applied across it, the current will not want to flow thru it so it is delayed until eventually it does. The result is the current is, as ELI above states, 90 degrees behind the voltage. Cosine of 90 degrees is 0, so power factor is 0. In reality all things have resistance and such so seldom is PF=0 in real life.
OK, our common induction motor. We put rated voltage at 60hz in USA into it and it runs at its nameplate base speed. Of course it draws current. It actually draws two separate currents at the same time: a current in phase with the voltage & this makes torque, & a current 90 degrees out of phase that makes the magnetic field inside the motor. At base speed, and below on a vfd, the magnetic field is a constant strength. So the motor draws a constant current at this 90 degree phase shift to make it and to make itself a motor.
But the torque producing in phase current varies depending on the load; no load and a 10hp or smaller motor is typically only drawing 1/2 amp to make torque enough to overcome bearing friction and windage to rotate. Load it up and this torque producing current increases LINEARLY with torque until the motor nameplate rated current is reached at rated load (torque). All this time, that magnetic field producing 90 degree shifted current is flowing at one constant value......
Taking a 10 hp motor as an example, 230v, 3ph, typically around 27 amps full load rating. That is 30#-ft torque (HP=N*T/5252). But at no load, it probably draws about .5 amps of in-phase current to make enough torque to overcome friction and rotate. The magnetic field is usually around 1/3 to 1/2 the nameplate amps, so lets assume 10 amps here. To see what a clamp on ammeter will read we have to vector add these 2 currents to get the sum. Luckily it is easy to do with Pythagorean theorem: C^2=A^2+B^2 or C=[A^2+B^2]^.5 where C is the resultant vector sum - what the ammeter reads - while A & B are the two currents 90 degrees apart.
So our clamp on ammeter will read [10*10+.5*.5]^2 or 10 amps on our 10hp motor with no load.
Likewise it will read 27 amps when fully loaded.. But wait, what part of the 27 amps is making torque? pythagorean to the rescue again: 27^2=10^2+x^2, so x=25 amps! Sounds like magic that the 10 amps there is STILL there too but only 'looks' like 2 amps until vector summed to the torque producing current.
Now that we have no load and full load current values, and know that the torque producing in-phase current increases linearly with load from .5 to 25 amps, and the magnetic field producing current is a constant at 10 amps, WE CAN DEFINE POWER FACTOR FOR THIS MOTOR.
Remember, power factor is the angle of the total current (vector sum of these two currents) that your ammeter will read at any given load. It is defined as the cosine of the angle of this resultant current from the voltage. Cosine is defined as adjacent/hypotenuse sides of a right angle triangle, which is what we have, since the magnetizing current is the opposite side, the torque producing current is the adjacent side, and the vector sum the ammeter reads is the hypotenuse.
So we can see the power factor of our 10hp motor is about .5/10 or .05 no load, and 25/27 or .93 fully loaded.
There is a lot of neat stuff we can do with this information once we understand PF, like understand why power factor capacitors work, understand why a 10 hp motor only draws about .5a*230v*1.73= 200 watts to spin unloaded, so our vfd with single phase input will only require us to pay the power company this 200/230v=.87amps to run it, even though our clammp on ammeter is reading 10 amps! Or how we can make that 10 amps not load go down to .5 amps with power factor caps, or how we can quickly see the magnetizing portion of our current, or how we can tell exactly what torque our motor is putting out at any given time by measuring the current, or, or....lots more neat stuff.