power factor

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domnic

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Will a motor with .80 pf and a motor with .90 pf both 2 hp cost the same to run?
 
Yes and no; other things being equal the PF alone will not affect the watts, just the VA.
But, other things may well not be equal since the construction difference that changes the PF may also affect the efficiency.
In short, you will not know.
 
Will a motor with .80 pf and a motor with .90 pf both 2 hp cost the same to run?
Obviously the higher pf machine will draw greater current so supply conductor losses will be greater.
For the motor, it depends on a number of factors as the Digger guy said.
 
Obviously the higher pf machine will draw greater current so supply conductor losses will be greater.
For the motor, it depends on a number of factors as the Digger guy said.

I'm confused, it is a 2Hp motor, the current is based on KVA; how will the full load current be higher with a lower power factor?
 
I'm confused, it is a 2Hp motor, the current is based on KVA; how will the full load current be higher with a lower power factor?
Power (watts in a 3 phase system, I'll skip the conversion to HP) is E (electromotive force, volts) * I (current, amps) * power factor * (square root of 3)

So at the same voltage, current with a pf=0.80 will be higher than with a pf=0.90.

Hard numbers, 2HP, 460V
0.8pf, 2.35A
0.9pf, 2.08A
 
Power (watts in a 3 phase system, I'll skip the conversion to HP) is E (electromotive force, volts) * I (current, amps) * power factor * (square root of 3)

So at the same voltage, current with a pf=0.80 will be higher than with a pf=0.90.

Hard numbers, 2HP, 460V
0.8pf, 2.35A
0.9pf, 2.08A

Thank you for the clarification ;)

There are many that do not realize that the rated nameplate Hp of a motor is the output power, NOT the electric input power. The electric input power must take into consideration the efficiency of the motor; and the FLA listed is for the input power. The listed power factor also relates to the rated KVA of the motor which from the formula is input.

kWrated = (Hp*745.7W/Hp)/efficiency = KVA*pf

Now, based on the OP; increasing the power factor will improve, i.e. lower the current, assuming the efficiency stays the same. However, in physical motor design it is expensive to have high PF and high efficiency; there are trade offs. As economic constraints dictate (cheaper sells more motors) the higher efficiency motors have to give up something and that is usually PF (lower).

Motors typically do not get matched to the load very well, and will not run at optimum performance and thus power factor is a poor way to lower operating costs. It is more economical to add low cost capacitors to improve PF, while purchasing high efficiency motors because there are more losses in the motor unrelated to current thus high efficiency gives you overall lower costs. Typically capacitors are not added to motors below 20Hp anyway because the effect on the total system is minimal.
 
... However, in physical motor design it is expensive to have high PF and high efficiency; there are trade offs. As economic constraints dictate (cheaper sells more motors) the higher efficiency motors have to give up something and that is usually PF (lower).
...
Out of all of that great response, that last part of this sentence is the only part I have a slight problem with, and mostly only in the impression it may leave with people who might see it as a "rule". So just indulge my expansion of the concept just to set the record straight(er).

The EPRI (Electric Power Research Institute) has (or had) some great white papers on techniques used to improve AC motor efficiency, and offered a class on it at one time that I was lucky enough to attend. Sacrificing PF was never mentioned as a strategy, although in some of the techniques, it may have been an unintended consequence, or rather, a consequence that was outweighd by other improvements that gain more efficiency than what was lost by having a lower PF. As a grossly over generalized rule, low PF represents current flowing in the circuit that is not producing useful work, so because all current flow results in heat losses at some level, lowering the PF of a motor while trying to get better efficiency from it is not something a motor mfr would likey pursue as a first line option. There is however a lot of "horse trading" that takes place in the process of improving overall efficiency, so there are no generalized assumptions that can be made.

For example, reducing the air gap by fractions of a millimeter to gain very tiny improvements in flux transfer can improve efficiency, which also IMPROVES the PF. But at the same time, it causes OTHER losses in the motor ("stray load losses") to increase, which might be worse, depending on the steel used, slot geometry, winding method etc. etc. So increasing the gap may actually end up improving overall efficiency more than reducing it in that design, which might result in a lower PF again. The point is, all of the causes and effects have to be weighed carefully in the process, but at no point does anyone purposely lower the PF. In fact the EPRI did a study of that in which, of the 5 EE motor designs tested, 2 showed lower PF, 2 showed higher PF, and 1 had no change in PF.

Back to the OP, the bottom line answer is that NO, your cost to run a 2HP motor will not be any different, in a significant way, whether the PF is .80 or .90 if that is the ONLY difference. But, to the earlier point, small motors like that have traditionally NEVER shown PF values at full load better than maybe .85, so the presumption made by Golddigger earlier is very valid; being the fact that if a small motor does show a .90 PF it would be a likely indicator that it is a newer energy efficient design, which will cost less to operate, just for reasons OTHER than the PF itself.
 
Obviously the higher pf machine will draw greater current so supply conductor losses will be greater.
For the motor, it depends on a number of factors as the Digger guy said.

Was in a hurry. Obviously lower power factor is higher current.
Apologies folks.
 
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