Power Factor for AC Pumps

[sphelps]

Originally Posted by mrhasan

I am not a die hard power guy (more in energy scenarios) but the basics is, the lesser the PF is, the more current it will draw from the outlet and hence more consumption for the device.

If you PF for RD6.5 was like 0.8 like albert said, it would be drawing only around 0.9A, meaning less power would be drawn out form the outlet and thus less consumption to run motor with the same power.

Maybe the motor became fault and hence PF is so low?

My understanding, as far as I know, is the power factor can be defined as how effectively a device uses the current available in the circuit. It's not really related to efficiency. AC pumps typically have a lower power factor than say DC pumps or other devices which follow basic V=I/R & P=VI. This relates to impedance which I don't want to get into but basically a high power factor of 1 means the device draws in X amount of amps and uses every last one. A power factor of 0.8 means it draws X amount but only uses 0.8*X while the remainder is fed back into the circuit so it doesn't actually use more power, it just draws more current.

[mrhasan]

Originally Posted by sphelps

My understanding, as far as I know, is the power factor can be defined as how effectively a device uses the current available in the circuit. It's not really related to efficiency. AC pumps typically have a lower power factor than say DC pumps or other devices which follow basic V=I/R & P=VI. This relates to impedance which I don't want to get into but basically a high power factor of 1 means the device draws in X amount of amps and uses every last one. A power factor of 0.8 means it draws X amount but only uses 0.8*X while the remainder is fed back into the circuit so it doesn't actually use more power, it just draws more current.

PF does talk about efficiency. In any AC systems, there are two sorts of powers: real and apparent. You are using the real power to drive the motor (82w) but you are actually taking in more power from the outlet.

Electric companies don't bill you on how much your devices consume but on how much you are consuming from the grid. In this case, you are withdrawing 1.85A from the grid. They don't care about what's the power factor of your device and how much you consuming, they care about how much you are withdrawing. In this case, its 115v and 1.85A resulting in 212.75W. I am not sure whether there is any PF accounted in billing (I will get back to you once I get to know about it from my colleague who is yet to come). Even if they do, it is bound to be above 0.9. So you are actually consuming 212.75W (under unitiy pf condition) for the motor while your motor is actually consuming 82W only.

As for DC, there's one huge reason why they are considered so efficient, they don't have any PF involved. PF is due to the phase difference between voltage and current which DC doesn't have and hence its PF is always 1. Rock solid 1. But since we plug in DC to AC, there is AC to DC conversion stage and power factor corrections are not very very effective in bringing pf to rock solid 1. And hence you are getting 0.97 for the DC motor.

[sphelps]

Ok now we're getting somewhere. My assumption is I'm getting billed in kWh which would include PFs so if my pump is using 82W I'm paying to 82W and not 212W. While the pump is pulling 1.8A it's not using it all, essentially most of it is going back to the grid so to speak. So it's not actually being used, hence the PF related to how effective not efficient. It would certainly be very valuable information for if you could confirm this.

[mrhasan]

Originally Posted by sphelps

Ok now we're getting somewhere. My assumption is I'm getting billed in kWh which would include PFs so if my pump is using 82W I'm paying to 82W and not 212W. It would certainly be very valuable information for if you could confirm this.

http://www.enmax.com/Power/Tariffs/P...or/default.htm

This might help you a bit for now. My colleague is still not here and I am not sure how Alberta is billed with PF like whether they consider any ratio or just unity pf in the grid. I will get the info from him and confirm the ratio.

[sphelps]

Originally Posted by mrhasan

http://www.enmax.com/Power/Tariffs/P...or/default.htm

This might help you a bit for now. My colleague is still not here and I am not sure how Alberta is billed with PF like whether they consider any ratio or just unity pf in the grid. I will get the info from him and confirm the ratio.

That sounds like its only applied to commercial use, being billed based on kVA and not kW.

[mrhasan]

Originally Posted by sphelps

That sounds like its only applied to commercial use, being billed based on kVA and not kW.

I am still confused about those components. Will get to you with the right stuff.

Electric companies will not be able to determine individual PF of all the devices and hence, even if they use any PF it is going to be an average of the whole house or just unity factor. I will get back to you after 12pm or something regarding this. My prof should be the best person to clear it up

[sphelps]

Thanks let me know

I can't tell you how ****ed I'll be if this pump is actually costing me over 200W of power.

[mike31154]

For the average home owner, the power company loses with respect to low PF, not the consumer. Pretty certain that large industrial customers are billed extra when their equipment includes many large inductive loads such as AC motors and they don't take steps to minimize a low PF.

[mrhasan]

Originally Posted by mike31154

For the average home owner, the power company loses with respect to PF, not the consumer. Pretty certain that large industrial customers are billed extra when their equipment includes many large inductive loads such as AC motors.

My logic is, even if the PF is included in their calculation, it got to be the pf of the whole house and not the individual motor. Generally, that gets to above 0.9.

[mrhasan]

Originally Posted by sphelps

While the pump is pulling 1.8A it's not using it all, essentially most of it is going back to the grid so to speak.

That's only applicable in some microgeneration programs. You are pulling out 1.8A through the pump and hence its yours. And more or less, the whole current is consumed by the motor other than some leakages.