Power Factor for AC Pumps

[sphelps]

OK so has anyone measured amps and watts on their AC pumps before? I've never looked too closely until today and I'm surprised at some of the numbers, take a look.

Voltage is 115V

Mag Pump - 0.3A , 22W (PF 0.64)
Red Dragon 6.5 - 1.85A , 82W (PF 0.39)
Red Dragon BK - 1.33A, 79W (PF 0.52)

Waveline DC10000 - 0.83A, 93W (PF 0.97)

I'm no electrical guy this makes no sense to me. Anyone care to educate me?

[albert_dao]

Something is off with your power factors on the rd6.5. I've measured them at 0.7-0.8 in the past. What device are you using?

[mrhasan]

Originally Posted by sphelps

OK so has anyone measured amps and watts on their AC pumps before? I've never looked too closely until today and I'm surprised at some of the numbers, take a look.

Voltage is 115V

Mag Pump - 0.3A , 22W (PF 0.64)
Red Dragon 6.5 - 1.85A , 82W (PF 0.39)
Red Dragon BK - 1.33A, 79W (PF 0.52)

Waveline DC10000 - 0.83A, 93W (PF 0.97)

I'm no electrical guy this makes no sense to me. Anyone care to educate me?

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?

[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.

[sphelps]

Originally Posted by albert_dao

Something is off with your power factors on the rd6.5. I've measured them at 0.7-0.8 in the past. What device are you using?

Looks off to me as well, but the current was measured with my profilux PAB and an energy monitor. Both read the same amount of current which is around 1.8A for the RD 6.5. Since both devices read the same I assume the number is accurate. I use the energy monitor to measure the wattage as well which is around 80W. This matches up with manufacture specs so I sort of assume it's right as well. The monitor seems to be accurate for everything else as well and while I expect so see a power factor for AC pumps I have a hard time understanding how the RD 6.5 can have one so low. Even the RD skimmer pump is really low. Even the mag pump as 0.6 seems low. I expected around 0.8 like you said so I just though maybe there was something I'm missing here. The power factors I listed is just a ratio from P = V * I * PF

[mike31154]

From what I gather with a little bit of reading on the subject, smaller AC motors generally have a lower power factor than larger ones. It also depends on the load being drawn. Varying load will cause the current/voltage lag to change and along with it, the power factor. It seems as if newer motors are designed with smarter controlling circuitry to improve the power factor.

[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.

[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.