Power Factor for AC Pumps
 

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? :biggrin:

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?

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.

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

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.

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.

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.

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.

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.

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.

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.

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 :D

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.

Easiest solution: move to DC motors. :P

I talked with a friend of mine from montreal and he said you are going to be billed for 1.85 * 115 * number of operating hours, theoretically. I will keep you updated :)

Well this is looking awesome, 65 Watt max stamped right on the $700 pump, meanwhile it's using over 200 Watts.

Hold on no need to get excited about it :razz: I haven't confirmed it yet. :)

Hurry up, the large hammer I'm holding above the pump is getting heavy

Nothing to confirm if you've measured consumption at 200 watts, other than what's stamped on the motor is evidently false info....

I talked with a electrical post doc over here and he said you are billed for how much current you are drawing and not how much your appliance is consuming. When he heard about the PF of your motor, he was like "its terrible". So in that case, its more than 200W and you are being billed for that.

I will talk with my prof and let you know the final verdict after few hours.

If you plug the pump into an energy monitor it reads 82W but it pulls 1.8A. So the question is as far as the power company goes are they billing me for 82W or 1.8Ax115V=207W

Where are you getting your power factor numbers from?

In any case put your watt meter to the pump, that will tell you the watts.

You're not billed kva in residential. And even if you were you need to look at what your whole system is drawing, not just one little part.

Watt meter on the pump reads 82W, same meter reads 1.8A & 115V

P = V * I * PF
So PF = P/(V*I) = 82W/(115V*1.8A) = 0.4

I'm just restating what has been said but:

PF= True Watts/ Volts*Amperes
True Watts= 82w
Volts= 115v
Amps= 1.80a

PF= 82/207 = 0.396

You're pump output is only 40% of the power it is drawing. Not good, not efficient! haha

Exactly.

There you go Steve. Another justification :) You better change the motor if you want the consumption to cut to half! DC motors FYI!

The RD is a DC motor ?

RD does have DC motor pumps but I don't think 6.5 is DC motors. Can't seem to find the exact motor specs of 6.5 but with that power factor, its impossible to be a DC motors pumps. DC motors don't have any pf, the only thing that will have some PF is the converter to run the pump (hence the 0.97 in Steve's DC motor).

Yeah not sure what I was thinking that pump is not DC :)

It's the fancy Laguna.

Yeap very fancy, want to buy it?

I'm still on the fence, everything I read states power companies bill consumers based on real power and not apparent.

Here's a little quote from a website that talked about power factor correction devices and how they don't work for residential.

Interesting! Well I hope thats how it is! I can't imagine a pump being out as much as yours apparently is so perhaps this is true.

I'm fairly confident at this point power factors really have nothing to do with a residential power bill, we're billed in kWh and our meters can't measure power factors, only real power in watts.

Some more talk about power factors and correction devices
http://michaelbluejay.com/electricity/powerfactor.html

It seems the only power losses from low power factors relate to the additional heat loss in the supplied cable due to the higher current. Here's a table comparing a power factor of 1 and 0.75 and the resulting power losses.

http://powerelectronics.com/site-fil...fc-Table02.jpg

If you live in Canada you are billed in kwh for a residential service, by law. Pf has no outcome on your bill.

Alright so I talked with my prof and he said you are billed for what you consume and its the amount of current that you are taking OUT from the power outlet, not the amount of power that the device is using. Because there is no way an utility meter can determine how much power each of the device is consuming separately. So in this case, since you motor is pulling out 1.8A through a 115V AC line, you meter is going to read 1.8A * 115V * number of operating hours. It cannot see the PF of the motor and the motor doesn't have any sort of feedback circuits that will tell the meter how much it is "actually" consuming. Meter will only note how much current you are sucking in the house for how long and at what voltage line.

So bottom line, your motor is "consuming" 207W from the grid but is using 82W for operation. Its how much you consume that gets billed on, not how much you actually use. Its like those phantom powers, you don't use it but you get billed for it.

I don't know about the calculation that someone did about less amp consumption but no impact on bill - its not theoretically viable. Less current = less consumption from the grid = less bill. Meter's don't look at the rating on the motors, it looks at how much "power" a.k.a V*I you are pulling in.

That goes against everything I read online, I just don't buy it. My understanding is still that it's not actually using that much current, that just the apparent power and not the actual. That extra 60% of the current in the circuit can't just disappear. A power factor of 0.4 doesn't mean 40% efficient.