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sphelps
03-07-2013, 01:47 AM
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:

albert_dao
03-07-2013, 06:09 AM
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
03-07-2013, 07:01 AM
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:

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
03-07-2013, 02:35 PM
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
03-07-2013, 02:42 PM
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
03-07-2013, 03:19 PM
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
03-07-2013, 03:51 PM
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
03-07-2013, 03:56 PM
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
03-07-2013, 03:58 PM
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/Power+Factor/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
03-07-2013, 04:00 PM
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
03-07-2013, 04:02 PM
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
03-07-2013, 04:03 PM
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.

sphelps
03-07-2013, 04:16 PM
http://www.enmax.com/Power/Tariffs/Power+Factor/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
03-07-2013, 04:23 PM
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

sphelps
03-07-2013, 04:32 PM
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.

mrhasan
03-07-2013, 04:35 PM
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 :)

sphelps
03-07-2013, 04:44 PM
Well this is looking awesome, 65 Watt max stamped right on the $700 pump, meanwhile it's using over 200 Watts.

mrhasan
03-07-2013, 04:46 PM
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. :)

sphelps
03-07-2013, 04:49 PM
Hurry up, the large hammer I'm holding above the pump is getting heavy

mike31154
03-07-2013, 05:01 PM
Nothing to confirm if you've measured consumption at 200 watts, other than what's stamped on the motor is evidently false info....

mrhasan
03-07-2013, 05:03 PM
Hurry up, the large hammer I'm holding above the pump is getting heavy

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.

sphelps
03-07-2013, 05:03 PM
Nothing to confirm if you've measured consumption at 200 watts, other than what's stamped on the motor is evidently false info....

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

Jeff000
03-07-2013, 05:08 PM
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.

sphelps
03-07-2013, 05:09 PM
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

wmcinnes
03-07-2013, 05:17 PM
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

wmcinnes
03-07-2013, 05:18 PM
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

Exactly.

mrhasan
03-07-2013, 05:21 PM
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

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

lastlight
03-07-2013, 05:23 PM
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 ?

sphelps
03-07-2013, 05:25 PM
http://www.boardsandrec.com/images/smilies/hulk.gif

mrhasan
03-07-2013, 05:29 PM
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).

lastlight
03-07-2013, 05:32 PM
Yeah not sure what I was thinking that pump is not DC :)

It's the fancy Laguna.

sphelps
03-07-2013, 05:35 PM
Yeah not sure what I was thinking that pump is not DC :)

It's the fancy Laguna.

Yeap very fancy, want to buy it?

sphelps
03-07-2013, 05:57 PM
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.

The power factor correction devices are said to improve the second half of the above equation, the Apparent Power. However you don't pay your utility for Apparent Power. You pay them for Real Power (Watts). Apparent Power is defined as the total power in an AC circuit, both dissipated AND returned! (scroll to the bottom of this link to view the power triangle and description of Apparent, Real and Reactive power). This means that if you currently have a poor power factor, your Apparent Power is higher, but all this means is that you are returning more unused electrons to the utility! But since they only charge you for used electrons (dissipated electrons = Real Power = Watts) you don't give a hoot about your Apparent Power!

Let's take an example of 2 completely identical motors sitting side by side. Both of these motors have the exact same efficiency and operate at 1.2 kW. The first motor doesn't have a power correcting device. The second motors does have PF correcting device.

Motor 1: 1.2 kW motor, connected to a 120 V circuit, PF = .7
Motor 2: 1.2 kW motor, connected to a 120 V circuit, PF = .999 (this has the Power Factor correction device, thus the excellent PF!)
Using the equation above we can show the amps (current) that will be dissipated in motor 1:

1.2 kW = .7 *120V * A → A= 14.29

And we can do the same thing for motor 2:

1.2 kW = .999*120V*A → A=10.01

But this doesn't mean you'll pay less to the utility! All this shows as that your power factor increases (gets better) your amperage decreases, but the Real Power (Watts = what the utility charges you) stays the same! Therefore no matter your power factor, in residential settings the utility is still going to show that you took the same amount of Real Power off of the power lines, so that is what you pay.

wmcinnes
03-07-2013, 06:07 PM
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.

sphelps
03-07-2013, 06:26 PM
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

Okay, so let's talk about "power factor". Without getting too technical, sometimes more power goes into a device than you'd expect, because of a special kind of inefficiency. The actual power used by your device is measured in kW, and that's what you're charged for. If your device uses only 80% of the power going into it, the power factor is 80%. Power factor is the Real Power (the amount your device actually uses) divided by the Apparent Power (the total going into it). For example, 80 kW (Real Power) 100 kVA (Apparent Power) = 80%. And again, residential customers are charged only for the Real Power, not the Apparent Power.

sphelps
03-07-2013, 06:38 PM
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-files/powerelectronics.com/files/archive/powerelectronics.com/images/0507pfc-Table02.jpg

Jeff000
03-07-2013, 07:01 PM
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

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

mrhasan
03-07-2013, 07:17 PM
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.

sphelps
03-07-2013, 07:26 PM
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.

Goatman
03-07-2013, 07:28 PM
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:

http://www.reefcentral.com/forums/showthread.php?t=1989362

sphelps
03-07-2013, 07:29 PM
http://www.reefcentral.com/forums/showthread.php?t=1989362

I read that thread already, it really isn't what we're talking about here.

Goatman
03-07-2013, 07:30 PM
I read that thread already, it really isn't what we're talking about here.
oops, sorry wrong paste...ignore :)

mrhasan
03-07-2013, 07:37 PM
I am not reading online since I am getting infos from electrical engineers. I have, out of curiosity, confirmed with another colleague of mine just now and is also saying that you get billed for the amount of voltage and current you are drawing out from the outlet; NOT how much your device is consuming. There's no such utility meter that can determine which device is consuming what. With the difference in calculation, he came to some conclusion:

1. The motor is really horrible
2. The readings are not taken from the correct point

I am not with number 2 because the DC pump is right on target regarding the readings; so I stick with the motor is horrible ;)

sphelps
03-07-2013, 07:49 PM
I cannot find anything to suggest a residential meter measure kVA, only watts.

There are three terms you will encounter when dealing with alternating (AC) power (as opposed to DC or batteries). The first is the kilowatt, abbreviated kW, and it represents real power. Real power can perform work. Power company utility meters on the side of your house measure this quantity and charge you for it.

The second term is reactive power, and it is measured in KVAR which is short for kilo volt amp reactive. Unlike kW, it cannot perform work. Residential customers do not pay for KVAR, and utility meters on houses do not record it.

The third term is apparent power, referred to as KVA. If you hook up two multimeters to measure current and voltage and then multiply the readings together, you get apparent power in volt amps, or VA. To distinguish it from real power, VA is used instead of W.

mrhasan
03-07-2013, 07:52 PM
I cannot find anything to suggest a residential meter measure kVA, only watts.

Yap residential meters record KWh, not KVA.

wmcinnes
03-07-2013, 07:53 PM
Yap residential meters record KWh, not KVA.

Thus your meter only counts the real power.


I'm with Steve and his research.

mrhasan
03-07-2013, 07:55 PM
I'm with Steve and his research.

Then my question is, how would each of the real power of individual components in the household be calculated by the utility meter?

wmcinnes
03-07-2013, 08:01 PM
Then my question is, how would each of the real power of individual components in the household be calculated by the utility meter?

They simply measure kW and not kVA.

They consist of a motor with the stator current being the current into the building and the rotor curreent being proportional to the voltage at the building input. The motor turns a disc (the disc you see rotating in the window) that passes through a magnetic field generating eddy currents in the disc and this regulates the speed at which the disc turns so that the meter can be adjusted to read correctly.

sphelps
03-07-2013, 08:03 PM
Then my question is, how would each of the real power of individual components in the household be calculated by the utility meter?

Because the utility meter measures the exact same thing as the energy monitor, it measures actual power in kWh. It doesn't measure apparent power or vars.

mrhasan
03-07-2013, 08:05 PM
They simply measure kW and not kVA.

They consist of a motor with the stator current being the current into the building and the rotor curreent being proportional to the voltage at the building input. The motor turns a disc (the disc you see rotating in the window) that passes through a magnetic field generating eddy currents in the disc and this regulates the speed at which the disc turns so that the meter can be adjusted to read correctly.

1. KW is not a direct unit so the meter actually does some conversions within itself

2. That is exactly how it is done. The rotor moves proportionally to the product of voltage and current to shows kWh consumption. No PF involved that's it. So when a outlet draws 1.85A @ 115V, that means the rotor is turning at a proportional rate directly to the product of those two.

The motor needs 82W to run and it is taking in 82W but due to its inefficient manner (age, water, etc.), the pump has to take in more power and hence 1.85*115V.

For this motor, you are bringing 1.85A "in" the house.

wmcinnes
03-07-2013, 08:12 PM
If the current and voltage are in phase, the motor speed is the product of voltage and current (PF=1). If they are not in phase the motor in the meter will run fast for part of a cycle and slower for the other part with the average speed being the product of the in-phase voltage and current. Thus measuring real power.

sphelps
03-07-2013, 08:15 PM
According the producers of the energy monitor:


In AC power measurements, there actually three separate power components that can be measured:

Apparent Power: Measured in VA (volt - amperes). This is measured by taking the RMS voltage and RMS current readings and multiplying them together. This is what the eMonitor reports for power. All electrical circuits must be sized to handle apparent power because instantaneously, this is the maximum amount of power that can be flowing in a circuit.

Real Power: Measured in watts. This is the actual power that is being consumed by the load. This is what the utility company measures on the meter, as is what customers see on their electrical bill.

Reactive Power: This is kind of like imaginary power. It is really the difference between the Apparent Power and the Real Power. It is power that flows back and forth while the voltage and current are out of phase and is caused by the inductive load. Your utility company does not charge for reactive power, but it must handle reactive power, and reactive power does cause real current to run in your wires, and can generate heat. Reactive power is measured in VAR (volt - amperes reactive).

mrhasan
03-07-2013, 08:16 PM
If the current and voltage are in phase, the motor speed is the product of voltage and current (PF=1). If they are not in phase the motor in the meter will run fast for part of a cycle and slower for the other part with the average speed being the product of the in-phase voltage and current. Thus measuring real power.

Yes. So how would the meter know which device has what amount of PF? It must do an equivalent resistance scenario for all the loads in the house with one common PF for the whole house and thus multiplying the voltage * current going into house * common PF. The meters can't judge individual PF of the devices.

mrhasan
03-07-2013, 08:23 PM
According the producers of the energy monitor:

Yap that's right. Like I said in the very beginning, the utility companies either consider pf=1 or maybe they just do some calculations to find a relative pf for the houses to multiply the RMS volt and current. The meter's simply can't find the individual PF of each appliances. Alongside, households doesn't generally have heavy PF lowering stuffs and hence its more or less above 0.9. Two or three pumps may have low PF in your house but in the end, it won't effect much to the pf of your whole house.

I have sent an email to enmax regarding this. The theory and the claims are just not going through and would love to clear it up too. :razz:

Another thing, if possible, if to measure the resistance of the motor (disconnect it and then measure) and then putting in the current (1.85) and resistance into the formula:
I^2/R.

sphelps
03-07-2013, 09:01 PM
Well my understanding of AC power isn't great, not my specialty by a long shot. However it seems obvious to me that the term reactive power (the difference between apparent and real) when referred to "imaginary power" isn't something you're going to billed for. It's not being used despite the fact it goes through the circuit. While I might find it amusing to see a line on my power bill that said "Imaginary power usage = 958 kWh (this is the power you didn't use)" I just can't see it.

It's why everything online states a utility meter that measures in kW measures real power and not apparent power. It's clear power factors have virtually zero impact on your power bill which is why all those power factor compensation devices are a scam, also well documented.

I've also sent an inquiry to Enmax and GE (meter manufacturer) for conformation.

mrhasan
03-07-2013, 09:02 PM
Alright Steve you don't have to break the motor. I came across an energystar article and they say that in NA grid system, if you have a low pf equipment in domestic facility, you draw in more current but you are only charged for the amount of power that you are consuming and the extra power due to the charge is actually put into the penalty of the industries who have to pay for pf corrections.

So I guess you are actually getting charged for 82W :P

Sorry for the confusions; sometimes theory and practical stuffs doesn't match :) I will still post the response I get from enmax.

mrhasan
03-07-2013, 09:05 PM
Well my understanding of AC power isn't great, not my specialty by a long shot. However it seems obvious to me that the term reactive power (the difference between apparent and real) when referred to "imaginary power" isn't something you're going to billed for. It's not being used despite the fact it goes through the circuit. While I might find it amusing to see a line on my power bill that said "Imaginary power usage = 958 kWh (this is the power you didn't use)" I just can't see it.

It's why everything online states a utility meter that measures in kW measures real power and not apparent power. It's clear power factors have virtually zero impact on your power bill which is why all those power factor compensation devices are a scam, also well documented.

I've also sent an inquiry to Enmax and GE (meter manufacturer) for conformation.

Please do post what they say. It will help me get my theories straight too.

World should revert back to DC grids like Edison's time. AC stuffs are just too complicated :twised:

sphelps
03-07-2013, 09:07 PM
Yap that's right. Like I said in the very beginning, the utility companies either consider pf=1 or maybe they just do some calculations to find a relative pf for the houses to multiply the RMS volt and current. The meter's simply can't find the individual PF of each appliances. Alongside, households doesn't generally have heavy PF lowering stuffs and hence its more or less above 0.9. Two or three pumps may have low PF in your house but in the end, it won't effect much to the pf of your whole house.

I have sent an email to enmax regarding this. The theory and the claims are just not going through and would love to clear it up too. :razz:

Another thing, if possible, if to measure the resistance of the motor (disconnect it and then measure) and then putting in the current (1.85) and resistance into the formula:
I^2/R.

That's really not what my quote said by the way, but at this point we'll have to agree to disagree until we get conformation either way.

sphelps
03-07-2013, 09:09 PM
Alright Steve you don't have to break the motor. I came across an energystar article and they say that in NA grid system, if you have a low pf equipment in domestic facility, you draw in more current but you are only charged for the amount of power that you are consuming and the extra power due to the charge is actually put into the penalty of the industries who have to pay for pf corrections.

So I guess you are actually getting charged for 82W :P

Sorry for the confusions; sometimes theory and practical stuffs doesn't match :) I will still post the response I get from enmax.

Good news so far, I'll wait for conformation from Enmax before celebrating though ;)

mrhasan
03-07-2013, 09:10 PM
That's really not what my quote said by the way, but at this point we'll have to agree to disagree until we get conformation either way.

I am convinced that utilities charge households just for amount consumed so you are getting charged for 82W. Maybe that's why household businesses are becoming popular now a days? ;)

mrhasan
03-07-2013, 09:11 PM
Good news so far, I'll wait for conformation from Enmax before celebrating though ;)

I am feeling bad for someone out there who is paying more because of your motor having low pf :lol:

sphelps
03-07-2013, 09:18 PM
I am feeling bad for someone out there who is paying more because of your motor having low pf :lol:

I still don't think it works that way, the extra current isn't lost. The only losses are from heat resulting from the difference in current that feeds through the lines plus some very minor losses in the pump which you would experience with most devices. The extra current flows back to the source. Large industries are sometimes penalized for low power factors due to the provider having to supply higher currents and accept higher losses due to cable inefficiency. In residential the losses are minimal and probably absorbed in general as delivery charges.

The real advantage of high power factors is the ability to design lower powered circuits, it really has very little to do with efficiency.

And before you curse me, measure your own pumps :lol: many home appliances such as washing machines are known to have very low power factors. And while I'm still really surpirsed at the number for these expensive pumps at the end of the day at least they don't cost that much to run and I'd still rank then significantly higher in quality over the wavelines I have as well.

sphelps
03-07-2013, 10:31 PM
Funny story, I just got a phone call from some direct energy provider wanting me to switch to them for electricity. I asked if they'll charge me for reactive power and she said she didn't know. I then asked how can you expect me to switch if you can't even tell me if I'll be billed for the imaginary power my pumps use. She hung up..

Madmak
03-07-2013, 10:48 PM
You've probably scared the crap out of her...

sphelps
03-07-2013, 10:56 PM
Well I thought it was a legit question, we've been talking about it all morning.

hillegom
03-07-2013, 11:12 PM
I like the fine print in a certain "other" thread

albert_dao
03-07-2013, 11:24 PM
What you SHOULD do is find out why your pump is running at a power factor completely out of line with my numbers. How old is it?

sphelps
03-07-2013, 11:29 PM
What you SHOULD do is find out why your pump is running at a power factor completely out of line with my numbers. How old is it?

About 4 years I think, in service for around 2.5. Everything seems good, runs like new and from my googling it seems a few users have discovered that RD pumps have low power factors. Also the bubble king 2000 is following a similar path. Doubtful I have two defective units.

Jeff000
03-08-2013, 12:15 AM
Funny story, I just got a phone call from some direct energy provider wanting me to switch to them for electricity. I asked if they'll charge me for reactive power and she said she didn't know. I then asked how can you expect me to switch if you can't even tell me if I'll be billed for the imaginary power my pumps use. She hung up..

You're asking some girl that has no technical training, only a script to read to sell you the service and answers to standard questions.
Had you said power factor and not reactive power you might have gotten an answer.

I'm a sparky myself, but not a power lineman. But I did call a buddy working for epcor, he's a meter tech for them, he said the meters only measure actual power use for residential and light commercial.

They aren't even allowed to have the smart meters monitor power factor for statistics.

sphelps
03-08-2013, 12:19 AM
Good stuff, thanks Jeff.

mrhasan
03-08-2013, 12:49 AM
Lol. She must have been like..."huh, imaginary power"......she chose a great day to call you ;)

Baldy
03-08-2013, 01:58 AM
pretty much everything you guys are asking about is all 2nd year electrician apprenticeship. Residental meters measure power, not apparent or reactive power. the meters used on larger commercial and industrial buildings are called demand meters and only charge for true power used until the buildings peak apparent power goes beyond a cirtain point, then they are charged heavily for the extra demand. even if your pump has a pf of 0.5, and pulls 120 va or apparent power (120v*1a), its only using 60w of power, and your only charged for 60w of power. Power companies dont charge residential buildings for apparent power because most of the devices we use have power factor correction to 0.9.

sphelps
03-08-2013, 02:43 AM
Yeah, yeah. Where were you this morning? ;)

mrhasan
03-08-2013, 02:45 AM
Yeah, yeah. Where were you this morning? ;)

I was actually thinking that too! So many words, searches and (umm...) interrogations could have been saved!

Baldy
03-08-2013, 02:49 AM
sorry i just saw the thread tonight. just for credits sake, i am a 4th year electrician, so ive had 2 school terms of the trig calculations to back this up. last year was all 3 phase which is quite a bit harder =P

mrhasan
03-08-2013, 02:55 AM
sorry i just saw the thread tonight. just for credits sake, i am a 4th year electrician, so ive had 2 school terms of the trig calculations to back this up. last year was all 3 phase which is quite a bit harder =P

Three phase is "quite" a bit harder? :neutral: You must be a genius when it comes to power. Being an electrical major myself, I feel crap when it comes to stuffs regarding AC and three phase Y-Delta craps. I am more of a DC person and my undergrad is not from here so never knew about how the grid works over here.

Thanks for clearing the confusions up. Learned a lot today :)

Cheers to Steve for starting this thread in the first place and keeping it warm! I wish I saw your powerhead add before; I need a power head badly :P

wmcinnes
03-08-2013, 02:59 AM
Us Power Engineers know everything there is about power generation but when it comes to how a residential power meter works.. Me feel stupid haha:lol:

mrhasan
03-08-2013, 03:01 AM
Us Power Engineers know everything there is about power generation but when it comes to how a residential power meter works.. Me feel stupid haha:lol:

In my defense, I am from a 3rd world country where we have more blackouts than on the moon :lol: Our generation works in a totally different way ;)

wmcinnes
03-08-2013, 03:05 AM
In my defense, I am from a 3rd world country where we have more blackouts than on the moon :lol: Our generation works in a totally different way ;)

Haha I can imagine it does!

Baldy
03-08-2013, 03:16 AM
Three phase is "quite" a bit harder? :neutral: You must be a genius when it comes to power. Being an electrical major myself, I feel crap when it comes to stuffs regarding AC and three phase Y-Delta craps. I am more of a DC person and my undergrad is not from here so never knew about how the grid works over here.

Thanks for clearing the confusions up. Learned a lot today :)

Cheers to Steve for starting this thread in the first place and keeping it warm! I wish I saw your powerhead add before; I need a power head badly :P

3 phase is like single phase, which is what is at the wall outlet, only x3. the phases are each 120 degrees out of phase with each other. 3 phase motors are far more efficient, smoother and quieter than single phase motors. I cant wait until one of these manufacturers comes out with a 3 phase pump and variable frequency drive.

electrician isnt really that difficult. we only learn what pertains to our trade. I know almost nothing about power generation =)

mark
03-08-2013, 03:31 AM
is your measurement equipment accurate?

sphelps
03-08-2013, 04:18 AM
is your measurement equipment accurate?

Seems to be. Amps matches another measurement device and while I haven't verified the wattage meter portion of it, it matches other device ratings so I assume its accurate. Either way it's unlikely the wattage used is much higher and if its lower the power factor only decreases more. So unless everything is way off the accuracy isn't that important here.

mrhasan
03-12-2013, 06:08 PM
This is what I got from Enmax:

Good morning Raied,

Thank you for contacting ENMAX. My name is Nancy and I appreciate the opportunity to respond to your billing inquiry.

Please accept our apologies for the delay in response. We have been experiencing higher than normal volumes which have impacted our service levels.

Raied, ENMAX meters measure wattage usage.

Should you have any further questions or concerns, please do not hesitate to reply to this email and I will be happy to assist you. Alternatively, you may contact our Customer Care Centre at 310-2010 (Toll Free in Alberta) Monday to Friday 8:00 am to 8:00 pm and Saturday 8:00 am to 4:30 pm.

Thank you for granting ENMAX the opportunity to be of service for you.

Yours truly,


Nancy C.
Customer Correspondence Team
ENMAX and City Utilities

sphelps
03-12-2013, 06:16 PM
Yeah she gave me a similar response today as well...


Good afternoon Steven,

Thank you for contacting ENMAX. My name is Nancy and I appreciate the opportunity to respond to your inquiry.

Your ENMAX electric meter measures the amount of power being utilized by your residence.

I did find an article online which may be of interest to you: http://lifeislikethat.com/?p=847.

Steven, should you have any further questions or concerns, please do not hesitate to reply to this email and I will be happy to assist you. Alternatively, you may contact our Customer Care Centre at 310-2010 (Toll Free in Alberta) Monday to Friday 8:00 am to 8:00 pm and Saturday 8:00 am to 4:30 pm.

Thank you for granting ENMAX the opportunity to be of service for you.

Yours truly,


Nancy C.
Customer Correspondence Team
ENMAX and City Utilities

lastlight
03-12-2013, 07:48 PM
Raied, ENMAX meters measure wattage usage

it took her a while to track someone down who knew what the deuce you guys were nattering about.

sphelps
03-12-2013, 08:20 PM
Not only that but she really didn't answer the question at all, honestly I think she has no idea what we're talking about. This is the inquiry I sent:

Hello,

As a current customer I'm investigating some techniques for energy saving in my home and I have question relating to power factors and whether a typical residential meter measures apparent power or real power.
For example if I have a load such as pump that has a power factor of 0.5 the current in the circuit will be twice that of what the actual device is using, if the measured current is 5A then the apparent power is 575W on a 115V circuit while the real power used by the device is 288W.
The question I have is does a residential meter measure and bill for real power or apparent power? Is it worth investing in new devices with higher power factors if the goal is reduce electricity cost?

Judging by the response it's pretty clear she has no idea what we're on about and relied on google to quickly find an article that somewhat discusses power factor.

mrhasan
03-12-2013, 08:41 PM
Exactly! Come on we know that the usage is based on the wattage meter but the question generally was how/what power does it read. Oh well :razz:

Delphinus
03-13-2013, 12:38 AM
I'm late to this party, holy moly my head is swimming after reading it from beginning to end. Anyhow, I always thought your billing was based, more or less, on the "loss of potential". I haven't seen that term thrown about in here so now it's in there.

And electrons. I'm pretty sure electricity has something to do with electrons.

And Shaw has little robots that live in your house's wires.

Word.

StirCrazy
03-13-2013, 12:48 AM
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.

nope this is not how it works
here is a simple explanation of PF I found

"Power factor is the percent of electrical power that does work. Resistive loads, such as lights and heater elements, always have unity (1.0) power factor; all power is used for work. Motors, because they are essentially large inductors, lag current and cause power factor issues. A motor with a .85 power factor uses 85 percent of the power for work. 15 percent is wasted.

For example, a 480VAC, 10HP motor with a 1.0 power factor uses 10.6 amps to run at 10HP. Lowering the power factor to .8 requires the motor to consume 13.2 amps to produce that same 10 HP."

I read up on this years and years ago and can talk about it all day (well maybe not all day) but it is hard to sit and type it out, hence the simple explanation I found. we used to be realy concerned with PF with lights to get the efficiency.


you if you have a pump that is rated at 120 watts but has a power factor of .6 in reality it will use 200 watts to do the work. the extra 80watts are lost through ineficient circuts, heat, ect... so when the power company bills you you are billed for the 200 watts. this is why the hydro companies have rebates for big business to upgrade to high PF equipment, it lowers the demand on the grid.

Steve

mrhasan
03-13-2013, 01:05 AM
nope this is not how it works
here is a simple explanation of PF I found

"Power factor is the percent of electrical power that does work. Resistive loads, such as lights and heater elements, always have unity (1.0) power factor; all power is used for work. Motors, because they are essentially large inductors, lag current and cause power factor issues. A motor with a .85 power factor uses 85 percent of the power for work. 15 percent is wasted.

For example, a 480VAC, 10HP motor with a 1.0 power factor uses 10.6 amps to run at 10HP. Lowering the power factor to .8 requires the motor to consume 13.2 amps to produce that same 10 HP."

I read up on this years and years ago and can talk about it all day (well maybe not all day) but it is hard to sit and type it out, hence the simple explanation I found. we used to be realy concerned with PF with lights to get the efficiency.


you if you have a pump that is rated at 120 watts but has a power factor of .6 in reality it will use 200 watts to do the work. the extra 80watts are lost through ineficient circuts, heat, ect... so when the power company bills you you are billed for the 200 watts. this is why the hydro companies have rebates for big business to upgrade to high PF equipment, it lowers the demand on the grid.

Steve

Not again!!! :cry:

That's what I thought but apparently, the meters in our houses only counts the amount of power used (consumed) by the devices and the industries, in return, are charged PF penalties.

mrhasan
03-13-2013, 01:06 AM
And electrons. I'm pretty sure electricity has something to do with electrons.


Ummm...electrons? wat's that? can you post a picture? doesn't sound like reef safe to me :neutral:

reefermadness
03-13-2013, 03:15 AM
Residence are only charged for REAL power, not APPARENT power (power factor).

Real power is measured in watts and apparent power is measured in VA (volt-amps).

You will notice on your bill you are being charged a measurement of watts which is real power.

wmcinnes
03-13-2013, 05:21 AM
Can we just make this thread die? I don't wanna debate over this anymore! :razz:

beefORchicken
03-13-2013, 05:34 AM
if you are worried about PF, capacitors can be used to cancel out the inductive load and bring PF back to ~1, probably more trouble than its worth though...

sphelps
03-13-2013, 05:37 AM
It's not a debate anymore, it's already been established and some people in the know have posted and confirmed. We don't pay for apparent power and power factors make virtually no difference when it comes to you power bill with the exception of some minor cable inefficiency losses. If you think otherwise you're wrong so consider yourself educated :)

wmcinnes
03-13-2013, 06:08 AM
ive agreed with you the whole time Steve!

Baldy
03-13-2013, 01:01 PM
nope this is not how it works
here is a simple explanation of PF I found

"Power factor is the percent of electrical power that does work. Resistive loads, such as lights and heater elements, always have unity (1.0) power factor; all power is used for work. Motors, because they are essentially large inductors, lag current and cause power factor issues. A motor with a .85 power factor uses 85 percent of the power for work. 15 percent is wasted.

For example, a 480VAC, 10HP motor with a 1.0 power factor uses 10.6 amps to run at 10HP. Lowering the power factor to .8 requires the motor to consume 13.2 amps to produce that same 10 HP."

I read up on this years and years ago and can talk about it all day (well maybe not all day) but it is hard to sit and type it out, hence the simple explanation I found. we used to be realy concerned with PF with lights to get the efficiency.


you if you have a pump that is rated at 120 watts but has a power factor of .6 in reality it will use 200 watts to do the work. the extra 80watts are lost through ineficient circuts, heat, ect... so when the power company bills you you are billed for the 200 watts. this is why the hydro companies have rebates for big business to upgrade to high PF equipment, it lowers the demand on the grid.

Steve

Ill chime in once more to clear up this misinformation. in your example, "200w" is actually 200VA. apparent power is measured in volt-amps. in a power triangle it is the hypotenuse. the 120w is correct, and is the horizontal side of the triangle. the one no ones talked about yet is the vertical side which is reactive power and its measured in VARs or volt-amp reactive. The power factor is a ratio between apparent power and true power. you are only charged for the true power used, which is the watts. that is the current that is in phase with the voltage. reactive power leads or lags the voltage, and isnt measured by residential meters. in this example, you will be charged for 120w NOT 200w.

StirCrazy
03-14-2013, 05:24 AM
in this example, you will be charged for 120w NOT 200w.

Unless your with bc hydro on a smart meter which can and do measure aparent power. A lot of people are pointing to this as the reason there power bill has increases after smart meter instalation. Other places who have smart meters in the us had the same problem.

Understand the reason the hydro companies want to be able to measure it, as they need larger capacity systems to compensate for aparent power. They may or may not be charging for it yet, i wont know till i see my next bill as mine was just installed.

Steve

sphelps
03-14-2013, 06:54 PM
Smart meters can measure apparent power and therefore a power factor for the whole house. Pretty sure any more modern digital meter probably can as well, smart or not. But they don't charge residential for apparent power, plain and simple. The meters are more accurate and measure all the brief high start up power generated by larger appliances which previous analog meters couldn't measure. This is one reason you'll see an increase. Other increases will result from peek time billing if applicable. Power companies will sometimes even list your house PF and apparent power in kVA on your bill but they won't charge you for it.

I'm sure a few people already using smart meters in BC can confirm this for you.