Power Factor for AC Pumps

[lastlight]

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

It's the fancy Laguna.

[sphelps]

Originally Posted by lastlight

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]

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]

Originally Posted by sphelps

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]

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]

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.

[Jeff000]

Originally Posted by sphelps

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]

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]

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]

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?

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