Power Factor for AC Pumps

[mrhasan]

Originally Posted by wmcinnes

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]

Originally Posted by mrhasan

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 ?

[mrhasan]

Originally Posted by lastlight

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]

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]