Power Factor for AC Pumps

[mrhasan]

Originally Posted by sphelps

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]

Originally Posted by mike31154

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]

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]

Originally Posted by Jeff000

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]

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]

Originally Posted by sphelps

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]

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 ?

[sphelps]

[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).