Sorry, Another Power Factor Question

Re: Sorry, Another Power Factor Question

jcheil wrote: »

I have a chest freezer, and running it thru the kill-a-watt meter it shows a power factor of .50
The voltage shows as 120v and the watts shown are 75 and the amps (I believe) shown were 1.2ish.
I don't remember what the va showed.

1) In any event. It is my understanding (correct me if I am wrong) that it means basically 1/2 of the power that is being consumed by this device is being "wasted" in some way?

Sort of yes and no. How is that for a clear answer?

2) So, does my inverter actually have to supply MORE power to it because of the poor power factor; thus draining the batteries faster? And if I added the proper motor run capacitor would that reduce the draw on the inverter/batteries?

Vector Math from Physics (high school). Work is actual "work done". If you have a swing pendulum, it switches between kinetic energy (high velocity at the bottom of the swing) and potential energy (no motion, but the weight is a few inches above the lowest point of the arc).

"We" (people) see work being done as the transfer of energy from one type to another. However, the physicists and engineers do not--From a black box point of view, there is no energy going into or out of the system (other than a little friction). So, no work is being done. Sort of what is happening with your electric motor... Some of the "energy" is sorted in the motor's coils (Inductors) and 1/2 of the cycle the wall outlet is "filling" the inductors with energy, and the other 1/2 of the cycle the inductors are pushing energy back into the outlet (no net work being done for that part of the current flow).

With your measurements, you see ~75 watts and (120 volts * 1.2 amps=) and ~144 VA of "stuff" being used by the motor.

• PF = Watts / VA = 75W / 144VA = 0.52 PF

We are used to seeing:

• Power = Volts * Current

And that is correct for "simple" systems (powering a filament lamp or electric heater). The "real" equation is:

• Power = Volts * Amps * Cos (phase angle between V and A vectors) = V*A*PF (PF=Cos phase angle, among other things)

An "induction motor" typically is very "Inductive" (like a coil vs a resistor or capacitor). And the current into the motor tends to "lag" the voltage sine wave:

• cos-1 (0.52) = 59 degrees (current lags voltage by 59 degrees)

One way to look at what this means to "us"... Think of standing in front of a car and pulling it with a rope. If you are dead in front of the car, 100% of your force goes into pulling the car forward. Cos 0 degrees = 1.00

If you stand of to the side of the car (say 59 degrees to the side), then Cos (59) = 0.52 of the force (52%) on the rope goes to pulling the car forward.

So--In terms of rope, you have to have the rope 1/0.52 = 1.92 times stronger to pull the car just as hard as you would if standing in front.

In terms of electricity... Your wiring to do 75 watts worth of work would have only required:

• P=V*I; I=P/V
• P=75W / 120V= 0.625 Amps

But when you have a Power Factor that does not equal one, then the device draws more current:

• P=V*I*PF; I = P/(V*PF)
• I = 75 Watts / (120V * 0.52PF) = 1.20 Amps

So, your wiring has to be almost 2x thicker to handle the additional amperage (due to power factor).

And any transformers will have to have a higher VA rating (1.20A * 120 VAC = 144 VA rated).

And, typically, most small AC inverters and gensets are rated in Watts. And their VA rating is the same as the Watt rating (i.e., when PF = 1.0; Watts=VA).

So, in this case, instead of a 75 Watt rated Inverter, you need a 144 VA (and Watt) rated inverter... (larger inverter/wiring to manage the higher current due to "poor" power factor).

Interestingly, on the DC side of the equation, the batteries will only need to supply the real power (75 Watt) amount of energy (neglecting losses).

• I=P/V = 75 watts / 12 volts = 6.25 Amps DC (at 12 volts)

There is more to discuss--But does this stuff make sense so far?

-Bill

Re: Sorry, Another Power Factor Question

The "front end" of the Iota chargers would need a complete redesign to "fix"... You cannot add a capacitor to "adjust" the PF of the Iotas.

The modern Inverter-Chargers, usually, have pretty good power factor and high overall efficiencies. As well as being much more programmable in their charging settings/options.

And the "magic" generator support (make your small generator look "larger" by adding battery power when starting/running heavy loads like well pumps).

-Bill

Re: Sorry, Another Power Factor Question

No--In my humble opinion--For the "standard" AC to DC power supply, it is the rectifying diodes and the "high voltage" (~380 volt nominal, typical) that converts from AC to DC that is the problem.

If you look up a couple of posts, you will see an oscilliscope picture that shows the Voltage Sine wave and the Current Peaks that don't look at all like Sine Waves. That is "poor" non-linear power factor.

A motor would simply have a sine wave current delayed by xx milliseconds which adding a capacitor can "fix".

-Bill