Correcting P.F. and sag, using caps, consumes power?

  • Thread starter Jeff57
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In summary: What I found was confusing. My voltage was of course 121, but the Power Factor (PF) was 0.01, the amps was 3.7 , the watts read was 6 but the VA read 450.So, am I paying a higher power bill because caps are drawing 6 watts 24/7 ? Do these capacitors really correct my power factor and sag or am I wasting my time?"You might be paying more because the power company charges by the voltage, not by the wattage. The power factor (PF) is a measure of how much of the power supplied to a load is used by the load, not lost in the supply lines. Your meter is probably reading 0
  • #1
Jeff57
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I live in the country about 15 miles from the nearest power utility capacitor bank.
I believe this 15x2= 30 miles (round trip) of wire causes a reactive inductance (or internal resistance) by the time it reaches my home. I think this contributes to micro-brownouts or sags in the current available to me.

The previous owner had the yard littered with burn out frigs and electronics.

So large AC motors like air compressors with large stall amps, use starting capacitors during start up.

I decided to add my own 80 mfd 370vac Capacitors, one on each side of the 220 v mains at the meter panel.

The caps are from Packard Inc part # PRC80. My hope was that these would lower the "internal resistance" of my utility company, and stop any sagging of the leading edge of the sine wave.

Here's my issue, I connected one identical 80mfd cap to a power cord and plugged it into a "kill-a-watt" meter ( http://www.p3international.com/products/special/P4400/P4400-CE.html )

, which displays the power consumption of any device that is plugged into it.

What I found was confusing. My voltage was of course 121, but the Power Factor (PF) was 0.01, the amps was 3.7 , the watts read was 6 but the VA read 450.

So, am I paying a higher power bill because caps are drawing 6 watts 24/7 ?
Do these capacitors really correct my power factor and sag or am I wasting my time?

How can I show a 3.7 amp draw and the watts be only 6 and not 120vx3.7a ?

So I have florescent lights that indicate a P.F. of .55 using the same meter. I think I pay a penalty for phantom power draws less than 1.0 PF. So does the 0.01 of the cap, help or hurt me?

Not even the power company rep can give me a clear answer.

Thanks for the reply's. I would like to understand the math and how these things work.
-Jeff57 10/17/10

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  • #2
Capacitors do not dissipate power and inductance is not resistance.
Your supply company meter (it will be either a rotating disc or electronic simulation of the same) would normally be the sort that measures the 'dot product' of V and I and will give you an answer VI cos(ϕ). There's no reason why the kill a watt meter would be any different, these days so you can probably believe what it tells you.
Any reactive components in the circuit can produce an increase in the magnitude of V or I. The reason you can lose more power than necessary through non-ideal power factor is that reactive components can cause excessive current to flow in the supply wires (more than with a pure resistive load). This causes resistive losses in the albeit fairly low resistance of the cables. But your meter won't measure that and you won't be charged for it. All the meter can do is to measure V.I 'at your end of things'. afaik, it is the supply company who demand a reasonable power factor as it is wasting 'their power' not yours.

btw, the reactance of your supply cable will be capacitive rather than inductive.

There's not much you can do about sag under heavy loads because the supply lines have a significant resistance. You could ask for the transmission line to be operated at a higher voltage and use step up step down transformers at each end but you would probably get a short answer to that one!

When you connected the capacitors across the meter output they must be passing enough in-phase current for the meter to register 6W, although their power factor, as a load is very low (as you'd expect from a reactance). If the company meter is reading that then you will be charged for it. The place to put the capacitors is on the device you are using so they are only connected when the device is operating. But I'm not sure why you would want to.

"So large AC motors like air compressors with large stall amps, use starting capacitors during start up." This reads as if you mean it's to stop the sag - it isn't, it's to introduce a phase difference in the motor windings so that it will start rotating. Single phase induction motors need help starting and, as with all motors, they take a lot of current at low speeds -causing sag.
 
  • #3
Here's the math

The cap impedance at 60 Hz is Z = +1/jωC = -j33 ohms

Current at 121 volts is V/R = 121/-j33 = +j3.7 amps, all reactive, so PF = 0 ideally, and zero watts.

Bob S
 
  • #4
Thank you both, that was real help, real fast.

What about noise. My radio is full of static on am & FM when my HID lights come on and they use digital ballasts that correct for pf.

I thought a cap close to the radio's ac socket would help, but no...
 
  • #5
I'd be more worried about the field created than the noise on the lines themselves. Proper decoupling in the amplifier would fix that, but I would be surprised if that was the issue...
 
  • #6
Jeff57 said:
Thank you both, that was real help, real fast.

What about noise. My radio is full of static on am & FM when my HID lights come on and they use digital ballasts that correct for pf.

I thought a cap close to the radio's ac socket would help, but no...

If the noise is being generated by the lights, you should place the capacitors near the lights.
Once the interference has traveled down the wires, it is radiated and picked up by the antenna of the radio, so it is too late to bypass the radio's power line.

If the lights are operating at mains voltage you probably should get an electrician to install any capacitors and also to check on your power factor efforts.
 

Related to Correcting P.F. and sag, using caps, consumes power?

1. How do capacitors help correct power factor and sag?

Capacitors store and release electrical energy in order to balance out the reactive power in an electrical system. This helps to correct power factor and sag, improving the efficiency of the system.

2. What is the relationship between correcting power factor and reducing power consumption?

Correcting power factor reduces the amount of reactive power in an electrical system, which in turn reduces power consumption. This is because reactive power does not contribute to useful work and only adds to the overall power load.

3. Can correcting power factor and sag lead to cost savings?

Yes, correcting power factor and sag can lead to cost savings in the long run. By reducing power consumption, utility bills can be lowered and equipment can operate more efficiently, resulting in lower maintenance and replacement costs.

4. How do you determine the right size and placement of capacitors for correcting power factor and sag?

The size and placement of capacitors for correcting power factor and sag depends on the specific electrical system and its needs. An analysis of the system's power factor and sag levels can help determine the appropriate size and placement of capacitors.

5. Are there any potential drawbacks to using capacitors to correct power factor and sag?

While using capacitors can be an effective way to correct power factor and sag, there are some potential drawbacks. Overcorrecting power factor can lead to overvoltage and other issues, and capacitors can also introduce harmonics into the system. It is important to carefully analyze the system and consult with experts before implementing capacitor correction.

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