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KVAR unit as energy saver
Old 07-07-2011, 11:20 PM   #1
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KVAR unit as energy saver

I recently had reason to revisit what the ‘buzz’ was behind KVAR units and my engineering side had me reading docs and understanding the concepts.
What I surmise is that some standard run capacitors locally (house side of the meter) handles power on surges that your motors (ac, refrig, pool pump) do.

Also as your motors run the brushes contacting on / off can also be “quieted” to a lower amperage draw through the meter.

Ok – so it’s a legal way to make your house loads or efficient.

An installed unit seems to start at $500 (many people pay more). Distilling the component parts as common run capacitors it seem to be sub $50 including case/surge protector… so doing the math, an installer coming over, a KVAR unit maker building a box, sales cycle…. At $500 its fair I guess.

Anyone install one and swear by it?
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Old 07-08-2011, 12:36 AM   #2
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Well, what that capacitor is doing is correcting the 'power factor', the odd inefficiency in electrical power use when the net load your house presents to the electrical grid looks like a big inductor, rather than being a purely resistive load.

The size of the capacitor needs to be matched to the inductive portion of the load your house presents. (Hint: if the installer/salesman isn't doing this, you may be being scammed...)

An inductive load like most electric motors, or older fluorescent light fixtures with magnetic ballasts will add a phase shift to the flow of power line current relative to voltage. Line voltage (in the US) oscillates at 60 times a second. With a purely resistive load, the voltage and current oscillate together, both peaking in one direction, at the same time every 60th of a second.



With a purely inductive or capacitive load, the voltage and current are 90 degrees out of phase, and no real power is consumed. This case is of concern in power distribution networks, because while that purely capacitive load uses no power, it draws real current through real wires, which have real resistance, and will consume power themselves. This reduces the remaining capacity for power carrying in the distribution network.


This is why electrical distribution networks will often have equipment in distribution substations to improve the power factor of their section of the grid.

A capacitor added at the mains entry to a home MAY improve power factor for the transmission line to the home. If you have a meter at the roadside, and a mile or two of transmission line to your home, with a consistent inductive load running in the home, one of these units might reduce your transmission line losses. It will not affect resistive losses on lines within the home to individual loads. You'll need to place capacitors, switched with each load, at the point of each load within the home.

Note that the resistive losses on the in-home wiring runs are very small.

Let's look at a washing machine. (Specifically, mine...) Out of a 41 minute cycle, the motor runs about 30 minutes, and has a power factor of 0.47.
  • 30 minutes
  • Power Factor: 0.47
  • Current: 8.94A
  • Real power: 513W
  • Apparent power: 1081VA

Impendence phase angle is arcos(True power / apparent power) = 62 degrees
Reactive power = Real power * atan(phase angle) = 423 VAR


If the power factor for the motor was 1.0, we could have done this wash with a current of 4.66A, instead of 8.94. That is, the power factor added a 'phantom' current of (8.94 - 4.66) or 4.27A. That extra current produced real resistive losses in the wiring between the washer and panel.


Lets say we have 40 feet, wired in a 15 amp circuit (code sez #14 copper wire). #14 has 2.6 ohms per 1000 feet, so our 40 foot run is 0.104 ohms. Toss in the breaker contacts, screw connections, and whatnot, and call it 0.11 ohms. Let's see how much power our phantom current costs us.



Power = (I^2)R
Power = (4.27A * 4.27A) * 0.11 ohms

Power = 2 Watts

We ran for 30 minutes, or 1/2 hour.

Energy = (2 Watts/(1000 Watts/Kilowatt)) * 1/2 hour
Energy = 0.001 KWh

Now we need to know how much your juice costs. Let's say you're in California, sucking maximum juice from PG&E, so you're in a really high bracket for that last watt, 29.385 cents per KWh.

Cost = KWh * (pennies per KWh)
Cost = 0.001 KWh * 29.385 pennies per KWh
Cost = 0.029385 penny.

You're going to have to do quite a few loads of laundry to pay off that capacitor you installed at the motor.

And once again, no, the whole house 'KVAR unit' won't have any effect on this.

The National Institute of Standards and Technology has some good information on this:

NIST Team Demystifies Utility of Power Factor Correction Devices

There's a more detailed technical report (PDF) at NIST Technical Note 1654, Regarding Electric Energy Savings, Power Factors, and Carbon Footprints: A Primer.
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Old 07-08-2011, 09:14 AM   #3
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Great explanation.

From way back I recall the caution never to correct an inductive load load to a PF of 1. .95 to .97 seemed to be the preferred range. Thus further reducing payback time. Can't remember the the details, it had to with resonance and harmonics. IIRC 3d harmonic was the major issue.

OTOH if in your junk box there happens to be the right size AC rated 900 to 1500 volt capacitor and want to hang it on on the motor side of the power switch, by all means go ahead.
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Old 07-08-2011, 09:24 AM   #4
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Thanks! - as i understand it the sale is mostly a one size fits all - "here's a unit size we have" and there is no analysis of your real home load.

Im compelled to know more about this because of an indirect customer project, but as an EE its a curious puzzle. I have an old oscilloscope in the garage which im sure can view the voltage waveform but im unsure at the moment how to peek at the current (a load resistor in line and scope across it?).
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Old 07-08-2011, 06:36 PM   #5
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Quote:
Originally Posted by nphx
Thanks! - as i understand it the sale is mostly a one size fits all - "here's a unit size we have" and there is no analysis of your real home load.

Im compelled to know more about this because of an indirect customer project, but as an EE its a curious puzzle. I have an old oscilloscope in the garage which im sure can view the voltage waveform but im unsure at the moment how to peek at the current (a load resistor in line and scope across it?).
Offhand, assuming the 'scope has fully isolated inputs (no common ground) I'd put line voltage on one input and the drop across that resistor on the other input. Set the gain so the vertical and horizontal deflections are the same. In phase, with a power factor of one, should make a clean diagonal line. Lower power factors should open up an ellipse, and a zero power factor should make a donut.

Mmmmm.. Donuts...

Caution: I'm away from my reference books, and so may be an unreliable narrator.
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Old 07-08-2011, 07:02 PM   #6
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I've never heard of these things, but here are some articles that say it's a scam:

NIST Team Demystifies Utility of Power Factor Correction Devices

KVAR Power Factor Correction in the Home is a Scam
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Old 07-08-2011, 07:27 PM   #7
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Using a "Kill A Watt", you can measure the Watt and VA.

Edit. Cross posted with TromboneAI which showed the use of the "Kill A Watt".
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Old 07-08-2011, 08:36 PM   #8
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Quote:
Originally Posted by KingB View Post
Using a "Kill A Watt", you can measure the Watt and VA.

Edit. Cross posted with TromboneAI which showed the use of the "Kill A Watt".

My "Kill A Watt" also displays Power Factor at the push of a button.
Can only use it on one leg of a single phase residential feed. Will not do 220 loads.
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Old 07-08-2011, 11:40 PM   #9
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Thank u paquette and al, interesting reading.
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Old 07-08-2011, 11:49 PM   #10
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even if you consider losses in your house, this device will not correct power factor from this device to washing machine but on utility supply side. So unless you have a mile between electric meter and your house.

Quote:
Originally Posted by M Paquette View Post
A capacitor added at the mains entry to a home MAY improve power factor for the transmission line to the home. If you have a meter at the roadside, and a mile or two of transmission line to your home, with a consistent inductive load running in the home, one of these units might reduce your transmission line losses. It will not affect resistive losses on lines within the home to individual loads. You'll need to place capacitors, switched with each load, at the point of each load within the home.

Note that the resistive losses on the in-home wiring runs are very small.

Let's look at a washing machine. (Specifically, mine...) Out of a 41 minute cycle, the motor runs about 30 minutes, and has a power factor of 0.47.
  • 30 minutes
  • Power Factor: 0.47
  • Current: 8.94A
  • Real power: 513W
  • Apparent power: 1081VA

Impendence phase angle is arcos(True power / apparent power) = 62 degrees
Reactive power = Real power * atan(phase angle) = 423 VAR


If the power factor for the motor was 1.0, we could have done this wash with a current of 4.66A, instead of 8.94. That is, the power factor added a 'phantom' current of (8.94 - 4.66) or 4.27A. That extra current produced real resistive losses in the wiring between the washer and panel.


Lets say we have 40 feet, wired in a 15 amp circuit (code sez #14 copper wire). #14 has 2.6 ohms per 1000 feet, so our 40 foot run is 0.104 ohms. Toss in the breaker contacts, screw connections, and whatnot, and call it 0.11 ohms. Let's see how much power our phantom current costs us.



Power = (I^2)R
Power = (4.27A * 4.27A) * 0.11 ohms

Power = 2 Watts

We ran for 30 minutes, or 1/2 hour.

Energy = (2 Watts/(1000 Watts/Kilowatt)) * 1/2 hour
Energy = 0.001 KWh

Now we need to know how much your juice costs. Let's say you're in California, sucking maximum juice from PG&E, so you're in a really high bracket for that last watt, 29.385 cents per KWh.

Cost = KWh * (pennies per KWh)
Cost = 0.001 KWh * 29.385 pennies per KWh
Cost = 0.029385 penny.

You're going to have to do quite a few loads of laundry to pay off that capacitor you installed at the motor.

And once again, no, the whole house 'KVAR unit' won't have any effect on this.

The National Institute of Standards and Technology has some good information on this:

NIST Team Demystifies Utility of Power Factor Correction Devices

There's a more detailed technical report (PDF) at NIST Technical Note 1654, Regarding Electric Energy Savings, Power Factors, and Carbon Footprints: A Primer.
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Old 07-10-2011, 01:13 AM   #11
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Quote:
Originally Posted by M Paquette View Post
The size of the capacitor needs to be matched to the inductive portion of the load your house presents. (Hint: if the installer/salesman isn't doing this, you may be being scammed...)
A couple days ago I dug up my old nuke power school EE notes and started going through the math, got distracted, and didn't come back to it until this afternoon. Then I thought "Let's check the board", and sure enough you guys saved me a lot more labor. Thanks!

And what they said.

In general, home improvements that very few people have ever purchased before are usually too good to be true.
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