Kvar energy saving device
- Thread starter mark480
- Start date
- Status
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
It's voodoo. Residential customers are generally not charged an extra fee for having a lower than unity power factor. Therefore, if you improve their power factor, you do not save them money. This topic has been discussed several times on this forum. I don't have the time right now to look up the earlier discussions, but if you do a word search, you might be able to find them.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080604-2153 EST
mark480:
The following quote is from
http://peswiki.com/index.php/Directory:pacifickvar#How_it_Works
My opinion without having a unit with which to experiment is that this statement is basically hogwash from the standpoint of the home owner.
Your home type watt-hour meter measures real power and reactive power is averaged to 0. Somewhere I discussed that sin t*sin t consists of DC and double frequency components. Averaging this you get the DC component. While sin t*cos t produces only a double frequency component and that averages to 0.
This Kvar device may be simply a capacitor to compensate your inductive loads. I do not see that they really define in detail how it works.
Changing power factor at the input to your home should have virtually no effect on your watt-hour meter reading. However, it is of benefit to the power company. But most homes are largely a resistive load. Inductive loads in most homes are motors and these are on-off devices with fairly short periods.
When I get time I will run some experiments on a different type of device than the Kvar unit. This other device should reduce wasted power in motors and thus run the motor cooler. This is basically a device that reduces the volt-time integral and thus reduces the peak magnetizing current. It is a type of voltage regulator.
My line voltage is typically 123 to 125 V. If this is effectively reduced to 110 or 105, then my refrigerator motor may actually use less power than at the higher voltage. The mechanical load on a refrigerator compressor should not change much with changing input voltage because motor speed is approximately constant relative to input voltage. However, the in phase to voltage input current will increase as the voltage drops. But if total input RMS current drops fast enough with dropping voltage, then because of the reduced magnetizing current there will be less motor loss.
.
mark480:
The following quote is from
http://peswiki.com/index.php/Directory:pacifickvar#How_it_Works
My opinion without having a unit with which to experiment is that this statement is basically hogwash from the standpoint of the home owner.
Your home type watt-hour meter measures real power and reactive power is averaged to 0. Somewhere I discussed that sin t*sin t consists of DC and double frequency components. Averaging this you get the DC component. While sin t*cos t produces only a double frequency component and that averages to 0.
This Kvar device may be simply a capacitor to compensate your inductive loads. I do not see that they really define in detail how it works.
Changing power factor at the input to your home should have virtually no effect on your watt-hour meter reading. However, it is of benefit to the power company. But most homes are largely a resistive load. Inductive loads in most homes are motors and these are on-off devices with fairly short periods.
When I get time I will run some experiments on a different type of device than the Kvar unit. This other device should reduce wasted power in motors and thus run the motor cooler. This is basically a device that reduces the volt-time integral and thus reduces the peak magnetizing current. It is a type of voltage regulator.
My line voltage is typically 123 to 125 V. If this is effectively reduced to 110 or 105, then my refrigerator motor may actually use less power than at the higher voltage. The mechanical load on a refrigerator compressor should not change much with changing input voltage because motor speed is approximately constant relative to input voltage. However, the in phase to voltage input current will increase as the voltage drops. But if total input RMS current drops fast enough with dropping voltage, then because of the reduced magnetizing current there will be less motor loss.
.
Last edited:
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080609-0946 EST
I have started some experiments on a device called GreenPlug. Very little information at this point.
My experiments yesterday were on a freezer. The freezer was warm meaning the compressor would run continuously during my experiment. My assumption is that under these conditions the mechanical load from the compressor will remain constant. May or may not be valid. The initial assumption is that at a fixed load the mechanical power will be constant. But in reality the slip for a given load will increase for a lower input voltage. I have no easy way to measure the motor RPM.
Since the GreenPlug is supposed to modulate the voltage to the load I need to know how this compressor responds to voltage variations.
I allowed the compressor to stablize for about 1/2 hour then made the following measurements. The instruments were --- Fluke RMS meter with Hall device current probe, Simpson 75 watt full scale wattmeter with a current transformer, a Fluke average reading meter for voltage, and a 7.5 A Powerstat for an adjustable voltage source.
The results were as follows:
If in fact the mechanical output power was constant, then the efficiency improved with lower voltage. I do not know if this is true or not but I doubt it.
Assuming this constant output power is correct, then maximum saving from running at 100 V instead of 130 is only about 30 W. Refrigeration systems do not normally ruin continuously.
As pointed out by somone in one of the other threads on this subject turning lights off will save vastly more energy. I can not find that thread right now.
With the above data I was ready to run the test with the GreenPlug device. Problem --- with that device in line this compressor could not get started. It tripped its thermal overload.
Later I will try the test again on a different motor.
.
I have started some experiments on a device called GreenPlug. Very little information at this point.
My experiments yesterday were on a freezer. The freezer was warm meaning the compressor would run continuously during my experiment. My assumption is that under these conditions the mechanical load from the compressor will remain constant. May or may not be valid. The initial assumption is that at a fixed load the mechanical power will be constant. But in reality the slip for a given load will increase for a lower input voltage. I have no easy way to measure the motor RPM.
Since the GreenPlug is supposed to modulate the voltage to the load I need to know how this compressor responds to voltage variations.
I allowed the compressor to stablize for about 1/2 hour then made the following measurements. The instruments were --- Fluke RMS meter with Hall device current probe, Simpson 75 watt full scale wattmeter with a current transformer, a Fluke average reading meter for voltage, and a 7.5 A Powerstat for an adjustable voltage source.
The results were as follows:
Code:
080608-1200 Power, Voltage, and Current measurements to a freezer compressor.
Voltage Current Power VA Power Factor
Volts Amperes Watts
100.1 3.74 319 374 0.853
100.2 3.68 319 368 0.867
109.9 3.62 323 397 0.814
120.1 3.55 332 426 0.779
120.2 3.56 332 427 0.778
130.3 3.8 349 495 0.705
130.3 3.7 344 482 0.714
These were not run in this order.If in fact the mechanical output power was constant, then the efficiency improved with lower voltage. I do not know if this is true or not but I doubt it.
Assuming this constant output power is correct, then maximum saving from running at 100 V instead of 130 is only about 30 W. Refrigeration systems do not normally ruin continuously.
As pointed out by somone in one of the other threads on this subject turning lights off will save vastly more energy. I can not find that thread right now.
With the above data I was ready to run the test with the GreenPlug device. Problem --- with that device in line this compressor could not get started. It tripped its thermal overload.
Later I will try the test again on a different motor.
.
barbeer
Senior Member
- Location
- Pinellas County, FL
Our local IAEI had a rep from KVAR at a recent meeting, and a rep from the POCO...........it was kinda funny! You could tell the salesman was just that as he could not explain how it all worked, we concluded that it was a capacitor set-up. He did however offer to set it up on someone's home for a free test? I'll check into the results.
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