Electricity Power Savers

[Donnyboy]

A guy gave a power saver device to one of my mates to trial for a month. He was told it would save up to 40% on your normal power bill. After a month my mate said it reduced alittle but with power bills they arent usually constant in consumption in his home so he gave it back as it didnt make much difference. Anyone got one and have success with them?

I think the cost was about 2000B. I just cant believe that they would work.

[Tywais]

There have been a few topics on this posted before. No, they don't work, at least in a standard residential environment.

[jaapfries]

There have been a few topics on this posted before. No, they don't work, at least in a standard residential environment.

Well; I beg to differ.

A year or so ago; I was also intro'd to this phenomena and f course I alsodid not believe the claims. I did 2 things to get to the truth:

1. I discussed this unit with an eletrotechnical expert; ex-Navy Captain, who opened-up the unit and had a look inside. From his nods I surmized he understood what he was looking at and then he explained to me that this circuit was designed to "take the dips & peaks" out of the power being circulated hrough the home.

Would it safe a significant amount of power ? - His reply was "The proof isin the eating" !

2. At the time, we had just finished building our second home, but we were not living in it yet. So; I noted down the meter's reading and did nt change the behaviour ofthe power-usage og this dwelling.[At this time,we had automatic lights-on/of system; fridge, freezer etc. running in the home; but nobody living in it].

After a precise 4 week period, I plugged-in the Power-Saver unit (I paid 4,200.-) and continued monitoring.

Conclussion: After this 2 x 1 month test, I simply compared the number of 'clicks' recorded on the meter and thus was able tom estabish a decrease in elecdtricityn used of almost 23%.

Hope my little contribution helps.

Cheers,

JGK/Pattaya

[chris11]

Hello

I would be careful with this. I had seen a report about this "Energy Saver" and they not save Energy they only let run the meter slower. So in clear words if you use it you steal electric. Chris

[PaulDee]

The Ask Andy column in 'Pattaya Today' had a similar question two weeks ago from a reader. Handy Andy's answer follows:

Promoters of power savers promise they will cut your energy bills by about 25 percent each month. However, the jury seems to be out on whether such devices are effective. My research indicates strong arguments on both sides of the question. Were I in your shoes, I would insist on a money-back guaranty that the device you purchase will indeed reduce your energy consumption as promised. If you get such assurance (in writing), try it out for two or three months and compare your power bills for those months against your power bills for the previous three months AND against the same three months one year ago.

[monty]

They work in theory, with one single load on them.

They work by lowering the power factor, so the electricity meter will run slower for any amount of watt deleivered (as the electricity meter registers Volt Amp and not really watts).

Now in a household you have a bunch of loads, all either capacitive or inductive and all with a different power factor.

Very few devices are neither inductive nor capacitive (a regular incandescent light bulb is one of the few, volts and amps are in perfect sync).

These power savers are either inductive or capacitive. Both will slow down the electricity meter when the load is a regular light bulb, by bringing the volts and the amps out of sync.

If your average load in the house is capacitive (i.e. the current lags the voltage), and the power saver uses an inductive device (coil), then your electricity meter will actually speed up, as it will bring current and voltage more in sync!

And if in your situation it does slow down the meter (by having a capacitive load and adding a capacitive device or vice versa), you are indeed stealing electricity from the company by tricking the meter.

Then realize, that in an average household the load can vary between capacitive and inductive non stop. A/C's starting and stopping, fridges kicking in, lights being turned on and of, PC's being turned on and of.

So IMO, on average the device will make your meter run faster sometimes, and slower on other times, depending on which electrical devices are used in your house. Savings? Who knows. One months 6 %, next month you pay 3% more.

In general electricity companies allow an average power factor of .95, meaning that you get up to 5% of your electricity for free.

Actually, most bigger consumers (small companies) will get an electronical meter which will record the power factor, and if it is too low (meaning the meter will run too slow for the power used) you will get charged for it.

And the charge is much higher then what the actual electricity would have been, mainly because a bad power factor sharply increases the losses in the electricity lines towards your house or business, making the cost towards the electricity company very high, much higher then the little extra electricity you are cheating out of them!

The only thing they "might" help on is in keeping regular lightbulbs from popping as much by softening the spikes in the lines.

[Crossy]

They work by lowering the power factor, so the electricity meter will run slower for any amount of watt deleivered (as the electricity meter registers Volt Amp and not really watts).

Sorry Monty, incorrect. All modern domestic electricity meters (and that includes the mechanical ones used in Thailand) are true Watt - Hour meters. Power Factor has no effect upon the reading unless it's really, really poor.

Much has been written on the various technical forums about these things. As yet I have not seen any of the manufacturers provide a proper demonstration of how these things work (the excuse is of course, that others would copy their design). The majority of them are as Monty says, just capacitors that 'improve' PF.

EDIT I have a Watt / VA / PF meter. If anyone would like to loan me their power saver (or I can come to them) we can do some proper quantitative tests (meter is plug-in so we can only check plug-in appliances). Anyone in northern BBK (Pathum Thani) with one of these things? Could be an interesting afternoon's activity.

[elkangorito]

Comments in blue.

They work in theory, with one single load on them.

They work by lowering the power factor, so the electricity meter will run slower for any amount of watt deleivered (as the electricity meter registers Volt Amp and not really watts).

Incorrect.

A good P.F. is high (above 0.8).

The typical electromechanical KWH meter does not & cannot measure VA. It can only measure kW & therefore the consumer is billed for "True Power".

Now in a household you have a bunch of loads, all either capacitive or inductive and all with a different power factor.

Many household loads are resistive. Very few are capacitive. Quite a few are inductive.

Very few devices are neither inductive nor capacitive (a regular incandescent light bulb is one of the few, volts and amps are in perfect sync).

These power savers are either inductive or capacitive. Both will slow down the electricity meter when the load is a regular light bulb, by bringing the volts and the amps out of sync.

Power Factor correction is always achieved by switching capacitance in & out of the circuit.

If your average load in the house is capacitive (i.e. the current lags leads the voltage), and the power saver uses an inductive device (coil) [these things never use induction to correct anything], then your electricity meter will actually speed up, as it will bring current and voltage more in sync!

Again, this is not true. See my first comment.

And if in your situation it does slow down the meter (by having a capacitive load and adding a capacitive device or vice versa), you are indeed stealing electricity from the company by tricking the meter.

Incorrect again.

Then realize, that in an average household the load can vary between capacitive and inductive non stop. A/C's starting and stopping, fridges kicking in, lights being turned on and of, PC's being turned on and of.

So IMO, on average the device will make your meter run faster sometimes, and slower on other times, depending on which electrical devices are used in your house. Savings? Who knows. One months 6 %, next month you pay 3% more.

<sigh>

In general electricity companies allow an average power factor of .95, meaning that you get up to 5% of your electricity for free.

Where do you get this faulty information from?

Actually, most bigger consumers (small companies) will get an electronical meter which will record the power factor, and if it is too low (meaning the meter will run too slow for the power used) you will get charged for it.

And the charge is much higher then what the actual electricity would have been, mainly because a bad power factor sharply increases the losses in the electricity lines towards your house or business, making the cost towards the electricity company very high, much higher then the little extra electricity you are cheating out of them!

The only thing they "might" help on is in keeping regular lightbulbs from popping as much by softening the spikes in the lines.

For my detailed answer, do a search for previous topics OR wait til I finish work this eveing OR ask Crossy.

[InterestedObserver]

Energy, kWh as measured by the meter, for free! George Westinghouse and Thomas Edison would die laughing at that joke. I especially like the one about capacitive current lagging voltage.

[jasreeve17]

It's getting very technical all of a sudden.

My missus got one. It didn't save money, but it did blow up and cause a small fire...

[InterestedObserver]

The typical electromechanical KWH meter does not & cannot measure VA. It can only measure kW & therefore the consumer is billed for "True Power".

You seem to be saying that if the power factor is 1.0 (unity) then a kWh meter will not measure anything; ie free energy. That's like saying a regular incandescent light bulb doesn't cost anything to operate; ie free watts.

[Tywais]

The typical electromechanical KWH meter does not & cannot measure VA. It can only measure kW & therefore the consumer is billed for "True Power".

You seem to be saying that if the power factor is 1.0 (unity) then a kWh meter will not measure anything; ie free energy.

No, the PFC at 1.0 means the current and voltage are in phase as you would get if it was a totally resistive load.

All the information can be found here, particularly the graphs. PFC.

[elkangorito]

Generally speaking, an electromechanical induction disc KWH meter consists of a voltage coil & two current coils. They are positioned in such a way that when the load PF is 1.0, the fluxes produced by the coils are effectively 90 degrees out of phase with each other & thus provides a maximum turning force to the disc.

As the PF decreases, the flux angle is reduced, which in turn reduces the turning force on the disc even though the same amount of current may be flowing through the current coils.

Edited September 10, 2009 by elkangorito

[InterestedObserver]

The typical electromechanical KWH meter does not & cannot measure VA. It can only measure kW & therefore the consumer is billed for "True Power".

You seem to be saying that if the power factor is 1.0 (unity) then a kWh meter will not measure anything; ie free energy.

No, the PFC at 1.0 means the current and voltage are in phase as you would get if it was a totally resistive load.

All the information can be found here, particularly the graphs. PFC.

Yes, with a power factor of 1.0 the voltage and current are in phase and the product of volts (V) times current (A) equals power (W) [VA=W]. Therefore, the statement that the typical electromechanical kWh meter does not and cannot measure VA is not correct. A 100 watt [VA=100] incandescent (resistive) light bulb will indeed cause the kWh meter disk to rotate.

Edited September 10, 2009 by InterestedObserver

[samran]

I just take out the plug from the wall when I'm not using something, seems to working in lowering costs.

[Crossy]

Despite the fact that 99% of these energy saving devices are snake oil, there is a class of device that can and does save significant amounts of energy, sadly they're not really applicable in Thailand unless you're in the unusual position of having a supply that's regularly high rather than low voltage.

Voltage Optimisation is a particularly effective means of saving energy in the UK, because there is a national problem of over-voltage. The declared electricity supply in the United Kingdom is now, as a result of European Harmonisation in 1995, 230V with a tolerance of +10% to -10%. This means that supply voltage can theoretically be anywhere between 207V and 253V depending on local conditions. However, the average voltage supplied from the national grid (in mainland UK) is 242V, compared to the (nominal) average European voltage of 220V. Therefore, most electrical equipment manufactured for Europe and the UK is rated at 220V and may operate satisfactorily at voltages down to 200V. By efficiently bringing supply voltages to the lower end of the statutory voltage range, voltage optimisation technology could yield average energy savings of around 13%.

AC induction motors are probably the most common type of motor load and are used in a variety of equipment including refrigeration, pumps, air conditioning etc. The de-rating effects of overvoltage on AC motors are well known. Overvoltage results in saturation of the iron core, wasting energy through eddy currents and increased hysteresis losses. Drawing excessive current results in excess heat output due to copper losses. The additional stress of overvoltage on motors will decrease motor lifetime. Avoiding overvoltage does not affect the motor speed since this is a function of the supply frequency and the number of poles in the motor provided the motor is correctly loaded. Nor does it reduce the efficiency of the motor and so substantial energy savings can be made through reducing iron and copper losses. This is especially apparent if the motor application means that it experiences a variety of loading conditions since the motor efficiency is further reduced with both overvoltage and less than full loading.

Now, one of the symptoms of overvoltage on an induction motor is a poor power factor, maybe this is why the device manufacturers believe 'fixing' the PF will improve efficiency. Sadly this is not the case, as in order to increase the efficiency you must reduce the supply voltage, a side effect of which is that the PF improves, fixing the symptom does not cure the disease

[pkrv]

Despite the fact that 99% of these energy saving devices are snake oil, there is a class of device that can and does save significant amounts of energy, sadly they're not really applicable in Thailand unless you're in the unusual position of having a supply that's regularly high rather than low voltage.

Voltage Optimisation is a particularly effective means of saving energy in the UK, because there is a national problem of over-voltage. The declared electricity supply in the United Kingdom is now, as a result of European Harmonisation in 1995, 230V with a tolerance of +10% to -10%. This means that supply voltage can theoretically be anywhere between 207V and 253V depending on local conditions. However, the average voltage supplied from the national grid (in mainland UK) is 242V, compared to the (nominal) average European voltage of 220V. Therefore, most electrical equipment manufactured for Europe and the UK is rated at 220V and may operate satisfactorily at voltages down to 200V. By efficiently bringing supply voltages to the lower end of the statutory voltage range, voltage optimisation technology could yield average energy savings of around 13%.

AC induction motors are probably the most common type of motor load and are used in a variety of equipment including refrigeration, pumps, air conditioning etc. The de-rating effects of overvoltage on AC motors are well known. Overvoltage results in saturation of the iron core, wasting energy through eddy currents and increased hysteresis losses. Drawing excessive current results in excess heat output due to copper losses. The additional stress of overvoltage on motors will decrease motor lifetime. Avoiding overvoltage does not affect the motor speed since this is a function of the supply frequency and the number of poles in the motor provided the motor is correctly loaded. Nor does it reduce the efficiency of the motor and so substantial energy savings can be made through reducing iron and copper losses. This is especially apparent if the motor application means that it experiences a variety of loading conditions since the motor efficiency is further reduced with both overvoltage and less than full loading.

Now, one of the symptoms of overvoltage on an induction motor is a poor power factor, maybe this is why the device manufacturers believe 'fixing' the PF will improve efficiency. Sadly this is not the case, as in order to increase the efficiency you must reduce the supply voltage, a side effect of which is that the PF improves, fixing the symptom does not cure the disease

I have to agree with you, long ago I developed 3 phase power supplies for Marconi International Marine. The first time I turned on my prototype incorporating a star and delta wound transformer (noise from switch mode power supplies and the phases etc was critical with transmitters as well as maintaining a smooth ish voltage) I took out the entire building's electricity. My lab bench was errrrr..... heavily upgraded after that.

Fixed with a simple (massive) relay and really really big resistors across the top.

Best bet I agree with samran - take out the plug - standby on our toys does cost money.

Edited September 10, 2009 by pkrv

[elkangorito]

Yes, with a power factor of 1.0 the voltage and current are in phase and the product of volts (V) times current (A) equals power (W) [VA=W]. Therefore, the statement that the typical electromechanical kWh meter does not and cannot measure VA is not correct. A 100 watt [VA=100] incandescent (resistive) light bulb will indeed cause the kWh meter disk to rotate.

Ahem!

For A.C. circuits that consist of resistance & reactance, the product of the line voltage & current does not equal the power consumed & therefore cannot be expressed in Watts. This is the Apparent power only (S).

Your supposition may be correct in the sense that Apparent Power (S) will equal True Power (P) when the Power Factor is equal to 1.0 (in this case, Apparent Power equals 0).

This, however, does not prove or mean that an electromechanical induction disc KWH meter can measure Apparent Power. It simply cannot do it by design.

I seriously hope you don't work in the electrical industry.

[trogers]

If I am to spend US$300+ on power saving, it would be on air-con units and refrigerator with inverter technology, rather than on such a device.

[Naam]

energy saving devices do exist! centuries ago i got married and flew to Germany to buy a house. an interesting feature of one of the houses i looked at was a high-tech power saving device consisting of a steel wire a few cm in length and a hole of 0.2mm drilled in the meter. pushing the wire in would slow down the rotating disk of the meter.

[Crossy]

energy saving devices do exist! centuries ago i got married and flew to Germany to buy a house. an interesting feature of one of the houses i looked at was a high-tech power saving device consisting of a steel wire a few cm in length and a hole of 0.2mm drilled in the meter. pushing the wire in would slow down the rotating disk of the meter.

555, should I give you a warning for discussing illegal practices Naam?

I can't believe the Thais haven't sussed this technique for energy saving. That said the positioning of some of the meters would make it hazardous and visible (our meter reader at the restaurant uses a boat and binoculars).

[Gary A]

My friend had the ultimate power saving device. He had a ZERO baht bill for several months. The power company finally told him they were going to disconnect his electricity at the pole unless he bought a new meter.

He offered to sell me his old meter but I declined.

[InterestedObserver]

Ahem!

Your supposition may be correct in the sense that Apparent Power (S) will equal True Power (P) when the Power Factor is equal to 1.0 (in this case, Apparent Power equals 0).

Explain please. If I have a 100 watt light bulb (power factor 1.0) drawing both volts and amps from the mains, how is it that Apparent Power (VA) equals zero. If VA=W=0 then the light bulb would not emit radiant energy.

[hhgz]

"A guy gave a power saver device to one of my mates to trial for a month. He was told it would save up to 40% on your normal power bill. After a month my mate said it reduced alittle but with power bills they arent usually constant in consumption in his home so he gave it back as it didnt make much difference. Anyone got one and have success with them? I think the cost was about 2000B. I just cant believe that they would work. "

You have answered your own question, and don't like your answer. Simply, the devices do not work. Your friend told you it doesn't work. You don't think it will work. Most of us told you it won't work. Still, you want to know if it works. PT Barnum is quoted, "A sucker is born every minute", and it certainly seems true.

[Crossy]

Ahem!

Your supposition may be correct in the sense that Apparent Power (S) will equal True Power (P) when the Power Factor is equal to 1.0 (in this case, Apparent Power equals 0).

Explain please. If I have a 100 watt light bulb (power factor 1.0) drawing both volts and amps from the mains, how is it that Apparent Power (VA) equals zero. If VA=W=0 then the light bulb would not emit radiant energy.

It should say Reactive Power (VAR) = 0

[InterestedObserver]

Ahem!

Your supposition may be correct in the sense that Apparent Power (S) will equal True Power (P) when the Power Factor is equal to 1.0 (in this case, Apparent Power equals 0).

Explain please. If I have a 100 watt light bulb (power factor 1.0) drawing both volts and amps from the mains, how is it that Apparent Power (VA) equals zero. If VA=W=0 then the light bulb would not emit radiant energy.

It should say Reactive Power (VAR) = 0

If the poster is trying to tell us that a kWh meter will not measure reactive energy, then I agree with him. However, he consistently refers to VA instead of VAR.

Edited September 11, 2009 by InterestedObserver

[elkangorito]

Ahem!

Your supposition may be correct in the sense that Apparent Power (S) will equal True Power (P) when the Power Factor is equal to 1.0 (in this case, Apparent Power equals 0).

Explain please. If I have a 100 watt light bulb (power factor 1.0) drawing both volts and amps from the mains, how is it that Apparent Power (VA) equals zero. If VA=W=0 then the light bulb would not emit radiant energy.

It should say Reactive Power (VAR) = 0

I guess I should've said that but I thought it would be obvious that if Reactive Power is 0, then Apparent Power will not be a part of the Power Triangle...only True Power remains.

[InterestedObserver]

I guess I should've said that but I thought it would be obvious that if Reactive Power is 0, then Apparent Power will not be a part of the Power Triangle...only True Power remains.

Try it this way. Power Factor (PF) can be defined as True Power (P) divided by Apparent Power (S) [PF=P/S]. A 100 watt light bulb with a power factor of 1.0 must have 100 volt-amperes of apparent power flowing to satisfy the equation. The so-called Power Triangle will contain equal values of P and S, but S is definitely not zero or completely absent. Reactive Power (Q) will indeed be zero.

Edited September 11, 2009 by InterestedObserver

[elkangorito]

I guess I should've said that but I thought it would be obvious that if Reactive Power is 0, then Apparent Power will not be a part of the Power Triangle...only True Power remains.

Try it this way. Power Factor (PF) can be defined as True Power (P) divided by Apparent Power (S) [PF=P/S]. A 100 watt light bulb with a power factor of 1.0 must have 100 volt-amperes of apparent power flowing to satisfy the equation. The so-called Power Triangle will contain equal values of P and S, but S is definitely not zero or completely absent. Reactive Power (Q) will indeed be zero.

Let's go back a bit.

I think I misunderstood you when you said VA=W. What I think you were trying to say was that the VA value was equal to the Watt value. If this is the case, I fully agree.

With regard to the "so called" Power Triangle, if Reactive Power is not zero, Apparent Power will also not be zero. As Reactive Power increases (decrease in Power Factor), so will the Apparent Power component increase.

Only in a purely resistive component (doesn't exist), Reactive Power (& thus Apparent Power) will not exist. Since all of this is theoretical, there are 3 power components - True Power (resistive - volts x amps but not called VA...it's called Watts).

Apparent Power (volts x amps and called VA because for circuits with resistance & reactance, the product of the voltage & the current does not equal the power consumed & can't be expressed in Watts.)

Reactive Power is defined by the product of the voltage & the proportion of the current that does not consume power.

So, what is the argument? Is it that VA is the same as Watts? If so, I disagree.

[Crossy]

I think some of us are mis-remembering college / uni (we all get older and it's years since I actually used any of this stuff).

This is the 'power triangle'

Apparent Power is the power that a device appears to draw (Volts x Amps, VA), it has two components Real Power (Watts) and Reactive Power (VAR).

Both Real and Reactive Power can tend to zero, but Apparent Power will always be non-zero if current is flowing. A purely resistive circuit will have zero Reactive Power (PF=1), a purely reactive (capacitive or inductive) circuit will have zero Real Power (PF=0). Note that all circuits will have some resistive and some reactive component so the notion of 'pure' resistance or reactance is illustrative only.

The Power Factor is Real Power/Apparent Power which also happens to be the cosine of the phase angle between V and I.

The electricity meter measures the 'real' component and that is what you pay for, however the circuit losses are proportional to the square of the current (I²R) which is a component of Apparent Power. For this reason the supply companies like to supply as little Reactive Power as possible, because nobody pays for it and because things like transformers and transmission lines are sized based upon the current they carry.