A horsepower is 550 foot-pounds of work per second or 33,000 per minute. A watt is one joule per second or one newton-meter per second. A horsepower is about 0.746 kilowatt. Torque is something completely different - it is a measure of twisting action and can be measured in pounds-feet or newton-meters. But- torque acting through an angle can be expressed as work. For example a newton-meter of torque acting around a revolution requires 6.28 (two pi radians) joules of work. A horsepower-hour is 2545 Btus of heat; a kilowatt hour is 3415 Btus. A Btu (British thermal unit) is the amount of heat required to heat a pound of water one degree Fahrenheit; a kilocalorie (which is what dietitians call a calorie) is the amount of heat required to heat a kilogram of water one degree Celsius and is equal to about 3.968 Btus. Above is from memory and Machinerys Handbook 18 Edition, 1969, by Oberg and Jones. I suppose the heat generated by the processor is the same heat generated by current flowing through a resistor. In motors, hysteresis is one thing that causes heat. If the voltage and current are out-of-phase the power factor is less than one and the volt-amps is greater than the watts.

---

Agreed, and that's what I took exception to. In physics, power is a measurement of work done or energy transfered per unit time. However, the original arguement implied that heat was the only form energy could take and that's just plain incorrect. Since any form of energy can be represented as an equivalent value of another form, this was an inappropriate "bending" of the semantics of heat.

There was one other comment which bothered me. This was the electrical examples offered.

First off, you are *not* billed for electricity based on current alone. If this was the case you would be charged on a per coulomb (electrical charge) basis, since amperes are defined as coulombs per second. You are billed in KWH precisely for the reason that it *is* a measurement of the "effort" or work it took the utility to generate the electricity you used. The effects of reactive power from the source's POV cannot be overlooked, which is why watthour meters typically have a correction factor builtin to compensate for this effect depending on their intended usage.

Second, the analogy of the A/C unit is completely ridiculous. For an ideal machine of a rated power, there is absolutely no difference from an efficiency POV to whether it runs at 1 volt, 115 volts, 230 volts, or even 1,000,000 volts.

The impact to a *real* machine is that as the rated power increases you gain in efficiency with a higher operating voltage due to a reduction in the I squared R losses in the machine, which in this case does manifest itself as heat, and waste heat at that.

Alinator

.......

James Watt, the Scottish inventor who defined the unit of heat in relation to the production of kinetic power

James Watt was British. He was was born in Scotland; a part of Great Britain. He lived and worked in Scotland and England i.e. Great Britain.

---

Speaking of cost...

The total work units that were invalid due whatever error (throughout my crunching history starting last year) and gotten zero credits, but time was spent nonetheless; the total time spent for invalid work units with errors is 570,519.47 seconds. That's for 165 WU.

At 21 cents (USD) per hour (I'm ignoring the 30 cents per hour on my latest bill), the total cost for errored out work units is $6.66. It gives ya the creeps, eh? LOL


Them bloody errors... costing
$6.66

The errors is ok, but the exact cost bothers me. AHHH!!! I'm just hoping the 21 cents per hour average is not correct. :( (It is an estimated average.)

I wasn't talking abt those little trickle charger panels...you can get 1 amp panels for not much more.

1 amp at 12 volts is not all that much -- about 12 watts (or if you prefer, 12 volt-amps). It'd take 5 of those to run a typical incandescent light-bulb.

As an aside, I just replaced the factory batteries in my UPS with something a little larger -- 75 amp-hours at 48 volts gives my server farm 6 hours on batteries. There are six machines, some of them slightly power hungry.

So, 300 amp/hours at 48v to cover one day. If you figure five hours of solid output from solar panels, you'd need panels good for 60 amps (at 48v), so take these and that looks like 40 panels, roughly 20" by 40"

Or, about 80 of those 1 watt panels to take one machine off grid.

---

Ned
I think your maths is about right, that was why I suggested if anybody is serious about using alternative power sources such as solar panels or small wind turbines I would suggest running the computer directly from DC so that the 'DC to AC' in inverter and AC to DC at i/p stage of computer power supply are eliminated along with all there losses, and potential faults.
I think you could then reduce your power requirements to about 60%.
I would also consider only using mobile cpu's and mobo's built with mobile chipsets, like pent M and Aopen i915Mm - HFS, which would use less than 60 VA to crunch, about the same as a light bulb.

Andy

.......
James Watt, the Scottish inventor who defined the unit of heat in relation to the production of kinetic power

James Watt was British. He was was born in Scotland; a part of Great Britain. He lived and worked in Scotland and England i.e. Great Britain.

You must be a Pom. I know a few Scotts & they prefer not to acknowledge the greater Britain.
:-)

---

Grant
Darwin NT

I wouldn't know how to do that WinterKnight, If and when I do have solar panels for the house($20,500 was the estimate), They would be making power and converting It into AC and feeding It to the grid so that I would have no bill or a very small one at the very least. Oh and there would be no batteries either.

You would use the DC output of the solar panels or wind geneartor to charge batteries, and connect you computer modified with a 12 or 24 volt DC input Power supply (DC/DC convertor), to the batteries. A google for '24v dc input ATX psu' will find companies that make/sell suitable power supplies, there are also companies that make computers with DC inputs usually ~18V, but the input spec is usually quite wide like 15 to 30 Volts.

typical conections

solar panel______batteries_________inverter___________computer psu
24V DC . . . . . . . . . . . . . . . . . 24V dc to 110V AC . . . . . . (normal)

or

Solar Panel______batteries__________24V i/p Computer psu

the efficiency of the inverter is about 70% and there are losses, 5 to 10 watts, at the input of the normal computer supply to convert to the AC input to DC.

So, first with inverter, for computer requiring 210 watts out of the psu, as average psu is 70% efficient and inverter is 70% efficient and batteries require 14 hour charge for quoted 10 hour output rating you solar panel would require a daily output of over 600 watts. Assuming 6 hrs/day with good output that would be rating of 2400 watts or 100amps at 24 volts. batteries would have to be of 1000 amperehour min. As charging period is only 6 hours the solar panels would be reuired to give about 150 amps output.

With 24 volt i/p psu, no inverter losses, solar panel output would go direct to psu for 6 hours/day, but would also have to charge batteries during this period. Power supply would require ~12 amps and about 70 amps would be required to charge the batteries.

Therefore the solar panel output could be cut to about 100 watts or 2/3 of inverter system and battery size could be cut by similar amount. But you would require a large battery charger for rainy days.

Thats my thoughts on the systems.

Andy

Hmmm, strictly speaking Watts are a unit of power. Heat is one way the power manifests itself in the physical world.

If all the energy was dissipated as heat, then incandescent lamps would give off no visible light, cars would not move when the engine was running, RF transmitters would not send a signal for the receiver to pick up, etc. IOW, some of the energy must go to the work the device was intended to perform.

Alinator

Incorrect. I have already stated the fact that watt is NOT being correctly used.

Yourstatement implies that because it's always been done that way, it is correct. That's argumentum ad antiquitutem. It's like saying because "slavery was done for hundreds of years, it is(was) right."

One could also demonstrably state your comments were dicto simpliciter.

What you see as light from your lights is rated in candelas, the amount of light given off.

RF transmitters are inductors, inductors do NOT generate heat. The Em the emit is rated in volt-amps (va) or kilo volt amps or higher.

You can easily verify that James Watt, whom created the unit "watt" never, ever had anything to do with electricity at all.

---

A horsepower is 550 foot-pounds of work per second or 33,000 per minute. A watt is one joule per second or one newton-meter per second. A horsepower is about 0.746 kilowatt.

Torque is something completely different - it is a measure of twisting action and can be measured in pounds-feet or newton-meters. But- torque acting through an angle can be expressed as work. For example a newton-meter of torque acting around a revolution requires 6.28 (two pi radians) joules of work. A horsepower-hour is 2545 Btus of heat; a kilowatt hour is 3415 Btus. A Btu (British thermal unit) is the amount of heat required to heat a pound of water one degree Fahrenheit; a kilocalorie (which is what dietitians call a calorie) is the amount of heat required to heat a kilogram of water one degree Celsius and is equal to about 3.968 Btus. Above is from memory and Machinerys Handbook 18 Edition, 1969, by Oberg and Jones.

Mostly correct, but their generalities are dicto simpliciter Your torque generated by a motor is different that the force generated by James Watt's steam engine which he rated in horsepower and how many units of heat were required to boil "x" amount of water in order to generate a linear force that would then be converted into a circular motion. It was the amount of force created along the drive-link shaft which he rated for linear force.

Where issues come into play is likesaying claories are equal to temperature. They're not. You can compare one another, but they simply aren't the same thing.

[qioute I suppose the heat generated by the processor is the same heat generated by current flowing through a resistor. In motors, hysteresis is one thing that causes heat.

You do have resistors on that chip you know. And, frequency increases the amount of heat, just as harmonics increase the amount of heat on a neutral.

If the voltage and current are out-of-phase the power factor is less than one and the volt-amps is greater than the watts.

Incorrect. When voltage and current are out of phase, you lose more to heat generated in the cinductor

---

Agreed, and that's what I took exception to. In physics, power is a measurement of work done or energy transfered per unit time. However, the original arguement implied that heat was the only form energy could take and that's just plain incorrect. Since any form of energy can be represented as an equivalent value of another form, this was an inappropriate "bending" of the semantics of heat.

There was one other comment which bothered me. This was the electrical examples offered.

First off, you are *not* billed for electricity based on current alone. If this was the case you would be charged on a per coulomb (electrical charge) basis, since amperes are defined as coulombs per second. You are billed in KWH precisely for the reason that it *is* a measurement of the "effort" or work it took the utility to generate the electricity you used. The effects of reactive power from the source's POV cannot be overlooked, which is why watthour meters typically have a correction factor builtin to compensate for this effect depending on their intended usage.

Second, the analogy of the A/C unit is completely ridiculous. For an ideal machine of a rated power, there is absolutely no difference from an efficiency POV to whether it runs at 1 volt, 115 volts, 230 volts, or even 1,000,000 volts.

The impact to a *real* machine is that as the rated power increases you gain in efficiency with a higher operating voltage due to a reduction in the I squared R losses in the machine, which in this case does manifest itself as heat, and waste heat at that.

Alinator

Again you are absolutely incorrect and I take exception to that. As I mentioned where you get charged for your voltage is when your home's pf is out of whack. Else why would a 240 volt air conditioner (motor load) be less to run that a 120 one? ASk your local power company. I run all of my lights (flourescents) on 240...and when I switched there..I saw another drop in my bill. Now if you were getting charged for voltage, you bill should remain the same, not drop.

Current is what you're really charged for.

And, like it or not wattage is an incorrectly used comparison term. Ask an EE. Oh wait, one IS telling you.

---

[quote]Agreed, and that's what I took exception to. In physics, power is a measurement of work done or energy transfered per unit time. However, the original arguement implied that heat was the only form energy could take and that's just plain incorrect. Since any form of energy can be represented as an equivalent value of another form, this was an inappropriate "bending" of the semantics of heat.

There was one other comment which bothered me. This was the electrical examples offered.

First off, you are *not* billed for electricity based on current alone. If this was the case you would be charged on a per coulomb (electrical charge) basis, since amperes are defined as coulombs per second. You are billed in KWH precisely for the reason that it *is* a measurement of the "effort" or work it took the utility to generate the electricity you used. The effects of reactive power from the source's POV cannot be overlooked, which is why watthour meters typically have a correction factor builtin to compensate for this effect depending on their intended usage.

Believe what you want, reality is different. You might try actually confirming oor denying your statements before making them.

No your meters do not. One of my summer jobs in the electrical field was calibrating replaced units. There isn't any factor built into them. I won't try confusing you with (verifiable) facts because you already have your mind made up.

One of your arguments is trying to apply physics jargon to electrical. What something means, a word, to a physicist, is often different from the electrical definition. To an electrician or an EE power and current are synonymous. To those that actually know what they're doing/talking about, watts means a loss.

That's incorrect: How many of the public would know how they're being billed, when a dicto simpliciter form of comparison exists? It doesn't make it correct but does make it srgumentum ad populum or numerum.

Second, the analogy of the A/C unit is completely ridiculous. For an ideal machine of a rated power, there is absolutely no difference from an efficiency POV to whether it runs at 1 volt, 115 volts, 230 volts, or even 1,000,000 volts.

Nice ad hominem, which still doesn't make you correct. In fact if you actually cared to do a little unbiased research, you'd learn you're wrong. Ever actually try pf correcting things and operating two identical (but diff motor voltage loads) items to find out? No I think not.

The impact to a *real* machine is that as the rated power increases you gain in efficiency with a higher operating voltage due to a reduction in the I squared R losses in the machine, which in this case does manifest itself as heat, and waste heat at that.

Alinator

You are correct only in the I^2R losses exist inb the conductors within the electrical device, partially due to amount of current flowing through the conductors inherant resistance/ft. That loss is measured in watts...heat dissapated.

Again you are absolutely incorrect and I take exception to that. As I mentioned where you get charged for your voltage is when your home's pf is out of whack. Else why would a 240 volt air conditioner (motor load) be less to run that a 120 one? ASk your local power company. I run all of my lights (flourescents) on 240...and when I switched there..I saw another drop in my bill. Now if you were getting charged for voltage, you bill should remain the same, not drop.

Current is what you're really charged for.

And, like it or not wattage is an incorrectly used comparison term. Ask an EE. Oh wait, one IS telling you.

---

... Current is what you're really charged for.

And, like it or not wattage is an incorrectly used comparison term. Ask an EE. Oh wait, one IS telling you.

Well, here in the UK our electricity suppliers measure and (bill for) domestic electricity usage using meters called "Watt-Hour" meters.

Note that (by Ohm's law) Watts = Volts * Amps

Look up how the meters work and what they measure. Then please tell us.

Aside, the way that the Vrms and Irms are measured can give an over-estimate if your power-factor is away from being balanced. So if you have a very reactive load, then you'll get billed for more than the energy you're actually using.


Don't know what you mean by "wattage is an incorrectly used comparison term". I guess Marketing people can distort science and physics any way they choose and reality is only an irrelevance. ;-(

"Watts" is a measure of rate of power conversion. In what way whatever energy is being converted does not matter. Steam power and electrical power can have the same Watts!


Happy crunchin',
Martin

[edit] Mmmm, and sorry about the many highly charged multiple puns in there! :-) [/edit]

---

See new freedom: Mageia Linux
Take a look for yourself: Linux Format
The Future is what We all make IT (GPLv3)

... Current is what you're really charged for.

And, like it or not wattage is an incorrectly used comparison term. Ask an EE. Oh wait, one IS telling you.

Well, here in the UK our electricity suppliers measure and (bill for) domestic electricity usage using meters called "Watt-Hour" meters.

Note that (by Ohm's law) Watts = Volts * Amps

Look up how the meters work and what they measure. Then please tell us.

Aside, the way that the Vrms and Irms are measured can give an over-estimate if your power-factor is away from being balanced. So if you have a very reactive load, then you'll get billed for more than the energy you're actually using.


Don't know what you mean by "wattage is an incorrectly used comparison term". I guess Marketing people can distort science and physics any way they choose and reality is only an irrelevance. ;-(

"Watts" is a measure of rate of power conversion. In what way whatever energy is being converted does not matter. Steam power and electrical power can have the same Watts!


Happy crunchin',
Martin

[edit] Mmmm, and sorry about the many highly charged multiple puns in there! :-) [/edit]

Do you have a big inductive load at your home? Most electricians do a pretty good job of having a balanced load for a home. Unless you have lots of Electronics and flourescent lights(More than a few computers) I really doubt balancing your load will do anything at all. You are talking about Harmonics and most houses will not have that problem.

---

Official Abuser of Boinc Buttons...
And no good credit hound!

Do you have a big inductive load at your home? Most electricians do a pretty good job of having a balanced load for a home. Unless you have lots of Electronics and flourescent lights(More than a few computers) I really doubt balancing your load will do anything at all. You are talking about Harmonics and most houses will not have that problem.

Nothing too fancy on the wiring here! :-)

Power-factor is at 0.90 at the moment. I've seen it as bad as 0.64. But then, my power usage isn't expensive enough to go chasing for that to get fixed. I suspect that the phase that my house is wired onto is getting distorted by a business unit nearby on the same substation phase. Or someone is doing Frankenstein experiments!

Most lights here are the 'energy saver' types, but there are not enough on to notice. The PC PSU is a pFC type... So... Who knows?


Or it mighht just be some nutter with a farm of freezers and aircon trying to counter Global Warming. It's hot here still!

Cheers,
Martin

---

See new freedom: Mageia Linux
Take a look for yourself: Linux Format
The Future is what We all make IT (GPLv3)

Those of us who crunch in California may want to be aware of new rates starting in August....

Baseline power covers basic needs -- it's 13 cents/kwh here (through Southern California Edison).

The new Tier 5 is 49 cents/kwh.

New rates start in August.

---

<snip>

And, like it or not wattage is an incorrectly used comparison term. Ask an EE. Oh wait, one IS telling you.

Ok, let's take some of your arguments one at a time:

1.) Your statement; "You can easily verify that James Watt, whom created the unit "watt" never, ever had anything to do with electricity at all."

Not quite. The fact is the unit of power representing 1 joule per second of energy transfer was named in honor of James Watt in 1889, seventy years or so after his death, and again in 1960. The only accurate part here is Watt apparently didn't have much interest in electricity, or at least in it's practical uses. *

He did however define the unit of power known as "Mechanical Horsepower" in 1782, and the motivation was so he could better market his improved steam engines to a broader spectrum of potential users. **

Strictly speaking in all electrical engineering terms, a joule is defined as the work required to move 1 coulomb of charge through a potential difference of 1 volt (gets rid of those nasty newton meters). Therefore 1 watt(electrical) can be defined as 1 coulomb volt per second.

My original beef resulted from this statement; "Wattage is merely a unit of HEAT disappation." While technically correct it is inaccurate since you CHOSE to obfuscate the sentence by emphasizing the word "heat". As written, most people would read this as saying in effect that power is the same thing as energy, which is false by definition. This is what I meant by bending the broad meaning of the word "heat".

A more correct and accurate syntax would be; "The wattage rating of many electrical devices is merely a measure of their heat dissipation capability.", a subtle but crucial difference, but of course you know that already.

My statement: "If all the energy was dissipated as heat, then incandescent lamps would give off no visible light..."

Your reply: "What you see as light from your lights is rated in candelas, the amount of light given off."

OK, then refute this; For an incandescent electric lamp:

E(electrical) = E(visible light) + E(heat)

I assume you will allow me to "lump" all other forms of energy except electricity and visible light as "heat" since that's your basic premise, or are you saying that visible light is not a form of energy?

I leave it to other readers to decide where the dicto simpliciter lies.

* http://en.wikipedia.org/wiki/Watt, etal
** http://en.wikipedia.org/wiki/Horsepower#History_of_the_term_.22horsepower.22, etal

2.) Your statement; "That's why a 240 volt ac unit is cheaper to operate than a 120."

The correct answer is; It depends...

We have to make an assumption here first, which is you were referring to a room air conditioner (RAC) in this statement. A reasonable assumption since I couldn't find an outdoor condensing unit (ie. Central A/C) which was designed to run on 120 volts, not that that means they don't exist.

Next, let's imagine one has an area which needs 12000 BTU/hr of cooling capacity to air condition. From a cost of operation viewpoint *of the machine only* which would be the best unit to purchase?

Given that, please refer to the following spec sheet:

http://www.fedders.com/catalog/appliances/roomac/fed_specs.pdf

Let's further say that due to installation concerns we have decided the choice boils down to one of the following 3 units:

A6Y12F2B 115 VAC, 11.0 A/cooling, 1230 watts/cooling, 9.8 EER
A7Y12F2B 115 VAC, 10.2 A/cooling, 1110 watts/cooling, 10.8 EER
A6Y12F7B 230 VAC, 5.6 A/cooling, 1230 watts/cooling, 9.8 EER

Well color me dumb possibly, but I'm going to go against your recommendation and buy the 115 volt A7Y12F2B. BTW, please note that A6Y12F2B outperforms the A6Y12F7B on a VA basis as well.

I specifically picked Fedders since they're one of the few manufacturers who provides easy online access to "full" nameplate specs for RAC's, but a quick survey of a number of other manufacturers who offer machines of the same cooling capacity in 115 and 230 volt models and using EER as the metric shows that 115 units can have the same or better energy performance than 230 volt models at the test conditions where EER is measured.

I'll leave the reasons why for the above phenomena up to the reader to figure out as a thought and/or research exercise. (Hint: They aren't rocket science, although some are pretty clever most are nothing "new" particularly.) :-)

I will grant you there is one aspect where your statement holds true, and that is 230 volt units will always have an advantage due to the resistive loss in the conductors feeding it. In some cases, this could be a significant problem due to heating of the wiring, especially in older buildings (we're talking 60-70 years and older).

So we will now expand the parameters to include infrastructure concerns as well. It just so happens I have an A7Y12F2B, so I decided to run a test to determine the infrastructure losses. I measured the no load voltage at the outlet with a true rms voltmeter and again when the unit was running with the fan at high speed and the compressor running. The indoor/outdoor temperatures were about 78/92 degrees respectively, reasonably close to the EER test conditions. Under these conditions I found there was a 3.0 volt drop with the machine in operation. For the sake of argument, let's assume the current drawn by the machine is the worst case value stated on the nameplate of 10.2 A. We can now calculate the dissipation in the line using P=VI, so in this case 30.6 W.

As a reality check, I then estimated the actual run length of the 14/2 AWG branch circuit to determine the expected series resistance of the cable using standard AWG tables. The reactive components of this transmission line are negligible and may be ignored.

Estimated Run Length: 50 feet load center to outlet (100 ft. equivalent conductor length)
Expected Resistance from Tables: 0.2575 ohms
Calculated Value: 0.2941 ohms

The results are close enough to be reasonably sure we're in the right ballpark, being off most likely due to overestimating the length of the run, limits of precision for the voltage measurement, and/or the initial assumption of the load current. Sorry, but I didn't have my current clamp handy and I learned to *never* go near the shop when on vacation a long time ago.

Now let's imagine we convert that run to 230 volts and perform the calculations again using the full load amp value for the A6Y12F7B and the calculated line resistance. The loss would now become 9.2 W, so as you said one could save 21.4 W in operational load with the 230 volt machine.

From that, let's assume you run the unit 24 hours a day for the 92 days from June 1st through August 31 and you pay $0.20/KWh, you could save $9.45 a year.

The catch is, from a Total Cost of Ownership POV, most people will need to call an electrician to run a new 230 volt branch circuit to the location, since losing the other outlets on the original 120 volt one won't make you too popular with other folks you share the building with (not to mention would probably be illegal in most places). Let's assume it would cost $250 to do that, you would have to run the machine for ~26.5 years to pay back the infrastructure upgrade and that doesn't take into account 230 volt units usually cost a little more initially in this range of machines. Since a fairly recent DOE study indicated the average service lifetime of a modern RAC is ~12.5 years this is starting to look like an economic loser. *

Therefore, my recommendation is; If a 230 volt circuit is already available or cost is no concern, then by all means go 230 for peace of mind (safety margin, code, etc.) and being an environmentally responsible property owner reasons (you will use less power), but if the budget is a problem leave it for the next guy and get the 115 volt model.

I think this pretty well covers the issue.

BTW, regarding any ad hominem; I did *not* call you ridiculous, I called your original statement and the implication of it ridiculous. There *is* a difference.

* http://www.eren.doe.gov/buildings/appliance_standards/residential/pdfs/tsdracv2.pdf, Chapter 2.2

3.) Regarding Electric Utility Company Billing and Metering Issues:

You are correct in stating a residential user is charged for reactive power, in that the utility builds a "factor" into their tariff schedule to account for it. However, for the vast majority if not all of US residential customers, you are not directly metered for it.

The reason; For the people who still have them, the classic electromechanical induction watthour meter does NOT respond to VARs to any significant degree by theory of operation, design, and calibration. For those who have modern electronic watthour meters, even though the meter is theoretically capable of measuring real and reactive power usage simultaneously, the models which are used for residential purposes do not mostly because the manufacturer charges a premium for their models which do. IOW, they put in a "dumbed" down version of the hardware and/or software which drives the meter and for residential use cheaper is better from the electric company's POV.

To verify this, one only needs to examine your utility companies tariff schedules (a matter of public record) and compare your residential rate to that charged to the various classes of commercial and industrial users.

Typically, one finds the rate for active power in KWH is roughly 1/4 to 1/3 less than what is charged to residential customers. On the other hand, they "hose" the Comm/Ind customer something on the order of 5 to 6 times their KWH rate for KVARH use when it exceeds a certain value stated in their service class. This is one of the reasons why these customers bend over backwards to make sure their power factor is as high as they can possibly (and safely) get it.

4.) Regarding Current as the Metric of Power Usage Measurement:

I'll stand by my statement for the simple reason that in order for it to be an accurate quantity to derive power usage from, the utility would be required to hold the voltage absolutely constant at all times (not very likely). This could lead to gross overbilling in brownout conditions and is one of the reasons why it was abandoned for such use many, many years ago.

This is due to the fact the metering device most likely would be some form of a current shunt integrating amp hour meter, calibrated to readout in KWH. * If ALL of your load is resistive, then there is no problem. However, in the real world this is not the case.

The problem is load devices like induction motors, fluorescent lamp ballasts, electronic power supplies, etc, which are essentially constant *output* power devices when operated within their design parameters, try to compensate for an input voltage sag by drawing more current from the source.

The catch is the metering device does not care one whit about what the line voltage is and the "load" impedance it is measuring is the shunt, not the actual load. Therefore the utility could increase the customer's indicated power usage by deliberately lowering the line voltage. Ignoring for the moment this would result in needless abuse of the customer's equipment as well as the utility's, it would be fraud from a billing POV.

* http://www.watthourmeters.com/westinghouse/shall-amp.html

5.) General References

http://en.wikipedia.org/wiki/Electricity_meter
http://en.wikipedia.org/wiki/Electricity_meter#Other_types_of_electricity_meter
http://www.themeterguy.com/
http://www.watthourmeters.com
http://www.aham.org/
http://www.portlandgeneral.com/business/manage_account/understanding_bill.asp?bhcp=1
http://www.uomschool.org/Meter_Book/Table%20of%20Contents/Meter%20Parts/Meter%20Parts.htm
http://www.themeterguy.com/Theory/watthour_meter.htm
http://www.usbr.gov/power/data/fist/fist3_10/vol3-10.pdf
http://www.aham.org/ht/a/GetDocumentAction/i/1792
http://207.140.180.12/dirsvc/aham.nsf/frmAboutRac?OpenPage
http://www.lmphotonics.com/m_control.htm
http://www.motorsanddrives.com/cowern/motorterms12.html
http://www.ecmweb.com/mag/electric_highs_lows_motor/index.html

6.) Summary

As for some of your other "shotgun" commentary, suffice to say some of it borders on technobabble (and I'm being generous), including the idea that physics doesn't apply to electrical engineering and it's "jargon".

I challenge our engineering student readers to put some of them down as answers to your next examination question where they might apply (term used loosely) and see how well they fly! Just make sure you can retake the test without penalty. :-)

So the question now arises in my mind, why would someone post such easily refuted commentary in a forum populated by such a high percentage of technically savvy folks like this one?

I refer the reader to this remark:

"Since I power-factor corrected at all inductive sources in my home, my bill has dropped considerably. One could over excite the field of a synchronous motor and cause it to act in a capacitive fashion...drive it off a frequency drive so the field is automatically adjusted to one's needs moment to moment...."

This leads me to the conclusion the whole purpose of the exercise was to direct the casual readers' attention to the plethora of quasi-snakeoil residential power factor correction device vendor(s).

My reasons are twofold:

1.) Power factor was not even mentioned in the this thread until you brought it up.

2.) PF correction for traditional inductive loads is not very effective when applied to electronic devices like DC power supplies, motor speed controls, electronic ballasts, etc. As hiamps brought up, harmonic distortion of the line current waveform is a far more prevalent problem in a commercial office setting nowadays and increasingly in the residential one as well.

Unfortunately, it is referred to in technical as well as popular literature as a "power factor" problem. In a sense I suppose it is from the viewpoint it leads to energy being wasted instead of performing useful work. However, traditional capacitor based bulk solutions have little effect on it. It can even make the problem worse by setting up a resonance situation in the power distribution system overall.

This does NOT mean in some cases a homeowner can't save a buck or two with these devices, it depends on your particular situation. I will go as far to say that if you are saving a bundle with them, then something was seriously screwy to start with and the root cause(s) needs to be further investigated (or you need to limit your Tesla coil experiments in time travel and/or warp drive).

I believe I have included enough base links so the reader can determine the validity of this post on their own by reading them, following the links provided on them, and creating appropriate Google keyword queries based on the information presented on them.

Alinator