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Old 11-30-2007, 12:46 PM   #31
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I am still waiting for the units of the power factor. Yea, multiplying and dividing is way too difficult for me.

You made a statement: "Power factor has units. If it was unitless, there would be no need for Power Factor at all." What is the units? Or is that too difficult.

Edit: And if power factor has units, then the cosine has units. And that's preposterous.


Last edited by spebby; 11-30-2007 at 12:53 PM.
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Old 11-30-2007, 01:05 PM   #32
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Originally Posted by spebby View Post
I am still waiting for the units of the power factor. Yea, multiplying and dividing is way too difficult for me.

You made a statement: "Power factor has units. If it was unitless, there would be no need for Power Factor at all." What is the units? Or is that too difficult.
BTW, I didn't say it had units but I did say that it was not unitless. Quite simply, Power Factor is Watts. But, what Watts? PF is never unitless. It is simply a matter of attaching the correct power to the correct unit. In your case, attaching the current to the voltage of the unit in question. The problem is the phase angle between the voltage & the current. This is the Power Factor. It can be corrected by a capacitor (if the phaser angle is lagging) but to what degree?

Spebby, earlier you argued that PF will not make a difference. Please tell me/us why & how.
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Old 11-30-2007, 01:23 PM   #33
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Quote:
Originally Posted by elkangorito View Post
BTW, I didn't say it had units but I did say that it was not unitless. Quite simply, Power Factor is Watts. But, what Watts? PF is never unitless. It is simply a matter of attaching the correct power to the correct unit. In your case, attaching the current to the voltage of the unit in question. The problem is the phase angle between the voltage & the current. This is the Power Factor. It can be corrected by a capacitor (if the phaser angle is lagging) but to what degree?

Spebby, earlier you argued that PF will not make a difference. Please tell me/us why & how.
Remember, you said, "If you have 1kw of work to be done, the meter will see 1 Kw if the power factor is 1. If the power factor is less than 1, then the meter will see more than 1 kw for 1 kw of work."
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Old 11-30-2007, 02:15 PM   #34
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elkangorito, I never said PF would not make a difference. Re-read the statement I made that you quoted:

"If you have 1kw of work to be done, the meter will see 1 Kw if the power factor is 1. If the power factor is less than 1, then the meter will see more than 1 kw for 1 kw of work."

Power Factor is a ratio of two different measures of power. And I am taking about sinusoidal voltages and currents. It much more complex for non-sinusoidal voltages and currents. A ratio of like measures is unitless. watts / watts is unitless which means it does not have a unit of measure. It's similar to a percentage, mathematically.

BTW, I have a B.S. degree in Electrical Engineering (years ago), although I have never worked as an electrical engineer. There is way more money in software.
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Old 11-30-2007, 11:11 PM   #35
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My comments in blue.

Quote:
Originally Posted by elkangorito View Post
So, how about the motor I posed earlier (in post 26)? Too difficult?
Answers a), b) & c) please.
I'm still waiting for the answers. See below for the questions.

Quote:
Originally Posted by spebby View Post
elkangorito, I never said PF would not make a difference. Re-read the statement I made that you quoted:

"If you have 1kw of work to be done, the meter will see 1 Kw if the power factor is 1. If the power factor is less than 1, then the meter will see more than 1 kw for 1 kw of work."

Yes, you are correct & I made a "slip of the tongue". But, the above statement is incorrect. This it what the statement should say,
"If you have 1kw of work to be done, the meter will see 1 Kw if the power factor is 1. If the power factor is less than 1, then the meter will see less than 1 kw for 1 kw of work."

The above is true providing that the values of voltage & current do not change. In the case of a motor for example, as the PF decreases, the current will increase to maintain the power input. The KWH meter will measure this as "true power". The power measured by the meter (consumed by the motor) will remain constant but the current through it will have increased.

Whilst you cast your mind back to university, I'll cast my mind back to High School:

True Power (single phase) = E x I x PF.



BTW, I have a B.S. degree in Electrical Engineering (years ago), although I have never worked as an electrical engineer. There is way more money in software.
Quote:
Originally Posted by elkangorito View Post

For example, a 1kW load is connected to a 250v AC supply (single phase). What is the current at the following Power Factors?

a] unity PF.
b] 0.8 PF.
c] 0.4 PF.

Can you please answer the questions for this 1 kW load?

This is a very simple multiplication and/or division calculation. Not difficult at all.
Since you have a "B.S." degree in electrical engineering, I can't imagine why you would think that a reduced PF would mean a reduced power input for any given appliance. Again, look at the above questions & at least prove that you can answer them. After you have answered them, it should become apparent why a reduced PF does not mean a reduced power input & therefore a reduced power consumption.
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Old 12-01-2007, 08:18 AM   #36
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The meter does not see the power factor of any load. All the meter sees is the increased current of a load with a power factor lower than 1 which means more watts. If I have a motor with a PF of .8 and replace it with a motor with a PF of .9 the meter will see less current and thus less power.

Edit: You are talking in circles. These statements contradict:

"If you have 1kw of work to be done, the meter will see 1 Kw if the power factor is 1. If the power factor is less than 1, then the meter will see less than 1 kw for 1 kw of work."

"it should become apparent why a reduced PF does not mean a reduced power input & therefore a reduced power consumption."

Last edited by spebby; 12-01-2007 at 08:21 AM.
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Old 12-02-2007, 03:27 AM   #37
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Originally Posted by spebby View Post
The meter does not see the power factor of any load. All the meter sees is the increased current of a load with a power factor lower than 1 which means more watts. If I have a motor with a PF of .8 and replace it with a motor with a PF of .9 the meter will see less current and thus less power.
This is not true. I shall explain but firstly, I'd like to clarify a point by way of example.
If a single phase 1kW load is connected to a 250v A.C. supply, what is the current flowing at power factors of;

1] unity.
2] 0.8.
3] 0.4.

Answers;
1] 4 Amps.
2] 5 Amps.
3] 10 Amps.

The point I'm trying to make here is that although the current changes according to the power factor, the load still absorbs 1 kW of True Power, which is exactly what induction disc kWH meters measure.

How do induction disc kWH meters work?
A single-phase kWH meter is essentially an induction motor, the speed of which is directly proportional to the voltage applied and the amount of current flowing through it. The phase displacement of the current, as well as the magnitude of the current, is automatically taken into account by the meter. In other words, the power factor influences the speed and the disk rotates with a speed proportional to true power. True Power does account for power factor, as in the above example with the motor. True Power = E x I x Cos phi. Please note that the True Power has remained constant whilst the current has increased at lower power factors.
For an induction disc kWH meter, if the power factor is decreased but the current remains the same (did not increase), less rotational torque would be produced onto the disc, which would result in less kW being measured.

These types of meters cannot measure power factor but they do account for variations in power factor up to certain limits.

The long & the short of this energy saving device is that although you will not see any savings at the kWH meter, you will reduce your line current & therefore allow equipment to operate more efficiently. This only applies if the kWH meter is the induction disc type.
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Last edited by elkangorito; 12-02-2007 at 03:32 AM.
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Old 12-02-2007, 08:50 AM   #38
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http://www.snapdrive.net/files/50779...ons%5B1%5D.pdf

This is a paper done by a couple of engineers about power quality solutions and energy savings.

Personally I am skeptical about many of the claims out there.
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Old 12-02-2007, 10:40 AM   #39
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Why are you assuming True power stays a constant? Why don't you assume apparent power stays a constant?
If you assume apparent power is the constant then the watts is lower with poor power factor.
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Old 12-03-2007, 03:40 AM   #40
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Why are you assuming True power stays a constant? Why don't you assume apparent power stays a constant?
If you assume apparent power is the constant then the watts is lower with poor power factor.
With regards to Apparent Power - for circuits that consist of combinations of resistance & reactance, the product of the line voltage & current do not equal the power consumed & therefore cannot be expressed in Watts. It is expressed in VA. This is for A.C. circuits only.

It must also be noted that Reactive Power (VAR) forms a part of this problem.

With regards to True Power remaining constant, it's simply a matter of Ohms Law as follows (series circuits);
Since a motor has a fixed resistance & a fixed inductive reactance (therefore a fixed Impedance), an applied voltage will cause a current to flow. The only thing that will change this current is a change in the phase angle between itself & the line voltage. In other words, the phasor sum of the voltage drops across the resistive component (copper wire) & the reactive component (the inductor), must equal the line voltage (since it remains constant). Changing the reactive component & therefore the voltage dropped across this component by using capacitors or inductors, will change the current but not the True Power. Thus the formulas for Ohms Law & True Power hold true. True Power; P = E x I x Cos phi, NOT E x I.
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Old 12-03-2007, 07:58 AM   #41
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I just had a thought...why not go through the list of FAQ's as listed in the Power Saver website. Here they are. My comments in blue;

Frequently Asked Questions.

How Does the Power-Save 1200™ Unit Work?
The Power-Save 1200™ Unit reduces the amount of power drawn from the utility by storing (in its capacitors) otherwise lost electricity (watts) caused by the inductive motors in your home. (Some examples of inductive motors are Air Conditioning units, refrigerators, freezers, washers, dryers, dishwashers, pool pumps, vacuum cleaners, furnace blower motors, fans etc.) The technology applied by the Power-Save 1200™ Unit supplies that stored electricity back to your inductive loads, thus causing you to decrease your demand from the utility. If you decrease your demand from the utility, your meter slows down, and you use less electricity. The thought is, you’ve already paid for that electricity, why pay for it and waste it when you can pay for it, store it, and reuse it again. This whole process is called power factor optimization.
This is a question I raised before. Are you billed by "demand" or by "kiloWatt hour"? From the responses so far, domestic residences are not billed on "demand".

What is Power Factor?
Power factor is the percentage of electricity that’s delivered to your house and used effectively, compared to what is wasted. For example, a 1.0 power factor means that all the electricity that’s being delivered to your home is being used effectively for its purpose. However, most homes in North America today have a .77 power factor or less. This means that 77% of the electricity that is coming thru your meter at your home or business is being used effectively, the other 23% is being wasted by your inductive load. With a low power factor, the utility has to deliver more electricity to do the same work. However, the Power-Save 1200™ increases that power factor in most cases to .97 or .98, thus increasing the effective use of your electricity and lowering your usage.
This is correct.

Does the Power-Save 1200™ work in any home?
Yes it does, as long as you have a circuit breaker panel with breaker switches and not the old screw in type fuses, the unit will work on any single-phase electric application for homes.
Generally, "wire" fuses have a Power Factor, albeit generally insignificant at most loads.

Will the Power-Save 1200™ affect any of my appliances and their normal use?
No, if anything, your motors will run about 10% cooler, which is good for a motor because heat is the enemy of a motor.
This is correct.

Is the Power-Save 1200™ tested and approved by independent labs?
Yes, the Power-Save 1200™ is UL listed and tested, and has also been tested by the University of Santa Clara California’s Electrical Engineering department, of which the results of that test are available on our website. Also, the technology is recognized by the U.S. Department of Energy.
The UL certification does not appear to relate to power savings. On the other hand, the university tests (whatever they are) may relate to power savings. Any university can equate a high Power Factor to the saving of "energy" (not necessarily money).

How much can I expect to save per month by using the Power-Save 1200™?
That depends on many factors. The size of your home, the amount of inductive motor load, and the amount you are paying per kilowatt-hour for electricity etc. However, generally speaking users of the product have seen up to 25% in reduced consumption, but the average savings is somewhere in the 15% to 20% range.
You can only save money if you;
1] have an electronic kWh meter that can measure Power Factor, and/or;
2] you are billed against your "demand", which is "current" (Amps) usage.

How long will it take for the Power-Save 1200™ to pay for itself?
Generally about 6-12 months, but again, the same factors above apply, some will see sooner (6 months), some will see later (12 months).
This depends upon how you are billed. See above.


Is the Power-Save 1200™ easy to install?
We recommend installation by an electrician. The unit comes with complete installation instructions. It installs in about 20-30 minutes.
This sounds correct.

How long will the Power-Save 1200™ last?
It has a predicted lifespan of up to 20 years.
This depends upon;
1] the quality of the supply.
2] the amount of load placed upon the Power Save unit.

Why haven’t I heard of this product until now?
That’s easy, two words “cost effectiveness”. Up until recently, electric rates throughout America were cheap, costing us 2, 3 or 4 cents per kilowatt-hour. Now, electric rates are 8, 10, 12, 14, and 19 and in some cases New York City is 22 cents per Kwh, and Hawaii is 33.5 cents per Kwh. At the cheaper rates the Power-Save 1200™ didn’t make sense, but at the current rates, it makes all the sense in the world.
A complete load of rubbish. Power Factor Correction has been around for a long time. The reason why domestic residences have not heard about it is because that domestic metering methods have excluded the need for PF correction. I.E. domestic metering methods did not allow for a reduced dollar value due to low Power Factor. In the early days, meters were easily made to run backwards by using a low PF load. These days, this is impossible. Not only this, all modern day induction disc meters will measure True Power accurately because they are the "cash register" for an energy supplier. This means that a low PF will not affect the "cash flow" entirely. As insurance, most energy suppliers impose a standard & fixed penalty to domestic consumers for an "assumed" Power Factor.

How much does the Power-Save 1200™ cost, and how do I receive it?
$299.95 plus $15 shipping. We ship the product directly to your door by regular ground shipment, and you should receive it in about 7-10 business days.
I could supply you with the same device for about $50.00 & still be making money. Installation would be extra.

What About Power-Save 1200™ Surge Protection?
The Power-Save 1200™ also protects the entire home against power surges. No longer a need for so many surge protectors in the home. The unit provides a broad range of protection for hardwired appliances and most home electronics such as televisions, satellite equipment, entertainment systems, etc. The unit protects from power line surges as well as spikes caused by internal wiring problems, loose connections and fluctuating demand from large motors such as appliances, vacuum cleaners, heating and cooling equipment, etc.
Unless this is a high quality MOV (Metal Oxide Varistor) protection, I wouldn't believe this. Also, this MOV must be able to operate within about 20 micro seconds & handle a minimum fault current of about 50 000 Amps to be of any use.

Is the Power-Save 1200™ Unit Warranted?
Yes, 5 year Manufacturers Warranty for full replacement. See our website for full details.
No, it is not "warranted". However, correcting the poor English, it is "warranteed" for 5 years.

Is there a “Money Back Guarantee”?
Yes, 60-day money back guarantee. If in 60 days, you don’t see reduction in usage on your electric bill, call us and let us know, and we’ll give you details on how to return the unit for a full refund of the purchase price. Installation cost will not be refunded.
I wonder how many people have returned this product?
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Old 12-03-2007, 05:36 PM   #42
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It's an AC circuit. Basic Ohms law does not apply. That why they call it VA(volt amps) instead of watts.
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Old 12-03-2007, 06:00 PM   #43
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Why can't this thread just die?
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Old 12-03-2007, 08:00 PM   #44
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The utility company charges based on total power consumed
true power is determined by W=ExIxPF. However, where there is evidence of a low power factor such as buildings supplying a lot of highly inductive loads, the power company imposes penalties in thier rate structure. So in some senses even with the powersave device, if you are being penalized you will still be penalized, even though the powersave is increasing your powerfactor, although your bill could still go down based on less energy consumed from the powersave storing the wasted energy and reusing it.
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Old 12-03-2007, 08:49 PM   #45
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The utility company charges based on total power consumed
true power is determined by W=ExIxPF. However, where there is evidence of a low power factor such as buildings supplying a lot of highly inductive loads, the power company imposes penalties in thier rate structure. So in some senses even with the powersave device, if you are being penalized you will still be penalized, even though the powersave is increasing your powerfactor, although your bill could still go down based on less energy consumed from the powersave storing the wasted energy and reusing it.
Incorrect. Again, your domestic power bill will not change because your kWh meter (induction type) will only measure True Power, which remains constant regardless of the PF. It is the VAR that increases which in turn, increases the VA.

Of course, if you have an electronic kWh meter that can measure the line PF & "demand", you may be billed extra.

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