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11-30-2007, 01:46 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 01:53 PM.

11-30-2007, 02:05 PM   #32
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Quote:
 Originally Posted by spebby 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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11-30-2007, 02:23 PM   #33
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Quote:
 Originally Posted by elkangorito 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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 11-30-2007, 03:15 PM #34 Member   Join Date: Oct 2007 Location: Oklahoma City Posts: 71 Rewards Points: 75 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.
12-01-2007, 12:11 AM   #35
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My comments in blue.

Quote:
 Originally Posted by elkangorito 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 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 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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 12-01-2007, 09:18 AM #36 Member   Join Date: Oct 2007 Location: Oklahoma City Posts: 71 Rewards Points: 75 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 09:21 AM.
12-02-2007, 04:27 AM   #37
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Quote:
 Originally Posted by spebby 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.

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 04:32 AM.

 12-02-2007, 09:50 AM #38 Master Electrician     Join Date: Mar 2007 Location: Baltimore, MD Posts: 332 Rewards Points: 250 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. __________________ John from Baltimore One Day at a Time "Experience is what you get when you were expecting something else" "The bitterness of low quality lingers long after the sweetness of low cost is forgotten"
 12-02-2007, 11:40 AM #39 Member     Join Date: Mar 2005 Location: Welland, Ontario Posts: 12,384 Rewards Points: 11,560 Blog Entries: 11 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.
12-03-2007, 04:40 AM   #40
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Quote:
 Originally Posted by joed 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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Last edited by elkangorito; 12-03-2007 at 04:43 AM.

 12-03-2007, 06:36 PM #42 Member     Join Date: Mar 2005 Location: Welland, Ontario Posts: 12,384 Rewards Points: 11,560 Blog Entries: 11 It's an AC circuit. Basic Ohms law does not apply. That why they call it VA(volt amps) instead of watts.
 12-03-2007, 07:00 PM #43 Member   Join Date: Nov 2007 Posts: 561 Rewards Points: 500 Why can't this thread just die?
 12-03-2007, 09:00 PM #44 Member   Join Date: Nov 2007 Location: Anywhere, USA Posts: 90 Rewards Points: 75 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.
12-03-2007, 09:49 PM   #45
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Quote:
 Originally Posted by RichyL 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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