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Old 11-27-2007, 07:25 PM   #16
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Adjusting power factor is a useless effort in a residential service since the POCO does not bill residential based on PF. You could have the worst power factor ever and the POCO would not penalize you for it. In fact it would save you money.

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Old 11-27-2007, 07:32 PM   #17
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But Joe how do you account for all the savings that guy had on his utility bill??...........
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Old 11-27-2007, 07:47 PM   #18
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Wow this guy saved over 1000 dollars in one month.......28% exceeding all expectations


I am seeing an average savings of at least 28% per month and I am very happy with the product. Extremely happy actually. As an example, prior to the installation of the Power-Save , my April 2006 electric bill was just over $2700.00. I just received my April 2007 bill in the amount of $1780.00. The product will literally pay for itself in only a few months."
- Mo Sandiq, Owner of Northwest Investment Properties LLC, Woodland, California
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Old 11-27-2007, 09:16 PM   #19
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Quote:
Originally Posted by joed View Post
Adjusting power factor is a useless effort in a residential service since the POCO does not bill residential based on PF. You could have the worst power factor ever and the POCO would not penalize you for it. In fact it would save you money.
Could you please explain how having a low PF can save you money.

Also, I noticed a "Stabilisation Deferral" rate on one of the customers electricity bills. What is this?

Without any further guessing about the metering equipment used at residences or billing methods, there can be no substantial savings by using such a device unless;

1] the energy supplier acknowledges the customers increased PF and/or,
2] the installation is large (high current) with a significant inductive load.
3] the device is able to automatically correct PF to a minimum of 0.8 & constantly maintain this level of correction. This only applies if 1] & 2] are apparent.
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Old 11-29-2007, 02:12 AM   #20
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But Joe how do you account for all the savings that guy had on his utility bill??...........
-------------------------------------------------------------------------
My guess..probably used way less energy for a month with sneakyness to show proof of his newly though up marketing scam, ie: the boob probably unplugged his hot water tank for a majority of the time !
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Old 11-29-2007, 06:07 PM   #21
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Could you please explain how having a low PF can save you money.
Because your meter only measure true power(watts). In residential billing there is no consideration or measurment of PF(at least not where I live). The further away for perfect PF(unity or 1) you are the less power the meter can read.
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Old 11-29-2007, 11:33 PM   #22
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Quote:
Originally Posted by joed View Post
Because your meter only measure true power(watts). In residential billing there is no consideration or measurment of PF(at least not where I live). The further away for perfect PF(unity or 1) you are the less power the meter can read.
Generally & for example, a 1kW inductive load will try to do 1kW of work. That is, if the PF decreases, the current will increase in an attempt to maintain the required level of work (Ohms Law).

So, a kWH meter will still see 1kW used regardless of the PF & therefore regardless of the line current.
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Old 11-30-2007, 09:53 AM   #23
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Quote:
Originally Posted by elkangorito View Post
Generally & for example, a 1kW inductive load will try to do 1kW of work. That is, if the PF decreases, the current will increase in an attempt to maintain the required level of work (Ohms Law).

So, a kWH meter will still see 1kW used regardless of the PF & therefore regardless of the line current.
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, 10:48 AM   #24
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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.
Really?

Prove it!......or can't you? I'll even give you time to do a google search.

BTW, "time" means up to 4 hours, so you have plenty of time to come up with an amazing answer.
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Last edited by elkangorito; 11-30-2007 at 11:30 AM.
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Old 11-30-2007, 11:29 AM   #25
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Power Factor is a unitless ratio of work (real power) / apparent power (volts * amps). For a resistive load the power factor is 1 and work = apparent power. Assuming work is a constant, and the power factor is less than 1, then the volts * amps must be greater than actual work. Since voltage is constant (assuming no voltage drop on feeders) the current increases. A single phase Kwh meter sees the volts * amps.
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Old 11-30-2007, 11:45 AM   #26
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Quote:
Originally Posted by spebby View Post
Power Factor is a unitless ratio of work (real power) / apparent power (volts * amps). It is not unitless. For a resistive load the power factor is 1 and work = apparent power & True Power. Assuming work is a constant and the power factor is less than 1, then the volts * amps must be greater than actual work. Since voltage is constant (assuming no voltage drop on feeders) the current increases. A single phase Kwh meter sees the volts * amps.
Dear oh deary me!

Let's see, for a given device;

P = E x I x Cos Phi. This formula excludes single phase motor efficiency but is for single phase machines.

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. Not difficult at all.
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Last edited by elkangorito; 11-30-2007 at 11:57 AM.
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Old 11-30-2007, 11:59 AM   #27
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My only comment is that power factor is a unitless ratio. Look it up. You divide watts by watts you get a unitless number.
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Old 11-30-2007, 12:11 PM   #28
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Quote:
Originally Posted by spebby View Post
My only comment is that power factor is a unitless ratio. Look it up. You divide watts by watts you get a unitless number.
Power factor is not unitless. If it was unitless, there would be no need for Power Factor at all. True Power & Apparent Power have different uses. Just because something uses lots of Apparent Power doesn't mean that it's wasteful. Then again, Apparent Power is another discussion, much the same same as Reactive Power.
Please make sure you can back up your statements before posting such stuff. In the meantime, please feel free to answer the question in the post above ie what is the current in each given PF?

I gave you the formula...simply apply it.
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Last edited by elkangorito; 11-30-2007 at 12:13 PM.
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Old 11-30-2007, 12:25 PM   #29
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What units is a power factor? Is it volts, amps, watts, joules, inches, feet, meters? You insist that it has units, so surely you know the units. What is the units of a cosine?

Yea, I can calculate the current at the meter in each case. And guess what, a) is less than b), and a) and b) are less than c. (assuming the voltage is constant). Exactly what I was saying.
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Old 11-30-2007, 12:37 PM   #30
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Whether you can calculate the current at each instantaneous phase angle or not, you still haven't answered my question about "current", which was your original question/assumption.

Power Factor is (all sinusoidal & RMS);

1] Volts/Amps.
2] True Power/Apparent Power.
3] Resistance/impedance.

So, how about the motor I posed earlier (in post 26)? Too difficult?
Answers a), b) & c) please.

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Last edited by elkangorito; 11-30-2007 at 12:42 PM.
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