Does this help.

Dear Don:
Did you answer my question regarding how much power a generator would be required to make to deliver the 205 watts
at the measured power factor? I must have missed that. Instead I think you may have used the opportunity take much pleasure in putting people down for asking legitimate questions and then not answering their questions. Is that the hallmark of a "real electrical engineer" or is that one of those infantile insult your mention.? You can do better than that!

Answer is it depends. Generators are specified in KW/KVA. The KVA rating is usually higher but who knows maybe not. Depends on the generator. A generator that could make 269 megawatts could.
Ignoring the requirement to have the KVA's is like thinking that just because you borrow money for your rent from the credit card company till the end of the billing cycle and pay it back with no interest that money did not exist. The credit card company had to have the money just like that generator had to have those -j174 amps.
The generator had to make some of that apparent power, and more fuel was required by the prime mover to create the torque that turned that generators shaft. Had the power factor been unity it would not. If this is multiplied thousands of thousands of times it could mean a power plant may not be needed, a peaker would not have to run or a new transmission line may not have to be built.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Do we deal with digital transmission lines?
Power engineering in a process environment like a water treatment plant, that has a megawatt emergency generator, and automatic transfer switch, variable speed drives, variable frequency drives, large pump motors, induction heaters, as well as air conditioning and its own FOFA transformer off the feeder involves lots of digital instrumentation which is communicated over digital transmission lines.
In power transmission systems switches, breakers, reclosers are all remotely monitored and controlled with digital networks and lots of digital instrumentation and SCADA backhaul to accomplish substation automation, switching management and load management. Some of the digital backhaul uses the high voltage wires in the transmission lines to relay this SCADA. The high voltage wires become digital transmission lines.
Most large MCCB's have digital trip control, remote trip control and remote reclosing all controlled by computers that talk directly to them commanding them as if they were PLC's . Real electrical engineers use digital transmission lines in power engineering.
Chances are someone you know has a demand meter on their house that reports your electricity usage back to their utility over the power lines or perhaps by digital radio. A real electrical engineer designed that product and another real electrical engineer designed the digital backhaul to get the information back to the utility.
Peace dawg
PS: Dont really care what people here say about whether or not I am a REAL engineer. When I need to prove it to the AHJ I whip out my stamp and sign my name on top, the secretary of state in two of the lower 48 states I practice in have my numbers. If I need to pick up another state the NCEES has my record on file Been a PE over 25 years.
I learned long ago I sleep better being an anonymous usenet user. You get exactly what you pay for on usenet.
Subject: Re: Residential & PF Date: Tuesday, September 09, 2008 2:18 AM
------- Yes there are real electrical engineers lurking on this group. Paul H is one, Daestrom is another, Ben Miller is one. I also am one. There are others and some definitely are not engineers- wecandoit, and the guys who spend their time in infantile insults are not.
Do we deal with digital transmission lines? - I don't and some of the others don't- power transmission and utilization is far more fun.
You got a good answer to your question from Paul and Martindale pointed out some facts. In considering residential service, the charges are based on energy as read by a KWH meter which actually measures the real power* time. Your KWH meter, at .38KWH over 1.62 hours indicates an average real power of 380/1.62 #0 Watts (rounded to 2 significant figures). This meter doesn't give a hoot about power factor Your killawatt indicates 205 Watts - The KWH meter is actually more accurate by a fairly large margin (money is involved).
In determining your costs- consider only the real power as that is all that the KWH meter cares about. The power factor as found from watts/Volt amps 5/269 =0.76 as opposed to the measured 0.78 . Your voltage times current indicates 259VA not 269VA so there are a few discrepancies but this is to be expected. The measured Volt-amps includes the effect of reactive volt amps which is a measure of energy put in during part of the cycle and returned in another part of the same cycle =average of 0 (in every 1/60 second) so it doesn't register in the watt reading. This is what gives the power factor which is the ratio between true power (watts) and the apparent power or volt-amperes. Using your measurements the reactive Volt amps 4VAR The power factor of different loads will depend on the load. A light bulb or heater will be unity power factor as they are pure resistance loads. A single phase motor will have a power factor at full load of the order of 0.7 or so at full load and about 0.5 at no load because it is inductive. Fluorescent lamps have inductive ballasts so they will also have a power factor less than unity.
All this tells us nothing about the efficiency of the generator, the transmission line or even the appliance itself. At this power level trying to correct power factor (at the appliance- not at the house service entrance) might save a couple of watts (line losses as Paul had indicated) but would cost more than it saved so why bother.
Does this help?
--

Don Kelly snipped-for-privacy@shawcross.ca
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----------------------------

--------- There is a considerable difference between a generator supplying 205MW and a load requiring 205 watts. I was dealing with the latter case and your question had nothing to do with that case. If I have offended, then I am deeply sorry- but after seeing the puerile comments of others, my fuse may have been shorter than it should be.
However:
Note that a generator of this size is generally rated in MVA and power factor. The MVA rating is the maximum continuous MVA available and the pf rating is set by the field current limitations (at the rated voltage). So at some point a 269MVA 0.8 pf the generator will be at its rating both field and stator current and the MW output at this point would be 215MW and an eyeball estimate of the capability curve of such a generator would indicate that 205MW at 076 pf lag would be a borderline case for continuous operation. If the rating was 269MVA at 0.9pf then there would be an field overload condition. You are correct in saying "it depends".

---------- Please note that the original situation was a residential service where only KWh measurements are taken. Except for small changes in losses and voltage drops in the wiring between the meter and the load, the power factor of the load isn't of concern in this situation as "apparent power" isn't being measured by either the utility or consumer. If, for commercial or industrial loads, there is demand metering, then there may or may not be a benefit from power factor correction- it depends where and how much. ---------

------- FALSE: The prime mover doesn't "see" the apparent power, only the real power (check any machines text). (except for the small change due to I^2R losses) is required for a given real power output at any power factor. I suggest that you go back to the fundamentals of AC circuit theory and start from the instantaneous power for v=Vm*cos(wt) and i=Im*cos(wt+a) and plot the instanteous power over a cycle. note the constant term and the double frequency term. Now average both of these and the double frequency term disappears leaving only the constant term (0.5)*Vm*Im*cos(a) =Vrms*Irms*cos(a) =Vrms*Irms*power factor This "real" power is what the prime mover sees. The reactive component that you see in the real, apparent, reactive triangle is a measure of the energy (and this is an artifact of going from the time to the frequency domain so we could use the advantages of phasor analysis) into and out from inductance and capacitance but the average of this energy over a cycle (or half cycle) is zero. This was recognized sometime more than 100 years ago.

------- And, in all these cases, the data sent is digital, not the transmission line. That was my point.

So? I have been a P.Eng over 50 years, and have been a seniour member of IEEE and have power system and academic experience during that time. I bet that your area is electronics and or digital systems while mine is power systems and machines. I won't argue with you with regard to the former but will with regard to the latter:).
--

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Added comments as to why I didn't consider you competent in the subject under discussion These comments do not, in any way, deal with your competence in other areas of electrical engineering. I suspect that in some areas of EE, you have knowledge that I don't have.
Quote:
"For safety reasons it is probably better that one does not go about modifying appliances to get closer to a resistive load but I understand where you are coming from. You are enlightened enough see your appliance is throwing 54 watt hours per hour to the "terrists" who provide the fuel for our polluting power plants.
Considering that the power delivered to you appliance by the electric utility is a fraction of the energy required to make it and pump it through the distribution system for your appliance to consume, correcting the power factor could save 100 watt hours per hour or close to a megawatt ( 24x365x100W) of prime mover energy per year from being consumed."
Unquote:
--
This statement of yours simply doesn't make sense - it appears to
compounding a fundamental conceptual error with respect to "power factor"
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ For the purpose of better understanding what is going on here and to move away from the Killa-Watt and my previous questionable blathering.................lets continue with this analysis. The question is how to calculate the I^2xR loss in the 14 gauge wire for the corrected and uncorrected power factor. Not that it matters other than how to calculate it. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given: 118VAC measured at the terminals of appliance 2.2A measured current draw .78 measured PF ( yeah from the killa-watt) 25' of number 14 between breaker and appliance 2.5756 Ohms per 1000' #14 at 77 degrees (room temp) $0.10 per KWH
Calculate VA 8VAC x 2.2 A = 259.6 Watts = VA x PF = 259.6 x .78 = 202.48 Phase Angle = PA =arcCOS PF8.73degrees VARS = Sin PA x VA = Sin 38.73 x 259.6 VA = 162.4 VARS Z = V/A 8V/2.2A = 53.63 ohms R= Watts/I^2 = 41.83 ohms XsubL = VARS/I^2 3.55 ohms reactive.
VA - Watts = 269.6-202.48 = 57.112 (What units VA?) What would you call this quantity? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ This is where it gets sketchy for me. Assuming ( Capital ASS here) this motor load is a resistor and an inductor in series. How would I figure out what affect the 33.55 ohms reactive has on the overall circuit? Is this a series or parallel circuit?
If correcting the phase angle at the load end would the capacitor bo across this series RL load? Just across the L (physically impossible?)
Not sure how to take this to the #14 wire.
How do you figure this ?
a) the difference between 259 (from V*I)and 205 TVA of which the bulk is

Thanks in Advance
peace dawg
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----------------------------

--------- So far so good. VA is volt-amperes but is commonly called "apparent power" -in this case 259 volt-amperes ---------

------- I'd call it garbage as it is meaningless. You have looked at the impedance triangle and you also have a "power triangle" VA=root(watts^2 +vars^2)

-------- If this is a small induction motor, then it would look like a T circuit with a speed dependent resistive load on the output side of the T. Essentially it is a transformer with a load resistance that depends on the speed. Unlike a normal transformer, it has a high (inductive) magnetizing current because of the air gap between rotor and stator. Now consider that this whole thing is in series with your #14. Capacitors at the motor can correct its power factor to unity but as the pf of the motor changes with load, the capacitors will under or overcorrect much of the time. For a motor of this size- the cost of the capacitors exceeds the benefits.
Now lets look at what is going on in the #14 wire with or without correction. 50ft (2 wires) of #14 =0.123 ohms
a) no correction, 2.2A (at 0.78pf) I^R=(2.2^2)0.123 =0.623 watts. For a year (8760 hours) this leads to 5.46 KWH or 55 cents
b) corrected to unity pf will result in a current of 2.2*0.78 =1.72A I^2R =(1.72^2)0.123=0.379 watts. For a year the energy is 3.32KWH and the cost is 33 cents
Saving by correction is 22 cents/year. ( my estimate was based on 50 ft to the load -double the losses or 50 cents/year).
Size of capacitor at 60Hz for an impedance of 33.6 ohms is about 80 microfarads and it should be rated for, say, 125VAC. Motor start capacitors are cheap (say <$5 but as they are normally designed for inermittent duty they might not be suitable (e.g. more money needed). So the breakeven period would be about 25 years. It's not worth the effort. If you are going to correct pf- choose a load where it makes a real difference.
--

Don Kelly snipped-for-privacy@shawcross.ca
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If the power factor was corrected would not the total power in the circuit now be 1.73A^2 x 41.83ohms = 125 watts?
Why is the power still 2.2A^2 x 41.83ohms = 202.48 watts.
What is going on here? Still not getting it.
peace dawg
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Or better yet why is the corrected circuit power now not E^2 x R 118 x 118/41.83 ohms = 332 watts?

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----------- First of all, I made an OOPS! I calculated the capacitance on the basis of the XsubL value which assumes a series connection which isn't the case as the capacitor would be added in parallel to the load. The capacitance that I gave would then be for a series capacitance and in that case -what you have written is absolutely correct except for the dilemma that you have encountered.
However, the current is correct as shown below.
The power is determined by the mechanical load and is still 202.5 watts give or take a small variation. That means that the apparent "resistance" will have changed accordingly. For good reasons, we don't want series resonance and capacitors are connected in parallel with the load. Since the voltage is constant, it is much easier to use an admittance model.
In fact the capacitor would be in parallel and the capacitor would have to supply -162.4VARs for correction. On that basis the capacitive admittance would be 162.4/(118^2) = 0.0117 Siemens (S) corresponding to 31 microfarads at 60 Hz. Then the complex power in would be 202.5 +j162.4 -j162.4 2.5 +j0 and the corresponding current would be 202.5/118=1.72A
Alternatively Y=1/Z =0.0186 @ -38.73 degrees =0.0145-j 0.0117 Siemens (mhos) and adding a parallel capacitive admittance of +0.0117 S as before. This results in a total admittance of 0.0145 -j0.0117 +j0.0117 = 0.0145 corresponding to 68.96 ohms and a current of 1.71 A at 118V.
I bypassed all this and simply used 2.2*0.78 =1.72 A at unity pf for the same power.
Note that, in general, power factor correction at the load terminals has the effect of improving the power factor which reduces losses, decreases feeder voltage drops and more importantly reduces the total KVA where there are demand charges based on KVA as well as energy charges- this is not a factor in residential loads but can be a factor for industrial loads. The balance that determines the amount of compensation is an optimization problem where the best point is where the incremental cost of poor power factor (losses and demand charges) = the incremental cost money and maintenance to correct the power factor. If the capacitor is at the incoming "panel" then the demand charges and some voltage improvement are the only benefits as loss improvements won't be obtained except by the upstream utility.
--

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Dear Don:
Would the new equivelent circuit look like this?
----line 118V--------------------- | | | | R | | | XsubL CAP | | _____________|__________|
or like this
----line 118V------------------------------- | | | | | | R | | | XsubL | | | CAP | | | _____________|_____ |___________|
R ~ 68 Xl ~ -j85 Xc ~ +j85
Regards Dawg
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----------------------------
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***By the way, your signs on Xc and XsubL are wrong. The convention is: inductive reactance is +jX and capacitive reactance is -jX .
A good basic review source would be Schaum's Outline on circuits (AC if there is one specific to that) as any problems that I see here are not so much power as circuit analysis oriented.
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If +JX = inductive reactance = lagging phase angle -JX = capacitive reactance = leading phase angle
Then
ELI the ICE man must be voltage kind of ice man? And The power vector points up when inductive and down when capacitive?
peace dawg
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----------------------------
watts and 300 Vars inductive and the pf is 0.8 lagging
P.S, I never used ELI the ICE man. Easier to remember the applicable relationships than the memnomic.
--

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