Does this help.

--------- 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:).

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:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 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 =118VAC x 2.2 A = 259.6 Watts = VA x PF = 259.6 x .78 = 202.48 Phase Angle = PA =arcCOS PF=38.73degrees VARS = Sin PA x VA = Sin 38.73 x 259.6 VA = 162.4 VARS Z = V/A =118V/2.2A = 53.63 ohms R= Watts/I^2 = 41.83 ohms XsubL = VARS/I^2 =33.55 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 =54VA of which the bulk is

Thanks in Advance

peace dawg

--------- 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

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

Or better yet why is the corrected circuit power now not E^2 x R

118 x 118/41.83 ohms = 332 watts?

----------- 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=202.5 +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.

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

----------------- The capacitor preferrably would be connected at the load terminals- not back at the panel as it appears in fig 1(and downstream of the KWH meter) where it would not reduce line losses or voltage drops in the line because it doesn't affect anything "downstream". The utility, being upstream of the panel, really wouldn't care less with regard to its position.

Now, whether the load is modelled as an R- L series circuit (fig1 ) (Z=53.6 @38.7 degrees =41.8+j33.6 ohms or as a parallel model Y=1/Z =1/R' -j(1/XsubL) =68.9 ohms in parallel with an inductive reactance of

+j85.7 ohms ***(as in fig 2) doesn't matter a damn as the two are equivalent as seen from the terminals where the measurements were taken. The presence or the absence of the capacitor doesn't change either model. (but for full equivalence, the capacitor in fig2 should be as shown in fig 3 below). Thinks of the circuit as something inside a black box and you have no idea of what is actually there except from these measurements at the electrical terminals at a given voltage. There is nothing else that is known including whether the R and L are voltage dependent or not or whether they depend on some other factor that wasn't observed. SO all we know is that the load draws 2.2A at 118V and 0.78 pf and the associated power and reactive etc that go with this data. 202.5 watts and 162.4VARS. We can then use any model, however complex which corresponds. The R-L series or the R_L parallel models do this. A monkey having a variable resistor in one hand and a variable inductor in the other hand and reacting to having it's tail twisted artistically could also do this- but is a bit harder to model.

Fig 3 below is what we have- nothing more

2.2A->

-----line----------------------------- 1.7A-> | | CAP BLACK BOX (118V 202.5W 162.5VAR) | |

--------------------------------------

On this basis, I would say that full pf compensation means that the capacitor should supply -162.4VAR at 118V (-j85.7 ohms = +j 0.0117 S). It then would draw j1.38A and is in parallel with the load drawing 1.72-j1.38 A resulting in a total current of 1.72+j0A.

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

***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.

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

The convention used in North America is inductive vars positive. This goes back a long long time - a typical load such as an induction motor required real power from the system and also required inductive vars from the system - both being considered positive. One had to try to increase the generator voltage to produce inductive vars just as i was necessary to try to increase speed to produce more real power).

This choice was established before the math of AC circuit analysis was well established and has been grandfathered in practice. Mathematically this corresponds to complex power S =P+jQ =VI* (* implying conjugate). Hence the complex power and impedance triangles match.

Suppose we have an inductive load consisting of Z=4+j3 ohms supplied at 50V @angle 0 Z=4+j3 =5 @+36.87 degrees The current is V/Z =(50 @0)/5 @36.87 =10 @ -36.87 degrees (lagging the voltage) The complex power is (50 @ 0)(10 @ -36.87)* =(50 @ 0)(10 @+36.87) =500 @

36.87 =400 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.