Can I Improve Power Factor Correction with a Parallel Component?

[KhalDirth]

I've been working on a mathematical problem for a few days now. I'm not sure where my math is falling apart.

I have an impedance connected in series with a source. In phasor notation, the impedance is 1.2649 at 71.565 degrees. In rectangular, this impedance is .4 + j1.2 . The power factor for this circuit is 0.32 lagging. I would like to make the circuit have a power factor of .98 leading, while retaining the original resistance (.4).

One solution is to place a component in parallel that provides reactive compensation.

I have attempted the solution in two ways. One is to convert the impedance into admittance, since parallel admittances add together (making the math simpler). The other way is to work through the math of two parallel impedances.

When I try the problem via admittance values, I get a final answer that is a capacitor. This makes sense, as any additional real resistance will change the real power dissipation of the total circuit (my goal here is to make sure real power dissipation stays the same, even with power factor correction).

When I try the problem via impedance values, my parallel component does not solve to the inverse value of the parallel admittance component I found in my first solution.

In fact, when I calculate what the total impedance should be (load in parallel with component), my answer does not equal the inverse of the admittance.

The total admittance I get from my first solution is .25 + j0.0508.

The total impedance I get from my second solution is .4 - j0.0812.

These two values are not inverse of each other. Why is that?

 

[Bob S]

The admittance corresponding to an impedance of 0.4+j1.2 ohms is 0.25-j0.75 mhos.

Bob S

 

[skeptic2]

Is this homework?

Is your question what value of reactance do I add to a 0.32 lagging power factor to get a 0.98 leading power factor or is it why is the admittance value not the inverse of the impedance value?

Instead of putting the reactance in parallel have you considered putting it in series?

 

[KhalDirth]

Is this homework?

Is your question what value of reactance do I add to a 0.32 lagging power factor to get a 0.98 leading power factor or is it why is the admittance value is not the inverse of the impedance value?

Instead of putting the reactance in parallel have you considered putting it in series?

This is not homework, or I'd have posted it in the homework section.

My question is not what value of reactance I need to place in parallel (I already have that answer). My question is: why do I not get the same answer when I evaluate the system as an impedance?

I will post a visual reply (shortly) to make more sense of this.

 

[skeptic2]

To answer that question we would need to see your calculations.

 

[KhalDirth]

To answer that question we would need to see you calculations.

I just finished drawing it up.

 

[KhalDirth]

To answer that question we would need to see you calculations.

You can see, I try to solve the problem two separate ways: one via impedance, one via admittance. I am able to find the right-handed solution, but not the left. Why are the two values for total impedance/admittance inverse of each other? They should be, since I took the load impedance and admittance values and calculated what the necessary total values should be for the same power factor.

 

[skeptic2]

When you add capacitance in parallel to an inductive impedance, you increase the value of R. In order to maintain R at the same value, you must add reactance in series. Your answer of 0.4 -j0.0812 is correct for a PF of 0.98. All you have to do is subtract the inductive reactance from that for the total capacitive reactance you need to add in series.

 

[KhalDirth]

When you add capacitance in parallel to an inductive impedance, you increase the value of R. In order to maintain R at the same value, you must add reactance in series. Your answer of 0.4 -j0.0812 is correct for a PF of 0.98. All you have to do is subtract the inductive reactance from that for the total capacitive reactance you need to add in series.

I'm still not sure that answers my question. It seems to me that the answer should be the same, regardless of whether or not I choose to model the system as an admittance or an impedance. Why then, do I not get the exact (but inverse) results for what my system should look like at the correct power factor?

Anyone else have any thoughts?

 

[skeptic2]

The reciprocal of Z should equal Y and vice versa. The fact that they don't means your arithmetic is wrong. You don't show enough of your work to determine where the problem is. Can you explain in more detail how you get the reciprocal of ZL or YL? How did you calculate the value of ZC and YC? How do you combine ZC with ZL to get ZT? The same for YC, YL and YT.

 

[KhalDirth]

The reciprocal of Z should equal Y and vice versa. The fact that they don't means your arithmetic is wrong. You don't show enough of your work to determine where the problem is. Can you explain in more detail how you get the reciprocal of ZL or YL? How did you calculate the value of ZC and YC? How do you combine ZC with ZL to get ZT? The same for YC, YL and YT.

Gladly! Here are the mathematics. I've highlighted my conundrum.

Note: Stating again, my problem is that those two values should be inverse of each other. The total impedance (Zt) (power factor corrected) should be the inverse of the total admittance (Yt). In the end, you're analyzing the same circuit, so why shouldn't they be inverse?

OR

Where am I going wrong with my math?

 

[KhalDirth]

The reciprocal of Z should equal Y and vice versa. The fact that they don't means your arithmetic is wrong. You don't show enough of your work to determine where the problem is. Can you explain in more detail how you get the reciprocal of ZL or YL? How did you calculate the value of ZC and YC? How do you combine ZC with ZL to get ZT? The same for YC, YL and YT.

Having a problem with the upload service. Will try again in a few.

 

[KhalDirth]

The reciprocal of Z should equal Y and vice versa. The fact that they don't means your arithmetic is wrong. You don't show enough of your work to determine where the problem is. Can you explain in more detail how you get the reciprocal of ZL or YL? How did you calculate the value of ZC and YC? How do you combine ZC with ZL to get ZT? The same for YC, YL and YT.

I had my file size too large. I've attached it as a .jpg. However, the quality might be too poor. So I've included a URL as well.

http://i993.photobucket.com/albums/af55/khaldirth/PFProblem2.jpg

 

[skeptic2]

Zl = 0.4 + j1.2

For a PF of 0.98 and a real part of 0.4, the imaginary part must be 0.4 * sin(acos(.98))/-0.98 = -j0.08122.

My way:
The reactance that must be added in series to get that value is -j0.08122 - j1.2 = -j1.28122.

Your way:
To find Zc given Zl and Zt
Zc = (Zl * Zt) / (Zl - Zt)

Zl mag = sqrt(0.4^2 + 1.2^2) = 1.264911
Zt mag = sqrt(0.4^2 + -0.08122^2) = 0.408163
(Zl * Zt) mag = Zl mag * Zt mag = 0.51629

Zl ang = atan(1.2 / 0.4) = 1.249046 radians
Zt ang = atan(-0.08122 / 0.4) = -0.20033 radians
(Zl * Zt) ang = Zl + Zt = 1.249 + -0.20033 = 1.0487 radians

Zl - Zt = (0.4 +j1.2) - (0.4 - j0.08122) = 0.8 + j1.28122
(Zl - Zt) mag = sqrt(0.8^2 + 1.28122^2) = 1.28122
(Zl - Zt) ang = atan(1.28122 / 08) = -1.5708

(Zl * Zt) / (Zl - Zt) mag = 0.51629 / 1.28122 = 0.402967
(Zl * Zt) / (Zl - Zt) ang = 1.0487 - 1.5708 = -0.52209

Zc real = 0.402967 * cos(-0.52209) = 0.349284
Zc imag = 0.402967 * sin(-0.52209) = -0.200955

There are two ways to get convert 0.4 + j1.2 to 0.4 - 0.08122. The first is to add 0 - j1.28122 in series. The second is to add 0.349284 - j0.200955 in parallel.

 

[KhalDirth]

Zl = 0.4 + j1.2

For a PF of 0.98 and a real part of 0.4, the imaginary part must be 0.4 * sin(acos(.98))/-0.98 = -j0.08122.

My way:
The reactance that must be added in series to get that value is -j0.08122 - j1.2 = -j1.28122.

Your way:
To find Zc given Zl and Zt
Zc = (Zl * Zt) / (Zl - Zt)

Zl mag = sqrt(0.4^2 + 1.2^2) = 1.264911
Zt mag = sqrt(0.4^2 + -0.08122^2) = 0.408163
(Zl * Zt) mag = Zl mag * Zt mag = 0.51629

Zl ang = atan(1.2 / 0.4) = 1.249046 radians
Zt ang = atan(-0.08122 / 0.4) = -0.20033 radians
(Zl * Zt) ang = Zl + Zt = 1.249 + -0.20033 = 1.0487 radians

Zl - Zt = (0.4 +j1.2) - (0.4 - j0.08122) = 0.8 + j1.28122
(Zl - Zt) mag = sqrt(0.8^2 + 1.28122^2) = 1.28122
(Zl - Zt) ang = atan(1.28122 / 08) = -1.5708

(Zl * Zt) / (Zl - Zt) mag = 0.51629 / 1.28122 = 0.402967
(Zl * Zt) / (Zl - Zt) ang = 1.0487 - 1.5708 = -0.52209

Zc real = 0.402967 * cos(-0.52209) = 0.349284
Zc imag = 0.402967 * sin(-0.52209) = -0.200955

There are two ways to get convert 0.4 + j1.2 to 0.4 - 0.08122. The first is to add 0 - j1.28122 in series. The second is to add 0.349284 - j0.200955 in parallel.

Alright. So you say that the second way is to add an impedance of 0.349284 - j0.200955 in parallel to my load impedance.

It is safe to say that this would be the same as adding an admittance of 1/(0.349284 - j0.200955) in parallel to my load admittance (.25 - j0.75)? The total admittance at this point would be (.25-j0.75) + (2.15099-j1.2375431), which equals 2.40099 + j.487543. This is not the same answer that I got in my admittance-based calculations (I determined that the final admittance would be .25 + j0.0508.

 

[KhalDirth]

Perhaps this touches on some concept of duality that I'm not getting?

 

[skeptic2]

Yes it is safe to say that this would be the same as adding an admittance of 1/(0.349284 - j0.200955) in parallel to the load admittance (.25 - j0.75). However, 1/(0.349284 - j0.200955) is not 2.15099 - j1.2375431 but 2.15099 + j1.2375431 which when added to 0.25 - j0.75 equals 2.400998 + j0.487543. 1/(2.400998 + j0.487543) = 0.4 - j0.081223.

 

[KhalDirth]

Yes it is safe to say that this would be the same as adding an admittance of 1/(0.349284 - j0.200955) in parallel to the load admittance (.25 - j0.75). However, 1/(0.349284 - j0.200955) is not 2.15099 - j1.2375431 but 2.15099 + j1.2375431 which when added to 0.25 - j0.75 equals 2.400998 + j0.487543. 1/(2.400998 + j0.487543) = 0.4 - j0.081223.

Alright, I understand your math. I agree that you can place 0.349284 - j0.200955 in parallel with the load to get the desired impedance. What I don't understand is why you can also add an admittance of j0.801 in parallel and get the same circuit. You see in my right-hand calculations, the Y_C is j0.801 in parallel with the load admittance. Do you agree that this is the same as placing a 1/(j0.801) impedance in parallel with the load? This would be an impedance of -j1.2489. So my right side calculations show that you can add an impedance of -j1.2489 in parallel with the load. Your parallel impedance calculations also show that you can add 0.349284 - j0.200955 in parallel to the load. To me, that's two different circuits, and that's why I don't understand how both answers are right. Like I posted earlier, perhaps there's a concept of duality I am missing here?

Also, thank you for entertaining the questions.

 

[skeptic2]

What I don't understand is why you can also add an admittance of j0.801 in parallel and get the same circuit. You see in my right-hand calculations, the Y_C is j0.801 in parallel with the load admittance. Do you agree that this is the same as placing a 1/(j0.801) impedance in parallel with the load?

No I don't agree. You cannot add a reactance to an impedance in parallel without affecting the value of the real part of the impedance. Complex arithmetic is not always intuitive. This is why I wanted to see your work, to see where the error was. It is also why I showed you my work, so you could follow the arithmetic.

 

[The Electrician]

Alright. So you say that the second way is to add an impedance of 0.349284 - j0.200955 in parallel to my load impedance.

It is safe to say that this would be the same as adding an admittance of 1/(0.349284 - j0.200955) in parallel to my load admittance (.25 - j0.75)? The total admittance at this point would be (.25-j0.75) + (2.15099-j1.2375431), which equals 2.40099 + j.487543. This is not the same answer that I got in my admittance-based calculations (I determined that the final admittance would be .25 + j0.0508.

The mistake you're making in all this is a false assumption that if you keep .25 as the real part of the final load expressed as an admittance, that when expressed as an impedance, its real part will be .4, which is your requirement.

 

[KhalDirth]

The mistake you're making in all this is a false assumption that if you keep .25 as the real part of the final load expressed as an admittance, that when expressed as an impedance, its real part will be .4, which is your requirement.

So, you're saying that if someone asked me to correct the power factor of a known load without changing the real impedance, I would not be able to do so by running the calculations via parallel admittance values (as I tried to do)?

 

[The Electrician]

So, you're saying that if someone asked me to correct the power factor of a known load without changing the real impedance, I would not be able to do so by running the calculations via parallel admittance values (as I tried to do)?

Nope, that's not what I said.

What I said is, if you want to do things on an admittance basis, you can't get the answer you want by keeping the real part of the admittance equal to .25, which is what you did so far.

Your goal is to keep the real part of the impedance equal to .4, and that's what you have to do, whether you do things on an admittance or impedance basis.

 

[KhalDirth]

Nope, that's not what I said.

What I said is, if you want to do things on an admittance basis, you can't get the answer you want by keeping the real part of the admittance equal to .25, which is what you did so far.

Your goal is to keep the real part of the impedance equal to .4, and that's what you have to do, whether you do things on an admittance or impedance basis.

Okay. So how would you run the calculation for parallel admittances, such that the final result would give you an impedance with a real part of .4?

 

[skeptic2]

Okay. So how would you run the calculation for parallel admittances, such that the final result would give you an impedance with a real part of .4?

Just as I showed you in post #14. As you know, the formula to calculate the parallel impedance given the two elements is Zt = (Z1 * Z2) / (Z1 + Z2). If you already know the resultant impedance and one of the two elements as in this case, the formula is Z2 = (Z1 * Zt) / (Z1 - Zt).

 

[KhalDirth]

Just as I showed you in post #14. As you know, the formula to calculate the parallel impedance given the two elements is Zt = (Z1 * Z2) / (Z1 + Z2). If you already know the resultant impedance and one of the two elements as in this case, the formula is Z2 = (Z1 * Zt) / (Z1 - Zt).

I'm saying, via parallel impedances method Yt = Yc + YL. I'm asking this, because it seems to me that the solution to this problem is determined by the way you solve it. In other words, can you do the math (Yt = Yc + Yl, as seen in my attachment) in such a way that the result yields a real impedance of .4 with a pf of .98 leading?

 

[skeptic2]

Yes, that will work too.

YL = 1/(0.4+j1.2) = 0.25-j0.75
YT = 1/(0.4-j0.0812) = 2.40+j0.487

YC = YT - YL = (2.40+j0.487) - (0.25-j0.75) = 2.151+j1.237
ZC = 1/YC = 0.349-j201

 

[KhalDirth]

Yes, that will work too.

YL = 1/(0.4+j1.2) = 0.25-j0.75
YT = 1/(0.4-j0.0812) = 2.40+j0.487

YC = YT - YL = (2.40+j0.487) - (0.25-j0.75) = 2.151+j1.237
ZC = 1/YC = 0.349-j201

Ah. This is interesting. You see, I found this problem in a text on power engineering (again, not HW, as I'm not in school). The authors did their calculations just as I did mine for admittance. You converted what you found should be your final impedance into an admittance, and then solved for Y_C. The authors simply converted Z_L into Y_L and THEN (this is the big part) they determined how they would have to adjust their admittance so that it would maintain ITS real portion at a .98 leading pf. They determined that the admittance would need to be .25 + j0.0508. Is their method flawed, as the Y_T they (and myself, in my attachment) derived maintained the real admittance portion from Y_L, instead of doing as you did, and deriving Z_T and then converting it to Y_T?

 

[skeptic2]

I suspect it is flawed. As I said complex arithmetic is not always intuitive. (I once stumped my 12th grade math teacher with this one. 1 = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1i * 1i = -1) What was the name and author of the text?

 

[KhalDirth]

I suspect it is flawed. As I said complex arithmetic is not always intuitive. (I once stumped my 12th grade math teacher with this one. 1 = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1i * 1i = -1) What was the name and author of the text?

Electric Circuit Analysis, Third Edition
David E. Johnson, Johnny R. Johnson, John L. Hilburn, Peter D. Scott
ISBN 0-471-36571-8

I emailed Peter Scott (no reply, yet).

I guess, reading the problem now, the authors never did say that they meant to maintain the real impedance. I just think it would be crazy to correct the pf of a circuit and totally change the amount of power it will actually absorb (or at least, not take into account that it would change). Is this normal practice for pf correction? I used this text twice in undergrad. I've been working in the power industry for about a year now, and I decided I would refresh myself on the math. That's when I came across this problem. I agree with you about the intuition.

Do you have your own intuition about this phenomenon? I mean, when working with resistance/conductance, you could work an equation from start to finish in the exact same manner using either perspective, and wind up with the same result. With reactivity/susceptivity, it seems this doesn't hold true (thus, you had to determine the initial/final impedances and THEN convert to admittance). If you have a way of thinking about this, I'd love to hear it. Or perhaps you just drink the coolaid?

I've attached scans of the problem and its referenced figure.

 

[KhalDirth]

I suspect it is flawed. As I said complex arithmetic is not always intuitive. (I once stumped my 12th grade math teacher with this one. 1 = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1i * 1i = -1) What was the name and author of the text?

Also, sqrt(-1*-1) = sqrt(-1) *sqrt(-1) ?

Is this some sort of mathematical weirdness? My calculator certainly doesn't like the idea
if I subtract one from the other, I get -2 or 2 (depending on which way I subtract). I must also be stumped.

 

[skeptic2]

Amazon lists the book for $137.75. There are three reviews of the book.


1 of 1 people found the following review helpful:
2.0 out of 5 stars Not for learning purposes, February 22, 2007
By the sultan - See all my reviews
This review is from: Electric Circuit Analysis (Hardcover)
This book skims over topics without a good insight over what is being discussed. The reading is quite difficult to grasp without a sufficient background in mathematics, like trigonometry or linear algebra. I found numerous mistakes and errors when doing problems, and the context of the book does not prepare the reader for some of the more complex problems that are given at the end of the chapter. The questions on the back do not start off with simple problems, and are usually found within the range of medium to very hard. I recommend buying another book, and coming back to this one just to do the problems.
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1 of 1 people found the following review helpful:
1.0 out of 5 stars this book is awful, December 12, 2006
By ShorD143 (TX) - See all my reviews
This review is from: Electric Circuit Analysis, Student Problem Set with Solutions (Paperback)
Imagine a scenario where you're completely confused in a class and you go to the book for help, only to realize that the book might have just confused you even more. Then you look at the examples to find that those examples given are way easier than any of the practice problems at the end of the section, which means the book is of no help whatsoever. I hate this book.
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2 of 3 people found the following review helpful:
2.0 out of 5 stars Poorly written., May 25, 2006
By Steve K. (California, USA) - See all my reviews
This review is from: Electric Circuit Analysis (Hardcover)
I (unfortunately) used this book in my first three undergraduate electrial engineering classes. This book should only be used as a reference for someone who is throughly familiar with the material. The basic concepts are poorly explained. Instead of writing thoughtful, well conceived explanations (or even using practical example of real world applications) the authors chose to use strictly mathmatical solutions as the explanations themselves. Although the reader may completely understand the math, the basic understanding (and use) of the concepts will remain unclear. Electric Circuits by Nilsson is considerably better written.

PS Many of my peers agree with this "tame" opinion.

 

[KhalDirth]

Also, sqrt(-1*-1) = sqrt(-1) *sqrt(-1) ?

Is this some sort of mathematical weirdness? My calculator certainly doesn't like the idea
if I subtract one from the other, I get -2 or 2 (depending on which way I subtract). I must also be stumped.

Ahh, I see. Imaginary numbers violate the properties of the square root.

 

[KhalDirth]

Amazon lists the book for $137.75. There are three reviews of the book.


1 of 1 people found the following review helpful:
2.0 out of 5 stars Not for learning purposes, February 22, 2007
By the sultan - See all my reviews
This review is from: Electric Circuit Analysis (Hardcover)
This book skims over topics without a good insight over what is being discussed. The reading is quite difficult to grasp without a sufficient background in mathematics, like trigonometry or linear algebra. I found numerous mistakes and errors when doing problems, and the context of the book does not prepare the reader for some of the more complex problems that are given at the end of the chapter. The questions on the back do not start off with simple problems, and are usually found within the range of medium to very hard. I recommend buying another book, and coming back to this one just to do the problems.
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Was this review helpful to you? Yes No


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Comment Comment



1 of 1 people found the following review helpful:
1.0 out of 5 stars this book is awful, December 12, 2006
By ShorD143 (TX) - See all my reviews
This review is from: Electric Circuit Analysis, Student Problem Set with Solutions (Paperback)
Imagine a scenario where you're completely confused in a class and you go to the book for help, only to realize that the book might have just confused you even more. Then you look at the examples to find that those examples given are way easier than any of the practice problems at the end of the section, which means the book is of no help whatsoever. I hate this book.
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2 of 3 people found the following review helpful:
2.0 out of 5 stars Poorly written., May 25, 2006
By Steve K. (California, USA) - See all my reviews
This review is from: Electric Circuit Analysis (Hardcover)
I (unfortunately) used this book in my first three undergraduate electrial engineering classes. This book should only be used as a reference for someone who is throughly familiar with the material. The basic concepts are poorly explained. Instead of writing thoughtful, well conceived explanations (or even using practical example of real world applications) the authors chose to use strictly mathmatical solutions as the explanations themselves. Although the reader may completely understand the math, the basic understanding (and use) of the concepts will remain unclear. Electric Circuits by Nilsson is considerably better written.

PS Many of my peers agree with this "tame" opinion.

I never liked it much myself. My professor swore by it. Said he looked at all the other books around. He was a very sharp guy, so I just took his word for it. Do you teach, as well?

 

[skeptic2]

No, I'm an RF engineer.

 

[uart]

I guess, reading the problem now, the authors never did say that they meant to maintain the real impedance. I just think it would be crazy to correct the pf of a circuit and totally change the amount of power it will actually absorb (or at least, not take into account that it would change). Is this normal practice for pf correction?

No, you've got it completely the wrong way around. You're the one who wants to change the real power!

Initially you've got a resistance and (say) some series inductive reactance, if you were to correct it by keeping the real component (resistance) constant and just cancelling out the reactance then you'd get a larger current (through the same resistance value) and therefore much more power. No it's not normally what you'd want, but that's what you've been wanting to do for two pages in this thread and no ones been able to talk you out of it.

If on the other hand you correct the PF with a parallel capacitor then as you've discovered the real component of the impedance increases, but the power remains unchanged (assuming of course that it's voltage driven). The power must stay constant as the capacitor can only change the reactive power, not the real power.