# power analysis concepts etc.

Discussion in 'Mathematics and Physics' started by PG1995, Sep 3, 2017.

1. ### PG1995Active Member

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Hi

It was written that P=Po(V/Vo)^alpha and Q=Qo(V/Vo)^beta. Later, it was said that if alpha=beta=2 then you get a constant impedance, and also if alpha=beta=1 then you get a constant load. I believe that 'Q' stands for reactive power and 'P' for real power.

I couldn't make any sense that how the two formulas above were reached at and how and why alpha=beta=2 results into a constant impedance and so on. Could you please help? Thank you.

2. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

It may help if in addition to P and Q you can explain what Po and Qo and V and Vo are.

c=a/b

you cant tell me too much about that other than a/b is a ratio, but if i state it like this:
R=V/I

then you might assume it is like Ohm's Law. If you dont know what a,b,c are though you cant understand it as well.

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3. ### steveBWell-Known MemberMost Helpful Member

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Hi PG,

Welcome back. I think you are correct that the P and Q are real and reactive power. Depending on the type of load (constant impedance, constant power, constant current etc), the P and Q formulas will be different. The alpha and beta seem to be a way to parameterize the formulas in terms of load type.

My recommendation is to derive formulas for P and Q for constant impedance, constant current and constant voltage loads. Then see if the resulting formulas match the form given by your instructor.

As an example lets do constant resistance, which is not quite the same as constant impedance.

Z=R+jX with X=0, hence Z=R

S=Vrms*Irms*cos(phi)=Vrms*Vrms/R =P +jQ

Hence P=V^2/r and Q=0 which look like they can fit the formulas you showed for constant impedance (alpha=beta=2)

Then if you let Po=Vo^2/R and Qo=0 you will get the correct formulas P=Po(V/Vo)^2 and Q=Qo(V/Vo)^2

Note that Vo can be an arbitrary voltage level, or a nominal value, or a rated value or even the peak voltage of the sine wave.

I verified that the procedure will work for X not equal to zero, but I leave it you to derive it. Then you have to figure out what alpha=beta=1 corresponds to. You say it is "constant load", but I don't know precisely what that means. Does that mean constant current? or constant power? or constant voltage etc?

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5. ### PG1995Active Member

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It means that constant current magnitude.

Po and Qo are the rated consumption when when V=Vo; Vo is nominal voltage.

I just wanted to let you know. I'll get back to it soon. Thank you.

Last edited: Sep 7, 2017
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6. ### PG1995Active Member

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So the formulae seem to be okay but one thing surprises me that why they are not mentioned anywhere. I tried to search them everywhere but couldn't find. They were just made up...?

I proved it for Z=R+jX although my head started spinning after seeing all that math and formulas again! Please see here.

I believe that the purpose of the formulae, at least the case α=β=2, is to be able to find the power for any other voltage_rms value other than nominal or rated voltage_rms value in a rather straightforward manner.

Does α=β=0 signify the case of constant power?

I'll try to prove α=β=1 case later today. Thank you.

Types of loads: constant current, constant power, constant impedance, constant voltage

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7. ### PG1995Active Member

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This is an addition to my previous post.

I tried to solve for alpha=beta=1 case without any success. I'm attaching the so-called solution which doesn't make any sense to me. Please have a look and guide me how to proceed in logical manner. Thank you for your help.

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8. ### steveBWell-Known MemberMost Helpful Member

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I'm not sure why your professor introduced those equations which are hard to find. Engineers often derive and use formulas that they did not look up. They could be formulas that are known in research journals that never made it into text books, or they may just be formulas that the professor decided were useful.

Maybe an important comment for me to make is that as engineers/scientist transition from students to professionals, they often use math/formulas/derivations like people use verbal or written language. The math provides a true language and can express various ideas. As an analogy, imagine that a music student goes to his professor and says, " Hey those musical phrases you told me to play and analyze are bothering me because i cant find them anywhere in any known composition. Beethoven didn't write them ... nor did Mozart. Where did they come from?" That music professor would be wondering why you don't know that music is an expressive language, capable of expressing original brand new ideas.

I think the R^2-X^2 should be R^2+X^2.

That seems reasonable. There can be many purposes, both intended and unintended. Personally, what I find interesting about the formula is that it is generalized with alpha and beta, thus allowing it to express different load types.

It would seem so.

Last edited: Sep 15, 2017
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9. ### steveBWell-Known MemberMost Helpful Member

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I'm not following your work, but it appears you took a very complicated method for what should be simple. In your previous work, the second line shows power as a function of voltage current and phase. If you look at that formula and assume current and phase is constant with only voltage varying, then it is obvious that power is proportional to voltage, which means alpha=beta=1.

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10. ### PG1995Active Member

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Thank you.

I had already corrected the mistake but forgot to fix it here.

I was able to found the formulae in few online books. I could put the formulae in proper context now and at the same I have few questions. I'll do it today.

It wasn't me. It was the instructor. By the way, I couldn't even understand his approach to solve the case α=β=2. Please see the attachment. Thank you.

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11. ### PG1995Active Member

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Let's focus on the exponential load model.

P=Po(V/Vo)^α
Q=Qo(V/Vo)^β

where Vo is the reference voltage, and the exponents α and β depend on the type of load (motor, heating, lighting, etc.). Note that Po and Qo are the active and reactive powers consumed under a voltage V equal to the reference Vo and relate to the amount of connected equipment. These have been called nominal load powers in contrast to the consumed powers P and Q under a voltage V equal to the reference V.

The following are the special cases:

Now I'm going to jump to something which had been bothering me. For example, for the case α=β=2 we wanted to reach the conclusion that somehow tells us directly or indirectly that the impedance Z does not vary with voltage.

I don't think that the result reached by us, P=Vrms²R/(R²+X²)) and Q=Vrms²X/(R²+X²), shows that Z does not vary with voltage; or, I might be missing something. On the other hand, the result reached by the instructor clearly shows that Z doesn't vary with the voltage; its value remains the same as it was for nominal reference voltage value Vo. The same goes for his result about the case α=β=2 which clearly shows that the current doesn't change as the voltage V varies from the reference voltage Vo.

By the way, did you want me to proceed like this for the constant current load case?

Thank you so much for your help!

Smart Grid Handbook, 3 Volume Set, Volume 1
edited by Chen-Ching Liu, Stephen McArthur, Seung-Jae Lee
Page 474

Voltage Stability of Electric Power Systems
By Thierry van Cutsem, Costas Vournas
Page 94

Power System Dynamics and Stability
By Jan Machowski, Janusz Bialek, James Richard Bumby
page 81

Smart Grids: Opportunities, Developments, and Trends
edited by A B M Shawkat Ali
page 6

pg 10 and 12:

6: https://cdn.selinc.com/assets/Literature/Publications/Technical Papers/6558_ExplorationDynamic_DJ_20120223_Web.pdf?v=20170306-110906
pg 2

7: http://www.iea.lth.se/publications/Theses/LTH-IEA-1034.pdf
pg 48

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12. ### steveBWell-Known MemberMost Helpful Member

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PG,

This information is very interesting. I never saw such a nice summary of load behavior for commonly used devices and equipment.

As far as how you proceed in deriving the constant current case, I think it is basically correct and showing the result that exponents of one correspond to constant current. But, I think the presentation is a little confusing. In a nutshell you are trying to show that the power is proportional to voltage, and that you can define a constant of proportionality in terms of nominal values of power and voltage.

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13. ### PG1995Active Member

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Thank you.

It's clear that as an impedance changes, "R" and "X" would change too. Do they change in proportion? For example, if R becomes doubled and X also becomes doubled then the impedance angle won't change. But I don't think that R and X need to change in proportion therefore as the magnitude of impedance changes so does the angle. Please have a look on the attachment. Thanks.

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14. ### PG1995Active Member

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This an addition to the previous post and I'm just trying to rephrase my question.

We need to keep the current, I, same thru an impedance Z with varying voltage Vrms. Obviously, to keep the current same, the impedance needs to dynamically vary. An impedance is Z=Z∠θ where Z is magnitude and θ is an impedance angle equal to θv-θi. So, how does the impedance vary? Does only its magnitude change or both magnitude and angle? Could both magnitude and angle change in such a way that the current remains the same? The context of the question is the formulae P=Po(V/Vo)^α and Q=Qo(V/Vo)^β with α=β=1. Thank you.

15. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

I have not been following this thread that closely, but if you mean the amplitude of the current has to stay constant by varying the load and the load is powered from the voltage source then we have:
sqrt(r^2+i^2)=E/I

where:
E is the magnitude of the source that may change,
I is the magnitude of the current to stay constant,
r is the real part of the impedance,
i is the imaginary part of the impedance.

Of course if you square both sides then:
r^2+i^2=(E/I)^2

16. ### PG1995Active Member

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Thank you, MrAl.

I'm not sure if it's just the amplitude or both amplitude and angle of the current. I think it might be both. You can refer to post #10 or we would need to wait for steveB to clarify this because he's been following this thread and understands its proper context.

Last edited: Sep 21, 2017
17. ### steveBWell-Known MemberMost Helpful Member

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So, my opinion is that one needs to make assumptions about the angle in some of the cases, when you are deriving. For example, the constant current case, we would say that the phase angle and the current are constant, to get to the desired formula. In general, I would think that the phase angle can change dynamically also. Honestly, I'm just guessing here and using common sense, as I've never needed to formally consider these questions.

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18. ### PG1995Active Member

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Thank you.

Are you saying that power factor angle doesn't change in the context of the formulae we were discussing? It'd also imply that only the magnitude of the impedance changes.

Last edited: Sep 21, 2017
19. ### steveBWell-Known MemberMost Helpful Member

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You have to ask yourself, what assumption did you make when you did the derivation. When I did the derivation (keep in mind I spent about 1 minute thinking about it) I felt I needed to assume that the angle was constant - not just the current magnitude.

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20. ### PG1995Active Member

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Hi,

The phase angle for current I21 is -127° . I cannot know if it's leading or lagging because this calculation was done with reference to voltage source V2 which has phase angle of 0 degree. I understand that it could be said V2 could be thought of as 585∠-180° and in this way current would lag the voltage. But I don't see the logical reason for proceeding like this. Could you please guide me? Thank you.

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