Why the formula P = VI COS φ is wrong.

Size: px

Start display at page:

Download "Why the formula P = VI COS φ is wrong."

Tobias O’Connor’
6 years ago
Views:

1 Why the formula P = VI COS φ is wrong. Note: his whole document deals with single-phase AC systems. For DC systems, there is no such concern as active or reactive power. Systems in transitions are borderline to those cases. Proves developed here stay true for 3-phases systems. For explanations purposes, we often consider being in the case of an ideal voltage source which is typically what is assumed for any system connected to the grid. Summary and Overview Objectives We all have learnt the importance of the formula P = VI COS φ, used to calculate the active power in AC systems. I am going to show you in this document that this formula is wrong.. Well, to be fair, it is right in the case of a pure sinusoidal system. However, in the real world, we seldom have a such a pure sinusoidal system. here is then another factor to take in account: harmonic distortion. So, I am going to explain you why, and what happens in non-pure sinusoidal AC systems. On the way, I will uncover for you the one big mystery of electricity: why reactive power is called reactive. More importantly, I will give you some hints of what it implies in the practical cases of electrical engineering, and how to use it. Please spare me the maths If you are not concerned with all the mathematical and physical demonstrations and ready to trust me with them, you may simply focus on reading the paragraphs highlighted in light blue (such as this one). In particular, you may go directly to the following paragraphs: Reactive power why is it called 'reactive'? You can ignore the equations and just read the explanation about the name 'reactive'. Implications for real systems page 1 / 0

2 able of Contents Summary and Overview...1 Objectives...1 Please spare me the maths...1 Origin of the formula P = VI COS φ...3 Back to the basis...3 Instantaneous power...3 Active power...3 Root mean square values...3 Apparent power...4 Some maths...4 basic trigonometry...4 Integrals of sinusoidal functions...5 General case of AC systems...5 RMS values of voltage and current...5 Apparent power...6 Active power...6 Reactive power why is it called 'reactive'?...8 So, what is wrong?...9 Pure sinusoidal signals vs. real life...9 Back to maths and trigonometry - Fourrier decomposition...9 Power factor...13 Definition of PF...13 Distorting power...13 Implications for real systems...14 Harmonics are bad...14 HD...14 How to measure the powers in your system?...14 Fundamental harmonic current...14 Active power...14 Reactive power...14 Distortion power...15 Apparent power...15 Power factor...15 Example - How bad is my CFL bad doctor?...15 How to get rid of harmonics...15 Passive Power Factor Correctors (PFC)...15 Active PFC page / 0

3 Origin of the formula P = VI COS φ Back to the basis Let me remind here some basic definitions, valid for all AC systems, sinusoidal or not. Instantaneous power It is a time defined function as the product of instantaneous voltage and current. p t =v t. i t (1) he instantaneous power is the power actually available at a given instant for a direct application (i.e.: mechanical work). Active power he active power is defined as the average of instantaneous power over a period. P= 1 p t dt () Which gives: where is a period of time (a cycle) P= 1 v t.i t dt (3) hus, in simple words, the active power is the usable power, available in average over a period of time. Root mean square values Root mean square (RMS) values were introduced as a mean of comparing DC and AC systems. For the voltage and current, they are: V = 1 I = 1 v t dt (4) i t dt (5) Which is simply a way of giving an average value voltage and current when they are varying constantly. page 3 / 0

4 Apparent power he apparent power is defined simply as the product of root mean square values of voltage and current: S=V. I (6) S is called apparent because it is the product those values. If you have high voltage and high current, you would expect to have high power, would you not? Moreover, for the person who builds the grid, he must size his equipments not according to active power, but to the RMS voltage and current, i.e. to the apparent power. Some maths basic trigonometry Let's now define our v(t) and i(t), as simple sinusoidal functions with a that famous φ phase difference, and see what comes out when we are writing again the equations: v t =V max cos t (7) i t =I max cos t page 4 / 0

5 I am sorry to inform you, some basic trigonometry is required to go further. Let us start with the basic: cos a.cosb= cos a b cos a b In the particular case of (a=b = ωt), this becomes Integrals of sinusoidal functions cos 1 cos ωt ωt= o calculate some RMS and power values, we will facing integrals of products of sinusoidal functions. he following results will therefore be useful later on. (9) (8) Let us calculate a value δ, integral of sinusoidal functions. = cos k 1 t 1. cos k t dt Using (8), we obtain: = 1 cos k 1 k t 1 cos k 1 k t 1 dt = 1 [ cos k 1 k t 1 dt cos k 1 k t 1 dt ] Now, the integral of any pure sinusoidal function over its period is null: cos k t dt=0 (10) for any integer value of k, except 0, if is the primitive period of this function ( ω= π ). So, if k 1 and k are positive integers (and one of them not null), the left term of the integral will always have a null result, while the right term will also be null, except in the case k 1 = k, in which =.cos 1 And in the case 1 =, this brings: = (11) page 5 / 0

6 General case of AC systems RMS values of voltage and current In the case of a sinusoidal signal (7), from the little trigonometric formula we had above (9), if we apply it to the definition of RMS value of voltage (4), we obtain: = V V max 1 cos t As the integral of the sinusoidal term over the period is null, as explained previously in (10), this becomes: dt = V V max Which is what you already knew, did you not? dt=v max (11) Of course, a similar result applies to the current: I = I max Apparent power And then the apparent power becomes: S=V. I= V max. I max (1) Active power If we now use the same trick of trigonometry to the definition of active power, we get: P= 1 I max cos t.v max cos t dt (13) P= V max. I max cos t cos dt Again, the term cos t. will have a value of 0 when integrated over, as explained in (10). We can ignore it, and that leaves us with the second term: P= V max. I max.. cos = V. I.cos max max P=S.cos =V. I. cos (14) page 6 / 0

7 Case of a bad COS Φ (example: induction oven) the instantaneous power is sometimes negative and thus the active power is quite low (around 50 W) Case of a good COS Φ (example: toaster) for the same voltage and current as previously, the instantaneous power is never negative and the active power is over 100 W. page 7 / 0

8 Reactive power why is it called 'reactive'? We just described what is, and calculated active power. Now, let us try to describe the concept of reactive power? In a common words definition, to react is the same as to act, but with a short delay - immediately after an initial triggering event finished. Since in most systems, we deal with ideal voltage source, the current is a consequence of the voltage. he current i acts according to what the voltage v commands at a given instant t: i t = f v t (where the function f will be defined by Ohm's Law) If you imagine now that the current reacts to what the voltage commands, it would mean it acts according to what the voltage commands, but with that delay d. i t d = f v t We said that the value of this delay is the amount of time so that the initial triggering event finish, i.e. enough time so it fade down from its paramount (peak value) to nothing (zero). In this case, this (re-)active power of this system - we choose to write it Q instead of P- would be: Q= 1 v t.i t d dt= 1 V max cos t. I max cos t d dt (15) Q=V. I. cos d In AC systems, for a fundamental sinusoidal signal, is the peak value is 1 (COS 0 = 1) and the following 'nothing' is ( cos =0 ). he delay d is then And, as you probably know: cos d =cos =sin Which brings us the the more traditional formula: Q=V. I. sin =S. sin (16) And as cos x sin x=1, we also get the well-known: Q =S P (17) page 8 / 0

9 So, what is wrong? Do you remember the hypothesis we added in the last section? We actually said that the voltage and the current were sinusoidal functions of time. Pure sinusoidal signals vs. real life Have you ever had a look at an oscilloscope connected to any CFL bulb during any lab work? If yes, then, you know it is not always true. Let's be optimistic and say our system have an ideal voltage source. his is often the assumption made when the power source is the national grid. It is also the assumption made when the the power source is in 'grid-forming' mode. We have then a sinusoidal voltage. v t =V max cos t However, the current is then given by the properties of the loads connected to it (via Ohm's Law), and there is nothing to guarantee that it be sinusoidal! Example of signals for an AC load in 'real life'. page 9 / 0

10 Back to maths and trigonometry - Fourrier decomposition If the load is a passive time-independent dipole, the current, if not sinusoidal, is still a timeperiodic function with the same period as the voltage. However, thanks to Joseph Fourier, we know that the function i(t) can be expressed with a negligible approximation as a sum of sinusoidal functions: We may notice: with i t = I k. cos k t (18) k =0 I k = I k max hat I 1 defines the fundamental harmonic. the RMS value of the kth harmonic. hat the different I k (for k>1) define all the other harmonics. hat I 0 is normally equals to 0, as the signal is symmetrical between positive and negative values (by definition of alternate current). Otherwise, it means the current also holds a DC component. We will consider I 0 null for the rest of the study. hat the value of φ is defined by the phase difference between the fundamental sinusoid and the voltage. page 10 / 0

11 It is then interesting to have a look at the RMS value of current : Which develops into: I = I = 1 k I = k k ' k. I k cos k t dt, I k cos k t. k ' I k ' cos k ' t dt I k. I k ' cos k t.cos k ' t dt From the proof of (11), we know that all the integral terms will be null, except when k=k', in which case it is worth /: I = k I k I = k=0 I k (19) page 11 / 0

12 Now, where this makes a difference, is in the calculation of active power (14) using the trigonometric formulas. Indeed, if we take a closer look, and separate in different integrals the different harmonics: P= 1 v t.i t dt= V max cos t. I k. cos k t dt k=0 P=V max I k cos t.cos k t dt (0) k=0 o have a clear understanding of what this formula means, it is more simple to depict the system as a sum of different parallel branches, with the current I k of a single harmonic passing through each one of them. You will notice that they all have the same voltage V. Each one of those branches has an active power P k, a reactive power Q k, and an apparent power S k. As you already know, in the case of parallel (or series) association, we have: P= P k (1) k=0 page 1 / 0

13 And of course, the same may be said from reactive power: Q= Q k (). However, please mind that this is not true of apparent power. Using again the formulas of trigonometry, we may look for the values of P k, Q k, and S k. If we use once more (8), we find : k=0 P k =V max. I k cos t.cos k t dt P k =V max. I cos k 1 t cos k 1 t k dt (3) For k=1, we find the usual result of P 1 = V max. I.cos =V. I 1 1.cos (4) and Q 1 =V. I 1.sin (5) For all other branches (k>1), the two terms of the integral are both null (as already explained in (10)), so P k and Q k are null! In other words, only the fundamental of the current signal carries active and reactive power! Regarding S k, from (3) (4) and (6), we already know that the apparent powers are the usual: similarly to what was found in (1). S k =V. I k (6) I will now bring your attention to the fact that since I I 1 (19, we end up with P=P 1 =V. I 1 cos V. I cos. page 13 / 0

14 Power factor Definition of PF he power factor was defined to take the place of the now not-adequate-any more COS φ : PF = P S Distorting power When we finished the Fourrier decomposition, the clever ones of you will have noticed that, though, for k>1, P k and Q k are null, S k may well not be null! Which implies that we are facing there another new type of power. he power of these harmonics is called distorting power, because this power is not active nor reactive and has for sole effect to distort the nice sinusoidal signal we would have if it was not there. It is noted with a D. For the fundamental harmonic, D 1 is null, and for the other branches (harmonics with k>1), we have: And for the whole system, we have: P=P 1 =V. I 1. cos Q=Q 1 =V. I 1.sin D=V. I k k 1 S = P Q D S k =D k =V. I k page 14 / 0

15 Implications for real systems Harmonics are bad Now, you should understand that harmonics are bad: they rise the amount of apparent power required to feed your system with a determined amount of active (or reactive) power. As a higher apparent power implies higher nominal currents in your power lines, it means more losses in transmissions, as well as having to size bigger your protections. In addition, harmonics have frequencies n times the frequency of the grid, which means different from for what the grid was designed. HD he otal Harmonic Distortion is a value which was introduced to summarize in a single figure the amount of distortion there is the signal. It may actually be either calculated for any periodic signal (i.e.: the voltage or the current). In our case, since we assumed we had an ideal sinusoidal voltage source, it should be for the current: HD i = I k k 1 k 1 I = k 1 k I I k = HD 1 I 1 i (7) I Where I is the root mean square value of the signal (19). he HD of a system must be controlled (different norms apply to different countries). How to measure the powers in your system? Well, in the end, what you actually want as an engineer facing a real system is not complicated equations, but an easy way to assess your system's performances. What you need to measure with your tools, are V, I, the COS φ, and the HDi. Fundamental harmonic current I 1 =I. 1 HD i Active power P=V. I 1 cos page 15 / 0

16 Reactive power Q=V. I 1 sin Distortion power D=S. HD i Apparent power S=V. I Power factor PF = P S = 1 HD i. cos Example - How bad is my CFL doctor? Let us take again the example of a CFL bulb light - a very bad one whom we know has plenty of dirty harmonics all over the place, as well as a bad COS φ - and compare it to a classical incandescent bulb light. he only data available is what we have from the catalogue: the rated active power, the COS φ, and the HDi. he following table shows how to proceed to find the different values. Value Incandescent CFL How to find it? Rated active power (W) Given Voltage ( V AC) Given COS φ 1 0,8 Given HDi 0% 0% Given I 1 Current (A) fundamental harmonic 0,43 0,11 = rated active power / (voltage x cos φ) Current (A) 0,43 0,11 = I1 / (1 HDi^)^0,5 Apparent power (VA) 100 5,5 = voltage x current Reactive power (VAr) 0 15 = Voltage x I 1 x SIN φ Distortion power (VA) 0 5,1 = S. ('HDi') Power factor 1 0,78 = (1-HDi^)^0,5 x cos φ How to get rid of harmonics You already know how to compensate your reactive power. So now, what can you do when you have those dirty harmonics? Use filters! page 16 / 0

17 Passive Power Factor Correctors (PFC) hose are electronics filters. hey are designed to pass only a chosen frequency (i.e. your fundamental harmonic, 50 or 60 Hz). All other harmonics will be damped. Ii is effective, but relies on the use of capacitors and inductances, which can be bulky and expensive. Active PFC hose are filters based on power electronics (i.e. with transistors). hey are more complex, smaller and often cheaper. heir principle of functioning is that they create opposite harmonics to compensate the harmonics of the system. page 17 / 0

18 One step further voltage harmonics So far, we had still assumed that the voltage was a pure sinusoidal signal. hough unusual, we may take the analysis one step further and have a glance at what happens when we have a non-sinusoidal AC signal. Resolution of the voltage signal Exactly like we resolved the current into a sum of sinusoidal currents, the same can be done with voltage, except the association will be in series and not in parallel. But of course, we did not drop the assumption that the voltage was not sinusoidal only to add back the assumption the current was. his means the resolution of current still holds also true, and if we combined them together, we have something like: page 18 / 0

19 Note that the same current I n goes through Z 1n, Z n, Z 3n, etc.. Power of voltage harmonic n If we refer to formula (3), we find that the active power in dipole Z kn, though which goes current I k and submitted to voltage v n is: Which is resolved similarly P kn = V n. I k max cos k n t cos k n t dt V For k=n, we find the usual result of max P kk = k. I.cos =V. I.cos k k k and Q kk =V k. I k. sin For all other branches ( k n ), the two terms of the integral are both null. P kn and Q kn are therefore null and all the apparent power is distorting power. page 19 / 0

20 Power of the all system Now, if you not only consider harmonic n, but the all system, you will find that every harmonic which is present in both the voltage and the current potentially carries active power. However, distorting power is also much higher. Acknowledgements Julien Herbreteau for reading and remarks. page 0 / 0

Lecture 05 Power in AC circuit

CA2627 Building Science Lecture 05 Power in AC circuit Instructor: Jiayu Chen Ph.D. Announcement 1. Makeup Midterm 2. Midterm grade Grade 25 20 15 10 5 0 10 15 20 25 30 35 40 Grade Jiayu Chen, Ph.D. 2