1 / 51

J. McCalley

J. McCalley. d-q transformation. Consider the DFIG as two sets of abc windings, one on the stator and one on the rotor. Machine model. ω m. θ m. 2. The voltage equation for each phase will have the form: That is, we can write them all in the following form:. Machine model.

bwinchester
Télécharger la présentation

J. McCalley

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. J. McCalley d-q transformation

  2. Consider the DFIG as two sets of abc windings, one on the stator and one on the rotor. Machine model ωm θm 2

  3. The voltage equation for each phase will have the form: That is, we can write them all in the following form: Machine model All rotor terms are given on the rotor side in these equations. We can write the flux terms as functions of the currents, via an equation for each flux of the form λ=ΣLkik, where the summation is over all six winding currents. However, we must take note that there are four kinds of terms in each summation. In considering the four kinds of terms in the flux linkage summations, we take note that induction machines are always of the smooth-rotor type, i.e., they do not have salient pole structure. Another way to say this is that they have uniform air-gap. We should also take note that the mutual inductance for each pair of windings may need some careful thought… 3

  4. Inductance, L: • L=λ/i (flux per unit ampere): More L sees more flux for a given current. • v=dλ/dt (Faraday’s Law)=L di/dt if L constant: More L creates more induced voltage • Self inductance is ability of current in a conductor to induce voltage in that conductor • Mutual inductance is ability of current in a conductor to induce voltage in another conductor. Inductance Which pair of coils has highest mutual inductance? Maximum Mutual Inductance Minimum Mutual Inductance Point: Mutual inductance depends on relative positions of coils 4

  5. Stator-stator terms: These are terms which relate a stator winding flux to a stator winding current. Because the positional relationship between any pair of stator windings does not change with rotor position, these inductances are not a function of rotor position; they are constants. • Rotor-rotor terms: These are terms which relate a rotor winding flux to a rotor winding current. As in stator-stator-terms, these are constants. • Rotor-stator terms: These are terms which relate a rotor winding flux to a stator winding current. As the rotor turns, the positional relationship between the rotor winding and the stator winding will change, and so the inductance will change. Therefore the inductance will be a function of rotor position, characterized by rotor angle θm. • Stator-rotor terms: These are terms which relate a stator winding flux to a rotor winding current. As described for the rotor-stator terms, the inductance will be a function of rotor position, characterized by rotor angle θm. Machine model 5

  6. There are two more comments to make about the flux-current relations: • Because the rotor motion is periodic, the functional dependence of each rotor-stator or stator-rotor inductance on θm is cosinusoidal. • Because θm changes with time as the rotor rotates, the inductances are functions of time. • We may now write down the flux equations for the stator and the rotor windings. Machine model Note here that all quantities are now referred to the stator. The effect of referring is straight-forward, given in the book by P. Krause, “Analysis of Electric Machinery,” 1995, IEEE Press, pp. 167-168. I will not go through it here. Each of the submatrices in the inductance matrix is a 3x3, as given on the next slide… 6

  7. - Diagonal elements are the self-inductance of each winding and include leakage plus mutual. - Off-diagonal elements are mutual inductances between windings and are negative because 120° axis offset between any pair of windings results in flux contributed by one winding to have negative component along the main axis of another winding – see below fig. - The “1/2” in off-diagonals results from definition of self and mutual inductances, e.g., see. Eqs. 1.5-16 and 1.5-21 and in Krause (pp. 52-53). - On this page, and in these notes, I have maintained distinct notation for the mutuals between stator windings, Lms, the mutuals between rotor windings Lmr, and the mutuals between stator and rotor windings Lsr. However, Krause in his book indicates they are equal, i.e., that Lms=Lmr=Lsr as a result of referring all quantities to the stator (see pp. 144-145 of his 2002 edition). I adopt this latter approach (that they are all equal) on slide 37 and also in my notes on (slide 8 of) torque and power. Machine model ωm θm -b axis 7

  8. Summarizing…. Machine model 8

  9. Combining…. Machine model It is here that we observe a difficulty – that the stator-rotor and rotor-stator submatrices, Lsr and Lrs, because they are functions of θm, and thus functions of time, will also need to be differentiated. So differentiation of fluxes results in expressions like Both terms result in time-varying coefficients on the variables (currents), and thus we end up with a set of ODEs with time varying coefficients. Such ODEs are well-known to be difficult to solve. 9

  10. θ d-axis q-axis ia a' iq id a This presents some significant difficulties, in terms of solution, that we would like to avoid. We look for a different approach. The different approach is based on the observation that our trouble comes from the inductances related to the stator-rotor mutual inductances that have time-varying inductances. In order to alleviate the trouble, we project the a-b-c currents onto a pair of axes which we will call the d and q axes or d-q axes. In making these projections, we want to obtain expressions for the components of the stator currents in phase with the d and q axes, respectively. Although we may specify the speed of these axes to be any speed that is convenient for us, we will generally specify it to be synchronous speed, ωs. Transformation One can visualize the projection by thinking of the a-b-c currents as having sinusoidal variation IN TIME along their respective axes (as we did in developing space vectors!). The picture below illustrates for the a-phase current. Decomposing the b-phase currents and the c-phase currents in the same way, and then adding them up, provides us with: Constants kq and kd are chosen so as to simplify the numerical coefficients in the generalized KVL equations we will get. 10

  11. We have transformed 3 variables ia, ib, and ic into two variables id and iq, as we did in the α-β transformation. In general, this yields an under-determined system, meaning • We can uniquely transform ia, ib, and ic to id and iq • We cannot uniquely transform id and iq to ia, ib, and ic (unless there is another constraint such as ia+ib+ic=0). • We will use as a third current the zero-sequence current: Transformation Recall our id and iq equations: We can write our transformation more compactly as 11

  12. Transformation A similar transformation resulted from the work done by Blondel (1923), Doherty and Nickle (1926), and Robert Park (1929, 1933), which is referred to as “Park’s transformation.” In 2000, Park’s 1929 paper was voted the second most important paper of the last 100 years (behind Fortescue’s paper on symmertical components). R, Park, “Two reaction theory of synchronous machines,” Transactions of the AIEE, v. 48, p. 716-730, 1929. G. Heydt, S. Venkata, and N. Balijepalli, “High impact papers in power engineering, 1900-1999, NAPS, 2000. See http://www.nap.edu/openbook.php?record_id=5427&page=175 for an interesting biography on Park, written by Charles Concordia. Park’s transformation uses a frame of reference on the rotor. In Park’s case, he derived this for a synchronous machine and so it is the same as a synchronous frame of reference. For induction motors, it is important to distinguish between a synchronous reference frame and a reference frame on the rotor. From a recent “PowerGlobe” discussion: “The real foundation of most of the synchronous machine theory talked today was laid in a paper by a French Engineer, Blondel, who was the first to propose "two reaction theory" in 1895. Then Doherty and Nickle published extensive analysis of synchronous machines using two reaction theory in a number of papers between 1923 and 1928. At the behest of Charlie Concordia (as told by Charlie himself), Park published three papers in 1928 to 1933 and organized the work of Doherty and Nickle in a matrix form and that is what is best known today in terms of Park's Transformation.  Concordia and Park were colleagues in GE at that time.” - OM Malik Robert H. Park, 1902-1994 12

  13. Transformation https://www.nap.edu/read/5427/chapter/34#175 13

  14. Transformation Here, the angle θ is given by where ɣ is a dummy variable of integration. The angular velocity ω associated with the change of variables is unspecified. It characterizes the frame of reference and may rotate at any constant or varying angular velocity or it may remain stationary. You will often hear of the “arbitrary reference frame.” The phrase “arbitrary” stems from the fact that the angular velocity of the transformation is unspecified and can be selected arbitrarily to expedite the solution of the equations or to satisfy the system constraints [Krause]. The constants k0, kq, and kd are chosen differently by different authors. One popular choice is 1/3, 2/3, and 2/3, respectively, which causes the magnitude of the d-q quantities to be equal to that of the three-phase quantities. However, it also causes a 3/2 multiplier in front of the power expression (Anderson & Fouad use k0=1/√3, kd=kq=√(2/3) to get a power invariant expression). 14

  15. The constants k0, kq, and kd are chosen differently by different authors. One popular choice is 1/3, 2/3, and 2/3, respectively, which causes the magnitude of the d-q quantities to be equal to that of the three-phase quantities. PROOF (iq equation only): Let ia=Acos(ωt); ib=Acos(ωt-120); ic=Acos(ωt-240) and substitute into iq equation: Transformation Now use trig identity: cos(u)cos(v)=(1/2)[ cos(u-v)+cos(u+v) ] Now collect terms in ωt-θ and place brackets around what is left: Observe that what is in the brackets is zero! Therefore: Observe that for 3kdA/2=A, we must have kd=2/3. 15

  16. Choosing constants k0, kq, and kd to be 1/3, 2/3, and 2/3, respectively, results in Transformation The inverse transformation becomes: 16

  17. Krause gives an insightful example in his book, where he specifies generic quantities fas, fbs, fcs to be a-b-c quantities varying with time on the stator according to: Note that these are not balanced quantities! Example The objective is to transform them into 0-d-q quantities, which he denotes as fqs, fds, f0s. 17

  18. This results in Example Now assume that θ(0)=-π/12 and ω=1 rad/sec. Evaluate the above for t= π/3 seconds. First, we need to obtain the angle θ corresponding to this time. We do that as follows: Now we can evaluate the above equations 3A-1, 3A-2, and 3A-3, as follows: 18

  19. Example 19

  20. Example Resolution of fas=cost into directions of fqs and fds for t=π/3 (θ=π/4). Resolution of fbs=t/2 into directions of fqs and fds for t=π/3 (θ=π/4). fqs=0.2357+0.0903+0.5576=0.8836 Composite of other 3 figures Resolution of fcs=-sint into directions of fqs and fds for t=π/3 (θ=π/4). fds=0.2357-0.3372-0.1494=-0.2509 20

  21. Comment Source: P. Krause, O. Wasynczuk, and S. Sudhoff, “Analysis of electric machinery and drive systems,” 2nd edition, IEEE Press, 2002, ch 3. 21

  22. The d-q transformation and its inverse transformation are given below. Inverse transformation It should be the case that KsKs-1=I, where I is the 3x3 identity matrix, i.e., 22

  23. Under balanced conditions, i0 is zero, and therefore it produces no flux at all. Under these conditions, we may write the d-q transformation as Balanced conditions 23

  24. Our d-q transformation is as follows: But, what, exactly, is θ? θ can be observed in the figure on the next slide as the angle between the rotating d-q reference frame and the a-axis, where the a-axis is fixed on the stator frame and is defined by the location of the phase-a winding. We expressed this angle analytically using Rotor circuit transformation where ω is the rotational speed of the d-q coordinate axes (and in our case, is synchronous speed). This transformation will allow us to operate on the stator circuit voltage equation and transform it to the q-d-0 coordinates. We now need to apply our transformation to the rotor a-b-c windings in order to obtain the rotor circuit voltage equation in q-d-0 coordinates. However, we must notice one thing: whereas the stator phase-a winding (and thus it’s a-axis) is fixed, the rotor phase-a winding (and thus it’s a-axis) rotates. If we apply the same transformation to the rotor, we will not account for its rotation, i.e., we will be treating it as if it were fixed. 24

  25. θ ωm d-axis q-axis θm β ω ia a' iq id a To understand how to handle this, consider the below figure where we show our familiar θ (see slide 10), the angle between the stator a-axis and the q-axis of the synchronously rotating reference frame. • We have also shown • θm, which is the angle between the stator a-axis and the rotor a-axis, and • β, which is the angle between the rotor a-axis and the q-axis of the synchronously rotating reference frame. Note β=θ-θm. • The stator a-axis is stationary, the q-d axis rotates at ω, and the rotor a-axis rotates at ωm. Rotor circuit transformation Consider the iar vector, in blue, which is coincident with the rotor a-axis. Observe that we may decompose it in the q-d reference frame, but we need to use β instead of θ. Conclusion: Use the exact same transformation, except substitute β for θ, and…. account for the fact that to the rotor windings, the q-d coordinate system appears to be moving at ω-ωm 25

  26. We compare our two transformations below. Rotor winding transformation, Kr Stator winding transformation, Ks Rotor circuit transformation We now augment our notation for the abc current vectors and for the qd0 current vectors to distinguish between quantities on the stator and quantities on the rotor: 26

  27. Basic strategy: • Write out voltage equations in terms of stator & rotor abc quantities. • Multiple through by transformation matrices Ks and Kr. • Manipulate resulting equations to see if we can write them only in terms of stator and rotor qd0 quantities, eliminating stator and rotor abc quantities. Transforming voltage equations Relations that we will heavily use in this process: Stator relations Rotor relations Transformation: Inverse Transformation: “x” can be voltages, currents, or flux linkages. 27

  28. Recall our voltage equations: Transforming voltage equations Let’s apply our d-q transformation to it…. 28

  29. Transforming voltage equations Let’s rewrite it in compact notation Now multiply through by our transformation matrices. 29

  30. Therefore: the voltage equation becomes Transforming voltage equations – Term 1 30

  31. What to do with the abc currents? We need q-d-0 currents! Transforming voltage equations – Term 2 Recall (slide 27): and substitute into above. Perform the matrix multiplication: Fact: KRK-1=R if R is diagonal having equal elements on the diagonal. Proof: KRK-1=KrUK-1=rKUK-1=rKK-1=rU=R. Therefore…. U is 3×3 identity matrix. 31

  32. Therefore: the voltage equation becomes Transforming voltage equations – Term 2 32

  33. Term 3 is: Transforming voltage equations – Term 3 Focusing on just the stator quantities, consider: Differentiate both sides Solve for Use λabcs =Ks-1λqd0s A similar process for the rotor quantities results in Substituting these last two expressions into the term 3 expression above results in Substitute this back into voltage equation… 33

  34. Transforming voltage equations – Term 3 So this is an achievement!!! We transformed both stator and rotor equations so that they are expressed in terms of qd0 quantities instead of abc quantities . One thing is left: we have currents and flux linkages as state variables. We want just one type of state variable. So… Let’s express flux linkages in terms of currents. 34

  35. Express the flux linkages in terms of currents by recalling that Transforming voltage equations – Term 3 and the flux linkage-current relations: Now write the abc currents in terms of the qd0 currents: Substitute the third equation into the second: Substitute the fourth equation into the first: 35

  36. Perform the first matrix multiplication: Transforming voltage equations – Term 3 and the next matrix multiplication: 36

  37. Now we need to go through each of these four matrix multiplications. I will here omit the details and just give the results (note also in what follows the definition of additional nomenclature for each of the four submatrices): Transforming voltage equations – Term 3 Krause indicates that LM=3/2(Lms)=3/2(Lmr), ch. 4.5, pp. 150 of his 2002 edition (I have renamed LM as “M”). I have adopted this convention here. Also, see last bullet in my comments on slide 7 of these notes. And since our inductance matrix is constant, we can write: Substitute the above expressions for flux linkage derivatives into our voltage equation: 37

  38. Substitute the above expressions for flux linkage derivatives into our voltage equation: Transforming voltage equations – Term 3 We still have the last term to obtain. To get this, we will do two things. Express individual qd0 elements of λqd0s and λqd0r in terms of qd0 currents. We will do this now but we will not actually use the results until slide 42-43 because waiting will allow us to make some interesting observations about expressions containing these terms. Obtain and 38

  39. Express individual q- and d- terms of λqd0s and λqd0r in terms of currents: Transforming voltage equations – Term 3 From the above, we observe (we will use these on slide 42-43): • We also observe • λ0s=Lσs i0s • λ0r=Lσr i0r • but we will see shortly that we will not need these. (Note that the above equations are given on p. 150 of Krause’s 2002 edition as eqs. 4.5-16, 4.5-17, 4.5-19, and 4.5-20, respectively.) 39

  40. Obtain and To get , we can obtain (via the fundamental theorem of calculus) Transforming voltage equations – Term 3 Therefore: Likewise, to get we can obtain: Therefore: 40

  41. Obtain Transforming voltage equations – Term 3 Obtain Substitute into voltage equations… 41

  42. Substitute into voltage equations… Transforming voltage equations – Term 3 Substitute the matrices into voltage equation. This results in: Note the “Speed voltages” in the first, second, fourth, and fifth equations. -ωλds ωλqs -(ω- ωm)λdr (ω- ωm) λqs 42

  43. Some comments on speed voltages: -ωλds, ωλqs, -(ω- ωm)λdr, (ω- ωm) λqs: • These speed voltages represent the fact that a rotating flux wave will create voltages in windings that are stationary relative to that flux wave. • Speed voltages are so named to contrast them from what may be called transformer voltages, which are induced as a result of a time varying magnetic field. • You may have run across the concept of “speed voltages” in Physics, where you computed a voltage induced in a coil of wire as it moved through a static magnetic field, in which case, you may have used the equation Blv where B is flux density, l is conductor length, and v is the component of the velocity of the moving conductor (or moving field) that is normal with respect to the field flux direction (or conductor). • The first speed voltage term, -ωλds, appears in the vqs equation. The second speed voltage term, ωλqs, appears in the vds equation. Thus, we see that the d-axis flux causes a speed voltage in the q-axis winding, and the q-axis flux causes a speed voltage in the d-axis winding. A similar thing is true for the rotor winding. Transforming voltage equations – Term 3 43

  44. Transforming voltage equations – Term 3 The voltage equation as we now have it: Let’s collapse the last matrix-vector product by performing the multiplication…. 44

  45. Transforming voltage equations – Term 3 Results In  Krause, on p. 151 of his 2002 edition, provides equivalent circuits that “fit” these equations, as follows: Here, LM=M, and Lls and Llr are Lσs and Lσr, respectively, and the primed notation indicates quantities referred to the stator, which we have assumed in the development of these notes. 45

  46. Transforming voltage equations – Term 3 From slide 39, we have the fluxes expressed as a function of currents  And then substitute these terms in: 46

  47. Transforming voltage equations – Term 3 Observe that the four non-zero elements in the last vector are multiplied by four currents from the current vector which multiplies the resistance matrix: ids, iqs, idr, iqr. So let’s now expand back out the last vector so that it is a product of a matrix and a current vector. Now change the sign on the last matrix and on all of its elements… 47

  48. Transforming voltage equations – Term 3 Notice that the resistance matrix and the last matrix multiply the same vector (the current vector); therefore, we can combine these two matrices. For example, element (1,2) in the last matrix will go into element (1,2) of the resistance matrix, as shown. This results in the expression on the next slide…. 48

  49. These are current derivatives. Final Model This is the complete transformed electric machine state-space model in “current form.” What is significant about it???? All coefficients are constant! So this is a set of ODEs with constant coefficients. Contrast to what we had in our model before the transformation (slide 9): 49

  50. Final Model 50

More Related