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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.

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J. McCalley

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  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. 3

  4. 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 θ. • 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 θ. Machine model 4

  5. 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 θ is cosinusoidal. • Because θ 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… 5

  6. 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. Machine model ωm θm 6

  7. Summarizing…. Machine model 7

  8. Combining…. Machine model It is here that we observe a difficulty – that the stator-rotor and rotor-stator terms, Lsr and Lrs, because they are functions of θr, and thus functions of time, will also need to be differentiated. Therefore differentiation of fluxes results in expressions like The differentiation with respect to L, dL/dt, will result in time-varying coefficients on the currents. This will make our set of state equations difficult to solve. 8

  9. 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 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 (a space vector!). The picture below illustrates for the a-phase. 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. 9

  10. We have transformed 3 variables ia, ib, and ic into two variables id and iq, as we did in the α-β transformation. This yields an undetermined 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. • 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 10

  11. 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. Park’s transformation uses a frame of reference on the rotor. In Parks 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. See http://www.nap.edu/openbook.php?record_id=5427&page=175 for an interesting biography on Park, written by Charles Concordia. Robert H. Park, 1902-1994 11

  12. 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). 12

  13. 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. 13

  14. Choosing constants k0, kq, and kd to be 1/3, 2/3, and 2/3, respectively, results in Transformation The inverse transformation becomes: 14

  15. 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. 15

  16. 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: 16

  17. This results in Example 17

  18. 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). Composite of other 3 figures Resolution of fcs=-sint into directions of fqs and fds for t=π/3 (θ=π/4). 18

  19. The d-q transformation and its inverse transformation is given below. Inverse transformation It should be the case that KsKs-1=I, where I is the 3x3 identity matrix, i.e., 19

  20. 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 20

  21. Our d-q transformation is as follows: But, what, exactly, is θ? θ can be observed in the below figure 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. 21

  22. 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, 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. • 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 space vector, in blue, which is coincident with the rotor a-axis. Observe that we may decompose it in the q-d reference frame only by using 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 22

  23. We compare our two transformations below. Rotor winding transformation, Kr Stator winding transformation, Ks Rotor circuit transformation We now augment our notation to distinguish between q-d-0 quantities from the stator and q-d-0 quantities from the rotor: 23

  24. Recall our voltage equations: Transforming voltage equations Let’s apply our d-q transformation to it…. 24

  25. Transforming voltage equations Let’s rewrite it in compact notation Now multiply through by our transformation matrices. Be careful with dimensionality. 25

  26. Therefore: the voltage equation becomes Transforming voltage equations – Term 1 26

  27. What to do with the abc currents? We need q-d-0 currents! Transforming voltage equations – Term 2 Recall: 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…. 27

  28. Therefore: the voltage equation becomes Transforming voltage equations – Term 2 28

  29. Term 3 is: Transforming voltage equations – Term 3 Focusing on just the stator quantities, consider: Differentiate both sides Solve for Use λabcs =K-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… 29

  30. Transforming voltage equations – Term 3 30

  31. Now let’s express the fluxes in terms of currents by recalling that Transforming voltage equations – Term 3 and the flux-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: 31

  32. Perform the first matrix multiplication: Transforming voltage equations – Term 3 and the next matrix multiplication: 32

  33. 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 And since our inductance matrix is constant, we can write: Substitute the above expression for flux derivatives into our voltage equation: 33

  34. Substitute the above expressions for flux & flux derivatives into our voltage equation: Transforming voltage equations – Term 3 We still have the last term to obtain. To get this, we need to do two things. Express individual q- and d- terms of λqd0s and λqd0r in terms of currents. Obtain and 34

  35. Express individual q- and d- terms of λqd0s and λqd0r in terms of currents: Transforming voltage equations – Term 3 From the above, we observe: 35

  36. Obtain and To get , we must consider: Transforming voltage equations – Term 3 Therefore: Likewise, to get , we must consider: Therefore: 36

  37. Obtain Transforming voltage equations – Term 3 Obtain Substitute into voltage equations… 37

  38. Substitute into voltage equations… Transforming voltage equations – Term 3 This results in: Note the “Speed voltages” in the first, second, fourth, and fifth equations. -ωλds ωλqs -(ω- ωm)λdr (ω- ωm) λqs 38

  39. 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 39

  40. Transforming voltage equations – Term 3 Substitute the matrices into voltage equation and then expand. This results in: Let’s collapse the last matrix-vector product by performing the multiplication…. 40

  41. Transforming voltage equations – Term 3 Results In  From slide 35, we have the fluxes expressed as a function of currents  And then substitute these terms in: 41

  42. Transforming voltage equations – Term 3 Observe that the four non-zero elements in the last vector are multiplied by two currents from the current vector which multiplies the resistance matrix. 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. 42

  43. Transforming voltage equations – Term 3 Notice that the resistance matrix and the last matrix multiply the same 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…. 43

  44. Final Model This is the complete transformed electric machine state-space model in “current form.” 44

  45. ids and iqs are currents in a fictitious pair of windings fixed on a synchronously rotating reference frame. • These currents produce the same flux as do the stator a,b,c currents. • For balanced steady-state operating conditions, we can use iqd0s = Ksiabcs to show that the currents in the d and q windings are dc! The implication of this is that: • The a,b,c currents fixed in space (on the stator), varying in time produce the same synchronously rotating magnetic field as • The ds,qs currents, varying in space at synchronous speed, fixed in time! Some comments about the transformation • idr and iqr are currents in a fictitious pair of windings fixed on a synchronously rotating reference frame. • These currents produce the same flux as do the rotor a,b,c currents. • For balanced steady-state operating conditions, we can use iqd0r = Kriabcr to show that the currents in the d and q windings are dc! The implication of this is that: • The a,b,c currents varying in space at slip speed sωs=(ωs- ωm) fixed on the rotor, varying in time produce the same synchronously rotating magnetic field as • The dr,qr currents, varying in space at synchronous speed, fixed in time! 45

  46. The electromagnetic torque of the DFIG may be evaluated according to where Wc is the co-energy of the coupling fields associated with the various windings. We are not considering saturation here, assuming the flux-current relations are linear, in which case the co-energy Wc of the coupling field equals its energy, Wf, so that: Torque in abc quantities We use electric rad/sec by substituting ϴm=θm/p where p is the number of pole pairs. • The stored energy is the sum of • The self inductances (less leakage) of each winding times one-half the square of its current and • All mutual inductances, each times the currents in the two windings coupled by the mutual inductance • Observe that the energy stored in the leakage inductances is not a part of the energy stored in the coupling field. Consider the abc inductance matrices given in slide 6. 46

  47. The stored energy is given by: Torque in abc quantities Applying the torque-energy relation to the above, and observing that dependence on θm only occurs in the middle term, we get So that But only Lsr depend on θm, so 47

  48. We may go through some analytical effort to show that the above evaluates to Torque in abc quantities Negative value for generation To complete our abc model we relate torque to rotor speed according to: J is inertia of the rotor in kg-m2 or joules-sec2 Inertial torque Mech torque (has negative value for generation) 48

  49. However, our real need is to express the torque in qd0 quantities so that we may complete our qd0 model. To this end, recall that we may write the abc quantities in terms of the qd0 quantities using our inverse transformation, according to: Torque in qd0 quantities Substitute the above into our torque expression: 49

  50. Torque in qd0 quantities I will not go through this differentiation but instead provide the result: 50

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