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J. McCalley

J. McCalley. Torque and Power. The electromagnetic torque of the DFIG may be evaluated according to. where W c is the co-energy of the coupling fields associated with the various windings; ϴ m is the angle in mechanical degrees between the main rotor axis and fixed reference.

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J. McCalley

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  1. J. McCalley Torque and Power

  2. 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; ϴm is the angle in mechanical degrees between the main rotor axis and fixed reference. 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 See Slides called “WE Generators – Coupling Fields” We use electric rad/sec by substituting ϴm=θm/p where p is the number of pole pairs. We learned in the above-mentioned slide-set (slide 35), that for a linear electromagnetic system with J electrical inputs (windings), the total field energy is given by: where Lpq is the winding’s self inductance when p=q and when p≠q, it is the mutual inductance between the two windings. Derivation: Given in pp 22-24 of Krause, Wasynchzuk, and Sudhoff, “Analysis of electric machinery,” 1995. 2

  3. Torque in abc quantities • 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 7 of “dq transformation” slides. * See pg. 178 of Krause. 3

  4. 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 4

  5. We may perform the above differentiation and associated matrix multiplication 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) 5

  6. The previous torque expression is a bit complicated; it is not obvious how to control abc stator and rotor currents in order to achieve a certain torque. So…let’s express torque in qd0 quantities, to see if there are any good control strategies that result. 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: 6

  7. Torque in qd0 quantities I will not go through this differentiation and matrix multiplication, but I will instead just provide the result (using M=(3/2)Lsr) which is surprisingly simple (see eq. 4.6-2 on pg. 154 of Krause 2002 edition): 7

  8. (1a) Eq. 3.16(a) on p. 59 in Wu’s book; Eq. 4.6-2 on p. 154 of Krause’s 2002 edition. By using the below expressions (see slide 39 of “dq transformations”), Torque in qd0 quantities making appropriate substitutions into (1a), and then performing algebraic manipulations, some other useful expressions may be derived from the above, as follows: (1b) Eq. 4.6-3 on p. 154 of Krause’s 2002 edition. (1c) Eq. 3.16(a) on p. 59 in Wu’s book; Eq. 4.6-4 on p. 154 of Krause’s 2002 edition (1d) Here, Wu defines on p. 56: Ls=Lls+Lm; Lr=Llr+Lm. Thus, the Lm here is what I have called Lms and Lmr (see slide 7 of notes on dq xfrmation.) I have also used M=(3/2)Lms. Thus, Lm here could be replaced by (2/3)M. Eq. 3.16(d) on p. 59 in Wu’s book. (1e) Eq. 8.56 on p. 257 in Wu’s book. 8

  9. Comment: We will work with the torque expressions on the previous slide to show that the electromagnetic torque Tem can be directly controlled by the rotor quadrature current iqr. At the same time, we will also show that the stator reactive power Qs can be directly controlled by the rotor direct-axis current idr. This will provide us the necessary means to control the rotor-side converter. As a last step before moving into the control, let’s obtain real and reactive powers in qd0 quantities. We do that next…. Torque in qd0 quantities 9

  10. We will develop power expressions using a space vector approach, taking the following steps: • Show that our space vectors have same magnitude as abc quantities. • Obtain power expressions for space vectors. • Substitute qd0 quantities into space-vector power expressions. Power in qd0 quantities (Alternative approach would be to express power in terms of abc quantities and then use transformation to convert to qd0 quantities.) 10

  11. Show that our space vectors have same magnitude as abc quantities. You will recall in our slides called “Machine Transformations,” in slide 8, we expressed our current space vector as: This space vector came about as a result of a scaling choice we made, of 2/3. Let’s see here if we can understand why we made that scaling choice. Power in qd0 quantities: step 1 Consider that the three phase currents may be expressed as: where Ip is the peak amplitude of the waveform. Substitution into the space vector expression above yields: Factor out the Ip and also multiply and divide the second term by 2 (to get it to a form recognizable as a trig identity) results in: Recalling that sinu sinv = (1/2)[cos(u-v)-cos(u+v)], we can write the above as: Taking magnitudes: Therefore, we can write: 11

  12. 2. Obtain power expressions for space vectors. Write down the familiar three-phase power expressions for P and for Q. where φ=θv-θi, and with θv=0 (the reference), φ=-θi. Change rms values to peak (Xrms=Xp/sqrt(2)); use Re, Im of rectangular expression Power in qd0 quantities: step 2 Take the √2√2=2 to the denominator out front; change the rectangular to exponential Bring the peak magnitudes inside the Re and Im operator; multiply by ejωt and by e-jωt Replace φ by –θi. Observe the space vectors inside the curly brackets. Therefore: (Note: The 3/2 is because our transformations are all amplitude invariant but not power invariant.) 12

  13. 3. Substitute qd0 quantities into space-vector power expressions. On slide 18 of our slides called “Machine Transformations,” we wrote that a space vector may be decomposed into our d-q transformation variables according to Power in qd0 quantities: step 3 Applying this to our v and i space vectors on the previous slide results in (assuming they are stator quantities) : Now substitute into our power expressions from the previous slide: Perform the algebra: Collect real and imaginary terms: Therefore: 13

  14. θ d-axis q-axis ia a' iq id a • We want to obtain our control relations. To do this, we make the following remarks: • Reference frame orientations: We have a choice to make in terms of how to align our d-q rotating reference frame. There are three questions about this which you may find helpful: • What does it mean to “align our d-q rotating reference frame”? •  The d-q coordinate reference frame rotates at synchronous speed. But at any given moment, it has a particular angular position. We align the d-q reference frame when we choose the d-q reference frame to have a particular angular position at a particular time (e.g., t=0), • Why do we have a choice about this? • The d-q reference frame is entirely “made-up.” There is nothing real about it. The d-q transformation is a mathematical decomposition or projection of real (rotating electric and magnetic) quantities. We are free to make that projection any way we like. Control concepts Put another way, we recall the definition of θ, which is given below and illustrated to the right. We are free to make θ(0) whatever we want.

  15. We want to obtain our control relations. To do this, we make the following remarks: • Reference frame orientations: We have a choice to make in terms of how to align our d-q rotating reference frame. There are three questions about this which you may find helpful: • What difference does the choice make? • Our stator-oriented space vectors rotate at synchronous speed which is the same rotational speed as our d-q coordinate reference frame. Thus, • if we choose to align our d-axis with a particular stator-oriented space vector, the q-axis component of that space vector will always be zero. • if we choose to align our q-axis with a particular stator-oriented space vector, the d-axis component of that space vector will always be zero. • Forcing certain d-axis or q-axis quantities to be permanently zero can simplify the control design. Control concepts

  16. Stator voltage-oriented control (SVOC): SVOC is a reference frame orientation very convenient for wind turbine DFIGs since the grid fixes the stator voltage. SVOC is achieved by aligning the d-axis of the synchronous reference frame with the stator voltage space vector vs. (We are free to align the reference frame however we like). Therefore the resultant d- and q-axis stator voltages are Control concepts The below figure illustrates (ref, pg. 256 of Wu). Observe the location of is indicates that the machine is operating as a generator with unity power factor. Also observe the location of Vr indicates it is operating supersynchronously (see slide 9 of DFIG Set #2).

  17. Steady-state: We would like to identify control values of torque and power that would result in appropriate steady-state operation. This means we can use steady-state relations, obtainable by setting derivatives to zero in our machine equations. From slide #44 in our slides called “dqTransformation” we saw: Control concepts All of this goes away in the steady-state, when the current derivatives are zero. And so our “steady-state” voltage equations become… 17

  18. Control relations – torque Extracting the relations for vqs and vds, we have: Solving for λds and λqs, respectively, we get Now substitute these relations for λds and λqs into our torque relation (1e) from slide 8: Use vqs=0 (SVOC): Electromagnetic torque is a function of d-axis rotor current and d-axis stator voltage only! Assume rs≈0: 18

  19. Return to our power relations of slide 13 (now we denote them as stator powers): Use vqs=0 (SVOC): Control relations – power From slide #39 of “dqTransformation” slides, we have that: Eqts. (*) Now Wu in his book (pg. 59) writes the above equations as: Eqts. (**) implying that Lm=M. However, he also writes (pg. 56) that: Eqts. (***) defining these previous two quantities (Ls and Lr) as the stator self inductance and the rotor self inductance, respectively. This is a contradiction, because the “M” in Eqts. (*) is 3/2 times Lms=Lmr=Lrs, yet the “Lm in Eqts. (***) is definitely the mutual inductance, which is the same as Lms=Lmr=Lrs. I will proceed here assuming that Eqts. (**) and (***) are correct (this needs review). This implies that the equations on next slide are correct. 19

  20. Control relations – power 20

  21. Solve the below for ids and iqs : Control relations – power Substitute into the power relations above: Now solve for idr and iqr, respectively: Recall our stator flux equations from slide 18 (but with vqs=0): Substitute them into our idr and iqr equations above: Assume rs≈0: For a given stator voltage, Ps and Qs can be controlled by the dq-axis rotor currents! 21

  22. Block diagram of DFIG wind turbine system w/ SVOC Note that the ωm/p converts T to Ps. Also, there should be minus sign here. Diagram from p. 260 of Wu. 22

  23. MPPT block generates the reference torque Tem* based on some method for max power point tracking. The simplest one to understand is use of the Turbine Power Profile, Fig. 1 below, but a better one is use of optimal torque control, Fig. 2 below. The asterisk in all below equations indicates that the quantity is a reference quantity. Conceptual control sequence of rotor-side converter Fig. 2 Fig. 1 The desired power or torque are identified by the MPPT scheme and then one of the below computations are made. • For a given stator reactive power reference Qs*, the q-axis rotor current reference iqr* is computed according to: • The reference currents idr* and iqr* are compared to the measured values, and the errors passed through PI controllers which are then transformed back to abc quantities for use in the PWM scheme to control the RSC (see previous slide). 23

  24. The figure shows that the control scheme requires the abc/qd0 transformation and the qd0/abc transformation. To get these for both rotor and stator quantities, we need to have the angles θ(t) and β(t), to perform the transformations. Obtaining the transformation angle Transformations for stator: Transformations for rotor: Observe the “θs calculator.” What is this? 24

  25. We want to align the d-axis with the stator space vector vs (as this will make vqs=0). From the figure below, angle θ is defined as the angle of the q-axis wrspt the a-axis stator reference. q-axis Obtaining the transformation angle d-axis • What if we wanted to align the q-axis with the stator space vector vs? • This means we would want to define • the location of θ, • to be the same as • the location of vs (θs). That is, we want θ=θs. But how to know θs? 25

  26. α-axis θs θs vs vs vβ a' a' a a vα β-axis Here is the vector we draw to represent vs at t=0. We want to obtain θs, which is the angle of vs. To get θs, we observe that we can obtain the horizontal and vertical components of vs via the α-β transformation. Obtaining the transformation angle The above calculation is very possible since it is easy to measure the instantaneous phase voltages. In fact, we need only to obtain two of the instantaneous phase voltages, as shown on the next slide. 26

  27. Under balanced conditions, we need measure only the a-phase and b-phase quantities, as shown below: Obtaining the transformation angle 27

  28. α-axis θs vs vβ a' a vα β-axis Obtaining the transformation angle When aligning the q-axis with vs, we set θ=θs. But what do we do if we want to align the d-axis with vs? Recalling that the d-axis is 90° behind the q-axis, then we want to have θ-90 positioned at θs, i.e., q-axis d-axis 28

  29. So the procedure for aligning the d-axis with vs is as follows: Measure the instantaneous phase voltages. Compute α-β components via Obtaining the transformation angle • Compute the angle of vs • Compute the initial angle θ(t=0) according to: 29

  30. θ ωm d-axis q-axis θm β ω ia a' iq id a We also need β for the rotor transformation, which is the angle between the rotor a-axis and the q-axis of the synchronously rotating reference frame. Recall from the below diagram (slide 25 of “dqTransformation” slides) that β=θ-θm. Obtaining the transformation angle 30

  31. Note the diagram on the right uses θsl instead of β. To understand this, from the below diagram, we observe that θsl is the angle between the rotor a-axis and the d-axis of the synchronously rotating reference frame, where θsl = θs -θr. Note this diagram uses θr which we have called θm so we have that θsl = θs-θm. From previous slide, we have that β=θ-θm and with SVOC, θ=θs+90, we have β=θs+90-θm. Therefore, we have θsl =β-90. So θsl and β differ by 90 deg. This is because the d-q transformation used in the book from which these diagrams came differs from the transformation used in these notes (see slide 34). Obtaining the transformation angle θm is measured by an encoder mounted on the shaft. (Reverting back to our own nomenclature), then β=θ-θm. 31

  32. So the procedure for aligning the d-axis with vs is as follows: Measure the instantaneous phase voltages. Compute α-β components via Obtaining the transformation angles These two measurements need to synchronize. • Compute the angle of vs • Compute the initial angle θ(t=0) according to: • The procedure for obtaining the angle β is • Measure the rotor angle θm • Compute: 32

  33. What happens to the d-q transformation with the angle θ=0? Comment 1 Implication: the α-β transformation is the d-q transformation if we permanently fix θ=0. So when we do the α-β transformation in step 2, we may also think about it as doing the d-q transformation with θ=0 (fixed on the stator a-phase axis). 33

  34. The transformation in Wu’s book, p. 52, uses angles wrspt to d-axis. • The transformation in Krause’s book, p. 135, uses angles wrspt q-axis. This also affects the Clarke transformation. This different results in slightly different transformation matrices (a sign change in row 2). Comment 2 Trnsfrmtns from Krause Trnsfrmtns from Wu The one I presented to you before (in notes called “Transformations”) The one I have used in these notes in order to maintain consistency with the d-q transformation we have used. 34

  35. We see on slide 22 that the rotor-side converter (RSC) functions control • the direct-axis rotor current in order to achieve a desired torque or stator power level; • the quadrature-axis rotor current in order to achieve a desired stator reactive power. Function of grid-side converter • The grid-side converter (GSC) performs two main functions: • It provides reactive power to the grid when required. The reactive power reference Q*GSC can be set to zero for unity power factor operation of the converter. • It keeps the DC link voltage vdc constant. • We observe in the figure the reference quantities v*dc and Q*GSC being fed to the GSC. 35

  36. Function of grid-side converter 36

  37. To get GSC reference quantities, we write real & reactive power equations for the GSC just as we wrote them in slide 13 for the RSC. The subscript “g” is used here to indicate these equations are for the grid-side of the GSC. This means that our GSC voltages are the same as we used for our RSC analysis (vs, vqs, vds) , but currents are not, i.e., the currents are out of the GSC rotor circuit. This results in: Function of grid-side converter – reactive power We deploy SVOC here just as we did for the RSC (we use the same θ(t=0)), and we obtain vqg=0. In this case, the above power expressions become: From the Qg equation (noting that vdg is the grid-side voltage in our SVOC scheme), we see we control the q-axis current, iqg, to achieve a desired Qg, according to: Q*g can be set to 0 for unity power factor operation of the GSC, so that the overall power factor of the DFIG is controlled by the RSC through its reference Q*s. Then: • a negative value of Q*s is used in the iqr equation (slide 21) to supply vars to grid • a positive value of Q*s is used in the iqr equation (slide 21) to absorb vars from grid. 37

  38. We already saw on the previous slide that The DC power crossing the DC link must equal the active power flowing between the grid and the GSC, as shown below. Therefore: Function of grid-side converter – DC link voltage iDCg iDCr + iDCc vDC - With the capacitance of C, then we have: Substituting for iDCg, and solving for dvDC/dt, yields: In the steady-state, the left-hand-side is zero, so: So specification on vDC translates to a requirement on idg. 38

  39. Rotor-side converter (RSC) is controlled so that it provides independent control of Tem and Qs. Wind turbine control levels Grid-side converter (GSC) is controlled so that it provides independent control of Vbus and Qg. Level I: Regulates power flow between grid and generator. Level II: Controls the amount of energy extracted from the wind by wind turbine rotor. Level III: Responds to wind-farm or grid-central control commands for MW dispatch, voltage, or frequency control. 39

  40. S. Achilles, N. Miller, E. Larsen, and J. MacDowell, “Voltage and reactive power control, presentation to NERC ERSTF, June 11-12, 2014, available at www.nerc.com/comm/Other/essntlrlbltysrvcstskfrcDL/VoltVarControl_Weaksys%20ERSTF%20JMM%20GE_0612.pdf. Reactive Power at No Wind “6. The network of variable speed wind turbine generator systems of claim 1 wherein the individual generators operate as static VAR regulators under any wind condition.” https://www.google.com/patents/US6924565 40

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