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J. McCalley

J. McCalley. Double-fed electric machines – steady state analysis Set 2.

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J. McCalley

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  1. J. McCalley Double-fed electric machines – steady state analysis Set 2

  2. The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set 1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a 50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power (j) Total real power generated (k) Tem using (l) Pmech using Ps+Pr-Pr,loss-Ps,loss & using (e) Rotor current (f) Rotor flux (g) Rotor voltage (h) Rotor real power (a) Synchronous speed (b) Line-to-neutral voltage (c) Line current (d) Stator flux Example Problem (a) Synchronous speed: Alternatively, the synchronous speed was given (table) as 1500 rpm, therefore:  (b) Line-to-neutral voltage: (c) Line current: (d) Stator flux 2

  3. The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set 1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a 50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power (j) Total real power generated (k) Tem using (l) Pmech using Ps+Pr-Pr,loss-Ps,loss & using (e) Rotor current (f) Rotor flux (g) Rotor voltage (h) Rotor real power (a) Synchronous speed (b) Line-to-neutral voltage (c) Line current (d) Stator flux Example Problem (e) Rotor current This is the referred rotor current! We can obtain actual rotor current from a (or u) =0.34: This phasor is at rotor frqncy, of fr=sfs=-0.25(50)= -12.5 Hz. Our development (slide 8 of DFIG data set#1) accounts for the rotor circuit being at ωr=sωs by dividing out the “s”, resulting in an effective rotor frequency (for purposes of circuit analysis) of ωs. The negative frequency indicates rotating magnetic field from rotor is in opposite direction to rotational direction of rotor. (f) Rotor flux This term, with λr represents the drop across Lm and Lσr as observed in (g) Rotor voltage Actual rotor voltage: 3

  4. The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set 1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a 50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power (j) Total real power generated (k) Tem using (l) Pmech using Ps+Pr-Pr,loss-Ps,loss & using (e) Rotor current (f) Rotor flux (g) Rotor voltage (h) Rotor real power (a) Synchronous speed (b) Line-to-neutral voltage (c) Line current (d) Stator flux Example Problem (h) Rotor real power (i) Rotor reactive power (j) Total real power generated • Comments: • Pm must be larger in magnitude to supply losses • The slip, at -0.25, is large; this is consistent with the stator winding being at its 2MW rating. • Qs=0, Qr=23.4kVARMagnetization is via rotor 4

  5. The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set 1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a 50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power (j) Total real power generated (k) Tem using (l) Pmech using Ps+Pr-Pr,loss-Ps,loss & using (e) Rotor current (f) Rotor flux (g) Rotor voltage (h) Rotor real power (a) Synchronous speed (b) Line-to-neutral voltage (c) Line current (d) Stator flux Example Problem (k) Tem (l) Pmech Alternatively, under a slight rearrangement (for reasons that will be clear on next slide) 5

  6. The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set 1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a 50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power (j) Total real power generated (k) Tem using (l) Pmech using Ps+Pr-Pr,loss-Ps,loss & using (e) Rotor current (f) Rotor flux (g) Rotor voltage (h) Rotor real power (a) Synchronous speed (b) Line-to-neutral voltage (c) Line current (d) Stator flux Example Problem (l) Pmech Now let’s compute it using Relate this calculation to the one on the previous slide: From previous slide, Pslip=-0.50542 MW. So: 6

  7. Consider the circuit below, which is analogous to our stator winding circuit. At any operating condition, we may characterize the circuit as an impedance Z=R+jX=Z/_θ, as indicated. Then we may express the current according to Machine Question: How to know quadrant of Is? Observe that current angle is always negative of impedance angle, θi=-θ Z Z V V I I Lag Lag I I V V Z Z Lead Lead 7

  8. We have developed the following (per-unit) relations: (1) Stator voltage equation (2) Rotor voltage equation (3) Stator winding flux equation Phasor diagrams for generator operation (4) Rotor winding flux equation Draw phasor diagram per below (CCW rotation is pos angle): Step 1: Draw Vs as reference (0°). Step 2: For gen Qs>0, lag; for gen Qs<0, lead. Draw Is phasor (in this case, we assume lag). Step 3: Use (1) to draw the stator flux phasor λs: Step 4: Use (3) to draw the rotor current phasor Ir: Step 5: Use (4) to draw the rotor flux phasor λr: Step 6: …. Vs - Isrs -Isrs Vs Is Ir=λs/lm – lsIs/lm lmIs λs/lm – lsIs/lm λr=lmIs+lrIr 8 λs= -j(Vs – Isrs) lrIr

  9. Draw phasor diagram per below (CCW rotation is pos angle): Step 1: Draw Vs as reference (0°). Step 2: For gen, Qs>0, lag; for gen, Qs<0, lead. Draw Is phasor. Step 3: Use (1) to draw the stator flux phasor λs: Step 4: Use (3) to draw the rotor current phasor Ir: Step 5: Use (4) to draw the rotor flux phasor λr: Step 6: Use (2) to draw the rotor voltage phasor Vr: Phasor diagrams for generator operation jsλr, s>0, sub-sync Vr=Irrr+jsλr, s<0 super-syn Vr=Irrr+jsλr, s>0 Vs - Isrs Isrs Vs Irrr Is Ir=λs/lm – lsIs/lm lmIs jsλr, s<0 λs/lm – lsIs/Lm Observe that the angle of Vr is heavily influenced by the sign of s. λr=lmIs+LrIr λs= -j(Vs – Isrs) lrIr 9

  10. Rotor-side converter (RSC) is controlled so that it provides independent control of Tem and Qs. We will study the steady-state actions of this particular control function. Wind turbine control levels 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 2 computes reference quantities for Level 1. Level 2 performs MPPT for conventional designs, in which case Level 3 is absent. Level 3 provides the ability of the WT to provide ancillary services. Level III: Responds to wind-farm or grid-central control commands for MW dispatch, voltage, frequency, or inertial control. 10 Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.

  11. 2. MPPT with optimal tip speed ratio: “In this method, the maximum power operation of the wind turbine is achieved by keeping the tip speed ratio to its optimal value λT,opt. The generator speed ωm is controlled by the power converters and will be equal to its reference in steady state, at which the MPPT is achieved.” Level II: MPPT Approach #2 Concept The block diagram above shows that if we have a measured vw, then we can compute the optimal speed ωm* from knowledge of the optimal tip speed ratio λT,opt. The optimal tip speed ratio λT,opt is obtained for a given pitch from the CP plot (performance coefficient) to the left, as a function of pitch. The particular value of λT,opt shown is for a pitch of θ=3 degrees. λT,opt 11

  12. Level II: MPPT #3 Concept Popt=Koptωm3 since Popt=Toptωm Topt=Koptωm2 Source: O. Anaya-Lara, N. Jenkins, J. Ekanayake, P. Cartwright, and M. Hughes, “Wind energy generation: modeling and control,” Wiley, 2009. 12

  13. Level II: MPPT #3 Concept Source: O. Anaya-Lara, N. Jenkins, J. Ekanayake, P. Cartwright, and M. Hughes, “Wind energy generation: modeling and control,” Wiley, 2009.c 13

  14. This leads to three possible maximum power point tracking (MPPT) schemes. 1. MPPT with turbine power profile: “The wind speed is measured in real-time by a wind speed sensor. According to the MPPT profile provided by the manufacturer, the power reference Pm* is generated and sent to the generator control system, which compares the power reference with the measured power Pm from the generator to produce the control signals for the power converters. Through the control of power converters and generator, the mechanical power Pm of the generator will be equal to its reference in steady state, at which the maximum power operation is achieved. It is noted that the power losses of the gearbox and drive train in the above analysis are neglected and, therefore, the mechanical power of the generator Pm is equal to the mechanical power PM produced by the turbine.” Level II: Summary of MPPT #1, 2, 3 Concepts 2. MPPT with optimal tip speed ratio: “In this method, the maximum power operation of the wind turbine is achieved by keeping the tip speed ratio to its optimal value λT,opt. The generator speed ωm is controlled by the power converters and will be equal to its reference in steady state, at which the MPPT is achieved.” 3. MPPT with optimal torque control: “The maximum power operation can also be achieved with optimal torque control according to equation (2.8), where the turbine mechanical torque TM is a quadratic function of the turbine speed ωM. For a given gear ratio and with the mechanical power losses of the gearbox and drive train neglected, the turbine mechanical torque TM and speed ωM can be easily converted to the generator mechanical torque Tm and speed ωm respectively. Figure 2-21 shows the principle of the MPPT scheme with optimal torque control, where the generator speed ωm is measured and used to compute the desired torque reference Tm*. The coefficient for the optimal torque Kopt can be calculated according to the rated parameters of the generator. Through the feedback control, the generator toque Tm will be equal to its reference Tm*in steady state, and the MPPT is realized. It is noted that there is no need to use the wind speed sensors in this scheme. ” rT: rotor radius (blade length) Source:B. Wu, Y. Lang, N. Zargari, and S. Kouro, “Power conversion and control of wind energy systems,” Wiley, 2011. 14

  15. 2012 Tutorial paper on WT Control for APC ACTIVE POWER CONTROL (APC) A focus on level 3 control (but it also does good job covering basics) http://www.nrel.gov/docs/fy12osti/54605.pdf Review section II for exam 1. 15

  16. Level 1 control DC bus voltage is controlled by grid-side converter (GSC) to a pre-determined value for proper operation of both GSC and RSC. We achieve Tem and Qs control objectives by controlling rotor-side voltage. We control rotor voltage to achieve a specified torque and stator reactive power. 16 Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.

  17. Our objective here is, for a fixed stator voltage (fixed by the grid), and a desired torque Tem,ref and a desired stator reactive power Qs,ref, we want to determine the rotor voltage to make it so. We will also examine the stator flux, stator current, rotor current, rotor flux, and stator real power, as shown in the diagram below. Level 1 control A given. What results from our control. The control variable to achieve our targets. So our problem is this: Given desired targets T*em and Q*s, compute the required Vr. Targets that we want to achieve. 17 Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.

  18. We draw the phasor diagram with stator flux as reference (0 degrees). Here, the stator flux, denoted by ψs (instead of λs), is specified as the reference. We have identified particular angles in this diagram. It is operating as a motor (stator current is almost in phase with stator voltage), and stator is absorbing reactive power (Is has a negative angle relative to Vs, so Zmotor=Vs/Is has a positive angle, indicating it is inductive and therefore absorbing). Level 1 control Note that ɣi is the angle by which Is leads λs. 18 Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.

  19. From voltage equation (slide 40 of DFIG Set #1): If we neglect drop across the stator resistance (it is typically very small), then: Level 1 control: Qs equation Substitute into the stator reactive power equation: Use Im(ja)=Re(a): From previous slide, note that ɣi is the angle by which Is leads λs , i.e., Substituting: Final equation for Qs: 19 Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.

  20. From HW3 (see slide 42 of DFIG Set #1): Again (from phasor diagram), note that ɣi is the angle by which Is leads λs , i.e., Level 1 control: Tem equation Substituting: Final torque equation: 20

  21. From phasor diagram: But recall our Qs and Tem equations: Level 1 control: Is equation Substituting into Is equation: Recall from slide 19:  Substituting into Is equation: So… given the targeted torque and stator reactive power, we know the necessary stator current. But what we need (because we can control it) is the rotor voltage… 21

  22. From slide 22 of DFIG Set #1: Solve for Is, Ir in terms of λs, λr. Level 1 control: λr equation Using these relations, together with: we may derive: Suggest to do this on your own before Exam1. 22

  23. Recall the rotor flux equation derived on the previous slide together with the rotor voltage equation (slide 40 of DFIG Set #1): Level 1 control: λr equation Neglecting the voltage drop in the rotor resistance, and equating: Solving for Vr , switching order of terms, and simplifying: Then, using 23

  24. Level 1 control: summary Also, we have stator and rotor powers as a function of Tem (see slide 28 of DFIG Set #1):  24

  25. + Level 1 control: magnetization Recalling that the stator voltage is at 90 degrees, we know that the inductive part of Im is the real part. This is consistent with the fact that the imaginary part of the above equation is almost zero, since LS=Lm+Lσs , and Lσs is small, so that Ls≈Lm. Thus, the magnetizing current is provided by the real part of the above. There are three “extreme” situations of interest: Magnetized by rotor: When Qs=0, Im is provided by Vs/(ωsLm). Magnetized by stator: When Qs>0 and Vs/(ωsLm)=QsLs/(3VsLm), Im is provided by Qs/3Vs. Capacitive operation: Here the real term in the above Im expression must be negative. This occurs when Qs<0 and: In this case, rotor current is increased. BUT….it is possible to do it with DFIG!!!! (SCIG cannot). 25

  26. Level 1 control: comments Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011. 26

  27. Level 1 control: magnitudes And this shows that these terms are functions of our desired reference quantities. The last two relations are given as a function of ωr, but it may be more intuitive to express them as a function of slip, where we can use ωr =sωs 27

  28. Level 1 control: magnitudes You can think of the rotor speed using ωm=(1-s) ωs which shows that for low positive slips, rotor speed is just below synchronous speed, and for low negative slips, rotor speed is just above synchronous speed. 28

  29. Level 1 control Fixed Qs=0 Fixed Tem=-1 • Is is independent of ωm but increases with |Tem| and with |Qs| • Is is the same independent of whether machine is absorbing or supplying vars. • Above equation indicates Is should be the same for Tem=1, Tem=-1. However, above equation neglected stator resistance Rs. Assuming fixed Vs, in motor mode (Tem=1), Rs causes voltage across rotor circuit to be less, and so Ir (and thus Is) must be greater to deliver same torque. In gen mode, Rs causes voltage across rotor circuit to be more, and so Ir must be less to deliver same torque. 29

  30. Level 1 control Fixed torque implies fixed rotor current if stator flux is fixed. Therefore fixed torque implies fixed rotor current if λssin(θλs – θIr) is fixed. Fixed Tem=-1 If IS (last slide) and Ir are both constant, how can Pmech change? (Tem=Pmechp/ωm, Pmech must decrease as ωm increases). Answer: Vr is changingSee next slide. Last comment here: Ir is independent of ωm for fixed torque but increases as Qs moves from + (absorbing) to – (supplying). This is consistent with our observation on slide 59, which indicated rotor current would increase for capacitive (supplying) operation. 30

  31. Level 1 control 0.5 0 -0.5 Fixed Tem=-1 Fixed Qs=0 • Vr is decreasing with ωm to ωm=ωs and then increases with ωm. • Vr depends mainly on speed of machine. • Vr does not change much with Tem or with Qs because VsLr/ωsLm tends to dominate. 31

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