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J. McCalley

J. McCalley. Double-fed electric machines – steady state analysis Set 1. Four configurations. We will study this one, the DFIG. 2. Power Grid. DFIG. Rotor. DC Link. DC AC. AC DC. Basic concepts. Rotor is wound: it has 3 windings. Stator has three windings.

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J. McCalley

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

  2. Four configurations We will study this one, the DFIG. 2

  3. Power Grid DFIG Rotor DC Link DC AC AC DC Basic concepts Rotor is wound: it has 3 windings. Stator has three windings. Induction machine looks like a transformer with a rotating secondary (rotor). In DFIG, we inject a voltage signal via the converter to control it. 3

  4. # of pole pairs Basic concepts Mechanical rad/sec We can manipulate to get: The induced rotor voltages have frequency of : Substitution into slip expression above yields: Two modes of operation: ωm< ωsωr>0s>0Subsynchronous operation ωm>ωsωr<0s<0Supersynchronous operation 4

  5. The underlining of a parameter indicates that it is a phasor. We neglect magnetizing inductance Lm in this figure, for simplicity, but it will be added back in later. It is modeled in parallel with Es. Per-phase steady-state model This is major departure from SCIG where V’’r=0. STATOR VOLTAGE EQUATION: at fs =stator voltage with frequency fs These quantities are referred to stator side. = emf in the stator windings with frequency fs = stator current with frequency fs =stator resistance =stator leakage reactance at fr ROTOR VOLTAGE EQUATION: These quantities are referred to rotor side, as indicated by double-prime notation. =rotor voltage with frequency fr =induced emf in the rotor windings with frequency fr =induced rotor current with frequency fr =rotor resistance =rotor leakage reactance= 5 Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.

  6. We know (see *) the coils on the armature and rotor will see a flux linkage proportional to the magnetizing flux φmand the number of turns N. Application of Faraday’s law results in the stator emf and the induced rotor voltage being expressed as: Ks, Kr: stator and rotor winding factors, respectively, which combine the pitch and distribution factors. Ns, Nr: number of turns of stator & rotor, respectively. fs, fr, frequency of stator & rotor quantities, respectively φm : magnetizing flux Referring quantities Solve both relations for φm and equate: But recall: The ratio Ks/Kr is normally very close to 1, therefore (A) Define the effective turns ratio: * A. Fitzgerald, C. Kingsley, and A. Kusko, “Electric Machinery,” 3rd edition, McGraw-Hill, 1971, pg. 148. 6

  7. We just derived that: (A) At a locked rotor condition (s=1), the device is simply a static transformer, and we have: Referring quantities This tells us (for locked rotor) if we want to move a voltage from rotor side to stator side, we multiply it by a=Ns/Nr. We can obtain similar relationships for currents and impedances, and so we define the rotor quantities referred to the stator according to: Rotor quantities are referred to the stator-side, indicated by unprimed quantities (we could also use single prime quantities as notation here). Rr Rs jωrLσr Is jωsLσs Ir This is locked rotor condition (s=1), therefore ωr=ωs and Ers=Es Es 3 3 3 3 Ers Vs Vr 3 3 3 3 We can account for other slip conditions using ωr=sωs and from (A), Ers=aE’’rs=sEs. 7

  8. From slide 5, the stator-side voltage equation (referred to stator) is: Rr Rs jsωsLσr Is jωsLσs Ir Es Referring quantities Ers=sEs Vs Vr or using the nomenclature of this figure Now write the rotor-side voltage equation (referred to stator): Divide by s and we get the following circuit: Rr/s The voltage on both sides of the xfmr is the same, therefore, we may eliminate the xfmr. At the same time, we model the magnetizing inductance jωsLm. Rs jωsLσr Is jωsLσs Ir 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 Es Es Vs Vr/s 8

  9. Rr/s Rs Is jωsLσs jωsLσr Ir Referring quantities Es jωsLm Vs Vr/s 3 3 3 3 3 3 9

  10. Rr/s Rs Is jωsLσs jωsLσr Ir Power relations Es jωsLm Vs Vr/s Modify the above circuit to separate slip-dependent and slip-independent terms: 3 3 3 3 3 3 Change the circuit accordingly…. 10

  11. Veq=Vr(1-s)/s Rr Rs Is jωsLσs jωsLσr Ir + - Power relations Req= Rr(1-s)/s Es jωsLm Vr Vs The mechanical power out of the machine is the power into the machine less the stator and rotor copper losses (as indicated in the figure at the bottom), i.e., Power balance relation: where Ps and Pr are powers entering the machine through the stator & rotor windings, respectively, and Ploss,s and Ploss,r are the stator and rotor winding losses, respectively. Pmech>0 is motor; <0 is gen 3 3 3 Here, the first two terms are given by: Ps DFIM Pr 3 3 3 Ps>0 it receives power via stator Ps<0 it delivers power via stator Ploss,s+Ploss,r Pr>0 it receives power via rotor Pr<0 it delivers power via rotor Pmech 11

  12. Veq=Vr(1-s)/s Rr Rs Is jωsLσs jωsLσr Ir + - Power relations Req= Rr(1-s)/s Es jωsLm Vr Vs The mechanical power out of the machine is the power into the machine less the stator and rotor copper losses, i.e., Power balance relation: where Ps and Pr are powers entering the machine through the stator & rotor windings, respectively, and Ploss,s and Ploss,r are the stator and rotor winding losses, respectively. 3 3 3 From a cct analysis point of view, if we include remaining terms, we should get 0, i.e., 3 3 3 This shows the slip-dependent terms are those responsible for mechanical power. 12

  13. Pmech Introduction of voltage source in rotor circuit provides ability to affect Pmech. Veq= Vr(1-s)/s Rr Rs Is jωsLσs jωsLσr Ir + - Ploss,r Ploss,s Req= Rr(1-s)/s Ps Es Pr jωsLm Vr Vs Power relations Ps Ps Pmech DFIM Pr Pr Ploss,s+Ploss,r Pmech 3 3 3 3 3 3 13

  14. Pmech Veq= Vr(1-s)/s Rr Rs Is jωsLσs jωsLσr Ir + - Req= Rr(1-s)/s Es jωsLm Vr Vs Power relations With Pmech>0 (see below), motor convention implies a positive value of Req represents a positive mechanical load (as it did for SCIM), so Req term should be positive on RHS. For Veq term, rotor current (Ir) direction is out of positive side of voltage source; therefore voltage source supplies power to circuit and should be subtracted from the mechanical loading represented by the Req term. If Pmech>0the machine is delivering power through the shaft: MOTOR! If Pmech<0the machine is receiving power through the shaft: GEN! 3 3 3 3 3 3 Terms are always same sign; but Pmech takes their difference If 0<s<1(sub)Req term is pos; Veq term is pos. If 0>s>-1(super)Req term is neg; Veq term is neg. So we observe that in either mode (sub or super), machine may motor or generate, depending on relative magnitudes of Req and Veq terms. 14

  15. Pmech Veq= Vr(1-s)/s Rr Rs Is jωsLσs jωsLσr Ir + - Power relations Req= Rr(1-s)/s Es jωsLm Vr Vs Is Pslip equal to the power associated with the elements in the dotted box? No! Pmech is equal to the power associated with the elements in the dotted box, and we see from the derived equation that Pmech and Pslip differ by –(1-s)/s. If the power at the rotor terminals does not equal Pmech, what makes up the difference? 3 3 3 3 3 3 Power from the stator that crosses the airgap! We will return to this! What is Pslip? It is Pr-Ploss,r = what is provided at the rotor terminals not lost in the winding losses. Where does it go? To or from the shaft! 15

  16. Veq= Vr(1-s)/s Rr Rs Is jωsLσs jωsLσr Ir + - Req= Rr(1-s)/s A first torque expression Es jωsLm Vr Vs (p: # of pole pairs) Recall from slide 4: 3 3 3 3 3 3 and Therefore: 16

  17. Veq= Vr(1-s)/s Rr Rs Is jωsLσs jωsLσr Ir + - Req= Rr(1-s)/s A second (equivalent) torque expression Es jωsLm Vr Vs Idea here is to compute Pmech from: using stator & rotor voltage eqts. Stator power: Stator voltage: Substitute Vs into Ps: 3 3 3 The middle two terms are purely imaginary, therefore: 3 3 3 First term is purely real, only the second term contains real and imaginary, therefore: 17

  18. Veq= Vr(1-s)/s Rr Rs Is jωsLσs jωsLσr Ir + - Req= Rr(1-s)/s A second (equivalent) torque expression Es jωsLm Vr Vs Rotor power: Rotor voltage: Substitute Vr into Pr: 3 3 3 3 3 3 The middle two terms are purely imaginary, therefore: First term is purely real, only the second term contains real and imaginary, therefore: 18

  19. Now substitute Ps and Pr into the power balance equation: A second (equivalent) torque expression Observe we have loss terms added and subtracted in the above, so they go away. Now consider what happens when you take the real part of a vector multiplied by j (or rotated by 90 degrees): ja a Observe that Re(ja) = - Im(a) Im(a) Re(ja) Therefore: 19

  20. Let’s consider another vector identity: taking imaginary part of a conjugated vector: Observe that Im(a*) = - Im(a) A second (equivalent) torque expression a Im(a) Im(a*) a* Therefore (applying to first term only): Recall: Recall: Therefore: 20

  21. Two equivalent torque expressions Torque expression #1: Need ωr, rotor voltage and rotor current Torque expression #2: Need stator current and rotor current A third set of equivalent torque expressions follow…. 21

  22. This will be very similar to HW3, problem 1. If we assume the magnetic core of the stator and rotor is linear, then we may express flux linkage phasors of each winding (stator winding and rotor winding, respectively): Mutual inductances Additional equivalent torque expressions Rotor winding Stator winding Self inductances ASIDE: Recall L=λ/ieach self inductance is comprised of mutual and leakage according to: The flux seen by the stator winding from the stator current is the same as the flux seen by the rotor winding from the stator current plus the flux from the stator current that leaks. Therefore: From stator winding equation: From rotor winding equation: Choose one of these equations and substitute into torque expression #2…. 22

  23. From stator winding equation: From rotor winding equation: Substitute into torque expression #2…. Additional equivalent torque expressions Using stator winding equation: Using rotor winding equation: Purely real Purely real 23

  24. On slides 17 and 18, we derived the following relations for the power into the stator and rotor respectively: Subtracting losses from both sides, we obtain: Airgap and slip power This quantity is power flowing from stator terminals to rotor shaft (neg for generator operation). In other words, it is power across the airgap (rotor losses accounted for in slip power). Therefore: This quantity is power transferred from the grid to the rotor shaft through the converter (neg when it is into the grid). It is called the slip power. Therefore: Bring out front the “s” in the slip power expression and use Re{ja}=-Im(a) (both): Use Im(a*) = -Im(a) on slip expression: The term -3Im{} in the slip power expression is Pairgap. Therefore: 24

  25. Slip power (the power flowing to the shaft from the rotor circuit) is –s times the airgap power (the power flowing to the shaft from the stator circuit). Therefore: • In subsynchronous mode (0<s<1), Pairgap and Pslip have opposite signs (only one of them is into the rotor shaft). Because |s| is always small, |Pslip|=|s||Pairgap| is much smaller than |Pairgap|. So the sign of Pairgap determines the operation (Pairgap>0 is motor, Pairgap<0 is generator). • In supersynchronous mode (-1<s<0), Pairgap and Pslip have same signs - either both are positive and therefore into the rotor shaft (motor operation) or both are negative and therefore out of the rotor shaft (generator operation). Airgap and slip power 25

  26. So we just proved that: where Torque from airgap and slip power Our power balance relation states: Therefore: Substituting we obtain (*) Recall:  Substituting: (**)  Aside: From (*) and (**), we may derive the below relation, which was also derived on slide 13.  26

  27. On slides 17 and 18, we derived the following relations for the power into the stator and rotor respectively: If we neglect the stator losses (3RSIs2) and rotor losses (3RrIr2): Approximate relations between active powers Bring out front the “s” in the rotor power expression and use Re{ja}=-Im(a) (both): Use Im(a*) = - Im(a) on the rotor power expression The term 3Im{} in the rotor power expression is PS. Therefore: Recall the power balance relation: Neglecting losses: Substituting Pr expression:   Recall:   27

  28. Both Approximate Exact Active power relations - summary 28

  29. Approximate Exact Active power relations - summary • Mechanical power is the stator power • less the rotor power, in subsynchronous mode (when Ps and Pr are opposite sign); • plus the rotor power, in supersynchronous mode (when Ps and Pr are same sign). • Mechanical power is the airgap power • less the slip power, in subsynchronous mode (when Pairgap and Pslip are opposite sign); • plus the slip power, in supersynchronous mode (when Pairgap and Pslip are same sign). 29

  30. Without losses With losses Power balance Pslip Pslip Pairgap Pairgap Ps Ps Pgrid Pr Pgrid Pr Ploss,r Ploss,s Pmech Pmech These figures assume proper sign convention (power flowing to the rotor is positive). 30

  31. Yellow is supersynchronism; red is subsynchronism, bold box is generator operation. Generator modes Operation (motor/generator) is determined by the sign of Tem or equivalently, the sign of Ps≈Pairgap (see slide 25). For each mode, we may use the two relations to track the signs of Ps and Pr from the signs of Tem and s. For example, for mode 2, Tem<0Ps<0; Ps<0, s<0Pr<0 Focusing on the generator modes, we observe the standard induction machine generating mode, supersynchronism, where ωm>ωs (mode 2). We also observe a subsynchronous mode (mode 3), where ωm<ωs. 31

  32. Recall the approximate relation Generator modes Operation must have |s|<1, so rotor power is always smaller than stator power. Mode 2 Pm= Pmech In fact, DFIGS always run within about -0.3<s<0.3. Therefore, the rating of the PE converter circuit need be only about 30% of the stator winding rating. Mode 3 These figures show actual flow direction for generator operation. They also neglect losses. 32

  33. What is the fundamental difference between DFIG & SCIG that allows DFIG to operate as a gen in both sub-synch & super-synch? Generator modes Control over rotor power! Mode 2 Unlike the SCIG, we can make the rotor cct inject or withdraw as much or as little power as we like, indepen-dent of conditions. Thus, Pm= Pmech • when power supplied by turbine is small, inject power via rotor into shaft – subsynchronous! Mode 3 • When power supplied by turbine is large, withdraw power via rotor from the shaft – supersynchronous! These figures show actual flow direction for generator operation. They also neglect losses. 33

  34. This figure assumes proper sign convention (power flowing to the rotor or into the stator is positive). Without losses Pslip Pairgap Ps Pgrid Pr Assume an operating condition such that Pmech=PWTrating. Then A question on rating Pmech For example, consider Pmech=PWTrating=-2 MW. In supersynchronous mode, with s=-0.3, Therefore stator winding must be rated for 1.5385 MW. But in the subsynchronous mode, s=+0.3, then Question: Does this mean that the stator of a 2 MW turbine must be rated for 2.8571? Answer: No. In subsynchronous mode, the mechanical power from the generator shaft is lower that that in the supersynchronous mode. If Pmech increases beyond a certain level, then machine speed increases into the supersynchronous mode. So above situation never occurs. The maximum power in subsynchronous mode at 30% slip is: 34

  35. Question: Since stator losses (3RSIs2) and rotor losses (3RrIr2) are always positive, and since we get sign changes with the numerical values of Pmech, Ps, and (sometimes) Pr, do the loss terms in the above equation need to have different signs for motor operation than for generator operation? That is, do we need to do the following? Question on sign of losses Motor operation: Generator operation: Answer: No. Our original equation applies for both motor & generator operation. Remember: Pmech is positive for motor operation; Ps, and Pr are positive when flowing into the device from the grid. It may help to think about the equation in two different, but equivalent forms. Motor operation: Generator operation: 50 = 45 +10 - 3 - 2 - 50 = - 55 + 3 + 2 35

  36. In general, per-unitization enables inclusion of DFIGs within a system model. • It also facilitates identification of inappropriate data. Finally, a per-unitized voltage provides the ability to know how far it is from its nominal value (usually also the “normal” value) without knowing that nominal value. • The procedure is to choose three base quantities and compute other necessary base quantities. We will choose our base quantities as • rated rms line-to-neutral stator voltage, Vbase=|Vs|rated (rms volts); • rated rms stator line current, Ibase=|Is|rated (rms amperes) • rated stator synchronous frequency, ωbase= ωs,rated (rad/sec)) Per-unitization Then we compute: • Base impedance: • Base inductance: • Base flux: • Base speed: • Three-phase power base: • Base torque: 36

  37. Once all base quantities are obtained, then per-unitization is easy: • Stator voltage in pu: Per-unitization – stator quantities • Stator current in pu: • Stator flux in pu: • Stator active power in pu: • Stator reactive power in pu: As usual, only the magnitude is transformed (angle remains unchanged). 37

  38. Rotor voltage in pu: Per-unitization – rotor quantities For rotor quantities, we use the same base quantities as for the stator quantities (with actual rotor quantities referred to the stator side). • Rotor current in pu: • Rotor flux in pu: • Rotor active power in pu: • Rotor reactive power in pu: As usual, only the magnitude is transformed (angle remains unchanged). 38

  39. Torque in pu: Per-unitization – torque, speed, R, L For rotor quantities, we use the same base quantities as for the stator quantities (with actual rotor quantities referred to the stator side). • Speed in pu: • Resistances in pu: Note that the resistances and inductances when expressed in pu are lower case. • Inductance in pu: 39

  40. From slides 17, 18, we obtain voltage equations for stator and rotor circuits: which we rearrange by collecting terms in jωs: Voltage equations expressed in per unit From slide 22, we obtain the equations for stator and rotor flux linkages: (*) We recognize the flux linkage expressions in the voltage equations. Therefore: Replace voltages, currents, flux linkages, w/ product of pu value & base quantity: Divide by base quantities (note from slide 36 that Vbase=ZbaseIbase=λbaseωbase and so this last step is a division by same value) ωs,pu=1 40

  41. Now consider the flux linkage equations: Voltage equations expressed in per unit Replace currents and flux linkages with the product of their pu value and their base quantity, then base quantities are used to per-unitize inductances to obtain: Per-unitize one of the torque equations (#2) as follows:  Per-unitize the power expressions to obtain: Notation Police!!!! . I have used θv,s, θi,s, θv,r, and θi,r previously (see slides 11 and 16). Here I am using γv, γi,, φv, and φi, respectively. 41

  42. Homework #3: This homework is due Monday September 26, 2016. It is also posted to the website. A. Using previous relations provided in the “Set 1” slides, derive the following torque expressions. Homework #3 (also posted to website) (and identify σ) B. Use Q = 3Im{VI*} and the equivalent circuit to derive reactive power expressions, in terms of Is and Ir for • The stator, Qs • The rotor, Qr • C. For each DFIG condition below, compute Pairgap and Pslip and draw the power flows similar to slide 30 in the “Set 1” slides. • Pmech=-1 MW with s=+0.30 (subsynchronous operation). • Pmech=-1MW with s=-0.30 (supersynchronous operation). D. Complete the table on the next slide (the boxed section) by computing the per-unit values of the indicated five resistances/inductances for the 2 MW machine. 42

  43. Homework u (or a) Rs Lσs Lm R’r Lσr Rr Lσr Ls Lr Vbase Ibase rs lσs lm rr lσr Per-unit values 43

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