Doublefed electric machines steady state analysis J Mc

Double-fed electric machines – steady state analysis J. Mc. Calley

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

Basic concepts Power Grid DFIG Rotor DC Link AC DC DC AC 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 will inject a voltage control signal via that converter. 3

Basic Concepts rotor Balanced voltages applied to stator windings provides a rotating magnetic field of speed which induces an emf in the rotor windings according to • eind=induced emf in one conductor of rotor • v=velocity of conductor relative to stator flux rotation • B=stator magnetic flux density vector • L=length of conductor in direction of wire (fs: 60 Hz, p: # of pole pairs) 4

Basic concepts We can manipulate to get: Mechanical rad/sec The induced rotor voltages have frequency of : Substitution into slip expression above yields: Observe three modes of operation: ωm< ωs ωr>0 s>0 Subsynchronous operation ωm= ωs ωr=0 s=0 Synchronous operation ωm>ωs ωr<0 s<0 Supersynchronous operation 5

Per-phase steady-state model STATOR VOLTAGE EQUATION: =stator voltage with frequency f at fs s 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 ROTOR VOLTAGE EQUATION: =rotor voltage with frequency fr =induced emf in the rotor windings with frequency fr =induced rotor current with frequency fs =rotor resistance =rotor leakage reactance= at fr These quantities are referred to rotor side, indicated by prime notation. 6

Referring quantities Application of Faraday’s Law allows the stator back emf and the induced rotor voltage to be 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, frequency of stator & rotor quantities, respectively φm : magnetizing flux Solve both relations for φm and equate: But recall: The ratio Ks/Kr is normally very close to 1, therefore Define the effective turns ratio: Define the induced rotor voltage referred to the stator side: 7

Referring quantities We just derived that: (*) At a locked rotor condition (s=1), the device is simply a static transformer, and we have: This tells us it 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: jωs. Lσs jωr. Lσr 3 3 Vs Rs 3 3 Is Es 3 3 Ers Rr Ir Rotor quantities are referred to the stator-side, indicated by unprimed quantities. Vr This is locked rotor condition (s=1), therefore ωr=ωs and Ers=Es We can account for other slip conditions using ωr=sωs and from (*), a. E’rs=s. Es. 8

Referring quantities Rs Is jsωs. Lσr jωs. Lσs Ir 3 3 Es Vs Rr 3 3 Ers=s. Es Vr Now write the rotor-side voltage equation (referred to stator): Divide by s Is Rs and we get the following circuit: jωs. Lσr jωs. Lσs Es Ir 3 3 Vs Rr/s 3 3 Es Vr/s The voltage on both sides of the xfmr is the same, therefore, we may eliminate the xfmr. . We represent a magnetizing inductance jωs. Lm in its place. 9

Referring quantities Is Rs jωs. Lσr Es Ir 3 3 Vs Rr/s 3 3 jωs. Lm Vr/s 10

Power relations Is Rs jωs. Lσr Es Ir 3 3 Vs Rr/s 3 3 jωs. Lm Vr/s We modify the above circuit slightly in order to clearly separate slip-dependent terms from loss terms: Change the circuit accordingly…. 11

Power relations Is Rs jωs. Lσr 3 3 Vs Es 3 3 jωs. Lm Rr Vr(1 -s)/s +Rr(1 -s)/s Ir Vr It is possible to prove that the mechanical power out of the machine is the power associated with the slip-dependent terms R 2(1 -s)/s and Vr(1 -s)/s. To do so, use: 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. Expressing the right-hand-terms of the power balance relation in terms of the above circuit parameters leads one to identify the slip-dependent terms as Pmech. Knowing that the slip-dependent terms are those responsible for mechanical power, we may obtain the power expressions from the circuit, as on the next slide. 12

Power relations Is Rs jωs. Lσr 3 3 Vs Es 3 3 jωs. Lm Pmech Veq= Vr(1 -s)/s I Rr r +Req= Rr(1 -s)/s Vr Rotor current (Ir) direction is out of positive side of voltage source; therefore it supplies power to circuit. But a normal (positive) resistance Req always consumes power. So these two terms should be opposite sign. Defining Pmech>0 (see below) as motor mode implies Req term should be added and Veq term should be subtracted. If Pmech>0 the machine is delivering power through the shaft: MOTOR! If Pmech<0 the machine is receiving power through the shaft: GEN! If 0<s<1 Req term is positive Veq term is positive Supplying P to cct If 0>s>-1 Req term is negative Veq term is negative Consuming P from cct. 13

A first torque expression Is Rs jωs. Lσr 3 3 Vs Es 3 3 jωs. Lm Veq= Vr(1 -s)/s I Rr r +Req= Rr(1 -s)/s Vr (p: # of pole pairs) Recall from slide 5: and Therefore: 14

A second (equivalent) torque expression Is Rs jωs. Lσr 3 3 Vs Es 3 3 jωs. Lm Veq= Vr(1 -s)/s I Rr r +Req= Rr(1 -s)/s Vr Stator power: Stator voltage: Substitute Vs into Ps: The middle two terms are purely imaginary, therefore: First term is purely real, only the second term contains real and imaginary, therefore: 15

A second (equivalent) torque expression Is Rs jωs. Lσr 3 3 Vs Es 3 3 jωs. Lm Veq= Vr(1 -s)/s I Rr r +Req= Rr(1 -s)/s Vr Rotor power: Rotor voltage: Substitute Vr into Pr: The middle two terms are purely imaginary, therefore: First term is purely real, only the second term contains real and imaginary, therefore: 16

A second (equivalent) torque expression Now substitute Ps and Pr into the power balance equation: 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 ja (or rotated by 90 degrees): Im(a) a Observe that Re(ja) = - Im(a) Re(ja) Therefore: 17

A second (equivalent) torque expression Let’s consider another vector identity: taking imaginary part of a conjugated vector: Observe that Im(a*) = - Im(a) a Im(a) Im(a*) a* Therefore: Recall: Therefore: 18

Two equivalent torque expressions Torque expression #1: Need rotor speed, rotor voltage and rotor current Torque expression #2: Need stator current and rotor current A third set of equivalent torque expressions follow…. 19

Additional equivalent torque expressions 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 Rotor winding Stator winding Self inductances ASIDE: Each self inductance is comprised of mutual and leakage according to: Therefore: From stator winding equation: From rotor winding equation: Choose one of these equations and substitute into torque expression #2…. 20

Additional equivalent torque expressions From stator winding equation: From rotor winding equation: Substitute into torque expression #2…. Using stator winding equation: Purely real Using rotor winding equation: Purely real 21

Airgap and slip power On slides 15 and 16, we derived the following relations for the power into the stator and rotor respectively: Subtracting losses from both sides, we obtain: This quantity is the power that flows from the stator terminals to the rotor (negative for generator operation). In other words, it is the power across the airgap. Therefore: This quantity is the power that is transferred from the grid to the rotor through the converter (negative 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 3 Im{} in the slip power expression is Pairgap. Therefore: 22

Airgap and slip power So we just proved that: where Our power balance relation states: Therefore: Substituting Recall: we obtain Substituting: 23

Approximate relations between active powers On slides 15 and 16, we derived the following relations for the power into the stator and rotor respectively: If we neglect the stator losses (3 RSIs 2) and rotor losses (3 Rr. Ir 2): 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 3 Im{} in the rotor power expression is PS. Therefore: Recall the power balance relation: Neglecting losses: Substituting Pr expression: Recall: 24

Active power relations - summary Exact Both Approximate 25

Power balance Without losses With losses Pgrid Ps Pairgap Ploss, s Pslip Pmech Pr Pgrid Ps Pairgap Ploss, r Pslip Pr Pmech These figures assume proper sign convention (power flowing to the rotor is positive). 26

Generator modes Mode Slip and speed Pmech Ps Pr 1. Motor (Tem>0) s<0, ωm>ωs (suprsynchrnsm) >0 (mch delivers mech pwr) >0 (mch receives power via stator) >0 (mch receives power via rotor) 2. Generator (Tem<0) s<0, ωm>ωs (suprsynchrnsm) <0 (mch receives mech pwr) <0 (mch delivers power via stator) <0 (mch delivers power via rotor) 3. Generator (Tem<0) s>0, ωm<ωs (subsynchrnsm) <0 (mch receives mech pwr) <0 (mch delivers power via stator) >0 (mch receives power via rotor) 4. Motor (Tem>0) s>0, ωm<ωs (subsynchrnsm) >0 (mch delivers mech pwr) >0 (mch receives power via stator) <0 (mch delivers power via rotor) For each mode, we may use three relations to track the sign Ps, ωr, and Pr from the signs of Tem and s. For example, for mode 2, Tem<0 Ps<0 and Tem<0, s<0 ωr<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, which is available to the DGIG as a 27 result of the machine receiving power from the grid via the rotor circuit.

Generator modes Recall the approximate relation Mode 2 Pm= Pmech In fact, DFIGS always run within about -0. 3<s<0. 3. Mode 3 These figures show actual flow direction for generator operation. They also neglect losses. Operation must have |s|<1, so rotor power is always smaller than stator power. Therefore, the rating of the PE converter circuit need be only about 30% of the stator winding rating. 28

This figure assumes proper sign convention (power flowing to the rotor or into the stator is positive). A question on rating Pgrid Assume an operating condition such that Pmech=PWTrating. Then Ps Without losses Pairgap Pslip Pr 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 in the supersynchronous mode. If Pmech increases beyond a certain level, then machine speed increases into the supersynchronous mode. So above situation never occurs. We can obtain the maximum power in subsynchronous mode as: 29

Question on sign of losses Question: Since stator losses (3 RSIs 2) and rotor losses (3 Rr. Ir 2) are always positive, and since we get sign changes with the numerical values of P mech, 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? 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: 50 = 45 +10 - 3 - 2 Generator operation: - 50 = - 55 + 3 + 2 30

Per-unitization 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)) Then we compute: • Base impedance: • Base inductance: • Base flux: • Base speed: • Three-phase power base: • Base torque: 31

Per-unitization – stator side Once all base quantities are obtained, then per-unitization is easy: • Stator voltage in pu: • 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). 32

Per-unitization – rotor side • Rotor voltage in pu: • Rotor current in pu: • Rotor flux in pu: For the rotor side, we use the same base quantities as on the stator side (with actual quantities referred to the stator side). • Rotor active power in pu: • Rotor reactive power in pu: As usual, only the magnitude is transformed (angle remains unchanged). 33

Per-unitization – torque, speed, R, L • Torque in pu: • Speed in pu: • Resistances in pu: • Inductance in pu: On the rotor side, we use the same base quantities as on the stator side (with actual quantities referred to the stator side). Note that the resistances and inductances when expressed in pu are lower case. As usual, only the magnitude is transformed (angle remains unchanged). 34

Voltage equations expressed in per unit From slides 15, 16, we obtain voltage equations for stator and rotor circuits: which we rearrange by collecting terms in jωs: From slide 20, we obtain the equations for stator and rotor flux linkages: (*) We recognize the flux linkage expressions in the voltage equations. Therefore: Now we can replace voltages, currents, and flux linkages with the product of their per-unit value and their base quantity, then the base quantities can be used to perunitize the resistances and frequency to obtain: 35

Voltage equations expressed in per unit Replace voltages, currents, flux linkages with the product of their pu value and their base quantity, then base quantities are used to per-unitize resistances and frequency to obtain: Now consider the flux linkage equations: 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: 36

Homework #3: This homework is due Monday, March 26. A. Using previous relations provided in these slides, derive the following torque expressions. (and identify σ) B. Use Q = 3 Im{V I*} and the equivalent circuit to derive reactive power expressions, in terms of Is and Ir for 1. The stator, Qs 2. The rotor, Qr C. For each DFIG condition below, compute Pairgap and Pslip and draw the power flows similar to slide 28. 1. Pmech=-1 MW with s=+0. 30 (subsynchronous operation). 2. Pmech=-1 MW 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. 37

Homework u (or a) Rs Lσs Lm R’r Lσr Rr Lσr Ls Lr Vbase Ibase Rs lσs lm rr lσr 38

Phasor diagrams for generator operation We have developed the following relations: (1) Stator voltage equation (2) Rotor voltage equation (3) Stator winding flux equation (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. 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 Vs lm Is Is Ir=λs/lm – ls Is/lm λr=lm Is+lr Ir λs= -j(Vs – Isrs) lr Ir 39

Phasor diagrams for generator operation 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: Vr=Irrr+jsλr, s<0 super-syn lm Is Is Vr=Irrr+jsλr, s>0 I r rr Ir=λs/lm – ls Is/lm jsλr, s<0 λs/lm Observe that the angle of Vr is heavily influenced by the sign of s. jsλr, s>0, sub-sync Vs - Isrs Vs – ls Is/Lm λr=lm Is+Lr Ir λs= -j(Vs – Isrs) lr Ir 40

Question: How to know quadrant of Is? Consider the circuit below, which is analogous to our stator winding circuit. At any operating condition, we may characterize the circuit as an impedance Machine Z=R+j. X=Z/_θ, as indicated. Then we may express the current according to Observe that current angle is always negative of impedance angle, θi=-θ Real pwr Reactive pwr P>0 motor R>0 Q>0 absorbing X>0 Real pwr Reactive pwr P>0 motor R>0 Q<0 supplying X<0 Z Real pwr Reactive pwr V I Lag I P<0 gen R<0 Q>0 absorbing X>0 Z Lead V I Lag Real pwr Reactive pwr V Z P<0 gen R<0 Q<0 supplying X<0 I V Z Lead 41

Example Problem The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The slip is s=-0. 25 (super-synchronous). Compute: (h) Rotor real power (d) Stator flux (a) Synchronous speed (e) Rotor current (i) Rotor reactive power (j) Total real power generated (b) Line-to-neutral voltage (f) Rotor flux (c) Line current (g) Rotor voltage (k) Tem (a) Synchronous speed: Alternatively, the synchronous speed was given as 1500 rpm, therefore: (b) Line-to-neutral voltage: (c) Line current: (d) Stator flux 42

Example Problem The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The slip is s=-0. 25 (super-synchronous). Compute: (h) Rotor real power (d) Stator flux (a) Synchronous speed (e) Rotor current (i) Rotor reactive power (j) Total real power generated (b) Line-to-neutral voltage (f) Rotor flux (c) Line current (g) Rotor voltage (k) Tem (e) Rotor current This is the referred rotor current! We can obtain the actual rotor current from a (or u) =0. 34: This phasor is at the rotor frequency, of fr=sfs=-0. 25(50)=-12. 5 Hz (f) Rotor flux (g) Rotor voltage Actual rotor voltage: 43

Example Problem The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The slip is s=-0. 25 (super-synchronous). Compute: (h) Rotor real power (d) Stator flux (a) Synchronous speed (e) Rotor current (i) Rotor reactive power (j) Total real power generated (b) Line-to-neutral voltage (f) Rotor flux (c) Line current (g) Rotor voltage (k) Tem (h) Rotor real power (i) Rotor reactive power (j) Total real power generated Comments: 1. Pm must be larger in magnitude to supply losses 2. This wind turbine’s rating should be 2. 55 MW. 3. The DFIG stator winding is rated for 2 MW. 44

Example Problem The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The slip is s=-0. 25 (super-synchronous). Compute: (h) Rotor real power (d) Stator flux (a) Synchronous speed (e) Rotor current (i) Rotor reactive power (j) Total real power generated (b) Line-to-neutral voltage (f) Rotor flux (c) Line current (g) Rotor voltage (k) Tem 45

Wind turbine control levels Rotor-side converter (RSC) is controlled so that it provides independent control of Tem and Qs. Let’s study the steady -state actions of this particular control function. 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 windfarm or grid-central control commands for MW dispatch, voltage, frequency, or inertial 46 control.

Level 1 control We achieve control objectives by controlling rotor-side voltage. This (openloop) control not heavily used for DFIGs Assume DC bus voltage is controlled by grid-side converter (GSC) to a predetermined value for properation of both GSC and RSC. We control rotor voltage to achieve a specified torque and stator reactive power. 47

Level 1 control 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 are also interested in the stator flux, stator current, rotor current, and rotor flux, and stator real power, as shown in the diagram below. 48

Level 1 control We draw the phasor diagram with stator flux as the 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 phasor diagram. It is operating as a motor (current is almost in phase with voltage), and the 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. 49

Level 1 control: Qs equation From voltage equation (slide 35): If we neglect drop across the stator resistance (it is typically very small), then: 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: 50

Level 1 control: Tem equation From HW 3 (see slide 37): Again (from phasor diagram), note that ɣi is the angle by which Is leads λs , i. e. , Substituting: Final torque equation: 51

Level 1 control: Is equation From phasor diagram: But recall our Qs and Tem equations: Substituting into Is equation: Recall from slide 50: Substituting into Is equation: 52

Level 1 control: λr equation From slide 20: Using these relations, together with: we may derive: 53

Level 1 control: λr equation Now use the rotor flux equation derived on the previous slide together with the rotor voltage equation (slide 35): Neglecting the voltage drop in the rotor resistance, we may derive: 54

Level 1 control: summary Also, we have stator and rotor powers as a function of Tem: 55

Level 1 control: magnitudes Magnitudes are attractive because then we can plot them. And this shows that these terms are functions of our desired reference quantities. The above relations are given as a function of ωr, but it may be more intuitive to plot them as a function of rotor speed, ωm, where we can compute ωr =sωm/(1 -s). You can think of the rotor speed as 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. 56

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 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. 57

Level 1 control Fixed torque implies fixed rotor current if stator flux is fixed. Because Tem=Pmechp/ωm, Pmech must decrease as ωm increases. Fixed Tem=-1 • Ir is independent of ωm for fixed torque but increases as Qs moves from + (absorbing) to – (supplying). 58

Level 1 control Both rotor current and stator current equations have real part determined by Q s and imaginary part determined by Tem (Vs is at 90° so real part of currents is in quadrature with Vs) Add them to obtain magnetizing current Magnetizing component. Very close to zero since Ls~Lm. Qs=0 (no stator reactive power): Magnetized from rotor current 0<Qs<3 Vs 2/Lsωs (reactive power into stator, abs) Magnetized from both currents. Qs=3 Vs 2/Lsωs (reactive power into stator, abs) Magnetized from stator current. Qs<0 (reactive power from stator, sup): Magnetized from both currents. 59

Level 1 control Fixed Qs=0 Fixed Tem=-1 • Pr linearly decreases w/ ωm for –Tem (gen) and linearly increases w/ ωm for +Tem(mot). • Pr is independent of whether machine is absorbing or supplying vars. Remember: ωm=(1 -s)ωs, ωr=sωs. 60

Level 1 control Fixed Qs=0 Fixed Tem=-1 • Vr is linearly decreasing with ωm to ωm=ωs and then linearly increasing with ωm. • Vr depends mainly on speed of machine. • Vr does not change much with Tem or with Qs because Vs. Lr/ωs. Lm tends to dominate. Remember: ωm=(1 -s)ωs, ωr=sωs. 61

Level 1 control Fixed Qs=0 Fixed Tem=-1 • Efficiency increases with ωm under all conditions (see next slide): • In the subsynchronous mode, stator windings carry |Pmech|+|Pr|. • In the supersynchronous mode, stator windings carry |Pmech|-|Pr|. • Efficiency decreases as |Qs| increases (most efficient for unity power factor). • More efficient when absorbing (magnetized from stator) than supplying (magnetized from rotor) 62

Generator modes Mode 2 Pm= Pmech Mode 3 63

Representing RSC with impedance It can be convenient in analyzing the steady-state performance of the DFIG to represent the RSC as an equivalent impedance, as indicated in the below figure. We can follow our earlier development (see slide 9), but with our RSC equivalent impedance represented: Rr Rs jsωs. Lσr Is jωs. Lσs Ir 3 3 Es Vs In slide 9: Divide by s 3 3 Ers=s. Es Req Vr 3 jωr. Leq 3 =jsωs Leq Now: Divide by s 64

Representing RSC with impedance Is Rs jωs. Lσs Rr/s Ir 3 3 Vs jωs. Lσr Vm 3 3 jωs. Lm Req/s Vr/s 3 jωs Leq 3 Equivalent RSC impedance is: Represent it in the circuit with: Let’s assume the DFIG operates at unity power factor. Then Qs=0, and for Vs=Vs/_0°, Question: Do we need to specify motor or generator operation in the above equation? Answer: Not for the relation Pairgap= Ps-Ploss, s (see slide 30). For motor op, Ps>0 and losses subtract so that Pairgap is smaller than Ps, consistent with the fact that power flows from stator to rotor. For gen op, Ps<0 and losses add so that Pairgap is larger than Ps, consistent with the fact that power flows from rotor to stator. However, the relation on the right assumes that Is is a magnitude (positive), and so it is correct for motor op. For gen op, we must use a negative magnitude to get the sign of Vs. Is correct. We could correct this by writing the RHS as Vs. Is-Rs. Is 2= (Vs-Rs. Is)Is, i. e. use phasor notation for the current instead of just magnitude. 65

Representing RSC with impedance Is Rs jωs. Lσr Vm Ir 3 3 Vs Rr/s 3 3 jωs. Lm Req/s Vr/s 3 jωs Leq 3 From slide 25, we know for the model (with losses) that Equating the two airgap expressions: Rewriting, we find a quadratic in Is: Obtain roots: Could be positive (motor) or negative (generator) With stator current calculated, we can use the circuit to find Vr and Ir…. 66

Representing RSC with impedance Is Rs jωs. Lσr Rr/s Ir 3 3 Im Vs Vm 3 3 jωs. Lm From KVL we can compute Vm: Req/s Vr/s 3 jωs Leq=j. Xeq/s 3 (Xeq= ωr. Leq) Then compute the magnetizing current Im: Then compute the rotor current Ir: Then compute the rotor voltage Vr: We can now obtain Zeq/s or Zeq: where Ir is computed from above relations. 67

Representing RSC with impedance Tem is increasing here. Req<0 converter transfers active power to rotor. Req>0 rotor delivers active power to the converter. 68

Homework #4 Consider a 1. 5 MW, 690 v, 50 Hz 1750 rpm DFIG wind energy system. The parameters of the generator are given on the next slide. The generator operates with a maximum power point tracking (MPPT) system so that its mechanical torque Tem is proportional to the square of the rotor speed. The stator power factor is unity. For each of the following speeds: 1750, 1650, 1500, 1350, and 1200 rpm, compute: • Slip • Tem (k. N-m) • Vr (volts) • Ir (amps) • Req (ohms) • Xeq (ohms) What kind of machine is this at 1500 rpm? 69

Homework #4 70

Homework #4 Converter equivalent impedance at 1500 rpm: So 1500 rpm is synchronous speed! 71

Homework #4 There is another solution which has very large current and is clearly not realistic. Be careful here because this solution assumed the direction of current Ir opposite to what we have assumed. Observe that slip=0. This implies that a DC current flows through the rotor circuit from the converter and the rotor leakage reactance and equivalent reactance are zero. The DFIG is operating like a synchronous machine where the rotor flux is produced by a DC current through a DC exciter. 72

SCIG Torque-slip characteristic You may recall, from EE 303 or your undergraduate course on electric machines that the torque-slip characteristic of the squirrel-cage induction generator (SCIG) appears as below. One observes that the SCIG operates as a generator only when it is in supersynchronous mode and a motor only when it is in subsynchronous mode. Motoring Generating Subsynchronous Supersynchronous Let’s see how we obtain this curve for SCIG, and let’s also compare what we do to what we need to do to obtain the analogous curves for the DFIG. 73

Comparison of equivalent circuits: SCIG vs DFIG Is Rs jωs. Lσr Rr/s Ir 3 3 Im Vm 3 3 jωs. Lm Vs SCIG The difference between the machines in terms of steady-state models is the ability to electrically absorb or supply complex power S via the rotor. Is Rs jωs. Lσr Rr/s Ir 3 3 Im Vs Vm 3 3 jωs. Lm Req/s Vr/s 3 jωs Leq=j. Xeq/s 3 DFIG Where do we see rotor losses in these circuits? … (next slide) (Xeq= ωr. Leq) 74

Comparison of equivalent circuits: SCIG vs DFIG Is Rs jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Im Vm 3 3 jωs. Lm Vs SCIG Split up the R/s terms in each circuit as R+R(1 -s)/s and the rotor losses become immediately apparent. Is Rs jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm DFIG Vr/s Req 3 jωs Leq=j. Xeq/s 3 (Xeq= ωr. Leq) Where do we see mechanical power in these circuits? … (next slide) 75

Comparison of equivalent circuits: SCIG vs DFIG Is Rs jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Im Vm 3 3 jωs. Lm Vs SCIG The mechanical power is represented by the slip-dependent resistances. Is Rs jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm DFIG Vr/s Req 3 jωs Leq=j. Xeq/s 3 (Xeq= ωr. Leq) But what do the other two terms in the DFIG circuit represent? … (next slide) 76

Comparison of equivalent circuits: SCIG vs DFIG Is Rs jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Im Vm 3 3 jωs. Lm Vs SCIG These terms represent the real and reactive power exchange between the rotor and the RSC. As we saw on slide 68, these terms, Req and Xeq can be pos (rotor transfers power to RSC) or neg (RSC transfers power to rotor). Is Rs jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm DFIG Vr/s Req 3 jωs Leq=j. Xeq/s 3 (Xeq= ωr. Leq) How to compute torque in for these machines? … (next two slides) 77

Comparison of equivalent circuits: SCIG vs DFIG 78

Torque equation for SCIG Is Rs jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Im Vs Vm 3 3 jωs. Lm SCIG Note that the “s” on the denominator provides that Pmech is positive for s>0, motor action, and negative for s<0, generator action. How to obtain Ir? …. (next slide) 79

Torque equation for SCIG Zs=Rs+j. Xσs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Im Vm 3 3 Zm=jωs. Lm Vs Find Thevenin looking in here. Is Zth jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Comment: Zm>>ZS, so Vth≈Vs, Zth=Zs is not a bad approximation. Vth 80

SCIG Torque-slip characteristic You may recall, from EE 303 or your undergraduate course on electric machines that the torque-slip characteristic of the squirrel-cage induction generator (SCIG) appears as below. One observes that the SCIG operates as a generator only when it is in supersynchronous mode and a motor only when it is in subsynchronous mode. Motoring Generating Subsynchronous Supersynchronous Now let’s take a look at the torque-speed curves for the DFIG…. (next slide) 81

Torque equation for DFIG Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm How to obtain Ir? …. (next slide) Vr/s Req 3 jωs Leq=j. Xeq/s 3 82

Comparison of equivalent circuits: SCIG vs DFIG Rs Is jωs. Lσr Rr 3 3 Req(1 -s)/s Im Vm 3 3 Zm=jωs. Lm Vs Ir Rr(1 -s)/s Vr/s Req 3 jωs Leq=j. Xeq/s 3 Find Thevenin looking in here. Is Zth jωs. Lσr Ir Rr(1 -s)/s 3 3 Vth Rr Comment: Zm>>ZS, so Vth≈Vs, Zth=Zs is not a bad approximation. Req(1 -s)/s Vr/s Req 3 jωs Leq=j. Xeq/s 3 83

Torque-slip characteristic for DFIG So how do we obtain the torque-slip characteristic for the DFIG? 1. Develop values of Zeq for various values of torque-speed control point (slides 66 -67): Aside: The above points result from the turbine control characteristic. This characteristic originates from the maximum power extracted from the wind, which is given by the power curve, described by Pmech~ωm 3. But Pmech=Temωm therefore Tem~ ωm 2. 2. For each value of Zeq, express Tem as a function of s (or ωm= ωs(1 -s)) for various values of s. torque-speed control point (slides 66 -67): 84

Torque-slip characteristic for DFIG The sign of Req and Xeq are for rotor current direction defined out of the rotor. These signs reverse for rotor current direction into the rotor as we have done. 85

Efficiency Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs jωs. Lσs Ir Rr(1 -s)/s 3 3 Vs jωs. Lσr Im Vm 3 3 Zm=jωs. Lm Req(1 -s)/s Vr/s Req 3 jωs Leq=j. Xeq/s 3 At 1750, the slip is s=(1500 -1750)/1500=-0. 1667 From your homework, you should compute that Is=1068. 2 amperes Ir=1125. 6 amperes, Req=0. 05375 ohms, Xeq=0. 02751 ohms. The mechanical power supplied to the generator 86

Efficiency Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs jωs. Lσs Ir Rr(1 -s)/s 3 3 Vs jωs. Lσr Im Vm 3 3 Zm=jωs. Lm Req(1 -s)/s Vr/s Req 3 jωs Leq=j. Xeq/s 3 From your homework, you should compute that Is=1068. 2 amperes Ir=1125. 6 amperes, Req=-0. 05375 ohms, Xeq=-0. 02751 ohms. The rotor power is This power is negative (because Req is negative); it is supersynchronous, therefore it is flowing out of the rotor to the RSC. 87

Efficiency Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs jωs. Lσs Ir Rr(1 -s)/s 3 3 Vs jωs. Lσr Im Vm 3 3 Zm=jωs. Lm Req(1 -s)/s Vr/s Req 3 jωs Leq=j. Xeq/s 3 From your homework, you should compute that Is=1068. 2 amperes Ir=1125. 6 amperes, Req=0. 05375 ohms, Xeq=0. 02751 ohms. The rotor and stator winding losses are 88

Efficiency Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs jωs. Lσs Ir Rr(1 -s)/s 3 3 Vs jωs. Lσr Im Vm 3 3 Zm=jωs. Lm Req(1 -s)/s Vr/s Req 3 jωs Leq=j. Xeq/s 3 From your homework, you should compute that Is=1068. 2 amperes Ir=1125. 6 amperes, Req=0. 05375 ohms, Xeq=0. 02751 ohms. The stator active power is 89

Efficiency Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs jωs. Lσs Ir Rr(1 -s)/s 3 3 Vs jωs. Lσr Im Vm 3 3 Zm=jωs. Lm Req(1 -s)/s Vr/s Req 3 jωs Leq=j. Xeq/s 3 From your homework, you should compute that Is=1068. 2 amperes Ir=1125. 6 amperes, Req=0. 05375 ohms, Xeq=0. 02751 ohms. The total power delivered to the grid is The difference between Pm and Pg is the losses on the stator and rotor windings: Efficiency is: 90

DFIG for non-unity power factor “FERC 661 -A [1] specifies that large wind farms must maintain a power factor within the range of 0. 95 leading to 0. 95 lagging, measured at the POI as defined in the Large Generator Interconnect Agreement (LGIA) if the Transmission Provider shows, in the system impact study that they are needed to ensure the safety or reliability of the transmission system. . ” [1] Order for Wind Energy, Order No. 661 -A, 18 CFR Part 35 (December 12, 2005). See also Interconnection for Wind Energy, Order No. 661, 70 FR 34993 (June 16, 2005), FERC Stats. & Regs. ¶ 31, 186 (2005) (Final Rule); see also Order Granting Extension of Effective Date and Extending Compliance Date, 70 FR 47093 (Aug. 12, 2005), 112 FERC ¶ 61, 173 (2005). “The Electrical System Operator (IESO) of Ontario essentially requires reactive power capabilities for large wind farms that are equivalent to that for synchronous generators, taking into consideration an equivalent impedance between the generator terminals and the POI [2]. The requirements include: … Supplying full active power continuously while operating at a generator terminal voltage ranging from 0. 95 pu to 1. 05 pu of the generator’s rated terminal voltage. ” “The Alberta Electric System Operator’s requirements [4] include: The wind farm’s continuous reactive capability shall meet or exceed 0. 9 power factor (pf) lagging to 0. 95 pf leading at the collector bus based on the wind farm aggregated MW output. ” E. Camm and C. Edwards, “Reactive Compensation Systems for Large Wind Farms, ” IEEE Transmission and Distribution Conference and Exposition, 2008. 91

DFIG for non-unity power factor E. Camm and C. Edwards, “Reactive Compensation Systems for Large Wind Farms, ” IEEE Transmission and Distribution Conference and Exposition, 2008. 92

DFIG for non-unity power factor “Along with the evolution of wind turbine technology, technical standards of wind generation interconnections become more restrictive. For example, unity power factor has been required for wind generation interconnections in many utilities or control areas in earlier years. Recently, the more restrict requirement with 0. 95 lead and lag power factor has been under discussion since the DFIG and full converter wind turbine technology has become mainstream of wind generation interconnection requests. ” I. Green and Y. Zhang, “California ISO experience with wind farm modeling, ” IEEE Power and Energy Society General Meeting, 2011. 93

DFIG for non-unity power factor Rs Is jωs. Lσs Rr jωs. Lσr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 Define: φ as power factor angle: -180<φ<180 Identify the current phasor as Ps is negative for gen; then cosφ is also negative; Ps is positive for motor; then cosφ is also positive; so Is is always positive. Therefore: Recalling , we may write We have just made the numerator positive for all values of φ. 94

DFIG for non-unity power factor Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 The sign of Ps determines the sign of the real part of the current. Ps is negative if machine is in generating mode (supplying real power). In this case, cosφ is negative because φ is in quadrant 2 or 3. If machine is supplying Q, then sign of Qs should be negative, sign of Im{Is*} should be negative, and therefore sign of Im{Is} should be positive. Given cos φ is negative: If machine is absorbing Q, then sign of Qs should be positive, sign of Im{Is*} should be positive, and therefore sign of Im{Is} should be negative. Given cos φ is negative: 95

DFIG for non-unity power factor Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 Define the magnetizing current factor: From the circuit, KCL requires: But the magnetizing current is entirely imaginary: or Substitution of the Im expression into the rotor current expression yields: If the machine is absorbing Q, then . Substituting into Ir: 96

DFIG for non-unity power factor Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 Assume the machine is operated at rated power, Ps, rated, and recall Recall from slide 94: and the substitute into previous expression : Factor out the Ps, rated/3 Vs…. (next slide): 97

DFIG for non-unity power factor Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 Factor out the -Ps, rated/3 Vs Combine terms with “j” Simplify 98

DFIG for non-unity power factor Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 So this is for absorbing (underexcited operation) If we repeat the exercise for supplying (overexcited operation), we will obtain this: The difference in sign on the square root term indicates higher rotor current is required for overexcited operation than for underexcited operation. No big surprise there! And so the rotor winding should be rated for the overexcited operation, at rated stator active power output. This would be…. (next slide) 99

DFIG for non-unity power factor Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 Rotor current for rated stator active power and reactive power generation It is interesting to see the relative magnitude between Ir and Is. Again, from slide 94: 100

DFIG for non-unity power factor Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 Rotor current for rated stator active power and reactive power generation Rotor current for rated stator active power and stator unity power factor Ratio of rotor current required for a given stator power factor when supplying Q and that required for unity stator power factor, all at rated stator active power 101

DFIG for non-unity power factor Rs Is jωs. Lσr Rr Ir Rr(1 -s)/s 3 3 Req(1 -s)/s Im Vs Vm 3 3 jωs. Lm Vr/s Req 3 jωs Leq=j. Xeq/s 3 Krs factor for a given stator power factor at rated stator active power. Krs factor for unity stator power factor at rated stator active power. Ratio of Krs factor for a given power factor to Krs factor for unity power factor, for rated stator active power. 102

DFIG for non-unity power factor Krs factor for a given stator power factor at rated stator active power. The prime notation on the Krs at the top of the graph indicates the values have been referred to the rotor for an “a” of about 0. 3. 103

DFIG for non-unity power factor Ratio of Krs factor for a given power factor to Krs factor for unity power factor, for rated stator active power. 104