1 / 60

J. McCalley

J. McCalley. Double-fed electric machines – steady state analysis Set 3. From slide 5 of DFIG Set 1, the stator-side voltage equation (referred to stator) is:. R r. R s. js ω s L σ r. I s. j ω s L σ s. I r. E s. Recall Slide #10 from DFIG Set #1. E rs =s E s. V s. V r.

mfields
Télécharger la présentation

J. McCalley

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. J. McCalley Double-fed electric machines – steady state analysis Set 3

  2. From slide 5 of DFIG Set 1, the stator-side voltage equation (referred to stator) is: Rr Rs jsωsLσr Is jωsLσs Ir Es Recall Slide #10 from DFIG Set #1 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 2

  3. 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. The equivalent impedance can be integrated into the conventional steady-state equivalent circuit of the DFIG, based on which an analysis of the DFIG can be performed under any operating conditions. We can follow our earlier development (slide 8 of DFIG Set #1), but with our RSC equivalent impedance represented. Representing RSC with impedance Rr Rs jsωsLσr Is jωsLσs Ir Req Es Ers=sEs Vs Vr jωrLeq =jsωs Leq In slide 8 of DFIG Set #1: Now: Divide by s Divide by s 3 3 3 3 3 3 3 3 3 3 3

  4. Rr/s Rs Is jωsLσs jωsLσr Ir Req/s Vm jωsLm Vs Vr/s Representing RSC with impedance jωs Leq 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°, No need for concern of conjugation here because angle is either 0 or 180. 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 35 of DFIG Set#1). 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 subtract so that |Pairgap| is larger than |Ps|, consistent with the fact that power flows from rotor to stator. 3 3 3 3 3 3 3 3 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 VsIs correct. We could correct this by writing the RHS as VsIs cosθ-RsIs2= (Vscosθ-RsIs)Is; we could also use: (Vs -RsIs)Is, where Is=+Is or -Is 4

  5. Rr/s Rs Is jωsLσs jωsLσr Ir Req/s Vm jωsLm Vs Vr/s Representing RSC with impedance jωs Leq From slide 28 of DFIGSet#1, we know for the model (w/losses) that: Equating the two airgap expressions: Rewriting, we find a quadratic in Is: Could be positive (motor, 0°) or negative (generator, 180°); no other angle is allowed since Qs=0. Obtain roots: 3 3 3 3 3 3 3 3 With stator current calculated, we can use the circuit to find Vr and Ir…. 5

  6. Rr/s Rs Is jωsLσs jωsLσr Ir Im Req/s Vm jωsLm Vs Vr/s Representing RSC with impedance jωs Leq=jXeq/s (Xeq= ωrLeq) From KVL we can compute Vm: Then compute the magnetizing current Im: Then compute the rotor current Ir: Then compute the rotor voltage Vr/s or Vr: We can now obtain Zeq/s or Zeq: 3 3 3 3 3 3 3 3 Observe that the expressions for Zeq use a -Ir in the denominator because of the defined directionality of this current. where Ir is computed from above relations. 6

  7. Tem is increasing here. Representing RSC with impedance Observe that Vr reverses polarity in moving from subsyn operation to supersyn operation. Req>0rotor cct consumes MW, implying the rotor delivers active power to the converter. Req<0rotor cct provides MW, implying the converter transfers active power to rotor. 7

  8. 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 (kN-m) • Vr (volts) • Ir (amps) • Req(ohms) • Xeq (ohms) Example (*) What kind of machine is this at 1500 rpm? 8 (*) B. Wu, Y. Lang, N. Zargari, & S. Kouro, “Power conversion and control of wind energy systems,” Wiley, 2011, pp. 244-245.

  9. Example 9

  10. Converter equivalent impedance at 1500 rpm: Example So 1500 rpm is synchronous speed! 10

  11. 8.1851 comes from rated torque value given on slide 9. The other solution (using addition) has very large current of 151,100A and is clearly not realistic. Example (1500 rpm) Here, this solution assumed the direction of current Ir opposite to what we have assumed and therefore used Ir=Is-Im instead of Ir=Im-Is. Then Vr equation and Zeq equation must be changed accordingly. But results are same. At 1500rpm, 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. 11

  12. Matlab code was developed, below: Lls=0.0001687; Llr=0.0001337; Lm=0.0054749; pp=2; Vs=Vsll/sqrt(3); n=1750; omega_m=pp*n*2*pi/60; omega_s=2*pi*50; Tem=-8185.1*(n/1750)^2; s=(omega_s-omega_m)/omega_s; Isroot=sqrt(Vs^2-4*Rs*Tem*omega_s/(3*pp)); Isplus=(Vs+Isroot)/(2*Rs); Isminus=(Vs-Isroot)/(2*Rs); Vm=Vs-Isminus*(Rs+i*omega_s*Lls); Im=Vm/(i*omega_s*Lm); Ir=Im-Isminus; Vr=s*Vm+Ir*(Rr+i*s*omega_s*Llr); Zeq=Vr/(-1*Ir); Example Note that, unlike the calculations on the previous slide, this code was developed using our standard directionality for Ir (into the stator circuit from the rotor circuit) and thus needs a negative sign in computing the impedance. 12

  13. The above code solves the problem for n=1750rpm, giving the following results: Vs = 398.3717 omega_m = 366.5191 omega_s = 314.1593 Tem = -8.1851e+003 s = -0.1667 Is = -1.0682e+003 Vm = 4.0120e+002 +5.6614e+001i Im = 3.2915e+001 -2.3326e+002i Ir = 1.1011e+003 -2.3326e+002i Vr = -65.6040 -17.7576i Zeq = 0.0538 + 0.0275i Example 13

  14. Solution to the problem for n=1650 rpm is as follows: Vs = 398.3717 omega_m = 345.5752 omega_s = 314.1593 Tem = -7.2764e+003 s = -0.1000 Is = -950.3619 Vm = 4.0089e+002 +5.0368e+001i Im = 2.9284e+001 -2.3308e+002i Ir = 9.7965e+002 -2.3308e+002i Vr = -38.4915 - 9.7646i Zeq = 0.0349 + 0.0183i Example 14

  15. Solution to the problem for n=1500 rpm is as follows: Vs = 398.3717 omega_m = 314.1593 omega_s = 314.1593 Tem = -6.0135e+003 s = 0 Is = -786.2759 Vm = 4.0046e+002 +4.1672e+001i Im = 2.4228e+001 -2.3282e+002i Ir = 8.1050e+002 -2.3282e+002i Vr = 2.1316 - 0.6123i Zeq = -0.0026 Example COMPARE TO SOLUTION GIVEN ON SLIDE 11 15

  16. Solution to the problem for n=1350 rpm is as follows: Vs = 398.3717 omega_m = 282.7433 omega_s = 314.1593 Tem = -4.8710e+003 s = 0.1000 Is = -637.5111 Vm = 4.0006e+002 +3.3787e+001i Im = 1.9644e+001 -2.3259e+002i Ir = 6.5715e+002 -2.3259e+002i Vr = 42.7114 + 5.5273i Zeq = -0.0551 - 0.0279i Example 16

  17. Example Solution to the problem for n=1200 rpm is as follows: Vs = 398.3717 omega_m = 251.3274 omega_s = 314.1593 Tem = -3.8487e+003 s = 0.2000 Is = -504.1578 Vm = 3.9971e+002 +2.6720e+001i Im = 1.5535e+001 -2.3239e+002i Ir = 5.1969e+002 -2.3239e+002i Vr = 83.2605 + 9.0985i Zeq = -0.1270 - 0.0743i 17

  18. A summary of results are in the table below; however, the results in the table were computed with Ir having opposite direction to what we have assumed (out of the rotor), therefore Ir has opposite sign to what is given above. Example The following results were computed with Ir having direction of into the rotor, consistent with our defined directionality.. 18

  19. Example Comparing my plots (on the right) to those we saw on slide 7, the |Vr| and |Ir| plots are very similar. The angle plot /_Vr is also very similar, but the angle plot /_Ir is shifted by 180° and the angle plot /_Vr-/_Ir is also shifted. This is due to the defined direction for Ir. 19

  20. In the next eight slides, we will develop the torque equation for the DFIG from the circuit, so that we can look at the torque-slip (or torque-speed) characteristic in the same way that we looked at it for the SCIG. Before doing that, let’s remind ourselves of the steps we took to obtain the torque-speed characteristic of a SCIG. Those steps were based on the steady-state circuit model of the SCIG. So it will be useful to compare the steady-state circuit model of the SCIG to the steady-state circuit model of the DFIG, which we will also do. Torque equation for DFIG 20

  21. Rr Rs Is jωsLσs jωsLσr Rr(1-s)/s Ir Im Vm Recall Torque equation for SCIG jωsLm Vs 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. 3 3 3 3 3 3 How to obtain Ir? …. (next slide) 21

  22. Rr Zs=Rs+jXσs Is jωsLσr Rr(1-s)/s Ir Im Vm Recall Torque equation for SCIG Zm=jωsLm Vs Find Thevenin looking in here. Comment: Zm>>ZS, so Vth≈Vs, Zth=Zs is not a bad approximation. Rr Zth Is jωsLσr Rr(1-s)/s Ir Vth 3 3 3 3 3 3 3 3 3 3 22

  23. We then found the torque-slip characteristic of the squirrel-cage induction generator (SCIG) 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. Recall SCIG Torque-slip characteristic Motoring Generating Subsynchronous Supersynchronous Now let’s take a look at the torque-speed curves for the DFIG…. (next slide) 23

  24. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm Req jωsLm Vr/s Vs Torque equation for DFIG jωs Leq=jXeq/s 3 3 3 3 3 3 3 3 24 How to obtain Ir? …. (next slide)

  25. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm Zm=jωsLm Req Vs Vr/s Torque equation for DFIG jωs Leq=jXeq/s Find Thevenin looking in here. Comment: Zm>>ZS, so Vth≈Vs, Zth=Zs is not a bad approximation. Rr Zth Is Rr(1-s)/s jωsLσr Ir Req(1-s)/s Req Vr/s Vth jωs Leq=jXeq/s 3 3 3 3 3 3 3 3 3 3 3 3 3 3 25

  26. Recall that we want to compute torque as a function of slip. That is, we put in a value of slip and get out a value of torque, which is what we did with the SCIG. But we have a problem here. What is it? Torque-slip characteristic for DFIM The problem here is that Zeq depends on torque. See slide 11 where the first thing we did was obtain torque from knowledge of speed – this followed from our assumption of a maximum power point tracking (MPPT) system that controlled torque Tem proportional to the square of the rotor speed. Why is this a problem? Because our objective is to compute torque as a function of slip. To get torque, based on the previous slide, we need Zeq. But to get Zeq, we need torque. 26

  27. 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 5-6): Torque-slip characteristic for DFIG 2. For each value of Zeq, express Tem as a function of s (or ωm= ωs(1-s)) for various values of s (see slide 25). 27

  28. The below curves result from this procedure. Observe Curve 4, which characterizes the torque-slip characteristic imposed by the MPPT control. Some comments: • The values of Req and Xeq on the lower graph result from this control, as we have seen in our previous example (slide 72). For example, point B is the point corresponding to Zeq2=-0.072-j0.033 in the subsynchronous mode. • The non-linear curves are those that result from fixing Zeq according to the MPPT control and then computing torque as a function of slip as done in step 2 on the previous slide (slide 91). For example, curve 2 results from fixing Zeq to Zeq2. Torque-slip characteristic for DFIG • With the MPPT in place, points on the nonlinear curves cannot be achieved. They are drawn to show the conceptual relationship between the torque-slip characteristics of the SCIG and the DFIG. • The DFIG always operates in the generator mode, in both subsynchronous and supersynchronous operation, due to the rotor voltage control imposed by the MPPT. • At s=0, without the MPPT, the DFIG is a synchronous machine and the torque is entirely set by the wind (curve 3). • For comparison, curve 5 shows the torque-slip characteristic when the rotor is short-circuited (resulting in a SCIG). The maximum torque of the DFIG is significantly higher! 28

  29. “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). DFIG for non-unity power factor “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. 29

  30. 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. 30

  31. DFIG for non-unity power factor A. Ellis, Robert Nelson, E. Von Engeln, R. Walling, J. McDowell, L. Casey, E. Seymour, W. Peter, C. Barker, and B. Kirby, “Reactive Power Interconnection Requirements for PV and Wind Plants – Recommendations to NERC,” Sandia Report SAND2012-1089, February, 2012, available at http://energy.sandia.gov/wp/wp-content/gallery/uploads/Reactive-Power-Requirements-for-PV-and-Wind-SAND2012-1098.pdf. 31

  32. “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.” DFIG for non-unity power factor I. Green and Y. Zhang, “California ISO experience with wind farm modeling,” IEEE Power and Energy Society General Meeting, 2011. A. Ellis, Robert Nelson, E. Von Engeln, R. Walling, J. McDowell, L. Casey, E. Seymour, W. Peter, C. Barker, and B. Kirby, “Reactive Power Interconnection Requirements for PV and Wind Plants – Recommendations to NERC,” Sandia Report SAND2012-1089, February, 2012, available at http://energy.sandia.gov/wp/wp-content/gallery/uploads/Reactive-Power-Requirements-for-PV-and-Wind-SAND2012-1098.pdf. 32

  33. DFIG for non-unity power factor 33

  34. Let’s first take a look at our equivalent impedance approach, similar to what we did for unity power operation. DFIG for non-unity power factor 34

  35. Rr/s Rs Is jωsLσs jωsLσr Ir Req/s Vm jωsLm Vs Vr/s Representing RSC with impedance jωs Leq Equivalent RSC impedance is: Represent it in the circuit with: Let’s assume the DFIG operates at non-unity power factor. Then for Vs=Vs/_0°, Alternatively, assume RS=0; then this becomes: Equating the two airgap expressions: Rewriting, we find a quadratic in Is: 3 3 3 3 3 3 3 3 An approximate expression 35

  36. Consider the same 1.5 MW DFIG analyzed under unity power factor (data is repeated on the next slide). Once again, assume 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. • (1) Assume the stator power factor is 0.95 leading. For each of the following speeds: 1750, 1650, 1500, 1350, and 1200 rpm, compute: • Slip • Tem (kN-m) • Is (use exact expression, i.e., with Rs) • Vr (volts) • Ir (amps) • Req(ohms) • Xeq (ohms) • (2) Repeat (1) except assume the stator power factor is 0.95 lagging. • (3) Repeat (1) except use approximate expression to obtain Is. Homework #4 36

  37. Example 37

  38. DFIG for non-unity power factor In the following slides on non-unity power factor operation of the DFIG, we will derive some relations between stator current, magnetization current, and rotor current which allow us to observe some design issues of the DFIG if it is to be operated under non-unity power factor. 38

  39. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm DFIG for non-unity power factor Req jωsLm Vs Vr/s jωs Leq=jXeq/s Define: φ as power factor angle: -180<φ<180 (see slide 9 of DFIG Set#2) as angle between Vs & Is). Ps is negative for gen; then cosφ is also negative; Ps is positive for motor; then cosφ is also positive; so here, Is is always positive, mag notation is right. Identify the current phasor as Therefore: 3 3 3 3 Recalling , we may write 3 3 3 3 The “±” is needed because the square root sign makes the numerator positive for all values of φ, but it needs to be able to take on both positive and negative values as does sinφ. 39

  40. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm DFIG for non-unity power factor Req jωsLm Vs Vr/s jωs Leq=jXeq/s 2 1 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. 3 4 If machine is supplying Q, then sign of Qs should be negative (according to defined directionality), sign of Im{Is*} should be negative, and therefore sign of Im{Is} should be positive (leading). Given cos φ is negative (generating): 3 3 3 3 3 3 3 3 If machine is absorbing Q, then sign of Qs should be positive (according to defined directionality), sign of Im{Is*} should be positive, and therefore sign of Im{Is} should be negative (lagging). Given cos φ is negative (generating): 40

  41. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm Req jωsLm Vr/s Vs DFIG for non-unity power factor jωs Leq=jXeq/s Define the magnetizing current factor: Is,rated is the stator current corresponding to the highest active power generated by the wind turbine (which occurs at maximum speed since Pmech=(ωm/p)Tem and Tem=kωm2). From the circuit, KCL requires: But Im is almost entirely imaginary: or (“Almost” because, although Im is 90° out of phase with Vm, it is not with Vs, which is the ref.) Substitution of the Im expression into the rotor current expression yields: 3 3 3 3 If the machine is absorbing Q, then, from the previous slide: 3 3 3 3 (We will also do the “supplying” case shortly) Substituting into Ir: 41

  42. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm DFIG for non-unity power factor Req jωsLm Vs Vr/s jωs Leq=jXeq/s Assume the machine is operated at rated power, Ps,rated. Therefore, the above becomes: Recall from slide 39: and then substitute into previous expression: 3 3 3 3 3 3 3 3 42

  43. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm DFIG for non-unity power factor Req jωsLm Vs Vr/s jωs Leq=jXeq/s  Factor out the -Ps,rated/3Vs Combine terms with “j”  3 3 3 3 3 3 3 3 Simplify  43

  44. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm DFIG for non-unity power factor Req jωsLm Vs Vr/s jωs Leq=jXeq/s So this is for absorbing (underexcited operation) If we repeat the exercise for supplying (overexcited operation), we will obtain this: 3 3 3 3 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! 3 3 3 3 And so the rotor winding should be rated for the overexcited operation, at rated stator active power output. This would be as follows…. (next slide) 44

  45. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm DFIG for non-unity power factor Req jωsLm Vs Vr/s jωs Leq=jXeq/s Rotor current for rated stator active power and reactive power generation (overexcited case). You would identify the rated rotor current from the minimum power factor required by industry standards. Assume this is pf=cosφ’. It might be 0.95. At some point in the future, it might be 0.80! Then: 3 3 3 3 3 3 3 3 It is important to understand how wind turbines were designed to operate before requiring them to operate in some certain fashion. 45

  46. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm DFIG for non-unity power factor Req jωsLm Vs Vr/s jωs Leq=jXeq/s Rotor current for rated stator active power and reactive power generation (overexcited case). Let’s see what design requirements are of the rotor relative to the stator. This requires inspecting the relative magnitude between Ir and Is,rated. Again, from slide 39: 3 3 3 3 3 3 3 3 Here, remember that Krs can give Ir,rated/Is,rated by simply using the minimum pf=cosφ’ in the formula. 46

  47. Krs factor for a given stator power factor at rated stator active power. Km factor at rated stator active power: Decreasing Nr Increasing rotor conductor rating DFIG for non-unity power factor In the plot, the prime notation on the Krs at the top of the graph indicates the values have been referred to the rotor for an a=Ns/Nr=Ir/Is of about 0.3 (so actual stator currents are multiplied by 0.3 in computing Krs’ ). 1. The Km factor is a proxy for amount of current Im necessary to achieve required magnetization level. For a given induced rotor voltate, magnetization increases with Im and Nr. Thus we save money by decreasing Nr if we also increase Km. 2. For a given pf=cosφ, the rotor winding current rating, indicated by K’rs, gets higher as the Km increases (and the number of turns decreases). So savings in number of turns might be offset by increased rotor conductor current rating. 3. For a given Km, the rotor current rating increases as required power factor decreases. 4. Because Ir,rated also is the DFIG converter current rating, Krs provides an indication of the DFIG converter rating to the converter rating necessary for a full converter interface. 47

  48. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm Req jωsLm Vr/s Vs DFIG for non-unity power factor jωs Leq=jXeq/s Now let’s look at the effect on the rotor current rating of lowering the required pf. The previous slide did this in terms of Ir/Is,rated. Now we want this in terms of Ir/Irpf=1.0. Rotor current for rated stator active power and reactive power generation (overexcited case). Rotor current for rated stator active power and stator unity power factor 3 3 3 3 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 3 3 3 3 48

  49. Rr Rs Is jωsLσs Rr(1-s)/s jωsLσr Ir Req(1-s)/s Im Vm Req jωsLm Vr/s Vs DFIG for non-unity power factor jωs Leq=jXeq/s Let’s take a look at the ratio of Krs/Krspf=1.0. 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. 3 3 3 3 3 3 3 3 From last slide: 49 Ir/Irpf=1.0 is almost the same as the ratio of Krs/Krspf=1.0.

  50. And so we observe from this plot that, changing the requirement from unity pf to 0.8, the Krsratio increases by 15% if high magnetization current is needed (low number of turns) and by about 6% if lower magnetization current is needed (high number of turns). DFIG for non-unity power factor Dividing the values on the vertical axis by 0.8, we obtain the Ir/Irpf=1.0 which will be 1.15/.8=1.4375 and 1.06/0.8=1.325, indicating when changing the requirement from unity pf to 0.8, rotor current increase by 44% if high magnetization current is needed (low number of turns) and by 33% if lower magnetization current is used (high number of turns). 50

More Related