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EE 559 2016 Iowa State University

EE 559 2016 Iowa State University. Overview of Wind Generators. James D. McCalley. Electric Generators. Type 1 Conventional Induction Generator (fixed speed).

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EE 559 2016 Iowa State University

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  1. EE 5592016Iowa State University Overview of Wind Generators James D. McCalley

  2. Electric Generators Type 1 Conventional Induction Generator (fixed speed) Types 1-3 are induction generators and must have gearboxes to change “slow” rotor rotational speed (5-20rpm) to “fast” generator rotational speed (750-3600rpm) needed by the 60 Hz grid freq seen in stator winding. Type 2 Wound-rotor Induction Generator w/variable rotor resistance Type 3 Doubly-Fed Induction Generator (variable speed) Type 4 maintains generator at “slow” rotor rotational speed producing low freq in stator winding which is transformed by converter to grid. Also called “direct-drive.” Usually uses a PM synchronous generator. Type 4 Full-converter interface Plant Feeders ac dc generator to to dc ac full power

  3. Two kinds of AC generators Note: DC gens can be used (and sometimes are for small wind turbines), but are not used for utility-scale wind turbines because their brush/commutator configuration is high-maintenance. Permanent magnet synchronous machine For type 4 For Types 2, 3 Steam & Gas Turbines Hydro-gens For Type 1 R. Gasch and J. Twele, “Wind Power Plants: Fundamentals, Design, Construction & Operation,” Chapter 11, Springer-Verlag, Berlin, 2012.

  4. Two kinds of AC generators for wind Synchronous gen, type 1: *stator freq=k×rotor speed. Induction gen, types 1-3: stator freq fixed by rotor speed and rotor freq. Connects to grid via converter Connects directly to grid (60Hz) Stator winding (AC) DC, supplied by external circuit or rotor is permanent magnet AC, induced via SLIP between rotor and stator mag field Rotor winding None (fixed speed) or Full converter interface (variable speed) Electric control DFIG (variable speed) ns is rpm, p is pole pairs. 4

  5. Synchronous Machine Structure Phase A STATOR (armature winding) ROTOR (field Winding or perm magnet) + 120° N Phase B One pole pair (p=1) + DC Voltage S The negative terminal for each phase is 180 degrees from the corresponding positive terminal. + Current is supplied to rotor windings externally, or the rotor can carry a permanent magnet. The two magnetic fields (one from stator and one from rotor) interact to produce torque. Phase C Air Gap

  6. Induction Machine Structure Phase A STATOR (armature winding) ROTOR (Squirrel cage) + 120° Phase B + Voltage is induced within rotor windings by relative motion (SLIP) between those windings and magnetic field produced by stator (Faraday’s Law). Then two magnetic fields (one from stator and one from rotor) interact to produce torque. The negative terminal for each phase is 180 degrees from the corresponding positive terminal. + Phase C Air Gap

  7. Some fundamental machine concepts:a. Rotating magnetic field b. Faraday’s lawc. Torque production

  8. Basic Concepts rotor Balanced voltages applied to stator windings provides a rotating magnetic field of speed ns. (fs: 60 Hz, p: # of pole pairs) Let’s look at that rotating magnetic field from the stator; the principles apply equally well to the induction machine and to the synchronous machine. 8

  9. There are 3 stator windings, separated in space by 120°, with each carrying AC, separated in time by ω0t=120°. • For cnvntnl sync gen, windings directly connected to 60 Hz grid. • For types 1-3 WT (ind gen), windings directly connected to 60Hz grid. • For type 4 WT (PMSG), windings connected to grid via converter. • Each of these three currents creates a magnetic field in the air gap of the machine. Let’s look at only the a-phase: Rotating magnetic field • Ba, in webers/m2, is flux density from the a-phase current. • /_Ia is there because it is the current that produces the field. • α is the spatial angle along the air gap. • For any time t, α=0,180 are spatial maxima (absolute value of flux is maximum at these points).

  10. Let’s fix α=0 and see what happens at 4 sequential times: 1. ω0t1, such that ω0t1+∟Ia is just less than π/2 2. ω0t2= ω0t1+90, 3. ω0t3= ω0t1+180, 4. ω0t4= ω0t1+270 Rotating magnetic field: fixed in space Positive rotation assumed in the counterclockwise direction.

  11. Fix t=t2: observe field densities at α=0, α=45, α=90, α=135, α=180, α=225, α=270, α=315. Fix t=t3: observe field densities at α=0, α=45, α=90, α=135, α=180, α=225, α=270, α=315. Rotating magnetic field: fixed in time • The magnetic field is sinusoidally distributed around the airgap. • The spatial maxima are always at the same location. • The above two attributes are consistent with the equation, when the first sinusoid is fixed.

  12. Now let’s consider the magnetic field from all three windings simultaneously. (1) Rotating magnetic field (2) (3) • Add them up, then perform trig manipulation to obtain: (4) Notice that each of (1), (2), and (3) has a spatial maximum which is fixed in space, as dictated by the second term of each the 3 expressions. But the spatial maximum of (4) rotates. This is a characteristic of a rotating magnetic field. One can observe this using the following: http://educypedia.karadimov.info/library/rotating_field.swf www.youtube.com/watch?v=eQk0OznWTjM The shape of the individual winding fields throughout the airgap are spatially fixed, but their amplitudes pulsate up and down. In contrast, the amplitude of the composite is fixed in time, but it rotates in space. What you will see in the video are just the variation of the maximum field point.

  13. Induction Machine – Faraday’s Law We have established that the existence of three phase currents in the three stator windings creates a rotating magnetic field in the air gap. When the stator windings are initially energized, the rotor is stationary, and so the rotor windings see this rotating magnetic field. What happens? The rotor windings are seeing a time-varying magnetic field, and by Faraday’s law, a voltage will be established in them. This voltage causes currents to flow, which in turn creates a rotating magnetic field in the air gap. A key point is that the rotor must rotate at a different speed than the rotating magnetic field of the stator, otherwise there is no relative motion between the rotor windings and the magnetic field the rotor windings seeSLIP.

  14. Induction Machine - Slip • s is “slip” which quantifies how much the rotor “slips” behind the stator-produced rotating magnetic field. More precisely, slip is • the amount (in electrical radians/sec) by which • the rotational speed of the stator rotating magnetic field exceeds • the rotational speed of the rotor • normalized by rotational speed of stator rotating magnetic field • Can slip be 0 in the steady-state? • No, because if it were, there would be no induced Vno I flow in rotor windings. • Now, denote rotor velocity in mechanical rad/sec as Ωm, and • p=number of pole pairs on the rotor, then: • Notes: • Some treatments define p=number of poles, in which case • The number of poles on rotor must be the same as the number of poles on stator.

  15. Induction Machine - Speeds We may also write this in terms of mechanical revolutions per minute (rpm), denoted by n, according to: Transform Ω to n: Or, since Ωm=ωm/p, Ωs=ωs/p, transform ω to n: Or, since ω=2πf, transform fs to ns: (If p=number of poles, then )

  16. Induction Machine - Speeds Summary: And finally:

  17. Induction Machine - Example Example: Consider a four-pole 60Hz induction machine. Compute slip, Ωm, and ωm for the following values of rpm rotor speed: 0, 500, 1000, 1750, 1800, 1850, 2600, 3600. Solution: To make these computations, we first compute synchronous speed: or Then we can use the following:

  18. To produce constant torque, two rotating magnetic fields must have the same rotational velocity. In the case of generator operation, we can think of bar A (the rotor field) as pushing bar B (the armature field), as in Fig. a (N pole repels N pole). In the case of motor operation, we can think of bar B (the armature field) as pulling bar A (the rotor field), as in Fig. b (N pole attracts S pole). Torque production Note: Analytic treatment of torque production is done using the coupling field approach. We will treat this a little later.

  19. Induction Machine – Freq of Rotor Currents Fact 1: The rotor must have speed (ωm) which differs from speed of rotating magnetic field (ωs) from stator. Otherwise, no voltage is induced in rotor windings. Fact 2: Constant torque only exists if the two rotating magnetic fields (one from stator currents and one from rotor currents) are at the same speed. How can Fact 1 and Fact 2 both be true? Speed of rotating magnetic field from rotor= rotor speed + {rotating magnetic field speed relative to rotor} Speed of RMF from rotor, relative to rotor Rotor speed Speed of RMF from stator Frequency (rad/sec) of rotor currents.

  20. Induction Machine – Clarification ωr is the frequency (rad/sec) of the electrical quantities in the rotor winding. But your added statement may need some further clarification, “due to rotor being in rotating stator magnetic field.” ωr occurs due to the motion (speed) of the magnetic field and the motion (speed) of the rotor: ωr= ωs-ωm. From another view, it is the speed of the rotating magnetic field from the rotor referenced to the rotor. ωm would be the speed of the rotating magnetic field from the rotor if the rotor currents were DC. It is the difference between • the speed of the rotating magnetic field from the stator and • the speed of the rotating magnetic field from the rotor referenced to the rotor ωm= ωs-ωr The little omegas (ωr, ωs, ωm) are related to their corresponding big omegas (Ωr, Ωs, Ωm) by ω=pΩ where p is the number of pole pairs. When p=1, they are equal. When p=2, the little omegas double the big omegas, etc. This shows that the little omega quantities reflect electrical “rotation” that is easily visualized as the sinusoid associated with the corresponding electrical quantities (currents and voltages); it is also visualized as the rotating B-field. If there are 2 poles, one full mechanical rotation takes the corresponding electrical quantities through one full electrical rotation. If there are 4 poles, one full mechanical rotation takes the corresponding electrical quantities through two full electrical rotations, etc.

  21. Transformer Per-phase Equivalent Circuit An impedance …An induction machine may be thought of as a transformer with a rotating secondary.

  22. Induction Machine Per-phase Equivalent Circuit Notation issue: In these slides, we are using the prime notation to indicate the quantity is referred from rotor to stator (in the DFIG notes, we use primed notation to indicate rotor quantity is referred to rotor). I’2 For transformer, this is an impedance The main difference between the transformer equivalent cct and the induction machine equivalent cct is the loading: the transformer load is an actual impedance whereas the induction machine load is a variable resistance that depends on “s”.

  23. Induction Machine Per-phase Equivalent Circuit I’2 What does the induction mach load represent? It represents the power provided to or from the shaft. Thus the induction mach load is a purely “real” load since energy transfer by mechanical means must be MW only (this is not the case for the transformer).

  24. Induction Machine Per-phase Equivalent Circuit Denote load resistance as: I’2 Req= R2’ is the rotor resistance referred to the stator (as indicated by the prime – see slide 20). The variable “s” is the slip, as defined earlier. “Referred to”: ohmic value of impedances differ in numerical value, according to Zprim/stat=(Nprim/stat/Nsec/rot)2 Zsec/rot based on whether they are represented on the primary/stator side or on the secondary/rotor side of the transformer/machine, where N is the number of turns on the indicated side of the device.

  25. Induction Machine - Power What about real power? I’2 Req= P1 PD

  26. Induction Machine Per-phase Equivalent Circuit Power crossing air gap? Includes mechanical load & rotor winding losses I’2 PG Req= P1 PD

  27. Induction Machine Per-phase Equivalent Circuit I’2 PG Req= P1 PD Compare PG to PD: Conclusion:

  28. Induction Machine - Torque Torque expression: Mechanical radians/sec Electrical radians/sec But recall that: Substitution yields: Assuming we know the applied voltage and speed (or slip), then computing torque requires getting I’2. We can do this using circuit analysis.

  29. Induction Machine – Obtaining I’2 I’2 Recall: When 2 impedances are in parallel, we obtain their equivalent using: Req One method: Given slip, compute load resistance, Req. Then Ohm’s Law provides current as applied voltage over equivalent impedance Then, by current division, we obtain: Now, for various values of slip, from s=-1 to s=+1, compute I’2 and subsequently T. Then plot T vs. s.

  30. Induction Machine – Obtaining I’2 I’2 Example: Consider a 6-pole induction generator with line-line voltage of 220 volts, and the below data.. R1=0.294 Ω X’2=0.209 Ω X1=0.503 Ω R’2=0.061 Ω RC=1000 Ω Xm=13.25 Ω Compute the torque for f=60 Hz, for Ωm=130 rad/sec. We need to compute: Req Therefore… Compute the line-to-neutral voltage, V1=220/sqrt(3)=127.017volts. Compute Ωs=2(pi)f/p=2(pi)(60)/3=125.664rad/sec. Compute slip: s=(Ωs-Ωm)/Ωs=(125.664-130)/125.664=-0.0345 Compute Req=R2’(1-s)/s=0.061(1- -0.0345)/(-0.0345)=-1.8289 Now obtain: Now obtain: Compute:

  31. Induction Machine – Obtaining I’2 I’2 Example: Consider a 6-pole induction generator with line-line voltage of 220 volts, and the below data.. R1=0.294 Ω X’2=0.209 Ω X1=0.503 Ω R’2=0.061 Ω RC=1000 Ω Xm=13.25 Ω Compute the torque for f=60 Hz, for Ωm=130 rad/sec. From previous slide: Req So: Then, by current division, we obtain: Ntn-meters

  32. I’2 Equivalent circuit by Thevenin Req An alternative way to obtain I’2 is by use of Thevenin, looking into the terminals as shown.

  33. Za=R1+jX1 By voltage division: Equivalent circuit by Thevenin Zb= Rc//jXm Vth V1 Note that V1 is the line-to-neutral voltage, given by V1=VLL/sqrt(3). Za=R1+jX1 The two impedances are in parallel: Zb= Rc//jXm

  34. I’2 Zth jX’2 R’2 DFIG equivalent circuit Thevenin Vth R’2 (1-s)/s

  35. Induction Machine – Plotting T-s characteristic Example: Consider a 6-pole induction generator with line-line voltage of 220 volts, and the below data. Plot the torque-speed characteristic for f=60 Hz. R1=0.294 Ω X’2=0.209 Ω X1=0.503 Ω R’2=0.061 Ω RC=1000 Ω Xm=13.25 Ω Solution: The below calculations were carried out for this machine for a range of slips.

  36. Type 1 Induction Generator – TD vs. s But it has a wide torque range, dictated by the wind. And so wind gusts are passed on directly to the grid. Motor Region of operation Generator Torque Range It is referred to as “fixed speed” because of the narrow band of speed operation shown here. Slip 1 0 -1 Supersynchronous operation Subsynchronous operation 36

  37. Type 1 Induction Generator – TD vs. s Torque-speed characteristics associated with wind turbines are not simply a function of a mechanical load but also reflect the aerodynamic features of the wind-blade interaction, and as a result, appear as shown in the figure, where the numbers 4-10 correspond to wind speed (m/sec). We observe that as the wind speed increases, the electromagnetic torque increases significantly, but the speed of the machine changes very little. For this reason, this kind of wind turbine configuration, using an induction machine with a squirrel cage rotor, is referred to as a fixed-speed machine. Slip 1 0 -1 Supersynchronous operation Subsynchronous operation 37

  38. Type 1 Induction Generator – TD, PD, I1, I’2 It is also interesting to look at the real power consumed and the currents, given by the below expressions.

  39. Type 1 Induction Generator In the top plot, P is what we called PD on the previous slide (power developed) since, as the bottom figure indicates, it goes to 0 when I’2 goes to 0. Also, T is what we called TD. Observe that when TD, PD, and I2 are 0, I1 is not zero and has a value of Im, which is the magnetizing current. One can understand this by considering the equivalent circuit, see next slide. In this plot, I2 is what we called I’2

  40. Type 1 Induction Generator I’2 Req Note from our example data (slide 30) that RC=1000 Ω and Xm=13.25 Ω. This is quite typical in that RC>>XM. Thus, for analysis of currents, we can neglect RC. Therefore, the current flowing into the shunt leg is just the current flowing into XM, which is the magnetizing current, IM. Now consider the operating condition s=0, and considering s=(ωs-ωm)/ωs, we have that ωs=ωm, i.e., the rotor is at synchronous speed. This means that Req=(1-s)/s=∞, i.e., the rotor side of our model is open circuited. In this case, under the condition that RC>>XM, we have that I1=IM, as indicated by the previous diagram. The operating condition s=0 in an induction machine corresponds to the “open-circuit” (infinite load impedance) condition for a transformer.

  41. Type 1 Induction Generator Let’s also look at Q1 and QG. I’2 QG Q1

  42. Type 1 Induction Generator This is PD. Observe that it is 0 for s=0 Positive is absorbing This is Q1. Observe that it is just a little greater than 0 for s=0. This is due to the magnetizing current IM flowing through X1 and XM. Positive is absorbing Sources: 1. “Chapter 6: Electrical Generator Machines in Wind-Energy Systems.” 2. T. Burton, et. al, “Wind Energy Handbook,” 2nd edition, Wiley, pp 439-440.

  43. Type 1 Induction Generator – Q1, PD We can also compute Q1 and PD for the same value of s and then plot Q1 vs. PD.

  44. Type 1 Induction Generator - Q1, PD Sources: 1. “Chapter 6: Electrical Generator Machines in Wind-Energy Systems.” 2. T. Burton, et. al, “Wind Energy Handbook,” 2nd edition, Wiley, pp 439-440. Positive is absorbing So Type 1 machine always absorbs reactive power from the network independent of whether it is gen or motor. Windfarms withType 1 machines must have reactive compensation at the substation. Generally, it is supplied via capacitor banks.

  45. Type 1 Induction Generator – Voltage dips When an induction generator experiences a voltage sag on the network, the generator speed increases. If it is not disconnected, it can accelerate to an unstable condition. This is illustrated on the next slide, where, following a voltage drop, the machine moves from point X to point Y, where the machine speeds up because the mechanical torque (a constant) is higher than the electrical torque (which has decreased as indicated by the difference in torque values between point X and Y). If the acceleration moves beyond the pull-out torque, then further speed increases cause further decrease in electrical torque, resulting in more acceleration, an unstable condition. Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.

  46. Type 1 Induction Generator – Voltage dips Mechanical torque stays relatively constant. At point X, Tmech=Te. At point Y, Tmech>Te and so machine accelerates. If Te exceeds “pull-out” torque (the hump), then further acceleration (e.g., point Z) will decrease Te and further accelerate the machine which decreases Te which…. UNSTABLE! But this behavior (tripping) will violate voltage-ride-through requirements (see next slide). Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.

  47. Type 1 Induction Generator – Voltage dips To avoid violation of low-voltage ride-through requirements, windfarms using fixed-speed induction generators (type 1) may need to deploy dynamic var capabilities (e.g., SVC or STATCOM) at the substation.

  48. Type 2 Induction Generator – speed cntrl Why do we want to provide speed control? Because it enables us to get closer to the maximum power point of the turbine, as indicated by the below diagrams.

  49. Type 2 Induction Generator – with R control Some control may be achieved via use of a resistor in series with the rotor, increasing R’2. This means that the effective rotor resistance becomes: This means that we must substitute R’2,ext for R’2 in the Req expression. Then the total rotor circuit resistance can be written as:

  50. Type 2 Induction Generator – with R control The circuit model is shown below. Note that we have combined the equivalent resistance with the rotor resistance. If you want to compute PD, then you need to use Req,ext, as shown on the next slide. I’2 R’2/s Rext/s

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