Answer each questionin a paragraph in 12hr

EE 499-002 WIND POWER #3 WIND GENERATORS Dr. Venkata Yaramasu Assistant Professor of Electrical Engineering Director of Advanced Motors, Power Electronics, and Renewable Energy (AMPERE) Laboratory School of Informatics, Computing, and Cyber Systems (SICCS) Northern Arizona University Phone: +1-928-523-6092 E-Mail: [email protected] Office Hours:

MoWeFr 12.30-1.30 p.m. in #69-210 By appointment in #90-112 Lecture:

MoWe 11.30 a.m. to 12.20 p.m. in #69-224 Lab:

Friday 11.30 a.m. to 2.00 p.m. in #69-234 Slides Credit:

Dr. Bin Wu, Ryerson University, Canada. Spring 2017Photo Courtesy:

Nordex Topics 2 1.Introduction 2.Reference Frame Transformation 3.Induction Generators 4.Synchronous Generators 1. Introduction Wind Generator Classification 3 Fig. 3-1 Classification of commonly used electric generators in large wind turbines Figure # in the textbook 1. Introduction Large Wind Generators 4 Generator Ty p eDoubly Fed Induction Generator (DFIG)Squirrel Cage Induction Generator (SCIG)Wound Rotor Synchronous Generator (WRSG)Permanent Magnet Synchronous Generator (PMSG) Rated Voltage (line-to-line) 690V 690V 400V 700V 3000V Rated Power 2MW 2.3MW 2.3MW 2.75MW 5.32MW Rated Stator Frequency50Hz 50Hz na na 19.6Hz 900 – 1900rpm600 – 1600rpm6 – 21.5rpm 6 – 18rpm58.6 – 146.9rpm Number of Poles 4 4 72 120 28 Brand/Model Gamesa G90Siemens SWT-2.3-101Enercon E-70Avantis AV928Multibrid M5000 2. Reference Frame Transformation Space Vector 5 3 2  3 2  abcframe 3 2  x Fig. 3-2 Space vector and its three-phase variablesx a,x b andx c. 2. Reference Frame Transformation abc/dq Transformation: dq arbitrary frame 6 3 2  3 2  3 2                               c ba q dx x x x x ) 3 / 4 sin( ) 3 / 2 sin( sin) 3 / 4 cos( ) 3 / 2 cos( cos 3 2                                           q d c bax x x x x ) 3 / 4 sin( ) 3 / 4 cos() 3 / 2 sin( ) 3 / 2 cos(sin cos         For a balanced system, the inverse transformation: (3.1) (3.3) Fig. 3-3 Equation # in the textbook 2. Reference Frame Transformation abc/dq Transformation: dq rotating reference frame 7 Space vector rotates the same speed as that of the dqframe.

x  3 2  3 2  dqxx 1 tan   3 2  x   is the synchronous speed of an induction or synchronous generator. Fig. 3-4 Decomposition of space vector into the dqrotating frame 2. Reference Frame Transformation abc/αβTransformation: Stationary Frame 8                          c bax x x xx 2 / 3 2 / 3 02 / 1 2 / 1 1 3 2                                   xx x x x c ba 2 / 3 2 / 12 / 3 2 / 10 1 For a balanced system, the inverse transformation: (3.6) (3.7) θin (3.1) = 0 3. Induction Generators Construction 9 - An induction machine can be used as a motor or a generator - The construction for the induction generator and motor is the same. Fig. 3-5 Cross-sectional view of and SCIG 3. Induction Generators Construction 10 SCIG Courtesy of ABB DFIG - Doubly fed induction generator SCIG - Squirrel cage induction generator DFIG 3. Induction Generators Induction Machine dq-axis Model 11 r s R R, – stator and rotor winding resistance lr ls L L, m L – stator and rotor leakage inductances – magnetizing inductance Fig. 3-8 Induction motormodel in the arbitrary reference frame v ds ,v qs –dq-axis stator voltage v dr ,v qr –dq-axis rotor voltage ids ,i qs –dq-axis stator current idr ,i qr –dq-axis rotor current ω –speed of the arbitrary reference frame ω r –speed of the synchronous reference frame qs  sR dsi lsL dmi mL lrL qr r   ) (  dsv drv rR dri ds p dr p ds  sR qsi lsL qmi mL lrL dr r   ) (  qsv qrv rR qri qs p qr p 3. Induction Generators Induction Machine dq-axis Model (Review) 12 Induction motor dq-axis model:

• Widely accepted • Extensively used qs  sR dsi lsL dmi mL lrL qr r   )(  dsv drv rR dri ds p dr p ds  sR qsi lsL qmi mL lrL dr r   )(  qsv qrv rR qri qs p qr p Fig. 3-8 Q: Can we use it for IG?

A: Q: How to operate an induction machine as a generator?

A: 3. Induction Generators Induction Machine dq-axis Model (Review) 13 Impor tant note:

When the model is used for an induction generator, the reference direction for the stator and rotor currents remain unchanged.

As a result, all the equations remains unchanged. qs  sR dsi lsL dmi mL lrL qr r    ) ( dsv drv rR dri ds p dr p ds  sR qsi lsL qmi mL lrL dr r   ) ( qsv qrv rR qri qs p qr p 3. Induction Generators Voltage and Flux Linkage Equations (Review) 14                   dr r qr qr r qrqr r dr dr r drds qs qs s qsqs ds ds s ds p i R vp i R vp i R vp i R v         ) () ( 1) Voltage equations: p = d/dt (3.14) qs  sR dsi lsL dmi mL lrL qr r   ) (  dsv drv rR dri ds p dr p ds  sR qsi lsL qmi mL lrL dr r   ) (  qsv qrv rR qri qs p qr p Three important equations for induction machine model 3. Induction Generators Voltage and Flux Linkage Equations (Review) 15                         qs m qr r qs m qr m lr qrds m dr r ds m dr m lr drqr m qs s qr m qs m ls qsdr m ds s dr m ds m ls dsi L i L i L i L Li L i L i L i L Li L i L i L i L Li L i L i L i L L ) () () () (     m ls s L L L  m lr r L L L  - Stator self inductance - Rotorself inductance(3.15) qs  sR dsi lsL dmi mL lrL qr r   ) (  dsv drv rR dri ds p dr p ds  sR qsi lsL qmi mL lrL dr r   ) (  qsv qrv rR qri qs p qr p 2) Flux linkage equations: 3. Induction Generators Torque Equation 16 3) Electromagnetic torque: ) ( 2 3 qsdsdsqs e ii P T    Note: Pis the number of pole pairs (3.16) qs  sR dsi lsL dmi mL lrL qr r   )(  dsv drv rR dri ds p dr p ds  sR qsi lsL qmi mL lrL dr r   )(  qsv qrv rR qri qs p qr p Q: The electromagnetic torque can be calculated by different equations, but the above one is extensively used. Why? A: 3. Induction Generators Simulink Model (Review) 17 From the voltage equations:                          S i R vS i R vS i R vS i R v dr r qr r qr qrqr r dr r dr drds qs s gs qsqs ds s ds ds // / /                            dr r qr qr r qrqr r dr dr r drds qs qs s qsqs ds ds s ds p i R vp i R vp i R vp i R v         ) () ( From the flux linkage equations:                         qs m qr r qs m qr m lr qrds m dr r ds m dr m lr drqr m qs s qr m qs m ls qsdr m ds s dr m ds m ls dsi L i L i L i L Li L i L i L i L Li L i L i L i L Li L i L i L i L L ) () () () (                                               qr dr qs ds r mr mm sm s qr dr qs dsi i i i L LL LL LL L 0 00 00 00 0     (3.17) (3.18) 3. Induction Generators Simulink Model (Review) 18 ] [ ] [ ] [ ] ][ [ ] [ ] [ ] [ ] ][ [ ] [ 1 1 1           L i i L L L i L Manipulation of flux linkage equations:                                                qr dr qs ds s ms mm rm r qr dr qs ds L LL LL LL L D i i i i     0 00 00 00 0 1 1 2 1m r s L L L D  where(3.20) (3.21) Motion equation:Important note:

For generator operation, bothT e and shaft mechanical torqueT m < 0             (b) 2 3(a) qs ds ds qs em e r i i P TT T JSP    3. Induction Generators Simulink Model (Review) 19 Fig. 3-9 3. Induction Generators Simulation Block Diagram 20 Circuit Breaker (Stator Frame) vas vbs vcs (vds ) ids iqs ias ibs ics T e T m r abc (Eq. 3.6) abc (Eq. 3.8) vβs vαs (vqs )IG Model in Arbitrary Reference Frame (Fig. 3-9)( iαs ) ( iβs ) Grid vdr = 0 vqr = 0 (for SCIG ) Fig. 3-10 Block diagram for dynamic simulation of SCIG with direct grid connection 3. Induction Generators Per Unit System 21 Table A-1, Appendix A 3. Induction Generators Per Unit System 22 Example Generator ratings: 690V, 50Hz, 2.59MVA Base voltage: 690/ ; Base current: 2168A; Base impedance: 0.1838 Ω Stator resistance: 0.011 Ω→ 0.006 pu 3 3. Induction Generators Case Study 3-1: Direct Start-up of SCIG 23 Purpose Investigate the dynamic performance of the SCIG wind energy system during startup with direct grid connection Induction generator 2.3MW / 690V / 50Hz / 1512rpm SCIG GB Grid CB System configuration 3. Induction Generators Case Study 3-1: Direct Start-up of SCIG 24 3 Generator Parameters (Table B-1, Appendix B) 3. Induction Generators Case Study 3-1: Direct Start-up of SCIG 25 Generator Parameters (Table B-1, Appendix B) 3. Induction Generators Case Study 3-1: Direct Start-up of SCIG 26 Simulation Conditions Starting procedures • The mechanical safety brake for the wind turbine is released •The blades are pitched slightly into the wind •The blades start to rotate → the rotor of the generator starts to rotate as well.

•The circuit breaker is closed when the rotor speed reaches 1450rpm (0.959pu) Note • During the startup, the mechanical input torque is very low since the blades are pitched slightly into the wind.

•After the startup, the pitch angle is adjusted to its optimal value to harvest the maximum power from the wind. 3. Induction Generators Case Study 3-1: Direct Start-up of SCIG 27 Q:

High inrush current. Why? Simulated waveforms –transient stator current 3. Induction Generators Case Study 3-1: Direct Start-up of SCIG 28 Note:

• High torque oscillations during the startup •The rated rotor speed (1512rpm, 1pu) is used as the base speed → synchronous speed ω s(1500rpm, 0.992 pu) is lower than the rated rotor speed. 3. Induction Generators Generator Steady State Operation 29 IEEE Recommended equivalent circuit Note:

•The steady state equivalent circuit can be derived from the dq-axis IG model •Pay attention to the current direction of I s andI r Is jX ls jX lr R r jX m V s Ir R s ss R r Im P ag P m Fig. 3-15 3. Induction Generators Torque-Speed Curve Derivation 30 m m m T P                 r r r r r m m R ss I P R ss I T1 3 / 1 1 3 1 2 2   P P s R I P T sag r r s m / 3 / 1 2           s R I P r r ag2 3  r r m R ss I P) 1 ( 3 2   Mechanical power derived from steady-state circuit:

Replacing in the definition of mechanical power leads to:

Using the definition of slip (1 ‐s) = ω r/ω s yields:

where P ag is the airgap power given by (3.28) (3.34) (3.35) (3.36) 3. Induction Generators Torque-Speed Curve Derivation 31  2 2 lr ls r ss r X X s R RV I            2 22 3 lr ls r ss r s m X X s R RV s R P T              Neglecting the magnetizing branch in the steady state circuit simplifies the rotor current calculation:

Finally, replacing I r in the torque equation of previous slide It relates T m with slip sfor a given stator voltage V s and stator frequency  s (3.37) (3.38) 3. Induction Generators Torque-Speed Curve 32 Generator Operation Slip < 0 T m < 0 Motor Operation Slip >0 T m > 0 Fig. 3-15 3. Induction Generators Power Flow 33 s cu r cu rot in s P P P P P , ,     For induction generator: Fig. 3-14 Power flow and losses in an induction generator 3. Induction Generators Case Study 3-2: Power & Efficiency Analysis of IG 34 Given:

•Induction generator: 2.3MW 690V, 50Hz, 1512rpm, squirrel cage (Table B-1, Appendix C) •Operating conditions: for a give wind speed, the rotor speed is 1506rpm, rotational losses are 23KW, and IG is grid connected.

Find:

•Stator and rotor currents (rms) •Stator power, mechanical power, input power, power factor •Stator and rotor winding losses, and generator efficiency 3. Induction Generators Case Study 3-2: Power & Efficiency Analysis of IG 35 Solution:

Q:Why negative slip? V V s      0 4 . 398 0 3 / 690 004 . 0 1500 / ) 1506 1500 (     s Rated phase voltage:

Slip:

Total Impedance:        3 . 145 330 . 0 ) //( lr r m ls s s jX s R jX jX R Z Stator and rotor currents :                        A 8 . 173 0 . 1030 /A 3 . 145 9 . 1206 3 . 145 330 . 00 3 / 690 lr r ms m rs s s jX s R jXI jX IZ V I Stator circuit Rotor circuitAirgap Is jX ls jX lr Rr jX m Vs Ir Rs ss 1 R r Im s Z 3. Induction Generators Case Study 3-2: Power & Efficiency Analysis of IG 36 Stator circuit Rotor circuitAirgap Is jX ls jX lr Rr jX m Vs Ir Rs ss 1 R r Im Solution (Continued) Mechanical power:

Mechanical torque: Stator power factor angle and power factor:                   822 . 0 cos3 . 145 3 . 145 0 s ss s s PFI V   kW 2 . 1186 ) 822 . 0 ( 9 . 1206 3 / 690 3 cos 3        s s s s I V P  Stator Active Power: kW 78 . 1195 / ) 1 ( 3 2    s s R I P r r m m kN 7.58 ) 60 / 2 1506 (10 8 . 1195 3           m m m P T 3. Induction Generators Case Study 3-2: Power & Efficiency Analysis of IG 37 Stator power to the grid: Solution (Continued) Stator and rotor winding losses:         kW 4.76 3kW 4.82 3 2 ,2 , r r r cus s s cuR I PR I P kW 2 . 1186 , ,     r cu s cu m s P P P P Generator efficiency: kW 8 . 1218    rot m inP P P Total input power: % 33 . 97 /  in s P P  3. Induction Generators Summary of Operation 38 Table 3-2 Operation of induction machine as a motor/generator 4. Synchronous Generators Construction 39 Fig. 3-16 Salient-pole wound rotor synchronous generator 4. Synchronous Generators Construction 40 Stator Rotor Courtesy of Enercon Wound rotor synchronous generator 4. Synchronous Generators Construction 41 Surface Mounted Shaft S S s S S N SN S N SN N N N N Shaft Inset permanent magnet Stator Stator winding slot Rotor Air gap Fig. 3-17 Permanent magnet synchronous generator (PMSG) Inset Magnets 4. Synchronous Generators Construction 42 High pole number PMSG Courtesy of Avantis Permanent Magnet Synchronous Generators (PMSG) Low pole number PMSG Courtesy of ABB High pole number PMSG Permanent Magnet Synchronous Generators (PMSG) Low pole number PMSG Courtesy of ABB 4. Synchronous Generators SG Dynamic Model 43 qs r  dsv     sR lsL ds p dsi dmL dr p dmi f ds r  qsv   sR lsL qs p qsi qmL qmi             qs ds r qs s qsds qs r ds s ds p i R vp i R v      The same as that for IG except the stator current direction Rotor:

Wound rotor excited by field current I f Note: SG model is in the rotor synchronous frame (ω= ω r ) Fig. 3-19 General model for SG 4. Synchronous Generators SG Dynamic Model 44 qs r  dsv     s R lsL ds p dsi dmL dr p dmi f ds r  qsv   s R lsL qs p  qsi qmL qmi Stator flux linkages:          qm ls qdm ls df dm r L L LL L LI L                         qs q qs qm ls qsr ds d f dm ds dm lsds f dm ds ls dsi L i L Li L I L i L Li I L i L ) () (   Rotor flux and self inductances: (3.51) (3.52) 4. Synchronous Generators SG Dynamic Model 45        qs q qsr ds d dsi Li L               qs ds r qs s qsds qs r ds s ds p i R vp i R v                   qs q r r ds d r qs s qsds d qs q r ds s ds pi L i L i R vpi L i L i R v     And assuming dλ r /dt = 0 due to constant If: (3.53) (3.51) Substituting λds and λqs into the following equations(From previous slide) 4. Synchronous Generators SG Dynamic Model 46              qs q r r ds d r qs s qsds d qs q r ds s ds pi L i L i R vpi L i L i R v     Fig. 3-20 Simplified SG model(3.53) (From previous slide) 4. Synchronous Generators SG Dynamic Model 47  ds qs qs ds ei i P T     2 3    qs ds q d qs r ei i L L i P T    2 3                       qs q qs qm ls qsr ds d f dm ds dm lsds f dm ds ls dsi L i L Li L I L i L Li I L i L ) () (   Torque equation:  ds qs qs ds ei i P T     2 3 Another torque expression:

For non-salient pole SG: L q = L d For salient pole SG: L q > L d The SG model can used for both non-salient and salient SG. (3.54) (3.55) 4. Synchronous Generators SG Dynamic Model 48 PMSG:λ r is produced by permanent magnet WRSG:λ r is produced by the rotor field winding Note:The SG model can used for both PMSG and WRSG. 4. Synchronous Generators SG Dynamic Model 49       qsqrrdsdrqssqs dsdqsqrdssds piL iLiRv piLiLiRv   Q:

How to build Simulink Model? A:    qsdsqdqsr eiiLLi P T    2 3  ine r TT J P dt d   4. Synchronous Generators SG Steady-State Model 50       rrdsdrqssqs qsqrdssds iLiRv iLiRv          qsqrrdsdrqssqs dsdqsqrdssds piL iLiRv piLiLiRv   Steady state model:

L(di/ dt) = 0 Q: Why L(di/dt) = 0?

A: 4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load 51 Rated Shaft Input Power 2.5 MW Rated Line-to-line Voltage 3950V (rms) Rated Stator Current 485A (rms) 1.0 pu Rated Apparent Power 3.32MVA 1.0 pu Rated Rotor Speed 400 rpm 1.0 pu Rated Torque 59.66 kN.m 1.0 pu Rated Stator Frequencyf s 40Hz 1.0 pu Number of Pole pairsP6 Rated Rotor Flux Linkageλ r 6.774Wb (peak) 0.528pu Base Impedance 4.70Ω1.0 pu Stator ResistanceR s 0.035Ω0.007pu d-axis Synchronous Inductance L d 10.7mH 0.574pu q-axis Synchronous InductanceL q 23.9mH 1.276pu Q:Salient or non-salient pole SG? 4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load 52 mH 26 . 8  L L  23 . 4 L R Given:

PMSG, standalone operation, operating at rated rotor speed, RL load.

Find:

Stator current, stator voltage, output power, output power factor Solution: ) ( ) ( ) )( ( ds L r qs L qs L r ds LL r L qs ds qs dsi L i R j i L i RL j R ji i jv v                        ds L qs L ds L r qs L qsqs L ds L qs L r ds L dsi X i R i L i R vi X i R i L i R v   (3.60) (3.61) 4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load 53 From (a): Substituting to (b):         ds L qs L qsqs L ds L dsi X i R vi X i R v (b)(a)              ds L r qs L r r ds d r qs sqs L r ds L qs q r ds si L i R i L i Ri L i R i L i R      qs s Lq L r dsi R RL L i  ) (  A 85 . 141 ) )( ( ) () ( 2 2       q L d L r s Ls L r r qs L L L L R RR R i    (3.61) (3.62) (3.64) (From previous slide) 4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load 54 Stator Voltage: A 0 . 249 ) (    qs s Lq L r dsi R RL L i  A 7 . 202 2 / 2 2    qs ds si i I              V 7 . 1124 V 9 . 772 r r ds d r qs s qsqs q r ds s ds i L i R vi L i R v     V 0 . 965 2 / 2 2    qs ds sv v V (3.65) (3.66) (3.67) (3.68) 4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load 55 Mechanical Power P – number of pole pairs Stator Winding Loss Output Power   m . kN 7 . 12 2 3     qs ds q d qs r ei i L L i P T  kW 0 . 531 /   P T T P r e m m m   kW 0 . 3 3 2 ,   s s s cu R I P kW 0 . 528 ,    s cu m L P P P (3.69) (3.71) (3.72) Electromagnetic Torque 4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load 56 Power Factor Alternative Way to Calculate PF (Method 1)             8 . 25 tan 1 LL r L RL   9 . 0 ) cos(   L L PF            kVA 7 . 255 ) ( 5 . 1kW 0 . 28 5 ) ( 5 . 1 qs ds ds qs Lqs qs ds ds Li v i v Qi v i v P 9 . 0 2 2    L LL L Q PP PF      8 . 25 9 . 0 cos 1 L  (3.74) (3.75) (3.76) 4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load 57 Alternative Way to Calculate PF (Method 2) Active load power can also be calculated as Considering rotational losses of the generator of 0.5% (12.5 kW), the efficiency is                              7 . 29 tan5 . 55 tan 11 ds qs ids qs v i iv v           9 . 0 cos 8 . 25 L Li v L PF     kW 0 . 528 cos 3   L s s L I V P  0.972 5 . 12 0 . 5310 . 528      rot mL P PP  (3.80) (3.82) Problems 58 Try to solve the following problems given in Appendix C of the textbook:

3-3, 3-4, 3-5, 3-6, and 3-8.