Answer each questionin a paragraph in 12hr

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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 2017

Photo Courtesy: Nordex

Topics

1. Introduction

2. Reference Frame Transformation

3. Induction Generators

4. Synchronous Generators

1. Introduction Wind Generator Classification

Fig. 3-1 Classification of commonly used electric generators in large wind turbines

Figure # in the textbook

1. Introduction Large Wind Generators

Generator Type

Doubly 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 Frequency

50Hz 50Hz na na 19.6Hz

900 – 1900rpm

600 – 1600rpm

6 – 21.5rpm 6 – 18rpm 58.6 –

146.9rpm Number of

Poles 4 4 72 120 28

Brand/Model Gamesa G90 Siemens

SWT-2.3-101 Enercon E-70

Avantis AV928

Multibrid M5000

2. Reference Frame Transformation Space Vector

3 2

ab c

fr am

e 3

2

x 

Fig. 3-2 Space vector and its three-phase variables xa, xb and xc.

2. Reference Frame Transformation abc/dq Transformation: dq arbitrary frame

3 2

  





  



 



  

  

 

  



x x x

x x

)3/4sin()3/2sin(sin )3/4cos()3/2cos(cos

3 2

 

 

  

 

  





  





 

 

  





  





x x

x x x

)3/4sin()3/4cos( )3/2sin()3/2cos(

sincos

  

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

Space vector rotates the same speed as that of the dq frame.

x 

3 2

x x1tan 

3 2

x 

 is the synchronous speed of an induction or synchronous generator.

Fig. 3-4 Decomposition of space vector into the dq rotating frame

2. Reference Frame Transformation abc/αβ Transformation: Stationary Frame

  





  



 



  

  

 

  



x x x

x x

2/32/30 2/12/11

3 2





 

  

    





  





 

  





  









x x

x x x

2/32/1 2/32/1

For a balanced system, the inverse transformation:

(3.6)

(3.7)

θ in (3.1) = 0

3. Induction Generators Construction

- 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

SCIGCourtesy of ABB

DFIG - Doubly fed induction generator SCIG - Squirrel cage induction generator

DFIG

3. Induction Generators Induction Machine dq-axis Model

rs RR , – stator and rotor winding resistance

lrls LL ,

– stator and rotor leakage inductances

– magnetizing inductance

Fig. 3-8 Induction motor model in the arbitrary reference frame

vds , vqs – dq-axis stator voltage

vdr , vqr – dq-axis rotor voltage

ids , iqs – dq-axis stator current

idr , iqr – dq-axis rotor current ω – speed of the arbitrary

reference frame

ωr – speed of the synchronous reference frame

qs sRdsi lsL

dmi

lrL qrr  )( 

dsv drv

rR dri

dsp drp

ds sRqsi lsL

qmi

lrL drr  )( 

qsv qrv

rR qri

qsp qrp

3. Induction Generators Induction Machine dq-axis Model (Review)

Induction motor dq-axis model: • Widely accepted • Extensively used

qs sRdsi lsL

dmi

lrL qrr  )( 

dsv drv

rR dri

dsp drp

ds sRqsi lsL

qmi

lrL drr  )( 

qsv qrv

rR qri

qsp qrp

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)

Important 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 sRdsi lsL

dmi

lrL qrr  )( 

dsv drv

rR dri

dsp drp

ds sRqsi lsL

qmi

lrL drr  )( 

qsv qrv

rR qri

qsp qrp

3. Induction Generators Voltage and Flux Linkage Equations (Review)

 



 













drrqrqrrqr

qrrdrdrrdr

dsqsqssqs

qsdsdssds

piRv





)(

1) Voltage equations:

p = d/dt

(3.14)

qs sRdsi lsL

dmi

lrL qrr  )( 

dsv drv

rR dri

dsp drp

ds sRqsi lsL

qmi

lrL drr  )( 

qsv qrv

rR qri

qsp qrp

Three important equations for induction machine model

3. Induction Generators Voltage and Flux Linkage Equations (Review)

 



 





 

qsmqrrqsmqrmlrqr

dsmdrrdsmdrmlrdr

qrmqssqrmqsmlsqs

drmdssdrmdsmlsds

iLiLiLiLL iLiLiLiLL

)( )(

 

mlss LLL 

mlrr LLL  - Stator self inductance

- Rotorself inductance

(3.15)

qs sRdsi lsL

dmi

lrL qrr  )( 

dsv drv

rR dri

dsp drp

ds sRqsi lsL

qmi

lrL drr  )( 

qsv qrv

rR qri

qsp qrp

2) Flux linkage equations:

3. Induction Generators Torque Equation

3) Electromagnetic torque:

)( 2

3 qsdsdsqse ii

P T  

Note: P is the number of pole pairs

(3.16)

qs sRdsi lsL

dmi

lrL qrr  )( 

dsv drv

rR dri

dsp drp

ds sRqsi lsL

qmi

lrL drr  )( 

qsv qrv

rR qri

qsp qrp

Q: The electromagnetic torque can be calculated by different equations, but the above one is extensively used. Why?

3. Induction Generators Simulink Model (Review)

From the voltage equations:

   

     





 













SiRv

drrqrrqrqr

qrrdrrdrdr

dsqssgsqs

qsdssdsds





 



 













drrqrqrrqr

qrrdrdrrdr

dsqsqssqs

qsdsdssds

piRv





)(

From the flux linkage equations:

 



 





 

qsmqrrqsmqrmlrqr

dsmdrrdsmdrmlrdr

qrmqssqrmqsmlsqs

drmdssdrmdsmlsds

iLiLiLiLL iLiLiLiLL

)( )(

 

    





    







   





   







    





    





i i

LL LL

00 00

 

(3.17)

(3.18)

3. Induction Generators Simulink Model (Review)

][][][]][[][][][]][[][ 111    LiiLLLiL Manipulation of flux linkage equations:

    





    







    





    





 



    





    





LL LL

D i i

i i

 

00 00

2 1 mrs LLLD where

(3.20)

(3.21)

Motion equation: Important note: For generator operation, both Te and shaft mechanical torque Tm < 0 

 

 





(b) 2

(a)

qsdsdsqse

mer

ii P

TT JS P





3. Induction Generators Simulink Model (Review)

Fig. 3-9

3. Induction Generators Simulation Block Diagram

Circuit Breaker

(Stator Frame)

vas

vbs

vcs

(vds) ids

iqs

ias

ibs

ics

Tm 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

Table A-1, Appendix A

3. Induction Generators Per Unit System

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. Induction Generators Case Study 3-1: Direct Start-up of SCIG

Purpose Investigate the dynamic performance of the SCIG wind energy system during startup with direct grid connection

Induction generator 2.3MW / 690V / 50Hz / 1512rpm

SCIGGB Grid

System configuration

3. Induction Generators Case Study 3-1: Direct Start-up of SCIG

Generator Parameters (Table B-1, Appendix B)

3. Induction Generators Case Study 3-1: Direct Start-up of SCIG

Generator Parameters (Table B-1, Appendix B)

3. Induction Generators Case Study 3-1: Direct Start-up of SCIG

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

Q: High inrush current. Why?

Simulated waveforms – transient stator current

3. Induction Generators Case Study 3-1: Direct Start-up of SCIG

Note: • High torque oscillations during the startup • The rated rotor speed (1512rpm, 1pu) is used as the base speed → synchronous speed ωs (1500rpm, 0.992 pu) is lower than the rated rotor speed.

3. Induction Generators Generator Steady State Operation

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 Is and Ir

Is jXls jXlr Rr

jXmVs

IrRs

s s Rr

Pag Pm

Fig. 3-15

3. Induction Generators Torque-Speed Curve Derivation

mmm TP 

  

   

  

   

 rr r

rr m

m Rs s

I P

R s s

IT 1

3 /

11 3

1 22 

P P

s R

I P

T s

agr r

s m /

3 /

1 2 

  

  



s R

IP rrag 23

rrm Rs s

IP )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) = ωr /ωs yields:

where Pag is the airgap power given by

(3.28)

(3.34)

(3.35)

(3.36)

3. Induction Generators Torque-Speed Curve Derivation

 2 2

lrls r

s r

XX s R

V I

  

  





 2 2

lrls r

s m

XX s R

V s RP

T 

 

  



 

Neglecting the magnetizing branch in the steady state circuit simplifies the rotor current calculation:

Finally, replacing Ir in the torque equation of previous slide

It relates Tm with slip s for a given stator voltage Vs and stator frequency s

(3.37)

(3.38)

3. Induction Generators Torque-Speed Curve

Generator Operation Slip < 0 Tm < 0

Motor Operation Slip >0 Tm > 0

Fig. 3-15

3. Induction Generators Power Flow

scurcurotins PPPPP ,,  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

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

Solution:

Q: Why negative slip?

VVs  04.39803/690

004.01500/)15061500( s

Rated phase voltage:

Slip:

Total Impedance:  3.145330.0)//( lr r

mlsss jXs R

jXjXRZ

Stator and rotor currents :

   

  



 



  



A8.1730.1030 /

A3.1459.1206 3.145330.0

03/690

lrrm

sm r

s s

jXsRjX IjX

Z V

Stator circuit Rotor circuitAirgap

Is jXls jXlr Rr

jXm Vs

IrRs

s s1 Rr

3. Induction Generators Case Study 3-2: Power & Efficiency Analysis of IG

Stator circuit Rotor circuitAirgap

Is jXls jXlr Rr

jXmVs

IrRs

s s1 Rr

ImSolution (Continued)

Mechanical power:

Mechanical torque:

Stator power factor angle and power factor:

  

 

822.0cos 3.1453.1450

sss

PF IV

 

kW 2.1186)822.0(9.12063/6903cos3  ssss IVP  Stator Active Power:

kW 78.1195/)1(3 2  ssRIP rrm

mkN 7.58 )60/21506(

108.1195 3 

 

 m

m m

P T

3. Induction Generators Case Study 3-2: Power & Efficiency Analysis of IG

Stator power to the grid:

Solution (Continued)

Stator and rotor winding losses:



  



kW 4.76 3

kW 4.82 3 2

2 ,

rrrcu

ssscu

RIP

kW 2.1186,,  rcuscums PPPP Generator efficiency:

kW 8.1218 rotmin PPP Total input power:

%33.97/  ins PP

3. Induction Generators Summary of Operation

Table 3-2 Operation of induction machine as a motor/generator

4. Synchronous Generators Construction

Fig. 3-16 Salient-pole wound rotor synchronous generator

4. Synchronous Generators Construction

Stator Rotor Courtesy of Enercon

Wound rotor synchronous generator

4. Synchronous Generators Construction

Surface Mounted

Shaft

S S

s S

N S

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

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

qsr

dsv



 

 sR lsL

dsp

dsi

dmL drp

dmi

dsr

qsv



sR lsL

qsp

qsi

qmL

qmi



  





qsdsrqssqs

dsqsrdssds

piRv



The same as that for IG except the stator current direction

Rotor: Wound rotor excited by field current If

Note: SG model is in the rotor synchronous frame (ω = ωr )

Fig. 3-19 General model for SG

4. Synchronous Generators SG Dynamic Model

qsr

dsv



 

 sR lsL

dsp

dsi

dmL drp

dmi

dsr

qsv



sR lsL

qsp

qsi

qmL

qmi

Stator flux linkages:

 

 



 



qmlsq

dmlsd

fdmr

LLL LLL

IL 

  

 









qsqqsqmlsqs

rdsdfdmdsdmls

dsfdmdslsds

iLiLL

iLILiLL

iILiL

)(



Rotor flux and self inductances:

(3.51) (3.52)

4. Synchronous Generators SG Dynamic Model

  

 

qsqqs

rdsdds

iL iL

 



  





qsdsrqssqs

dsqsrdssds

piRv





  





qsqrrdsdrqssqs

dsdqsqrdssds

piLiLiRv





And assuming dλr /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



  





qsqrrdsdrqssqs

dsdqsqrdssds

piLiLiRv





Fig. 3-20 Simplified SG model

(3.53) (From previous slide)

4. Synchronous Generators SG Dynamic Model

 dsqsqsdse iiPT   2 3

  qsdsqdqsre iiLLiPT  2 3

 

  

 









qsqqsqmlsqs

rdsdfdmdsdmls

dsfdmdslsds

iLiLL

iLILiLL

iILiL

)(



Torque equation:

 dsqsqsdse iiPT   2 3

Another torque expression:

For non-salient pole SG: Lq = Ld For salient pole SG: Lq > Ld

The SG model can used for both non-salient and salient SG.

(3.54)

(3.55)

4. Synchronous Generators SG Dynamic Model

PMSG: λr is produced by permanent magnet

WRSG: λr is produced by the rotor field winding

Note: The SG model can used for both PMSG and WRSG.

4. Synchronous Generators SG Dynamic Model



  





qsqrrdsdrqssqs

dsdqsqrdssds

piLiLiRv





Q: How to build Simulink Model?

  qsdsqdqsre iiLLiPT  2 3

 iner TTJ P

dt d

 

4. Synchronous Generators SG Steady-State Model



  





rrdsdrqssqs

qsqrdssds

iLiRv







  





qsqrrdsdrqssqs

dsdqsqrdssds

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

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 Frequency fs 40Hz 1.0 pu Number of Pole pairs P 6 Rated Rotor Flux Linkage λr 6.774Wb (peak) 0.528pu Base Impedance 4.70Ω 1.0 pu Stator Resistance Rs 0.035Ω 0.007pu d-axis Synchronous Inductance Ld 10.7mH 0.574pu q-axis Synchronous Inductance Lq 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

mH26.8LL 23.4LR

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

Find: Stator current, stator voltage, output power, output power factor Solution:

)()(

))((

dsLrqsLqsLrdsL

LrLqsdsqsds

iLiRjiLiR

LjRjiijvv











  





dsLqsLdsLrqsLqs

qsLdsLqsLrdsLds

iXiRiLiRv



(3.60)

(3.61)

4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load

From (a): Substituting to (b):



  





dsLqsLqs

qsLdsLds

iXiRv

(b) (a)



  





dsLrqsLrrdsdrqss

qsLrdsLqsqrdss

iLiRiLiR





qs sL

qLr ds iRR

LL i



 

)(

A 85.141 ))(()(

)( 22

 

 

qLdLrsL

sLrr qs

LLLLRR RR

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

Stator Voltage:

A 0.249 )(

 

  qs

qLr ds iRR

LL i



A7.2022/22  qsdss iiI



  





V 7.1124

V 9.772

rrdsdrqssqs

qsqrdssds

iLiRv





V0.9652/22  qsdss vvV

(3.65)

(3.66)

(3.67)

(3.68)

4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load

Mechanical Power P – number of pole pairs

Stator Winding Loss

Output Power

   m.kN 7.12 2 3

 qsdsqdqsre iiLLiPT 

kW 0.531/  PTTP remmm 

kW 0.33 2,  ssscu RIP

kW 0.528,  scumL PPP

(3.69)

(3.71)

(3.72)

Electromagnetic Torque

4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load

Power Factor

Alternative Way to Calculate PF (Method 1)

 

  

   8.25tan 1

Lr L R

L  9.0)cos(  LLPF 



  





kVA 7.255)(5.1

kW 0.285)(5.1

qsdsdsqsL

qsqsdsdsL

ivivQ

ivivP

9.0 22 

 

L L

P PF     8.259.0cos 1L

(3.74)

(3.75)

(3.76)

4. Synchronous Generators Case study 3-4: Steady-State Analysis of SG with RL Load

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

 



 





  

   

 

  

   

 



7.29tan

5.55tan

qs i

qs v

i i

v v



  

 

9.0cos 8.25

ivL

PF  

kW 0.528cos3  LssL IVP 

0.972 5.120.531

0.528 

 

rotm

PP P



(3.80)

(3.82)

Problems

Try to solve the following problems given in Appendix C of the textbook:

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