Topic7_DFIG_WECS_2012

1

ELE806 Alternative Energy Systems

Topic 7 Doubly Fed Induction Generator WECS

Bin Wu Professor

Dept of Electrical & Computer Eng.Ryerson University

Contact InfoTel: (416) 979-5000 ext: 6484Tel: (416) 979-5000 ext: 6484Email: [email protected]

Nordex N100, 2.5MW DFIG

1. Introduction

DFIG WECS Main Topics

2. Super‐ and Sub‐synchronous Operation

3. Steady State Model of DFIG

4. Unity Power Factor Operation

5. Leading and Lagging Power Factor Operation

2

6. Stator Voltage Oriented Control of DFIG WECS

7. DFIGWECS Start‐up and Experiments

2

IntroductionSystem Configuration

2MW (example)

0.6MW0.6MW

3

Block diagram of DFIG WECS

Note: Converter power rating: 0.3 x generator rated power

2MW

IntroductionSystem Configuration

690V

DFIG

GB

Grid

0.6MW0.6MW

RSC GSC

DFIG WECS with 2‐level VSC

4

3

‐ Rotor speed range: (0.6 ~ 1.2) ωs or (0.7 ~ 1.3) ωs

‐ωs is the synchronous speed of the generator

IntroductionPower-speed Characteristics

Sub‐synchronous speed operation:

ωr < ωs

Super‐synchronous speed ti

3rmP

operation:

ωr > ωs

5

Assumptions:Power losses in DFIG and converters are ignored.

IntroductionPower Flow in DFIG WECS

Pm – mechanical power

Ps – stator power

Pr – rotor power

Pg – grid power

Note: Power flow in the rotor and converters are bidirectional

Q:

Is it too much for the stator?

6

4

Is jXls jXlr Rr/s IrRs RSC

Equivalent Impedance of RSCDFIG Steady‐state Equivalent Circuit

DFIG Steady State Model

jXmVs

Im

jXeq/s

Vr/s

Req/s

Vm Zeq/sPag

The rotor‐side converter can be

modeled by an equivalent impedance Zeq

Q: How to find Req and Xeq?

7

Is jXls jXlr Rr/s IrRs RSC

Q: How to find Reqand Xeq?

DFIG Steady State ModelEquivalent Impedance of RSC

jXmVs

Im

jXeq/s

Vr/s

Req/s

Vm Zeq/sPag

Step 1 Find I given V ω and T (or P )

Note:The stator is connected to the grid.

Step 1 – Find Is given Vs , ωs and Tm (or Pm )

Step 2 – Find Vm and Im

Step 3 – Find Ir and Vr

Step 4 – Find Req and Xeq

8

5

The equivalent impedance of the converter is defined by

Dividing by the slip yields:

eqsleqeqeqeq LjRjXRZ


Dividing by the slip yields:

where

Step 1 – Find the stator current Is

Assuming that the stator operates at a unity power factor, the air‐gap power

is given by

eqseqeqsleqeq LjsRsLjsRsZ ////

ssl s

is given by

The air‐gap power can also be defined by

Combining both:

sssssmag IRIVIVP )(33

P

TP ms

ag

ssssms IRIV

P

T)(3

03

2 P

TIVIR ms

ssss

(from the equivalent circuit)

(from the machine theory)

9

Solving for Is:

03

2 P

TIVIR ms

ssss

mssss

sP

TRVV

I 3

42


Step 2 – Find Vm and Im

The voltage across the magnetizing branch is

3Psssss

s R2

lsssssm LjRIVV

where Note: The stator current angle of 180°

is due to the generating mode and unity power factor operation.

The magnetizing current is

180 and 0 ssss IIVV

ms

mm

Lj

VI

10

6

msr III

Step 3 – Find Ir and Vr


lrs

rrmr Lj

s

RIVsV / lrsrrmr LjsRIVsV

Is jXls jXlr Rr/s IrRs

Im

RSC

Req/s

jXmVs

jXeq/s

Vr/s

q

Vm Zeq/sPag

11

jXRsV r

///

Step 4 – Find Req and Xeq


from which

sjXsRI

eqeqr

r//

r

reqeq I

VjXR

12

7


Q: Physical meaning of positive and

negative Req from the power flow point of view ?

A:

Req > 0 → rotor delivers power to the grid, super‐synch mode;

Req > 0

Req < 0 → rotor absorbs power from the grid, sub‐synch mode;

See Fig. 8.3‐6 on Slide 70

13

Req < 0

Case Study 1 – Equivalent Impedance of RSCConsider:

− DFIG: 1.5MW/690V/1068A/50Hz/1750rpmQ: Number of pole pairs?


− Parameters given in Table B‐5 of Appendix B

− Generator operates with an MPPT scheme

−Mechanical torque Tm is proportional to the square of the rotor speed

− Unity stator power factor operation

Investigate relationship between the rotor voltage, rotor current, and the

Q: Number of pole pairs?A: P = 2

equivalent impedance of the rotor‐side converter when

a) DFIG operates at super‐synchronous speed of 1750rpm

b) DFIG operates at synchronous speed of 1500rpm, and

c) DFIG operates at sub‐synchronous speed of 1200rpm.

14

8

a) DFIG operating at super‐synchronous speed (1750 rpm, rated)

Stator current Q: Synchronous speed?A: 1500rpm

Case Study 1 – Equivalent Impedance of RSC


where

Voltage across the magnetizing branch

omitted)A 10×1.514 solution other (theA 2.10682

3

4

5

2

ss

smsss

s IR

P

TRVV

I

.2 and mΩ 25.6 ,rad/sec 502 N.m, 1.8185 ,V 3/690 PRTV ssms

A: 1500rpm.

Rated torque due to the rated rotor speed of 1750rpm

g g g

3 690 / 3 0 1068.2 180 (0.00625 100 0.1687 10 )

401.2 56.6 405.2 8 V

sm s s s lsV V I R j L

j

j

15

Magnetizing current

A0826235262339232 jV

I mm



The rotor current

The rotor voltage

A0.826.23526.23392.32 jLj

Ims

m

A0.1686.112526.2331.1101 jIII msr

9.16497.67lrsrrmr LjsRIVsV

where the slip is

Finally, the equivalent impedance can be calculated by

1667.0/)( srss

2751.0 05375.0/ jIVZ rreq

02751.0

05375.0

eq

eq

X

R

16

9

b) DFIG operating at the synchronous speed (1500 rpm)

Stator current



where


omitted)A 10×1.511 solution other (theA 3.7862

3

4

5

2

ss

smsss

s IR

P

TRVV

I

kN.m 0135.61851.8)1750/1500( 2 mT

Rated values

(with MPPT)


9.56.4027.415.400 jLjRIVV lsssssm

17

Magnetizing current

A1.841.23482.23223.24 jLj

VI

ms

mm



The rotor current

The rotor voltage

where

j ms

A0.16428.84382.23250.810 jIII msr

V 162178.2 rrlrsrrmr RILjsRIVsV 0/)( srss

Finally, the equivalent impedance is

0.000263.0 180 00263.0/ jIVZ rreq

0

00263.0

eqsleq

req

LX

RR

Q: Main differences between a) and b)? A: Pay attention to the calculation of Tm and slip.

18

10

c) DFIG operating at the sub‐synchronous speed of 1200rpm

Calculated results for the three operating modes:



Rotor Speed (rpm) 1200 1500 1750 (rated)

Slip 0.2 0 -0.1667 (rated)

Tm [kN.m] -3.849 -6.014 -8.185

rV [V] 2.6756.83 0.16218.2 9.164965.67

rI [A] 9.155285.569 0.164281.843 0.168566.1125

Sub‐synch Synchronous Super‐synch

Req [Ω] -0.126989 -0.00263 0.053751

Xeq [Ω] -0.074293 0 0.027513

19

Torque – Speed characteristic of DFIG To obtain torque‐speed

curves, use the same

d i T i 3


procedure as in Topic 3

(generators), but include

the RSC equivalent

impedance Zeq

Note:

Different curves areDifferent curves are

obtained for different

values of Reqand Xeq

20

11

Steady-state analysisConsider the steady‐state equivalent circuit


Neglecting rotational losses, the transferred and dissipated powers are

eqrrm ssRRIP /)1)((3 2

ssss

ssscu

rrrcu

eqrr

eqrrm

IVP

RIP

RIP

RIP

cos3

3

3

3

))((

2,

2,

2

mode synchronou-subfor

mode synchronou-superfor

rs

rsg

PP

PPP

Power transferred to the grid

21

Steady-state analysisPower flow diagrams


22

12

Consider the 1.5MW/690V DFIG operating under the same conditions as those in Case

Study 1. Investigate the super‐synchronous, synchronous and sub‐synchronous modes

of operation.

Case Study 2 – Steady-state Analysis with PFs = 1

Unity Power Factor Operation

p

a) Operation at the rotor speed of 1750rpm (super‐synchronous mode)

With rotor current calculated in Case Study 1, the mechanical power of the DFIG can be calculated by

16670/166701263000053750611253

/13

2

2

ssRRIP reqrm (from the equivalent circuit)

Alternatively,

kW 1500

1667.0/1667.0126300.005375.06.11253

kW 1500/60217508185.1 mmm TP (from the machine theory)

23

The rotor power

kW 29.20405375.06.11253)(3 2 eqrr RIP

Unity Power Factor OperationCase Study 2 – Steady-state Analysis with PFs = 1

The rotor and stator winding losses

The stator active power

kW 0.10)(3 2, rrrcu RIP

kW 07.9)(3 2, ssscu RIP

kW64.1276)180cos(2.10683/690cos3 ssss IVP

Generating mode with unity PF

Total power delivered to the grid

)(ssss

kW 1480.93 rsg PPP %7.981500/93.1480/ mg PP

Efficiency:

24

13

b) Operation of the DFIG at synchronous and sub‐synchronous modes

Following the same procedures in a), the calculated results are given by


Operating Mode Sub-synchronous

Operation Synchronous

Operation Super-synchronous

Operation ωm [rpm] 1200 1500 1750 (rated) s Slip 0.2 0 -0.1667 (rated)

|Tm| [kN.m] 3.849 6.014 8.1851 Req [Ω] -0.126989 -0.002630 0.053751 Xeq [Ω] -0.074293 0 0.027513

Is [A] 504.16 786.28 1068.22 Ir [A] 569.29 843.28 1125.57 Vr [V] 83.76 2.22 67.97

|Pm| [kW] 483.64 944.61 1500.0 |Pr| [kW] 123.47 5.61 204.29 Pcu,r [kW] 2.56 5.61 10.0 Pcu,s [kW] 2.02 4.92 9.07 |Ps| [kW] 602.53 939.69 1276.64

|Pg| [kW] 479.06 934.08 1480.93

25

Power versus speed curves of the DFIG operating at super‐synchronous, synchronous and sub‐synchronous modes


26

14

c) Simplified calculations

In large wind generators, the stator resistance is small (< 0.01pu)

Wh i i h i ( ) f h i b


When operating with unity (stator) power factor, the air‐gap power can be

calculated by

The stator current can the be calculated by

ssssssag IVIRIVP 3)(3

s

sms V

PTI

3

/

With Is calculated, all other analysis can be performed.

Note: This is a much simpler method, which can be used to facilitate

analysis of non‐unity power factor operation as well.

s

27

Comparison between normal and simplified calculations

Operating Mode Sub-synchronous

Operation Super-synchronous

Operation 1200 rpm 1750 rpm


ωm [rpm] 1200 rpm 1750 rpm

Method 1 Method 2 Error |%| Method 1 Method 2 Error |%| s Slip 0.2 0.2 N/A -0.1667 -0.1667 N/A

|Tm| [kN.m] 3.849 3.849 N/A 8.185 8.185 N/A Req [Ω] -0.12699 -0.12671 0.22 0.05375 0.05339 0.26 Xeq [Ω] -0.07429 -0.07398 0.31 0.02751 0.02735 0.57 Is [A] 504.2 505.8 0.32 1068.2 1075.8 0.71 Ir [A] 569.3 570.9 0.28 1125.6 1133.2 0.68

Vr [V] 83.6 83.7 0.12 67.9 68.0 0.15 |P | [kW] 483 6 485 3 0 35 1500 1510 7 0 71

Summary: Errors caused by simplified calculations are very small.

|Pm| [kW] 483.6 485.3 0.35 1500 1510.7 0.71 |Pr| [kW] 123.5 123.9 0.34 204.3 205.7 0.68

Pcu,r [kW] 2.557 2.571 0.55 9.996 10.13 1.3 Pcu,s [kW] 2.020 2.034 0.69 9.071 9.201 1.4 |Ps| [kW] 602.5 604.5 0.34 1276.6 1285.7 0.71

|Pg| [kW] 479.0 480.6 0.33 1480.9 1491.4 0.77

28

15

d) Calculations with leading and lagging power factor

With leading or lagging PF, the stator current can be calculated by

sms

PTI

/

Case Study 2 – Steady-state Analysis

Leading and Lagging PF Operation

Following the same procedures, the results for sub‐, super‐ and synchronous

modes with leading and lagging PF are as follows:

sss

V cos3

Stator PF ωm (rpm) 1200 1500 1750

s (Slip) 0.2 0 -0.1667

Leading PF (0.95)

8161

rV [V] 7.589.86 6.31514.2 3.16567.73

rI [A] 2.1423.659 4.1489.955 7.1513.1259 8.161s

Req [Ω] -0.0957 -0.00263 0.04277 Xeq [Ω] -0.0906 0 0.03992

Lagging PF (0.95)

8.161s

rV [V] 9.665.80 5.2146.2 1.16430.62

rI [A] 3.1732.525 5.1779.815 3.1732.1117

Req [Ω] -0.1493 -0.00263 0.0550 Xeq [Ω] -0.0360 0 0.0089

29



Torque‐slip curves of DFIG operating with a leading stator PF of 0.95

30

16



Tm (pu)

-2.0

Tm – s curve(R X )

Tm – s curve

Req, Xeq (pu)

00.1 -0.1 -0.20.20.3 s (slip)

-1.5

-1.0

-0.5

0

Req = 0Xeq = 0

with Req2, Xeq2 (Req1,Xeq1)

31

Torque‐slip curves of DFIG operating with a lagging stator PF of 0.9500.1 -0.1 -0.20.20.3

Req, Xeq (pu)

Req

Xeq

s (slip)-1.0

-0.5

0

0.5

1.0Req2 = -0.165pu

Xeq2 = -0.029pu

Req1 = 0.098pu

Xeq1 = 0.024puReq

Xeq

Principles of SVOC‐ The stator of the DFIG is directly connected to the grid.

‐ Since stator voltage is fixed by the grid, it is convenient to use voltage oriented

control (VOC) instead of field oriented control (FOC)

Stator Voltage Oriented Control

sv

dri

qri

( ) ( )

‐ Space vector diagram of DFIG with SVOC in the super‐synchronous mode

Stator voltage vector aligned with the d‐axis of the synch frame:

0qsv

sds vv

ri

qrv

drvrv

si

Speed of the rotating frame:

Slip angle:

ss f 2

θsl = θs – θr

32

17

DFIG

GB

RSC

Grid

~GSC

vs

Stator Voltage Oriented ControlPrinciples of SVOC – system block diagram

PWM PWM

VOCController

PI

dq/abc

PIidriqr

iar, ibr

vdc

θs

θsCalculator

abc/dq

θr

θs

θsl

θsl

vdr* vqr

*

var, vbr, vcr* * *

vas, vbs

vdc* Q*

GSC

vds

33

Q

iqr*idr

*

GSC

MPPTωm Te

dsm

ss

vPL

L

3

2

ms

ds

L

v

*

mds

s

Lv

L

3

2*s

Principles of SVOCElectromagnetic torque is given by

)(2

3qsdsdsqse ii

PT (8.53) in textbook


The stator flux linkages are

from which

qrmqssqs

drmdssds

iLiL

iLiL

s

drmdsds L

iLi

(8 55)

Substituting (8.5‐6) into (8.5‐4) yields

s

qrmqsqs L

iLi

)(2

3qsdrdsqr

s

me ii

L

PLT (8.56)

(8.55)

34

18

Principles of SVOCThe dq‐axis stator flux linkages can also be written as

qssqsds

iRv


Substituting (8.59) to (8.56), and assuming Rs=0 and vqs=0 forSVOC, we have

s

dssdsqs

sds

iRv

PL3

(8.59)

dsdrss

me vi

L

PLT

2

3

Note: With a fixed stator voltage, torque can be controlled by d‐axis rotor current.

(8.62)

35

Principles of SVOCBy definition, the active and reactive power are

)(2

3qsqsdsdss ivivP


With for SVOC,

Substituting (8.55) into (8.64) yields

)(

2

3qsdsdsqss ivivQ

qsdss

dsdss

ivQ

ivP

2

32

3

0qsv (8.64)

g ( ) ( ) y

s

qrmqsdss

s

drmdsdss

L

iLvQ

L

iLvP

2

3

2

3

(8.65)

36

19

Principles of SVOCFrom (8.65)

dss

d

sdr L

PLv

Li 1

3

2


The above equation can be rewritten as

qsm

smds

sqr

mmds

LQ

Lv

Li

LLv

1

3

2

3

(a) 3

2 s

mds

sdr P

Lv

Li

(8.66)

(b) 3

2

ms

dss

mds

sqr

mds

L

vQ

Lv

Li

Note: The stator active and reactive power can be controlled through rotor currents and voltages!

(8.68)

37

DFIG

GB

RSC

Grid

~GSC

vs

Control Scheme


PWM PWM

VOCController

PI

dq/abc

PIidriqr

iar, ibr

vdc

θs

θsCalculator

abc/dq

θr

θs

θsl

θsl

vdr* vqr

*

var, vbr, vcr* * *

vas, vbs

vdc* Q*

GSC

vds

Q

iqr*idr

*

GSC

MPPTωm Te

dsm

ss

vPL

L

3

2

ms

ds

L

v

*

mds

s

Lv

L

3

2*s

(8.68b)(8.62)

38

20

Control SchemeDFIG

GB

RSC

Grid

~GSC

vs


PWM

RSC

PWM

VOCControllerdq/abc

idriqr

iar, ibr

vdc

θs

θsCalculator

abc/dq

θr

θs

θsl

θsl

vdr* vqr

*

var, vbr, vcr* * *

GSC

vas, vbs

* *

vds

39

PI PI

Q

iqr*idr

*

vdc Q*GSC

MPPTωm Te

dsm

ss

vPL

L

3

2

ms

ds

L

v

*

mds

s

Lv

L

3

2*s

v

vs

1tan

cs

bs

as

v

v

v

v

v

2/32/30

2/12/11

3

2

(8.68b)(8.62)

Control Scheme DFIG

GB

RSC

Grid

~GSC

vs


PWM

RSC

PWM

VOCControllerdq/abc

idriqr

iar, ibr

vdc

θ

θsCalculator

abc/dq

θr

θs

θsl

θsl

vdr* vqr

*

var, vbr, vcr* * *

GSC

vas, vbs

* * vd

(8.68b)(8.62)

PI PI θs

Q

iqr*idr

*

vdc Q*GSC

MPPTωm Te

dsm

ss

vPL

L

3

2

ms

ds

L

v

vds

*

mds

s

Lv

L

3

2*s

‐ Keep Vdc constant

‐ Grid side reactive power control

VOC Controller: ‐ Refer to Topic 4 for details

40

21

Case Study 3 – Dynamic performance of SVOCTransients from super‐synchronous to sub‐synchronous operation

Consider:‐ 1.5MW/690V DFIG


‐ Step change in wind speed are investigated from 1.0 pu to 0.7 pu ‐ Rotational losses are neglected (Te=Tm in steady‐state)

System parameters:

Induction generator Generator ratings: 1.5MW/690V/50Hz/1750rpm/8.185kN.m

Generator parameters: Table B-5 (Appendix B)

Torque reference at t = 0 pu) (1.0 kN.m 185.8* eT

System input variables Reactive Power Reference for GSC 0* GSCQ

DC link voltage reference *dcv = 436.2 V

Control scheme Stator Voltage Oriented Control (SVOC) Fig. 8-14

41

System parameters (continued)

Converter type Two-level VSC

Modulation scheme Space vector modulation

Stator Voltage Oriented ControlCase Study 3 – Dynamic performance of SVOC

Rotor-side converter

p

Switching sequence Seven segment; Table 4-4

Switching frequency 4 kHz

AC harmonic filter NO

dc link filter Capacitor 130.73 mF

Grid-side converter The same as RSC

Voltage/Frequency 690V/50HzElectric Grid

Voltage/Frequency 690V/50Hz

Line inductance (harmonic filter) 0.0775 mH

Moment of inertia Reduced for short simulation time 98.26 kg.m2 (H=1.1 sec)

Reference Frame Transformation

abc/dq transformation Eq. (3.1), Chapter 3

dq/abc transformation Eq. (3.2), Chapter 3

42

22

Torque versus speed characteristics

Stator Voltage Oriented ControlCase Study 3 – Dynamic performance of SVOC

43

Power versus speed characteristics

Case Study 3 – Dynamic performance of SVOC


44

23

Speed and torque dynamic response



‐ At t = 0.5 s, mechanical torque goes from A to B due to step change in wind speed‐ Electromagnetic torque cannot change abruptly since it depends on MPPT which changes according to speed (speed changes slowly due to inertia)

45

Speed and torque dynamic response



‐ Speed decreases since mechanical torque has been reduced‐ Speed reaches steady state when Te is equal to Tm.

46

24

Rotor voltage, rotor current, and stator current

‐ Rotor frequency varies withspeed



p

‐ Around t = 1s systemcrosses synchronous speed‐ Rotor frequency is 0‐ Rotor current is DC‐ Rotor voltage is zero sincethere is no drop in Leq and Llr

‐ Rotor phase‐a current leadsh b iphase‐b current in super‐

synchronous mode, but thisreverses in the sub‐synchronous mode‐ Values of stator current will be checked in steady‐state analysis.

47

Case Study 4 – Steady-state analysis of SVOC

With the same parameters of Case Study 3 and at 0.7pu wind speed

The mechanical torque of the generator is

N m740108185)70()( 22 TT


The stator current can be calculated by

where

N.m7.40108185)7.0()( ,pu, Rmmm TT

omitted)A 10×1.509 (A 3.5252

3

4

5

2

ss

smsss

s IR

P

TRVV

I

2andmΩ256rad/sec502V3/690 PRV

The voltage of the magnetizing branch is

.2and mΩ25.6,rad/sec502 ,V 3/690 PRV sss

3 690 / 3 0 525.3 180 (0.00625 100 0.1687 10 )

399.8 27.8 400.7 4.0 V

sm s s s lsV V I R j L

j

j

48

25

The magnetizing current is calculated by

0.860.23342.23219.16 jLj

VI

mm



The rotor current is

The rotor voltage is determined by

Lj ms

8.15627.58942.23250.541 jIII msr

5.699.76lrsrrmr LjsRIVsV

where

Then the slip frequency at the 0.7 pu rotor speed is

rad/sec 58.575021833.0 ssl s

1833.0/)( srss

(9.165Hz)

49

The peak values of the rotor voltage and rotor current are

pu) (0.664A 25.83327.58922 rr Ii



The rotor power factor angle is

The equivalent impedance for the rotor‐side converter is

pu) (0.273 V 9.10899.7622 rr Vv

3.1508.1565.6r

06472.0 11351.0/ jIVZ rreq

from which

06472.0

11351.0

eq

eq

X

R

50

26

The mechanical power of the generator is

/13

2

2 ssRRIP reqrm



Alternatively,

The rotor power is

The rotor and stator winding losses are

kW 50.5141833.0/1833.0126300.011351.027.5893 2

kW 50.514/6027.017507.4010 mmm TP

kW 24.118)11351.0(27.5893)(3 22 eqrr RIP

The rotor and stator winding losses are

The stator active power is

,kW 74.2)(3 2, rrrcu RIP kW 19.2)(3 2

, srscu RIP

kW 81.627)180cos( 89.275 3/690cos3 ssss IVP

Note: the DFIG operates as a generator with unity PF

51

The total power delivered to the grid is

kW 509.57 24.18181.276 rsg PPP



The efficiency of the DFIG is then

The grid current is calculated by

The peak grid current is

%0.9950.514/57.509/ mg PP

A 37.4264.3983

106.509

3

3

g

gg V

PI

Q: Why is the grid current lower than stator current?The peak grid current is

The peak stator current is

pu) (0.48A 0.60337.42622 gg Ii

(0.59pu)A 9.7423.52522 ss Ii

lower than stator current?

A: Because the grid power is equal to the stator minus the rotor power in the sub‐synch mode

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Verifying steady‐state values by simulationCase Study 4 – Steady-state analysis of SVOC


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Case Study 5 – Analysis based on phasor diagram

‐ Same parameters as Case Study 4. The DFIG is operated at 0.7 pu rotor speed (sub‐synchronous mode) and unity stator power factor

‐ Steady‐state dq‐frame phasor diagram


‐ d‐axis aligned with the stator voltage phasor sV

Note: dq axes in this diagram are stationary for steady‐state analysis

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The peak values of the rotor voltage and current are calculated by

pu) (0.273 V 9.10899.7622 rr Vv



where Vr and Ir are the rms values calculated in Case Study 4

Based on the phasor diagram, the rotor dq‐axis voltages are

pu)(0.664A 4.8333.58922 rr Ii

pu)(0.031 V33.12)5.6sin(9.108

pu) (0.272 V 2.108)5.6cos(9.108

qr

dr

v

v

The rotor dq‐axis currents

pu)(0.03V33.)5.6s (9.08qrv

pu) (0.262A 3.328)8.156sin(4.833

pu) (-0.61A 0.766)8.156cos(4.833

qr

dr

i

i

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Verification with simulation results



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Startup Process

DFIG Startup and Experiments

System block diagram

Step 1 – Initial parking state‐Wind peed is below cut‐in speed or above cut‐out speed‐ Switches SW1 and SW2 are open (usually electromechanical switches)‐ The turbine blades are pitched out of the wind‐ No torque is generated by the turbine

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Step 2 – Turbine acceleration and stator voltage generated

‐Wind speed surpasses cut‐in speed or reduces below cut‐out speed

Startup Process


‐ The turbine blades are pitched to produce starting torque (turbine rotates)

‐ SW2 closes and power converters are energized

‐ GSC controls dc‐link voltage to rated value

‐ RSC provides excitation to the rotor, producing a rotating magnetic flux

‐ A 3‐phase balanced voltage is induced in the stator

‐ The stator voltage is monitored for synchronization with the grid

‐ Torque reference is set to zero

‐ No power is generated

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Step 3 – Synchronization of the voltage/frequency with the grid

‐ During the rotor speed acceleration, both stator voltage and

Startup Process


frequency are fully controlled by the rotor‐side converter

‐ When the generator reaches a speed consistent with the MPPT of

the measured wind speed, the stator voltage, frequency and

phase angle are adjusted to match those of the grid for

synchronizationsynchronization

‐ When the synchronization is achieved, SW1 is closed

‐ The DFIG WECS is connected to the grid

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Step 4 – Power generation and optimal pitch angle

‐ The torque or power reference is increased from zero to a value

Startup Process


generated from the MPPT algorithm according to the measured

wind speed

‐ The blade pitch angle is also adjusted to its optimal value

‐ Maximumwind energy conversion efficiency is achieved

‐ The start‐up process is completed.

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Startup Process – Experimental Results‐ Startup process with rotor speed at 0.8 pu (Steps 1 and 2 achieved)

‐ Results show Step 3 at which SW1 is closed


Stator current does not increase due to perfect synchronization

Ripple appears in stator current due to switching harmonics

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Startup Process – Experimental Results

Rotor currents during transition from super‐ to sub‐synchronous speed

Super synch S b h


Super-synch Sub-synch

iaribr iar ibr

The rotor currents are DC when crossing synchronous speed

The Phase‐a rotor current goes from lagging to leading the phase‐brotor current like seen in Case Study 4

500ms/div

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Angles of stator voltage vector, rotor position and slip

Startup Process – Experimental Results


Note: The rotor frequency is slower than the stator frequency; The DFIG is in sub‐sync mode of operation.

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Problems

DFIG unity power operation: 8‐1, 8‐2

DFIG leading/lagging power operation: 8‐13, 8‐14

DFIG steady state analysis of dq axis voltages/currents : 8 19 8 20

DFIG stator voltage oriented control (SVOC): 8‐22, 8‐23

DFIG steady state analysis of dq‐axis voltages/currents : 8‐19, 8‐20

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Topic7_DFIG_WECS_2012

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