ECE 576 – Power System Dynamics and Stability

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

1

Lecture 10: Synchronous Machines Models

Announcements

• Homework 2 is due now• Homework 3 is on the website and is due on Feb 27• Read Chapters 6 and then 4

2

etc

de d ed

de d ep

se s e

X X X

R R R

Single Machine, Infinite Bus System (SMIB)

Book introduces new variables by combining machinevalues with line values

Usually infinite busangle, qvs, is zero

3

s

ss

sst

HT

T

1

2

“Transient Speed”

Mechanical time constant

A small parameter

Introduce New Constants

4

We are ignoring the exciter and governor for now; they will be covered in much more detail later

1 sin

1 cos

dese d t qe s vs

s

qese q t de s vs

s

oese o

dR I V

dt T

dR I V

dt T

dR I

dt

Stator Flux Differential Equations

5

1 sin

1 cos

det qe s vs

s

qet de s vs

s

dV

dt T

dV

dt T

An exact integral manifold (for any sized ε):

cos

sin

de s vs

qe s vs

V

V

( : )s t

dNote T

dt

Special Case of Zero Resistance

Without resistancethis is just anoscillator

6

12

11

qdo q d d

d dd d d s d q fd

d s

ddo d q d s d

dET E X X

dt

X XI X X I E E

X X

dT E X X I

dt

Direct Axis Equations

7

22

22

dqo d q q

q qq q q s q d

q s

qqo q d q s q

dET E X X

dt

X XI X X I E

X X

dT E X X I

dt

Quadrature Axis Equations

8

s2

(recall and T = )s t t s ss

ts M de q qe d FW

d HT Tdt

dT T I I Tdt

Swing Equations

9

These are equivalent to the more traditional swing expressions

2

s

M de q qe d FWs

d

dtH d

T I I Tdt

1

2

d s d dde de d q d

d s d s

q s q qqe qe q d q

q s q s

oe oe o

X X X XX I E

X X X X

X X X XX I E

X X X X

X I

10

Stator Flux Expressions

2 2

1 sin

1 cos

t d q

edd e d t eq s vs

s

eqq e q t ed s vs

s

ed ep d

eq ep q

V V V

dV R I V

T dt

dV R I V

T dt

X I

X I

Network Expressions

11

3 fast dynamic states

, ,de qe oe

6 not so fast dynamic states

1 2, , , , ,q d d q tE E

8 algebraic states

, , , , , , ,d q o d q t ed eqI I I V V V

Machine Variable Summary

12

We'll getto the exciterand governorshortly;for nowEfd is fixed

Elimination of Stator Transients

• If we assume the stator flux equations are much faster than the remaining equations, then letting go to zero creates an integral manifold with

13

0 sin

0 cos

0

se d qe s vs

se q de s vs

se o

R I V

R I V

R I

Impact on Studies

14

Image Source: P. Kundur, Power System Stability and Control, EPRI, McGraw-Hill, 1994

1

2

d s d dde de d q d

d s d s

q s q qqe qe q d q

q s q s

oe oe o

X X X XX I E

X X X X

X X X XX I E

X X X X

X I

15

Stator Flux Expressions

2

1

0

sin

0

cos

q qd sse d qe q d q

q s q s

s vs

d s d dse q de d q d

d s d s

s vs

X XX XR I X I E

X X X X

V

X X X XR I X I E

X X X X

V

16

Network Constraints

2

2

1

2

2vs

q s q qd q q d q

q s q s

jd s d d

q dd s d s

js d d q

j je ep d q s

X X X XE X X I

X X X X

eX X X Xj EX X X X

R jX I jI e

R jX I jI e V e

17

"Interesting" Dynamic Circuit

vssdepqeq

vssqepded

VIXIRV

VIXIRV

cos

sin

These last two equations can be written as one complex equation.

vsjs

jqdepe

jqd

eV

ejIIjXRejVV

22

18

"Interesting" Dynamic Circuit

2

21

q s q qd q q d q

q s q s

jd s d dq d

d s d s

X X X XE X X I

X X X X

X X X Xj E eX X X X

E

E

19

Subtransient Algebraic Circuit

2

often neglected

21

E E X X Id q q d q

q

jj E eq d

d

20

Subtransient Algebraic Circuit

Subtransientsaliency useto be ignored(i.e., assumingX"q=X"d). However thatis increasinglyno longerthe case

Simplified Machine Models

• Often more simplified models were used to represent synchronous machines

• These simplifications are becoming much less common• Next several slides go through how these models can be

simplified, then we'll cover the standard industrial models

21

Two-Axis Model

• If we assume the damper winding dynamics are sufficiently fast, then T"do and T"qo go to zero, so there is an integral manifold for their dynamic states

22

1

2

d q q s d

q d q s q

E X X I

E X X I

Two-Axis Model

• Then

23

11

12

0ddo d q d s d

qdo q d d

d dd d d s d q fd

d s

qdo q d d d fd

dT E X X I

dtdE

T E X Xdt

X XI X X I E E

X X

dET E X X I E

dt

Two-Axis Model

• And

24

22

22

0qqo q d q s q

dqo d q q

q qq q q s q d

q s

dqo d q q q

dT E X X I

dtdE

T E X Xdt

X XI X X I E

X X

dET E X X I

dt

vssdqepqdes VEIXXIRR sin0

vssqdepdqes VEIXXIRR cos0

Two-Axis Model

25

FWqddqqqddMs

s

qqqdd

qo

fddddqq

do

TIIXXIEIETdtdH

dtd

IXXEdtEd

T

EIXXEdt

EdT

2

Two-Axis Model

26

No saturationeffects areincludedwith thismodel

2 2

0 sin

0 cos

sin

cos

s e d q ep q d s vs

s e q d ep d q s vs

d e d ep q s vs

q e q ep d s vs

t d q

R R I X X I E V

R R I X X I E V

V R I X I V

V R I X I V

V V V

Two-Axis Model

27

Flux Decay Model

• If we assume T'qo is sufficiently fast then

28

dqo d q q q

qdo q d d d fd

s

M d d q q q d d q FWs

M q q q d q q q d d q FW

M q q q d d q FW

dET E X X I 0

dtdE

T E X X I Edtd

dt2H d

T E I E I X X I I Tdt

T X X I I E I X X I I T

T E I X X I I T

Flux Decay Model

This model is no longer common

29

Classical Model

• Has been widely used, but most difficult to justify• From flux decay model

• Or go back to the two-axis model and assume

30

0 0

q d do

q

X X T

E E

( const const)q d do qo

q d

X X T T

E E

2 20 0

00 1

0tan 2

q d

q

d

E E E

E

E

Or, argue that an integral manifold exists for

Rffddq VREEE ,,,, such that const.qE

const qdqd IXXE

2 20 0 0 0

0 1tan 2

d q d q qE E X X I E

Classical Model

31

j δ-π 2d qI + jI e

32

Classical Model

dt sd

0

0

0

2sins

M vs FWd ep

H d E VT T

dt X X

This is a pendulum model

a) Full model with stator transients

b) Sub-transient model c) Two-axis model

d) One-axis model

e) Classical model (const. E behind

0

1

s

0 doqo TT

0qoT

dX )

Summary of Five Book Models

33

Damping Torques

34

• Friction and windage– Usually small

• Stator currents (load)– Usually represented in the load models

• Damper windings– Directly included in the detailed machine models

– Can be added to classical model as D(w-ws)

Industrial Models

• There are just a handful of synchronous machine models used in North America–GENSAL

• Salient pole model–GENROU

• Round rotor model that has X"d = X"q

–GENTPF

• Round or salient pole model that allows X"d <> X"q

–GENTPJ• Just a slight variation on GENTPF

• We'll briefly cover each one

35

Network Reference Frame

• In transient stability the initial generator values are set from a power flow solution, which has the terminal voltage and power injection–Current injection is just conjugate of Power/Voltage

• These values are on the network reference frame, with the angle given by the slack bus angle

• Voltages at bus j converted to d-q reference by

36

, ,

, ,

sin cos

cos sind j r j

q j i j

V V

V V

, ,

, ,

sin cos

cos sinr j d j

i j q j

V V

V V

Similar for current; see book 7.24, 7.25

, , In book j r j i j i Di QiV V jV V V jV

Network Reference Frame

• Issue of calculating , which is key, will be considered for each model

• Starting point is the per unit stator voltages (3.215 and 3.216 from the book)

• Sometimes the scaling of the flux by the speed is neglected, but this can have a major impact on the solution

37

d q d qEquivalently, V +jV +jI

d q s d

q d s q

s q d

V R I

V R I

R I j

Two-Axis Model

• We'll start with the PowerWorld two-axis model (two-axis models are not common commercially, but they match the book on 6.110 to 6.113

• Represented by two algebraic equations and four differential equations

38

q q s q d d

d d s d q q

E V R I X I

E V R I X I

1 1( ) , ( )

2, ( )

q dq d d d fd d q q q

do qo

s M d d q q q d d q FWs

dE dEE X X I E E X X I

dt T dt T

d H dT E I E I X X I I T

dt dt

The bus number subscript is omitted since it is not usedin commercial block diagrams

Two-Axis Model

• Value of is determined from (3.229 from book)

• Once is determined then we can directly solve for E'q and E'd

39

s qE V R jX I Sign convention oncurrent is out of thegenerator is positive

Example (Used for All Models)

• Below example will be used with all models. Assume a 100 MVA base, with gen supplying 1.0+j0.3286 power into infinite bus with unity voltage through network impedance of j0.22–Gives current of 1.0-j0.3286 and generator terminal voltage

of 1.072+j0.22 = 1.0946 11.59

40

Infinite Bus

slack

X12 = 0.20

X13 = 0.10 X23 = 0.20

XTR = 0.10

Bus 1 Bus 2

Bus 3

0.00 Deg 6.59 Deg

Bus 4

1.0463 pu

11.59 Deg

1.0000 pu

1.0946 pu -100.00 MW-32.86 Mvar

100.00 MW57.24 Mvar

Two-Axis Example

• For the two-axis model assume H = 3.0 per unit-seconds, Rs=0, Xd = 2.1, Xq = 2.0, X'd= 0.3, X'q = 0.5, T'do = 7.0, T'qo = 0.75 per unit using the 100 MVA base.

• Solving we get

41

1.0946 11.59 2.0 1.052 18.2 2.814 52.1

52.1

E j

0.7889 0.6146 1.0723 0.7107

0.6146 0.7889 0.220 0.8326d

q

V

V

0.7889 0.6146 1.000 0.9909

0.6146 0.7889 0.3287 0.3553d

q

I

I

Two-Axis Example

• And

42

0.8326 0.3 0.9909 1.1299

0.7107 (0.5)(0.3553) 0.5330

1.1299 (2.1 0.3)(0.9909) 2.9135

q

d

fd

E

E

E

Saved as case B4_TwoAxis

Subtransient Models

• The two-axis model is a transient model• Essentially all commercial studies now use subtransient

models• First models considered are GENSAL and GENROU,

which require X"d=X"q

• This allows the internal, subtransient voltage to be represented as

43

( )sE V R jX I

d q q dE jE j

Subtransient Models

• Usually represented by a Norton Injection with

• May also be shown as

44

q dd qd q

s s

jE jEI jI

R jX R jX

q d d q

d q q ds s

j j jj I jI I jI

R jX R jX

In steady-state w = 1.0

GENSAL

• The GENSAL model has been widely used to model salient pole synchronous generators– In the 2010 WECC cases about 1/3 of machine models were

GENSAL; in 2013 essentially none are, being replaced by GENTPF or GENTPJ

• In salient pole models saturation is only assumed to affect the d-axis

45

GENSAL Block Diagram (PSLF)

46

A quadratic saturation function is used. Forinitialization it only impacts the Efd value

GENSAL Initialization

• To initialize this model 1. Use S(1.0) and S(1.2) to solve for the saturation coefficients

2. Determine the initial value of d with

3. Transform current into dq reference frame, giving id and iq

4. Calculate the internal subtransient voltage as

5. Convert to dq reference, giving P"d+jP"q="d+ "q

6. Determine remaining elements from block diagram by recognizing in steady-state input to integrators must be zero

47

s qE V R jX I

( )sE V R jX I

GENSAL Example

• Assume same system as before, but with the generator parameters as H=3.0, D=0, Ra = 0.01, Xd = 1.1, Xq = 0.82, X'd = 0.5, X"d=X"q=0.28, Xl = 0.13, T'do = 8.2, T"do = 0.073, T"qo =0.07, S(1.0) = 0.05, and S(1.2) = 0.2.

• Same terminal conditions as before• Current of 1.0-j0.3286 and generator terminal voltage of

1.072+j0.22 = 1.0946 11.59 • Use same equation to get initial

48

1.072 0.22 (0.01 0.82)(1.0 0.3286)

1.35 1.037 1.70 37.5

s qE V R jX I

j j j

j

GENSAL Example

• Then

And

49

( )

1.072 0.22 (0.01 0.28)(1.0 0.3286)

1.174 0.497

sV R jX I

j j j

j

sin cos

cos sin

0.609 0.793 1.0 0.869

0.793 0.609 0.3286 0.593

d r

q i

I I

I I

GENSAL Example

• Giving the initial fluxes (with = 1.0)

• To get the remaining variables set the differential equations equal to zero, e.g.,

50

0.609 0.793 1.174 0.321

0.793 0.609 0.497 1.233q

d

0.82 0.28 0.593 0.321

1.425, 1.104

q q q q

q d

X X I

E

Solving the d-axis requires solving two linearequations for two unknowns

GENSAL Example

• Once E'q has been determined, the initial field current (and hence field voltage) are easily determined by recognizing in steady-state the E'q is zero

51Saved as case B4_GENSAL

2

2

1 ( )

1.425 1 1.1 0.5 (0.869)

1.425 1 1.25 1.425 0.8 0.521 2.64

fd q q d d D

q

E E Sat E X X I

B E A

Saturationcoefficientswere determinedfrom the twoinitial values