ece 576 – power system dynamics and stability prof. tom overbye dept. of electrical and computer...
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ECE 576 – Power System Dynamics and Stability
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
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