ECE 576 – Power System
Dynamics and Stability
Lecture 7: Synchronous Machine Modeling
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
1
Announcements
•
•
•
•
•
Read Chapter 3, skip 3.7 for now
Skim Chapter 4; we'll be returning to this in greater
detail later
Homework 2 is posted on the web; it is due on Tuesday
Feb 16
Midterm exam is moved to March 31 (in class) because
of the ECE 431 field trip on March 17
We will have extra classes on Mondays Feb 8, 15 and
March 7 from 1:30 to 3pm in 5070 ECEB
– No class on Tuesday Feb 23, Tuesday March 1 and Thursday
March 3
2
Load Modeling
•
•
•
The load model used in transient stability can have a
significant impact on the results
By default PowerWorld uses constant impedance models
but makes it very easy to add more complex loads.
The default (global) models are specified on the Options,
Power System Model page.
These models
are used only
when no other
models are
specified.
3
Load Modeling
•
•
•
More detailed models are added by selecting “Stability
Case Info” from the ribbon, then Case Information,
Load Characteristics Models.
Models can be specified for the entire case (system), or
individual areas, zones, owners, buses or loads.
To insert a load model click right click and select insert
to display the Load Characteristic Information dialog.
Right click
here to get
local menu and
select insert.
4
Dynamic Load Models
•
•
•
•
•
Loads can either be static or dynamic, with dynamic
models often used to represent induction motors
Some load models include a mixture of different types
of loads; one example is the CLOD model represents a
mixture of static and dynamic models
Loads models/changed in PowerWorld using the Load
Characteristic Information Dialog
Next slide shows voltage results for static versus
dynamic load models
Case Name: WSCC_9Bus_Load
5
WSCC Case Without/With
Complex Load Models
•
Below graphs compare the voltage response following a
fault with a static impedance load (left) and the CLOD
model, which includes induction motors (right)
1.05
1
0.95
0.9
0.85
0.8
0.75
0.7
0.65
0.6
0.55
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
g
b
c
d
e
f
b
c
d
e
f
g
b
c
d
e
f
g
3
4
5
6
7
8
V (pu)_Bus Bus 2 f
g
b
c
d
e
V (pu)_Bus Bus 5 g
b
c
d
e
f
V (pu)_Bus Bus 3 f
g
b
c
d
e
V (pu)_Bus Bus 6 g
b
c
d
e
f
V (pu)_Bus Bus 4
V (pu)_Bus Bus 8 g
b
c
d
e
f
V (pu)_Bus Bus 9 g
b
c
d
e
f
V (pu)_Bus Bus1
V (pu)_Bus Bus 7
9
10
0
1
2
g
b
c
d
e
f
b
c
d
e
f
g
b
c
d
e
f
g
3
4
5
6
7
8
V (pu)_Bus Bus 2 f
g
b
c
d
e
V (pu)_Bus Bus 5 g
b
c
d
e
f
V (pu)_Bus Bus 3 f
g
b
c
d
e
V (pu)_Bus Bus 6 g
b
c
d
e
f
V (pu)_Bus Bus 4
V (pu)_Bus Bus 8 g
b
c
d
e
f
V (pu)_Bus Bus 9 g
b
c
d
e
f
V (pu)_Bus Bus1
9
10
V (pu)_Bus Bus 7
6
Under-Voltage Motor Tripping
•
In the PowerWorld CLOD model, under-voltage motor
tripping may be set by the following parameters
– Vi = voltage at which trip will occur (default = 0.75 pu)
– Ti (cycles) = length of time voltage needs to be below Vi
•
•
before trip will occur (default = 60 cycles, or 1 second)
In this example change the tripping values to 0.8 pu and
30 cycles and you will see the motors tripping out on
buses 5, 6, and 8 (the load buses) – this is especially
visible on the bus voltages plot. These trips allow the
clearing time to be a bit longer than would otherwise be
the case.
Set Vi = 0 in this model to turn off motor tripping.
7
37 Bus System
•
Next we consider a slightly larger, 9 generator, 37 bus
system. To view this system open case GOS_37Bus.
The system one-line is shown below.
SLACK345
A
MVA
A
MVA
1.03 pu
slack
75 MW
49 Mvar
RAY345
A
1.03 pu
TIM345
SLACK138
A
A
MVA
MVA
MVA
1.02 pu
A
RAY138
A
A
MVA
TIM138
1.01 pu
33 MW
13 Mvar
A
A
MVA
18 MW
5 Mvar
16.0 Mvar
A
1.02 pu MVA
MVA
MVA
1.02 pu
RAY69
74 MW
A
TIM69
PAI69
1.01 pu
1.01 pu
MVA
GROSS69
17 MW
3 Mvar
A
A
23 MW
7 Mvar
27 Mvar
MVA
A
A
1.03 pu
1.02 pu
MVA
MVA
A
FERNA69
MVA
MVA
MVA
23 MW
7 Mvar
A
MORO138
HISKY69
MVA
A
12 MW
5 Mvar
A
20 MW
6 Mvar
1.00 pu
1.00 pu
A
PETE69
DEMAR69
MVA
HANNAH69
45 MW
12 Mvar
A
18.2 Mvar
UIUC69
1.00 pu
MVA
1.01 pu
A
1.00 pu
140 MW
45 Mvar
A
56 MW
MVA
58 MW
36 Mvar
1.00 pu
A
A
1.01 pu
7.3 Mvar
HALE69
MVA
15 MW
MVA
5 Mvar
75 MW
35 Mvar
A
MVA
A
60 MW
12 Mvar
36 MW
10 Mvar
MVA
A
A
7.3 Mvar
1.01 pu
0.0 Mvar
MVA
MVA
1.00 pu
1.01 pu
PATTEN69
A
MVA
1.01 pu
LAUF69
1.02 pu
150 MW
-14 Mvar
1.01 pu
A
A
MVA
MVA
23 MW
6 Mvar
23 MW
3 Mvar
WEBER69
22 MW
15 Mvar
10 MW
5 Mvar
1.01 pu
BUCKY138
A
MVA
A
MVA
SAVOY69
A
1.02 pu
38 MW
4 Mvar
JO138
JO345
MVA
A
MVA
A
MVA
ROGER69
28 MW
6 Mvar
LAUF138
1.01 pu
A
BLT69
1.01 pu
MVA
A
MVA
SHIMKO69
MVA
A
LYNN138
1.02 pu
4 Mvar
MVA
MVA
1.01 pu
A
14 MW
3 Mvar
BLT138
1.00 pu
MVA
AMANDA69
14 MW
A
A
27 MW
3 Mvar
MVA
13 Mvar
16 MW
-14 Mvar
MVA
MVA
BOB69
MVA
12.8 Mvar
A
A
MVA
A
A
A
MVA
23 MW
6 Mvar
A
1.01 pu
MVA
MVA
BOB138
MVA
58 MW
40 Mvar
29.1 Mvar
HOMER69
A
4.8 Mvar
MVA
1.01 pu
WOLEN69
1.00 pu
1.01 pu
A
MVA
SAVOY138
A
A
150 MW
-2 Mvar
MVA
MVA
150 MW
-2 Mvar
A
MVA
1.02 pu
A
MVA
1.03 pu
To see summary
listings of the
transient stability
models in this case
select “Stability
Case Info” from the
ribbon, and then
either “TS Generator
Summary” or “TS
Case Summary”
8
Transient Stability Case and Model
Summary Displays
Right click on a line
and select “Show
Dialog” for more
information.
9
37 Bus Case Solution
60
59.98
59.96
59.94
59.92
59.9
59.88
59.86
59.84
59.82
59.8
59.78
59.76
59.74
59.72
59.7
59.68
59.66
59.64
59.62
59.6
0
1
2
3
4
g
b
c
d
e
f
b
c
d
e
f
g
b
c
d
e
f
g
b
c
d
e
f
g
b
c
d
e
f
g
5
6
7
8
9
10
11
12
13
14
15
16
Speed_Gen WEBER69 #1 f
g
b
c
d
e
Speed_Gen JO345 #2
b
c
d
e
f
g
Speed_Gen JO345 #1
g
b
c
d
e
f
Speed_Gen ROGER69 #1 g
b
c
d
e
f
Speed_Gen BOB69 #1
Speed_Gen LAUF69 #1
17
18
19
20
Graph
shows the
generator
frequency
response
following
the loss
of one
generator
Speed_Gen SLACK345 #1
Speed_Gen BLT138 #1
Speed_Gen BLT69 #1
10
Stepping Through a Solution
•
Simulator provides functionality to make it easy to see
what is occurring during a solution. This functionality
is accessed on the States/Manual Control Page
Transfer results
to Power Flow
to view using
standard
PowerWorld
displays and
one-lines
Run a Specified Number of Timesteps or Run
Until a Specified Time, then Pause.
See detailed results
at the Paused Time
11
Dynamics Scenario
Demo shows
an interactive,
42 bus dynamics
scenario simulated
using the
PowerWorld
Dynamics Studio
12
Synchronous Machine Modeling
•
Electric machines are used to convert mechanical
energy into electrical energy (generators) and from
electrical energy into mechanical energy (motors)
– Many devices can operate in either mode, but are usually
•
•
customized for one or the other
Vast majority of electricity is generated using
synchronous generators and some is consumed using
synchronous motors, so that is where we'll start
Much literature on subject, and sometimes overly
confusing with the use of different conventions and
nominclature
13
Synchronous Machine Modeling
3 bal. windings (a,b,c) – stator
Field winding (fd) on rotor
Damper in “d” axis
(1d) on rotor
2 dampers in “q” axis
(1q, 2q) on rotor
14
Dq0 Reference Frame
•
•
•
Stator is stationary,rotor is rotating at synchronous speed
Rotor values need to be transformed to fixed reference
frame for analysis
Done using Park's transformation into what is known as
the dq0 reference frame (direct, quadrature, zero)
– Parks’ 1929 paper voted 2nd most important power paper of
•
20th century (1st was Fortescue’s sym. components paper)
Convention used here is the q-axis leads the d-axis
(which is the IEEE standard)
– Others (such as Anderson and Fouad) use a q-axis lagging
convention
15
Fundamental Laws
Kirchhoff’s Voltage Law, Ohm’s Law, Faraday’s
Law, Newton’s Second Law
Stator
d a
va  ia rs 
dt
d b
vb  ib rs 
dt
d c
vc  ic rs 
dt
Rotor
v fd  i fd r fd 
v1d  i1d r1d 
v1q  i1q r1q 
v2 q  i2 q r2 q 
d  fd
dt
d 1d
dt
d 1q
Shaft
d shaft 2
 
dt
P
2 d
J
 Tm  Te  T f 
P dt
dt
d 2 q
dt
16
Dq0 transformations
vd 
va 
 
v 
T

v
dqo  b 
 q
v 
vc 
 o
or i, 
vd 
va 
v   T 1 v 
 b  dqo  q 
v 
vc 
 o
17
Dq0 transformations
Tdqo

P
sin
 shaft

2

2
P
 cos  shaft
3
2

1


2

2 
2  
P
P
sin   shaft 
sin


shaft



3 
3 
2
2

2 
2  
P
P
cos   shaft 
cos


shaft



3 
3 
2
2

1
1


2
2

with the inverse,


P
P
sin

cos

1
shaft
shaft


2
2

P
2 
P
2  


1
Tdqo  sin   shaft   cos  shaft   1
3 
3  
2
 2
2 
2  
 P
 P
sin


cos

 shaft
 1
  2 shaft 3 
3  
2
18
Dq0 transformations
Note: This transformation is not power invariant. This
means that some unusual things will happen when we
use it.
Example: If the magnetic circuit is assumed to be
linear
abc  Liabc

(symmetric)
Tabc  TLT 1idqo
1
dqo  TLT
 idqo
Not symmetric if T is not power invariant.
19
Transformed System
Stator
d
vd  rsid  q  d
dt
d q
vq  rsiq  d 
dt
d
vo  rsio  o
dt
Rotor
v fd  r fd i fd 
v1d  r1d i1d 
v1q  r1qi1q 
v2 q  r2 qi2 q 
Shaft
d  fd
dt
d 1d
dt
d 1q
d shaft

2

P
dt
2 d
J
 Tm  Te  T f 
P dt
dt
d 2 q
dt
20
Electrical & Mechanical Relationships
Electrical system:
d
v  iR 
dt
(voltage)
d
vi  i R  i
dt
2
Mechanical system:
 2  d
J 
 Tm  Te  T f 
 P  dt
(power)
(torque)
2
2
 2  d 2
J  
 Tm  Te  T f 
dt P
P
P
P
2
(power)
21
Derive Torque
•
•
•
Torque is derived by looking at the overall energy
balance in the system
Three systems: electrical, mechanical and the coupling
magnetic field
– Electrical system losses in form of resistance
– Mechanical system losses in the form of friction
Coupling field is assumed to be lossless, hence we can
track how energy moves between the electrical and
mechanical systems
22
Energy Conversion
Look at the instantaneous power:
3
3
vaia  vbib  vcic  vd id  vqiq  3voio
2
2
23
Change to Conservation of Power
Pin
 vaia  vbib  vcic  v fd i fd  v1d i1d  v1qi1q
elect
 v2qi2q


Plost  rs ia2  ib2  ic2  r fd i 2fd  r1d i12d  r1qi12q  r2qi22q
elect
d fd
da
db
dc
d1d
Ptrans  ia
 ib
 ic
 i fd
 i1d
dt
dt
dt
dt
dt
elect
 i1q
d1q
dt
 i2q
d2q
dt
24
With the Transformed Variables
3
3
Pin  vd id  vqiq  3voio  v fd i fd  v1d i1d
2
elect 2
 v1qi1q  v2qi2q
3 2 3 2
Plost  rsid  rsiq  3rsio2  r fd i 2fd  r1d i12d
2
elect 2
 r1qi12q  r2qi22q
25
With the Transformed Variables
3 P d shaft
3 dd 3 P d shaft
Ptrans  
qid  id

d iq
2 2 dt
2 dt 2 2 dt
elect
d fd
3 dq
do
d1d
 iq
 3io
 i fd
 i1d
2 dt
dt
dt
dt
d1q
d2q
 i1q
 i2q
dt
dt
26
Change in Coupling Field Energy
dW f
2
 Te
dt
P
d
da
db
 ia
 ib
dt
dt
dt
d fd
dc
d1d
 i1d
 ic
 i fd
dt
dt
dt
 i1q
d1q
dt
 i2 q
d2 q
dt
This requires the lossless coupling field
assumption
27
Change in Coupling Field Energy
For independent states , a, b, c, fd, 1d, 1q, 2q
dW f
dt

W f

d W f

a
dt
da W f

b
dt
W f dc W f


c dt
 fd
d fd
dt

db
dt
W f
1d
d1d
dt
W f d1q
W f d2 q


1q dt
2 q dt
28
Equate the Coefficients
2 W f
Te 
P

ia 
W f
a
etc.
There are eight such “reciprocity conditions for
this model.
These are key conditions – i.e. the first one gives
an expression for the torque in terms of the
coupling field energy.
29
Equate the Coefficients
W f


3P
d iq  qid  Te

 shaft 2 2
W f
3
 id ,
d 2
W f
 fd
 i fd ,
W f
3
 iq ,
q 2
W f
1d
 i1d ,
W f
o
W f
1q
 3io
 i1q ,
W f
2 q
 i2 q
These are key conditions – i.e. the first one gives an
expression for the torque in terms of the coupling field energy.
30
Coupling Field Energy
•
The coupling field energy is calculated using a path
independent integration
– For integral to be path independent, the partial derivatives of
•
all integrands with respect to the other states must be equal
i fd
3 id
For example,

2  fd d
Since integration is path independent, choose a
convenient path
– Start with a de-energized system so all variables are zero
– Integrate shaft position while other variables are zero, hence
no energy
– Integrate sources in sequence with shaft at final shaft value
31
Do the Integration
W f  W fo 
d
 shaft

o
 shaft

3 P ˆ
ˆ i

i


 2 2 d q
qd
q

 ˆ
d shaft
o
3
3
ˆ
  id d d   iq d ˆq   3io d ˆo
2
2
o
o
d
q
oo
 fd
1d
1q
2q
  i fd d ˆ fd   i1d d ˆ1d   i1q d ˆ1q   i2q d ˆ2q
 ofd
1od
1oq
2oq
32
Torque
•
•
•
Assume: iq, id, io, ifd, i1d, i1q, i2q are independent of shaft
(current/flux linkage relationship is independent of
shaft)
Then Wf will be independent of shaft as well
Since we have
W f
 shaft
3P

d iq  q id   Te  0

22
3P

Te  
d iq  qid 
22
33
Define Unscaled Variables
P
   shaft   st
2
s is the rated
synchronous speed
d plays an important role!
d d
 rsid  q  vd
dt
d q
 rsiq  d  vq
dt
d o
 rsio  vo
dt
d  fd
 r fd i fd  v fd
dt
d 1d
 r1d i1d  v1d
dt
d 1q
dt
d 2 q
dt
  r1qi1q  v1q
  r2 qi2 q  v2 q
d
   s
dt
2 d
 3  P 
J
 Tm     d iq  qid  T f 
p dt
 2  2 


34
Convert to Per Unit
•
As with power flow, values are usually expressed in per
unit, here on the machine power rating
VBase I Base  PBase
•
•
Two common sign conventions for current: motor has
positive currents into machine, generator has positive
out of the machine
Modify the flux linkage current relationship to account
for the non power invariant “dqo” transformation
35
Convert to Per Unit
Va 
Ia 
a 
va
VBABC
ia
I BABC
Vb 
,
Ib 
,
a
 BABC
,
vb
VBABC
ib
I BABC
b 
Vc 
,
Ic 
,
b
 BABC
,
vc
VBABC
,
ic
I BABC
c 
c
 BABC
where VBABC is rated RMS line-to-neutral stator
voltage and
I BABC
PB

,
3VBABC
 BABC 
VBABC
B
36
Convert to Per Unit
Vd 
vd
VBDQ
Vq 
,
id
,
Id 
I BDQ
d 
d
 BDQ
Iq 
,
vq
VBDQ
iq
I BDQ
q 
Vo 
,
Io 
,
q
 BDQ
,
vo
VBDQ
,
io
I BDQ
o 
o
 BDQ
where VBDQ is rated peak line-to-neutral stator voltage
and
VBDQ
2 PB
I BDQ 
,  BDQ 
3VBDQ
B
37
Convert to Per Unit
V fd
v fd
v1q
v2 q
v1d

, V1d 
, V1q 
, V2 q 
VBFD
VB1D
VB1Q
VB 2Q
I fd 
 fd 
i fd
I BFD
, I1d 
 fd
 BFD
i1d
I B1D
, 1d 
, I1q 
1d
 B1D
i1q
I B1Q
, 1q 
, I 2q 
1q
 B1Q
i2 q
I B 2Q
,  2q 
2 q
 B 2Q
Hence the  variables are just normalized
flux linkages
38
Convert to Per Unit
Where the rotor circuit base voltages are
PB
PB
VBFD 
, VB1D 
,
I BFD
I B1D
VB1Q 
PB
I B1Q
VB 2Q 
,
PB
I B 2Q
And the rotor circuit base flux linkages are
 BFD 
VBFD
 B1Q 
VB1Q
B
B
,
,
 B1D 
VB1D
 B 2Q 
VB 2Q
B
B
,
39
Convert to Per Unit
Rs 
rs
Z BDQ
,
R fd 
r fd
Z BFD
,
r1q
R1q 
,
Z B1Q
Z BDQ 
VBDQ
I BDQ
,
Z BFD
VBFD

,
I BFD
Z B1Q 
VB1Q
I B1Q
,
r1d
R1d 
,
Z B1D
R2 q 
Z B1D
r2 q
Z B 2Q
,
VB1D

,
I B1D
Z B 2Q 
VB 2Q
I B 2Q
40
Convert to Per Unit
•
Almost done with the per unit conversions! Finally
define inertia constants and torque
2
H
TM
1
2
J (B )
2
P , M  2H
SB
s
Tm
, TELEC
TB
Te
, TFW
TB
T fw
TB
, TB
SB
2
B
P
41
Synchronous Machine Equations
1 d d

 Rs I d   q  Vd
 s dt
s
1 d q

 Rs I q   d  Vq
 s dt
s
1 d o
 Rs I o  Vo
 s dt
1 d fd
  R fd I fd  V fd
 s dt
1 d1d
  R1d I1d  V1d
 s dt
1 d 1q
  R1q I1q  V1q
s dt
1 d 2 q
  R2 q I 2  V2 q
s dt
d
   s
dt
2 H d
 TM   d I q   q I d  TFW
 s dt


42