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TRANSCRIPT
ECEN 667 Power System Stability
Lecture 6: Synchronous Machine Modeling
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
Texas A&M University
1
Announcements
• Read Chapter 3
• Homework 1 is due today
• Homework 2 is due on Thursday September 19
2
37 Bus System
• Next we consider a slightly larger, ten generator, 37
bus system. To view this system open case
AGL37_TS. The system one-line is shown below.
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”
slack
Aggieland Power and LightSLA CK345
SLA CK138
HOWDY345
HOWDY138
HOWDY69
12MA N69 63%
A
MVA
GIGEM69
KYLE69
KYLE138
WEB138
WEB69
BONFIRE69
FISH69
RING69
T REE69
CENT URY69
REVEILLE69
T EXA S69
T EXA S138
T EXA S345
BA T T 69
NORT HGA T E69
MA ROON69
SPIRIT 69
YELL69
RELLIS69
WHIT E138
RELLIS138
BUSH69
MSC69
RUDDER69
HULLA BA LOO138
REED69
REED138
A GGIE138 A GGIE345
12%A
MVA
25%A
MVA
41%A
MVA
24%A
MVA
50%A
MVA
31%A
MVA
55%A
MVA
54%A
MVA
33%A
MVA
28%A
MVA
44%A
MVA
24%A
MVA
34%A
MVA
72%A
MVA
27%A
MVA
50%A
MVA
44%A
MVA
29%A
MVA
37%A
MVA
37%A
MVA
37%A
MVA
46%A
MVA
37%A
MVA
55%A
MVA
34%A
MVA
19%A
MVA
68%A
MVA
72%A
MVA
56%A
MVA
48%A
MVA
55%A
MVA
43%A
MVA
67%A
MVA
25%A
MVA
50%A
MVA
28%A
MVA
38%A
MVA
38%A
MVA
A
MVA
A
MVA
21%A
MVA
51%A
MVA
52%A
MVA
41%A
MVA
61%A
MVA
70%A
MVA
54%A
MVA
54%A
MVA
1.02 pu
1 .01 pu
1 .02 pu
1 .03 pu
0 .98 pu
1 .00 pu1 .01 pu
0 .99 pu
1 .00 pu
1 .02 pu
1 .00 pu
1 .01 pu1 .01 pu
1 .01 pu
1 .02 pu
1 .02 pu
0 .98 pu
0 .98 pu
1 .04 pu
1 .008 pu
1 .01 pu
1 .01 pu
1 .02 pu
0 .99 pu
0 .99 pu
1 .00 pu
1 .00 pu 1 .00 pu
1 .00 pu
1 .00 pu1 .00 pu
79%A
MVA
1.02 pu
73%A
MVA
34%A
MVA
PLUM138
A
MVA
1.01 pu
19%A
MVA
1.01 pu
57%A
MVA
270 MW
34 MW
0 Mvar
59 MW
17 Mvar
MW 250
100 MW
30 Mvar
20 MW
8 Mvar
MW 500
100 MW
30 Mvar
61 MW 17 Mvar
59 MW
6 Mvar
70 MW
0 Mvar
93 MW
58 Mvar
58 MW
17 Mvar
MW 0 36 MW
24 Mvar
96 MW
20 Mvar
MW 60
37 MW
14 Mvar
53 MW
21 Mvar
16.2 Mvar 29 MW
8 Mvar
93 MW
65 Mvar 82 MW
27 Mvar
20.3 Mvar
35 MW
11 Mvar
25 MW
10 Mvar
38 MW
10 Mvar 22 MW
0 Mvar
31.2 Mvar
22.2 Mvar
12.7 Mvar
29.4 Mvar
31.2 Mvar
20.8 Mvar
MW 80
31 MW
13 Mvar
MW 92
27 MW
4 Mvar
MW 12
49 MW
17 Mvar
MW 80
MW 100
deg 0
tap1.0875
tap1.0125
tap1.0000
3
Transient Stability Case and Model Summary Displays
Right click on a line
and select “Show
Dialog” for more
information.
4
37 Bus Case Solution
Graph
shows the
rotor angles
following
a line fault
5
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
Run a Specified Number of Timesteps or Run
Until a Specified Time, then Pause.
See detailed results
at the Paused Time
Transfer results
to Power Flow
to view using
standard
PowerWorld
displays and
one-lines
6
Physical Structure Power System Components
Generator
P, Q
Network
Network
control
Loads
Load
control
Fuel
Source
Supply
control
Furnace
and Boiler
Pressure
control
Turbine
Speed
control
V, ITorqueSteamFuel
Electrical SystemMechanical System
Voltage
Control
P. Sauer and M. Pai, Power System Dynamics and Stability, Stipes Publishing, 2006.
7
Dynamic Models in the Physical Structure
Machine
Governor
Exciter
Load
Char.
Load
RelayLine
Relay
Stabilizer
Generator
P, Q
Network
Network
control
Loads
Load
control
Fuel
Source
Supply
control
Furnace
and Boiler
Pressure
control
Turbine
Speed
control
V, ITorqueSteamFuel
Electrical SystemMechanical System
Voltage
Control
P. Sauer and M. Pai, Power System Dynamics and Stability, Stipes Publishing, 2006.
8
Generator Models
• Generators can
have several
classes of models
assigned to them
– Machine Models
– Exciter
– Governors
– Stabilizers
• Others also available
– Excitation limiters, voltage compensation, turbine load
controllers, and generator relay model
9
Generator Models
10
Machine Models
11
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 we'll start there
• There is much literature on subject, and sometimes it is
overly confusing with the use of different conventions
and nomenclature
12
Synchronous Machine Modeling
3 bal. windings (a,b,c) – stator
Field winding (fd) on rotor
Damper in “d” axis
(1d) on rotor
Two dampers in “q” axis
(1q, 2q) on rotor
13
Two Main Types of Synchronous Machines
• Round Rotor
– Air-gap is constant, used with higher speed machines
• Salient Rotor (often called Salient Pole)
– Air-gap varies circumferentially
– Used with many pole, slower machines such as hydro
– Narrowest part of gap in the d-axis and the widest along
the q-axis
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)
• Convention used here is the q-axis leads the d-axis
(which is the IEEE standard)
15
Synchronous Machine Stator
Image Source: Glover/Overbye/Sarma Book, Sixth Edition, Beginning of Chapter 8 Photo
16
Synchronous Machine Rotors
• Rotors are essentially electromagnets
Image Source: Dr. Gleb Tcheslavski, ee.lamar.edu/gleb/teaching.htm
Two pole (P)
round rotor
Six pole salient
rotor
17
Synchronous Machine Rotor
Image Source: Dr. Gleb Tcheslavski, ee.lamar.edu/gleb/teaching.htm
High pole
salient
rotor
Shaft
Part of exciter,
which is used
to control the
field current
18
Fundamental Laws
• Kirchhoff’s Voltage Law, Ohm’s Law, Faraday’s
Law, Newton’s Second Law
11 1 1
11 1 1
22 2 2
fdfd fd fd
dd d d
qq q q
qq q q
dv i r
dt
dv i r
dt
dv i r
dt
dv i r
dt
aa a s
bb b s
cc c s
dv i r
dt
dv i r
dt
dv i r
dt
shaft 2
2m e f
d
dt P
dJ T T T
P dt
Stator Rotor Shaft
19
Dq0 Transformations
1
or ,
d a
q dqo b
co
a d
b dqo q
c o
v v
v T v i
vv
v v
v T v
v v
In the next few slides
we’ll quickly go
through how these
basic equations are
transformed into the
standard machine
models. The point
is to show the physical
basis for the models.
And there is NO quiz
at the end!!
20
Dq0 Transformations
2 2sin sin sin
2 2 3 2 3
2 2 2cos cos cos
3 2 2 3 2 3
1 1 1
2 2 2
shaft shaft shaft
dqo shaft shaft shaft
P P P
P P PT
with the inverse,
13
2
2cos
3
2
2sin
13
2
2cos
3
2
2sin
12
cos2
sin
1
shaftshaft
shaftshaft
shaftshaft
dqo
PP
PP
PP
T
Note that the
transformation
depends on the
shaft angle.
21
Transformed System
11 1 1
11 1 1
22 2 2
fdfd fd fd
dd d d
qq q q
qq q q
dv r i
dt
dv r i
dt
dv r i
dt
dv r i
dt
2
2
shaft
m e f
d
dt P
dJ T T T
P dt
dd s d q
qq s q d
oo s o
dv r i
dt
dv r i
dt
dv r i
dt
Stator Rotor Shaft
22
Electrical & Mechanical Relationships
Electrical system:
2
(voltage)
(power)
dv iR
dt
dvi i R i
dt
Mechanical system:
2
2(torque)
2 2 2 2(power)
m e fw
m e fw
dJ T T T
P dt
dJ T T T
P dt P P P
P is the
number of
poles (e.g.,
2,4,6); Tfw
is the friction
and windage
torque
23
Torque Derivation
• 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 are in the form of resistance
– Mechanical system losses are 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
24
Energy Conversion
ooqqddccbbaa iviviviviviv 32
3
2
3
Look at the instantaneous power:
25
Change to Conservation of Power
dt
di
dt
di
dt
di
dt
di
dt
di
dt
di
dt
diP
iririririiirP
iv
ivivivivivivP
dd
fdfd
cc
bb
aa
electtrans
qqqqddfdfdcbaselectlost
qqddfdfdccbbaaelectin
22
11
11
222
211
211
2222
22
1111
We are using
v = d/dt here
26
With the Transformed Variables
222
211
211
2222
2211
11
32
3
2
3
32
3
2
3
qqqq
ddfdfdosqsdselectlost
qqqq
ddfdfdooqqddelectin
irir
iririririrP
iviv
ivivivivivP
27
With the Transformed Variables
dt
di
dt
di
dt
di
dt
di
dt
di
dt
di
idt
dP
dt
dii
dt
dPP
dd
fdfd
oo
qdshaftd
ddqshaft
electtrans
22
11
113
2
3
22
3
2
3
22
3
28
Change in Coupling Field Energy
fdW
dt
PTe
2
dt
dai
dt
d abi
dt
d b
ci dt
d cfdi
dt
d fddi1
dt
d d1
qi1 dt
d q1qi2
dt
d q2
This requires the lossless coupling field assumption
First term on
right is what is
going on
mechanically,
other terms are
what is going
on electrically
29
Change in Coupling Field Energy
For independent states , a, b, c, fd, 1d, 1q, 2q
fdW
dt
fW
dt
d f
a
W
dt
d a f
b
W
dt
d b
f
c
W
dt
d c f
fd
W
dt
d fd
1
f
d
W
dt
d d1
1
f
q
W
dt
d q1
2
f
q
W
dt
d q2
30
Equate the Coefficients
2 f fe a
a
W WT i
P
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.
31
Equate the Coefficients
shaft
fW
3
2 2d q q d e
Pi i T
dd
fi
W
2
3
3, , 3
2
f fq o
q o
W Wi i
fdfd
fi
W
1 1 2
1 1 2
, , ,f f f
d q qd q q
W W Wi i i
These are key conditions – i.e. the first one gives an expression for
the torque in terms of the coupling field energy.
32
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
• Since integration is path independent, choose a
convenient path
– Start with a de-energized system so variables are zero
– Integrate shaft position while other variables are zero
– Integrate sources in sequence with shaft at final value
3For example,
2
fdd
fd d
ii
33
Define Unscaled Variables
11 1 1
fdfd fd fd
dd d d
dr i v
dt
dr i v
dt
2shaft s
Pt
ds d q d
qs q d q
os o o
dr i v
dt
dr i v
dt
dr i v
dt
11 1 1
22 2 2
qq q q
qq q q
dr i v
dt
dr i v
dt
2 3
2 2
s
m d q q d f
d
dt
d PJ T i i T
p dt
s is the rated
synchronous speed
plays an important role!
34
Synchronous Machine Equations in Per Unit
1
1
1
ds d q d
s s
qs q d q
s s
os o o
s
dR I V
dt
dR I V
dt
dR I V
dt
11 1 1
1
1
fdfd fd fd
s
dd d d
s
dR I V
dt
dR I V
dt
11 1 1
22 2
1
12
qq q q
s
qq q
s
dR I V
dt
dR I V
dt
2
s
M d q q d FWs
d
dt
H dT I I T
dt
Units of H are
seconds
35
Sinusoidal Steady-State
3
2cos2
3
2cos2
cos2
3
2cos2
3
2cos2
cos2
isssc
isssb
isssa
vsssc
vsssb
vsssa
tII
tII
tII
tVV
tVV
tVVHere we consider the
application to balanced,
sinusoidal conditions
36
Simplifying Using
• Define
• Hence
• These algebraic
equations can be
written as complex
equations
issq
issd
vssq
vssd
II
II
VV
VV
cos
sin
cos
sin
2shaft s
Pt
The conclusion is
if we know , then
we can easily relate
the phase to the dq
values!
/ 2
/ 2
jj vsV jV e V ed q s
jj isI jI e I ed q s
37
Summary So Far
• The model as developed so far has been derived
using the following assumptions
– The stator has three coils in a balanced configuration,
spaced 120 electrical degrees apart
– Rotor has four coils in a balanced configuration located
90 electrical degrees apart
– Relationship between the flux linkages and currents must
reflect a conservative coupling field
– The relationships between the flux linkages and currents
must be independent of shaft when expressed in the dq0
coordinate system
38
Assuming a Linear Magnetic Circuit
• If the flux linkages are assumed to be a linear
function of the currents then we can write
The rotor
self-
inductance
matrix
Lrr is
independent
of shaft
1 1
1 1
2 2
a a
b bss shaft sr shaft
c c
fd fd
d d
rs shaft rr shaftq q
q q
i
iL L
i
i
iL L
i
i
39
Conversion to dq0 for Angle Independence
1
1 11
1 1
2 2
d d
q q
o odqo srdqo ss dqo
fd fd
d d
rrrs dqoq q
q q
i
i
iT LT L Ti
iLL T
i
i
40
Conversion to dq0 for Angle Independence
1 1
1 1
1 1 1 1 1 1
3
2
3
2
d s md d sfd fd s d d
fd sfd d fdfd fd fd d d
d s d d fd d fd d d d
L L i L i L i
L i L i L i
L i L i L i
1 1 2 2
1 1 1 1 1 1 2 2
2 2 1 2 1 2 2 2
3
2
3
2
q s mq q s q q s q q
q s q q q q q q q q
q s q q q q q q q q
L L i L i L i
L i L i L i
L i L i L i
o s oL i
,md A B
mq A B
3L L L
2
3L L L
2
For a round rotor
machine LB is small
and hence Lmd is
close to Lmq. For a
salient pole machine
Lmd is substantially
larger
41
Convert to Normalized at f = s
• Convert to per unit, and assume frequency of s
• Then define new per unit reactance variables
, ,
, ,
, ,
,
,
s mqs s s mds md mq
BDQ BDQ BDQ
s fdfd s fd 1d sfds 1d 1dfd 1d fd 1d
BFD B1D BFD s1d
s 1q1q s 2q2q s 1q2q s1q
1q 2q 1q2q
B1Q B2Q B1Q s2q
fd fd md 1d 1d md
1q 1q mq 2q 2
LL LX X X
Z Z Z
L L LLX X X
Z Z Z L
L L L LX X X
Z Z Z L
X X X X X X
X X X X X
,
q mq
d s md q s mq
X
X X X X X X
42
Key Simulation Parameters
• The key parameters that occur in most models can
then be defined as 2
2
1
1
1
1
1
1 1
1
1 1
,
mdd s d
fd
md fd
mqq s q
q
mq q
qdo qo
s fd s q
XX X X
X
X X
XX X X
X
X X
XXfdT T
R R
These values
will be used in
all the
synchronous
machine models
In a salient rotor machine
Xmq is small so Xq = X'q;
also X1q is small so
T'q0 is small
43
Key Simulation Parameters
• And the subtransient parameters
These values
will be used in the
subtransient machine
models. It is common
to assume X"d = X"q
1
1 2
1 21 2
1 1
1
1 1 1
1
1 1 1
1 1 1 1,
1 1 1 1
d s
md fd d
q s
mq q q
do d qo qs d s q
md d mq q
X X
X X X
X X
X X X
T X T XR R
X X X X
44
Example Xd/Xq Ratios for a WECC Case
45
Example X'q/Xq Ratios for a WECC Case
About 75% are Clearly Salient Pole Machines!
46
Internal Variables
• Define the following variables, which are quite
important in subsequent models
Hence E'q and E'd are
scaled flux linkages
and Efd is the scaled
field voltage
11
XmdE
q fdXfd
Xmq
Ed qX
q
XmdE V
fd fdRfd
47
Dynamic Model Development
• In developing the dynamic model not all of the
currents and fluxes are independent
– In this formulation only seven out of fourteen are
independent
• Approach is to eliminate the rotor currents,
retaining the terminal currents (Id, Iq, I0) for
matching the network boundary conditions
48
Rotor Currents
• Use new variables to solve for the rotor currents
1
1
1 12
1
d s d dd d d q d
d s d s
fd q d d d dmd
d dd d d s d q
d s
X X X XX I E
X X X X
I E X X I IX
X XI X X I E
X X
49
Rotor Currents
2
1 2
2 22
1
q s q q
q q q d q
q s q s
q d q q q qmq
q qq q q s q d
q s
o s o
X X X XX I E
X X X X
I E X X I IX
X XI X X I E
X X
X I
50
Final Complete Model
These first three equations
define what are known
as the stator transients; we
will shortly approximate
them as algebraic constraints
12
q d ddo q d d d d d s d q fd
d s
dE X XT E X X I X X I E E
dt X X
22
q qdqo d q q q q q s q d
q s
X XdET E X X I X X I E
dt X X
51
Final Complete Model
11
22
2
ddo d q d s d
qqo q d q s q
s
M d q q d FWs
dT E X X I
dt
dT E X X I
dt
d
dt
H dT I I T
dt
TFW is the friction
and windage
component
1
2
d s d sd d d q d
d s d s
q s q q
q q q d q
q s q s
o s o
X X X XX I E
X X X X
X X X XX I E
X X X X
X I
52
Single-Machine Steady-State
The key variable
we need to
determine the
initial conditions
is actually , which
doesn't appear
explicitly in these
equations!
1
0
0
0
0
0
s d q d
s q d q
s o o
q d d d fd
d q d s d
R I V
R I V
R I V
E X X I E
E X X I
s d q d d
q q q d
o s o
E X I
X I E
X I
2
0
0
0
0
d q q q
q d q s q
s
m d q q d FW
E X X I
E X X I
T I I T
53
Field Current
• The field current, Ifd, is defined in steady-state as
• However, what is usually used in transient stability
simulations for the field current is the product
• So the value of Xmd is not needed
/fd fd mdI E X
fd mdI X
54
Single-Machine Steady-State
• Previous derivation was done assuming a linear
magnetic circuit
• We'll consider the nonlinear magnetic circuit later but
will first do the steady-state condition (3.6)
• In steady-state the speed is constant (equal to s), is
constant, and all the derivatives are zero
• Initial values are determined from the terminal
conditions: voltage magnitude, voltage angle, real and
reactive power injection
55
Determining without Saturation
• In order to get the initial values for the variables we
need to determine
• We'll eventually consider two approaches: the simple
one when there is no saturation, and then later a
general approach for models with saturation
• To derive the simple approach we have
d s d d q q
q s q q d d
V R I E X I
V R I E X I
56
Determining without Saturation
• In terms of the terminal values
/2Since
j
jq d d q
j e
E X X I E e
( )as s q asE V R jX I
The angle on E