ece 576 – power system dynamics and stability prof. tom overbye dept. of electrical and computer...
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![Page 1: ECE 576 – Power System Dynamics and Stability Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign](https://reader035.vdocuments.net/reader035/viewer/2022081420/551a17cb55034646638b463e/html5/thumbnails/1.jpg)
ECE 576 – Power System Dynamics and Stability
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
1
Lecture 6: Synchronous Machine Modeling
Special Guest: TA Soobae Kim
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Announcements
• Read Chapter 3
2
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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
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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
4
Synchronous Machine Modeling
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Dq0 Reference Frame
• Stator is stationary and rotor is rotating at synchronous speed
• Rotor values need to be transformed to fixed reference frame for analysis
• This is done using Park's transformation into what is known as the dq0 reference frame (direct, quadrature, zero)
• 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
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11 1 1
11 1 1
22 2 2
fdfd fd fd
dd d d
qq q q
qq q q
dv i r
dtd
v i rdt
dv i r
dtd
v i rdt
Kirchhoff’s Voltage Law, Ohm’s Law, Faraday’s Law, Newton’s Second Law
aa a s
bb b s
cc c s
dv i r
dtd
v i rdt
dv i r
dt
shaft 2
2m e f
d
dt Pd
J T T TP dt
Stator Rotor Shaft
6
Fundamental Laws
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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
7
Dq0 transformations
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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
22
cos3
22
sin
13
22
cos3
22
sin
12
cos2
sin
1
shaftshaft
shaftshaft
shaftshaft
dqo
PP
PP
PP
T
8
Dq0 transformations
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abcabc Li
(symmetric)
dqodqo
dqoabc
iTLT
iTLTT
1
1
Not symmetric if T is not power invariant.
Example: If the magnetic circuit is assumed to be linear
Note: This transformation is not power invariant. This means that some unusual things will happen when we use it.
9
Dq0 transformations
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11 1 1
11 1 1
22 2 2
fdfd fd fd
dd d d
qq q q
qq q q
dv r i
dtd
v r idt
dv r i
dtd
v r idt
2
2
shaft
m e f
d
dt Pd
J T T TP dt
dd s d q
qq s q d
oo s o
dv r i
dtd
v r idt
dv r i
dt
Stator Rotor Shaft
10
Transformed System
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Electrical system:
2
(voltage)
(power)
dv iR
dtd
vi i R idt
Mechanical system:
2
2(torque)
2 2 2 2(power)
m e f
m e f
dJ T T T
P dt
dJ T T T
P dt P P P
Electrical & Mechanical Relationships
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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
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Energy Conversion
ooqqddccbbaa iviviviviviv 323
23
Look at the instantaneous power:
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dt
di
dt
di
dtd
idt
di
dtd
idt
di
dtd
iP
iririririiirP
iv
ivivivivivivP
dd
fdfd
cc
bb
aa
electtrans
qqqqddfdfdcbaselectlost
qqddfdfdccbbaaelectin
22
11
11
222
211
211
2222
22
1111
14
Change to Conservation of Power
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222
211
211
2222
2211
11
323
23
323
23
qqqq
ddfdfdosqsdselectlost
qqqq
ddfdfdooqqddelectin
irir
iririririrP
iviv
ivivivivivP
15
With the Transformed Variables
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dt
di
dt
di
dtd
idt
di
dtd
idt
di
idt
dPdt
dii
dt
dPP
dd
fdfd
oo
qdshaftd
ddqshaft
electtrans
22
11
113
23
223
23
223
16
With the Transformed Variables
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fdW
dt
PTe
2 dtd
ai 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
17
Change in Coupling Field Energy
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For independent states , a, b, c, fd, 1d, 1q, 2q
fdW
dt fW
dtd 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
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Change in Coupling Field Energy
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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.
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Equate the Coefficients
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shaft
fW
3
2 2 d q q d eP
i i T
dd
f iW
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3, , 3
2f f
q oq o
W Wi i
fdfd
f iW
1 1 2
1 1 2
, , ,f f fd q q
d q q
W W Wi i i
20
Equate the Coefficients
These are key conditions – i.e. the first one gives an expression for the torque in terms of the coupling field energy.
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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 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
21
3For example,
2fdd
fd d
ii
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3 ˆ ˆ ˆ2 2
3 3ˆ ˆ ˆ32 2
1 21ˆ ˆ ˆ ˆ
1 1 1 1 2 2
1 1 2
shaft PoW W i i df f d q q d shaftoshaft
qd oi d i d i dd d q q o o
o o od q o
fd q qdi d i d i d i dfd fd d d q q q q
o o o ofd d q q
22
Do the Integration
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Torque
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• 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
dqqde iiP
T 22
3
fd q q d e
shaft
W 3 Pi i T 0
2 2
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11 1 1
fdfd fd fd
dd d d
dr i v
dtd
r i vdt
2 shaft sP
t
ds d q d
qs q d q
os o o
dr i v
dtd
r i vdt
dr i v
dt
11 1 1
22 2 2
qq q q
qq q q
dr i v
dtd
r i vdt
2 3
2 2
s
m d q q d f
d
dtd P
J T i i Tp dt
Define Unscaled Variables
s is the ratedsynchronous speedd plays an important role!
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Convert to Per Unit
• As with power flow, values are usually expressed in per unit, here on the machine power rating
• 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
25
BaseBaseBase PIV
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, , ,
, ,
, ,
a b ca b c
BABC BABC BABC
a b ca b c
BABC BABC BABC
a b ca b c
BABC BABC BABC
v v vV V V
V V V
i i iI I I
I I I
where VBABC is rated RMS line-to-neutral stator voltage and
,3
B BABCBABC BABC
BABC B
P VI
V
Convert to Per Unit
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, , ,
, ,
, ,
qd od q o
BDQ BDQ BDQ
qd od q o
BDQ BDQ BDQ
qd od q o
BDQ BDQ BDQ
vv vV V V
V V V
ii iI I I
I I I
where VBDQ is rated peak line-to-neutral stator voltage and
2,
3BDQB
BDQ BDQBDQ B
VPI
V
27
Convert to Per Unit
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1 211 1 2
1 1 2
1 211 1 2
1 1 2
1 211 1 2
1 1 2
, , ,
, , ,
, , ,
fd q qdfd d q q
BFD B D B Q B Q
fd q qdfd d q q
BFD B D B Q B Q
fd q qdfd d q q
BFD B D B Q B Q
v v vvV V V V
V V V V
i i iiI I I I
I I I I
Convert to Per Unit
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Hence the variables are just normalizedflux linkages
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Where the rotor circuit base voltages are
11
1 21 2
, ,
,
B BBFD B D
BFD B D
B BB Q B Q
B Q B Q
P PV V
I I
P PV V
I I
And the rotor circuit base flux linkages are
11
1 21 2
, ,
,
BFD B DBFD B D
B B
B Q B QB Q B Q
B B
V V
V V
Convert to Per Unit
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11
1
21 2
1 2
11
1
1 21 2
1 2
, , ,
1, ,
, , ,
,
fds ds fd d
BDQ BFD B D
qq q
B Q B Q
BDQ BFD B DBDQ BFD B D
BDQ BFD B D
B Q B QB Q B Q
B Q B Q
rr rR R R
Z Z Z
rr qR R
Z Z
V V VZ Z Z
I I I
V VZ Z
I I
Convert to Per Unit
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Convert to Per Unit
• Almost done with the per unit conversions! Finally define inertia constants and torque
31
( ),
, , ,
2
B
B s
fwm e BM ELEC FW B
B B BB
1 2J 2H2 PH M
S
TT T ST T T T
2T T TP
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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
dtH d
T I I Tdt
Synchronous Machine Equations
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32
cos2
32
cos2
cos2
32
cos2
32
cos2
cos2
isssc
isssb
isssa
vsssc
vsssb
vsssa
tII
tII
tII
tVV
tVV
tVV
Sinusoidal Steady-State
33
Here we consider the applicationto balanced, sinusoidal conditions
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0
2cos
2
2sin
2
o
vssshaftBDQ
BABCsq
vssshaftBDQ
BABCsd
V
tP
VVV
V
tP
VVV
V
2sin
2
2cos
2
0
s BABCd shaft s is
BDQ
s BABCq shaft s is
BDQ
o
I I PI t
I
I I PI t
I
I
Transforming to dq0
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issq
issd
vssq
vssd
II
II
VV
VV
cos
sin
cos
sin
/ 2
/ 2
jj vsV jV e V ed q s
jj isI jI e I ed q s
2 shaft sP
t
Simplifying Using d
• Recall that
• Hence
• These algebraic equations can be written as complex equations,
35
The conclusion is if we know d, thenwe can easily relatethe phase to the dq values!
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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
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Assuming a Linear Magnetic Circuit
• If the flux linkages are assumed to be a linear function of the currents then we can write
37
1 1
1 1
2 2
a a
b bss shaft sr shaft
c c
fd fd
d drs shaft rr shaft
q q
q q
i
iL Li
i
iL L
i
i
The rotorself-inductancematrix Lrr is independentof qshaft
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L12 = 0 L12 = + maximum
L12 = - maximum
Inductive Dependence on Shaft Angle
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Stator Inductances
• The self inductance for each stator winding has a portion that is due to the leakage flux which does not cross the air gap, Lls
• The other portion of the self inductance is due to flux crossing the air gap and can be modeled for phase a as
• Mutual inductance between the stator windings is modeled as
39
cos( )A B shaftL L P
cos( )A B shaft
1L L P offset
2
The offset angleis either 2 /3 or-2 /3
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1
1 11
1 1
2 2
d d
q q
o odqo srdqo ss dqo
fd fd
d drrrs dqo
q q
q q
i
i
iT LT L Ti
iLL T
i
i
40
Conversion to dq0 for Angle Independence
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1 1
1 1
1 1 1 1 1 1
3
23
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
23
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
Conversion to dq0 for Angle Independence
41
,md A B
mq A B
3L L L
33
L L L3
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Convert to Normalized at f = ws
• Convert to per unit, and assume frequency of ws
• Then define new per unit reactance variables
42
, ,
, ,
, ,
,
,
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 s1q1q 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
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1
1
1 1 1
1
d d d md fd md d
fd md d fd fd d md d
d md d d md fd d d
d
X I X I X I
X I X I c X I
X I c X I X I
c
1 2
1 1 1 2
2 1 2 2
1
q q q mq q mq q
q mq q q q q mq q
q mq q q mq q q q
q
X I X I X I
X I X I c X I
X I c X I X I
c
o s oX I
Normalized Equations
1 1 2,fd d q qd q
md mq
X Xc c
X X
43