ECE 576 – Power System
Dynamics and Stability
Lecture 11:Commercial Machine Models
and Exciters
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
1
GENROU Initialization
•
•
Then solve five nonlinear equations from five
unknowns
– The five unknowns are d, E'q, E'd,  'q, and  'd
Five equations come from the terminal power flow
constraints (giving voltage and current) and from the
differential equations initially evaluating to zero
– Two differential equations for the q-axis, one for the d-axis
(the other equation is used to set the field voltage
2
Announcements
•
•
•
•
•
Read Chapters 4 and 6
Homework 2 is due today
Homework 3 is posted on the web; it is due on
Thursday Feb 25
Midterm exam is moved to March 31 (in class) because
of the ECE 431 field trip on March 17
We will have an extra class on March 7 from 1:30 to
3pm in 5070 ECEB
– No class on Tuesday Feb 23, Tuesday March 1 and Thursday
March 3
3
Implementing Saturation Models
•
•
•
When implementing saturation models in code, it is
important to recognize that the function is meant to be
positive, so negative values are not allowed
In large cases one is almost guaranteed to have special
cases, sometimes caused by user typos
– What to do if Se(1.2) < Se(1.0)?
– What to do if Se(1.0) = 0 and Se(1.2) <> 0
– What to do if Se(1.0) = Se(1.2) <> 0
Exponential saturation models have also been used
4
GENSAL Example with Saturation
•
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
E fd  Eq 1  Sat ( Eq )    X d  X d  I D


 1.1298 1  3.82 1.1298  0.838    1.784  3.28
 1.1298 1  B 1.1298  A    2.1  0.3 (0.9909)
2
2
Saturation
coefficients
were
determined
from the two
initial values
Saved as case B4_GENSAL_SAT
5
GENROU
•
•
The GENROU model has been widely used to model
round rotor machines
Saturation is assumed to occur on both the d-axis and
the q-axis, making initialization slightly more difficult
6
GENROU Block Diagram
The d-axis is
similar to that
of the
GENSAL; the
q-axis is now
similar to the
d-axis. Note
that saturation
now affects
both axes.
7
GENROU Initialization
•
•
Because saturation impacts both axes, the simple
approach will no longer work
Key insight for determining initial d is that the
magnitude of the saturation depends upon the
magnitude of ", which is independent of d
   V  ( Rs  jX ) I
•
This point is
crucial!
Solving for d requires an iterative approach; first get a
guess of d using 3.229 from the book
E d  V   Rs  jX q  I
8
GENROU Initialization
•
•
Then solve five nonlinear equations from five
unknowns
– The five unknowns are d, E'q, E'd,  'q, and  'd
Five equations come from the terminal power flow
constraints (giving voltage and current) and from the
differential equations initially evaluating to zero
– Two differential equations for the q-axis, one for the d-axis
(the other equation is used to set the field voltage
9
GENROU Initialization
•
•
Use dq transform to express terminal current as
 I d   sin d
I   
 q  cos d
 cos d   I r 
sin d   I i 
These values will change during
the iteration as d changes
Get expressions for  "q and  "d in terms of the initial
terminal voltage and d
– Use dq transform to express terminal voltage as
Vd   sin d
V   
 q  cos d
– Then from
 cos d  Vr 
sin d  Vi 
Recall Xd"=Xq"=X"
and w=1 (in steady-state)
 q  j d  Vd  jVq   ( Rs  jX )  I d  jI q 
 q  Vd  Rs I d  X I q Expressing complex
 d  Vq  Rs I a  X I d
equation as two real
equations
10
GENROU Initialization Example
•
•
Extend the two-axis example
–
For two-axis 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.
– For subtransient fields assume X"d=X"q=0.28, Xl = 0.13,
T"do = 0.073, T"qo =0.07
– for comparison we'll initially assume no saturation
From two-axis get a guess of d
E  1.094611.59   j 2.0 1.052  18.2   2.81452.1
 d  52.1
11
GENROU Initialization Example
•
And the network current and voltage in dq reference
Vd  0.7889 0.6146 1.0723 0.7107 

V   




0.6146
0.7889
0.220
0.8326
q






 
 I d  0.7889 0.6146  1.000  0.9909

I   




 q  0.6146 0.7889   0.3287   0.3553
•
Which gives initial subtransient fluxes (with Rs=0),
    j   w  V
q
d
d
 jVq   ( Rs  jX )  I d  jI q 
 q w  Vd  Rs I d  X I q  0.7107  0.28  0.3553  0.611
 d w  Vq  Rs I a  X I d  0.8326  0.28  0.9909  1.110
12
GENROU Initialization Example
•
•
•
Without saturation this is the exact solution
Initial values are:
d = 52.1 ,
E'q=1.1298,
E'd=0.533,
 'q =0.6645,
and  'd=0.9614
Efd=2.9133
Saved as case
B4_GENROU_NoSat
13
Two-Axis versus GENROU Response
Figure compares rotor angle for bus 3 fault, cleared at
t=1.1 seconds
14
GENROU with Saturation
•
•
•
•
Nonlinear approach is needed in common situation in
which there is saturation
Assume previous GENROU model with S(1.0) = 0.05,
and S(1.2) = 0.2.
Initial values are: d = 49.2 , E'q=1.1591, E'd=0.4646,  'q
=0.6146, and  'd=0.9940
Efd=3.2186
Saved as case
B4_GENROU_Sat
15
Two-Axis versus GENROU Response
16
GENTPF and GENTPJ Models
•
These models were introduced into PSLF in 2009 to
provide a better match between simulated and actual
system results for salient pole machines
– Desire was to duplicate functionality from old BPA TS code
– Allows for subtransient saliency (X"d <> X"q)
– Can also be used with round rotor, replacing GENSAL and
•
GENROU
Useful reference is available at below link; includes all
the equations, and saturation details
https://www.wecc.biz/Reliability/gentpj-typej-definition.pdf
17
GENSAL Results
Chief Joseph
disturbance
playback
GENSAL
BLUE = MODEL
RED = ACTUAL
Image source :https://www.wecc.biz/library/WECC%20Documents/Documents%20for
%20Generators/Generator%20Testing%20Program/gentpj%20and%20gensal%20morel.pdf
18
GENTPJ Results
Chief Joseph
disturbance
playback
GENTPJ
BLUE = MODEL
RED = ACTUAL
19
GENTPF and GENTPJ Models
•
•
GENTPF/J d-axis block diagram
GENTPJ allows saturation function to include a
component that depends on the stator current
Se = 1 + fsat( ag + Kis*It)
Most of
WECC
machine
models
are now
GENTPF
or
GENTPJ
If nonzero, Kis typically ranges from 0.02 to 0.12
20
Theoretical Justification for
GENTPF and GENTPJ
•
In the GENROU and GENSAL models saturation shows
up purely as an additive term of Eq’ and Ed’
– Saturation does not come into play in the network interface
•
equations and thus with the assumption of Xq”=Xd” a simple
circuit model can be used
The advantage of the GENTPF/J models is saturation
really affects the entire model, and in this model it is
applied to all the inductance terms simultaneously
– This complicates the network boundary equations, but since
these models are designed for Xq”≠ Xd” there is no increase in
complexity
21
GENROU/GENTPJ Comparison
Easy Paper Suggestion!
22
GenRou, GenTPF, GenTPJ
Figure compares gen 4 reactive power output for the
0.1 second fault
23
Voltage and Speed Control
P, 
Q,V 
24
Exciters, Including AVR
•
•
•
Exciters are used to control the synchronous machine
field voltage and current
– Usually modeled with automatic voltage regulator included
A useful reference is IEEE Std 421.5-2005
– Covers the major types of exciters used in transient stability
simulations
– Continuation of standard designs started with "Computer
Representation of Excitation Systems," IEEE Trans. Power
App. and Syst., vol. pas-87, pp. 1460-1464, June 1968
Another reference is P. Kundur, Power System Stability
and Control, EPRI, McGraw-Hill, 1994
– Exciters are covered in Chapter 8 as are block diagram basics
25
Functional Block Diagram
Image source: Fig 8.1 of Kundur, Power System Stability and Control
26
Types of Exciters
•
•
•
•
None, which would be the case for a permanent magnet
generator
– primarily used with wind turbines with ac-dc-ac converters
DC: Utilize a dc generator as the source of the field
voltage through slip rings
AC: Use an ac generator on the generator shaft, with
output rectified to produce the dc field voltage;
brushless with a rotating rectifier system
Static: Exciter is static, with field current supplied
through slip rings
27
Brief Review of DC Machines
•
•
•
•
Prior to widespread use of machine drives, dc motors
had a important advantage of easy speed control
On the stator a dc machine has either a permanent
magnet or a single concentrated winding
Rotor (armature) currents are supplied through brushes
and commutator
The f subscript refers to the field, the
Equations are
a to the armature; w is the machine's
v f  if Rf  Lf
di f
dt
dia
va  ia Ra  La
 Gwmi f
dt
speed, G is a constant. In a
permanent magnet machine the field
flux is constant, the field equation
goes away, and the field impact is
embedded in a equivalent constant
to Gif
Taken mostly from ECE 330 book, M.A. Pai, Power Circuits and Electromechanics
28
Types of DC Machines
•
If there is a field winding (i.e., not a permanent magnet
machine) then the machine can be connected in the
following ways
– Separately-excited: Field and armature windings are
connected to separate power sources
• For an exciter, control is provided by varying the field
current (which is stationary), which changes the armature
voltage
– Series-excited: Field and armature windings are in series
– Shunt-excited: Field and armature windings are in parallel
29
Separately Excited DC Exciter
(to sync
mach)
ein1  r f 1iin1  N f 1
a1 
1
1
f1
d f 1
dt
1 is coefficient of dispersion,
modeling the flux leakage
30
Separately Excited DC Exciter
•
Relate the input voltage, ein1, to vfd
f 1
v fd  K a1w1a1  K a1w1
1
N f 1 1
N f 1 f 1 
v fd
K a1w1
d f 1 N f 1 1 dv fd
Nf1

dt
K a1w1 dt
N f 1 1 dv fd
ein  iin rf 1 
K a1w1 dt
1
Assuming a constant
speed w1
1
31
Separately Excited DC Exciter
•
If it was a linear magnetic circuit, then vfd would be
proportional to in1; for a real system we need to account
for saturation
v fd
iin1 
 f sat v fd v fd
K g1
Without saturation we
can write
Kg1
K a1w1

L f 1us
N f 1 1
Where L f 1us is the
unsaturated field inductance
32
Separately Excited DC Exciter
ein  r f 1iin1  N f 1
1
d f 1
dt
Can be written as
ein 
1
rf 1
K g1
 
v fd  r f 1 f sat v fd v fd 
L f 1us dv fd
K g1
dt
This equation is then scaled based on the synchronous
machine base values
X md
X md v fd
E fd 
V fd 
R fd
R fd VBFD
33
Separately Excited Scaled Values
KE

sep
rf 1
K g1
L f 1us
TE 
K g1
X md
VR 
ein1
R fd VBFD
 
 VBFD R fd

S E E fd  r f 1 f sat 
E fd 
 X

md


Thus we have
TE
dE fd
dt
 


  KE
 S E E fd  E fd  VR


sep


Vr is the scaled
output of the
voltage
regulator
amplifier
34
The Self-Excited Exciter
•
When the exciter is self-excited, the amplifier voltage
appears in series with the exciter field
TE
dE fd
dt
 


  KE
 S E E fd  E fd  VR  E fd


sep


Note the
additional
Efd term on
the end
35
Self and Separated Excited Exciters
•
The same model can be used for both by just modifying
the value of KE
TE
dE fd
dt


  K E  S E  E fd  E fd  VR


KE
 KE
 1  typically K E
 .01


self
sep
self


36
Saturation
•
•
A number of different functions can be used to
represent the saturation
The quadratic approach is now quite common
S E ( E fd )  B ( E fd  A) 2
An alternative model is S E ( E fd ) 
•
B ( E fd  A) 2
E fd
Exponential function could also be used
S E  E fd   Ax e
Bx E fd
37
Exponential Saturation
KE  1
S E E fd   0.1e
0.5 E fd
.5 E fd

Steady state VR  1  .1e

 E
 fd
38
Exponential Saturation Example
Given:
K E  .05

S E  E fd
max


  0.27


S E  .75 E fd
max

VR
 1.0
max
Find:

  0.074

Ax , Bx and E fd max
S E  Axe
Bx E fd
E fd max  4.6
Ax  .0015
Bx  1.14
39
Voltage Regulator Model
Amplifier
dVR
TA
 VR  K AVin
dt
VRmin  VR  VRmax
VR
Vref  Vt  Vin 
KA
K A  Vt  Vref
In steady state
Big
Modeled
as a first
order
differential
equation
There is often a droop in regulation
40
Feedback
•
•
This control system can often exhibit instabilities, so
some type of feedback is used
One approach is a stabilizing transformer
N2
dIt1
Ltm
Large Lt2 so It2  0 VF 
N1
dt
41
Feedback
dIt1
E fd  Rt1It1   Lt1  Ltm 
dt

dVF
Rt1
N 2 Ltm dE fd 
  VF 


 Lt1  Ltm  
dt
N1 Rt1 dt 

1
TF

KF
42
IEEE T1 Exciter
•
This model was standardized in the 1968 IEEE
Committee Paper with Fig 1 shown below
43
IEEE T1 Evolution
•
This model has been subsequently modified over the
years, called the DC1 in a 1981 IEEE paper (modeled
as the EXDC1 in stability packages)
Note, KE in the
feedback is the
same as
the 1968
approach
Image Source: Fig 3 of "Excitation System Models for Power Stability Studies,"
IEEE Trans. Power App. and Syst., vol. PAS-100, pp. 494-509, February 1981
44
IEEE T1 Evolution
•
In 1992 IEEE Std 421.5-1992 slightly modified it,
calling it the DC1A (modeled as ESDC1A)
Same model is in 421.5-2005
Image Source: Fig 3 of IEEE Std 421.5-1992
VUEL is a
signal
from an
underexcitation
limiter,
which
we'll
cover
later
45
Initialization and Coding: Block
Diagram Basics
•
•
•
To simulate a model represented as a block diagram, the
equations need to be represented as a set of first order
differential equations
Also the initial state variable and reference values need
to be determined
Next several slides quickly cover the standard block
diagram elements
46
Integrator Block
u
•
KI
s
y
Equation for an integrator with u as an input and y as an
output is
dy
 KI u
dt
•
In steady-state with an initial output of y0, the initial
state is y0 and the initial input is zero
47
First Order Lag Block
u
•
K
1  Ts
Input
y
Output of Lag Block
Equation with u as an input and y as an output is
dy 1
  Ku  y 
dt T
•
•
In steady-state with an initial output of y0, the initial
state is y0 and the initial input is y0/K
Commonly used for measurement delay (e.g., TR block
with IEEE T1)
48
Derivative Block
u
•
•
•
KDs
1  sTD
y
Block takes the derivative of the input, with scaling KD
and a first order lag with TD
– Physically we can't take the derivative without some lag
In steady-state the output of the block is zero
State equations require a more general approach
49
State Equations for More
Complicated Functions
•
•
There is not a unique way of obtaining state equations
for more complicated functions with a general form
du
0 u  1 
dt
dy
0 y  1 
dt
d mu
 m m 
dt
d n 1 y d n y
  n 1 n 1  n
dt
dt
To be physically realizable we need n >= m
50
General Block Diagram Approach
•
One integration approach is illustrated in the below
block diagram
Image source: W.L. Brogan, Modern Control Theory,
Prentice Hall, 1991, Figure 3.7
51
Derivative Example
•
•
•
Write in form
KD
s
TD
1 TD  s
Hence 0=0, 1=KD/TD, 0=1/TD
Define single state variable x, then
dx
y
 0 u   0 y  
dt
TD
KD
y  x  1u  x 
u
TD
Initial value of
x is found by recognizing
y is zero so x = -1u
52
Lead-Lag Block
u
•
•
•
1  sTA
1  sTB
input
y
Output of Lead/Lag
In exciters such as the EXDC1 the lead-lag block is
used to model time constants inherent in the exciter; the
values are often zero (or equivalently equal)
In steady-state the input is equal to the output
To get equations write
TA
1
s
in form with 0=1/TB, 1=TA/TB, 1  sT
TB
TB
A

0=1/TB
1  sTB
1 TB  s
53
Lead-Lag Block
•
The equations are with
0=1/TB, 1=TA/TB,
0=1/TB
then
dx
1
 0 u   0 y   u  y 
dt
TB
TA
y  x  1u  x  u
TB
The steady-state
requirement
that u = y is
readily apparent
54