Systems
Each Multi-machine synchronous machine is represented by a constant
voltage source behind the direct axis transient reactance
The input powers are assumed to remain constant
Using the pre-fault voltage, all loads are converted to
equivalent admittances to ground and are assumed to
remain constant
Damping or asynchronous powers are ignored
The mechanical rotor angle of each machine coincides
with the electrical angle of the excitation voltage source
Power Systems I
Modeling Steps
Solve the initial load flow and obtain the initial bus
voltage magnitude and phase angle
Calculate the machine currents prior to the disturbance
I mach i
*
Smach i
=*
Vmach i
Obtain the voltages behind the transient reactances
+
+
Emach i = Vmach i + j X d I mach i
Convert all loads to equivalent admittances
yi 0 =
Power Systems I
Si*
Vi
2
=
Pi j Qi
Vi
2
Modeling Steps
Combine the generator models with the networks bus
admittance matrix with converted loads
0 Ynn
I = YT
mach n mach
Yn mach Vn
E+
Ymach mach mach
Use kron reduction (matrix form) to remove the network
buses from the matrix
1
reduced
Ybus
= Ymach mach YnTmach [Ynn ] Yn mach
reduced
I mach = Ybus E+
mach
Power Systems I
Modeling Steps
Express in terms of the machines excitation voltages, the
power output
*
+*
Smach i = Emach i I mach i
[
+*
Pmach i = Emach i I mach i
]
m
+
I mach i = Emach jYij
j =1
m
+
+
Pmach i = Emach i Emach j Yij cos( ij i + j )
j =1
Power Systems I
Multi-Machine Transient Stability
Set conditions
m
+
+
P initial mach i = Emach i Emach j Y1ij cos( ij i + j )
1
j =1
Pei (
) = P1machi
Pm i = Pei ( ) t =0
Set machine inertia constants on a common system
power base
SG i
Hi =
H G mach i
SB
Power Systems I
Multi-Machine Transient Stability
Solve the system of ODEs of the faulted network
m
H i d 2 i
= Pmi Ei+ E +j Y2 ij cos( 2ij i + j )
2
f 0 dt
j =1
Solve the system of ODEs of the post-fault network
m
H i d 2 i
= Pmi Ei+ E +j Y3ij cos( 3ij i + j )
2
f 0 dt
j =1
Power Systems I
Example
Consider the 3 machine system below
select generator #1 as the swing machine with a constant angle
of 0 degrees
determine the system stability when a fault on the line 5-6 near
bus 6 is cleared in 0.4 and 0.5 seconds
H = 20
Xd = j0.2 1
4
5
6
H=5
Xd = j0.25 3
Power Systems I
2 H=4
Xd = j0.15
Example
Load Data
1
2
3
4
5
6
P
0
0
0
1.0
0.9
1.6
Q
0
0
0
0.7
0.3
1.1
Generator Data
1
2
3
V
1.06
1.04
1.03
Power Systems I
P
--1.5
1.0
Line Data
1
1
1
2
3
4
5
4
5
6
4
5
6
6
R
0.035
0.025
0.040
0.000
0.000
0.028
0.026
X
0.225
0.105
0.215
0.035
0.042
0.125
0.175
B
0.0065
0.0045
0.0055
0.0000
0.0000
0.0035
0.0300
Example
Phase angle difference (fault cleared at 0.4s)
150
Delta, degree
Power angle / time
fault clearing in 0.4 s
100
50
0
-50
-100
Power Systems I
0
0.5
1
t, sec
1.5
Example
Phase angle difference (fault cleared at 0.5s)
1400
Power angle / time 1200
fault clearing in 0.5 s
Delta, degree
1000
800
600
400
200
0
-200
Power Systems I
0
0.5
1
t, sec
1.5