Before solving the differential equations to determine the transient stability or instability of the system, it is necessary to compute the initial conditions.
Let us consider an ‘ n ’ bus power system with ‘ m ’ generators ( m
< n ). Without loss of generality, it is assumed that the ‘ m ’ generators are located at first ‘ m ’ buses of the system. Towards computation of initial conditions, initially the load flow solution of the system is computed. From the load flow solution, following quantities are availbale: a.
V i
∠
θ i b.
P
Li
, Q
Li c.
P
Li
, Q
Li
, P
Gi
, Q
Gi for i
=
1 , 2 ,
⋯⋯ n for i
= m
+
1 , m
+
2 ,
⋯⋯ n for i
=
1 , 2 ,
⋯⋯ m
P
Gi
In the above, P
Li and Q
Li denote the real and reactive power load at bus ‘ i ’ respectively. Similarly, and Q
Gi denote the real and reactive power generation at bus ‘ i ’ respectively. Further, the quantities V i and θ i denote the voltage magnitude and angle of i t h bus respectively. With these information, following calculations are carried out:
(i) At any bus ‘ i ’, the loads are converted to equivalent admittance as;
¯
Li
=
P
Li
− jQ
Li
V i
2 for i
=
1 , 2 ,
⋯⋯ n (6.16)
(ii) Augment the Y
BUS matrix of the system as;
BUS
( i, i
) =
old
BUS
( i, i
) +
¯
Li for i
=
1 , 2 ,
⋯⋯ n (6.17) where,
¯ old
BUS is the original Y
BUS matrix of the system used in load flow calculation.
(iii) At the generator buses, the generators are represented as equivalent voltage sources behind the direct axis transient reactances as,
E i
= ¯ i
+ jx
∕ di
I
¯ i
= jx
∕ di
P
Gi
− jQ
Gi
V i
∗
=
E i
∠
δ i for i
=
1 , 2 ,
⋯⋯ m (6.18)
In equation (6.18), the quantity x
∕ di denotes the direct axis transient reactance of the
While performing the transient stability analysis, the magnitude i th machine.
∣
E i
∣ is held constant. The equivalent diagram of the ‘ n ’ bus power system is shown in Fig. 6.3
.
With the initial conditions computed as above, we are now ready to solve the transient stability problem. Basically, the transient stability problem is solved by two techniques: i) partition explicit
(PE) method and ii) simultaneous implicit (SI) method. In the PE method, the network algebraic equations and the generator differential equations are solved separately. In the SI method, these
256

Figure 6.3: Equivalent representation of generators for transient stability analysis algebraic and differential equations are solved together. In this course, however, we are going to discuss PE method only. Before, discussing the PE method, it is necessary to describe the network algebraic equations.
The network algebraic equations are represented as;
I
BUS
= ¯
BUS
V
BUS
In equation (6.19), Y
BUS is the bus impedance matrix of the system and
I
BUS
V
BUS
= [
I
¯
1
= [ ¯
1
I
¯
2
V
2
⋯⋯
I
¯ m
⋯⋯ ¯ m
I
¯ m
+
1
V m
+
1
⋯⋯
I
¯ n
] T
⋯⋯ ¯ n
] T
257
(6.19)
(6.20)

The voltage current relationship of the generator reactance is given by,
¯ k
= jx
∕ di i k
⇒ i k
=
¯ k v k
(6.21) respectively and y k
(
In equation (6.21),
¯ k
− ¯ k
) as ( ¯
¯ k k
= i k
= ¯ k and v k are the current through and voltage across the generator reactance
1 jx
−
∕ di
V
= − j
/ k
). Or, x
∕ di
. Now, at the generator terminals,
I
¯ k
+
¯ k
V k
=
¯ k
¯ k
I
¯ k
= ¯ k
=
¯ k v k
; or, I
¯ k
=
(6.22)
From equations (6.22) and (6.23) we get,
( ¯
11
Y
21
+
¯
1
V
1
Y m 1
⋮
¯
( m
+
1
)
, 1
Y n 1
⋮
V
1
¯
1
) ¯
1
V
1
+
+ (
Y
22
Y
12
+
¯
Y m 2
⋮
¯
V
2
+
+
+
+
+ ¯
( m
+
1
)
, 2
⋮
Y n 2
V
2
2
2
) ¯
2
V
2
+ ⋯ +
+ ⋯ +
+ ⋯ +
+ ⋯ + (
Y
Y
Y mm
1 m
2 m
+ ⋯ +
+ ⋯ +
+ ⋯ +
V
V m m
+
⋮ y m
⋯⋯⋯
⋮
⋯⋯⋯
) ¯ m
+ ⋯ +
Y
1 n
+ ⋯ +
Y
2 n
+ ⋯ +
+ ⋯ +
Y mn
⋮
V n
V n
V n
+ ⋯ + ¯
( m
+
1
)
,n
+ ⋯ +
+ ⋯ + ¯ nn
⋮
V n
¯ n
=
¯
1
=
¯
2
= ⋮
= y m
E
E
=
0
= ⋮
=
0
1
2
E m
E
2
(6.23)
From equation (6.23), the voltages V
1
, V
2
⋯⋯ ¯ n can be solved, for known values of E
1
⋯⋯ ¯ n
. With these known terminal voltages, the electrical power output of each generator
, can be calculated as;
P ei
=
Re
( ¯
I
I
¯ i
∗ ) for i
=
1 , 2 ,
⋯ m (6.24)
⎪⎪⎪⎪⎪
⎫⎪⎪⎪⎪
⎪⎪⎪⎪⎪
⎪⎪⎪⎪⎪
Where,
I
¯ i
=
¯ i
( ¯ i
− ¯ i
) for i
=
1 , 2 ,
⋯ m
With the above equations, we are now in a position to discuss the PE method.
(6.25)
In the PE method, the numerical integration of the generator differential equations are carried out separately from the solution of the network algebraic equations. For numerical integration of differential equations, the total simulation time ( t
T of duration ∆ t seconds. Thus, ∆ t
= t
T
N problem, with PE method, are as follows.
) is divided into N intervals, each interval being
. Now, the major steps for solving the transient stability
1.
Obtain the load flow solution of the given system. Thereafter, compute the internal voltages of all the generators (
¯ i
, for i
=
1 , 2 ,
⋯⋯ m ) using equations (6.16) - (6.18). Please note that the magnitudes E i would be kept constant at these calculated values throughout the simulation.
2.
With the values of E i obtained above, solve the equation set (6.23) to obtain the terminal voltages at all the buses ( the generators (
¯ ei
, for
¯ i
, for i
=
1 , 2 ,
⋯⋯ n ). Subsequently, the electrical power output of all i
=
1 , 2 ,
⋯⋯ m ) are computed from equations (6.24) - (6.25).
258

3.
Under steady state condition, mechanical power input to each generator is equal to its generator electrical output power (neglecting losses). Therefore, P mi
=
P ei
, for i
=
1 , 2 ,
⋯⋯ m .
Also, under steady state, all the generators are assumed to operate at synchronous speed ( ω s
).
ω i
, P mi
4.
Thus, after the above three steps, at for i
=
1 , 2 ,
⋯⋯ m t
=
0 , the variables pertaining to generators ( E i
) and the network bus voltages (
¯ i
, for
, δ i
=
1 , 2 ,
⋯⋯ n ) are all known.
i
5.
The simulation process advances to t
=
∆ t . At this instant, first the network equations
, given in the equation set (6.23) are solved to compute the bus voltages and subsequently, the output electrical power of each generator is calculated by using equations (6.24) - (6.25). Now, if the steady state condition is still maintained, i.e. if there is no change in the network (as compared to the network condition at t
=
0 ), then the calculated values of P ei would be again equal to the corresponding value of P mi
.
6.
With the solution of the network equations at hand, we should now solve the solve the generator differential equations for calculating the values of δ i and ω i at t
=
∆ t . Towards that end, let us first re-write the swing equations of i th machine for convenience in equations (6.26) - (6.27) below.
2 H i dω i
ω s dt
=
P mi
−
P ei
(6.26) dδ i dt
=
ω i
−
ω s
(6.27)
Now, from equation (6.26), as P mi
=
P ei
, dω i =
0 . In other words, there is no change in the
δ i dt speed of the generator and hence angle
ω i
=
ω s
. As a result, from equation (6.27), would also be maintained at the value calculated at dδ i =
0 and hence, the t
=
0 dt
. Therefore, under steady state condition, at t
=
∆ t , both δ i and ω i would be maintained at the values calculated at
7.
Assume that the steady state condition continues from t
=
0 .
t
=
0 to t
= ( k
−
1
)
∆ t , where, k is a positive integer. Following the arguments described at steps 5 and 6, it can be easily seen that at the end of t
=
0 .
t
= ( k
−
1
)
∆ t sec., both δ i and ω i would still be maintained at the values calculated at
8.
Now, let us assume that a three phase to ground short circuit fault occurs in the system at t
= t o
= k ∆ t at the ` th bus. To accommodate this fault condition in the network equations, the element Y
`` is increased manyfold to reflect very high admittance from bus ‘ ` ’ to ground. With this imposed condition, the network bus voltages are calculated from equation set (6.23). Subsequently, the output electrical power of each generator is calculated by using equations (6.24) - (6.25) corresponding to time t
= t o
.
9.
With the values of to obtain the values of δ i
P and ei
ω i calculated in step 8, the swing equations (6.26) - (6.27) are integrated at t
= t o
. Now, for integrating the swing equations, initial values of
δ i and ω i are required. These initial values are taken to be equal to the values of δ i and ω i obtained at the end of denoted as as
¯ i
=
E i
δ i
∠
δ i t
= ( k
−
1
)
∆ t . For brevity, the value of δ
( t o
( t o
( t o
)
. With this value of δ i
( i
=
1 , 2 ,
⋯⋯ m ) calculated at t
= t o is i
)
, the voltage behind the transient reactance is updated
)
, for i
=
1 , 2 ,
⋯⋯ m , where the magnitude E i has been taken to be equal to the constant value calculated at step 1.
259

10.
The simulation advances to t
= t o
+
∆ t . If the fault still persists, the values of P ei calculated as described in step 8. However, for this purpose, the values of E i are in equation set (6.23) are taken to be equal to the latest values calculated at the end of
ω i
, δ i
11.
With the values of and E i at the end of equations, the latest values of
P t
= t o
ω ei i calculated at and δ i t
= t o
+
∆ t , step 9 is repeated to update the variables
+
∆ t . Again, please note that for integrating the differential
(calculated at t
= t o t
= t o sec.
) have been taken as the initial values.
12.
Steps 10 and 11 are repeated to update the variables at
( t cl t
= t o
+
2∆ t, t o
+
3∆ t, t o
+
4∆ t,
⋯⋯ till the fault clears.
13.
At t
= t cl
= p ∆ t , p being a positive integer and p
> k ), the fault clears. This condition is imposed in the network equations by restoring Y
`` to its original pre-fault value and subsequently, steps 8 - 12 are repeated to obtain the variations of δ i and ω i
. By observing the variation of δ i
, the stability of the system is assessed.
In step 9, the generator differential equations are numerically integrated. In the next lecture, we will look into two such numerical integration techniques, namely, i) modified Euler’s method and ii)
4 th order Runga-Kutta technique .
260