Download 6.3 Transient stability analysis

6.3
Transient stability analysis
Before solving the differential equations to determine the transient stability or instability of the
system, it is necessary to compute the initial conditions.
6.3.1
Computation of 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. Vi ∠θi
for i = 1, 2, ⋯⋯ n
b. PLi , QLi
for i = m + 1, m + 2, ⋯⋯ n
c. PLi , QLi , PGi , QGi
for i = 1, 2, ⋯⋯ m
In the above, PLi and QLi denote the real and reactive power load at bus ‘i’ respectively. Similarly,
PGi and QGi denote the real and reactive power generation at bus ‘i’ respectively. Further, the
quantities Vi and θi denote the voltage magnitude and angle of it h bus respectively. With these
information, following calculations are carried out:
(i) At any bus ‘i’, the loads are converted to equivalent admittance as;
ȳLi =
PLi − jQLi
Vi 2
for i = 1, 2, ⋯⋯ n
(6.16)
(ii) Augment the YBUS matrix of the system as;
old
ȲBUS (i, i) = ȲBUS
(i, i) + ȳLi
for i = 1, 2, ⋯⋯ n
(6.17)
old
where, ȲBUS
is the original ȲBUS 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,
Ēi = V̄i + jx∕di I¯i = jx∕di
PGi − jQGi
= Ei ∠δi
V̄i∗
for i = 1, 2, ⋯⋯ m
(6.18)
In equation (6.18), the quantity x∕di denotes the direct axis transient reactance of the ith machine.
While performing the transient stability analysis, the magnitude ∣Ei ∣ 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
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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.
6.3.2
Network algebraic equations
The network algebraic equations are represented as;
ĪBUS = ȲBUS V̄BUS
(6.19)
In equation (6.19), ȲBUS is the bus impedance matrix of the system and
T
ĪBUS = [I¯1 I¯2 ⋯⋯ I¯m I¯m+1 ⋯⋯ I¯n ]
V̄BUS = [V̄1 V̄2 ⋯⋯ V̄m V̄m+1 ⋯⋯ V̄n ]
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T
(6.20)
The voltage current relationship of the generator reactance is given by,
v̄k = jx∕di īk
⇒
īk = ȳk v̄k
(6.21)
In equation (6.21), īk and v̄k are the current through and voltage across the generator reactance
1
= −j/x∕di . Now, at the generator terminals, I¯k = īk = ȳk v̄k ; or, I¯k =
jx∕di
ȳk (Ēk − V̄k ) as (v̄k = Ēk − V̄k ). Or,
I¯k + ȳk V̄k = ȳk Ēk
(6.22)
respectively and ȳk =
From equations (6.22) and (6.23) we get,
(Ȳ11 + ȳ1 ) V̄1
Ȳ21 V̄1
⋮
Ȳm1 V̄1
Ȳ(m+1),1 V̄1
⋮
Ȳn1 V̄1
+
Ȳ12 V̄2
+ (Ȳ22 + ȳ2 ) V̄2
+
⋮
+
Ȳm2 V̄2
+ Ȳ(m+1),2 V̄2
+
⋮
+
Ȳn2 V̄2
+
+
+
+
+
+
+
⋯
⋯
⋯
⋯
⋯
⋯
⋯
+
Ȳ1m V̄m
+
Ȳ2m V̄m
+
⋮
+ (Ȳmm + ȳm ) V̄m
+
⋯⋯⋯
+
⋮
+
⋯⋯⋯
+
+
+
+
+
+
+
⋯
⋯
⋯
⋯
⋯
⋯
⋯
+
Y1n V̄n
+
Y2n V̄n
+
⋮
+
Ymn V̄n
+ Ȳ(m+1),n V̄n
+
⋮
+
Ȳnn V̄n
= ȳ1 Ē1
= ȳ2 Ē2
=
⋮
= ȳm Ēm
=
0
=
⋮
=
0
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
(6.23)
From equation (6.23), the voltages V̄1 , V̄2 ⋯⋯ V̄n can be solved, for known values of Ē1 ,
Ē2 ⋯⋯ Ēn . With these known terminal voltages, the electrical power output of each generator
can be calculated as;
for i = 1, 2, ⋯ m
Pei = Re (ĒI I¯i∗ )
(6.24)
Where,
I¯i = ȳi (Ēi − V̄i )
for i = 1, 2, ⋯ m
(6.25)
With the above equations, we are now in a position to discuss the PE method.
6.3.3
Partition explicit solution scheme
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 (tT ) is divided into N intervals, each interval being
tT
of duration ∆t seconds. Thus, ∆t =
. Now, the major steps for solving the transient stability
N
problem, with PE method, are as follows.
1. Obtain the load flow solution of the given system. Thereafter, compute the internal
voltages of all the generators (Ēi , for i = 1, 2, ⋯⋯ m) using equations (6.16) - (6.18). Please note
that the magnitudes Ei would be kept constant at these calculated values throughout the simulation.
2. With the values of Ēi obtained above, solve the equation set (6.23) to obtain the terminal
voltages at all the buses (V̄i , for i = 1, 2, ⋯⋯ n). Subsequently, the electrical power output of all
the generators (P̄ei , for i = 1, 2, ⋯⋯ m) are computed from equations (6.24) - (6.25).
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3. Under steady state condition, mechanical power input to each generator is equal to its
generator electrical output power (neglecting losses). Therefore, Pmi = Pei , for i = 1, 2, ⋯⋯ m.
Also, under steady state, all the generators are assumed to operate at synchronous speed (ωs ).
4. Thus, after the above three steps, at t = 0, the variables pertaining to generators (Ei , δi ,
ωi , Pmi for i = 1, 2, ⋯⋯ m) and the network bus voltages (V̄i , for i = 1, 2, ⋯⋯ n) are all known.
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 Pei would be again equal to the
corresponding value of Pmi .
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 δi and ωi at t = ∆t. Towards that end,
let us first re-write the swing equations of ith machine for convenience in equations (6.26) - (6.27)
below.
2Hi dωi
= Pmi − Pei
(6.26)
ωs dt
dδi
= ωi − ωs
(6.27)
dt
dωi
Now, from equation (6.26), as Pmi = Pei ,
= 0. In other words, there is no change in the
dt
dδi
speed of the generator and hence ωi = ωs . As a result, from equation (6.27),
= 0 and hence, the
dt
angle δi would also be maintained at the value calculated at t = 0. Therefore, under steady state
condition, at t = ∆t, both δi and ωi would be maintained at the values calculated at t = 0.
7. Assume that the steady state condition continues from 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 = (k − 1)∆t sec., both δi and ωi would still be maintained at the values calculated at
t = 0.
8. Now, let us assume that a three phase to ground short circuit fault occurs in the system
at t = to = k∆t at the `th bus. To accommodate this fault condition in the network equations,
the element Ȳ`` 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 = to .
9. With the values of Pei calculated in step 8, the swing equations (6.26) - (6.27) are integrated
to obtain the values of δi and ωi at t = to . Now, for integrating the swing equations, initial values of
δi and ωi are required. These initial values are taken to be equal to the values of δi and ωi obtained
at the end of t = (k − 1)∆t. For brevity, the value of δi (i = 1, 2, ⋯⋯ m) calculated at t = to is
denoted as δi (to ). With this value of δi (to ), the voltage behind the transient reactance is updated
as Ēi = Ei ∠δi (to ), for i = 1, 2, ⋯⋯ m, where the magnitude Ei has been taken to be equal to the
constant value calculated at step 1.
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10. The simulation advances to t = to + ∆t. If the fault still persists, the values of Pei are
calculated as described in step 8. However, for this purpose, the values of Ēi in equation set (6.23)
are taken to be equal to the latest values calculated at the end of t = to sec.
11. With the values of Pei calculated at t = to + ∆t, step 9 is repeated to update the variables
ωi , δi and Ēi at the end of t = to + ∆t. Again, please note that for integrating the differential
equations, the latest values of ωi and δi (calculated at t = to ) have been taken as the initial values.
12. Steps 10 and 11 are repeated to update the variables at t = to + 2∆t, to + 3∆t, to +
4∆t, ⋯⋯ till the fault clears.
13. At t = tcl (tcl = p∆t, p being a positive integer and p > k ), the fault clears. This condition
is imposed in the network equations by restoring Ȳ`` to its original pre-fault value and subsequently,
steps 8 - 12 are repeated to obtain the variations of δi and ωi . By observing the variation of δi , 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)
4th order Runga-Kutta technique.
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