power system stability

1. Power system stability

2. What is Stability?

3. Power system stability
This is the ability of a power system (having multiple synchronous
machines i.e. generators and motors interconnected through a
transmission network) to attain the original or a new steady state
after it experiences respectively a small disturbance or a large
disturbance.
This ability mainly stems from the adjustment capability of
synchronous machines by adjusting their rotor angles in the event
of any disturbance

4. What are disturbances?
Small Disturbances
 Incremental changes in load
 Incremental changes in generation
2 Large Disturbances
 Loss of a large generator or load
 Faults on transmission lines

5. Types of stability
 The stability arising from small disturbance (e.g.
change in AVR gain parameter, gradual or slow
variation in load or generation etc.) is termed
steady state stability
 The stability arising from a large disturbance (e.g. fault, line
switching, sudden large load or generation change) is termed
transient stability

6. Classification of power system stability
Power system
stability
Rotor angle
stability
Small
disturbance
angle stability
Short term
Transient
stability
Short term
Frequency
stability
Short term Long term
Voltage stability
Large
disturbance
voltage stability
Short term Long term
Small
disturbance
voltage stability
Short term Long term

7. New categories of stability
Ref :
Hatziargyriou, Nikos, et al. "Definition and classification of power system stability–revisited &
extended." IEEE Transactions on Power Systems 36.4 (2020): 3271-3281.

8. Rotor angle stability
 Rotor angle stability is the ability of the interconnected synchronous machines
running in the power system to remain in the state of synchronism.
Small disturbance (small signal) stability
 Ability to maintain synchronism under small disturbances.
 Since disturbances are small, nonlinear differential equations can be linearized.
 It is easy to solve.
2 Large disturbance (Transient) stability
 Ability to maintain synchronism under large disturbances.
 Since disturbances are large, nonlinear differential equations can not be linearized.
 It has to be solved numerically. It is difficult.However, we can use a graphical approach called Equal
Area Criterion
for analyzing the stability of a single machine connected to an infinite bus using the classical model.

9. Rotor angle and synchronous machine

10. Swing equation
 Swing equation describes the rotor dynamics for a synchronous machine
Consider a synchronous generator developing an electromagnetic torque Te(and
a corresponding electromagnetic power Pe) while operating at the synchronous
speed ws. If the input torque provided by the prime mover.

11. Swing equation
𝜏 = 𝐼𝛼
𝐽
𝑑2
𝜃𝑚
𝑑𝑡2 = 𝑇𝑎 = 𝑇𝑚 − 𝑇𝑒
𝜃𝑚 = 𝜔𝑠𝑚𝑡 + 𝛿𝑚
𝑑𝜃𝑚
𝑑𝑡
= 𝜔𝑠𝑚 +
𝑑𝛿𝑚
𝑑𝑡
𝑑2
𝜃𝑚
𝑑𝑡2
=
𝑑2
𝛿𝑚
𝑑𝑡2
𝐽
𝑑2𝛿𝑚
𝑑𝑡2 = 𝑇𝑎 = 𝑇𝑚 − 𝑇𝑒

12. Swing equation (cont.)
 𝐽𝜔𝑚
𝑑2𝛿𝑚
𝑑𝑡2 = 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒 Multiplying both sides by ωm and substituting P =
T ωm )
 M
𝑑2𝛿𝑚
𝑑𝑡2 = 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒
M= Jωm is inertia constant or angular
momentum of rotor at sync. speed
M
𝑑2𝛿𝑚
𝑑𝑡2 = 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒
2𝐻𝑆𝑚𝑎𝑐ℎ
𝜔𝑠𝑚
𝑑2
𝛿𝑚
𝑑𝑡2
= 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒
M=
2𝐻𝑆𝑚𝑎𝑐ℎ
𝜔𝑠𝑚
MJ/mech rad
H is another inertia constant

13. Swing equation (cont.)
2𝐻𝑆𝑚𝑎𝑐ℎ
𝜔𝑠𝑚
𝑑2𝛿𝑚
𝑑𝑡2
= 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒
2𝐻
𝜔𝑠𝑚
𝑑2
𝛿𝑚
𝑑𝑡2
=
𝑃𝑎
𝑆𝑚𝑎𝑐ℎ
=
𝑃𝑚 − 𝑃𝑒
𝑆𝑚𝑎𝑐ℎ
2𝐻
𝜔𝑠
𝑑2𝛿
𝑑𝑡2 = 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒 per unit

14. Swing equation (cont.)
𝐻
𝜋𝑓
𝑑2𝛿
𝑑𝑡2= 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒 per unit

𝐻
180𝑓
𝑑2𝛿
𝑑𝑡2= 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒 per unit
𝛿 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑎 𝑠𝑦𝑛𝑐ℎ𝑟𝑜𝑛𝑜𝑢𝑠𝑙𝑦 𝑟𝑜𝑡𝑎𝑡𝑖𝑛𝑔 𝑟𝑒𝑓𝑒𝑟𝑛𝑐𝑒 𝑎𝑥𝑖𝑠,
𝑑𝛿
𝑑𝑡
𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑦𝑛𝑐ℎ𝑟𝑜𝑛𝑜𝑢𝑠 𝑠𝑝𝑒𝑒𝑑
If swing equation is referred to system base in multi machine system, then all powers in pu and H
needs to be converted as follows
𝐻𝑠𝑦𝑠𝑡𝑒𝑚 = 𝐻𝑚𝑎𝑐ℎ
𝑆𝑚𝑎𝑐 ℎ
𝑆𝑠𝑦𝑠𝑡𝑒𝑚

15. Coherently swinging machine
 Machine 1
2𝐻1
𝜔𝑠
𝑑2𝛿1
𝑑𝑡2 = 𝑃𝑎1 = 𝑃𝑚1 − 𝑃𝑒1 per unit---------------------(1)
2𝐻2
𝜔𝑠
𝑑2𝛿2
𝑑𝑡2 = 𝑃𝑎2 = 𝑃𝑚2 − 𝑃𝑒2 per unit-----------------(2)
Machine 2
Since δ1 = δ2= δ, Adding (1) and (2)
2𝐻
𝜔𝑠
𝑑2𝛿
𝑑𝑡2 = 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒 per unit
Where H= H1+H2
(
2𝐻1
𝜔𝑠
+
2𝐻2
𝜔𝑠
)
𝑑2𝛿
𝑑𝑡2= (𝑃𝑚1+ 𝑃𝑚2 )- (𝑃𝑒1+ 𝑃𝑒2 )

16. example
Calculate equivalent H constant.
Machines which swing
together are called
coherent machines

17. Non coherent machine
Since δ1 ≠ δ2. And Power flow from the machine with
greater δ to lower δ. Lets assume δ1 > δ2

18. Non coherent machine (cont)
 A noteworthy application of these equations concerns a two- machine system having only one
generator (machine one) and a synchronous motor (machine two) connected by a network of pu
reactances. Whatever change occurs in the generator output is thus absorbed by the motor, and
we can write
In order to be stable the angular differences betweenmachines must
decrease after the final switching operation-such as the opening of circuit
breaker to clear a fault .

19. Power angle equation
𝑃𝑘 + 𝑗𝑄𝑘 = 𝑉𝑘𝐼𝑘 ∗
𝐼𝑏𝑢𝑠 = 𝑌𝑏𝑢𝑠𝑉𝑏𝑢𝑠
𝑃1 + 𝑗 𝑄1 = 𝐸1
′
𝑌11 𝐸1
′
< (𝛿1−𝛿1 − 𝜃11)+ 𝐸1
′
𝑌12 𝐸2
′
< (𝛿1−𝛿2 − 𝜃12)
𝑃1 = | 𝐸1
′
|
2
𝑌11 cos(𝜃11)+ 𝐸1
′
𝑌12 𝐸2
′
𝑐𝑜𝑠(𝛿1−𝛿2 − 𝜃12)

20. Power angle equation (cont…)

25. Kron Reduction
When a bus k between bus I and j is
equivalenced
𝑌′𝑖𝑗 = 𝑌𝑖𝑗 −
𝑌𝑖𝑘𝑌𝑘𝑗
𝑌𝑘𝑘

26. Kron reduction

 𝑌11
′
= 𝑌11 −
𝑌13∗𝑌31
𝑌33
= −𝑗3.333 −
𝑗3.333∗𝑗3.333
−𝑗10.8333
= −𝑗2.308
 𝑌12
′
= 𝑌12 −
𝑌13∗𝑌32
𝑌33
= 0 −
𝑗3.333∗𝑗2.50
−𝑗10.833
= 𝑗0.769

29. Synchronising power coefficient
 A commonsense requirement for an acceptable operating point is that the generator should not lose synchronism when
small temporary changes occur in the electrical power output from the machine. To examine this requirement for fixed
mechanical input power Pm consider small incremental changes in the operating point parameters; that is consider
𝛿 = 𝛿0 + 𝛿∆
𝑃𝑒 = 𝑃𝑒0 + 𝑃𝑒∆
 we obtain the power-angle equation for the general two-machine system in the form
𝑃𝑒 = 𝑃𝑒0 + 𝑃𝑒∆=𝑃𝑚𝑎𝑥sin(𝛿0 + 𝛿∆)
= 𝑃𝑚𝑎𝑥(sin 𝛿0cos 𝛿∆+cos 𝛿0sin 𝛿∆)----------(1)
𝛿∆ is a small incremental displacement from 𝛿0
𝑠𝑖𝑛𝛿∆ ≅ 𝛿∆ and 𝑐𝑜𝑠𝛿∆ ≅1---------(2)
From (1) and (2)
𝑃𝑒0 + 𝑃𝑒∆ = 𝑃𝑚𝑎𝑥(sin 𝛿0+ 𝛿∆ cos 𝛿0)----------(3)

30. Synchronising power coefficient
𝑁𝑜𝑤, 𝑤𝑟𝑖𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑤𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (
2𝐻
𝜔𝑠
𝑑2𝛿
𝑑𝑡2 = 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒)

31. Synchronising power coefficient
When Sp is positive, the solution of 𝛿∆(t)
corresponds to that of simple harmonic motion. ; such
motion is represented by the oscillations of an undamped
swinging pendulum. When Sp is negative, the solution
𝛿∆(t) increases
exponentially without lim it

32.  Swing equation of a machine connected to infinite bus
Equal Area Criterion

33. The equal area criterion states
that the kinetic energy added to
the rotor following a fault must
be removed
after the fault in order to restore
the rotor to synchronous speed .

34. Equal Area Criterion
• The generator is operating initially at synchronous speed with a rotor angle of 𝛿0 and the input
mechanical power Pm equals the output electrical power Pe , as shown .
• When the fault occurs at t=0 the electrical power output is suddenly zero while the input mechanical
power is
unaltered

35. Equal Area Criterion prove

36. Equal area criterion (cont---)
 If we denote the time t o clear the fault by 𝑡𝑐 then the acceleration is constant for
time t less than 𝑡𝑐 and is given by

37. Equal area criterion (cont---)
 At the instant of fault clearing the increase in rotor speed and the angle
separation between the generator and the infinite bus are given by
In a system where one machine is swinging with respect to
an infinite bus we may use this principle of equality of
areas, called the equal-area criterion , to
determine the stability of the system under transient
conditions without solving the swing equation.

38. Critical clearing angle & Critical clearing time
 there is a critical angle for clearing the fault in order to satisfy the requirements
of the equal-area criterion for stability. This angle, called the critical clearing
angle 𝛿𝑐𝑟
 The corresponding critical time for removing the fault is called the critical
clearing time tcr . Thus, the critical clearing time is the maximum elapsed time
from the initiation of the fault until its isolation such that the power system is
transiently stable.

40. Critical clearing time

42. Application of EAC to a more generalized case and finding CCA
when output or transferred electrical power during a fault is not
zero
.
P

44. STEP-BY-STEP SOLUTION OF THE SWING
CURVE :
M
𝑑2𝛿𝑚
𝑑𝑡2 = 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒
 When fault occurs 𝑃𝑎 ≠ 0 ,
Rotor accelerates , since
𝑑2𝛿𝑚
𝑑𝑡2 ≠ 0 ,
 Speed ω increases, δ
increases too
 Since 𝑃𝑒=𝑃𝑚𝑎𝑥𝑠𝑖𝑛𝛿 , 𝑃𝑎
decreases with increasing δ
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝛼 =
𝑑2𝛿𝑚
𝑑𝑡2 =
𝑑𝜔
𝑑𝑡
= 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝜔
 Since 𝑃𝑎 decreases with increasing δ
𝑑2𝛿𝑚
𝑑𝑡2 =
𝑑𝜔
𝑑𝑡
= 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝜔 decreases.
So the curve has a decreasing slope

45. STEP-BY-STEP SOLUTION OF THE
SWING CURVE : assumptions
 The accelerating power Pa computed at the beginning of an interval is constant
from the middle of the preceding interval to the middle of the interval
considered.
 Throughout any interval the angular velocity is constant at the value computed
for the middle of the interval.

46. STEP-BY-STEP SOLUTION OF THE
SWING CURVE cont
The accelerating power Pa computed at the beginning of
an interval is constant from the middle of the preceding
interval to the middle of the interval considered.
Throughout any interval the angular velocity
is constant at the value computed for the
middle of the interval.

47. STEP-BY-STEP SOLUTION OF THE
SWING CURVE

𝐻
180𝑓
𝑑2𝛿
𝑑𝑡2= 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒----------(1)
 𝛼 =
𝑑2𝛿𝑚
𝑑𝑡2 =
𝑑𝜔
𝑑𝑡

𝑑𝜔
𝑑𝑡
can be written as lim
∆𝑡→0
∆𝜔
∆𝑡
 Now
∆𝜔
∆𝑡
=
𝜔𝑟,𝑛−1/2−𝜔𝑟,𝑛−3/2
∆𝑡
---------(2)
 From (1)
∆𝜔
∆𝑡
= 𝑃𝑎,𝑛−1 ×
180𝑓
𝐻
---------(3)
 Fromm (2) and (3) we get

48. Step by step solution cont..
∆𝛿𝑛
∆𝑡
= 𝜔𝑟,𝑛−1/2
∆𝛿𝑛−1
∆𝑡
= 𝜔𝑟,𝑛−3/2
-----------(6)
-----------(5)
-----------(4)
Substituting from equation 4 and 5 to eqn 6
𝜔𝑟,𝑛−1/2 − 𝜔𝑟,𝑛−3/2=
∆𝛿𝑛
∆𝑡
−
∆𝛿𝑛−1
∆𝑡
= 𝑃𝑎,𝑛−1 ×
180𝑓
𝐻
∆𝑡
∆𝛿𝑛 − ∆𝛿𝑛−1= 𝑃𝑎,𝑛−1 ×
180𝑓
𝐻
∆𝑡2

49.  ∆𝛿𝑛 − ∆𝛿𝑛−1= 𝑃𝑎,𝑛−1 ×
180𝑓
𝐻
∆𝑡2
 ∆𝛿𝑛= ∆𝛿𝑛−1 + 𝑃𝑎,𝑛−1 ×
180𝑓
𝐻
∆𝑡2
 Where ∆𝛿𝑛=𝛿𝑛 − 𝛿𝑛−1 𝑎𝑛𝑑 ∆𝛿𝑛−1=𝛿𝑛−1 − 𝛿𝑛−2
 therefore 𝛿𝑛= ∆𝛿𝑛+ 𝛿𝑛−1
 Where ∆𝛿𝑛= ∆𝛿𝑛−1 + 𝑘𝑃𝑎,𝑛−1
Type equation here.

50. STEP-BY-STEP SOLUTION OF THE SWING
CURVE

51. Multimachine stability studies
 The mechanical power input to each machine remains constant during the
entire period of the swing curve computation.
 Damping power is negligible.
 Each machine may be represented by a constant transient reactance in series
with a constant transient internal voltage.
 The mechanical rotor angle of each machine coincides with δ , the electrical
phase angle of the transient internal voltage.
 All loads may be considered as shunt impedances to ground with values
determined by conditions prevailing immediately prior to the transient
conditions.
The system stability model based on these assumptions is
called the classical stability model, and studies which use
this model are called classical stability studies

52. Steps to solve multimachine stability
 By forming the pre fault bus admittance matrix and adding at the diagonal
positions the admittances obtained by converting loads (as follows) , the
augmented admittance matrix is formed
 Then the during fault condition is reflected appropriately in this augmented Ybus and
it is reduced to nxn matrix eliminating non generator buses when n is the number of
generators
 Similarly the after fault condition is reflected appropriately in this augmented Ybus and
it is reduced to nxn matrix eliminating non generator buses when n is the number of
generators
.

53. Steps to solve multimachine stability
 The Pei in the swing equation of a generator ‘i’ is evaluated using an equation where
the admittances between generator buses i and j are taken from the reduced matrix
corresponding to during or after fault condition whichever is needed. e.g., in a 3
generator system this eqn.

56. In load flow analysis the Ybus is formed without considering
machine nodes. But for stability analysis since the Ybus matrix
needs to be reduced among machines’ nodes, the machine
terminal bus is ‘moved’ to internal node so that machine
reactance is also added with transformer reactance in Ybus.
e.g. Bus 1 is now ‘behind’ Xd1’ instead of XT1

58.  A three-phase fault at bus 4 so it is
same as reference bus and hence
reflected by simply deleting 4th row
and 4th column
 Then reducing 4x4 matrix by using
kron’s formula

60. Post fault swing equation
 Sincefault is cleared by removing line 4-5
so that in prefault Ybus the elements Y44, Y55 need to
be recomputed by subtracting from them y45
(=1/(0.018+j0.11)
and half of corresponding line charging susceptance
(j0.226/2).
 Also Y45= Y54 =0.0+j0.0
After Kron reduction

64.  Since the method assumes discontinuity in Pa at the middle of an interval so that if an
actual discontinuity occurs at the beginning of an interval it must be considered by
taking average of the two values.
𝛿𝑛= ∆𝛿𝑛+ 𝛿𝑛−1
Where ∆𝛿𝑛= ∆𝛿𝑛−1 + 𝑘𝑃𝑎,𝑛−1

67.  Fault cleared at 0.225sec

68. Swing curves

70. Factors effecting transient stability
 An analysis of the set of equations shows that
• A) acceleration (due to a fault) or deceleration (due to increase in electrical load) of a synchronous
machine is less and hence stability is more if
1. H is higher i.e. machine rotor size is higher.
2. The difference between mechanical input power (Pm) and electrical power transfer (Pe) is less. Pe can be
increased by increasing E i.e. excitation and decreasing reactance X.
• B) The critical clearing time increases i.e. a machine can retain stability for a longer time if
3. H is higher
4. Pre-fault power (equal to Pm neglecting loss) is lower.

71. Means to improve stability
Operational strategies:
i) Increase machine excitation so that E and hence Pe increases during a fault; this is performed by what is
known as PSS (Power System Stabilizer)
ii) Fast valve control of turbines i.e. during a fault the turbine valve will close faster while during a sudden
increase in electrical load demand the valve will open faster so that the difference between Pm and Pe
and hence acceleration or deceleration reduces.
iii) Faster fault clearing (much less than critical clearing time) using high speed relays and CBs.
iv) Selective pole operation of CBs to clear a fault i.e. healthy phases remain intact so that Pe after fault
increases and hence rotor acceleration becomes less.
v) In case generation i.e. mechanical power output from turbine cannot be increased when needed, the
last resort is to shed electrical demand (load) using under frequency auto load shed relays installed at
substations, to reduce the difference between Pm and Pe .
Design strategies:
i) Choice of machines with high H and low Xd’ where possible.
ii) Reduction of transfer reactance X by parallel lines, series compensation, decreasing transformer
reactance etc. so that Pe increases during a fault.
Use of HVDC line to interconnect the areas within a system or to interconnect two systems so that
oscillation of machines in one area /system will not spread to other area/system