Unit V -Stability Analysis

1. EE8501 – POWER SYSTEM ANALYSIS
Unit – V – STABILITY ANALYSIS
By
Mr. A. Arun Kumar,
Assistant Professor,
Department of Electrical and Electronics Engineering.
Email: arunkumar@ritrjpm.ac.in
Cell: 98430 80689Power System Analy sis / Unit - I 1

2. POWER SYSTEM STABILITY
Power system stability may be broadly defined as that property that
enables it to remain in state of operating equilibrium (a state in
which opposing forces or influences are balanced) under normal
operating conditions and to regain an acceptable state of
equilibrium after being subjected to a disturbance.
Power system Stability studies which evaluate the impact of
disturbances on the electromechanical dynamic behavior of the
power system.
Traditionally, Stability problem has been one of maintaining the
synchronous operation

3. FACTORS DECIDING POWER
SYSTEM STABILITY PHENOMENON
Power system stability recognized as an
important study to maintain system security
for secured system operation. The factors are
• Initial operating conditions
• Network topology
• Nature of disturbances such as type of fault,
location of the fault, severity of the
disturbance.

4. FUNDAMENTAL ASSUMPTION IN
STABILITY STUDIES
• Only synchronous frequency currents and
voltages are considered in the stator windings
and the power system. Consequently, dc offset
current and harmonic components are
neglected.
• Symmetrical components are used in the
representation of unbalanced faults.
• Generated voltage is considered unaffected by
machine speed variation.

6. ROTOR ANGLE STABILITY
Rotor angle stability is the ability of
interconnected synchronous machines of a
power system to remain in synchronism.
A fundamental factor in this problem is that
manner in which the power outputs of
synchronous machines vary as their rotors
oscillate.
Basic Concept of “Synchronous machine”.
Contd(…)

7.  In general, the term "field windings"
applies to the windings that produces the main
magnetic field in a machine, and the term
"armature windings" applies to the windings
where the main voltage is induced.
 Synchronous generators are by definition
synchronous, meaning that the electrical
frequency produced is locked in or
synchronized with the mechanical rate of
rotation of the generator.

8. Contd(…)

9. Contd(…)
(a) SMALL SIGNAL STABILITY
It is the ability of the power system to maintain
synchronism under small disturbances.
 Local Modes
 Interarea modes
 Control modes
 Torsional Modes
(b) Transient Stability
It is the ability of the power system to maintain
synchronism when subjected to a severe
transient disturbances.

10. VOLTAGE STABILITY
 Voltage stability is the ability of a power system to
maintain steady acceptable voltages at all buses in the
system under normal operating conditions and after
being subjected to a disturbance.
 The main factor causing instability is the inability of the
power system to meet the demand for reactive
power.
 The heart of the problem is usually the voltage drop
that occurs when active power and reactive power flow
through inductive reactances associated with the
transmission network.

11. SWING EQUATION
 The equation describing the relative motion of the rotor in
terms of load angle (δ) with respect to the stator field as a
function of time is known as swing equation.
 It is second order non linear differential equation.
 Swing equation governing rotor motion of a synchronous
machine is based on the elementary principle in dynamics
which states the accelerating torque is the product of the
moment of inertia of the rotor times its angular
acceleration.

12. J – Total moment of inertia of the rotor masses in
kg-m2
𝜃 𝑚 - Angular displacement of the rotor w.r.t a
stationary axis in mechanical rad
t – Time in Seconds
Tm – Mechanical torque or shaft torque by the
prime mover in N-m
Te – Net electrical torque in N-m
Ta – Net accelerating torque in N-m
𝐽
𝑑2 𝜃 𝑚
𝑑𝑡2 = 𝑇𝑎 ….. (1)
𝑇𝑎 = 𝑇 𝑚 − 𝑇𝑒 𝑁 − 𝑚 ….. (2)

13. 𝜃 𝑚 = 𝜔𝑠𝑚 𝑡 + 𝛿 𝑚 ….. (3)
𝜔𝑠𝑚 - synchronous speed of the machine
𝛿 𝑚 Angular displacement of the rotor in mechanical radians
𝑑𝜃 𝑚
𝑑𝑡
= 𝜔𝑠𝑚 +
𝑑𝛿 𝑚
𝑑𝑡
….. (4)
𝑑2 𝜃 𝑚
𝑑𝑡
= 𝜔𝑠𝑚 +
𝑑𝛿 𝑚
𝑑𝑡
…..(5)
𝑑2 𝜃 𝑚
𝑑𝑡
=
𝑑2 𝛿 𝑚
𝑑𝑡2 ….. (6)

14. 𝐽
𝑑2 𝛿 𝑚
𝑑𝑡2 = 𝑇𝑎 𝑁 − 𝑚 ….. (7)
𝐽𝜔 𝑚
𝑑2 𝛿 𝑚
𝑑𝑡2 = 𝑃𝑎 𝑖𝑛 𝑤 …..(8)
𝑀
𝑑2 𝛿 𝑚
𝑑𝑡2 = 𝑃𝑎 𝑖𝑛 𝑤 …...(9)
𝑀 – Inertia Constant
𝐻 =
𝑆𝑡𝑜𝑟𝑒𝑑 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 𝑚𝑒𝑔𝑎 𝑗𝑜𝑢𝑙𝑒𝑠 𝑎𝑡 𝑠𝑦𝑛𝑐ℎ𝑟𝑜𝑛𝑜𝑢𝑠 𝑠𝑝𝑒𝑒𝑑
𝑚𝑎𝑐ℎ𝑖𝑛𝑒 𝑟𝑎𝑡𝑖𝑛𝑔 𝑖𝑛 𝑀𝑉𝐴
𝐻 =
1
2
𝐽𝜔2
𝑠𝑚
𝑆 𝑚𝑎𝑐ℎ
…..(10)
𝐻 =
1
2
𝑀𝜔 𝑠𝑚
𝑆 𝑚𝑎𝑐ℎ
𝑀𝐽
𝑀𝑉𝐴
……(11)

15. 𝑀 =
2𝐻
𝜔 𝑠𝑚
𝑆 𝑚𝑎𝑐ℎ
𝑀𝐽
𝑚𝑒𝑐ℎ 𝑟𝑎𝑑
…..(12)
2𝐻
𝜔 𝑠𝑚
𝑑2 𝛿
𝑑𝑡2 =
𝑃 𝑎
𝑆 𝑚𝑎𝑐ℎ
𝑖𝑛 𝑝. 𝑢 …… (13)
2𝐻
𝜔 𝑠
𝑑2 𝛿
𝑑𝑡2 =
𝑃 𝑎
𝑆 𝑚𝑎𝑐ℎ
𝑖𝑛 𝑝. 𝑢 …… (14)
𝐻
𝜋𝑓
𝑑2 𝛿
𝑑𝑡2 = 𝑃𝑎 𝑖𝑛 𝑝. 𝑢 …… (15)
𝛿 𝑖𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
𝐻
180𝑓
𝑑2 𝛿
𝑑𝑡2 = 𝑃𝑎 𝑖𝑛 𝑝. 𝑢 …… (16)
𝛿 𝑖𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑑𝑒𝑔𝑟𝑒𝑒𝑠

16. ema
m
PPPwhere
WinPa
dt
d
M

2
2

ema
m
PPPwhere
p.uinPa
dt
d
f
H

2
2
*180


17. POWER ANGLE EQUATION
 Mechanical power input to the machine remains
constant during the period of electromechanical
transient of interest. In other words, it means that the
effect of the turbine governing loop is ignored, being
much slower than the speed of transient.
 Rotor speed changes are insignificant.
 Effect of voltage regulating loop during the transient is
ignored. As a consequence the generated machine emf
remains constant.

19. FACTORS INFLUENCING TRANSIENT STABILITY
 How heavily the generator is loaded.
 The generator output during the fault. This depends on the
fault location and type.
 The fault-clearing time.
 The postfault transmission system reactance.
 The generator reactance. A lower reactance increases peak
power and reduces initial rotor angle.
 The generator inertia. The higher the inertia, the slower the
rate of change in angle. This reduces the kinetic energy
gained during the fault.
 The generator internal voltage magnitude. This depends on
the field excitation.
 The infinite bus voltage magnitude.

20. ENHANCEMENT OF TRANSIENT STABILITY
Methods of improving transient stability try to
achieve one or more of the following effects:
I. Reduction in the disturbing influence by
minimizing the fault severity and duration.
II. Increase of restoring synchronizing forces.
III. Reduction of the accelerating torque
through control of prime-mover
mechanical power.
IV. Reduction of the accelerating torque by
applying artificial loads

21. The following are various methods of achieving objectives:
1) High speed fault clearing
2) Reduction in transmission system reactance
3) Regulated shunt compensation
4) Dynamic braking
5) Reactor switching
6) Independent-pole operation of circuit breakers
7) Single pole switching
8) Steam turbine Fast-valving
9) Generator Tripping
10) Controlled system separation and Load shedding
11) High speed excitation systems
12) Discontinuous excitation control
13) Control of HVDC transmission links

22. EQUAL AREA CRITERION
 The principle by which stability under transient
conditions is determined without solving the swing
equation is called equal area criterion stability.
 A method used for quick prediction of stability.
 It is based on the graphical interpretation of the energy
stored on rotating mass as an aid to determine if the
machine maintains its stability after a disturbance.
 This method is only applicable to a single machine
infinite bus or two-machine system.

23. ASSUMPTIONS MADE IN EQUAL AREA
CRITERION
 Constant input over the time interval
being considered.
 Damping effect is neglected.
 Constant voltage behind transient
reactance.

24. CRITICAL CLEARING ANGLE AND TIME
The maximum allowable value of the
clearing time and angle for the system to
remain stable are known respectively as
critical clearing time and angle.
why?
0
2


  crmcr tP
2H
f
m
cr
cr
Pf
H
t

 )(2 0



25. DIFFERENT CASES IN EQUAL AREA CRITERION
Case 1 : Step Change in mechanical input
Case 2 : Sudden loss of parallel lines.
Case 3 : Sudden short circuit on one of parallel
lines at end of one line.
Case 4 : Sudden short circuit on one of parallel
lines at the middle of one line
2max3max
3max02max0
coscos)(
cos
PP
PPP mmm
cr






26. RESPONSE TO A STEP CHANGE IN
MECHANICAL POWER

28. SOLUTION OF SWING EQUATION BY MODIFIED
EULER’S METHOD
ALGORITHM:
Step 1 : The appropriate Equal Area Criterion
method is to be chosen for the given
system and the values are to be found
out at different conditions.
Step 2 : The values of Pe1, pe2 and Pe3 are to be
found out.
Step 3 : Then the solution of swing equation using
modified euler’s method are to be applied.
Step 4 : compute
Step 5 : Compute the first iteration
t;, 00 

29. Step 6 : Predicted values are
Step 7 : Derivatives at the end of first iteration
]sin[
)(
2
2max
00






PP
H
f
dt
d
f
dt
d
m
0
i
i




t
dt
d
t
dt
d
i
i
p
i
i
i
p
i







*
)(
*
1
1






p
i
p
i
p
i
H
Paf
dt
d
dt
d
0
p
i
11
1
)(
1




 






30. Step 8 : The final corrected values are
t
dt
d
dt
d
p
ii
i
c
i 













 
 *
2
1
1



t
dt
d
dt
d
p
ii
i
c
i 











 


 
 *
2
1
1




31. SOLUTION OF SWING EQUATION BY
RK METHOD
Step 1 : Obtain the prefault conditions and
determine the generator internal
voltages behind transient reactances.
Step 2 : Calculate Pe1 during prefault conditions,
Pe2 during fault conditions andPe3
during postfault conditions.
Step 3 : Initialise iteration count; K=0 and
Step 4:Determine the stability by finding the
constants.
t

32.  
 
 
 
  tPP
H
f
l
tlK
tPP
H
f
l
t
l
K
tPP
H
f
l
t
l
K
tPP
H
f
t
dt
d
l
tt
dt
d
K
Kem1
0
k4
K
em1
0
3
k3
K
em1
0
2
k2
em1
0
1
k
k
k
k
k
k
k






















)(4
3
)
2
(
2
)
2
(
1
)(
1
3
2
1
2
2
















33. Step 5 : calculate the change in state variable
Step 6 : The new state variables are
Step 7 : If (fault clearing time) is reached
stop it, else repeat the iterations.
Set k=k+1
Step 8 : Stop
 
 4321
4321
22
6
1
22
6
1
lKllll
KKKKK
k
k


kkk
kkk
l
K






1
1
t

34. Problems in H and M
Problem 1:
The moment of inertia 8 pole, 75 MVA, 11
kV, 3 phase, 0.6 p.f, 50 Hz turbo alternator is
20,000 kg-m2. Calculate H and M.
Inertial Constant 𝑯 =
𝑾 𝒌
𝑺 𝒃𝒂𝒔𝒆
𝑴 =
𝑺 𝒃𝒂𝒔𝒆 ∗ 𝑯
𝟏𝟖𝟎 ∗ 𝒇

35. Answers:
H=0.8225 MJ/MVA
M=0.0069 MJ sec/elec. deree

36. Problem 2:
A 50 Hz, 8 pole turbo alternator rated at 100
MVA, 13.2 kV has an inertia constant of 9
MJ/MVA. Find
(a) Energy stored in the rotor at synchronous speed
(b)The rotor accleration if the mechanical input
raised to 90 Mw for an electrical load of 60 MW.
𝑾 𝒌 =
𝟏
𝟐
J𝝎 𝟐
𝒔𝒎 or Inertial Constant 𝑯 =
𝑾 𝒌
𝑺 𝒃𝒂𝒔𝒆
𝑯
𝟏𝟖𝟎 × 𝒇
𝒅 𝟐 𝜹
𝒅𝒕 𝟐
= 𝑷𝒎 − 𝑷𝒆

37. Answers:
W= 900 MJ
Acceleration = 300 electrical degrees/sec2

38. END