NERC Standard
The bulk power system will achieve an adequate
level of reliability when it is planned and operated
such that:
Introduction to Power System
Stability
1.
2.
Mohamed A. El-Sharkawi
Department of Electrical Engineering
University of Washington
Seattle, WA 98195
http://SmartEnergyLab.com
Email: elsharkawi@ee.washington.edu
3.
4.
5.
©Mohamed El-Sharkawi, University of Washington
1
NERC Standard
The System remains within acceptable limits;
The System performs acceptably after credible
contingencies;
The System contains (limit) instability and
cascading outages;
The System’s facilities are protected from severe
damage; and
The System’s integrity can be restored if it is lost.
©Mohamed El-Sharkawi, University of Washington
2
System Operating Limit (SOL)
• “Reliable Operation means operating the elements of
the Bulk-Power System within equipment and electric
system thermal, voltage, and stability limits so that
instability, uncontrolled separation, or cascading
failures of such system will not occur as a result of
sudden disturbance, including a Cybersecurity
Incident, or unanticipated failure of system elements.”
• MW, MVar, Amperes, Frequency or Volts
that satisfies the most limiting of the
operating criteria for a specified system
configuration to ensure operation within
acceptable reliability criteria
• This is a security measure
• NERC standard is for dynamic performance
requirements
– Security
– Stability
©Mohamed El-Sharkawi, University of Washington
3
Interconnection Reliability
Operating Limit (IROL)
4
SOL and IROL
• Following a contingency or other system
event that cause the system to operate
outside set reliability boundaries,
Transmission Operators (TO) are
obligated to return its transmission system
to within SOL or IROL as soon as
possible.
• IROL is a SOL that, if violated, could lead
to instability, uncontrolled separation, or
cascading outages that adversely impact
the reliability of the bulk power system.
• This is a stability measure
©Mohamed El-Sharkawi, University of Washington
©Mohamed El-Sharkawi, University of Washington
5
©Mohamed El-Sharkawi, University of Washington
6
1
What is stability?
Stability
• During fault, and right after the fault is
cleared, the power system stability is
determined by the capabilities of its
generators to
• Is the ability of the system to achieve
steady state operating condition after a
disturbance
– All oscillations are damped out
– maintain connected to the grid
– provide the extra reactive power
– provide fast ramping down of real power
f
• Generators are permitted to trip off line only
in the case of a permanent fault on a
directly connected circuit
60Hz
Time
©Mohamed El-Sharkawi, University of Washington
7
©Mohamed El-Sharkawi, University of Washington
Balance of power in
Generators
What is security?
Pm
• If the system is stable after a disturbance,
the system is secure if all its key
components are operating within their
design limits
9
Turbine Speed and Imbalance of
Power
Pe
Pm
G
Pm  Pe  ~
if Pm  Pe ;
dn
0
dt
G
Mechanical
Power
Controlled at
the power plant
At steady state
©Mohamed El-Sharkawi, University of Washington
8
dn
dt
n  ns ; constant rotor speed
Pe
Electrical
Power
Controlled by
the customers
Pm = Pe
©Mohamed El-Sharkawi, University of Washington
10
Turbine Speed and System
Frequency
f 
p
n
120
f: The frequency of the terminal voltage of the generator
p: The number of poles of the generator
n: The speed of the generator (turbine)
if Pm  Pe ; n  ns
if Pm  Pe ; n  ns
©Mohamed El-Sharkawi, University of Washington
11
©Mohamed El-Sharkawi, University of Washington
12
2
If Pm>Pe
Pm
G
If Pm<Pe
Pe
Pm
• The surplus energy is stored in the rotating
mass of the generator in the form of kinetic
energy
– The machine speeds up (frequency
increases)
– The current inside the machine increases
– Over-speed and over-current protections will
eventually trip the machine
©Mohamed El-Sharkawi, University of Washington
Pe
• The deficit in energy is drawn from the
kinetic energy of the rotating mass of the
generator
– The machine initially slows down (frequency
decreases)
– If not corrected, the machine could operate as a
motor
– Machine is tripped to prevent mechanical damages
13
©Mohamed El-Sharkawi, University of Washington
Transient Stability Analysis
14
Unstable System
• Transient stability analysis determines
whether the generator, after a disturbance,
reaches a new stable operating point.
– The input mechanical power of the generator
is equal to the output electric power, and
– The frequency of the generator is the same as
the frequency of the system before the
disturbance (i.e. the generator speed is the
same as the prefault synchronous speed)
©Mohamed El-Sharkawi, University of Washington
G
15
n (f)
ns=60Hz
©Mohamed El-Sharkawi, University of Washington
Unstable System
16
Stable System
n (f)
n (f)
ns=60Hz
ns=60Hz
Time
©Mohamed El-Sharkawi, University of Washington
Time
17
©Mohamed El-Sharkawi, University of Washington
18
3
Power Generation
• 99+ % of all power are generated by the
synchronous generators
• Synchronous machines can operate as
generators or motors
Synchronous Generator
El-Sharkawi@University of Washington
El-Sharkawi@University of Washington
21
El-Sharkawi@University of Washington
20
22
Small Synchronous Machine
El-Sharkawi@University of Washington
23
El-Sharkawi@University of Washington
24
4
a
a
b
X
c
f
b
N
X
N
X
c
Vaa’
Vbb’
Vcc’
X
Time
c
X
25
El-Sharkawi@University of Washington
If
b
X
Rotor
Stator
f
c
S
b
S
El-Sharkawi@University of Washington
26
Equivalent Circuit
Open Stator
R
Xs
N
s
Vf
Ef
Vt
Ef
S
Ef ~
E f is directly proportional to the excitation current
df
dt
X s  Synchronous Re ac tan ce
R  Armature Re sis tan ce
If
The frequency of E f is proportional to the synchronous speed
R  X s
s
El-Sharkawi@University of Washington
27
El-Sharkawi@University of Washington
Generator Equivalent Circuit
Generator Equivalent Circuit
Xs
Xs
Ia
Ia
Ef
Ef

Vt
Ia
Vt is Fixed (infinite Bus)
Ef is function of If
Magnitude and phase of Ia
are dependant variables
E f  Vt  I a X s
El-Sharkawi@University of Washington
Ia Xs

Vt
Vt
Ef
28
29
E f  Vt  I a X s
El-Sharkawi@University of Washington
30
5
Power equations
f
Xs
Power equation
t
Ef
Ia
Ia Xs

Ef
Vt
Ia
I a cos  
Vt

I a X s cos   E f sin 
Ef


E f sin 
Xs
Vt

Ia
I a Xs
P  3 Vt I a cos 
P  3 Vt I a cos 
P
Vt and Ef are phase quantities
Qt  3 Vt I a sin 
31
El-Sharkawi@University of Washington
3 Vt E f
sin 
Xs
32
El-Sharkawi@University of Washington
Power Characteristics
of Generator
X
s
Reactive Power equations
Ef
Ia
Ia Xs

Ef
Xs
t
Vt
Ef
P

Ia
P
3 Vt E f
sin 
Xs
Pmax 

90 o
Ia
Vt
3 Vt E f
I a Xs
Vt
Ia

Vt
Qt  3 Vt I a sin 
I a X s sin   E f cos   Vt
Xs
El-Sharkawi@University of Washington


Ef
P max
l
f
33
Vt and Ef are phase quantities
El-Sharkawi@University of Washington
34
A Synchronous Generator
Connected to Large System
Qt  3 Vt I a sin 
I a X s sin   E f cos   Vt
Qt  3 Vt I a sin  
3Vt
E f cos   Vt 
Xs
If Ef cos  > Vt ;
Qt is positive and Current is lagging
If Ef cos  < Vt ;
Qt is negative and Current is leading
If Ef cos  = Vt ;
Qt is zero and Current is in phase
El-Sharkawi@University of Washington
Ef
Vt
G
V
Transmission line
Terminal bus
V and f may vary
35
©Mohamed El-Sharkawi, University of Washington
Infinite bus
V and f cannot vary
36
6
Power Equation
X = Xs+Xline
P
I
Pmax
Ef
Ef
P

V
V Ef
sin 
X
Pm
V
P
Power Characteristics of
Generator
Ef V
sin 
Real Power
X
V
Q  E f cos   V 
Reactive Power
X

©Mohamed El-Sharkawi, University of Washington
37

90o
©Mohamed El-Sharkawi, University of Washington
Maximum Power
38
Operating Point
P
P
P
Pmax
Pm
Ef
Pm

Ef V
X
sin 
Operating point
V
l
90
o

Phasor Diagram at
Pmax
©Mohamed El-Sharkawi, University of Washington

39


©Mohamed El-Sharkawi, University of Washington
40
Generation Limit
• The generator must generate less than the Pmax
(called pull-out power).
• The difference between Pmax and the actual
power generated is the generation margin.
• The generation margin must be positive and
large enough to ensure the dynamic stability of
the system.
©Mohamed El-Sharkawi, University of Washington
Dynamic Stability Assessment
(DSA)
41
©Mohamed El-Sharkawi, University of Washington
42
7
Phasor diagram
Objective of DSA
X
• Dynamic stability analysis determines whether
the system, after a disturbance, reaches a new
stable operating point.
I
– The input mechanical power of the generator is equal
to the output electric power
– The frequencies of the generator is the same as the
frequency of the system (i.e. the generator speed is
the synchronous speed)
– All voltages and currents are within the allowable
limits
©Mohamed El-Sharkawi, University of Washington
43
Rotation of Phasor Diagram
fg is the frequency of
the generator
Rotor
Ef
Ef
Ef

©Mohamed El-Sharkawi, University of Washington
fg 
V
f is the frequency at
the infinite bus
©Mohamed El-Sharkawi, University of Washington
45
©Mohamed El-Sharkawi, University of Washington
Pm
• If the frequency of the generator (fg) is 60 Hz (or
50 Hz), the speed of the generator’s shaft is called
synchronous speed (ns).
©Mohamed El-Sharkawi, University of Washington
46
Power Control
Frequency/Speed Relationship
p
ns
120
p
50 
ns
120
p
n
120
p is the number of magnetic poles of the generator
Infinite bus
60 
44
• The frequency (fg) of the generator’s voltage
is proportional to the speed of the
generator’s shaft (n).
f  60 or 50 Hz
fg 
V
I
Frequency/Speed Relationship
fg

I X

V
p
n
120
G
if Pm  P;
dn
0
dt
P
Pm  P  ~
dn
dt
n  ns ; constant rotor speed
if Pm  P; n increases
if Pm  P; n decreases
in 60 Hz system
in 50 Hz system
47
©Mohamed El-Sharkawi, University of Washington
48
8
Power Control
• When more power is
needed, the generator
speed increases (n>ns)
by increasing the
mechanical power
into the generator
– The frequency of the
generator increases
(fg>f )
– The angle  increases
– The current of the machine
increases
fg  f
Swings due to Sudden increase in Mechanical Power
P
at t2
3
2
Pm2
I2 X
Pm1
fg  f
Ef
2
1
at t1
I1 X
Ef
f
1
– Hence, the power increases

V
©Mohamed El-Sharkawi, University of Washington
49
Swings due to Sudden increase in Mechanical Power
2
3

©Mohamed El-Sharkawi, University of Washington
50
Swings due to Sudden increase in Mechanical Power
P
P
2
Pm2
Pm1
1
Pm1


1

2

Operating point
Powers
Acceleratio
n
Rotor
Speed
Power angle

Operating point
Powers
Acceleratio
n
Rotor
Speed
Power angle

1
Before disturbance
Pm1= Pe
0
n = ns
1
1 to 2
After increasing Pm
Pm2 > Pe
+ (n↑)
n > ns
↑
©Mohamed El-Sharkawi, University of Washington
51
Swings due to Sudden increase in Mechanical Power
©Mohamed El-Sharkawi, University of Washington
Swings due to Sudden increase in Mechanical Power
P
P
2
2
Pm2
Pm1
52
Pm2
1
Pm1

2
Operating point
Powers
Acceleratio
n
2
Pm2 = Pe
0
©Mohamed El-Sharkawi, University of Washington

1

2

Rotor
Speed
Power angle

Operating point
Powers
Acceleratio
n
Rotor
Speed
Power angle

n > ns
2
After 2
Pm2 < Pe
- (n↓)
n > ns
↑
53
©Mohamed El-Sharkawi, University of Washington
54
9
Swings due to Sudden increase in Mechanical Power
Swings due to Sudden increase in Mechanical Power
P
P
3
3
2
2
Pm2
Pm1
Pm2
1
Pm1

2
3
Operating point
Powers
Acceleratio
n
3
Pm2 < Pe
- (n↓)

1

Rotor
Speed
Power angle

Operating point
n = ns
3
3 to 2
©Mohamed El-Sharkawi, University of Washington
55
Swings due to Sudden increase in Mechanical Power
2

Powers
Rotor
Speed
Power angle

Pm2 < Pe
- (n↓)
n < ns
↓
©Mohamed El-Sharkawi, University of Washington
56
Swings due to Sudden increase in Mechanical Power
P
P
3
3
2
2
Pm2
Pm1
3
Acceleratio
n
Pm2
1
Pm1

2
3

1

2
3

Operating point
Powers
Acceleratio
n
Rotor
Speed
Power angle

Operating point
Powers
Acceleratio
n
Rotor
Speed
Power angle

2
Pm2 = Pe
0
n < ns
2
2 to 1
Pm2 > Pe
+ (n↑)
n < ns
↓
©Mohamed El-Sharkawi, University of Washington
57
Swings due to Sudden increase in Mechanical Power
©Mohamed El-Sharkawi, University of Washington
58
Swings due to Sudden increase in Mechanical Power
P
P
3
2
2
Pm2
Pm1
Pm2
1

Pm1
2
3
Operating point
Powers
Acceleratio
n
1
Pm2 > Pe
+ (n↑)
©Mohamed El-Sharkawi, University of Washington

Rotor
Speed
n = ns
1

2

Power angle

Operating point
Powers
Acceleratio
n
Rotor
Speed
Power angle

1
1 to 2
Pm2 > Pe
+ (n↑)
n > ns
↑
59
©Mohamed El-Sharkawi, University of Washington
60
10
Summary
Operating point
Powers
Acceleration
Rotor
Speed
Power angle

1
Before disturbance
Pm1= Pe
0
n = ns
1
Pm2 > Pe
+ (n↑)
n > ns
↑
2
Pm2 = Pe
0
n > ns
2
2 to 3
Pm2 < Pe
- (n↓)
n > ns
↑
3
Pm2 < Pe
- (n↓)
n = ns
3
3 to 2
Pm2 < Pe
- (n↓)
n < ns
↓
2
Pm2 = Pe
0
n < ns
2
2 to 1
Pm2 > Pe
+ (n↑)
n < ns
↓
1
Pm2 > Pe
+ (n↑)
n = ns
1 to 2
After Pm increased
Swing Angle

3
2
1
dt
1
©Mohamed El-Sharkawi, University of Washington
Time
Pm  P  ~ dn
61
©Mohamed El-Sharkawi, University of Washington
Damped Oscillations (Stable)
62
Damped Oscillations (Stable)

3
• Damping is due to several factors such
as
2
– Resistances of the various power system
components
– Controllers installed in the power plants
1
Pm  P   M dn  D n
Time
dt
©Mohamed El-Sharkawi, University of Washington
63
©Mohamed El-Sharkawi, University of Washington
Undamped Oscillation (Unstable)
64
Effect of Excitation

P
E f2
3
E f1
2
Pm
1
P
E f2 > E f1
Pm  P   M dn  D n
Time
2
When D is negative
dt
©Mohamed El-Sharkawi, University of Washington
65
1
©Mohamed El-Sharkawi, University of Washington
V Ef
X
sin 

66
11
Effect of Increasing Excitation
Increase transmission Capacity
P
• The maximum power that CAN be
delivered increases
• The real power is unchanged (Pm
unchanged)
• The power angle decreases
Xs
Pm
Vt
V Ef
sin 
X
Vo
Xl
G
Terminal bus
©Mohamed El-Sharkawi, University of Washington
67
Infinite bus
©Mohamed El-Sharkawi, University of Washington
Xs
Pm
68
Vt
V
Xl
G
Xl1
Xs
Xl2
Ia
Ef
©Mohamed El-Sharkawi, University of Washington
P
69
P1 
Pm
1 2
©Mohamed El-Sharkawi, University of Washington
V Ef
X
70
Dynamic Stability Assessment
(DSA)
• DSA Methods
V Ef
sin
( X s  Xl )
P
V
©Mohamed El-Sharkawi, University of Washington
V Ef
sin
( X s  0.5 X l )
P2 
Vt
sin 
•
•
•
•
•
Time-domain solution
Energy Margin
Equal area
Eigenvalues
Pattern Recognition

71
©Mohamed El-Sharkawi, University of Washington
72
12
• Time domain methods seek to set up and solve
a set of differential equations that describe the
motion of the machines connected to the system
• Advantage:
Frequency
Time Domain Solution
Frequency
Time Domain Solution
– Direct numerical integration can provide accurate
information on the stability of the system.
• Disadvantage:
Time
Time
– The numerical integration is performed in each time
interval; time consuming and slow
©Mohamed El-Sharkawi, University of Washington
Unstable System
73
Time Domain Solution
©Mohamed El-Sharkawi, University of Washington
74
Equal Area Criterion
Frequency
• A DSA method that represents the kinetic
energy gained or lost due to oscillations.
• If the deceleration energy is equal or larger than
the acceleration energy, the system is stable.
Time
Stable System
©Mohamed El-Sharkawi, University of Washington
75
Equal Area Criterion
• Assume a sudden increase in the mechanical
energy to the generator
– The speed of the machines increases from n1 to n2
KE ~ n 2
P
3
Change in Kinetic Energy KE
2
Pm2
Pm1
KE ~ n
76
Equal Area Criterion
• Kinetic Energy KE

©Mohamed El-Sharkawi, University of Washington
1
2

©Mohamed El-Sharkawi, University of Washington
77
©Mohamed El-Sharkawi, University of Washington
2
3

78
13
Stability Condition
Analysis of Equal Area Criterion
• Acceleration Kinetic Energy KEa

KE a  KE 2  KE1  ~ n22  n12

Decelerated Kinetic Energy KEd

KE d  KE 2  KE 3  ~ n22  n32


n
2
2
P
3
2
 
 n32  n22  n12
Pm1

Pm1
n3  n1
1

2
3
1
©Mohamed El-Sharkawi, University of Washington
79

Since
n1  ns
Hence
n3  ns

KEd  KEa
2
3
©Mohamed El-Sharkawi, University of Washington
Equal Area Criterion

80
Equal Area Criterion
Since
Since
n  ns 
Pm  P   power unbalance
Then
d
 n
dt
Then

1 2
Pm 2  P  d  KEa
n 1
t2
 Pm 2  P  dt  KEa
t1

t3
1 3
P  Pm 2  d  KEd
n 2
 P  Pm 2  dt  KEd
t2
©Mohamed El-Sharkawi, University of Washington
81
©Mohamed El-Sharkawi, University of Washington
P
Since
n  constant
Aa
3
2
Pm2
KE a  KE d
For stable System
Aa = Ad
Ad
2
3
1
2
Pm1
 Pm 2  P  d   P  Pm 2  d
Acceleration area  Deceleration area
1

Aa  Ad
©Mohamed El-Sharkawi, University of Washington
82
Representation of Aa and Ad
Stability Condition
Then
3
2
Pm2
Pm2

Condition for stable system
P
KEd  KEa
83
2
©Mohamed El-Sharkawi, University of Washington
3

84
14
Unstable System
P
Aa
3-max
2
Pm2
Ad(max)
Pm1
General Stability Condition
If at 3-max
n3 > ns
Then
Aa > Ad(max)
KE a  KE d (maximum)
1
2
 3 max
1
2
 Pm 2  P  d   P  Pm 2  d

3 max
2
Aa  Ad (max)

©Mohamed El-Sharkawi, University of Washington
85
©Mohamed El-Sharkawi, University of Washington
Example: Opened Breaker
General Stability Condition
KEa  KEd (maximum)
CB
G
2
 3 max
1
2
86
TL
Breaker
 Pm 2  P  d   P  Pm 2  d
Assume that the CB is opened for a short time
Aa  Ad (max)
©Mohamed El-Sharkawi, University of Washington
87
©Mohamed El-Sharkawi, University of Washington
Analysis of Opened Breaker
P
Pm
Aa
2
3
c
Clearing angle
©Mohamed El-Sharkawi, University of Washington
3
Ad min
P
the system is
stable
1

Critical Clearing Angle
If n3 = ns
Or
Aa = Ad,
Ad
88
Pm
Aa max
3-max
1


2
cr
Critical Clearing angle
89
©Mohamed El-Sharkawi, University of Washington
The critical
clearing angle (cr)
is the maximum
angle for a stable
system, i.e. when
Aa max = Ad min
3 max 
90
15
Critical Clearing Angle
Critical Clearing Angle
For stable system
Aa max 
 cr
 P
 P  d 
m
1
Ad min 
 cr
 P
m
Aa max  Ad min
 0  d  Pm  cr  1 
Pm  cr  1   Pmax cos  cr  cos 1   Pm  cr  1  
1
  1
 P
m
 P  d 
 cr
Then
  1
 P
m
 Pmax sin   d
cos  cr    21  sin 1  cos 1
 cr
 Pmax cos  cr  cos 1   Pm   1   cr 
©Mohamed El-Sharkawi, University of Washington
10.91
Energy Margin
• Calculate the transient energy at the
instant the disturbance is cleared.
• Determine the critical energy for the
current disturbance.
• Calculate the transient energy margin.
93
Eigenvalues Method
94
• The system is represented by

Eigenvalues
Root locus plots
Nyquist criteria
Routh-Hurwitz criteria
• Inaccuracies resulting from representing a highly
nonlinear system by a set of linear equations.
©Mohamed El-Sharkawi, University of Washington
©Mohamed El-Sharkawi, University of Washington
Eigenvalues Method
• The power system is converted into a set of
linear equations.
• The dynamic behavior can be analyzed by any
of the linear techniques
–
–
–
–
10.92
Energy Margin: Steps
• To predict the transient behavior of a power system
without having to conduct a complete time domain
simulation.
• The values of an energy function is calculated and
compared with the critical value to determine the
stability.
• Accurate only when the operating point is within the
region of energy function.
• Because of the approximation in the energy function,
results are often pessimistic
©Mohamed El-Sharkawi, University of Washington
©Mohamed El-Sharkawi, University of Washington
95
xa xd
dx
a x d
dt
Where d is a disturbance
©Mohamed El-Sharkawi, University of Washington
96
16
If  is Negative (Stable System)
Justification
n
• The solution of the equation is
x  d e t
where
  j 
ns
Imaginary component
Root (eigenvalue)
Real component
Time
©Mohamed El-Sharkawi, University of Washington
97
If  is Positive (Unstable System)
©Mohamed El-Sharkawi, University of Washington
98
Eigenvalues Method
n
• A linear analysis of the system
• Requires the detailed model of the system and
the knowledge of all its parameters
ns
Time
©Mohamed El-Sharkawi, University of Washington
99
©Mohamed El-Sharkawi, University of Washington
100
17