POWER GENERATION AND TRANSMISSION
EEE 420
STABILITY
Power system stability may be broadly defined
as that property of a power system that enables
it to remain in a state of operating equilibrium
under normal operating conditions and to regain
an acceptable state of equilibrium after being
subjected to a disturbance.
Transient stability studies are needed to ensure
that the system can withstand the transient
condition following a major disturbance.
Stability studies are conducted when planning
for
1. new generation
2. transmitting facilities
IMPORTANCE OF STABILITY STUDIES
The studies are helpful in determining such
things as
the nature of the relaying system required
 critical clearing time of circuit breakers
voltage level of the system
transfer capability between systems
CLASSIFICATION OF POWER SYSTEM
STABILITY PROBLEMS
1. Rotor Angle Stability
I. Steady State (Small Signal) Stability
II. Transient Stability
1.
Voltage Stability
Rotor Angle Stability
• Rotor angle stability is the ability of
interconnected synchronous machines of a
power system to remain in synchronism.
Steady State Stability
Steady state (small-disturbance) stability is the
ability of a power system to maintain
synchronism when subjected to small
disturbances as occur continually in normal
operation due to small variations in consumption
and generation.
• Steady state stability
Instability may be of two forms
 Steady increase in rotor angle due to the lack
of sufficient synchronizing torque.
 Rotor oscillations of increasing amplitude
due to lack of sufficient damping torque.
Swing Equation
The motion of the rotor of a synchronous generator is
described by the equation
d θm
J 2 = Ti − Te
dt
2
(1)
where
J is rotor moment of inertia
Ti is shaft torque developed by turbine
Te is electromagnetic torque developed by synchronous
machine
Θm is angular displacement of rotor in mechanical radians
From equation (1), Swing equation can be derived
as
2H d δ
= Pi − Pe
2
ω 0 dt
2
(2)
where
Pi is shaft input power in per unit
Pe is electromagnetic power in per unit
δ is the load angle, power angle, torque angle or
internal angle.
H is inertia constant defined as the stored kinetic energy at
synchronous speed in MJ or MW-sec per unit MVA of
machine rating.
Jω 0
× 10 6 MJ
H =
MVA
2 Pr
(3)
Pr = rating of mechanice in MVA
ω 0 = mechanical or rotor synchronous speed in radians
Power Transfer
Consider a single synchronous machine
connected to an infinite bus through an external
impedance or reactance
E∠δ
Synchronous
machine
V∠0o
Ze = Re + jX e I
R
Infinite bus
The complex power injected into the infinite
bus(or the complex power at its receiving end) if
R = 0 is given by
S R = V∠0 × I R
*
(4)
The complex power injected is
PR
VE sin δ
=
Xe
(5)
Steady State Stability Limit
The steady state stability limit of the simplest electrical
system is defined as the greatest possible power at its
receiving end under a given condition of operation and
excitation in the presence of small disturbances.
E∠δ
Xd
Xe
V∠0o
The electromagnetic power is given as
EV
sin δ
Pe =
Xd + Xe
(6)
Suppose a small disturbance causes the rotor
angle to vary by ∆δ
Thus, changes from δ 0 to (δ 0 + ∆δ)
where subscript ‘0’ denotes steady state
condition.
Then from electromagnetic power equation
 ∂Pe 
∆Pe = 
 ∆δ = c∆δ
 ∂δ  0
(7)
Suppose that Pi is constant because the governor
is slow to act compared to the speed of energy
dynamics.
Substituting equation (7) into the equation (2)swing equation, we obtain
 2H  d (δ 0 + ∆δ)


= Pi − (Pe 0 + ∆Pe )
2
dt
 ω0 
2
2H d 2 ∆δ
= − ∆Pe
2
ω 0 dt
d 2 ∆δ
M 2 = −C∆δ
dt
for Pe 0 = Pi
2H
where M =
ω0
(8)
(9)
(10)
d 2 ∆δ
+ C∆δ = 0
M
2
dt
The characteristic equation is
Mp + C = 0
2
and its roots are
C
p=± −
M
Natural frequency of oscillations = p/ 2π
(11)
Case 1
If C > 0, the roots are pure imaginary and any
small disturbance appearing in the system will
result in continuous oscillations.
Line resistance and damper windings of
machines ignored in the analysis cause the
system oscillations to decay. The system is
therefore stable for a small disturbance so long
as
∂P
 e

 >0
 ∂δ  0
Case 2
If C < 0, both roots are real and one of them is
positive. In this case, any small disturbance
results in a periodic rise of the torque angle, and
synchronism is soon lost.
Thus the system is unstable if
 ∂Pe 
 <0

 ∂δ  0
EV
At δ 0 = 90 , C = X cos δ 0 = 0
o
The angle δ 0 = 90 o therefore determines the steady
state stability limit Pm
Thus,
EV
EV
o
Pm =
sin 90 =
X
X
(11)
Example 1
For the system where X d = 1.20p.u., V = 1.0p.u., X e = 0.60p.u., E = 1.20p.u.
• H = 4 MW − sec MVA and the system frequency
calculate the frequency of natural oscillations if the
generator is loaded to (a) 50% and (b) 80% of its
maximum power limit.
Solution
(a) For 50% loading
Pe 0
sin δ 0 =
= 0.5 ⇒ δ 0 = 30 o
Pm
1.2 × 1
EV
cos 30 o = 0.577 pu / electrical rad
C=
cos δ 0 =
1.8
X
M=
2H
2H
2× 4
=
=
= 0.0255p.u. / electrical rad
ω0 2πf 0 2 × π × 50
p=± −
C
0.577
=±j
= ± j 4.757
M
0.0255
Natural frequency of oscillations = 4.757rad / sec = 4.757 = 0.757Hz
2π
Cont’d
(b) For 80% loading
sin δ 0 =
C=
Pe 0
= 0.8 ⇒ δ 0 = 53.1o
Pm
1.2 × 1
cos 53.1o = 0.4pu / electrical rad
1.8
M = 0.0255p.u. / electrical rad
C
0.4
p=± −
= ±j
= ± j3.961
M
0.0255
3.961
= 0.637 Hz
Natural frequency of oscillations = 3.961rad / sec =
2π
Example 2
For the system shown in the figure below,
calculate the limit of steady state power with
and without reactor switch closed.
X t = 0.1 pu
X dg = 1 pu
Eg = 1.2 pu
X L = 0.25 pu
X t = 0.1 pu
X c = 1 pu
M
X mg = 1 pu
Em = 1.0 pu
Solution
(a) Reactor switch is open
Total reactance between generator and motor
X = X dg + X t + X L + X t + X mg
= 1 + 0.1 + 0.25 + 0.1 + 1 = 2.45 p .u .
Therefore,
Pm =
EgEm
X
1.2 × 1
=
= 0.49p.u.
2.45
Cont’d
(b) Reactor switch is closed.
The equivalent circuit is as follows
a
j1 pu
j 0.1 pu
j 0.25 pu
j 0.1 pu
− j1 pu
Eg = 1.2 pu
j1 pu
b
M E = 1.0 pu
m
c
This can be reduced to (by star-to-delta conversion)
X
Eg = 1.2 pu
M E = 1.0 pu
m
Cont’d
This gives
Xa Xb
X = Xa + Xb +
Xc
Where X a = j(1 + 0.1 + 0.25) = j1.35
X b = j(1 + 0.1) = j1.1
X c = − j1
Therefore
X = j1.35 + j1.1 +
( j1.35)( j1.1) = j2.45 − j1.485 = j0.965p.u.
− j1
1.2 × 1
= 1.244 p .u .
Steady state power limit, Pm =
0.965