CHAPTER TWO

POWER SYSTEM STABILITY

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 Oscillations in Two Area System
A mechanical analogy to this phenomenon can be visualized
in figure below. Suppose that there is a set of balls of different
sizes connected to each other by a set of strings. The balls
represent generators having a specific mechanical
characteristic (that is, inertia). The strings represent the
transmission line interconnecting the generators.

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Mechanical Analogy of Transient Stability
Now suppose that there is a disturbance in which one of the
balls is struck with a cue. The ball now begins to swing, and as
a result, the string connected to the ball also oscillates. In
addition, the other strings to which this string is connected
are also affected, and this in turn affects the other balls
connected to these strings. As a result, the entire
interconnected system of balls is affected, and the system
experiences oscillations in the strings and motion of the balls.
If these oscillations in the strings become large, one of the
strings may break away from the rest, resulting instability. On
the other hand if the oscillation dies down and the entire
system comes back to rest as in the situation prior to the ball
being struck. This condition is analogous to a power system
being “transiently stable”.
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The Swing Equation

The motion of the synchronous generator’s rotor is determined by

Newton's second law, which is given as:

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inertia constant, or “H” constant

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The power output of the machine is given by

The steady state stability limit of a generator or system can be defined

as the maximum power that can be transmitted for a slow change in
load.
If the change in excitation is assumed to be initiated immediately
following the change in load (as happens with automatic voltage
regulator action), the stability limit under such conditions may be
termed dynamic stability limit. The dynamic limit will, in general, be
higher than the steady state stability limit as defined above.
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 𝑥1 0
x 0 =
𝑥2 0
∆𝛿𝑜
∆𝛿𝑜
= =
𝑑∆𝛿𝑜
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𝑑𝑡
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 A 50Hz synchronous generator having inertia constant
H=9.94 MJ/MVA and a transient reactance Xd`=0.3pu is
connected to an infinite bus through the following network.
The generator is delivering 1.0pu to the infinite bus at a
voltage 1 pu. Determine the power angle equation and swing
equation for the system applicable to the operating
conditions. Assume the damping power coefficient is D=0.138
pu. Consider a small disturbance of 10o or 0.1745 radians.
Obtain equations of rotor and generator frequency motion.

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Solution
The series reactance between the terminal voltage and the infinite
bus is

And therefore; the 1.0 pu power output of the generator is determined by

the power angle equation

So that the terminal voltage is

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Hence, the desired equation is

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 The power angle equation is plotted. Note that the mechanical input power
Pm is constant and intersects the sinusoidal power angle curve at operating angle
o = 36.378o. The swing equation for the machine may be written

Under the operating condition

which corresponds exactly to the mechanical power input Pm and the

acceleration is zero.

To obtain the equations of rotor and generator frequency motion

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Hence, the equations of rotor and generator frequency motion

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Example:
A synchronous generator, capable of developing 500MW
of power, initially operates at a power angle of 8o. By how much
can the input shaft power be increased suddenly without loss of
stability?

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Solution:
Initially, at

Let be the power angle to which the rotor can swing before losing
synchronism. Then the equal-area criterion requires that the equation

which yields,

must be satisfied.
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 From figure shown

Substituting

By trial and error method, yields

Hence, without loss of stability, the system can accommodate a sudden

increase of

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The critical clearing time ,denoted tc ,is the longest fault
duration allowable for stability.
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Example
A 50 Hz generator is supplying 550 MW to an
infinite bus (with 1.0 per unit voltage and unity p.f)
through two parallel transmission lines. Determine
the critical clearing time for a temporary three-
phase-to-ground fault which shown in figure below.
H = 20 seconds.

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Solution

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EXAMPLE:
The synchronous generator shown in figure is initially
operating in the steady-state condition, when a temporary
three-phase-to-ground bolted short circuit occurs on line 1-3
at bus 1, shown as point F in figure. Three cycles later the
fault extinguishes by itself. Due to a relay misoperation, all
circuit breakers remain closed. Determine whether stability is
or is not maintained and determine the maximum power
angle. The inertia constant of the generating unit is 3 per
unit-seconds on the system base. Assume Pm = 1 p.u and
remains constant throughout the disturbance. o = 23.95o =
0.4179 radian and f = 60 Hz. Also assume the infinite bus
receive 1 p.u real power at 0.95 p.f lagging.

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SOLUTION:
Plots of Pe and Pm versus  are shown in figure below.

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When the short circuit occurs, Pe instantaneously drops to
zero and remains at zero during the fault since power
cannot be transferred past faulted bus 1.

At t=3 cycles = 3/60 = 0.05 second,

Integrating twice with initial condition

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The accelerating area A1, shaded in figure is

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At t = 0.05 s the fault extinguishes and Pe instantaneously
increases from zero to the sinusoidal curve in figure.
 continues to increase until the decelerating area A2 equals
A1. That is,

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The nonlinear algebraic equation can be solved iteratively to
obtain

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Solution:
𝑃𝑒 = 𝑃𝑚𝑎𝑥 sin(𝛿)

𝑃𝑚𝑒𝑐 = 𝑃𝑒 𝑎𝑡 𝛿𝑜

1 = 𝑃𝑚𝑎𝑥 sin(23.95)

𝑃𝑚𝑎𝑥 = 2.4638 pu
𝛿𝑚𝑎𝑥 = π − 23.95 = 156.050 = 2.7222 𝑟𝑎𝑑

1
𝛿𝑐 = 𝑐𝑜𝑠 −1 [ (2.7222-0.4179)+cos(156.05)]
2.4638

𝛿𝑐 = 88.776𝑜 𝑜𝑟 1.5494 𝑟𝑎𝑑

2∗3∗(1.5494−0.4179)
𝑡𝑐 = = 0.19 sec
𝜋∗60∗1
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 3
𝑡(3 𝑐𝑦𝑐𝑙𝑒) = = 0.05 𝑠𝑒𝑐
60

∴ 𝑡𝑐 > 𝑡(3 𝑐𝑦𝑐𝑙𝑒)

Stability is maintained

Note: after 12 cycle , stability is losing from this system

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 Transient Stability Analysis

 For transient stability analysis we need to

consider three systems
1. Pre fault - before the fault occurs the
system is assumed to be at an equilibrium
point
2. During faulted - the fault changes the
system equations, moving the system away
from its equilibrium point
3. Post fault - after fault is cleared the system
hopefully returns to a new operating point
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Example
A 60 Hz generator is supplying 550 MW to an
infinite bus (with 1.0 per unit voltage and unity p.f)
through two parallel transmission lines. Determine
the critical clearing time for a fault midway down
one of the lines. H = 20 seconds.

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Solution

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 Further Considerations of the Swing Equation

In a stability study of a power system with many synchronous

machines only one MVA base common to all parts of the
system can be chosen. Since the right-hand side of the swing
equation for each machine must be expressed in per unit on
this common system base, it is clear that H on the left-hand
side of each swing equation must also be consistent with the
system base.
𝐻𝑠𝑦𝑠𝑡𝑒𝑚
𝑆𝑚𝑎𝑐ℎ
= 𝐻𝑚𝑎𝑐ℎ
𝑆𝑠𝑦𝑠𝑡𝑒𝑚

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In a stability study for a large system with many machines
geographically dispersed over a wide area it is desirable to
minimize the number of swing equations to be solved.

Machines which swing together are called coherent

machines.
Consider a power plant with two generators connected to the
same bus which is electrically remote from the network
disturbances. The swing equations on the common system
base are

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Adding the equations together and denoting 1 and 2 by 
since the rotors swing together, we obtain

where H = ( H1 + H2 ), Pm = ( Pm1 + Pm2 ), and Pe = ( Pe1 + Pe2).

This fact is often used in stability studies involving many
machines in order to reduce the number of swing equations
which need to be solved.

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For any pair of noncoherent machines in a system swing
equations

Multiplying each side by H1 H2 / ( H1 + H2 ) and rearranging,

we find that

which also may be written more simply in the form of the

basic swing equation as follows:

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where δ12 = δ1 - δ2

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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 pure reactances. Whatever change occurs in the
generator output is thus absorbed by the motor, and we can
write

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Under these conditions, Pm12 = Pm and Pel2 = Pe , so
equation reduce to

Let us assume that the transmission line is loss less. Then in

the simple case where the power from machine 1 flows to
machine 2, we get

Now since the system is lossless, will also imply that Pm1 = -
Pm2 . This means that in the steady state, the power
generated at machine 1 is absorbed through machine 2, we
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Let us now assume that H1 = H2 = H , V1 = V2 = 1.0 per unit
and Pm1 = 0. We then get from

where the oscillation frequency ω is given by

Thus the weighted difference of angles will approximate
simple harmonic motion for small changes in δ12 and the
frequency will decrease for an increase in inertia H or
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Design methods for improving transient stability
There are a number of design measures that can be
implemented to improve the transient stability of a power system:

1. Improve the maximum power transfer capability

Improving the maximum power transfer capability of a system means
that power can be transferred through alternative un-faulted portions of the
network when a fault occurs. The effect of a fault on the system will not be as
extreme. The maximum transfer capability of a system can be improved by the
following methods:
a. Implement and use higher system voltage levels (system losses will
decrease as current flows will be lower, especially important in cases
where line distances are large)
b. Install additional transmission lines.
c. Install lines and transformers with smaller reactance values
d. Install series capacitive transmission line compensation to reduce the
overall reactance of lines
e. Install static VAR compensators and flexible AC transmission systems
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2. Implement high speed fault clearing
It is vital to clear faults before the critical clearing time is reached
so the quicker a fault is cleared the better.

3. Implement high speed re-closure of circuit breakers

As the majority of transmission line short circuits are temporary,
re-closure post fault can be beneficial in providing better power transfer
capability. Care must be taken in this case to ensure that the re-closing
on a permanent fault and any subsequent re-opening will not adversely
affect the stability of the system.
4. Implement single pole switching
The majority of short circuits are single line to ground and the
independent switch out of only the faulted phase means that some
power flow can continue across the faulted line. Studies have shown
that single line to ground faults are self-clearing even when only the
faulted phase is de-energized.
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5. Use generators with larger machine inertia and with
lower transient reactance
A larger synchronous machine inertia constant (H) results in a
reduction in angular acceleration and therefore a slowing down of
angular swings. The critical clearing time is increased.
Reducing the machine transient reactance increases the power transfer
capability during faults and in the periods post fault.

6. Use fast responding, high gain exciters

Modern excitation systems can be designed to act quickly and
with high gain in the event of sensing a low terminal voltage during
faults. The effect is to increase the generator output during the fault and
post fault periods. Critical clearing times are increased.

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7. Implement fast valve
Some steam turbines are equipped with fast valve to divert
steam flows and rapidly decrease the mechanical output. When a fault
occurs near to the generator the electrical power output is reduced and
the fast valve acts to balance the mechanical and electrical powers. This
provides reduced acceleration and longer critical clearing times.

8. Breaking Resistors
In power systems, areas of generation can be temporarily
separated from the load areas. When the separation occurs the
breaking resister can be inserted into the generation area for a second
or two in order to slow the acceleration.

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