POWER SYSTEM

STABILITY
Getachew Biru (Dr.-Ing.)
 Introduction
• Power System Stability is the ability of the system to
bring back its operation to steady state condition
within a minimum possible time after having
undergone any transience or disturbance.
• The stability limit defines the maximum power
permissible to flow through a particular part of the
system for which it is subjected to line disturbances.
 Introduction

• The power system stability or synchronous stability

of a power system can be of several types
depending upon the nature of the disturbance, it
can be classified into the following three types as
shown below:
 Introduction
• Steady State Stability: It refers to the ability of the
system to regain its synchronism (speed &
frequency of the network) after slow and small
disturbance which occurs due to gradual power
changes.
 Introduction
• Transient Stability: Transient Stability is the ability
of the power system to maintain its stability after
large, major and sudden disturbances. For example,
occurrence of faults, sudden load changes, loss of
generating unit, line switching.
 Introduction

• Dynamic stability refers to the power system's

ability to maintain operational stability following a
small or large disturbance with the aid of automatic
regulation and control devices.
 Power Angle Curve
• Consider a synchronous machine connected to an
infinite bus through a transmission line of reactance Xl
shown in a figure below. Let us assume that the
resistance and capacitance are neglected.
 Power Angle Curve
Equivalent diagram of synchronous machine
connected to an infinite bus through a transmission
line of series reactance Xl is shown below:

V = V<0⁰ – voltage of infinite bus

E = E<δ – voltage behind the direct axis synchronous
reactance of the machine.
Xd = synchronous reactance of the machine
 Power Angle Curve
• The complex power delivered by the generator to the
system is S= VI*
• The armature current (V= V < 0. 𝐸ത = 𝐸 < 𝛿) E

ത 𝑉
𝐸− ഥ 𝐸 cos 𝛿−𝑉+𝑗 E sin 𝛿 IX
• I= = 
𝑗𝑋 𝑗𝑋
• The apparent power V
𝐸 cos 𝛿−𝑉−𝑗 E sin 𝛿
S= P+j Q = V I*= 𝑉 −𝑗𝑋
𝑉 𝐸 sin 𝛿 𝑉 (𝐸 cos 𝛿− 𝑉) 𝑉 𝐸 sin 𝛿 𝐸 𝑉 cos 𝛿− 𝑉 2
S= +𝑗 = + 𝑗
𝑋 𝑋 𝑋 𝑋

𝑉 𝐸 sin 𝛿
P=
𝑋
𝐸 𝑉 cos 𝛿− 𝑉 2
Q=
𝑋
 Power Angle Curve
• Active power transferred to the system

• The reactive power transferred to the system

 Power Angle Curve
• The maximum steady-state power transfers occur
when δ = 0
 Power Angle Curve
• The graphical representation of Pe and the load
angle δ is called the power angle curve.
 Power Angle Curve

Steady state stability

 Power Angle Curve

• Conditions for Steady State Stability:

 Power Angle Curve
• Maximum power transfer is given by

• To increase the steady state stability limit.

➢ A system can never be operated at higher than its
steady state stability limit.
➢ By reducing the X (reactance) or by raising the |E|
or by increasing the |V|, the improvement of
steady state stability limit of the system is possible.
➢ Stability limit is increased by quick excitation
voltage control.
➢ To reduce the X in the transmission, parallel line
are employed.
 Power Angle Curve
Example of Steady State Instability
Assume that the system load starts to grow. As the system
load grows, system voltages decline. Since the generator is
in manual excitation the generator terminal voltage also falls.
 Power Angle Curve
Example of Transient Stability
 Power Angle Curve
Transient Stability: The
transient environment is
characterized by a
system that undergoes a
sudden, severe
disturbance. In contrast
to the steady state
environment where
changes occur gradually,
the transient
environment involves
rapid changes.
 Power Angle Curve
Transient Stability
 Power Angle Curve
Transient Stability
• Immediately following the line opening the operating
point shifted from point “A” on the pre-disturbance
curve to point “B” on the post disturbance curve. This
movement represented an immediate reduction in
active power transfer due to the opening of the line.
• The angle spread then increased from “B” through
“C” and on to “D”. The angle spread stopped
increasing at “D”.
• As the operating point moved, there were two areas
created bounded by the points “A-B-C” and “C-D-E”.
These two areas are shaded and labeled “1” and “2”
in the Figure.
 DYNAMICS OF A SYNCHRONOUS MACHINE
• The kinetic energy of the rotor at synchronous
machine is

• where, J = rotor moment of inertia in kg-m2

• sm = synchronous speed in rad (mech)/s

• P = number of machine poles

Angular momentum in MJ-s/elect rad

DYNAMICS OF A SYNCHRONOUS MACHINE
• The inertia constant H is defined as follows:

where, G = machine rating (base) in MVA (3-phase)

H = inertia constant in MJ/MVA or MW-s/MVA: defined as kinetic energy
per apparent power unit

M is called the angular momentum

Taking G as base, the inertia constant in pu is
 DYNAMICS OF A SYNCHRONOUS MACHINE
• The inertia constant H has a characteristic value or
a range of values for each class of machines.
• H varies from 1-10 depending on the type of
generators.
• The value of H is considerably higher for steam
turbo-generator than for water wheel generator.
DYNAMICS OF A SYNCHRONOUS MACHINE
The Swing Equation

• The differential equation governing the rotor

dynamics can be written as:
DYNAMICS OF A SYNCHRONOUS MACHINE
The Swing Equation
• The Equation can therefore be converted into its
more convenient power form by assuming the
rotor speed to remain constant at the synchronous
speed (ωsm)
• Multiplying both sides of Eq.
DYNAMICS OF A SYNCHRONOUS MACHINE
The Swing Equation

• Rewriting the Eq.

DYNAMICS OF A SYNCHRONOUS MACHINE
The Swing Equation
• It is more convenient to measure the angular
position of the rotor with respect to a
synchronously rotating frame of reference. Let

rotor angular displacement from synchronously

rotating reference frame (called torque angle or
power angle)
DYNAMICS OF A SYNCHRONOUS MACHINE
The Swing Equation
• Hence Eq. can be written in terms of  as
the swing equation

• Dividing throughout by G, the MVA rating of the

machine,

in pu of machine rating as base

DYNAMICS OF A SYNCHRONOUS MACHINE

The Swing Equation

• The swing equation can then be written as

is a non-linear second-order differential

equation with no damping
 Multi-machine System
The Swing Equation
• In a multi-machine system a common system base
must be chosen.
• Equation can then be written as
 Multi-machine System
Machines Swinging Coherently
• Consider the swing equations of two machines on
a common system base.

• Since the machine rotors swing together

(coherently or in unison)
• Adding the above Eqs
where
 Multi-machine System
Machines Swinging Coherently
• The equivalent inertia

• The above results are easily extendable to any

number of machines swinging coherently.
 Multi-machine System
Exercise
• Two 50 Hz generating units operate in parallel
within the same plant, with the following ratings:
• Unit 1: 500 MVA, PF=0.8, 13.2 kV, 3600rpm, H=4MJ/MVA
• Unit : 1000 MVA, PF=0.9, 13.2 kV, 1800rpm, H=5MJ/MVA
• Calculate the equivalent H constant
 Equal Area Criterion
• The equal area criterion is a simple graphical
method for concluding the transient stability of
two-machine systems or a single machine against
an infinite bus.
• Starting with swing equation
 Equal Area Criterion
• Over a lossless line, the real power transmitted will
be

• Consider a fault occurs in a synchronous machine

which was operating in steady state. Here, the
power delivered is given by

• For clearing a fault, the circuit breaker in the

faulted section should have to be opened up. This
process takes 5/6 cycles and the successive post-
fault transient will take an additional few cycles.
 Equal Area Criterion
• The prime mover which is giving the input power
is driven with a turbine. For turbine mass system,
the time constant is in the order of few seconds and
for the electrical system, it is in milliseconds.
• Thus, while the electric transients take place, the
mechanical power remains stable.
• The transient study mainly looks into the capability
of the power system to retrieve from the fault and
to give the stable power with a new probable load
angle (δ).
Equal Area Criterion
 Equal Area Criterion
• The power angle curve is considered which is
shown in the fig.
• Imagine a system delivering ‘Pm’ power on an
angle of δo is working in a steady state. When a
fault occurs; the circuit breakers opened and the
real power is decreased to zero. But the Pm will be
stable. As a result, accelerating power,
 Equal Area Criterion
• The power differences will result in rate of change
of kinetic energy stored within the rotor masses.
Therefore, due to the accelerating power, the rotor
will accelerate. Consequently, the load angle (δ)
will increase.
 Equal Area Criterion
• Now, we can consider an angle δc at which the
circuit breaker re-closes. The power will then come
back to the usual operating curve. At this moment,
the electrical power will be higher than the
mechanical power. But, the accelerating power (Pa)
will be negative.
• Therefore, the machine will get decelerate. The load
power angle will still continue to increase because
of the inertia in the rotor masses.
• This increase in load power angle will stop in due
course and rotor of the machine will start to
decelerate or else the synchronization of the system
will get lose.
Equal Area Criterion
 Equal Area Criterion
• The equal area criterion is a simple graphical
method for concluding the transient stability of
two-machine systems or a single machine against
an infinite bus.
• Starting with swing equation

• where, M = Angular Momentum

Pe = Electrical Power
Pm = Mechanical Power (Pm)
Pa=accelerating power
δm= Load Angle
 Equal Area Criterion
• From the above equation,

• Multiplying both side by

• Integrating both side,

 Equal Area Criterion
• OR

• The above Eq. gives the relative speed of the

machine with respect to the synchronously
revolving reference frame. For stability, this speed
must be zero at sometime after the disturbance.
Therefore the stability criterion,
 Equal Area Criterion
• Consider the machine operating at the equilibrium
point δ0, corresponding to the mechanical power
input Pmo=Pe0 as shown in figure.
• Consider a sudden increase in input power
represented by a horizontal line Pm. Since PmoPe0 ,
the acceleration power on the rotor is positive and
the power angle increases.
• The energy stored in the rotor during the initial
acceleration is,
Equal Area Criterion
 Equal Area Criterion
• The area A1 represents the kinetic energy stored by
the rotor during acceleration, and A2 represents the
kinetic energy given up by the rotor to the system,
and when it is all given up, the machine has
returned to its original speed.
• The area under the curve should be zero, which is
possible only when PA has both accelerating and
decelerating powers, i.e., for a part of the curve Pm>
Pe and for the other Pe> Pm.
• For a generation action, Pm> PE for the positive area
and Pe> Pm for negative areas A2 for stable
operation. Hence the name equal area criterion.
 Equal Area Criterion
• With increase in, the electrical power increases and
δ = δ1, the electrical power matches the new input
power Pm1. Even though the accelerating power is
zero at this point, the rotor is running above
synchronous speed; hence δ and the electrical
power Pe continue to increase.
• Now Pm<Pe, causing the rotor decelerates toward
synchronous speed until δ = δmax.
• The energy given as the rotor decelerates back to
synchronous speed is,
 Equal Area Criterion
• The result is that the rotor swings to point b and
the angle δmax, at which point

• This is known as the equal area criterion. The rotor

angle would then oscillate back and forth between
δ0 and δmax at its natural frequency.
• The damping present in the machine will cause
these oscillations to subside and the new steady
state operation would be established at point b.
 Critical Clearing Angle
• The critical clearing angle is defined as the
maximum change in the load angle curve before
clearing the fault without loss of synchronism.
• In other words, when the fault occurs in the system
the load angle curve begin to increase, and the
system becomes unstable.
• The maximum angle at which the fault can be
cleared and the system becomes stable is called
critical clearing angle.
 Critical Clearing Angle
• Consider a simple case
– A three-phase fault at the sending end
– Pe, during fault=0 if all resistances are neglected
 Critical Clearing Angle
• Solve Critical Clearing Angle

• Integrating both sides:

 Critical Clearing Angle
• Solve the CCT from the CCA:
• During the fault (Pe, during fault=0):

c
DYNAMICS OF A SYNCHRONOUS MACHINE
The Swing Equation

With damper winding:

Dm is the damping coefficient