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MODULE – I
POWER SYSTEM REPRESENTATION
Main components of power system
Generation, transmission and distribution are the main components of power system. The
equivalent circuits of all the components of the power system are drawn and they are
interconnected to form equivalent circuit model of power system. If the shunt arm of transformer
is omitted from equivalent circuit, the resulting diagram is called impedance diagram model of
power system. If we omit static loads, resistance, shunt arm of each transformer and line
charging capacitance of the transmission line from the equivalent circuit model, the resulting
diagram is called reactance diagram model of power system. A simplified diagram by omitting
the completed circuit through the neutral and by indicating the components of the power system
by standard symbols rather than by their equivalent circuit is used for representing a power
system. This is called single line diagram model.
Advantages and disadvantages of interconnected power system
Advantages:
1. Less no. of generators are required as reserve for operation at peak loads. Hence the
reserve capacity of the generating station gets reduced.
2. Less no. of generators which are running without load are required for meeting the
sudden unexpected increase in load.
3. It allows the use of most economical sources of power depending on time.
Disadvantages:
1. It increase the amount of current which flows when a short circuit occurs on a system and
thereby requires the installation of breakers which are able to interrupt a larger current.
2. Synchronism must be maintained between of all the interconnected systems.
Single Line Diagram
One-line diagram or single-line diagram (SLD) is a simplified notation for representing a
three-phase power system. The one-line diagram has its largest application in power flow
studies. Electrical elements such as circuit breakers, transformers, capacitors, bus bars, and
conductors are shown by standardized schematic symbols. Instead of representing each of
three phases with a separate line or terminal, only one conductor is represented. It is a form
of block diagram graphically depicting the paths for power flow between entities of the
system. Elements on the diagram do not represent the physical size or location of the
electrical equipment, but it is a common convention to organize the diagram with the same
left-to-right, top-to-bottom sequence as the switchgear or other apparatus represented.
Symbols used in SLD
SLD of a typical power system
Need for base values
The components of power system may operate at different voltage and power levels. It
will be convenient for analysis of power system if the voltage, power, current ratings of
the components of the power system is expressed with reference to a common value
called base value. The base values of voltage, current, power and impedance are required
to represent the power system by reactance diagram. Selection of base values for any two
of them determines the base values of the remaining two. Normally base MVA is selected
for the entire power system and any one side voltage of power system can be selected as
base KV and the base KV for remaining sides can be calculated using transformation
ratio of the transformers. Base value for current is base KVA divided by base KV. Base
value for impedance is base voltage divided by base current.
Per unit representation
Per unit of any quantity is defined as the ratio of the quantity to its base value and it is
expressed as a decimal.
𝑃𝑒𝑟 𝑢𝑛𝑖𝑡 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝐵𝑎𝑠𝑒 𝑣𝑎𝑙𝑢𝑒
Advantages of per unit representation
(1) The per unit impedance referred to either side of a single phase transformer is the
same.
(2) The per unit impedance referred to either side of a three phase transformer is the
same regardless of the three phase connections whether they are Y-Y, Δ-Δ or Δ-Y
(3) The chance of confusion between the line and phase quantities in a three phase
balanced system is greatly reduced.
(4) The manufacturers usually provide the impedance values in per unit.
(5) The computational effort in power system is very much reduced with the use of
per unit quantities.
Base current, Ib and base impedance of a three phase system
The equation for base current Ib is,
The equation for base impedance is,
Also,
Where,
Ib
= Line value of base current.
kVAb = 3-phase base KVA
kVb
= line to line base kV
Zb
= Base impedance per phase.
Equation for converting the per unit impedance expressed in one base to another
Equation for transforming base kV on LV side to HV side of a transformer and vice versa
Impedance diagram:
Approximations

The neutral reactance is neglected.

The shunt branches in equivalent circuit of induction motor are neglected.
Reactance diagram:
Approximations

The neutral reactance is neglected.

The resistances are neglected.

All static loads and induction motors are neglected.

Shunt arms of transformer and transmission line are neglected.
Load impedance and load admittance per phase of a balanced star connected load
Where,
P = Three phase active power of star connected load in watts.
Q = Three phase reactive power of star connected load in VARs.
VL = Line voltage of load
Load impedance and load admittance per phase of a balanced delta connected load
Where,
P = Three phase active power of delta connected load in watts.
Q = Three phase reactive power of delta connected load in VARs.
VL = Line voltage of load.
Bus admittance matrix (Y-bus)
The matrix consisting of the self and mutual admittance of the power system network is
called bus admittance matrix. It is given by the admittance matrix Y in the node basis matrix
equation of a power system and it is denoted as Ybus. The diagonal elements of bus admittance
matrix are called self admittances of the buses and off diagonal elements are called mutual
admittances of the buses. Bus admittance matrix is a symmetrical matrix.
The equation for bus admittance matrix is,
YbusV = I
where
Ybus = Bus admittance matrix of order (n x n )
V = Bus voltage matrix of order (n x1)
I = Current source matrix of order (n x1)
n = Number of independent buses in the system
Advantages of bus admittance matrix, Ybus.
i) Data preparation is simple.
ii) Formation and modification is easy.
iii) Since the bus admittance matrix is sparse matrix (i.e., most of its elements are zero), the
computer memory requirements are less
Z-bus
The methods to determine the Y-bus and Z-bus matrices are primitive network, network
graph theory and incidence matrix. Y-bus can be determined by inspection method, if there is no
mutual coupling between lines and Z-bus can be formed either by inverting Y-bus or by using
bus building algorithm which has the following steps.
1. Adding a branch of impedances from a new bus p to the reference bus.
2. Adding a branch of impedance Zb from a new bus-p to an existing bus.
3. Adding a branch of impedance Zb from an existing bus-q to the reference bus.
4. Adding a branch of impedance Zb between two existing buses h and q.
Primitive network
Primitive network is a set of unconnected elements which provides information regarding
the characteristics of individual elements only. Graph shows the geometrical interconnection of
the elements of a network. A sub graph is any subset of elements of a graph. Path is a sub graph
of connected elements with not more than two elements connected to any one side. A graph is
connected if and only if there is a path b/n every pairs of nodes. If each element of the connected
graph is assigned a direction it is called oriented graph. Tree is a sub graph connecting all the
nodes of the oriented graph. Tree is a connected sub graph. A cutest is the minimum set of
elements in the graph, which when removed, divides a connected graph into two connected sub
graph.
The performance equations of primitive network are given below.
V + E = ZI (In Impedance form)
I + J = YV (In Admittance form)
where V and I are the element voltage and current vectors respectively.
J and E are source vectors.
Z and Y are the primitive Impedance and Admittance matrices respectively.
Symmetrical components
An unbalanced system of N related vectors can be resolved into N systems of balanced
vectors. The N sets of balanced vectors are called symmetrical components.
1. Positive sequence components
2. Negative sequence components.
3. Zero sequence components.
The positive sequence components of a three phase unbalanced vectors consists of three vectors
of equal magnitude, displaced from each other by 120 in phase and having the same phase
sequence as the original vectors.
The negative sequence components of a three phase unbalanced vectors consists of three vectors
of equal magnitude displaced from each other by 120 degree in phase and having the phase
sequence opposite to that of the original vectors.
The zero sequence components of a three phase unbalanced vectors consists of 3 vectors of equal
magnitude and with zero phase displacement from each other.
Sequence impedance and Sequence networks
The sequence impedances are the impedances offered by the devices for the like
sequence component of the current. The single phase equivalent circuit of a power system
consists of impedances to current of any one sequence is called sequence network. The
impedance of the circuit element for positive, negative and zero sequence component currents
are called positive, negative and zero sequence impedances respectively. The reactance diagram
of a power system, when formed using positive, negative and zero sequence reactance’s are
called positive, negative and zero sequence reactance diagram respectively.
Types of busses in power system
The meeting point of various components in a power system is called a bus. The bus is a
conductor made of copper having negligible resistance. The buses are considered as points of
constant voltage in a system. Each bus in a power system is associated with four quantities ie.
real power, reactive power, magnitude of voltage, and phase angle of voltage. The different types
of buses are
1) Load bus
2) Generator bus
3) Slack bus.
A generator bus is called voltage controlled bus as the magnitude of voltage and real
power are specified for it. In a voltage controlled bus the magnitude of the voltage is not allowed
to change and it is also called PV bus.
A load bus is called PQ bus because real and reactive components of power are specified
for the bus. In a load bus the voltage is allowed to vary within permissible limits.
A slack bus is called swing bus as the magnitude and phase of the bus voltage are
specified for it. The swing bus is the reference bus for load flow solution and it is required for
accounting line losses. Usually one of the generator buses is selected as swing bus. The phase
angle of the slack bus voltage is zero.
Need for slack bus
A slack bus is called swing bus as the magnitude and phase of the bus voltage are
specified for it. The swing bus is the reference bus for load flow solution and it is required for
accounting line losses. Usually one of the generator buses is selected as swing bus. The phase
angle of the slack bus voltage is zero.
The slack bus is needed to account for transmission line losses. In a power system the
Total power generated will be equal to sum of power consumed by loads and losses. In a power
system only the generated power and load power are specified for buses. The slack bus is
assumed to generate the power required for losses. Since the losses are unknown the real and
reactive power are not specified for slack bus.
Load flow analysis
The study of various methods of solution to power system network is referred to as load flow
study. The solution provides the voltages at various buses, power flowing in various lines and
line losses. The information obtained from a load flow study is the magnitude and phase of bus
voltages, real and reactive power flowing in each line and the line losses. The load flow solution
also gives the initial conditions of the system when the transient behavior of the system to be
studied. The load flow study of a power system is essential to decide the best operation existing
system and for planning the future expansion of the system. It is also essential for designing the
power system. The operating constraints imposed in load flow studies are reactive power limits
for generator buses and allowable change in magnitude of voltage for load buses. The Gauss
seidal method and Newton Raphson method are the two iterative methods used for load flow
analysis.
The following has to be performed for a load flow study.
a. Representation of the system by single line diagram.
b. Formation of impedance diagram using the information in single line diagram.
c. Formulation of network equations
d. Solution of network equations
Guass seidal method
The load flow equation of Gauss method is given by
Advantages of Guass seidal method

Calculations are simple and so the programming task is less.

The memory requirement is less.

This method is applicable for only small power system with less number of busses.

Computations per iteration are less.
Disadvantages of Gauss seidal method

Requires large number of iterations to reach converge.

Not suitable for large systems.

Convergence time increases with size of the system
Load flow equation of Newton-Raphson method
Jacobian matrix
The matrix formed from the first derivatives of load flow equations is called Jacobian
matrix. The elements of Jacobian matrix will change in all iterations. The elements of the
jacobian matrix are obtained by partially differentiating the load flow equations with respect to
unknown variable and then evaluating the first derivatives using the solution of previous
iteration.
Advantages in Newton Raphson method
In N-R method, the set of nonlinear simultaneous equations are approximated to a set of linear
simultaneous equations using tailors series expansion and the terms are limited to first order
approximation.

The N-R method is faster, more reliable and the results are accurate.

Requires less no. of iterations for convergence.

Suitable for large size system

Number of iterations does not depend on the size of the system.
Disadvantages of N-R method

The programming is more complex.

The memory requirement is more.

The number of computations per iteration is more.

Requires inversion of large Jacobian matrix in every iterations.
Comparison of G-S method and N-R methods
MODULE – II
ECONOMIC LOAD DISPATCH
Economic dispatch
Economic dispatch is the short-term determination of the optimal output of a number of
electricity generation facilities, to meet the system load, at the lowest possible cost, while
serving power to the public in a robust and reliable manner. The Economic Dispatch
Problem is solved by specialized computer software which should honour the operational
and system constraints of the available resources and corresponding transmission
capabilities. Economic dispatch has online economic load dispatch and unit commitment.
The optimum allocation of generating units at each generating station at various station
load levels is termed as unit commitment. The online determination of the amount of
power that has to be generated by the committed units to meet the instantaneous demand
at minimum production cost is online economic load dispatch.
Constraints to be considered for economic dispatch
Equality constraint: PG = Ploss + PD and QG = Qloss + QD
Inequality constraint:

Generator constrains which is the maximum and minimum real and reactive
power generation limit.

Voltage constrain which is the maximum and minimum allowable voltage
magnitude and angle.

Running spare capacity constrain which gives the reserve capacity needed to meet
outages and unexpected load demands.

Transformer tap setting constrain which is the maximum and minimum tap ratio.

Transmission line constrain which is the maximum transmission capacity limited
by thermal capacity of line.

Network security constrain which indicates that all constrains should be satisfied
without outages and with one outage.
Input-output curve, heat rate curve and incremental fuel cost curve
Input-output curve is the curve drawn between input energy rate in MKal/hr or fuel cost
in Rs/hr and output power in MW.
Heat rate curve is the curve drawn between heat rate in MKcal/MWH and output power
in MW.
Incremental fuel cost curve is the curve drawn between incremental fuel cost in Rs/hr and
output power in MW. Incremental fuel cost is the first order differential of cost function
with respect to PG.
Heat rate function, fuel input function, cost function and incremental fuel cost function
Heat rate function, Hi(PGi) = Ci1/PGi + bi1 + ai1PGi in MKcal/MWH
Fuel input function is got by multiplying heat rate function by P Gi
Fi(PGi) = Ci1 + bi1 PGi + ai1PGi2 in MKcal/hr
Cost function is got by multiplying fuel input function by cost of fuel (K) in Rs/MKcal.
Ci(PGi) = KCi1 +K bi1 PGi + Kai1PGi2 = Ci + bi PGi + ai PGi2 in Rs./hr
Incremental fuel cost is the first order differential of cost function with respect to PG.
ICi = 2aiPGi + bi in Rs./MWH
Also, PGi = αi + βi(ICi) + γi(ICi)2 in MW
Automatic load dispatching
The economic load dispatch means to distribute the load among the generating units actually
paralleled with the system in such a manner as to minimize the cost of supplying the minute to
minute requirements of the system. In large interconnected system, it is humanly impossible to
calculate and adjust such generation and hence the help of digital computer system along with
analogue devices is sought and the whole process is carried out automatically , hence the term
automatic load dispatching is used for the purpose.
The block diagram of an automatic control system used for load dispatching is shown in figure.
Operator
Station
operator
Computer
Console
Machine
Control
Genr.
Controller
Computer: The computer predicts the load and suggests economic loading. It transmits
information to machine controllers.
Data input: The computer receives a lot of essential data from the telemetering system
and
from the paper tape. Telemetering data comes to the computer either as analogue signals
representing line power flows, plant outputs or as signal bits indicating switch or isolator
positions. The system is entirely automatic, paper tape stores all the basic data required, and
finally asking the station to switch the machine controllers to automatic control.
Console: The console is the component through which the operator can converse with the
computer. He can obtain certain information required for some action to be taken under
emergency condition should such a condition arise in the system or he can put data into into it if
needed. the instruction being issued and state of energy system can be examined on cathode ray
display associated with the console.
The console has the facilities of security checking and load flows for the network calculations.
such a facility can be called manually at almost any instant of time and the result at least one of
the remaining branches will be overloaded.
Machine controller: The computer sends instruction regarding the optimal generation to the
machine controller at regular interval which in turn implements them. control on each machine is
applied by a closed loop system which uses a measure of action power generated and may be
recorded automatically after a specified interval of time, if required. The security check reveals
and displays that if one or more than one particular branches suffer outage, which operates
through the conventional speeder motor. These are referred to as controller power loops.
In the power frequency loop an error signal proportional to the difference between the derived
and actual frequency and power is developed. A summed error signal is formed from these two
components and is converted in the motor controller to a frame of pulses that are applied to a
speed governor reference setting motor called the speeder motor. the duration and amplitude of
these pulses are fixed but the pulse rate is made proportional to the summed error signal. The
pulses are applied as raise or lower command to the speeder motor in accordance with the error
signal and thus the output of the generators is increased or decreased accordingly.
Economic load dispatch without losses
The economic dispatch problem is defined as
n
Min FT =
 Fn
n
subject to PD =
n 1
 Pn
n 1
Where FT is total fuel input to the system, Fn the fuel input to the nth unit, PD the total load
demand and Pn the generation of nth unit.
By making use of Lagrangian multiplier the auxiliary function is obtained as
F = FT +  (PD -
n
P
n 1
n
)
Where  is the Lagrangian multiplier.
Differentiating F with respect to the generation Pn and equating to zero gives the condition for
optimal operation of the system.∂
FT
F
=
+  (0-1) = 0
Pn Pn
FT
-  =0
Pn
=
Since FT = F1 +F2 +............+Fn

FT
F
=
= 
Pn
Pn
And therefore the condition for optimum operation is
dF1
dF2
dFn
=
=....................
= 
dPn
dP1
dP2
Here
dFn
= incremental production cost of plant n in Rs per MWhr.
dPn
The incremental production cost of a given plant over a limited range is represented by
dFn
= FnnPn + fn
dPn
fn = intercept of incremental production cost curve.
Fnn = slope of incremental production cost curve.
Economic load dispatch including transmission losses
When the energy is to be transported over large distance, transmission losses in some cases may
amount to 20 to 30% of the total load & it then become necessary to take these losses also into
account, when developing an economic dispatch strategy.
consider two generators having equal incremental production cost. If generator 2 has a local load,
according to equal incremental production cost criteria total load must be shared by the 2, but
commonsense tells that it is more economical to let generator 2 supply most of the local load
because generator 1 has to supply in addition to the load, the transmission losses also. So the
equal incremental criteria not hold good under such situation, here we have to consider the
transmission losses also.
The optimal load dispatch problem including transmission losses is defined as
n
Min FT =
F
n 1
n
n
Subject to PD +PL -
P
n 1
n
=0
Where PL is the total system loss which is assumed to be a function of generation and the other
term have their usual significance.
Making use of Lagrangian multiplier  , the auxiliary function is given by
F= FT +  (PD+PL-  Pn )
The partial differential of this expression when equated to zero gives the condition for
optimal load dispatch ie.
 P

F
F
= T +   L  1  0
Pn
Pn
 Pn

Or
dFn
P
+  L =  ................................(1)
dPn
Pn
PL
is known as the incremental cost of received power in Rs.per MWhr.
Pn
Here the term
The above equation is known as coordination equation because they coordinate the incremental
transmission losses with the incremental cost of production.
To solve these equations the loss formula equation is expressed in terms of generations and is
approximately expressed as
PL =
 PmBmnPn ..............................(2)
m
n
Where Pm and Pn are the source loadings, Bmn the transmission loss coefficients.
The solution of coordination equation
dFn
P
+  L =  requires the calculation of
dPn
Pn
PL
which is obtained from equation (2), as
Pn
PL
= 2  BmnPm ......................(3)
Pn
m
Also
dFn
= FnnPn +fn .....................(4)
dPn
 the coordination equations can be written as
FnnPn + fn +   2BmnPm = 
Collecting all coefficients of Pn we obtain


Pn(Fnn+2  Bnn) = -    2 BmnPm   fn  
 m n

Solving for Pn we obtain
Pn = 1-
fn

 2BmnPm
m n
Fnn

 2Bnn
.......................(5)
MODULE – III
AUTOMATIC GENERATION AND VOLTAGE CONTROL
Power systems are large and complex electrical networks. In any power system, generations are
located at few selected points and loads are distributed throughout the network. In the power
system, the system load keeps changing from time to time. The delivered power must meet
certain minimum requirements with regards to the quality of the supply. The following determine
the quality of the power supply:
i) The system frequency must be kept around the specified 50 Hz with a variation of ± 0.05
Hz.
ii) The magnitude of bus voltages are maintained within narrow prescribed limits around the
normal value. Generally voltage variation should be limited to ±5%.
Voltage and frequency controls are necessary for the effective operation of power systems.
Necessity of maintaining frequency constant:
BASIC CONTROL LOOPS [P- f and Q-V CONTROL LOOPS]
In order to perform voltage and frequency control, a basic generator will have two control loops
namely:
1) Automatic load frequency control loop(P-f control loop)
2) Automatic voltage regulator loop (Q-V control loop)
Fig. 1 shows the two control loops, AVR loop and ALFC loop.
1) Automatic load frequency control loop(P-f control loop)
The automatic load frequency control (ALFC) loop regulates the real power output of the
generator and its frequency (speed). This loop responds to a frequency (speed) changes via the
speed governor and the steam flow is regulated with the aim of matching the real power
generation to relatively fast load fluctuations that take place in one to several seconds.
The automatic load frequency control (ALFC) loop controls the magnitude of terminal
frequency, f. Terminal frequency is continuously sensed and rectified. The measured system
frequency f is compared with a rated reference frequency fref. The resulting “error frequency”,
serves as input to the steam valve controller which controls the required amount of steam to the
turbine generator set, so that the generator terminal frequency f reaches the value fref.
2) Automatic voltage regulator loop (Q-V control loop)
The automatic voltage regulator (AVR) loop controls the magnitude of terminal voltage, |V|.
Terminal voltage is continuously sensed, rectified and smoothed. The strength of this dc
signal, being proportional to |V|, is compared with a dc reference |V|ref. The resulting “error
voltage” after amplification and signal shaping, serves as input to the exciter, which applies
the required voltage to the generator field winding, so the generator terminal voltage |V|
reaches the value |V|ref.


AVR loop affects the magnitude of the generator emf E. As the internal emf determines
the magnitude of the real power, it is clear that changes in the AVR loop will be felt in
the ALFC loop.
AVR loop is much faster than the ALFC loop and hence AVR dynamics may settle
before they can make themselves felt in the slower load-frequency control channel.
 LOAD FREQUENCY CONTROL (SINGLE AREA):
The control method used to maintain the system frequency constant is commonly known as
AUTOMATIC LOAD FREQUENCY CONTROL (ALFC). The ALFC loop will maintain
control only during small and slow changes in load and frequency.
The basic role of ALFC is:
1. To maintain the desired megawatt output power of a generator matching with the changing
load.
2. To assist in controlling the frequency of larger interconnection.
3. To keep the net interchange power between neighboring areas, at the predetermined values.
 MODELING SINGLE TURBO-GENERATOR SYSTEM SUPPLYING POWER TO A
LOCAL SERVICE AREA:
To understand the load frequency control problem, consider a single turbo-generator system
supplying an isolated load. The study of ALFC is done by modeling the single turbo-generator
system supplying power to a local service area. It involves the model of the three main parts
given below:
1. MODEL OF SPEED GOVERNING SYSTEM
2. TURBINE MODEL
3. GENERATOR LOAD MODEL
1. SPEED GOVERNING SYSTEM:
The real power control mechanism of a generator is shown in Fig. 1. The main parts of the speed
governing system are:
1) Speed changer
2) Speed governor
3) Hydraulic amplifier
4) Linkage mechanism
Fig. 1 Flyball Speed Governing System
a) SPEED CHANGER:
- It provides a power output setting for the turbine
- Its upward movement opens the upper pilot valve, so that more steam is admitted to the
turbine, this give rise to higher power output
- The reverse happens if the speed changer moves downward
b) SPEED GOVERNOR (FLY BALL SPEED GOVERNOR):
- It is a purely mechanical, speed sensitive device coupled directly to the prime movers
(turbine) to adjust the control valve opening via. Linkage mechanism
- It senses a speed deviation or a power change command, and converts it into appropriate
valve action
c) HYDRAULIC AMPLIFIER:
- It is connected between the governor and the valve
- It consists of a pilot valve and the main piston
- In hydraulic amplification, a large mechanical force is necessary so that the steam valve
could be opened or closed against high- pressure inlet steam
- Using hydraulic amplifiers, hydraulic amplification is obtained by converting the movement
of low power pilot valve into movement of high power level main piston
d) LINKAGE MECHANISM:
- It transform the fly ball movement to the turbine valve(steam valve) through a hydraulic
amplifier
- ABC is a rigid link pivoted at point ‘B’ and CDE is another rigid link pivoted at point
‘D’
- With the help of linkage mechanism, the position of the pilot valve can be changed in the
following ways:
(i) Directly by the speed changer
(ii) Indirectly through feedback due to position changes in the linkage mechanism
resulting from a change in speed
MODEL OF SPEED GOVERNING SYSTEM:
Assume that the system is initially operating under steady conditions:
i.
Linkage mechanism is stationary
ii. Pilot valve is closed
iii. Steam valve is opened by a definite magnitude
iv.
Turbine is running at constant speed with turbine power output balancing the
generator load.
The movements are assumed positive in the directions of arrows. The generator output power
will increase due to the following two conditions:
a) Due to the “raise” command in the speed changerThe linkage movements will cause “A” to move downwards; “C” to move upwards; “D” to
move upwards; “E” to move downwards. This allows more steam or water flow into the
turbine resulting incremental increase in generator output power.
b) Due to decrease in speed of the generatorWhen the speed drops, linkage point “B” moves upwards and again generator output power
will increase.
This operation can be characterized by
f0 =system frequency (speed)
𝑃𝐺0 =generator output=turbine output (neglecting generator loss)
𝑦𝐸0 = steam valve setting
Due to a positive change (“raise” command) in the speed changer power setting, ΔP C, point A on
the linkage mechanism move downwards by a small amount Δ𝑦𝐴 . It is a command which causes
the turbine power output to change and can therefore be written as,
Δ𝒚𝑨 =kCΔ𝑷𝑪
(1)
Command ΔPC sets into motion a sequence of events,



point “C” to move upwards by Δ𝒚𝑪
point “D” to move upwards by Δ𝒚𝑫
point “E” to move downwards by Δ𝒚𝑬
These movements allow more steam or water flow into the turbine resulting incremental increase
in generator output power and causes frequency to increase. Let us model these events
mathematically,

Movement of point C by Δ𝒚𝑪 :
Two factors that contribute to movement of point C (Δ𝑦𝐶 ) are-
a) Δ𝑦𝐴 caused due to change in the speed changer power setting, ΔPC
Δ𝑦𝐶 ’= -k1Δ𝑦𝐴 =-k1kCΔ𝑃𝐶
(2)
Where k1=(l2/l1)
b) Δf caused due to change in the speed of the generator, as measured by Δ𝑦𝐵
Δ𝑦𝐶 ’’=k2Δf
(3)
Where k2=( l1 + l2 /l1)
The net movement of C, Δ𝑦𝐶 =Δ𝑦𝐶 ’+Δ𝑦𝐶 ’’
Δ𝒚𝑪 =-k1kCΔ𝑷𝑪 +k2Δf

(4)
Movement of point D by Δ𝒚𝑫 : It is the amount by which the pilot valve opens. It is
contributed by Δ𝑦𝐶 and Δ𝑦𝐸
Δ𝒚𝑫 =k3Δ𝒚𝑪 +k4Δ𝒚𝑬
(5)
where k3=l4 /( l3+ l4) and k4=l3 /( l3+ l4)
It is to be noted that for a positive Δ𝑦𝐶 , the change Δ𝑦𝐷 is positive.

Movement of point E by Δ𝒚𝑬 :The output of the hydraulic amplifier is Δ𝑦𝐸 . This
depends on the position of main piston, which in turn depends on the quantity of oil
flow in the piston. For a small change Δ𝑦𝐷 in the pilot valve position, we have
Δ𝑦𝐸 = -k5∫Δ𝑦𝐷 dt
(6)
The constant k5 depends on the orifice, cylinder geometries and fluid pressure.
For a positive Δ𝑦𝐸 more fuel is admitted, speed increases, linkage point B moves
downwards causing linkage points C and D to move downwards resulting the change
Δ𝑦𝐷 as negative.
Δ𝒚𝑬 =k5∫- Δ𝒚𝑫 dt
Thus,
(7)
Taking the Laplace transform of equations (4),(5) and (7), we get
Δ𝑌𝐶(𝑆) = −𝑘1 k C Δ𝑃𝐶(𝑆) + 𝑘2 Δ𝐹(𝑆)
(8)
Δ𝑌𝐷(𝑆) = 𝑘3 Δ𝑌𝐶(𝑆) + 𝑘4 Δ𝑌𝐸(𝑆)
(9)
Δ𝑌𝐸(𝑆) = −𝑘5
1
𝑆
Δ𝑌𝐷(𝑆)
(10)
Substituting equa.(8) and (10) in (9), we can write
Δ𝑌𝐸(𝑠) =
𝑘1 𝑘3 kC Δ𝑃𝐶(𝑆) − 𝑘2 𝑘3 Δ𝐹(𝑠)
(𝑘4 +
𝟏
𝑲𝒔𝒈
Thus, Δ𝒀𝑬(𝒔)= [ΔPC(s) - 𝑹 ΔF(s)] (𝟏+𝑻
𝒔𝒈 𝑺
where 𝑲𝒔𝒈=
R=
k1 kC
k2
k1 k3 kC
k4
(12)
𝑠
)
𝑘5
)
=gain of speed governor
= speed regulation of the governor
(13)
Tsg =k
𝟏
4 k5
=time constant of speed governor
Here the constant R has dimension hertz per MW and is referred as speed regulation
of the governor.
𝑲𝒔𝒈
𝟏 + 𝑻𝒔𝒈 𝑺
ΔPC(s)
Δ𝒀𝑬(𝒔)
-
𝟏
𝑹
ΔF(s)
2. TURBINE MODEL:
Turbine model is obtained by relating the dynamic response of a steam turbine in terms
of changes in power output (ΔPt)to changes in steam valve opening (Δ𝑦𝐸 ).The turbine
power increment ΔPt depends entirely upon the steam valve openingΔ𝑦𝐸 and the
characteristic of the turbine. Different type of turbines will have different
characteristics. Taking transfer function of a turbine with single time constant (𝑇𝑡 ), we
can write
𝑲
ΔPt(s) = (𝟏+𝑻𝒕 𝑺)Δ𝒀𝑬(𝒔)
𝒕
(14)
𝑲𝒕
𝟏 + 𝑻𝒕 𝑺
Δ𝒀𝑬(𝒔)
ΔPt(s)
3. GENERATOR LOAD MODEL:
In normal steady state, the turbine power Pt keeps balance with the generator power PG resulting
in zero acceleration and a constant speed and frequency.
Since the generator is supplying power to a load in its service area. We make these assumptions:
a) The system is originally running in its normal state with complete power balance, that is,
PG0 = PD0 + losses. If the losses are neglected
PG-PD=0
(15)
The frequency is at normal value f0. All rotating equipment represents a total kinetic energy
0
of 𝑊𝑘𝑒
in MW sec.
b) By connecting additional load, load demand increases by ΔPD. To meet the increment in load
demand the generation must immediately increase by ΔPG, so that ΔPG = ΔPD.
The increment in power input to the generator-load system is ∆PG-∆PD
This increment in power input to the system gets absorbed in two ways.
i)
By the change in the total kinetic K.E. ( 𝑖. 𝑒.
𝑑𝑊𝑘𝑒
𝑑𝑡
)
At scheduled frequency(𝑓 0 ) , the stored kinetic energy in the generator rotor is
0
Wke
= H x Pr MW
(16)
where Pr is the MW rating of the turbo-generator and H is its inertia constant
Since the Kinetic energy is proportional to the square of the speed(frequency), the kinetic
energy at s frequency of (𝑓 0 + ∆𝑓) is
0
𝑊𝑘𝑒 = 𝑊𝑘𝑒
(
𝑓0 +∆𝑓 2
)
𝑓0
MW sec
Neglecting the higher order
𝑊𝑘𝑒 = 𝐻 𝑃𝑟 (1 +
2∆𝑓
𝑓0
) MW sec
(17)
Rate of change of kinetic energy is therefore,
𝒅𝑊𝑘𝑒
𝒅𝒕
ii)
=2
𝐻𝑃𝑟 𝑑∆𝑓
(18)
𝑓 0 𝑑𝑡
By the change in the load, due to change in frequency(i. e.
Frequency dependency of load can be written as B=
𝝏𝑃𝐷
𝝏𝒇
𝜕𝑃𝐷
𝜕𝑓
)
MW/Hz
for small changes in frequency ∆𝑓, the above equation can be expressed as
𝝏𝑃𝐷
𝝏𝒇
∆𝑓 =B ∆𝑓
Where B is positive for a predominantly motor load
The power balance equation can be written as,
(19)
∆𝑃𝐺 − ∆𝑃𝐷 = 2
𝐻𝑃𝑟 𝑑∆𝑓
(
) + 𝐵∆𝑓
𝑓 0 𝑑𝑡
Dividing throughout by 𝑃𝑟 and rearranging, we get
𝐻
𝑑∆𝑓
∆𝑃𝐺(𝑝𝑢) − ∆𝑃𝐷(𝑝𝑢) = 2 𝑓0 ( 𝑑𝑡 ) + 𝐵(𝑝𝑢) ∆𝑓 (20)
Taking the laplace transform, we can write ∆𝐹(𝑆) as
∆𝐹(𝑆) =
∆𝑃𝐺 − ∆𝑃𝐷
2𝐻
𝐵 + ( 0 𝑆)
𝑓
𝑲𝒑𝒔
∆𝑭(𝑺) = [∆𝑷𝑮(𝒔) − ∆𝑷𝑫(𝒔) ] 𝐱 (𝟏+𝑻
𝒑𝒔 𝑺
1
Where 𝐾𝑝𝑠 = 𝐵= power system gain and
)
(21)
2𝐻
𝑇𝑝𝑠 =𝐵𝑓0= power system time constant
Equation (21) can be represented in block diagram form as shown below,
∆𝑷𝑫(𝒔)
𝑲𝒑𝒔
𝟏 + 𝑻𝒑𝒔 𝑺
∆𝑷𝑮(𝒔)
∆𝑭(𝑺)
The complete block diagram representation of an isolated power system with feedback loop is as
shown below,
𝑲𝒔𝒈
𝟏 + 𝑻𝒔𝒈 𝑺
-
𝟏
𝑹
𝑲𝒕
𝟏 + 𝑻𝒕 𝑺
𝑲𝒑𝒔
𝟏 + 𝑻𝒑𝒔 𝑺
 TWO-AREA SYSTEM LOAD FREQUENCY CONTROL
A control area is characterized by the same frequency throughout. In the single-area case we
could thus represent the frequency deviation by the single variable Δf. Having interconnected
them with a tie-line therefore leads us to the assumption that the frequency deviations in the two
areas can be represented by two variables Δf1 and Δf2 respectively.
The control objective is to regulate the frequency of each area and to simultaneously regulate the
tie line power as per inter area power contracts.
The complete block diagram representation of isolated control area1 with feedback loop is as
shown below,
𝑲𝒔𝒈𝟏
𝟏 + 𝑻𝒔𝒈𝟏 𝑺
𝑲𝒕𝟏
𝟏 + 𝑻𝒕𝟏 𝑺
𝑲𝒑𝒔𝟏
𝟏 + 𝑻𝒑𝒔𝟏 𝑺
-
𝟏
𝑹
Similarly, block diagram representation of isolated control area2 with feedback loop is as shown
below,
𝑲𝒔𝒈𝟐
𝟏 + 𝑻𝒔𝒈𝟐 𝑺
-
𝟏
𝑹
𝑲𝒕𝟐
𝟏 + 𝑻𝒕𝟐 𝑺
𝑲𝒑𝒔𝟐
𝟏 + 𝑻𝒑𝒔𝟐 𝑺
 MODELING THE TIE-LINE and BLOCK DIAGRAM FOR TWO-AREA SYSTEM
In normal operation the power flows in the tie-line connecting the areas 1 and 2 is given by
Ptie,1 =
|𝑉1 ||𝑉2 |
𝑋12
sin(δ10 – δ02 )
(1)
Where Ptie,1 is the power transported out of area 1
δ10 & δ02 are the power angles of equivalent machines of the two areas
For small incremental deviations in angles δ1 and δ2 the tie-line power changes by an amount
ΔPtie,1 ≈
|𝑉1 ||𝑉2 |
𝑋12
cos(δ10 – δ02 ) (Δδ1 – Δδ2 )
(2)
∆Ptie,1(pu) =𝑇12 (Δδ1 – Δδ2 )
Where 𝑇12 =
(3)
|𝑉1 ||𝑉2 |
𝑋12 𝑃𝑟1
cos(δ10 – δ02 ) = synchronizing coefficient
We like to have ΔPtie,1 in terms of frequency deviations Δf1 and Δf2.
𝜔
1 𝑑𝛿
We know f =2𝜋 = 2π 𝑑𝑡 and δ=2π∫f dt
Hence
ΔPtie,1 = 2 π 𝑇12 (∫ ∆𝑓1 𝑑𝑡 − ∫ ∆𝑓2 𝑑𝑡)
(4)
where ∆𝑓1 and ∆𝑓2 are incremental frequency changes in areas 1 and 2 respectively
The incremental power balance equation for area 1 can be written as,
∆𝑃𝐺1 − ∆𝑃𝐷1 = 2
𝐻1 𝑃𝑟1 𝑑∆𝑓1
𝑓10
(
𝑑𝑡
) + 𝐵1 ∆𝑓1 + ΔPtie,1
Dividing throughout by 𝑃𝑟1 and rearranging, we get
∆𝑃𝐺1 − ∆𝑃𝐷1 = 2
𝐻1
𝑓10
𝑑∆𝑓1
(
𝑑𝑡
) + 𝐵1 ∆𝑓1 + ΔPtie,1
Taking the Laplace transform of the above equation and reorganizing, we get
∆𝐹1(𝑠) = [∆𝑃𝐺1(𝑠) − ∆𝑃𝐷1(𝑠) − ΔPtie,1(s) ]
1
𝑲𝒑𝒔𝟏
𝟏+𝑻𝒑𝒔𝟏 𝑺
2𝐻1
Where 𝐾𝑝𝑠1 = 𝐵 = power system gain of area 1 and 𝑇𝑝𝑠1 =𝐵
1
1𝑓
0
(5)
= power system time constant of
area 1
Similarly the incremental tie line power out of area 2 is given by,
ΔPtie,2 = 2 π 𝑇21 (∫ ∆𝑓2 𝑑𝑡 − ∫ ∆𝑓1 𝑑𝑡)
Where 𝑇21 =
|𝑉2 ||𝑉1 |
𝑋21 𝑃𝑟2
(6)
P
cos(δ02 – δ10 )=(Pr1 ) 𝑇12 = 𝑎12 𝑇12
r2
Defining the tie-line power in direction 2 to 1 as ΔPtie,2
ΔPtie,2 = - ΔPtie,1
so,
ΔPtie,2 =- 2 π 𝑎12 𝑇12 (∫ ∆𝑓1 𝑑𝑡 − ∫ ∆𝑓2 𝑑𝑡)
(7)
(8)
On taking the laplace transform of the above equation,we have
ΔPtie,2 =- 2 π
𝑎12 𝑇12
𝑆
(∆𝐹1(𝑆) − ∆𝐹2(𝑆) )
(9)
MODULE – IV
SHORT CIRCUIT ANALYSIS
Short circuit studies or fault analysis
The short circuit studies are essential in order to design or develop the protective schemes for
various parts of the system. The protective scheme consists of current and voltage sensing
devices, protective relays and circuit breakers. The selection of these devices mainly depends on
various currents that may flow in the fault conditions.
What is the need for short circuit study?

In order to determine the current interrupting capacity of the circuit breakers so that the
faulted equipments can be isolated.

To establish the relay requirements and settings to detect the fault and cause the circuit
breaker to operate when the current flowing through it exceeds the maximum value.
Fault
A fault in a circuit is any failure which interferes with the normal flow of current. The faults are
associated with abnormal change in current, voltage and frequency of the power system. The
faults may cause damage to the equipments if it is allowed to persist for a long time. Hence every
part of the system has been protected by means of relays and circuit breakers to since the faults
to isolate the faulty part from the healthy part in the event of fault. The fault occur in a power
system due to insulation failure of equipments, flashover of lines initiated by a lightning stroke,
due to permanent damage to conductors and towers.
Classification of Faults
In one method, the faults are classified as,
1. Shunt faults - due to short circuits in conductors
2. Series faults - due to open conductors.
In another method,
1. Symmetrical faults - fault currents are equal in all the phases and can be analyzed on per phase
basis
2. Unsymmetrical faults – fault currents are unbalanced and so they can be analyzed only using
symmetrical components.
Types of shunt faults
1. Single line-to-ground fault
2. Line-to-line fault
3. Double line-to-ground fault
4. Three phase fault
Types of series faults
1. One open conductor fault
2. Two open conductor fault
Symmetrical fault
The fault is called symmetrical fault if the fault current is equal in all the phases. This fault
conditions are analyzed on per phase basis using Thevenin’s theorem or using bus impedance
matrix. The three-phase fault is the only symmetrical fault.
Fault calculations
The fault condition of a power system can be dived into sub transient, transient and steady state
periods. The currents in the various parts of the system and in the fault are different in these
periods. The estimation of these currents for various types of faults at various locations in the
system is commonly referred as fault calculations.
Fault current can be reduced,

By providing neutral reactance.

By introducing a large value of shunt reactance between buses.
Symmetrical fault current estimation






Calculate Base Values
Calculate p.u impedances
Calculate internal emfs of generators and pre-fault voltage at fault point using pre-fault
load current.
Draw reactance diagram under fault condition
Calculate the pu fault current at various parts of the system and in the fault.
The actual value of fault current is obtained by multiplying pu values by the respective
base current.
Synchronous reactance
The synchronous reactance is the ratio of induced emf and the steady state rms current (i.e., it is
the reactance of a synchronous machine under steady state condition). It is the sum of leakage
reactance and the reactance representing armature reaction. It is given by,
Xs = Xl + Xa
Where,
Xs = Synchronous reactance
Xl = Leakage reactance
Xa = Armature reaction reactance.
Subtransient reactance
The subtransient reactance is the ratio of induced emf on no-load and the subtransient
symmetrical rms current, (i.e., it is the reactance of a synchronous machine under subtransient
condition). It is given by,
Where
Xl = Leakage reactan Xa = Armature reaction reactance Xf = Field winding reactance
Transient reactance
The transient reactance is the ratio of induced emf on no-load and the transient symmetrical rms
current. (i.e.,it is the reactance of synchronous machine under transient condition). It is given by,
Where
Xl = Leakage reactan Xa = Armature reaction reactance Xf = Field winding reactance
Significance of subtransient reactance and transient reactance in short circuit studies
The subtransient reactance can be used to estimate the initial value of fault current
immediately on the occurrence of the fault. The maximum momentary short circuit current rating
of the circuit breaker used for protection or fault clearing should be less than this initial fault
current.
The transient reactance is used to estimate the transient state fault current. Most of the
circuit breakers open their contacts only during this period. Therefore for a circuit breaker used
for fault clearing (or protection), its interrupting short circuit current rating should be less than
the transient fault current.
Fault current in a generator
where
I = Steady state symmetrical fault current
I ' = Transient symmetrical fault current
Xd = Direct axis synchronous reactance
Xd’= Direct axis transient reactance
Eg = RMS voltage from one terminal to neutral at no load.
Subtransient and transient internal voltage of the generator
Subtransient and transient internal voltage of the motor
Xd ’ =Direct axis transient reactance
Symmetrical faults are analysis
The symmetrical faults are analyzed using per unit reactance diagram of the power system. Once
the reactance diagram is formed, then the fault is simulated by short circuit or by connecting the
fault impedance at the fault point. The currents and voltages at various parts of the system can be
estimated by any of the following methods.
1) Using Kirchoff’s laws
2) Using Thevenin’s theorem
3) By forming bus impedance matrix.
Bus impedance matrix in fault current estimation
Consider the circuit shown in Figure below
Network depicting a symmetrical fault at bus-4.
We assume that a symmetrical fault has occurred in bus-4 such that it is now connected to the
reference bus. Let us assume that the pre-fault voltage at this bus is Vf . To denote that bus-4 is
short circuit, we add two voltage sources Vf and - Vf together in series between bus-4 and the
reference bus. Also note that the sub-transient fault current If flows from bus-4 to the reference
bus. This implies that a current that is equal to - If is injected into bus-4. This current, which is
due to the source - Vf will flow through the various branches of the network and will cause a
change in the bus voltages. Assuming that the two sources and Vf are short circuited. Then - Vf is
the only source left in the network that injects a current - If into bus-4. The voltages of the
different nodes that are caused by the voltage - Vf and the current - If are then given by
where the prefix Δ indicates the changes in the bus voltages due to the current - If .
From the fourth row of above matrix equation we can write
Combining the above two equations, we get
We further assume that the system is unloaded before the fault occurs and that the magnitude and
phase angles of all the generator internal emfs are the same. Then there will be no current
circulating anywhere in the network and the bus voltages of all the nodes before the fault will be
same and equal to Vf . Then the new altered bus voltages due to the fault will be given as
Symmetrical components
An unbalanced system of N related vectors can be resolved into N systems of balanced vectors
called symmetrical components.
In a 3-phase system, the three unbalanced vectors (either current or voltage vectors) can
be resolved into three balanced system of vectors. They are,
1) Positive sequence components
2) Negative sequence components
3) Zero sequence components.
Negative sequence components consist of three phasors equal in magnitude, displaced from
each other by 120o in phase, and having the phase sequence opposite to that of the original
phasors. Va2, Vb2 and Vc2 are the negative sequence components of Va, Vb and vc.
Zero sequence components consist of three phasors equal in magnitude and with zero phase
displacement from each other. Vao, Vbo and Vco are the zero sequence components of Va, Vb
and Vc.
Operator ‘a’
An operator which causes a rotation of 120o in the anticlockwise direction is
known as operator ‘a’. The value of ‘a’ is 1∠120o.
The polar form and rectangular form of operator ‘a’ is given by,
a = 1∠120˚ --------polar form
= -0.5 + j0.806 ----------rectangular form
The polar form and rectangular form of operator ‘a2’ is given by,
a2 = 1∠240˚ --------polar form
= -0.5 - j0.806 ----------rectangular form
Unbalanced voltages Va, Vb and Vc in terms of symmetrical components Va1, Va2 and
Va0
The expression of unbalanced voltages in terms of symmetrical components are,
Symmetrical components Va1, Va2 and Va0 in terms of unbalanced vectors Va, Vb and Vc
The expression of symmetrical components in terms of unbalanced vectors are,
Connection of sequence network for LG fault
The Thevenin equivalent of the sequence network is shown in Fig..
Connection of sequence network for LL fault
Cconnection of sequence network for LLG fault
Selection of CB
A typical circuit breaker operating time is given in figure. Once the fault occurs, the
protective devices get activated. A certain amount of time elapses before the protective relays
determine that there is over-current in the circuit and initiate trip command. This time is called
the detection time. The contacts of the circuit breakers are held together by spring mechanism
and, with the trip command, the spring mechanism releases the contacts. When two current
carrying contacts part, a voltage instantly appears at the contacts and a large voltage gradient
appears in the medium between the two contacts. This voltage gradient ionizes the medium
thereby maintaining the flow of current. This current generates extreme heat and light that is
called electric arc. Different mechanisms are used for elongating the arc such that it can be
cooled and extinguished. Therefore the circuit breaker has to withstand fault current from the
instant of initiation of the fault to the time the arc is extinguished.
Typical circuit breaker operating time.
Two factors are of utmost importance for the selection of circuit breakers. These are:


The maximum instantaneous current that a breaker must withstand and
The total current when the breaker contacts part.
The instantaneous current following a fault will also contain the dc component. In a high power
circuit breaker selection, the sub-transient current is multiplied by a factor of 1.6 to determine the
rms value of the current the circuit breaker must withstand. This current is called the momentary
current. The interrupting current of a circuit breaker is lower than the momentary current and
will depend upon the speed of the circuit breaker. The interrupting current may be asymmetrical
since some dc component may still continue to decay.
Breakers are usually classified by their nominal voltage, continuous current rating, rated
maximum voltage, K -factor which is the voltage range factor, rated short circuit current at
maximum voltage and operating time. The K -factor is the ratio of rated maximum voltage to the
lower limit of the range of the operating voltage. The maximum symmetrical interrupting current
of a circuit breaker is given by K times the rated short circuit current.
MODULE – V
STABILITY
Power System stability
The stability of a system is defined as the ability of power system to return to a stable operation
in which various synchronous machines of the system remain in synchronism or ‘in step’ with
each other, when it is subjected to a disturbance.
The steady state stability is defined as the ability of a system to remain stable for small
disturbance like gradual change in load.
The transient stability is defined as the ability of a system to remain stable for large disturbance
like sudden change in load and fault.
The dynamic stability is defined as the ability of a system to remain stable for large disturbance
like sudden change in load and fault under the influence of the controller.
Stability limit
The maximum power that can be transferred to the receiving system without loss of
synchronism is called stability limit.
The steady state stability limit is the maximum power that can be transferred by a machine to a
receiving system without loss of synchronism for small disturbance like gradual change in load.
The transient stability limit is the maximum power that can be transferred by a machine to a
receiving system without loss of synchronism for large disturbance like sudden change in load
and fault.
The dynamic stability limit is the maximum power that can be transferred by a machine to a
receiving system without loss of synchronism for large disturbance like sudden change in load
and fault under the influence of the controller. The transient stability limit is always less than the
steady state stability limit.
Swing curve
The swing curve is the plot between the power angle and time. The power angle is defined as
the angular displacement of the rotor from synchronously rotating reference frame.
It is usually plotted for a transient state to study the nature of variation in angle for a sudden
large disturbance.
If the magnitude of power angle increases without damping, the system is said to be unstable.
Thus swing curve is used to analyze the stability of a power system.
Swing equation
Let us consider a three-phase synchronous alternator that is driven by a prime mover. The
equation of motion of the machine rotor is given by
where
J
is the total moment of inertia of the rotor mass in kgm2
is the mechanical torque supplied by the prime mover in
Tm
N-m
Te
is the electrical torque output of the alternator in N-m
θ
is the angular position of the rotor in rad
Neglecting the losses, the difference between the mechanical and electrical torque gives
the net accelerating torque Ta . In the steady state, the electrical torque is equal to the
mechanical torque, and hence the accelerating power will be zero. During this period the
rotor will move at synchronous speed ωs in rad/s.
The angular position θ is measured with a stationary reference frame. To represent it with
respect to the synchronously rotating frame, we define
where δ is the angular position in rad with respect to the synchronously rotating
reference frame. Taking the time derivative of the above equation we get
Defining the angular speed of the rotor as
we can write as
We can therefore conclude that the rotor angular speed is equal to the synchronous speed
only when dδ / dt is equal to zero. We can therefore term dδ / dt as the error in speed.
Taking derivative, we can then rewrite as
Multiplying both sides by ωm we get
where Pm , Pe and Pa respectively are the mechanical, electrical and accelerating power in
MW.
We now define a normalized inertia constant as
Substituting this, we get
In steady state, the machine angular speed is equal to the synchronous speed and hence
we can replace ωr in the above equation by ωs. Note that Pm , Pe and Pa are given in MW.
Therefore dividing them by the generator MVA rating Srated we can get these quantities in
per unit
per unit. Hence dividing both sides of equation by Srated we get
This equation describes the behavior of the rotor dynamics and hence is known as the
swing equation. The angle δ is the angle of the internal emf of the generator and it
dictates the amount of power that can be transferred. This angle is therefore called the
load angle.
Power angle equation
In this we shall consider this relation for a lumped parameter lossless transmission line. Consider
the single-machine-infinite-bus (SMIB) system shown in figure. In this the reactance X includes
the reactance of the transmission line and the synchronous reactance or the transient reactance of
the generator. The sending end voltage is then the internal emf of the generator. Let the sending
and receiving end voltages be given by
A single machine infinite bus system
Power angle diagram
We then have
The sending end real power and reactive power are then given by
This is simplified to
Since the line is loss less, the real power dispatched from the sending end is equal to the real
power received at the receiving end. We can therefore write
where Pmax = V1 V2 / X is the maximum power that can be transmitted over the transmission line.
Equal area criteria
The real power transmitted over a lossless line is given by power angle equation. Now
consider the situation in which the synchronous machine is operating in steady state delivering a
power Pe equal to Pm when there is a fault occurs in the system. Opening up of the circuit
breakers in the faulted section subsequently clears the fault. The circuit breakers take about 5/6
cycles to open and the subsequent post-fault transient last for another few cycles. The input
power, on the other hand, is supplied by a prime mover that is usually driven by a steam turbine.
The time constant of the turbine mass system is of the order of few seconds, while the electrical
system time constant is in milliseconds. Therefore, for all practical purpose, the mechanical
power is remains constant during this period when the electrical transients occur. The transient
stability study therefore concentrates on the ability of the power system to recover from the fault
and deliver the constant power Pm with a possible new load angle δ .
Consider the power angle curve shown in figure. Suppose the system is operating in the steady
state delivering a power of Pm at an angle of δ0 when due to malfunction of the line, circuit
breakers open reducing the real power transferred to zero. Since Pm remains constant, the
accelerating power Pa becomes equal to Pm . The difference in the power gives rise to the rate of
change of stored kinetic energy in the rotor masses. Thus the rotor will accelerate under the
constant influence of non-zero accelerating power and hence the load angle will increase. Now
suppose the circuit breaker re-closes at an angle δc. The power will then revert back to the
normal operating curve. At that point, the electrical power will be more than the mechanical
power and the accelerating power will be negative. This will cause the machine decelerate.
However, due to the inertia of the rotor masses, the load angle will still keep on increasing. The
increase in this angle may eventually stop and the rotor may start decelerating, otherwise the
system will lose synchronism.
Note that
Power-angle curve for equal area criterion
Hence multiplying both sides of (9.14) by
and rearranging we get
Multiplying both sides of the above equation by dt and then integrating between two arbitrary
angles δ0 and δc we get
Now suppose the generator is at rest at δ0. We then have dδ / dt = 0. Once a fault occurs, the
machine starts accelerating. Once the fault is cleared, the machine keeps on accelerating before it
reaches its peak at δc , at which point we again have dδ / dt = 0. Thus the area of accelerating is
given as
In a similar way, we can define the area of deceleration. In figure, the area of acceleration is
given by A1 while the area of deceleration is given by A2 . This is given by
Now consider the case when the line is reclosed at δc such that the area of acceleration is larger
than the area of deceleration, i.e., A1 > A2 . The generator load angle will then cross the point δm ,
beyond which the electrical power will be less than the mechanical power forcing the
accelerating power to be positive. The generator will therefore start accelerating before is slows
down completely and will eventually become unstable. If, on the other hand, A1 < A2 , i.e., the
decelerating area is larger than the accelerating area, the machine will decelerate completely
before accelerating again. The rotor inertia will force the subsequent acceleration and
deceleration areas to be smaller than the first ones and the machine will eventually attain the
steady state. If the two areas are equal, i.e., A1 = A2 , then the accelerating area is equal to
decelerating area and this is defines the boundary of the stability limit. The clearing angle δc for
this mode is called the Critical Clearing Angle and is denoted by δcr. We then get from figure by
substituting δc = δcr
Critical clearing time and critical clearing angle
The critical clearing angle is the maximum allowable change in the power angle before
clearing the fault, without loss of synchronism.
The critical clearing time can be defined as the maximum time delay that can be allowed to clear
a fault without loss of synchronism.
The equal area criterion for stability states that the system is stable if the area under power vs
angle curve reduces to zero at some value of angle.
Methods of improving transient stability

Increase system voltage using AVR.

Use of high speed excitation system.

Reduction in system transfer reactance by series compensation.

Use of high speed reclosing breaker.

Reducing conductor spacing.

Increasing conductor diameter.

Using HVDC links.

Using breaking resistors.

Using bypass valving.

Using full load rejection technique.
Multimachine stability
If a system has any number of machines, then each machine is listed for stability by advancing
the angular position, δ of its internal voltage and noting whether the electric power output of the
machine increases (or) decreases. If it increases, i.e if ∂Pn / ∂δn > 0 then machine n is stable. If
every machine is stable, then the system having any number of machine is stable.
Assumptions made in multimachine stability studies
The assumptions made are,
• The mechanical power input to each machine remains constant during the entire period of the
swing curve computation
• Damping power is negligible
• Each machine may be represented by a constant transient reactance in series with a constant
transient voltage.
• The mechanical rotor angle of each machine coincides with δ , the electrical phase angle of the
transient internal voltage.