This document provides an overview of power systems, including different types of power generation sources like thermal, hydroelectric, nuclear, gas turbine, and diesel power plants. It also describes the basic components of a power system such as generators, transformers, transmission lines, and loads. Additionally, it discusses the voltage structure of electric power systems including generating stations, transmission systems, and distribution systems. Finally, it introduces the need for system analysis in planning and operating power systems, and distinguishes between steady state and transient state stability analysis.

1. UNIT-I
INTRODUCTION
SYLLABUS
Need for system analysis in planning and operation of power system - distinction
between steady state and transient state - per phase analysis of symmetrical three-phase
system. General aspects relating to power flow - short circuit and stability analysis -
per unit representation.
BASIC COMPONENT OF POWER SYSTEMS:
 Generator
 Transformer
 Transmission line
 Loads
GENERATION OF POWER: -
There are different sources from which the power can be produced for
running the machines in industry. The prime source of Power is sun, but its energy cannot
be efficiently utilized. The conventional sources from which the is obtained are
 Fuel (coal, oil and gas)
 Water
 .Nuclear disintegration of Uranium
The energy so obtained is used for running either steam turbines or water turbines.
The mechanical power obtained from the turbines is then efficiently converted into
electrical energy, by the use of alternators.
Thermal Power Plants:
Thermal power stations produce electrical energy from the heat released by the
combustion of coal, oil or natural gas. In a steam power plant, steam is produced in the
boiler by utilizing the heat of coal combustion.

2. The steam is then expanded in the steam turbine and is condensed, in a condenser
and fed into the boiler again. The steam turbine drives the alternator which converts
mechanical energy of the turbine into electrical energy.
Hydro Electric Power Plants:
Hydro electric power plants utilize the potential energy of water at high
level to produce electrical energy. In a hydroelectric power plant, water head is created
by constructing a dam across a river or lake. From the bottom of the dam water is led to a
water turbine. The water turbine captures the energy in the falling, water and changes the
hydraulic energy into mechanical energy which in turn is converted into electrical energy
by generators.
The power that can be extracted from a waterfall depends on its height and rate of
flow. Therefore, the size. and physical location of hydroelectric power plant depends on
these two factors: The available power can be calculated from the equation
P= 9.8 QH KW
Where ,
P- Available power in KW
Q- Flow of Water in m3
/sec
H - Head of water in meters.
Nuclear Power Plants:
Nuclear power plants convert nuclear energy into electrical energy. In Nuclear
power station, heavy elements such as Uranium or Thorium are subjected. To nuclear
fission in a special apparatus known as reactor. The heat energy thus released is utilized
in raising steam at high temperature and pressure.

3. The steam runs the steam turbine which converts steam energy into mechanical
energy. The turbine drives the alternator which converts .mechanical energy into
electrical energy. The most important feature of a nuclear power station is that huge
amount of electrical energy can be produced from a relatively small amount of nuclear
fuel as compared to other conventional types of power stations.
Gas Turbine Plant:
In gas turbines, air is used as the working fluid. It is compressed by the
compressor and fed to the combustion chamber where it is heated by burning fuel in the
chamber to the gas turbine where it expands and delivers mechanical energy. The gas
turbine drives the alternator which converts mechanical energy into electrical energy. The
compressor, the gas turbine and the alternator are mounted on the same shaft so. that part
of turbine power is supplied to the compressor besides the alternator.
Diesel Power Plant:
A diesel engine is a prime mover which obtains its. energy from a liquid fuel
generally known as diesel oil. It converts this energy obtained into mechanical woi-k. An
alternator or a D.C. generator mechanically coupled to it converts the mechanical energy
developed, into electrical energy. Diesel engines are preferred with small generating
stations.
VOLTAGE STRUCTURE OF ELECTRIC POWER SYSTEM:
An electric power system consists of. three. principal divisions: .
1) Generating Stations
2) Transmission Lines
3) Distribution System
The electrical power is generated in bulk ‗at the generating‘ stations which are
also called as Power Stations. The network of conductors between the power station and
the consumers can be broadly divided into two parts via, transmission system and
distribution system. Each part can be further subdivided into two — primary transmission
& secondary transmission and primary & secondary distribution.

4. Generating Station:
It is the place where electric power is produced by alternators operating in
parallel. The usual generation - voltage is 11 KV. For economy in the transmission of
electric power, the generation voltage is stepped up to 132 K‘! at the generating station
with the help of transformers. The transmission of electric power at high voltages has
several advantages including the saving of conductor material and high transmission
efficiency.
Transmission Systems:
The transmission of electrical energy over greater distances was developed in the
beginning of the twentieth century, since then it has made a rapid progress in its design
and methods of operation which has resulted into a greater reliability and Continuity.
(a) Primary transmission
Generally the primary transmission is carried at 66KV, 132KV, 220KV, or
4O0KV.The electric power at this high voltage is transmitted by 3Φ, 3wire overhead
system to the outskirts of the city. This forms the primary
transmission.
(b) Secondary transmission
The Primary transmission line terminates at the receiving station which usually
lies at the outskirts of the city. At the receiving station, the voltage is reduced to 33KV by
step down transformers. From this station, electric power is transmitted at 33KV by 3,
3wire overhead system to various substations located at the strategic points in the city.
This forms the secondary transmission.
Distribution Systems:
(a) Primary distribution:
The secondary transmission line terminates at the sub station where voltage is
reduced from 33KV to 1 1KV, and transmitted by 3$, 3 wire system. The 11 1KV lines
run along the important road sides of the city This forms the primary distribution. Big
consumers are generally supplied power at 1 1KV for further handling with their own
substation.

5. (b) Secondary distribution:
The electric power from primary distribution line is delivered to distribution
substation. These substations are located near the consumer localities and step down the
voltage to 400V, 3$, 4Wire for secondary distribution. The voltage between any two
phases is 400V and between any phase and neutral is 230V.
Secondary distribution system consists of feeders, distributors and service mains
Feeders radiating from the distribution system Feeders supply power to the distributors.
No consumer is given direct connection from the feeders. Instead, the consumers are
connected to the distributors through their service mains.
NEED FOR SYSTEM ANALYSIS IN PLANNING AND OPERATION OF
POWER SYSTEMS
The engineering of the last two decades of the twentieth century of challenging
tasks, which he can meet only by keeping himself abreast of the recent scientific
advances and the latest techniques.
PLANNING:
The planning side ,we has to make decisions on how much electricity to generate
where, when and by/using what fuel. We has t be involved in construction tasks of great
magnitude both in generation and transmission. We has to solve the problems of planning
and coordinated operation of a vast and complex power network, so as to achieve a high
degree of economy and reliability. In a country like India, we has to additionally face the
perennial problem of power shortages and to evolve strategies for energy conservation
and load management.
OPERATION:
The improvement and expansion of a power system, a power system engineer
needs load flow studies, short circuit studies, and stability studies we has to know the
principles of economic load dispatch and load frequency control. The solution of these

6. problems and the enormous contribution made by digital computers to solve the planning
and operational problems of power systems are also investigated.
DISTINCTION BETWEEN STEADY STATE AND TRANSIENT STATE:
The stability of a physical system is referred as its capability to return to the
original or a new equilibrium state on the occurrence of a disturbance. During transient
period, following the disturbance, the system state may not be in equilibrium but as the
time passes and tends to infinity the system achieves an equilibrium state if the system is
stable. In the equilibrium state the physical variables do not change with time.
The power system has a number of synchronous machines operating in parallel.
The voltage in synchronous machines is generated by the flux produced by field windings
of the machines there for the phase difference between the generated voltages of the any
two machines is the same as the electrical angle between the corresponding machine
rotors. When the machines are running with synchronism, the rotor angles of all the
synchronous machines with reference to a synchronously rotating reference axis do not
change with time denoting that the power is equilibrium state. If the machines may be
running faster than the others and angular position of their rotor relative to the other
machines will continue to advance resulting in continuous advancement of their
generated voltage phase angle relative to the voltage angle slower machines. This denotes
that the system is not in equilibrium state. Therefore, the power system operates in
equilibrium state if the various synchronous machines in the system are running in
synchronism or in step with each other.
The stability of an interconnected power system is its ability to return to
equilibrium state, that is a state in which all the synchronous machines are running with
synchronism, after having been subjected to some form of disturbance. The disturbance
may be as simple as very small slow change in loads or as large or complex as sudden
change in loads, loss of generators, transmission facilities etc. the precise definition of
stability as published by American Institute of Electrical Engineers is as follows
Stability when used with reference to the power system, it that attribute of the
system, or part of the system, which enables it to develop restoring forces between the

7. elements therefore, equal to or grater than the disturbing forces so as to restore a state of
equilibrium between the elements.
STEADY STATE, TRANSIENT STATE STABILITY
Power System Stability
The power system stability can be classified into three types.
 Steady state stability
 Transient state stability
 Dynamic stability
This stability‘s are according to the magnitude and duration of disturbance and time
frame of study.
Stability:
The ability of the power system to remain in stable condition (or) The ability of
the generator to remain in synchronism.
Steady state stability:
The ability of the power system to remain in synchronism for small and slow
variations of load. If will not affected by the stability of the system.
Transient state stability:
The ability of the generator to remain in synchronism if it is subjected to a large
and sudden variations of loads.
Dynamic stability:
The ability of the generator to remain in stable in the period following the end of
the transient period. It is similar to steady state stability. The variation is slightly above
the rotor natural frequency.

8. Transients in simple circuits:
For analyzing circuits for transients we will make use of Laplace transform
technique which is more powerful and easy to handle the transient problems than the
differential equation technique. We will assume here lumped impedances only. The
transients will depend upon the driving source also, that is whether it is a d.c. source or an
ac source. We will begin with simple problems and then go to some complicated
problems.
ER PHASE ANALYSIS OF SYMMETRICAL THREE PHASE SYSTEM
A complete diagram of a power system representing all the three phases becomes
too complicated for a system of practical size, so much so that it may no longer convey
the information it is intended to convey. It is much more practical to represent a power
system by means of simple symbols for each component resulting in that is called as one-
line diagram.
Per system leads to great simplification of three phase networks involving
transformers. An impedance diagram draw on a per unit basis does not require ideal
transformers to be included in it.
An important element of a power system is the synchronous during both steady
state and transient state conditions. The synchronous machine model in steady state is
presented.
SINGLE PHASE SOLUTION BALANCED THREE PHASE NETWORKS
In 3Φ network the solution can be easily carried out solving a single phase
network corresponding to the reference phase circuits. Under balanced condition In is
zero. At that time generator and load neutral are same potential and also neutral
impedance Zn will not affect the network behaviors.
The solution of a three-phase network under balanced conditions is easily carried
out by; solving the sjgepta5e network corresponding to the reference phase. Figure (1)

9. shows a simple, balanced three-phase network. The generator and load neutrals are
therefore at the same potentials so that In=0. Thus the neutral impedance Z, does not
affect network behavior.
For the reference phase a
Ea= (ZG+ZL)Ia
Eb= (ZG+ZL)Ib
Ec= (ZG+ZL)Ic
The currents and voltages in the other phases have the same magnitude but are
progressively shifted in phase by 120о
. Equation corresponds to the single-phase network
of Fig(2) whose solution completely determines the solution of the three-phase network.
Fig (1) Balanced three phase network
ZG
ZG
ZG
+
+
N
Ea
Eb
ZL
ZL
Ia
a
ZL
Ec
+

10. Fig (2) single phase equivalent of a balanced 3Φ network
Consider now the case where a three-phase transformer forms, part of a three-
phase system. If the transformer is Y/Y connected as shown in Fig. (2) single equivalent
of the three-phase circuit it can be obviously represented by a single-phase transformer
with primary and secondary pertaining to phase ‗a‘ of the three phase transformer.
ONE LINE DIAGRAM REPRESENTATION OF SINGLE PHASE POWER
SYSTEMS
breakers need not be shown in a load flow study but are a must for a protection study.
Power system networks arc represented by one- line diagrams using suitable symbols
- for generators, motors, transformers and loads. It is a convenient practical way of
network representation rather than drawing the actual three-phase diagram which may
indeed be quite cumbersome and confusing for a practical size power network. Generator
and transformer connections—star, delta, and neutral grounding aft indicated by symbols
drawn by the side of the representation Of these elements. Circuit breakers are
represented as rectangular blocks. Figure 10 shows the one-line diagram of a simple
power system. The reactance data of the elements are given below the diagram.
N
+
Ea
ZL
ZG

11. Fig (1) :One line diagram representation of a simple power system
Generator No. 1: 30 MVA, 10.5 kV, X‘=1.6 Ω
Generator No.2: 15 MVA, 6.6 kV, X‘ = 1.2 Ω
Generator No. 3: 25 MVA, 6.6 kV, X‘ 0.56 Ω
Transformer T (3 Φ): 15 JyIVA, 33/11 kV, X — 15.2 ohms per phase on tension side
Transformer T, (3 Φ): 15 MVA, 33/6.2 kV, X — 16 ohms per phase on high-tension side
Transmission line: 20.5 Ω /phase
Load A: 15 MW, I kV 0.9 lagging power factor.
Load B: 40 MW, 6.6 kV, 0.85 lagging power factor.
No Generators are specified in three-phase MVA. line-to-line voltage and per
phase reactance (equivalent star), Transformers are specified in three-phase KVA, line-
to-line transform on ratio, and per phase (equivalent star) impedance on one side. Loads
ate specified in three-phase MW, line-to-line voltage and power factor.
IMPEDANCE DIAGRAM:
Single-phase transformer equivalents are shown as ideal transformers with
transformer impedances indicated on the appropriate side. Magnetizing reactances of the
transformers have been neglected. This is a fairly good approximation for most power

12. system studies, the generators are represented as voltage sources with series resistance
and inductive reactance The transmission line Is represented by a Π-model loads are
assumed to be passive (not involving rotating machines) and are represented by resistance
and inductive reactance in series, Neutral grounding impedances do not appear in the
diagram a balanced conditions are assumed.
The impedance of power system diagram is used for load flow studies. The
impedance diagram can be obtained from the single line diagram by replacing all the
components of the power system buy their single phase equivalent circuit.
The following approximations are made while forming impedance diagram
 The current limiting impedance connected between the generator neutral and
ground are neglected since under balanced condition no current flow through
neutral.
 Since the magnetizing current of a transformer is very low when compared to
load current the shunt branches in the equivalent circuit of the transformer can
be neglected.
 If the inductive reactance of a component is very high when compared to
resistance then the resistance can be omitted, which introduces a little error in
calculations.
REACTANCE DIAGRAM:
The reactance diagram is used for fault calculations.
The following approximations are made while forming reactance diagram
 The neutral to ground impedance of the generator is neglected for symmetrical
faults.
 Shunt branches in the equivalent circuits of transformer are neglected.
 The resistance in the equivalent circuit of various components of the system are
omitted.
 All static loads are neglected.

13.  Induction motors are neglected in computing fault current of few cycle after the
fault occurs, because the current contributed by an induction motor dies out very
quickly after the induction motor is short circuited.
 The capacitance of the transmission lines are neglected.
PER UNIT SYSTEM
The per unit values of any quantity is defined as the ratio of actual value of
quantity to the base value of the quantity.
Per unit = Actual Value/ Base Value
% Per unit value = (Actual Value/ Base Value) * 100
The power system requires the base values of four quantities and they are voltage,
power, current and impedance. Selection of base values for any two of them determines
the base values of the remaining two.
SINGLE PHASE SYSTEM
Let
KVAb = Base KVA
KVb = Base Voltage in KV
Ib = Base Current in Amp
Zb = Base Impedance in Ω
Let
Base current Ib = Base KVA b / Base Voltage
Ib = KVAb / KVb in Amps ……………………(1)
Base Impedance Zb = (KVb * 1000) / Ib in Ω …..……………….(2)
(or) Zb = MVAb / Ib in Ω
On substituting for Ib values from equation (1) in equation (2) we get,
Base Impedance Zb = (KVb * 1000) / (KVAb / KVb ) in Ω
Zb = (KVb
2
* 1000) / KVAb in Ω ……………(3)
(or) Zb = KVb
2
/ MVAb in Ω

14. Per unit impedance Zp.u = Actual impedance in Ω / Base impedance in Ω
Zp.u = Z / [(KVb
2
* 1000) / KVAb ]
Zp.u = (Z * KVAb) / (KVb
2
* 1000) ……………(4)
(or) Zp.u = (Z * KVAb) / MVAb
Zp.u(old) = (Z * KVAb(old)) / MVAb(old) ……………...(5)
Z = [Zp.u(old) * MVAb(old)] / KVAb(old) …….(6)
Zp.u(new) = (Z * KVAb(new)) / MVAb(new)…………….(7)
On substituting for Z values from equation (6) in equation (7) we get,
Zp.u(new) = Zp.u(old) * (MVAb
2
(new) / MVAb
2
(old) ) * (KVAb
2
(old) / KVAb
2
(new) )
THREE PHASE SYSTEM
Let
KVAb = 3Φ Base KVA
KVb = Line to Line Base Voltage in KV
Ib = Line values Base Current in Amp
Zb = Base Impedance in Ω
Now, KVAb = √3 * KVb * Ib …………………………………(1)
( In 3Φ systems, KVA = √3 * VL * IL *10-3
= √3 * KVL * IL)
From equation (1) we get,
Ib = Base KVA b / √3 * Base Voltage
Ib = KVAb / √3 * KVb in Amps …………………………(2)
Base Impedance Zb = [(KVb/ √3) * 1000] / Ib in Ω
Zb = (KVb * 1000) / √3 Ib in Ω ………………………....(3)
On substituting for Ib values from equation (2) in equation (3) we get,
Zb = (KVb
2
* 1000) / KVAb ………………………………(4)

15. Per unit impedance Zp.u = Actual impedance in Ω / Base impedance in Ω
Zp.u = Z / [(KVb
2
* 1000) / KVAb ]
Zp.u = (Z * KVAb) / (KVb
2
* 1000) ……………(5)
(or) Zp.u = (Z * KVAb) / MVAb
Zp.u(old) = (Z * KVAb(old)) / MVAb(old) ……………...(6)
Z = [Zp.u(old) * MVAb(old)] / KVAb(old) …….(7)
Zp.u(new) = (Z * KVAb(new)) / MVAb(new)…………….(8)
On substituting for Z values from equation (7) in equation (8) we get,
Zp.u(new) = Zp.u(old) * (MVAb
2
(new) / MVAb
2
(old) ) * (KVAb
2
(old) / KVAb
2
(new) )
GENERAL ASPECTS RELATING TO POWER FLOW, SHORT CIRCUIT AND
STABILITY ANALYSIS
The planning, design and operation of power systems require continuous and
comprehensive analysis to evaluate current system performance and to• ascertain the
effectiveness of alternative plans for system expansion. The problems. which are
generally encountered in power system analysis include Load flow study, short circuit-
analysis Transient stability study problem. This section provides an engineering
description of these problems and also the mathematical techniques that are required for
the computer solution of these problems.
Load Flow Study:
An interconnected power system represents an electric network with a multitude
of branches and nodes, where the transmission lines typically constitute the branches. In
power engineering terminology the nodes are referred to as ―Buses‖ At some of the buses
power is being injected into the network, whereas at most other buses it is being tapped
by the system loads. In between, the power will flow in the network meshes.

16. The load flow problem consists of the calculation of power flows and voltages of
a network for specified terminal. or bus conditions. This information is essential for the
continuous evaluation of the current performance of a power system and for analyzing
the effectiveness of alternative plans for system expansion to meet increased
load demand. .
Short Circuit Analysis:
A fault in a circuit is any failure which Interferes with the normal flow of current.
Most faults on transmission lines are caused by lightning which results in the flashover of
insulators. Most of the faults on the power system lead to a short circuit condition. When
such a condition occurs, a heavy current flows through the equipment causing
considerable damage to the equipment and interruption of service to the consumers.
The knowledge of the fault currents is necessary for selecting the circuit breakers
of adequate rating, designing the substation equipment, determining the relay settings, etc
The fault calculations provide the information about the fault currents and the voltages at
various parts of power system under different fault conditions.
Stability Study:
In practice, when a power system is subjected to abnormal conditions, they must
be rectified so as to avoid the possibility of damage to the equipment. It is extremely
important to ensure the stability of the system in such situations.
Stability studies provide information related to the capability of a power system to
remain in synchronism during major disturbances. Following the disturbance, the rotor
angular positions will experience transient deviations. If all the individual rotor angles
settle down to new post fault steady state values, the system is transient stable. Transient
stability analysis is performed by combining a solution of the algebraic equations
describing the network with a numerical solution of the differential equations, which
describe the dynamics of the individual generators.

17. PROBLEMS:
1) The base current and base voltage of 345 KV system or choose to be 3000 A and
300 KV. Determine the base impedance and per unit voltage of system.
Given:
Base voltage = 300 KV
Base Current = 3000 A
Actual voltage = 345 KV
Required:
P.U Voltage = ?
Solution:
Zb = Base Voltage / Base Current
= 300*10
-3
/ 3000
Zb =100 Ω
P.U Voltage = Actual Voltage / Base Voltage
= 345 / 300
= 1.15 p.u
2) The three phase with rating 1000 KVA, 33KV has its armature resistance and
synchronous reactance as 20 Ω / ph and 70 Ω /ph . Calculate per unit impedance
of generator.
Given:
KVb = 33 KV
KVAb = 1000 KVA
R = 20 Ω

18. X = 70 Ω
Required:
P.U impedance = ?
Solution:
Per unit impedance Zp.u = Actual impedance in Ω / Base impedance in Ω
Base impedance Zb = (KVb
2
* 1000) / KVAb
Zb = [(332
) *1000] / 1000
Zb = 1089 Ω (or) 1.089 KΩ
Actual impedance Z = R+jX
= 20+j70
Per unit impedance Zp.u = Z / [(KVb
2
* 1000) / KVAb ]
Zp.u = 20+j70 / 1089
Zp.u = 0.018 + j 0.064 p.u
3) A generator is rated at 500 MVA , 22 KV its star connected winding has a
reactance of 1.1 p.u . Find the value of the reactance of winding . If the generator
is working in a circuit for which bases are specified has 100 MVA , 20 KV then
find the per unit value of reactance of generator winding on specified base.
Given:
KVb = 22 KV
MVAb = 500 MVA
Zp.u = 1.1 p.u
KVb(new) = 20 KV
MVAb(new) = 100 MVA
Required:
Z = ?
Z (P.U) = ?
Solution:
i) Actual Reactance (Z)

19. Z = [Zp.u(old) * MVAb(old)] / KVAb(old)
= 1.1 * 222
/ 500
= 1.064 Ω / ph
ii) Z (P.U)
Zp.u(new) = Zp.u(old) * (MVAb
2
(new) / MVAb
2
(old) ) * (KVAb
2
(old) / KVAb
2
(new) )
= 1.1 *(100/500) * (222
/ 202
)
Zp.u(new) = 0.2662 p.u
4) The reactance of a generator designated X11
is given as 0.25 per unit based on
the generators name plate rating of 18KV, 500MVA.The base for calculation is 20
KV,100MVA.Find X11
on the new base.
GIVEN:
Z (p u) given =0.25 p.u
Base KV given = 18KV
Base MVA given = 500MVA
Base MVA new = 100 MVA
Base KV new = 20KV
TO FIND:
Z (p u) new
SOLUTION:
Z ( p u) new = Z p u, given
2







BaseKVnew
nBaseKVgive
enBaseMVAgiv
BaseMVAnew
= 0.25
2
20
18
500
100













Z p u ( new) =0.0405 per unit.
5) If the reactance in ohms is 15, find the per unit value for a base of 15KVA and
10 KV.
GIVEN:

20. Actual reactance = 15
KV Base=10KV
KVA Base = 15 KVA.
TO FIND: Z, p u.
SOLUTION:
Z p u =
nceBaseimpeda
ceActualreactan
Base impedance, Z b = 1000
2

KVA
KVB
=  67.66661000
15
102
Z p u = pu0022.0
67.6666
15

Z p.u = 0.0022 p u.
6) A generator rated at 30 MVA, 11KV, has a reactance of 20%. Calculate its per
unit reactance for a base of 50 MVA and 10KV. (Nov/Dec – 2004)
GIVEN:
Base MVA, given = 30 MVA
Base KV, given =11KV
Z p u, given = 0.2
Base MVA, new = 50MVA
Base KV, new = 10 KV
TO FIND :
Z p u ,new.
SOLUTION:
Z p u (new) = Z p u, given
2
,,
,







newBaseKV
nBaseKVgive
givenBaseMVA
newBaseMVA
= 0.2
2
10
11
30
50













Z p u (new) = 0.4033p .u.

21. 7) A  connected generator rated at 300MVA, 33KV has a reactance of 1.24 p.u .
Find the ohmic value of reactance.
GIVEN:
Base MVA = 300MVA
Base KV = 33 KV
Per unit reactance = 1.24 p .u.
TO FIND :
Actual reactance.
SOLUTION:
Base impedance, Z b = phase
MVAB
KVB
/63.3
300
33
.
. 22

Z p u = ZbZpuActualval
valBase
valAct

.
.
Reactance of generator = Z p u Zb
= 1.24 63.3
Actual value = 4.5012 phase/ .
8) The base KV and Base MVA of a 3  transmission line is 33KV and 10 MVA
respectively. Calculate the base current and base impedance.
GIVEN:
Base KV = 33KV
Base MVA = 10MVA
TO FIND :
Base current, I b
Base impedance, Z b
SOLUTION:
Base current, I b =
B
B
B
B
KV
MVA
KV
KV
3
1000
3



22. I b = A95.174
333
100010



Base impedance, Z b =
  phase
MVA
KV
B
B
/9.108
10
3322

I b = 174.95 A
Z b = 108.9 phase/ .
9) In the above problem if the reactances are given in p. u. Compute the p.u. quantities
based on a common base.
Transformer T1 = 0.209
T2 = 0.220
Generator G1 = 0.435
G2 = 0.413
G3 = 0.3214
SOLUTION:
All the above p.u values are given based on their own ratings of the equipments.
All these must be changed to common base.
Z p u (new) = Z p u (given) 
givenMVA
newMVA
B
B
,
,

newKV
givenKV
B
B
,
,
2
2
Generator G1:
Z p u (new) = 0.435 
30
30

2
11
5.10






= 0.3964
Z 1G (p u), new = 0.3964
Transformer T1:
Z p u (new) = 0.209 
15
30

2
33
33






= 0.418
Z 1T (p u), new = 0.418
Transformer T2:
Zpu (new) = 0.220 
15
30

2
33
33






= 0.44

23. Z 2T (pu),new = 0.44
Generator G2:
Zpu (new) = 0.413 
15
30

2
2.6
6.6






= 0.936
Z 2G (pu),new = 0.936
Generator G3:
Zpu (new) = 0.3214 
25
30

2
2.6
6.6






= 0.437
Z 3G (pu),new = 0.437
10) A 3 ,  transformer with rating 100 KVA, 11KV/400V has its primary
and secondary leakage reactance as 12/phase and 0.05/phase respectively.
Calculate the p. u reactance of transformer.
GIVEN:
KVA =100
Vp = 11KV
Vs = 400V
Xp = 12 1/ Xphase 
X s = 0.05 2/ Xphase
TO FIND:
Per unit reactance of transformer.
SOLUTION:
The high voltage winding (primary) ratings are chosen as base values.
KV B = 11KV
KVA B = 100KVA
Base impedance, Z 

 1210
100
100011
1000
22
B
B
B
KVA
KV
Transformer line voltage ratio, K= 0364.0
11000
400


24. Total leakage reactance referred to primary 211 XXXO  1
= X
2
2
1
K
X

= 12 + 2
)0364.0(
05.0
XO1 = 12 +37,736
XO1 = 49.736 phase/
p u reactance per phase Xpu =
nceBsaeimpeda
cegereacTotalleaka tan
X p u = pu
Z
XO
B
0411.0
1210
736.491

X p u =0.0411 p u
Case (2) ;
The low voltage winding (secondary) ratings are chosen as base values.
KV B =0.4 KV
KVA B = 100 KVA
Base impedance, Z
100
10004.0
1000)(
222


B
B
B
B
B
KVA
KV
OR
MVA
KV
Z  6.1B
Transformer line voltage ratio, K = .0364.0
11
4.0

Total leakage reactance referred to secondary,
XO 2
1
12 XX 
= K 21
2
XX 
= (0.0364) 2
05.012 
XO ./0659.02 phase
P .u reactance per phase, X p u.

25. X p u = 0411.0
6.1
0659.0tan 2

BZ
XO
nceBaseimpeda
cegereacTotalleaka
X p u = 0.0411 p u.
RESULT:
P .u reactance of a transformer referred to both primary and
secondary are same.
12) A 3  transformer is constructed using three identical 1 transformer of
rating 200 KVA, 63.51KV/11KV transformer. The impedances of primary and
secondary are 20+j 45  and 0.1 +j 0.2 respectively. Calculate the p .u impedance
of the transformer.
GIVEN:
KVA KVAB 6003200 
3for
63.51 KV11/3
 110/11 KV
Z p =20 + j45 1Z
Z s = 0.1 + j0.2 2Z
Case (1):
The high voltage winding (primary) ratings are chosen as base
values.
KV B =110KV
KVA B = 600 KVA.
Base impedance, Z 

 67.20166
600
1000110
1000
22
B
B
B
KVA
KV
Transformer line voltage ratio, K= 11/110 =0.1
Total impedance referred to primary: ZO1

26. Z01 =Z 22
2
1
1
21
1.0
2.01.0
)4520(
J
J
K
Z
ZZ


Z01 =20 +j 45 +10 +j 20 = 30 + j 65 phase/
Z01 =30 +j 65 phase/ .
P .u impedance per phase,
Z p u =
7.20166
6530lim j
nceBaseimpeda
pedanceTota 

Z p u = 1.487 33
10223.310 
 j
Z p u = 0.001487+j0.003223 p u.
Case (2) :
The low voltage winding (secondary) ratings are chosen as base.
KV KVB 11
KVA KVAB 600
Base impedance per phase,
Z 

 67.201
600
100011
1000
22
B
B
B
KVA
KV
Transformer line voltage ratio K =11/110 =0.1
Total impedance referred to secondary} Z 0 2
1
12 ZZ 
= K 21
2
ZZ 
= (0.1) )2.01.0()4520(2
jj 
= 0.2+ j 0.45+ 0.1+j0.2
Z 0 phasej /65.03.02 
P .u reactance per phase,
Z p u =
67.201
65.03.0lim j
nceBaseimpeda
pedanceTota 


27. Z p u = 1.487 33
10223.310 
 j .
Z p u =0.001487+j0.003223 p u.
RESULT:
The p .u impedance of transformer referred to both primary and
secondary are same.
13) The 3 ratings of a 3 winding transformer are
Primary:  connected , 110KV,20MVA
Secondary:  connected, 13.2KV,15MVA
Tertiary: - connected, 2.1KV, 0.5MVA.
Three short circuit tests performed on this transformer yielded the following results.
(1) Primary excited , secondary shorted : 2290V,52.5A
(2) Primary excited , tertiary shorted : 1785V, 52.5A
(3) Secondary excited ,tertiary shorted :148v, 328A
Find the p .u impedance of the star –connected 1 equivalent circuit for a base
of 20MVA, 110KV in the primary circuit .Neglect the resistance.
SOLUTION:
Test (1) & (2) are performed on primary winding, hence the p. u
impedances of Z p s and Z p t can be obtained directly using the
primary winding ratings as base values.
The test (3) is performed on secondary winding, hence the p .u
impedance Z s t is obtained using secondary winding rating as bases
and then it can be converted to primary winding base.
To find p. u value of Z p s and Z p t
KV B , primary = 110KV
MVA B , p y = 20MVA
Base impedance of primary c k t} Z b, p y =
20
11022

B
B
MVA
KV

28. Z b, p y = 605 phase/
(P y -  -connected)
Z p s in phasephase /1835.25
5.52
3/2290
/ 
P u value of z p s = up
pyZb
inZps
.0416.0
605
1835.25
,
,


(Z p s) )( pu = 0.0416 p u.
(P y -  connected)
Z p t in ./6299.19
5.52
3/1785
/ phasephase 
P u value of Z p t = 0324.0
605
6299.19
,
,


pyZb
inZpt
(Z p t ) p u =0.0324 p u.
To find p u value of Z s t
Base KV,sec = 13.2KV
MVA B ,sec= 15 MVA
Z b , sec = phase
MVA
KV
B
B
/616.11
15
2.13sec, 2

(Secondary-  -connected)
Z s t in phasephase /2605.0
328
3/148
/ 
P u value of Z s t, Z s t (p u) =
sec,
,
Zb
inZst 
=
616.11
2605.0
Z s t(p u) = 0.0224 p u.
P .u value of Z s t on secondary circuit base values can be converted to primary
circuit base using the following formula.

29. Z p u, new = Z p u given
2
,,
,, 












newKV
KV
givenMVA
MVA
B
givenB
B
newB
New refers to primary & given refers to secondary.
Secondary KV =13.2KV
New KV KVB 2.13
110
2.13
110 
Z 1
s t (p u) =0.0224 pu0299.0
2.13
2.13
15
20
2













Z 1
s t (p u) = 0.0299 p u.
To compute Z p , Zs1
and Z t1
Z p =
2
1
 tZsZptZps 1
 =
2
1
 0299.00324.00416.0 
Z p =0.0221 p u
.
Z    0324.00299.00416.0
2
1
2
1 11
 ZptZstZpss
Z 1
s = 0.0196 p u
Z t    0416.00299.00324.0
2
1
2
1 11
 ZpsZstZpt .
Z t1
=0.01035 p u.
14) A 120 MVA, 19.5 KV generator has a synchronous reactance of 0.15 p. u and it is
connected to a transmission line through a transformer rated 150MVA, 230/18KV
(  / ) with x =0.1 p .u.
a) Calculate the p .u reactance by taking generator rating as base values.
b) Calculate the p .u reactance by taking transformer rating as base values.
c) Calculate the p .u reactance’s for a base value of 100MVA and 220KVon HT side
of the transformer.

30. SOLUTION:
a) Generator rating as base values:
Base MVA, MVA B , new = 120 MVA
Base KV, KV B , new = 19.5KV
Z g (p u) ,given = 0.15 p u. Zt (p u)=0.1 pu.
For transformer
(Z t) p u, new = Z p u given
2













newKV
givenKV
givenMVA
newMVA
B
B
B
B
= 0.1 pu0682.0
5.19
18
150
120
2













Z t ( p u) = 0.0682 p u.
b) Transformer rating as base values :
KV B , new = 18 KV
MVA newB , = 150 MVA
Z t = 0.1 p u , Z g = 0.15 p u
Z g, new (p u ) =0.15 pu2200.0
18
5.19
120
150
2













Z g (p u) = 0.2200 p u.
c) KV B = 220KV on HT side
MVA B =100MVA
(Z t ) p u, new = 0.1 pu0729.0
220
230
150
100
2













(Z t ) p u = 0.0729 p u.
Generator is connected to LT side of t r

31. KV B  220 KV22.17
220
18 
Z g (p u) , new = 0.15
2
22.17
5.19
120
100












 = 0.1603 p u.
Z g ( p u) = 0.1603 p u.
RESULT:
a) Z p u, gen = 0.15 p u , Z p u , tr = 0.0682 p u .
b) Z p u, gen =0.22 p u, Z p u t r = 0.1 p u.
c) Z p u gen = 0.1603 pu , Z p u, tr = 0.0729pu .
15) A 50 KW, 3 ,  -connected load is fed by a 200KVA, transformer with voltage
rating 11 KV/ 400V through a feeder. The length of the feeder is 0.5 km and the
impedance of the feeder is 0.1+ j 0.2 /km, if the load pf is 0.8. Calculate the p.u
impedance of the load and feeder.
Given:
P =50KW
KVA B = 200KVA
Vpy = 11KV
Vsec =400V
Length of the feeder = 0.5km
Z feed = (0.1+j 0.2) km/
= (0.1+j 0.2) 0.5 =0.05 +j 0.1 ph/
pf = 0.8.
Solution:
Choose secondary values as base values.
KV KVB 4.0
KVA B =200KVA
Base impedance, Zb =
B
B
KV
KV 10002



 8.0
200
10004.0 2
Z act 1 = 0.05 + j 0.1 ph/

32. Z p u = puj
j
Zb
Zact
125.00625.0
8.0
1.005.0



Z p u = 0.0625+ j 0.125 p u.
To calculate p .u impedance of the load
P = 50KW, PF = COS = 0.8

 86.36)8.0(1
COS
sin  = 0.599
Reactive power, Q = 599.0
8.0
50
sin
cos
 

p
= 37.44 KVAR.
Load impedance Per phase , Z L = 3
22
10)44.3750(
400][


 jjQp
VL
Z L =
44.3750
10400 32
j
 
= 



83.3656.2
83.3646.62
160
ZL = 2.05 + j 1.535 ./ phase
p .u value of load impedance, Z 


8.0
535.105.2
,
j
Z
Z
pu
b
L
L
Z )(puL = 2.563+j 1.9188p u.

33. UNIT - II
NETWORK MODELLING
SYLLABUS
Primitive network and its matrices - Bus incidence matrix - Bus
admittance and bus impedance matrix formation - -equivalent circuit of
transformer with off-nominal-tap ratio - Modelling of generator, load, shunt
capacitor, transmission line, shunt reactor for short circuit, power flow and
stability studies.
INTRODUCTION
 The systematic computation of currents and voltage in power system networks
may be achieved by constructing a suitable mathematical model. A network
matrix equation provides a suitable and convenient mathematical model and
describes both the characteristics of the individual elements of the power system
network and their interconnections.
 For analysis of large power system network a primitive network is first
considered. The primitive network is a set of uncoupled elements of the power
system network.
 Generally data available from any power company is in the form of primitive
network matrix. This matrix does not provide any information‘s pertaining to the
network connections. So that the primitive network matrix must be transformed
into a network matrix that describes the performance of the interconnected power
system network.
 The performance of the network matrix used in the network equations can be in
Dos frame of reference or loop frame of reference. In the loop frame of reference

34. the variables are loop voltages and loop currents. In the bus frame of reference the
variables are nodal voltages and nodal currents.
 The network equations corresponding to the bus frame and loop frame of
references are very useful In short circuit studies, load flow studies and power
system stability.
BASICS OF NETWORK TERMINOLOGY
There are certain terms used in network solution which must be defined before the
network equations are developed. Every electrical network consists of circuit elements,
branches, nodes and loops.
(i) Circuit element : These are active elements and passive elements.
Active elements : Voltage and current sources.
Passive elements : Resistor (R), Inductor (L) and Capacitor (C).
(ii) Branch : A branch is a single element or group of elements connected in either series
or parallel to form a device with two terminals only, these terminals forming the only
connections to other branches. A branch may or may not contain active elements.
(iii) Node: This is a point in a network which forms the connection between two or more
branches. In a circuit diagram of the network a node may be a line representing the bus
bar or impedance less connections.

35. (iv) Loop: Any path through two or more branches which forms a closed circuit is called
a loop or mesh.
For- example, a sample network is shown.
Fig. (2.1)
Nodes are a, b, c, d and e
Loops are L1, L2 & L3
Voltage source :e (t)
Branches are 1, 2, 3, 4, 5, 6 & 7
PRIMITIVE NETWORK AND ITS MATRICEX
A set of unconnected branches of a given network is called a primitive network.
A branch may consist of active and passive elements.
These set of unconnected branches are known as primitive network. For any
general network we can represent any unconnected branch in impedance form or
admittance form.

36. EP Epq Eq
P - + q
ipq
Vpq = Ep - Eq
Fig(a) : Admittance Form
jpq
Ep Eq
p ipq q
ipq + jpq
Vpq = Ep - Eq
Fig(b) : Impedance Form
Fig : Representation of a network component
Network components represented both impedance form and admittance form are
shown in figure.
The performance of the components can be expressed using either form
The variables and parameters are
Vpq ------- voltage across the element pq
Epq ------- source voltage in series with element pq
ipq -------- current through element pq
jpq --------- source current in parallel with element pq
Zpq --------- self impedance of element pq
ZPq
Ypq

37. Ypq ------- self admittance of element pq
Each element has two variables. Vpq and ipq.
In steady state these variables and the parameters of the elements Zpq and Ypq are
real number for direct current circuit and complex number for alternating current circuits.
The performance equation of an element in impedance form is
)1(..............................pqpqpqpq iZEV 
or in admittance form is
)2..(..............................pqpqpqpq VYji 
the parallel source current in admittance form is related to the series source
voltage in impedance form by
)3...(..............................* pqpqpq EYj 
pq
pq
Z
Y
1

A set of unconnected element is defined as a primitive network. The performance
equation of a primitive network are given by
In the impedance form
V + E = Z*I …………………………(4)
(or) BusBusBus IZE *
In the admittance form
I + J = Y *V ………………… ……..(5)
(or) BusBusBus EYI *
where,
BusE ------- Vector of bus voltage measured to the reference bus
BusI -------- Vector of impressed bus currents
V and I are the element voltage and current vectors
J and E are source vectors
Z and Y are the primitive impedance and admittance matrices

38. BUS INCIDENCE MATRIX:
GRAPH
Incidence Matrix : A
n be the number of nodes
e be the number of elements
A is the incidence matrix whose n-rows correspond to ‗n‘ nodes and
e – columns correspond to ‗e‘ elements.
The matrix elements are
a ij = 1 If the jth
element is incident to but directed away from
node i.
a ij = -1 If the jth
element is incident to but directed towards the
node i.
a ij = 0 If the jth
does not incident
5
9
8
7
2
3
4
1
21
0 3
4
5

39. The dimension of this matrix is n x e
For the given graph, the incidence matrix is
6 x 9  n = 6 (including reference node)
e = 9
6 rows , 9 columns.
n 1 2 3 4 5 6 7 8 9
0 1 1 1
1 -1 1
2 -1 1 1
3 -1 -1 -1 1
4 -1 1 1
5 -1 -1 -1
Reduced incidence (or) Bus incidence Matrix –„A‟:
Any node of the connected graph can be selected as the reference node and
then the variables of the remaining (n-1) nodes are termed as buses and measured with
respect to the assigned node.
Te matrix ‗A‘ obtained from the incidence matrix A is by deleting the
reference row (corresponding to reference node) is termed as REDUCED OR BUS
INCIDENCE MATRIX.
Number of buses = n-1
Order of the matrix is (n-1) x e

40. For the given graph, (n-1) x e 5 x 9
Node - 0 is taken as the reference node.
e
n 1 2 3 4 5 6 7 8 9
1 -1 1
2 -1 1 1
3 -1 -1 -1 1
4 -1 1 1
5 -1 -1 -1
FORMATION OF BUS ADMITTANCE AND BUS IMPEDANCE
MATRICES:
The bus admittance matrix YBus can be obtained with the help of bus incidence
matrix [A] .
W.K.T  VYji  (1)
Premultiply the equation (1) by [A]
        VYAjAiA  (2)
By KCL,
  0iA
  BusIjA  (3)
ie. The algebraic sum of external current sources at each bus and is equal to the
impressed bus current IBus.
From (3), (2) becomes

41.    VYAIBus (4)
   VEA Bus
T
(5)
from (5), (4)      Bus
T
Bus EAYAVYAI 
BusBusBus EYI 
  T
Bus AYAY 
 Y primitive admittance matrix
  1
 BusBus YZ
Procedural steps for finding BusBus YZ &
Step 1: Draw the graph, Tree of the given network and identify the nodes and
assign the direction of currents.
Step 2: Formulate the primitive square admittance matrix YPR
Step 3: Formulate the bus incidence matrix [A] and determine [A]T
Step 4: The bus admittance matrix is determined using
Step 5: The bus impedance ZBus is found using
  T
prBus AYAy 
  1
 BusBus YZ
  T
prBus AYAY 

68. UNIT – III
SYLLABUS
SHORT CIRCUIT ANALYSIS
 Need for short circuit study.
 Approximations in modelling
 Calculation for radial networks.
 Symmetrical short circuit analysis
 Symmetrical component transformation
 Sequence impedances
 Z–bus in phase frame and in sequence frame fault matrices
 Unsymmetrical fault analysis.
Need For Short Circuit Study
Short circuit study helps in determining the correct current rating of circuit
breaker as well as the associated current transformers. So that the system is well
protected under all possible circumstances and at the same time the expenditure is
minimized.
Another purpose is grading the operating time of various circuit breaker. In such
a way that it will operate only if circuit breakers which are ahead of it towards the part do
not affect. This is called relay coordination.
In short circuit analysis bus impedance matrix plays a vital role.

69. Approximations in modeling :
We have already seen about the approximate model circuit for transmission lines
and synchronous generator and motors.
Symmetrical Short Circuit Analysis
A power network comprises synchronous generators, transformers, lines and loads.
Though the operating conditions at the time of fault are important, the loads can be
neglected during fault, as voltages dip very low so that currents drawn by loads can be
neglected in comparison to fault currents.
The synchronous generator during short circuit has a characteristic time varying
behavior. In the event of a short circuit, the flux per pole undergoes dynamic change with
associated transients in damper and field windings.
The reactance of circuit model of the machine changes in the first few cycles from
a low subtransient reactance to a higher transient value finally setting at a still higher
synchronous value. Depending upon the arc interruption time of circuit breakers, a
suitable reactance value is used for the circuit model of synchronous generators for short
circuit analysis.
Short Circuit Of A Synchronous Machine On No Load
The armature reaction of a synchronous generator produces a demagneting flux
under steady state short circuit condition. Due to this armature reaction, the reactance Xa
is in series with the induced emf.‘
This reactance Xa is combined with the leakage reactance X of the machine is
called synchronous reactance Xd‘.
In salient pole machines Xd is the direct axis synchronous reactance.
The steady state short circuit model of a synchronous machine is shown on per phase
basis.

70. Let us consider now the sudden short circuit of a synchronous generator is
initially operating under open circuit. The machine undergoes a transient in all the three
phases. Immediately in the event of short circuit, the symmetrical short circuit current is
limited only by the leakage reactance of the machine. At this time, to help the main flux
some flux will be produced in the field winding and damper winding due to the short
circuit current. These currents decay in accordance with the winding time constants.
The time constant of damper winding is less than that of field winding because the
damper winding has a low leakage inductance and the field winding has a high leakage
inductance.
So that the reactance‘s of field winding and damper windings are all considere).
also these two windings are in parallel and also parallel with armature reactanc >‗
The combination of these reactance‘s in series with the leakage reactance is called at
subtransient reactance Xd‖.

145. State the assumptions made In Fault Studies
 All shunt connections from system buses to the reference node (neutral ) can be
neglected in all equivalent circuits representing transmission lines and
transformers.
 Load impedances are much larger than those of network components and so they
can be neglected in system modeling.
 All buses of the system have rated / nominal voltage of 0
00.1  per unit so that no
prefault current flows in the network.
 The voltage source plus series impedance equivalent circuit can be transformed to
an equivalent current source plus shunt impedance model. Then the shunt
impedances of the machine model represent the only shunt connections to the
references node.

146. UNIT – IV
POWER FLOW ANALYSIS
SYLLABUS
Problem definition - Bus classification - Derivation of power flow
equation - Solution by Gauss Seidel and Newton Raphson methods -
P-V bus adjustments for both methods - Computation of slack bus
power -Transmission loss and line flow.
INTRODUCTION
The load flow studies involves the solution of the power system network under
steady state conditions. The solution is obtained by considering certain inequality
constraints imposed on node voltages and reactive power of generators.
The main information obtained from a load-flow study are the magnitude and
phase angle of the voltage at each bus and the real and reactive power flowing in each
line. The load flow solution also gives the initial conditions of the system when the
transient behavior of the system is to be studied.
The load flow study of a power system is essential to decide the best operation of
existing system and for planning the future expansion of the system. It is also essential
for designing a new power system.
The load flow solution of power system lo condition have been presented in this
chapter. For analysis of balanced system a single phase equivalent circuit of power
system such as impedance diagram is adequate.
A load flow study of a power system generally require the following steps.
(i) Representation of the system by single line diagram

147. (ii) Determining the impedance diagram using the information in single line
diagram.
(iii) Formulation of network equations
(iv) Solution of network equations.
Under steady state conditions the network equations will be in the form of simple
algebraic equation in system the load and hence generation are continuously changing,
but for calculation purpose we assume that the loads and generation are fixed at a
particular value for a suitable period of time, e.g., 15 minutes or 30 minutes.
TYPES OF BUSES:
In a power system the buses are meeting at various components. The generator
will be feed energy to buses and loads will draw energy from buses. In the network of a
power system the buses becomes nodes and so a voltage can be specified for each bus.
Therefore each bus in a power system, is associated with four quantities and they
are real power, reactive power, magnitude of voltage and phase angle of voltage. In a
load flow problem two quantities (out of four) are specified for each bus and the
remaining two quantities are obtained by solving the load flow equations.
The buses of a power system can be classified into following three types based on
the quantities being specified for the buses.
The table 2.1 shows the quantities specified and to be obtained for each type of
bus.
(i) Load bus (or PQ-bus)
(ii) Generator bus (or voltage controlled bus or PV bus)
(iii) Slack bus (or swing bus or reference bus).

148. The following table shows the quantities specified and to be obtained for various types of
buses
Bus type Quantities specified Quantities to be
obtained
Load bus P,Q ,V
Generator bus P, V Q,
Slack bus V , P,Q
 Load Bus:
The bus is called load bus, when real and reactive components of power are
specified for the bus. The load flow equations can be solved to find the magnitude and
phase of bus voltage. In a load bus the voltage is allowed to vary within permissible
limits, for example ±5%.
 Generator Bus:
The bus is called generator bus, when real power and magnitude of bus voltage
are specified for the bus. The load flow equation can be solved to find the reactive power
and phase of bus voltage. Usually for generator buses reactive power limits will be
specified.
 Slack Bus:
The bus is called slack bus if the magnitude and phase of bus volt are specified for
the bus. The slack bus is the reference bus for I flow solution and usually one of the
generator bus is selected as a slack bus.
NEED FOR SLACK BUS
Basically the power system has only two types of buses and they are load and
general buses. In these buses only power injected by generators and power drawn by
loads are specified but the power loss in transmission lines are not accounted. In a power

149. system the total power generated will be equal to sum of power consumed by loads and
losses.
Therefore in a power system,
Sum of complex Sum of complex Total (complex) power
Power of generator = power of loads + loss in transmission lines
The transmission line losses can be estimated only if the real and reactive power
of buses are known. The powers in the buses will be known only after solving the load th
equations. For these reasons, the real and reactive power of one of the generator bus is n
specified and this bus is called slack bus. It is assumed that the slack bus generates the
real ax reactive power required for transmission line losses. Hence for a slack bus, the
magnitude at phase of bus voltage are specified and real and reactive powers are obtained
through the load flow solution.
STATIC LOAD FLOW EQUATION (OR) SLFE :
Either the bus self and mutual admittances which compose the bus admittance
matrix YBus or the driving point and transfer impedance which compose ZBus may be
used in solving the power –flow problem.
The YiJ of N x N bus admittance matrix is given by
ijijij yy 

Bij
ijij
Gij
ijij SinyjCosy   (1)
The voltage at typical bus (i) of the system is given in polar coordinates by
 iiiiii jCosVVV  sin (2)
and voltage at another bus (j) can be written as
 jjjjjj jCosVVV  sin (3)

150. The net current injected into the network at bus (i) in terms of elements Yin of
YBus is given by the summation.
NiNiii VyVyVyI  ...................2211


N
n
nini VyI
1
(4)
Let Pi  be the net real power entering the network at the bus (i)
Qi  be the net reactive power entering the network at the bus (i)
The complex conjugate of the power injected at bus (i) is
(5)
ininniin
N
n
ii QVVyjQP   1
(6)
Expanding this equation and equating real and reactive parts, we get,
 ininniin
N
n
i CosVVyP   1
(7)
 ininniin
N
n
i sINVVyQ   1
(8)
(7) & (8) are called as SLFE  Static Load Flow Equations
(7) & (8) forms the polar form of the power flow equations . They provide calculated
values for the net real power Pi and reactive power Qi entering the network at typical bus
(i).


N
n
niniii VyVjQP
1
*

151. Let, Pgi  Scheduled power being generated at bus (i)
Pdi  Scheduled power demand of the load at that bus.
Then, is the net scheduled power being injected into the network at bus (i)
Notation for active power at typical bus (i) in power flow studies.
Pi, Calc  Calculated value of Pi
Mismatch calcPSchPP iii ,,  (9)
  calcPPPP idigii , (10)
Similarly for reactive power at bus (i)
calcQschQQ iii ,,  (11)
  calcQQQQ idigii  (12)
Notation for reactive power at bus (i) in power flows studies.
Mismatches occur in the course of solving a power flow problem when calculated
values of Pi and Qi do not coincide with the scheduled values.
If Pi, Calc = Pi, sch &
Qi, calc = Qi, sch
then mismatches 0 ii QP
The power balance equations are,
  0,1
 digiischiii PPPPPg (13)
  0,11
 digiischiii QQQQQg
For potentially unknown quantities associated with each bus (i) are Pi, Qi, voltage
angel i and voltage magnitude iV
The general practice in power flow studies is to identify three types of buses in
the network.
digii PPSchP 

152. At each bus (i) two of the four quantities iii PV ,, and Qi are specified and remaining
two are calculated.
LOAD FLOW SOLUTION (EQUATION) :
The assumptions and approximations made in the load flow equations.
   iCosVyVP ninn
N
n
inii   1
(1)
   iSinVyVQ ninn
N
n
inii   1
(2)
are given by
i. Line resistances being small are neglected. Active power loss PL of the
system is zero.
Thus in (1) & (2)  9090 iiin and 
ii.  ni   is small 






6

so that    niniSin  
ii. All buses other than the slack bus (bus (1)) are PV buses.
i.e. Voltage magnitudes at al the buses including the slack bus are specified
(1) & (2) are reduced as
   NinVyVP in
N
n
inii ......2,1,
1
 
 (3)
    
Ni
yVnCosVyVQ iiiin
N
in
n
inii
......2,1
2
1

 



(4)
Since iV ‘S are specified equation (3)
represents a set of linear algebraic equations in i ‗s which are (N-1) in number as 1 is
specified at slack bus  01  .
The Nth
equation corresponding to slack bus (N=1) is redundant as the real power
injected at this bus is now fully specified as,

153.  0;
22
1   
L
N
i
Gi
N
i
Di PPPP
Equation (3) can be solved explicitly for n .............., 32 which is substituted in (4)
yields Qi ‗s the reactive power bus injections.
FORMULATION OF LOAD FLOW EQUATIONS USING Y Bus
MATRIX
The load flow equations can be formed using either the mesh or node basis
equation of a power system. However, from the view point of computer time and
memory, the nod admittance formulation using the nodal voltages as the independent
variables is the mo economic.
The node basis matrix equation of a n-bus system given by
Ybus V=I …………………………(1)
where
Ybus ------- Bus admittance matrix of order (nxn)
V Bus (node) --- voltage matrix of order (n x 1)
I --------- Source current matrix of order (n x 1).
An separating the real and imaginary parts of eqn (1) we get.
………………. (2)
………………… (3)
…………………….. (4)
The equations (2.13), (2.14) and (2.15) are called load-flow equations of Newton-
Raphson method.

154. LOAD FLOW SOLUTION BY GAUSS-SEIDEL METHOD
The Gauss-Seidel method is an iterative algorithm for solving a set of non-linear
load flow equations. The non-linear load flow equations are given by equ (2),
when p = 1,2 n and this equation is presented here for convenience
V










  

 
1
1 1
*
)(
1 p
q
n
pq
qpqqpq
p
pp
pp
p VYVY
V
jQP
Y
……………. (5)
where P = 1,2,3 ……… n
The variables in the equations obtained from equ (5) for p = 1,2,3 ……. n are the
node voltages V1 ,V2 ,V3 ,……….Vn. In Gauss-Seidel method an initial value of voltages
are assumed and they are denoted as V1
‘
,V2‘ ,V3
‘
,……….Vn
‘
. On substituting these
initial values in equ (5) and by taking p = 1, the revised value of bus- 1 voltage V1
‘
is
computed. The revised value of bus voltage V1
‘
is replaced for initial value V0 and the
revised bus-2 voltage V2‘ is computed. Now replace the V1‘ for V1 and V2‘ for V2 and
perform the calculationforbus-3 and soon.
The process of computing all the bus voltages as explained above is called one
iteration. The iterative process is then repeated till the bus voltage converges within
prescribed accuracy. The convergence of bus voltage is quite sensitive to the initial
values assumed. Based on practical experience it is easier to get a set of initial voltages
very close to final solution.
In view of the above discussions the load flow equation [5] can be written in the
modified form as shown below.
V










  

 

1
1 1
1
*
1
)(
1 p
q
n
pq
K
qpq
K
qpqK
p
pp
pp
K
p VYVY
V
jQP
Y
…………………….(6)
where,
Vi
k
= kth
iteration value of bus voltage Vi
Vi
k+1
= (k+1)th
iteration value of bus voltage Vi
It is important to note that the slack bus is a reference bus and so its voltage will
not change. Therefore in each iteration the slack bus voltage is not modified

155. For a generator bus, the reactive power is not specified. Therefore in order to calculate
the phase of bus voltage of a generator bus using equ (6), we have to estimate the reactive
power from the bus voltages and admittances as shown below.
From equ we get.
……………………….(7)
From equ(7) the equation for complex power in bus-p during (k + i)th
iteration can be
obtained as shown in equ (8).
…………………(8)
The reactive power of bus-p during (k + i)th
iteration is given by imaginary part of equ (8)
………….. (9)
Also, for generator buses a lower and upper limit for reactive powers will be
specified. In each iteration, the reactive power of generator bus is calculated using equ(9)
and then checked with specified limits. If it violates the specified limits then the reactive
power of the bus is equated to the limited and it is treated as load bus. If it does not
violate the limits then the bus is treated as generator bus.
GAUSS SEIDEL METHOD USED IN POWER FOLW ANALYSIS:
Digital solutions of the power flow problems follow an iterative process
by assigning estimated values to the unknown bus voltages and by calculating a
new value for ach bus voltage from the estimated value at the other buses and the
real and reactive power specified.

156. A new set of values for the voltage at each bus is thus obtained and used
to calculate still another set of bus voltages.
Each calculation of a new set of voltages is called an ITERATION.
The iterative process is repeated until the changes at each bus are less than a
specified minimum value.
We derive equations for a four bus system with the slack bus designated as number 1
computations start with bus (2)
If P2, sch and Q2, sch are the scheduled real and reactive power, entering the
network at bus (2)
From the equation


N
n
niniii VyVjQP
1
*
(1)
with i = 2 and N = 4
424323222121*
2
22
VyVyVyVy
V
schjQschP


(2)
Solving for V2 gives
 







 424323121*
2
22
22
2
1
VyVyVy
V
schjQschP
y
V (3)
If suppose bus (3) and (4) are also load buses with real and reactive power
specified.
At bus (3)
 







 434232131*
3
33
33
3
1
VyVyVy
V
schjQschP
y
V (4)
Similarly at bus (4)
 







 343242141*
4
44
44
4
1
VyVyVy
V
schjQschP
y
V (5)

157. The solution proceeds by iteration based on scheduled real and reactive power at
buses (2), (3) and (4). The scheduled slack bus voltage is 111  VV and initial
voltage estimates      0
4
0
3
0
2 , VandVV at other buses
Solution of equation (3) gives the corrected voltage  1
2V calculated from
 
 
   
 







 0
424
0
323121*0
2
,2,2
22
1
2
1
VyVyVy
V
schQschP
y
V (6)
 All the quantities in the right hand side expression are either fixed
specifications or initial estimates.
The calculated value  1
2V and the estimated value  0
2V will not agree.
Agreement would be reached to a good degree of accuracy after several iterations.
This value would not be the solution for V2 for the specific power flow conditions,
however because the voltages on which this calculation for V2 depends are the
estimated values  0
3V and  0
4V at the other buses, and the actual voltages are not yet
known.
Substituting
 1
2V in 4th
equation we obtain the first calculated value at bus
(3)
 
 
   
 







 0
434
1
232131*0
3
,3,3
33
1
3
1
VyVyVy
V
schjQschP
y
V (7)
The process is repeated at bus (4)
 
 
   
 







 1
343
1
242141*0
4
,4,4
44
1
4
1
VyVyVy
V
schjQschP
y
V (8)
This completes the first iteration in which the calculated values are found for each
state variable.
Then, the entire process is carried out again and again until the amount of
correction in voltage at every bus is less than some predetermined precision index.
This process of solving the power – flow equations is known as the “GAUSS
– SEIDEL ITERATIVE METHOD”

158. It is common practice to set the initial estimates of the unknown voltages at all
load buses equal to 0
00.1  per unit.
Such initialization is called a FLAT START because of the uniform voltage
profile assumed.
For a system of N buses the general equation for the calculated voltage at any bus
(i) where P and Q are scheduled is
 
 
   








  




N
ij
k
jij
i
j
k
jijk
i
ii
ii
k
i VyVy
V
schjQschP
y
V
1
1
1
1
*1
,,,1
(9)
The superscript (K) denotes the number of iteration in which the voltage is currently
being calculated and (k-1) indicates the number of the preceding iteration.
Equation (9) applies only at load buses where real and reactive power are
specified.
An additional step is necessary at Voltage controlled buses where voltage
magnitude is to remain constant.
The number of iterations required may be reduced may be reduced considerably
if the connection in voltage at each bus is multiplied by some constant that increases the
amount of correction to bring the voltage closer to the value it is approaching.
The multiplier that accomplishes this improved convergence is called an
ACCELERATION FACTOR
The difference between the newly calculated voltage and the best previous
voltage at the bus is multiplied by the appropriate acceleration factor to obtain a better
correction to be added to the previous value
For example,
At bus (2) in the first iteration we have the accelerated value accV .1
2 defined by
 
     1
2
0
2
1
2 1, VVaccV  
       
 0
2
1
2
0
2
1
2 , VVVaccV   (10)
 acceleration factor.
Generally, for bus (i) during iteration K, the accelerated value is given by,
 
     k
i
k
acci
k
i VVaccV   1
,1,

159.        1
,
1
,, 
 k
acci
k
i
k
acci
k
i VVVaccV  (11)
In power flow studies  is generally set at about 1.6 and cannot exceed 2 if
convergence is to occur.
Voltage controlled buses (or) PV – buses:
When voltage magnitude rather than the reactive power is specified at bus (i), the
real and imaginary components of the voltages for each iteration are found by first
computing a value for the reactive power.
From
n
N
n
iniii VyVjQP 

1
*






 
j
N
j
ijimi VyVIQ
1
*
(12)
Equivalent algorithmic expression
       














 




 1
1
1
1
*1 k
j
N
j
ij
k
j
i
j
ij
k
im
k
i VyVyVIQ (13)
Im Imaginary par of
 k
iQ is substituted in equation (9) to find a new value of
 k
iV
The components of the new
 k
iV are then multiplied by the ratio of the
specified constant magnitude iV to the magnitude of
 k
iV from (9)th
equation.
In 4 – bus system, if bus (4) is voltage controlled.
Eqn (13) becomes,
         
  0
444
1
343
1
242141
*0
4
1
4 ,, VyaccVyaccVyVyVIQ m  (14)
 The calculated voltages of buses (2) and (3) are accelerated values of the
first iteration.
Substitute  1
4Q for Q4,sch in (9) for bus (4) yields.

160.  
 
 
   








 accVyVyVy
V
jQschP
y
V 1
343
1
242141*0
4
1
44
44
1
4
,1
(15)
 all the quantities on the right hand side are known.
 Since 4V is specified,
We correct the magnitude of  1
4V as
 
 
 1
4
1
4
4
1
4 ,
V
V
VcorrV  (16)
and proceed to the next step with stored value  
corrV ,1
4 of bus (4) voltage having the
specified magnitude in the remaining calculations of the iteration.
The reactive power Qg must be within definite limits.
naxg QQQ min
Advantages of Gauss seidel method
 Calculations are simple and so the programming task is lesser.
 The memory requirement is less
 Useful for small size system.
Disadvantages:
 Requires large number of iterations to reach convergence.
 Not suitable for large systems
 Convergence time increases with size of the system.

161. FLOWCHART FOR LOAD FLOW SOLUTION BY GAUSS-SEIDEL METHOD

162. NEWTON – RAPHSON METHOD FOR LOAD FLOW PROBLEM
To apply the Newton – Raphson method to the solution of the power – flow
equations, we express the bus voltages and line admittances in polar form.
From,
 inin
N
n
niini CosVVyP   1
(1)
 inin
N
n
niini SinVVyQ   1
when n = i in the above equations and separating the terms by summations,
  inin
N
n
inniiiii CosyVVGVP   1
2
(2)
 inin
N
in
n
inniiiii SinyVVBVQ   

1
2
(3)
Gii and Bii 

ijij B
ij
G
ijijij ijSinyjijCosyy  
(4)
jBijGijy ijij 
Assume all buses, (except the slack bus) as load buses with known demands Pdi and
Qdi.
The slack bus has specified values for 1 and 1V ,
For other buses in the network, the two static variable s i and iV are to be
calculated in the power – flow solution .
The power mismatches for the typical load bus (i)
calcPischPiPi ,, 
(5)
calcQischQiQi ,, 
For real power Pi,

163. 4
4
3
3
2
2
4
4
3
3
2
2
V
V
P
V
V
P
V
V
PPPP
P iiiiii
i 

















 





(6)
The last three terms can be multiplied and divided by their respective voltage
magnitudes without altering their values.
3
3
3
3
2
2
2
24
4
3
3
2
2 V
V
V
P
V
V
V
V
P
V
PPP
P iiiii
i
















 





4
4
4
4
V
V
V
P
V i



 (7)
A mismatch equation can be written for reactive power Qi.
3
3
3
3
2
2
2
24
4
3
3
2
2 V
V
V
Q
V
V
V
V
Q
V
QQQ
Q iiiii
i
















 





4
4
4
4
V
V
V
Q
V i



 (8)
Each nonslack bus of he system has two equations like those for pi and Qi
Collecting all the mismatch equations into vector – matrix form yields.
 The solution of (9) is found by the following iteration.
 Estimate values
 0
i and
 0
iV for static variables.
 Use the estimates to calculate
 
calcPi
0
, and
 
calcQi
0
, ,from (2) and (3)
mismatches
 0
iP and
 0
iQ from (5) and the partial derivative elements of the
Jacobian J.
 Solve (9) for the initial corrections
 0
i and
 
 0
0
i
i
V
V
 Add the solved corrections to the initial estimates to obtain.

164.      001
iii   (10)
     001
iii VVV 
 
 
  







 0
0
0
1
i
i
i
V
V
V (11)
 Use the new values
 1
i and
 1
iV as starting values for iteration 2 and
continue.
The general formulas are
     k
i
k
i
k
i  
1
(12)
     k
i
k
i
k
i VVV 
1
 
 
  






 
 k
i
k
ik
i
V
V
V 1 (13)
For the four bus system sub matrix J11 has the form





































4
4
3
4
2
4
4
3
3
3
2
3
4
2
3
2
2
2
11



ppp
ppp
ppp
J (14)
Expressions for the elements of this equation are easily found by differentiating the
appropriate number of terms in (2)
The off diagonal element of J11
 ijijijji
j
i
iSinyVV
P





(15)
The typical diagonal element of J11
 inininni
N
in
ni
i
iSinyVV
P







1

165. 

 


N
in
n n
iP
1  (16)
By comparing (16) & (3)
iiii
j
i
BVQ
P 2




(17)
The formulas for the elements of sub matrix J21 is given by
The off diagonal element of J21 is
 ijijijji
j
i
iCosyVV
Q





(18)
The main diagonal element of J21 are
 inininni
N
in
ni
i
iCosyVV
Q







1


 


N
in
n n
Qi
1  (19)
Comparing (19) with (2) for Pi
]
The off diagonal elements of J12 are simply the negatives of elements in
J21  This is obtained by multiplying
jV
Pi


 by jV
The diagonal elements of J12 are fond by
 
















inin
N
in
n
inniiii
i
i
i CosyVGVV
V
P
V 
1
2 (21)
iiii
j
i
BVQ
P 2




iiii GVP
i
Qi 2





166. Comparing (21) with (19) & (20)
iiiiiii
i
i
i
i
i GVPGV
Q
V
P
V
22
2 





 (22)
The off diagonal elements of submatrix J22 of the Jacobian are found using
 
i
i
ijijijij
j
i
j
P
SinjVV
V
Q
V







(23)
The main diagonal elements are given by
iiiiiij
i
i
i
i
i BVQBV
P
V
Q
V
22
2 





 (24)
Generally
The off diagonal elements, ji 
j
i
j
j
i
ij
V
Q
V
P
M







(25)
j
i
j
j
i
ij
V
P
V
Q
N






  (26)
Diagonal elements, j = i
iiii
i
i
i
i
i
ii BViM
V
Q
V
P
M
2
2






 
(27)
iiii
i
i
i
i
i
ij GViN
V
P
V
Q
N
2
2






 
(28)
Using equation (9)


























44
2
4444342444342
3433
2
33332343332
242322
2
222242322
44
2
4444342444342
3433
2
33332343332
242322
2
222242322
2
2
2
2
2
2
BVMNMMNN
MBVMMNNN
MMBVMNNN
GVMNNMMM
NGVNNMMM
NNGVNMMM

167. 

























































4
3
2
4
3
2
4
4
3
3
22
4
3
2
/
Q
Q
Q
P
P
P
V
V
V
V
VV



(29)
when voltage controlled buses are given
For example if bus (4) is voltage controlled,
Then 4V has a specified constant value and voltage correction 0
4
4


V
V
So 6th
column of (29) is multiplied by zero and so it may be removed .
 Q4 is not specified.
4Q cannot be defined.
So sixth row of (29) is removed.
In general,
If there are Ng voltage controlled buses, (except slack bus)
Will have (2N-Ng-2) rows and columns.
Advantages of Newton – Raphson method
 Faster, more reliable and results are accurate
 Requires less number of iterations for convergence
 Suitable for large systems.

168. FLOWCHART FOR NEWTON RAPHSON POWER FLOW METHOD
PQ bus
PV bus
Slack bus
start
Read the system data
Form the Bus
admittance matrix YBus
Set iteration count k = 0
Convert the YBus in
complex form to polar
form
Set Bus count i = 0
Test whether
ith
bus is PV
or PQ or
Slack bus
Calculate
)sin(
1
)(
ijijijj
n
j
i
k
i YVVQ   
i = i+1
A
B
D

169. Is
max.
)(
i
k
i QQ 
Is
max.
)(
i
k
i QQ 
Set Qi
(k)
= Qi,max Set Qi
(k)
= Qi,min
)()( k
i
sch
i
k
i QQQ 
)()( k
i
sch
i
k
i QQQ 
Calculate
)cos(
1
)(
ijijijj
n
j
i
k
i YVVP   
)()( k
i
sch
i
k
i PPP 
i ≥ ni = i+1B
A
Yes
NO
NO NO
YesYes

170. Is
Max
)(k
iP ≤ ε
&
Max
)(k
iQ ≤ ε
Initialize the jacobian
matrix J =1
Calculate the jacobian matrix
J1,J2,J3 and J4 using the
formula
Form the power mismatch
equation






















VJJ
JJ
Q
P 
43
21
Calculate























Q
P
JJ
JJ
V
1
43
21
1
C
NO Yes

171. Set bus count i =1
Is
ith
bus is PV
or PQ or
Slack bus
Calculate
)()()1( k
i
k
i
k
i VVV 

Calculate
)()()1( k
i
k
i
k
i  

1
i ≥ n
i = i+1
k = k+1
No
Yes
PQ bus
PV bus
i = i+1
Slack bus
D

172. C
Set bus count i = 0
Is
ith
bus is PV
or Slack bus


n
j
ijijijji YVVQ
1
1 )sin( 
i ≥ n
Find the line current ,line flows and
line losses
Display the result
Stop


n
j
ijijijji YVVP
1
1 )cos( 
i = i+1
PV bus
Slack bus
NO
Yes

173. Disadvantages:
 Programming logic is more complex than Gauss –seidel method.
 Memory requirement is more
 Number of calculations per iterations are higher than G.S method.
COMPARISON OF THE G-S AND N-R METHODS OF LOAD FLOW SOLUTIONS.
PROBLEM
Consider the three bus system. Each of the three lines has a series impedance of
0.02 + j0.08pu and a total shunt admittance of j0.02 pu. The specified quantities at the
buses are tabulated
Bus Real load
demand PD
Reactive
load
demand QD
Real power
generation
PG
Reactive
power
generation
QG
Voltage
specification.
1 2.0 1.0 Unspecified Unspecified V1 = 1.04 +
j0(Slack bus)
2 0.0 0.0 0.5 1.0 Unspecified (PG
Bus)
3 1.5 0.6 0.0 QG3=? )(04.13 PVbusV 
Controllable reactive power sources of available at bus 3 with the constraint
puQG 5.10 3 
G-S method N-R method
1. Variables are expressed in
rectangular co-ordinates.
2. Computation time per
iteration is less.
3. It has linear convergence
characteristics.
4. The number of iterations
required for convergence
increases with size of the
system.
5. The choice of slack bus is
critical.
Variables are expressed in polar
co-ordinates.
Computation time per iteration is
more.
It has quadratic convergence
characteristics.
The number of iterations are
independent of the size of the
system.
The choice of slack bus is arbitrary.

174. Find the load flow solution using the NR method. Use a tolerance of 0.01 for power
mismatch.
Solution:
For each line
764.11941.2
08.002.0
1
j
j
y 


0
96.7513.12 
Each off diagonal element
0
04.10413.12764.11941.2  j
Self term  764.11941.22 j =5.882 – j23.528= 0
95.7523.24 














000
000
000
95.7523.2404.10413.1204.10413.12
04.10413.1295.7523.2404.10413.12
04.10413.1204.10413.1295.7523.24
Busy
To start iteration chose from (1)
 
010
2 jdV 
 
00
3 
(1)
 inin
N
n
niini CosVVYP   1
 inin
N
n
niini SinVVyQ   1
  2222
2
2212121122  CosyVCosyVVP 
 23232332   CosyVV
   32323223313131133   CosyVVCosyVVP
3333
2
3 CosyV
  2222
2
2212121122  SinyVSinyVVQ 
32232332   SinyVV
Substituting the given and assumed values
 
puP 23.00
2 
 
puP 12.00
3 
 
puQ 96.00
2 

175. from
calcPschPP iii ,, 
calcQschQQ iii ,, 
   
calcpspePP i
0
2
0
2 ,  = 0.5 - (-0.23)
 
73.00
2 P
 
  62.112.05.10
3 P
 
  96.196.010
2 Q
The changes in variables at the end of the first iteration are obtained as


















































2
2
3
2
2
2
2
3
2
2
2
3
2
2
3
2
2
2
2
3
2
V
QQQ
V
PPP
V
PPP
Q
P
P
















2
3
2
V


Jacobian elements can be evaluated by differentiating the expressions given above
for P2, P3, Q2 with respect to 232 ,&, V and substituting the given and assumed values
at the start of iteration.
The changes in variables are obtained as
 
 
 
1
1
2
1
3
1
2
54.2205.311.6
05.395.2423.12
64.523.1247.24




























V













96.1
62.1
73.0













089.0
0654.0
023.0
 
 
 
 
 
 
 
 
 

































































089.0
0654.0
023.0
0
0
0
1
2
1
3
1
2
1
2
0
3
1
2
1
2
1
3
1
2
VVV






 
 
 

























089.1
0654.0
023.0
1
2
1
3
1
2
V


using iQ equation
 
4677.01
3 Q

176.    
0677.16.04677.03
1
3
1
3  DG QQQ
which is with in limits.
The final results are
radV 024.0081.12 
radV 0655.0041.13 
45.06.015.03 GQ (with in limits)
 791.0031.11  jS
1005.02 jS 
15.05.13 jS 
Transmission loss = 0.031 pu.

177. UNIT – V
STABILITY ANALYSIS
SYLLABUS
Swing equation in state space form - Equal area criterion - Stability
analysis of single machine connected to infinite bus by modified
Euler’s method using classical machine model - Critical clearing angle
and time - Multi-machine stability analysis using classical machines
model and constant Admittance load representation using Runge-
Kutta method - Causes of Voltage instability - Voltage stability
proximity indices for two-bus system.
INTRODUCTION
STABILITY
The stability of a system is defined as the ability of power system to return to
stable (synchronous) operation when it is subjected disturbance.
The stability studies are classified depending on the nature of the disturbance.
Depending on the nature of disturbance the stability studies can be classified into
the following three types.
a. Steady state stability
b. Dynamic stability
c. Transient stability

178. Steady State Stability
The steady state stability is defined as the ability of a power system to
remain stable i.e., without loosing synchronism for small disturbances.
P max =
x
EV
Dynamic stability
The ability of the generator to remain in stable in the period following the end of
the transient period. It is similar to steady state stability. The variation is slightly above
the rotor natural frequency.
Transient Stability
The transient stability is defined as the ability of a power system to remain
stable (i.e., without loosing synchronism) for large disturbances.
Assumptions Made upon Transient Stability
The three assumptions upon transient suability are,
a. Rotor speed is assumed to be synchronous. Infact it varies
insignificantly during the course of the stability transient.
b. Shunt capacitances are not difficult to account for in a stability study.
c. Loads are modelled as constant admittances.
d.
Methods of Maintaining Stability
The methods of maintaining stability are, HVDC links, braking resistors, short
circuit limiters and turbine fast valving or by pass valving and full load rejection
technique.

179. STABILITY LIMITS:
Steady State Stability Limit
The steady state stability limit is the maximum power that can be transmitted by a
machine (or transmitting system) to a receiving system without loss of synchronism.
In steady state the power transferred by synchronous machine (a power system) is
always less than the steady state stability limit.
Transient Stability Limit
The transient stability limits is the maximum power that is transmitted by
a machine (or transmitting system) to a fault or a receiving system during a
transient state without loss of synchronism. The transient stability limit is always
less than the steady state stability limit.
TO IMPROVE THE TRANSIENT STABILITY LIMIT OF POWER SYSTEM
The transient stability limit of power system can be improved by,
a. Increase of system voltage
b. Use of high speed excitation systems
c. Reduction in system transfer reactance
d. Use of high speed reclosing breakers
STABILITY STUDY
The procedure of determining the stability of a system upon occurrence of a
disturbance followed by various switching off and switching on actions is called a
stability study.

180. SWING EQUATION OF A SYNCHRONOUS MACHINE
The equation governing rotor motion of a synchronous machine is given by
J 2
2
dt
d m
Ta = Tm – Te N-m  (1)
J – The total moment of inertia of the rotor masses, in Kg –m2
.
m - Angular displacement of the rotor in the respect to a stationary axis in (rad)
t- Time in seconds
Tm –mechanical (or) shaft torque in N-m
Te – electrical (or) electromagnetic torque in n-m
Ta – Net accelerating torque in N-m
 Tm and Te are positive for synchronous generator.
 Tm is the resultant shaft torque which tends to accelerate the rotor in the positive
m direction.
Under steady state operation of generator Tm = Te  Ta = 0
Mechanical torque Tm and Electrical torque Te are considered positive for
synchronous generator i.e., Tm is the resultant torque which accelerates the rotor
in positive m direction of rotation
Ta = 0  In this case there is no acceleration or deceleration of the rotor masses
and the resultant constant speed is a synchronous speed
To prime
mover
Generator
Tm ωm
Te
Electrical
network
Pe
Bus Bar

181. m  Measure of rotor angle
m Increases with time even at constant synchronous speed
m = wsmt +  m  (2)
wsm - Synchronous speed of the machine in mechanical radians per second
 m - Angular displacement of the rotor in mechanical radians per second
differentiate (2) with respect to time

dt
d m
wsm +
dt
d m
 (3) and
2
2
dt
d m
2
2
dt
d m
 (4)
(3)  shows
dt
d m
is constant and equals synchronous speed i.e.,
dt
d m
= wsm
when
dt
d m
= 0
dt
d m
is the deviation of the rotor speed from synchronism and units in mechanical
radians per second
(4)  Rotor acceleration measured in mechanical radians per second. Squared
Substitute (4) in (1)
 J 2
2
dt
d m
= Ta = Tm – Te N-m  (5)
 wm =
dt
d m
 (6)
Multiply (5) by wm on both sides
power = torque x angular velocity
Jwm 2
2
dt
d m
= Pa = Pm-Pe watts  (7)
Pm  shaft power input  power supplied by prime mover
Pe  electrical power output
Pa  accelerating power
Jwm  angular momentum of rotor at synchronous speed wsm

182. Jwm is denoted by M called as inertia constant
M  unit – Joules – second / mech.radian
(7)  M 2
2
dt
d m
= Pa = Pm-Pe Watts  (8)
In machine data  constant related to inertia is used i.e., H
H =
inginMVAmachinerat
dronousspeenMJatsynchticenergyistoredkine
H =
mach
sm
s
Jw2
)2/1(
=
mach
sm
s
Mw)2/1(
MJ/MVA  (9) (where , smJw = M)
From (9)
M=
smw
H2
machs MJ/mech.rad  (10)
Substitute (10) in (8)
smw
H2
machs . 2
2
dt
d m
= Pa = Pm-Pe
smw
H2
2
2
dt
d m
=
mach
a
s
P
=
mach
em
s
PP 
(11)
m  in mech.radians
wsm  in mech.radians / second
(11) 
sw
H2
2
2
dt
d 
= Pa = Pm-Pe per unit (12)
(12)  sw and  should be in same unit
either mech (or) electrical degrees (or) radians
Subscript m in wsm and  m denotes that they are in mech.units , otherwise
electrical units are implied
(12)  subs ws = 2 f
(12) 
f
H
 2
2
dt
d 
= Pa = Pm-Pe per unit (13)
(12)  Swing Equation

183. (13)   is in electrical radians
when  is in electrical radians (rad = degx( / 180))
(13) 
f
H
180 2
2
dt
d 
= Pa = Pm-Pe per unit (14)
(12)  second order differential equation .
This can be written as the 2 first order differential equations
sw
H2
.
dt
dw
= Pm-Pe per unit (15)
dt
d
= w- ws (16) w, ws and   involve electrical radians (or) electrical
degrees.
Multi machine system
Common system base is chosen
machs = machine rating (base)
ssystem = system base
(13) 
sys
mach
s
s
[
f
Hmach
 2
2
dt
d 
] = Pm-Pe perunit in sys.base  (17)
 Hsystem = Hmach
system
mach
s
s
 (18)
machine inertia constant in system base
Machines Swinging Coherently
Consider the swing equations of two machine on a common system base
f
H

1
2
1
2
dt
d 
= Pm1 – Pe1 pu  (19)
f
H

2
2
2
2
dt
d 
= Pm2 – Pe2 pu  (20)
since machines rotor swing together ( 1 =  2 =  )
Add (19) and (20)

184. f
Heq
 2
2
dt
d 
= Pm - Pe  (21)
where, Pm = Pm1 + Pm2
Pe = Pe1 + Pe2
Heq = H1+H2
Heq = H1
system
mach
s
s1
+ H2
system
mach
s
s2  (22)
(21) can be extended to any number of machines swinging coherently
PROBLEM
A 50HZ, four pole turbo generator rated 100MVA; 11KV has an inertia
constant of 8.0 MJ/MVA. Find (a). the stored energy in the rotor at
synchronous speed (b). If the mechanical input is suddenly raised to 80MW
for an electrical load of 50MW, find rotor acceleration, neglecting mechanical
and electrical losses.(c). If the acceleration calculated in part (b) is maintained
for 10cycles, find the change in torque angle and rotor speed in rev.per min at
the end of this period.[UQ]
Solution :
(a). Stored energy = SH (MVA rating of M/C*H) = 100X8 = 800MJ
(b). Pa = Pm – Pe = 80 -50 = 30MW
Pa = M
 
 2
2
dt
d 
M =
f
SH
180
=
5180
800
X
=
45
4
MJ = s/elec.deg.
45
4  
 2
2
dt
d 
=
4
4530X
= 337.5 elec.deg/S2
( c). 10cycles =
50
10
= 0.2 sec(t =
f
1
)

185. Change in  =
2
1
 X t2
=
2
1
(337.5)X(0.2)2
= 6.75 ele.deg.
= 60 X 
3602
5.337
X
= 28.125 rpm/sec.
Rotor speed at the end of 10 cycles = )(
120
t
p
f

=
4
50120X
+ (28.125 X0.2)
= 1505.625rpm.
A 2pole 50HZ, 11KV turbo alternator has a ratio of 100MW,pf – 0.85 lagging. The
rotor has a moment of inertia of 10,000 Kgm2
.calculate H and M.[UQ]
Given:
P=2, f =50HZ pf = 0.85lag
V = 11KV, P = 100MW, J = 10,000Kgm2
Solution:
M in pu =
 
bMVA
eletricalsMinMJ /
H = PufM
Syn.speed in rad/sec,Ws = 2 ns
ns = syn.speed in rpm = 






2
5022 X
p
f
= 50rps.
Ws = 2 ns = 2 X50 = 314.16 ele.rad/sec.
Inertia constant M = J sW
p
2
2






X10-6
in MJ –s /electrical.
M = 10,000 16.314
2
2
2
X





X10-6
= 3.146 in MJ –s /electrical.

186. MVA rating of M/C , S=
85.
100
.

fP
P
= 117.675MVA
KVb = 11KV and MVAb = 117.675MVA
H in pu =
65.117
1416.3
= 0.0267p.u
H =  f Mpu =  X50X0.0267 = 4.194MW-s/MVA.
Machine connected to infinite bus
Xe = Line reactance
XTransfer = ed XX '
Pe =  sinsin max
'
P
X
VE
Transfer









Two machine system:
Pm1 = -Pm2 = Pm
Pe1 = -Pe2 = Pe
Swing equations for 2 machines





 





 






11
11
2
1
2
H
PP
f
H
PP
f
dt
d emem


-----(1)





 





 






221
22
2
21
2
H
PP
f
H
PP
f
dt
d emem


-----(2)
em PP
HH
HH
f
dt
d





 





 
(
)(
21
21
2
21
2


)
2
2
dt
d
f
Heq 

= Pm - Pe
where 1  - 2
Heq =
21
21
HH
HH

Pe = sin'
2
'
1
'
2
'
1








 ded XXX
EE

187. EQUAL AREA CRITERION
In a system where one machine is swinging with respect to an
infinite bus, it is possible to study transient stability by means of a simple
criterion, without resorting to the numerical solution of a swing equation.
Consider the swing equation,
f
H
 2
2
dt
d 
= Pa = Pm-Pe
Plot of  Vs t for stable & unstable systems
0
If a system is unstable  continues to increase indefinitely with time and the machine
loses synchronism. On the other hand, if the system is stable ,  (t) performs oscillations
(nonsinusoidal) whose amplitude decreases in actual practice because of damping
terms(not included in the swing equation) .This fact can be stated as a stability
criterian,that the system is stable if at some time
0
dt
d
And is unstable, if 0
dt
d
For a sufficiently long time (more than 1 s will generally do)
The condition for the stability can be written as
Area 1
Area 2
P
δδmδo
Pm
Constant input
line

188. 0
0
 


dPa
The condition of stability can therefore be stated as: the system is stable if the area under
(accelerating power) -  curve reduces to zero at some value of  . In other words
,the positive (accelerating )area under Pa -  curve must equal the negative
(decelerating) area and hence the name ‗equal area‘ criterion of stability.
To illustrate the equal area criterion of stability, we now consider several types of
disturbance that may occur in a single machine infinite bus bar system.
Sudden change in Mechanical Input:
The transient model of a single machine tied to infinite bus bar is given. The electrical
power transmitted is given by
Pe =  sinsin max'
'
P
XX
VE
ed










Under steady operating condition
Pm0 = Pe0 = Pmax sin 
Pe - diagram for sudden increase in mechanical input to generator.
Let the mechanical input to the rotor be suddenly increased to Pm1 (by opening the steam
valve). The accelerating power Pa = Pm1 –Pe causes the rotor speed to increase ( s  )
and so does the rotor angle. at angle 1 , Pa = Pm1 –Pe (=Pmax sin 1 ) = 0 (state point at
b)but the rotor angle continues to increase as s  .Pa now becomes negative
(decelerating), the rotor speed begins to reduces but the angle continues to increase till at
angle 2 , s  once again(state point at c), the decelerating area A2 equals the
accelerating area A1 (areas are shaded, 0
2
0
 


dPa .since the rotor is decelerating, the
speed reduces below s and the rotor angle begins to reduce. the state point now
traverses the Pe - curve in the opposite direction as indicated by arrows in fig(). It is
easily seen that the system oscillates about the new steady state point b ( )1  with

189. angle excursion up to 0 and 2 on the 2 sides. these oscillations are similar to the simple
harmonic motion of an inertia-spring system except that are not sinusoidal.
As the oscillations decay out because of inherent system damping (not modeled),the
system settles to the new steady state where
Pm1 = Pe = Pmax sin 1
Areas A1 and A2 are given by
A1 = 


dPP em )(
1
0
1 
A2 = 
2
1
(


eP -Pm1) d
For the system to be stable, it should be possible to find angle 2 such that A1 = A2. as
Pm1 line as shown in fig().under this condition, 2 acquires the maximum value such that
2 = 





 
max
11
1max sin
P
Pm

Limiting cases of transient stability with mechanical input suddenly increased.
0
Any further increase in Pm1 means that the area available for A2 is less than A1, so
that the excess kinetic energy causes  to increase beyond point c and the decelerating
n
Pmax
δo
Area 1
Area 2P
δδm
Pm
Constant input
line
P0
δm
Pm
δcr
Pa=Pm sin δ

190. power change over to accelerating power, with the system consequently becoming
unstable. It has thus been shown by use of the equal area criterion that there is an upper
limit to sudden increases in mechanical input (Pm1 –Pm0), for the system in question to
remain stable.
It may also be noted from Fig () that the system will remain stable even though
the rotor may oscillate beyond  = 90o,
so long as the equal area criterion is met. The
condition of
 = 90o
is meat for use in steady state stability only and does not apply to the transient
stability case.
Effect of Clearing time on Stability
 Machine operating with mechanical input Pm at steady angle  0(Pm=Pe)
 If a 3 phase fault occurs at ‗P‘ of outgoing radial line, the electrical output is
decreases to zero Pe = 0, a to b.
 Accelerating area A1 is increases from b to c.
 A t time tc (clearing time) corresponding angle is  c(clearing angle) the faulted
line is cleared by opening of the line circuit breaker.
Pmax
Pm
Constant input
line
Pm
δ0 δc δm δπ/2
Curve X (Before the
fault)
Curve Y (During the
fault)
Curve Z (After the
fault)

191.  The system once again becomes healthy and Pe = Pmaxsin point moves to d.
 Decelerating area starts and point moves along decreases.
 System finally settles to ‗a‘ in an oscillatory manner.
 If clearing of faulty line is delayed A1 and  1 to  max.
 For clearing time or angle larger than this value, the system should be unstable as
A2<A1.
Critical clearing time (tcr)and Critical clearing angle( cr):
Maximum allowable value of clearing time and angle for the system to remain stable.
To find tcr and  cr
Pe =0 during fault.
There fore  max = 0  -----(1)
Pm = Pmaxsin 0 --------(2)
A1 =   )(0 0
0




cr
crmm PdP -------(3)
A2 =   


dPP
cr
m 
max
sinmax
= Pmax(cos cr -cos max)-Pm( max- cr)------(4)
For the system to be stable , A1 = A2 , (3) = (4)
Pe1 = )1(sin
21
'
'



IIXXX
VE
d
Line 2 switched off,
Pe2 = )1(sin
1
'
'



XX
VE
d
X1= Pmax2
Pmax2<Pmax1
Because (X 1
'
1
'
() XXXX dd  II X2)

192.  Line 2 is switched off. electrical operating point shift to curve2 (at b)
 A1 accelerating area 1  decelerating area A2
Stable A1= A2
Before fault,
Pe1 =  sinsin max
21
'
'
P
IIXXX
VE
d


During fault, Pe2 = 0
The rotor accelerators , therefore  increases.
 Synchronism remains unless the fault is cleared in time.
 During clearing time tc, the fault line is disconnected
 Power flows through healthy line
Pe3=  sinsin 3max
1
'
'
P
XX
VE
d


Pmax3<Pmax1
crc  
A1 =   
cr
dPPm



0
sin2max
A2 =   
max
sin3max



cr
dPP m
 max = 1
sin 
 





3maxP
Pm
A1 = A2 for stability
cos cr =
 
2max3max
max3max02maxmax coscos
180
PP
PPpm

 

Pe3=  sinsin 3max
1
'
'
P
XX
VE
d

