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.

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.

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.

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.

(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

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

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.

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)

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

+

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

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

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.

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 = (KVb2 * 1000) / KVAb in Ω ……………(3)

(or) Zb = KVb2 / MVAb in Ω

Per unit impedance Zp.u = Actual impedance in Ω / Base impedance in Ω

Zp.u = Z / [(KVb2 * 1000) / KVAb ]

Zp.u = (Z * KVAb) / (KVb2 * 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) * (MVAb2

(new) / MVAb2

(old) ) * (KVAb2

(old) / KVAb2(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)

Per unit impedance Zp.u = Actual impedance in Ω / Base impedance in Ω

Zp.u = Z / [(KVb2 * 1000) / KVAb ]

Zp.u = (Z * KVAb) / (KVb2 * 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) * (MVAb2

(new) / MVAb2

(old) ) * (KVAb2

(old) / KVAb2(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.

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.

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

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 / [(KVb2 * 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)

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) * (MVAb2

(new) / MVAb2

(old) ) * (KVAb2

(old) / KVAb2

(new) )

= 1.1 *(100/500) * (222 / 20

2 )

Zp.u(new) = 0.2662 p.u

4) The reactance of a generator designated X 11 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 X 11 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:

Actual reactance = 15

KV Base=10KV

KVA Base = 15 KVA.

TO FIND: Z, p u.

SOLUTION:

Z p u = nceBaseimpeda

ceActualreac tan

Base impedance, Z b = 1000

2

KVA

KVB

= 67.6666100015

102

Z p u = pu0022.067.6666

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.

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 = phaseMVAB

KVB/63.3

300

33

.

. 22

Z p u = ZbZpuActualvalvalBase

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

I b = A95.174333

100010

Base impedance, Z b =

phaseMVA

KV

B

B /9.10810

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

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

1210100

1000111000

22

B

BB

KVA

KV

Transformer line voltage ratio, K= 0364.011000

400

Total leakage reactance referred to primary 211 XXXO 1

= X2

21

K

X

= 12 +2)0364.0(

05.0

XO 1 = 12 +37,736

XO 1 = 49.736 phase/

p u reactance per phase Xpu = nceBsaeimpeda

cegereacTotalleaka tan

X p u = puZ

XO

B

0411.01210

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, Z100

10004.01000)(

222

B

B

B

BB

KVA

KVOR

MVA

KV

Z 6.1B

Transformer line voltage ratio, K = .0364.011

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.

X p u = 0411.06.1

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

67.20166600

10001101000

22

B

BB

KVA

KV

Transformer line voltage ratio, K= 11/110 =0.1

Total impedance referred to primary: ZO1

Z01 =Z22

21

1

211.0

2.01.0)4520(

JJ

K

ZZZ

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

67.201600

1000111000

22

B

BB

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

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

110 22

B

B

MVA

KV

Z b, p y = 605 phase/

(P y - -connected)

Z p s in phasephase /1835.255.52

3/2290/

P u value of z p s = uppyZb

inZps.0416.0

605

1835.25

,

,

(Z p s) )( pu = 0.0416 p u.

(P y - connected)

Z p t in ./6299.195.52

3/1785/ phasephase

P u value of Z p t = 0324.0605

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 = phaseMVA

KV

B

B /616.1115

2.13sec, 2

(Secondary- -connected)

Z s t in phasephase /2605.0328

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.

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 2.13110

2.13110

Z 1 s t (p u) =0.0224 pu0299.02.13

2.13

15

202

Z 1 s t (p u) = 0.0299 p u.

To compute Z p , Zs 1 and Z t 1

Z p =2

1 tZsZptZps 1 = 2

1 0299.00324.00416.0

Z p =0.0221 p u

.

Z 0324.00299.00416.02

1

2

1 11 ZptZstZpss

Z1s = 0.0196 p u

Z t 0416.00299.00324.02

1

2

1 11 ZpsZstZpt .

Z t 1 =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.

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 pu0682.05.19

18

150

1202

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 pu2200.018

5.19

120

1502

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 pu0729.0220

230

150

1002

(Z t ) p u = 0.0729 p u.

Generator is connected to LT side of t r

KV B 220 KV22.17220

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/

Z p u = pujj

Zb

Zact125.00625.0

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(1COS

sin = 0.599

Reactive power, Q = 599.08.0

50sin

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

Zpu

b

LL

Z )( puL = 2.563+j 1.9188p u.

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

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.

(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.

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

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

pqZ

Y1

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

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

2 1

0 3

4

5

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

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, 0iA

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

VYAIBus (4)

VEA BusT

(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

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.

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.

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‖.

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 000.1 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.

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

(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).

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

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)

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 VyI1

(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 VyVjQP1

*

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

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

inn

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,

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.

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 VYVYV

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 VYVYV

jQP

Y …………………….(6)

where,

Vik = k

th iteration value of bus voltage Vi

Vik+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

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.

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 VyVjQP1

*

(1)

with i = 2 and N = 4

424323222121*

2

22 VyVyVyVyV

schjQschP

(2)

Solving for V2 gives

424323121*

2

22

22

2

1VyVyVy

V

schjQschP

yV (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

1VyVyVy

V

schjQschP

yV (4)

Similarly at bus (4)

343242141*

4

44

44

4

1VyVyVy

V

schjQschP

yV (5)

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

1VyVyVy

V

schQschP

yV (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

1VyVyVy

V

schjQschP

yV (7)

The process is repeated at bus (4)

1

343

1

242141*0

4

,4,4

44

1

4

1VyVyVy

V

schjQschP

yV (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”

It is common practice to set the initial estimates of the unknown voltages at all

load buses equal to 000.1 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 VyVyV

schjQschP

yV

1

11

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,

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 VyVIQ1

*

(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.

accVyVyVy

V

jQschP

yV 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

44

1

4 ,V

VVcorrV (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.

FLOWCHART FOR LOAD FLOW SOLUTION BY GAUSS-SEIDEL METHOD

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

inn

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,

4

4

3

3

2

2

4

4

3

3

2

2

VV

PV

V

PV

V

PPPPP 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

PV

V

V

V

PV

PPPP iiiii

i

4

4

4

4V

V

V

PV 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

QV

V

V

V

QV

QQQQ iiiii

i

4

4

4

4V

V

V

QV 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.

001

iii (10)

001

iii VVV

0

0

01

i

i

i

V

VV (11)

Use the new values 1i 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

VV 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 iSinyVVP

(15)

The typical diagonal element of J11

inininni

N

inni

i iSinyVVP

1

N

inn n

iP

1 (16)

By comparing (16) & (3)

iiii

j

i BVQP 2

(17)

The formulas for the elements of sub matrix J21 is given by

The off diagonal element of J21 is

ijijijji

j

i iCosyVVQ

(18)

The main diagonal element of J21 are

inininni

N

inni

i iCosyVVQ

1

N

inn 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

inn

inniiii

i

i

i CosyVGVVV

PV

1

2 (21)

iiii

j

i BVQP 2

iiii GVPi

Qi 2

Comparing (21) with (19) & (20)

iiiiiii

i

i

i

ii GVPGV

Q

V

PV

222

(22)

The off diagonal elements of submatrix J22 of the Jacobian are found using

i

iijijijij

j

ij

PSinjVV

V

QV

(23)

The main diagonal elements are given by

iiiiiij

i

i

i

ii BVQBV

P

V

QV

22

2

(24)

Generally

The off diagonal elements, ji

j

ij

j

iij

V

QV

PM

(25)

j

ij

j

iij

V

PV

QN

(26)

Diagonal elements, j = i

iiii

i

ii

i

iii BViM

V

QV

PM

22

(27)

iiii

i

ii

i

iij GViN

V

PV

QN

22

(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

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 04

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.

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

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 ≥ n i = i+1 B

A

Yes

NO

NO NO

Yes Yes

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

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

C

Set bus count i = 0

Is

ith

bus is PV

or Slack bus

n

j

ijijijji YVVQ1

1 )sin(

i ≥ n

Find the line current ,line flows and

line losses

Display the result

Stop

n

j

ijijijji YVVP1

1 )cos(

i = i+1

PV bus

Slack bus

NO

Yes

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.

Find the load flow solution using the NR method. Use a tolerance of 0.01 for power

mismatch.

Solution:

For each line

764.11941.208.002.0

1j

jy

096.7513.12

Each off diagonal element

004.10413.12764.11941.2 j

Self term 764.11941.22 j =5.882 – j23.528= 095.7523.24

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

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

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.

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

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.

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.

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

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 m2

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

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

H2machs MJ/mech.rad (10)

Substitute (10) in (8)

smw

H2machs .

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

(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)

f

H eq

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/S

2

( c). 10cycles = 50

10 = 0.2 sec(t =

f

1)

Change in = 2

1 X t

2 =

2

1(337.5)X(0.2)

2 = 6.75 ele.deg.

= 60 X3602

5.337

X= 28.125 rpm/sec.

Rotor speed at the end of 10 cycles = )(120

tp

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 sWp

2

2

X10

-6 in MJ –s /electrical.

M = 10,000 16.3142

22

X

X10

-6 = 3.146 in MJ –s /electrical.

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

'

PX

VE

Transfer

Two machine system:

Pm1 = -Pm2 = Pm

Pe1 = -Pe2 = Pe

Swing equations for 2 machines

11

11

2

1

2

H

PPf

H

PPf

dt

d emem

-----(1)

221

22

2

21

2

H

PPf

H

PPf

dt

d emem

-----(2)

em PPHH

HHf

dt

d

(

)(

21

21

2

21

2

)

2

2

dt

d

f

H eq

= Pm - Pe

where 1 - 2

Heq = 21

21

HH

HH

Pe = sin'

2

'

1

'

2

'

1

ded XXX

EE

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

0dt

d

And is unstable, if 0dt

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

δ δm δo

Pm

Constant input

line

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'

'

PXX

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, 02

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

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 sinP

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 2 P

δ δm

Pm

Constant input

line

P0

δm

Pm

δcr

Pa=Pm sin δ

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 δ π/2

Curve X (Before the

fault)

Curve Y (During the

fault)

Curve Z (After the

fault)

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(sin21

'

'

IIXXX

VE

d

Line 2 switched off,

Pe2 = )1(sin1

'

'

XX

VE

d

X1= Pmax2

Pmax2<Pmax1

Because (X 1

'

1

' () XXXX dd II X2)

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

'

'

PIIXXX

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

'

'

PXX

VE

d

Pmax3<Pmax1

crc

A1 = cr

dPPm

0

sin2max

A2 = max

sin3max

cr

dPP m

max = 1sin

3maxP

Pm

A1 = A2 for stability

cos cr =

2max3max

max3max02maxmax coscos180

PP

PPp m

Pe3= sinsin 3max

1

'

'

PXX

VE

d

Case C: Reclosure

If circuit breaker of line 2 are reclosed

power transfer becomes

Pe4 = Pe1 = Pmax1sin

Reclosure time trc = tcr +

= time between clearing and reclosure.

Mentioning the assumptions clearly and developing necessary

equations describe the step by step solution of swing bus.[UQ]

Most powerful methods

1. Modified Euler‘s method.

2. Runge kutta method.

Another method to solve swing equation is point by point method (or)step

by step method.

Point by point method for one m/c tied to infinite bus bar.

This procedure can also be applied to multimachine system.

String equationM

P

M

PP

dt

d am

)sin( max

2

2

The solution (t) is obtained at discrete intervals of time with interval spread of t

Uniform throughout.

Accelerating power and change in speed which are continuous functions of time

are discredited as below

f

HM

GHM

Or input system

The accelerating power Pa computed at the beginning of the interval is

assumed to remain constant from the middle of the preceding interval to

the middle of the interval being considered )2

1()

2

3( nton

The angular rotor velocity

dt

d is assumed constant throughout any

interval.(n-2 ton-1)

Point by point solution of swing equation

Discrete solution

Continuous

solution

n-2 n-1 n

Pa (n-2)

Pa (n-1)

Pa (n)

Pa

n n-2

n-3

2

n-1

2 n-1

n-3

2

n-1

2

n

n-2 n-1 n

n-1

n-2

t t

n

n-1

t

t

t

t

t

t

At the end of the (n-1)th interval the acceleration power is

n-1 has been previously calculated.

The change in velocity from tnton )2

1()

2

3(

The change in during the (n-1)th interval is

During nth

interval

Subtract(3)from(4) using(2)

Similarly Pa(n),n+1andn+1ca be calculated.

Normal time interval is 0.5seconds.

Greater accuracy is achieved by reducing the time duration of intervals.

Fault or switching causes a disconnectivity in Pa.if such thing happens

then Pa=average value of the values of Pa before and after discontinuity.

The increment of angle occurring during the first interval after a fault is

applied at t=0.

Equation (5) becomes

1sin)1( 1max nma PPnP

2)1(2

3

2

1

nPM

ta

nn

32

3211

n

nnn t

42

11

n

nnn t

6

5)1(

2

1

ninn

nnn PaM

t

2

02

1

aP

M

t

Pa0+Pa immediately after occurrence of fault.

Before fault→in steady state Pa0-=0

If the fault is cleared at the beginning of nth

interval

)1(nPa Immediately before clearing fault

)1(nPa After clearing fault.

Multimachine stability

Multimachine stability studiesclassical representation

To reduce the complexity of system modeling the following assumptions are made in

transient stability studies

The mechanical power input to each machine remains constant during the

entire period of the string curve computation.

Damping power is negligible.

Each machine may be represented by a constant transient reactance in

series with a constant transient internal voltage.

The Mechanical rotor angle of each machine coincides with , the

electrical phase angle of the transient internal voltage.

All loads may be considered as shunt impedances to ground with values

determined by conditions prevailing immediately prior to the transient

conditions.

The system stability model based on these assumptions is called THE CLASSICAL

STABILITY MODEL and studies which use this model are called CLASSICAL

STABILITY STUDIES.

)1()1(2

1))1(()( nPnPnpnP aaaa

COMPUTATIONAL ALGORITHM FOR SOLVING SWING CURVES

USING MODIFIED EULERS METHOD.

State variable formulation of swing equations:

The swing equation for the Kth

generator is 2

2

dt

d =

kH

f = (

GKP - GKP )

(K = 1,2,……….. m)-(1)

For multi machine case, it is more convenient to organize the equation in state

variable form.

X1k = K = '

KE

X2k = K

KX1 = X2k

KX 2 =

kH

f = (

GKP - GKP ) (K = 1,2,……….. m)-(2)

Initial state vector (upon the occurrence of fault) is

KX1 = K = 0

KE

KX 2 = 0

The state form of swing equation (2) can be solved by the many available

integration algorithms.

Among many modified Euler‘s method is a convenient method.

Computational algorithm for obtaining swing curves using modified Euler‟s

methods

(1)Carry out a load flow study (prior to disturbance) using specified voltage and powers.

(2)Compute the voltage behind transient reactance‘s to generators ( KE )using

'

GiE = Vi +jGiGi IX '

This fixes generator emf magnitudes and initial rotor angle.

Reference is slack bus voltage 1V .

(3). Compare YBus during fault, post fault and line reload.

(4). Set time count r =0

(5). Compute generator power outputs using appropriate YBus with the help of the general

form

P1 = '

1E2 G11+ '

1E '

2E 12Y cos ( 1 - 2 - 12 )

This gives r

GKP for t = t(r)

After the occurrence of the fault, the period is divided into uniform discrete time

intervals ( t ) so that time is counted as t(0)

, t(1)

……….( t = 0.05 seconds)

(6). Compute mKXX r

k

r

K ............2,1,, )(0

2

)(

1

from (2)nd

equation.

(7) Compute the first state estimates for t = t(r+1)

as

tXXX r

K

r

K

r

K )(

1

)(

1

)1(

1

K = 1,2, 3,……. M

tXXX r

K

r

K

r

K )(

2

)(

2

)1(

2

(8). Compute the first estimates of )1( r

KK

)1( r

KE = KE ( cos )1(

1

r

KX + j sin )1(

1

r

KX )

(9). Compute )1( r

GKP

calculate YBus and ( nnn 1 )

(10). Compute [ )1(

1

r

kX , )1(

2

r

kX , K = 1,2 …. m) from equation(2).

(11). Compute the average values of state derive

)(

,1

r

avgkX = )1(

1

)(

12

1 r

K

r

K XX K = 1,2……m

)(

,2

r

avgkX = )1(

2

)(

22

1 r

K

r

K XX

(12). Compute the final estimate for EK at t = t(r+1)

)1(

1

r

kX = )(

1

r

kX + )(

1

r

kX average t

)1(

2

r

kX = )(

2

r

kX + )(

2

r

kX K = 1,2 ………m

(13). Compute the final estimate for EK at t = t(r+1)

using )1( r

KE = 0

KE ( cos )1(

1

r

KX + j sin )1(

1

r

KX )

(14). Print ( )1(

1

r

KX , )1(

2

r

KX ), K = 1,2…..m

Test for time limit (time for which swing curve is to be plotted)

Check if r>r final. If not r = r +1 and repeat from step 5.other wise print results and

stop.

The swing curve of all the machine are plotted.

If the rotor angle of a machine (or a group of machines) with respect to other

machines . Increases with out bound, it is an unstable state and machine or

machines falls out of step.

SOLUTION OF SOLVING SWING EQUATION BY RUNGE – KUTTA

METHOD

The modified Euler method and Runge- Kuttan method are the two popular

methods used for the solution of swing equation.

Runge – Kuttan method is self starting and gives an accuracy of the order of 5t

This method uses Taylor‘s series expansion, truncated after the fourth term.

This methods is also known as Runge – Kutta fourth order approximation

The swing equation is a second order differential equation.

Consider two first order differential equations in two variables x and y such

that

yxfdt

dx,

yxgdt

dy,

with known initial conditions x0 and y

0 and a time interval t , the following eight

constants are calculated as

tyxfK 00

1

0

1

tyxgl 000

1 ,

tlyKxfK 0

1

00

1

00

2 5.0,5.0

tlyKxgl 0

1

00

1

00

2 5.0,5.0

tlyKxfK 0

2

00

2

00

3 5.0,5.0

tlyKxgl 0

2

00

2

00

3 5.0,5.0

tlyKxfK 0

2

00

3

00

4 5.0,

tlyKxgl 0

3

00

3

00

4 ,

We use the above eight constants to compute the changes in x and y as follows.

0

4

0

3

0

2

0

1

0 2226

1KKKKx

0

4

0

3

0

2

0

1

0 226

1lKllly

The we update the values of t, x and y

tt 01

001 xxx

001 yyy

Then we replace x0 and y

0 by x

1 and y

1 and recalculate the K‘s , l‘s , x and y .

In general

tnt n 11

nnn xxx 1

nnn yyy 1

For using the above method for solving the swing equation of one machine

connected to infinite bus.,

We put,

x =

dt

dy

Then

,fdt

d

and

PePiM

gdt

d

1,

where, f

GJM

Mega – Joule – sec / elect . rod.

The initial value 0 is calculated using the prefault values

0max SinPPP ei

Finax

PiSin 1

0

Te values of Pe used in equation

pePiM

gdt

d

1,

For, 0<t< tc

For, t>tc ,

tc clearing time.

critical clearing time.

The expression for the critical clearing time t cr is given by

tms

ocrcr

P

H

)(4

where,

H is the constant

cr is the critical clearing angle

o is the rotor angle

P m is the mechanical power

s is the synchronous speed.

SinPPPe mbe 1

SinPPPe mce 11

Problem:

A 50 Hz generator is delivering 50% of the power that it is capable of delivering

through a transmission line to an infinite bus. A fault occurs that increases the

reactance between the generator and the infinite bus to 500% of the value before the

fault, when the fault is isolated the maximum power that can be delivered is 75% of the

original maximum value. Determine the critical clearing angle for the condition

described.[UQ]

Solution:

Prefault

0

1

'Sin

X

VEPci prefault

01 max SinPpe

1max01max 5.0 pSinP

= 0.52359 radians

During fault the reactance is 500% of the value before fault.

During fault

Pmax2 pu2.05

1

%100

%500

Post fault

Pmax3 = 0.75

`max

3max

1

P

pSin m

5.0max mpP

0

0 30

75.0

5.01Sin

333.11 Sin

7297.0

00 18.13881.41180 m

02max0max CosxppCos mcr

23

max3

maxmax

max

pp

Cosp

052359.04118.25.0

2.075.0

4118.275.052359.02.0

CosCos

55.0

5589.01732.09441.01

Coscr

55.0

212.01Coscr

radcr 175.1

Voltage Stability

VOLTAGE INSTABILITY:

At lower voltage very high current is taken to produce the power.

The thermal ratings of the line are apparent for loads of PF less than unity. The

possibility exists that before the thermal rating is reached, the operating power may

be on that part of the characteristics where small changes in load cause large

voltage changes and hence voltage instability occurs.

radiansm 4118.2

radcr 033.67

In power system network

Nominal voltage value is 1pu.

Many phenomena tend to make the actual voltage different from the nominal

value.

An unacceptable voltage level means Voltage instability.

If the voltage departs too much from the nominal value, the phenomenon is

known as VOLTAGE COLLAPSE.

Generally voltage in stability occurs during and after large disturbances.

VOLTAGE SECURITY:

The ability of the system to maintain the voltage at load points, at acceptable

level is known as voltage security.

Voltage stability problem arise mainly in the event of faults.

Rotor angle stability problems arise during and after faults.

The voltage instability problem arises due to reactor power mismatch.

(1). The series inductance of the line absorbs lagging vars.

(2). The shunt capacitance absorbs leading vars

These two are equal only if the load is equal to natural load.

(3). Voltage stability is obtained by shunt reactors.

Shunt compensation involves the use of shunt capacitors and shunt reactors to

avoid voltage instability.

Series compensation is used on short lines to improve voltage stability.

A synchronous generator of reactance 1.20 pu is connected to an infinite bus bar

( V = 1.0 pu) through transformers and a line of total reactance of 0.60 pu. The

generator no load voltage is 1.20 pu and its inertia constant is H = 4 MWs/MVA.

The resistance and machine damping may be assumed negligible. The system

frequency is 50 Hz.

Calculate the frequency of natural oscillation if the generator is loaded to

(i)50%

And (ii) 80% of its maximum power limit.

Solution

(i) For 50% loading

Sin 0 = maxP

Pe = 0.5 or 0 = 30 0

030

eP

= 8.1

12.1 cos30 0

M(pu) = 50

H =

50

4

S 2 /elect rad

From characteristic equation

P = 2

1

30

/0

M

PJ e

= 2

1

4

50577.0

J = J4.76

Frequency of oscillations = 4.76 rad / sec

= 2

76.4 = 0.758 Hz

(ii) For 80% loading

Sin 0 = maxP

Pe = 0.8 or 0

0 1.53

Pe01.53 =

8.1

12.1 cos53.1 0

= 0.4 MW (pu)/elect rad

p = 2

1

4

504.0

J = J 3.96

Frequency of oscillations = 3.96 rad/sec

= 2

96.3 =0.63 Hz

A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a

transmission circuit in which resistances is ignored .A fault takes place reducing the

maximum power transferable to 0.5 pu where as before the fault, this power was 2.0pu

and after the clearance of the fault, it is 1.5 pu .By the use of equal area criterion,

determine the critical clearing angle.

Solution :

Here

P .5.1,5.0,0.2 maxmax puandPpuPpuIIIMAXIII

Initial loading P m =1.0 pu

radP

P

I

m 523.02

1sinsin 1

max

1

0

rad41.25.1

1sin 1

max

cos cr =

2max3max

max3max02maxmax coscos

PP

PPpm

Cos 337.05.05.1

41.2cos5.1523.0cos5.0)523.041.2(0.1

cr

Or

3.70cr .