Power System Analysis unit - I

EE8501 – POWER SYSTEM ANALYSIS
Unit – I – Power System
By
Mr. A. Arun Kumar,
Assistant Professor,
Department of Electrical and Electronics Engineering.
Email: arunkumar@ritrjpm.ac.in
Cell: 98430 80689Power System Analy sis / Unit - I 1

COMPONENTS OF POWER SYSTEM
 Generation
 Transmission
 Distribution
 Utilization
2Power System Analysis / Unit – I/Mr. A.Arun Kumar, AP/EEE

Power System
Power System Analysis / Unit - I 3

ADVANTAGES OF
INTERCONNECTED (GRID) POWER SYSTEM
• Interconnection of power systems increases the reliability of the
system.
• It reduces the Steady state frequency error.
• For operation at peak loads, less number of generators are required
as a reserve for operation. Hence reserve capacity of the generating
station gets reduced.
• Less number of generators which are running without load are
required for meeting the sudden unexpected increase in load. Hence
the spinning reserve of the generating station gets reduced.
• It allows the use of most economical sources of power.
4
Power System Analysis / Unit - I

SYMBOLS OF POWER SYSTEM
COMPONENTS
1. Synchronous Generator
2. Motor
3. Two winding Transformer
4. Short Transmission Line
5. Static Load
6. Circuit Breaker
7. Three Phase Connection 5Power System Analysis / Unit - I

IMPEDANCE DIAGRAM
ASSUMPTIONS MADE :
• The neutral reactances are neglected.
• The shunt branches in the equivalent circuits
of transformers are neglected.
6Power System Analysis / Unit - I

REACTANCE DIAGRAM
I. The neutral reactances are neglected.
II. The shunt branches in the equivalent circuits
of transformers are neglected.
III. The resistances are neglected.
IV. All static loads are neglected.
V. The capacitance of transmission lines is
neglected.

Problem- 1
• Draw an impedance and reactance diagram
for a given power system.

Problem- 2

Problem- 3

Power System restructuring – An
Overview
Restructuring of the power industry aims at abolishing the
monopoly in the generation and trading sectors, thereby,
introducing the competition at various levels wherever it is
possible.
Electricity sector restructuring, also popularly known as
deregulation, is expected to draw private investment, increase
efficiency, promote technical growth and improve customer
satisfaction as different parties compete each other to win their
market share and remain in business.

Typical structure of vertically integrated
electric utility

Typical structure of a deregulated
electric system

Definitions
Deregulation:
Deregulation is a restructuring of the rules and economic incentives that
government set up to control and drives the electric supply industries.
It is changing the existing monopoly franchise rule or other regulations of
regulated industry, that affect how electric companies do business, and how
customers may buy electric power and services.
The suitable word in re-regulation which creates changes to encourage
competition wherever it is possible.
Restructuring:
It is disassembly of the original structure and re-assembly into another form
for better efficiency and performance.
Restructuring of regulated power sector is to separate the functions of power
generation, transmission, distribution and electric supply to consumers

Reasons for Introducing Deregulation
1. High Tariffs
2. Encourage innovation
3. Better customer choice
4. Change in generation economics of sale
5. Improvement in managerial efficiencies
6. Better experience of other deregulated market
7. Pressure from financial institution
8. Lack of public resources for future development
9. Need for regulation changed.

Benefits of Deregulation
1. Cheaper electricity
2. Efficient capacity expansion planning
3. Cost minimization
4. More choice
5. Employment
6. Pricing is cost reflective rather than a set tariff

CONCEPT OF PER UNIT SYSTEM
Whose percentage is more?
Marks Scored By Student 1 Marks Scored By Student 2
450 720

ADVANTAGES OF PER UNIT SYSTEM
1) The per unit system is ideal for the computerized analysis
and simulation of complex power system problems.
2) The per unit impedance referred to either side of
transformer is same for single phase transformer and three
phase transformer as well.
3) The comparison characteristics of the various electrical
apparatuses of different types of ratings is facilitated by
expressing the value of reactance's in per unit based on their
ratings.
4) The computational effort in the power system is very much
reduced with the use of per unit quantities. Contd(…)

Contd(…)
5) The per unit system helps to find the relative magnitude
information.
6) The chance of confusion between line and phase quantities in
a three phase balanced system is greatly reduced.
7) The manufacturers usually provide the impedance values in
per unit.

PROBLEM 4
Three generators are rated as follows:
Generator 1 :100 MVA, 33 kV, reactance 10%
Determine the reactance of the generators
corresponding to base value of 200 MVA,35 kV.

ANSWERS :
Generator 1 : 0.1778 p.u

PROBLEM 5
A single phase transformer is rated at 110/440 V,
2.5 kVA and leakage reactance is measured from a low
voltage side is 0.06 ohm. Determine the leakage
reactance in per unit.

ANSWERS :
Low Voltage Side:
Xbase=4.84
Xp.u=0.01239
High Voltage Side:
Xbase=77.44
Xactual=0.96
Xp.u=0.01239

PROBLEM 6
Draw an reactance diagram, showing all impedances in p.u. Choose
100 MVA as a base and 20 kV as base voltage for generator.
G : 40 MVA, 20 kV, X”=20%
T1 : 60 MVA, 20/200 kV, X”=30%
T2 : 60 MVA, 200/20 kV, X”=30%
M : 20 MVA, 20 kV, X”=10%
Line : 200 kV, Z=150+j250 Ω

ANSWERS :
G : j0.5 p.u
T1 : j0.5 p.u
Line : 0.375+j0.625
T2 : j0.5 p.u
M : j0.5 p.u

PROBLEM 6
The one line diagram of a power system is shown in
figure. Draw the per unit per phase reactance
diagram
G :90 MVA, 11 kV, X”=18%
T1 :70 MVA, 11/110 kV, X”=15%
T2 :60 MVA, 110/11 kV, X”=10%
T3 :30 MVA, 11/220 kV, X”=9%
T4 :50 MVA, 220/11 kV, X”=12%
Line 1 :110 kV, Z=80Ω
Line 2 :220 kV, Z=120Ω
M :85 MVA, 11 kV, X”=13%

ANSWERS :
G : j0.2 p.u
T1 : j0.2143 p.u
Line 1 : j0.6612 p.u
T2 : j0.1667 p.u
T3 : j0.3 p.u
Line 2 : j0.2479 p.u
T4 : j0.24 p.u
M : j0.1529 p.u

Problem 7
Determine the bus admittance matrix of the
representative power system shown in figure
Bus code Admittance
1-2 2-j8.0
1-3 1-j4.0
2-3 0.666-j2.664
2-4 1-j4.0
3-4 2-j8.0

ANSWERS :
3-j12 -2+j8 -1+j4 0
-2+j8 3.666-j14.664 -0.666+j2.664 -1+j4
-1+j4 -0.666+j2.664 3.666-j14.664 -2+j8
0 -1+j4 -2+j8 3-j12

Problem 8
Determine the bus admittance matrix of the
representative power system shown in figure
Bus code i-k Impedance,
Zik
Line Charging
yij/2
1-2 0.02+j0.06 J0.03
1-3 0.08+j0.24 J0.025
2-3 0.06+j0.18 J0.020

ANSWERS :
6.25-j18.695 -5+j15 -1.25+j3.75
-5+j15 6.667-j19.95 -1.667+j5
-1.25+j3.75 -1.667+j5 2.917-j8.705

Singular Transformation Method
Graph:
A graph is a representation of a
network by a set of nodes and set of lines that
connect the nodes.
Nodes:
A node can be described as a
terminal of an element.

Edges:
A line segment which connects two nodes.
Oriented graph:
It is called as directed graph. It
can be defined as the edges of a graph which
directs from one node to the other node.

Problem 9
Form bus admittance matrix by singular
transformation method. Take (1) as reference
node.
Element No.
Self Impedance
Bus Code Impedance
1 1-2 (1) j0.35
2 1 – 3 j0.2
3 3 – 4 j0.2
4 1 – 2 (2) j0.2
5 2 - 4 j0.2

Ybus= j7.4404 0 j3.3333
0 -j10 j5
j3.333 j5 -j8.3333
Ybus= -j12.85 0 J5
0 -j10 j5
j5 j5 -J10

ZBUS MARTIX FORMATION

Problem-11
Construct the impedance matrix

ANSWER :
Z= j0.6 j0.5 j0.5
j0.5 j0.625 j0.625
j0.5 j0.625 j0.675

Problem-12
Obtain the bus impedance matrix using bus
building algorithm.

ANSWER :
Z= j0.2793 j0.2206 j0.2498
j0.2206 j0.2793 j0.2502
j0.2498 j0.2502 j0.3492

Power System Analysis unit - I

More Related Content

What's hot (20)

Similar to Power System Analysis unit - I (20)

Recently uploaded (20)

Power System Analysis unit - I