Course File On Power system-II By MS. M.MANISHA...

KG Reddy College of Engineering & Technology

(Approved by AICTE, New Delhi, Affiliated to JNTUH, Hyderabad)

Chilkur (Village), Moinabad (Mandal), R. R Dist, TS-501504

Accredited by NAAC

1

Course File On

Power system-II

By

MS. M.MANISHA

Assistant Professor,

Electrical & Electronics Engineering

K. G. Reddy College Of Engineering and Technology

2019-2020

HOD Principal

EEE KGRCET




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COURSE FILE Subject Name : Power systems -II

Faculty Name : M.Manisha

Designation : Assistant Professor

Regulation /Course Code : R16/ EE502PC

Year / Semester : III / Ist

Department : Electrical & Electronics

Engineering

Academic Year : 2019-20




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COURSE FILE CONTENTS

S.N. Topics Page No.

1 Vision, Mission, PEO’s, & PO’s, PSOs

2 Syllabus (University Copy)

3 Course Objectives, Course Outcomes And Topic Outcomes

4 Course Prerequisites

5 CO’s, PO’s Mapping

6 Course Information Sheet (CIS)

a). Course Description

b). Syllabus

c). Gaps in Syllabus

d). Topics beyond syllabus

e). Web Sources-References

f). Delivery / Instructional Methodologies

g). Assessment Methodologies-Direct

h). Assessment Methodologies –Indirect

i). Text books & Reference books

7 Micro Lesson Plan

8 Teaching References Plan

9 Lecture Notes -Unit Wise (Hard Copy)

10 OHD/LCD SHEETS /CDS/DVDS/PPT (Soft/Hard copies)

11 University Previous Question papers

12 MID exam Descriptive Question Papers

13 MID exam Objective Question papers

14 Assignment topics with materials

15 Tutorial topics and Questions

16 Unit wise-Question bank

1 Two marks question with answers 5 questions

2 Three marks question with answers 5 questions

3 Five marks question with answers 5 questions

4 Objective question with answers 10 questions

5 Fill in the blanks question with answers 10 questions

17 Beyond syllabus Topics with material

18 Result Analysis-Remedial/Corrective Action

19 Sample Students Descriptive Answer sheets

20 Sample Students Assignment Sheets

21 Record of Tutorial Classes

22 Record of Remedial Classes

23 Record of guest lecturers conducted




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PART-2

S.NO. Topics

1 Attendance Register/Teacher Log Book

2 Time Table

3 Academic Calendar

4 Continuous Evaluation-marks (Test, Assignments etc)

5 Status Request internal Exams and Syllabus coverage

6 Teaching Diary/Daily Delivery Record

7 Continuous Evaluation – MID marks

8 Assignment Evaluation- marks /Grades

9 Special Descriptive Tests Marks

10 Sample students descriptive answer sheets

11 Sample students assignment sheets




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1. VISION, MISSION, PROGRAM EDUCATIONAL OBJECTIVES

(A) VISION

To become a renowned department imparting both technical and non-technical skills to the students

by implementing new engineering pedagogy’s and research to produce competent new age electrical

engineers.

(B) MISSION

To transform the students into motivated and knowledgeable new age electrical engineers.

To advance the quality of education to produce world class technocrats with an ability to adapt to

the academically challenging environment.

To provide a progressive environment for learning through organized teaching methodologies,

contemporary curriculum and research in the thrust areas of electrical engineering.

(C) PROGRAM EDUCATIONAL OBJECTIVES

PEO 1: Apply knowledge and skills to provide solutions to Electrical and Electronics Engineering

problems in industry and governmental organizations or to enhance student learning in educational

institutions

PEO 2: Work as a team with a sense of ethics and professionalism, and communicate effectively to

manage cross-cultural and multidisciplinary teams

PEO 3: Update their knowledge continuously through lifelong learning that contributes to

personal, global and organizational growth

(D) PROGRAM OUTCOMES

PO 1: Engineering knowledge: Apply the knowledge of mathematics, science, engineering

fundamentals and an engineering specialization to the solution of complex engineering problems.

PO 2: Problem analysis: Identify, formulate, review research literature, and analyze complex

engineering problems reaching substantiated conclusions using first principles of mathematics,

natural science and engineering sciences.

PO 3: Design/development of solutions: design solutions for complex engineering problems and

design system components or processes that meet the specified needs with appropriate

consideration for the public health and safety, and the cultural, societal and environmental

considerations.




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PO 4: Conduct investigations of complex problems: use research based knowledge and research

methods including design of experiments, analysis and interpretation of data, and synthesis of the

information to provide valid conclusions.

PO 5: Modern tool usage: create, select and apply appropriate techniques, resources and modern

engineering and IT tools including prediction and modeling to complex engineering activities with

an understanding of the limitations.

PO 6: The engineer and society: apply reasoning informed by the contextual knowledge to

assess societal, health, safety, legal and cultural issues and the consequent responsibilities relevant

to the professional engineering practice.

PO 7: Environment sustainability: understand the impact of the professional engineering

solutions in the societal and environmental contexts, and demonstrate the knowledge of, and need

for sustainable development.

PO 8: Ethics: apply ethical principles and commit to professional ethics and responsibilities and

norms of the engineering practice.

PO 9: Individual and team work: function effectively as an individual and as a member or

leader in diverse teams, and in multidisciplinary settings.

PO 10: Communication: communicate effectively on complex engineering activities with the

engineering community and with society at large, such as, being able to comprehend and write

effective reports and design documentation, make effective presentations, and give and receive

clear instructions.

PO 11: Project management and finance: demonstrate knowledge and understanding of the

engineering and management principles and apply these to one’s own work, as a member and

leader in a team, to manage projects and in multidisciplinary environments.

PO 12: Lifelong learning: recognize the need for, and have the preparation and ability to engage

in independent and lifelong learning in the broader context of technological change.

(E) PROGRAM SPECIFIC OUTCOMES

PSO-1: Apply the engineering fundamental knowledge to identify, formulate, design and investigate

complex engineering problems of electric circuits, power electronics, electrical machines and power

systems and to succeed in competitive exams like GATE, IES, GRE, OEFL, GMAT, etc.

PSO-2: Apply appropriate techniques and modern engineering hardware and software tools in power

systems and power electronics to engage in life-long learning and to get an employment in the field of

Electrical and Electronics Engineering.




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PSO-3: Understand the impact of engineering solutions in societal and environmental context, commit

to professional ethics and communicate effectively.

2. SYLLABUS (UNIVERSITY COPY)

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD.

B.Tech. III Year I Sem. L T P C

4 1 0 4

(a) COURSE OBJECTIVES

To compute inductance and capacitance of different transmission lines.

To understand performance of short, medium and long transmission lines.

To examine the traveling wave performance and sag of transmission lines.

To design insulators for over head lines and understand cables for power

transmission.

(b) COURSE OUTCOMES

To compute inductance and capacitance for different configurations of transmission lines.

To analyze the performance of transmission lines can understand transient’s phenomenon

of transmission lines.

To Analyze the Skin and Proximity effects - Description and effect on Resistance of

Solid Conductors/

To Calculate sag and tension calculations

To compute THE overhead line insulators and underground cables.

UNIT-I

Transmission Line Parameters: Types of conductors - calculation of resistance for solid conductors

- Calculation of inductance for single phase and three phase, single and double circuit lines, concept

of GMR & GMD, symmetrical and asymmetrical conductor configuration with and without

transposition, Numerical Problems.

Calculation of capacitance for 2 wire and 3 wire systems, effect of ground on capacitance,

capacitance calculations for symmetrical and asymmetrical single and three phase, single and double

circuit lines, Numerical Problems.




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UNIT-II

Performance of Short and Medium Length Transmission Lines: Classification of Transmission

Lines - Short, medium and long line and their model representations - Nominal-T, Nominal-Pie and

A, B, C, D Constants for symmetrical & Asymmetrical Networks, Numerical Problems.

Mathematical Solutions to estimate regulation and efficiency of all types of lines - Numerical

Problems.

Performance of Long Transmission Lines: Long Transmission Line - Rigorous Solution,

evaluation of A,B,C,D Constants, Interpretation of the Long Line Equations, Incident, Reflected and

Refracted Waves -Surge Impedance and SIL of Long Lines, Wave Length and Velocity of

Propagation of Waves - Representation of Long Lines - Equivalent-T and Equivalent Pie network

models (numerical problems).

UNIT – III

Power System Transients: Types of System Transients - Travelling or Propagation of Surges -

Attenuation, Distortion, Reflection and Refraction Coefficients - Termination of lines with different

types of conditions - Open Circuited Line, Short Circuited Line, T- Junction, Lumped Reactive

Junctions (Numerical Problems), Bewley’s Lattice Diagrams (for all the cases mentioned with

numerical examples).

Various Factors Governing The Performance of Transmission Line: Skin and Proximity effects -

Description and effect on Resistance of Solid Conductors - Ferranti effect - Charging Current - Effect

on Regulation of the Transmission Line.

Corona - Description of the phenomenon, factors affecting corona, critical voltages and power loss,

Radio Interference.

UNIT-IV

Overhead Line Insulators: Types of Insulators, String efficiency and Methods for improvement,

Numerical Problems - voltage distribution, calculation of string efficiency, Capacitance grading and

Static Shielding.

Sag and Tension Calculations: Sag and Tension Calculations with equal and unequal heights of

towers, Effect of Wind and Ice on weight of Conductor, Numerical Problems - Stringing chart and

sag template and its applications.

UNIT-V

Underground Cables: Types of Cables, Construction, Types of Insulating materials, Calculation of

Insulation resistance and stress in insulation, Numerical Problems. Capacitance of Single and 3-




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Core belted cables, Numerical Problems. Grading of Cables - Capacitance grading - Numerical

Problems, Description of Inter-sheath grading - HV cables.

TEXT BOOKS:

“C. L. Wadhwa”, “Electrical power systems”, New Age International (P) Limited Publishers, 1998.

“Grainger and Stevenson”, “Power Systems Analysis”, Mc Graw Hill, 1st Edition 2003.

“M. L. Soni, P. V. Gupta, U.S. Bhatnagar and A. Chakrabarthy”, Power System Engineering,

Dhanpat Rai & Co Pvt. Ltd, 2009.

REFERENCE BOOKS:

“I. J. Nagarath & D. P Kothari” , “Power System Engineering”, TMH, 2nd Edition 2010

“B. R. Gupta”, “Power System Analysis and Design”, Wheeler Publishing, 1998.

“Abhijit Chakrabarti and Sunitha Halder”, “Power System Analysis Operation and control”, PHI, 3rd

Edition, 2010




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3. COURSE OBJECTIVES AND COURSE OUTCOMES

(a) COURSE OBJECTIVES

To compute inductance and capacitance of different transmission lines.

To explain the performance of short, medium and long transmission lines.

To examine the traveling wave performance and sag of transmission lines.

To design insulators for over head lines and understand cables for power transmission.

(b) COURSE OUTCOMES: After completion of this course, the student will able to

(CO1) Able to Compute inductance and capacitance for different configurations of

transmission lines.

(CO2) Analyze the performance of transmission lines Can understand transient’s phenomenon of

transmission lines. (CO3) Analyze the skin and Proximity effects - Description and effect on Resistance of Solid

Conductors

(CO4) Calculate sag and tension calculations.

(CO5) Explain overhead line insulators and underground cables.




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(c)TOPIC OUTCOMES

S.NO TOPIC TOPIC OUTCOME

UNIT-I At the end of the topic, the student will be able to

L1 Introduction To The Course Recollect the Electrical components and

devices

L2 Types Of Conductors - Calculation Of

Resistance For Solid Conductors

Classify the conductors

L3 Calculation Of Inductance For Single Phase And

Three Phase

Calculate the Inductance For Single

Phase And Three Phase

L4 Single And Double Circuit Lines, Concept Of

GMR & GMD

Estimate the Single And Double Circuit

Lines

L5 Symmetrical Conductor Configuration With And

Without Transposition

Apply the Symmetrical Conductor

Configuration With And Without

Transposition

L6 Asymmetrical Conductor Configuration With

And Without Transposition

Apply the Asymmetrical Conductor

Configuration With And Without

Transposition

L7 Numerical Problems. solve the problems

L8 Calculation Of Capacitance For 2 Wire And 3

Wire Systems

Calculate the Capacitance For 2 Wire

And 3 Wire Systems

L9 Effect Of Ground On Capacitance, Identify Effect Of Ground On

Capacitance,

L10 Capacitance Calculations For Symmetrical

Single And Three Phase

Analyse the Capacitance Calculations

For Symmetrical Single And Three

Phase

L11 Capacitance Calculations For Asymmetrical


Analyse the Capacitance Calculations

For Symmetrical Single And Three

Phase

L12 Single And Double Circuit Lines Identify Single And Double Circuit

Lines

L13 Numerical Problems Solve the problems

UNIT-II L14 Classification Of Transmission Lines - Short,

Medium And Long Line And Their Model

Representations

Classify the transmission lines

L15 Nominal-T, Nominal-Pie And A, B, C, D

Constants For Symmetrical & Networks

Apply Nominal-T, Nominal-Pie And A,

B, C, D Constants For Symmetrical &




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Networks


Constants For Asymmetrical Networks

Apply Nominal-T, Nominal-Pie And A,

B, C, D Constants For Symmetrical &

Networks

L17 Numerical problems Solve the problems

L18 Mathematical Solutions to estimate regulation

and efficiency of all types of lines

Apply Mathematical Solutions to

estimate regulation and efficiency of all

types of lines


L20 Interpretation of the Long Line Equations Derive Interpretation of the Long Line

Equations

L21 Incident, Reflected And Refracted Waves Identify the Incident, Reflected And

Refracted Waves

L22 Surge Impedance And Sil Of Long Lines Analyse the Surge Impedance And Sil

Of Long Lines

L23 Wave Length And Velocity Of Propagation Of

Waves

Analyse Wave Length And Velocity Of

Propagation Of Waves

L24 Representation Of Long Lines - Equivalent-T

And Equivalent Pie Network Models

Classify the network models

L25 Numerical Problems

UNIT-III L26 Types of System Transients Classify the types of system transients

L27 Travelling or Propagation of Surges Analyse the travelling surges

L28 Attenuation, Distortion, Reflection and

Refraction Coefficients

Attenuation, Distortion, Reflection and


L29 Termination of lines with different types of

conditions

Classify the transmission lines

L30 Open Circuited Line, Short Circuited Line Draw the open circuited and short

circuited lines

L31 T- Junction, Lumped Reactive Junctions (Numerical Problems)

Solve the problems

L32 Bewley’s Lattice Diagrams (for all the cases

mentioned with numerical examples).

Draw the Bewley’s Lattice diagram

L33 Skin and Proximity effects - Description and

effect on Resistance of Solid Conductors

Derive the effect on resistance of solid

conductors

L34 Ferranti effect - Charging Current - Effect


Derive the regulation of the

transmission lines

L35 Corona - Description of the phenomenon,

factors affecting corona

Identify the factors effecting corona

L36 Critical voltages and power loss, Radio

Interference.

Derive the critical voltages and power

loss ,radio interference




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UNIT-IV L37 Types of Insulators, String efficiency and

Methods for improvement

Classify the insulators


L39 voltage distribution Analyse voltage distribution

L40 calculation of string efficiency Calculate string efficiency

L41 Capacitance grading and Static Shielding. Analyse the capacitance grading and

static shielding.

L42 Sag and Tension Calculations with equal

heights of towers

Calculate the sag and tension

calculations with equal heights of

towers

L43 Sag and Tension Calculations with unequal

heights of towers

Calculate the sag and tension

calculations with equal heights of

towers

L44 Effect of Wind and Ice on weight of

Conductor

Derive the wind and ice on weight of

conductor

L45 Numerical Problems - Solve the problems

L46 Stringing chart and sag template and its

applications

Identify the Stringing chart and sag

template and its applications

UNIT-V L47 Types of Cables, Construction Classify the cables

L48 Types of Insulating materials, Calculation of

Insulation resistance and stress in

insulation

Classify the insulating materials

L49 Types of Insulating materials Classify the insulating materials

L50 Calculation of Insulation resistance and

stress in insulation.

Calculate the insulation resistance and

stress in insulation

L51 Numerical Problems. Solve the problems

L52 Capacitance of Single and 3-Core belted

cables

Derive the capacitance of single and 3-

core belted cables


L54 Grading of Cables Identify the cables

L55 Capacitance grading - Numerical Problems Solve the problems

L56 Description of Inter-sheath grading - HV

cables

Describe the inter sheath grading

L57 Revision Revise the types of insulating materials

L58 Revision Revise the types of cables




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4. COURSE PREREQUISITES

1. Power Systems –I

2. Electromagnetic field theory

5) CO’S, PO’S MAPPING:

CO&PO Mappings

PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12

CO1 - 3 2 - - - - - - - - -

CO2 - - 2 - 3 - - - - - - -

CO3 3 3 - - - - - - - - - -

CO4 2 - 2 - - - - - - - - -

CO5 3 - - - - 3 - - - - - -




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6. COURSE INFORMATION SHEET (CIS)

6. (a) Course description

PROGRAMME: B. Tech.

(Computer Science Engineering.)

DEGREE: B.TECH

COURSE: Power System II YEAR: III SEM: I CREDITS: 4

COURSE CODE: EE502PC

REGULATION: R16

COURSE TYPE: CORE

COURSEAREA/DOMAIN:

Architecture/organization

CONTACT HOURS: 4+0 (L+T)) hours/Week.

6. (b) Syllabus

Unit Details Hours

I Transmission Line Parameters:

Types of conductors - calculation of resistance for solid

conductors - Calculation of inductance for single phase and

three phase, single and double circuit lines, concept of GMR &

GMD, symmetrical and asymmetrical conductor configuration

with and without transposition, Numerical Problems.

Calculation of capacitance for 2 wire and 3 wire systems, effect

of ground on capacitance, capacitance calculations for

symmetrical and asymmetrical single and three phase, single

and double circuit lines, Numerical Problems.

13




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II Performance of Short and Medium Length Transmission

Lines:

Classification of Transmission Lines - Short, medium and long

line and their model representations - Nominal-T, Nominal-Pie

and A, B, C, D Constants for symmetrical & Asymmetrical

Networks, Numerical Problems. Mathematical Solutions to

estimate regulation and efficiency of all types of lines -

Numerical Problems.

Performance of Long Transmission Lines:

Long Transmission Line - Rigorous Solution, evaluation of

A,B,C,D Constants, Interpretation of the Long Line Equations,

Incident,

Reflected and Refracted Waves -Surge Impedance and SIL of

Long Lines, Wave Length and Velocity of Propagation of

Waves - Representation of Long Lines - Equivalent-T and

Equivalent Pie network models (numerical problems)

10

III Power System Transients:

Types of System Transients - Travelling or Propagation of

Surges - Attenuation, Distortion, Reflection and Refraction

Coefficients - Termination of lines with different types of

conditions - Open Circuited Line, Short Circuited Line, T-

Junction, Lumped Reactive Junctions (Numerical Problems),

Bewley’s Lattice Diagrams (for all the cases mentioned with

numerical examples).

Various Factors Governing The Performance of

Transmission Line:

Skin and Proximity effects - Description and effect on

Resistance of Solid Conductors - Ferranti effect - Charging

Current - Effect on Regulation of the Transmission Line.

Corona - Description of the phenomenon, factors affecting

corona, critical voltages and power loss, Radio Interference.

11




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IV Overhead Line Insulators:

Types of Insulators, String efficiency and Methods for

improvement, Numerical Problems - voltage distribution,

calculation of string efficiency, Capacitance grading and Static

Shielding.

Sag and Tension Calculations:

Sag and Tension Calculations with equal and unequal heights

of towers, Effect of Wind and Ice on weight of Conductor,

Numerical Problems - Stringing chart and sag template and its

applications.

10

V Underground Cables:

Types of Cables, Construction, Types of Insulating materials,

Calculation of Insulation resistance and stress in insulation,

Numerical Problems. Capacitance of Single and 3-Core belted

cables, Numerical Problems. Grading of Cables - Capacitance

grading - Numerical Problems, Description of Inter-sheath

grading - HV cables.

13

Contact classes for syllabus coverage 57

Lectures beyond syllabus 02

Tutorial classes 10

Classes for gaps & Add-on classes 02

Total No. of classes 71

6 (c) Gaps in syllabus

S.NO. DESCRIPTION PROPOSED ACTIONS

1 Equivalent-T and Equivalent Pie network models 1 period (Class Room)

Calculation of Insulation resistance and stress in

1 periods (Class Room)




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2 insulation

6(d) Topics beyond Syllabus

1 Stringing chart and sag template and its applications.

Guest Lecture

2 Rigorous Solution NPTEL

6 (e) Web Source References

Sl. No. Name of book/ website

a. https://www.youtube.com/watch?v=1Ym2OviN0XM

b. https://www.youtube.com/watch?v=tWSz7Xm8mMk

c. https://www.youtube.com/watch?v=AfaxqVq1x6c

6(f)Delivery / Instructional Methodologies:

CHALK & TALK STUD. ASSIGNMENT WEB RESOURCES

LCD/SMART

BOARDS

STUD. SEMINARS ☐ ADD-ON COURSES

6(g)Assessment Methodologies - Direct

Assignments Stud. Seminars Tests/Model

Exams

Univ. Examination

https://www.youtube.com/watch?v=1Ym2OviN0XM

https://www.youtube.com/watch?v=tWSz7Xm8mMk

https://www.youtube.com/watch?v=AfaxqVq1x6c




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Stud. Lab

Practices

Stud. Viva ☐ Mini/Major

Projects

☐ Certifications

☐ Add-On

Courses

☐ Others

6(h) Assessment Methodologies - Indirect

Assessment Of Course Outcomes

(By Feedback, Once)

Student Feedback On

Faculty (Twice)

☐Assessment Of Mini/Major Projects By

Ext. Experts

☐ Others

6(i) Text books and References

T/R BOOK TITLE/AUTHORS/PUBLICATION

Text Book “C. L. Wadhwa”, “Electrical power systems”, New Age International (P) Limited Publishers, 1998.

Text Book “Grainger and Stevenson”, “Power Systems Analysis”, Mc Graw Hill, 1st Edition 2003.

Text Book “M. L. Soni, P. V. Gupta, U.S. Bhatnagar and A. Chakrabarthy”, Power System Engineering, Dhanpat Rai & Co Pvt. Ltd, 2009.

Reference

Book “I. J. Nagarath & D. P Kothari” , “Power System Engineering”, TMH, 2nd Edition 2010

Reference

Book “B. R. Gupta”, “Power System Analysis and Design”, Wheeler Publishing, 1998.

Reference “Abhijit Chakrabarti and Sunitha Halder”, “Power System Analysis Operation




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Book and control”, PHI, 3rd Edition, 2010

7. MICRO LESSON PLAN

S.N. Topic Schedule data Actual Date

UNIT-1 At the end of the topic, the student will be able to

L1 Introduction To The Course L2 Types Of Conductors - Calculation Of

Resistance For Solid Conductors

L3 Calculation Of Inductance For Single Phase

And Three Phase

L4 Single And Double Circuit Lines, Concept Of

GMR & GMD

L5 Symmetrical Conductor Configuration With


L6 Asymmetrical Conductor Configuration With


L7 Numerical Problems. L8 Calculation Of Capacitance For 2 Wire And 3

Wire Systems

L9 Effect Of Ground On Capacitance, L10 Capacitance Calculations For Symmetrical


L11 Capacitance Calculations For Asymmetrical


L12 Single And Double Circuit Lines L13 Numerical Problems

UNIT-II




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L13 Classification Of Transmission Lines - Short,

Medium And Long Line And Their Model

Representations


Constants For Symmetrical & Networks


Constants For Asymmetrical Networks

L16 Numerical problems L17 Mathematical Solutions to estimate regulation

and efficiency of all types of lines


L19 Interpretation of the Long Line Equations L20 Incident, Reflected And Refracted Waves

L21 Surge Impedance And Sil Of Long Lines

L22 Wave Length And Velocity Of Propagation Of

Waves

L23 Representation Of Long Lines - Equivalent-T

And Equivalent Pie Network Models


UNIT-III

L25 Types of System Transients L26 Travelling or Propagation of Surges L27 Attenuation, Distortion, Reflection and


L28 Termination of lines with different types of

conditions

L29 Open Circuited Line, Short Circuited Line L30 T- Junction, Lumped Reactive Junctions (Numerical

Problems)

L31 Bewley’s Lattice Diagrams (for all the cases

mentioned with numerical examples).

L32 Skin and Proximity effects - Description and

effect on Resistance of Solid Conductors

L33 Ferranti effect - Charging Current - Effect


L34 Corona - Description of the phenomenon,

factors affecting corona




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L35 critical voltages and power loss, Radio

Interference.

UNIT-IV

L36 Types of Insulators, String efficiency and

Methods for improvement


L38 voltage distribution

L39 calculation of string efficiency

L40 Capacitance grading and Static Shielding.

L41 Sag and Tension Calculations with equal

heights of towers

L42 Sag and Tension Calculations with

unequal heights of towers

L43 Effect of Wind and Ice on weight of

Conductor

L44 Numerical Problems -

L45 Stringing chart and sag template and its

applications

UNIT-V

L46 Types of Cables, Construction

L47 Types of Insulating materials, Calculation

of Insulation resistance and stress in

insulation

L48 Types of Insulating materials

L49 Calculation of Insulation resistance and

stress in insulation.

L50 Numerical Problems.

L51 Capacitance of Single and 3-Core belted

cables


L53 Grading of Cables

L54 Capacitance grading - Numerical Problems

L55 Description of Inter-sheath grading - HV

cables

L56 Revision




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L57 Revision L58 Lecture beyond syllabus (Nptel)




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8) Teaching Schedule

Subject ELECTRICAL TECHNOLOGY

Text Books (to be purchased by the Students)

Book 1 “C. L. Wadhwa”, “Electrical power systems”, New Age International (P) Limited

Publishers, 1998.

Book 2 “M. L. Soni, P. V. Gupta, U.S. Bhatnagar and A. Chakrabarthy”, Power System

Engineering, Dhanpat Rai & Co Pvt. Ltd, 2009.

Reference Books Book 3 “I. J. Nagarath & D. P Kothari” , “Power System Engineering”, TMH, 2nd Edition 2010

Book 4 “B. R. Gupta”, “Power System Analysis and Design”, Wheeler Publishing, 1998.

Unit

Topic Chapters No’s No of

classes Book 1 Book 2 Book 3 Book 4

I Line constant calculations 2 2 - - 7

Capacitance of transmission lines 3 2 3 - 6

Wave propagation on transmission

lines - 13 - - 2

Characteristics and performance of

power transmission lines - - 5 - 3

II Performance of lines 4 3 - - 5

III Performance of lines 4 3 - - 3

Corona 6 6 - - 4

Transients in power system 12 - 13 - 4

IV

Over head line insulators - 5 - - 3

Mechanical design of transmission

lines 7 - - - 3

Insulation coordination and over 16 - - - 4




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voltage protection

V Underground cables - 9 - - 7

Insulated cables 9 - - - 6

Contact classes for syllabus coverage 57

Tutorial classes,

Lecture beyond

syllabus and gaps in

the syllabus

14

Total No. of classes 71




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11.University Previous Question papers




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12. MID exam Descriptive Question Papers

Q.NO QUESTION

CO

mapping

Blooms

taxonomy

level

1 1.What are the differences between nominal-T and

nominal-π methods?

CO1 Understanding

2 2. Define A, B, C and D constants of a transmission line?

What are their values in short lines?

CO1 Understanding

3

3. Find GMD, GMR for each circuit, inductance for each

circuit, and total inductance per meter for two circuits that run

parallel to each othar. One circuit consists of three 0.25 cm

radies conductors’. Tha second circuit consists of two 0.5 cm

radies conductor

CO2 Apply

4

4. A three phase 60 Hz line is arranged as shown. The

conductors are ACSR Drake. Find the capacitive reactance for

1 mile of the line.

CO2 Understanding

K. G. Reddy College of Engineering &Technology

(Approved by AICTE, Affiliated to JNTUH)

Chilkur (Vil), Moinabad (Mdl), RR District

College Code

QM

Name of the Exam: I Mid Examinations Marks: 10

Year-Sem &

Branch: III-I Duration: 60 Min

Subject: PS Date &

Session

Answer ANY TWO of the following Questions

2X5=10




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1.What are the differences between nominal-T and nominal-π methods?

b) The transmission line having its effective length more than 80 km but less than 250 km is

generally referred to as a medium transmission line. Due to the line length being considerably high,

admittance Y of the network does play a role in calculating the effective circuit parameters, unlike in

the case of short transmission lines. For this reason the modeling of a medium length transmission

line is done using lumped shunt admittance along with the lumped impedance in series to the

circuit.These lumped parameters of a medium length transmission line can be represented using three

different models, namely-

Nominal Π representation.

Nominal T representation.

End Condenser Method.

Let’s now go into the detailed discussion of these above mentioned models.

Nominal Π Representation of a Medium Transmission Line

In case of a nominal Π representation, the lumped series impedance is placed at the middle of the

circuit where as the shunt admittances are at the ends. As we can see from the diagram of the Π

network below, the total lumped shunt admittance is divided into 2 equal halves, and each half with

value Y ⁄ 2 is placed at both the sending and the receiving end while the entire circuit impedance is

between the two.The shape of the circuit so formed resembles that of a symbol Π, and for this reason

it is known as the nominal Π representation of a medium transmission line. It is mainly used for

determining the general circuit parameters and performing load flow analysis.

As we can see here,

VS and VR is the supply and receiving end voltages respectively, and Is is the current flowing through

the supply end. IR is the current flowing through the receiving end of the circuit. I1 and I3 are the

values of currents flowing through the admittances. And I2 is the current through the impedance Z.

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Now applying KCL, at node P, we get. Similarly applying KCL, to

node Q. Now substituting equation (2) to equation (1)

Now by applying KVL to the circuit,

Comparing equation (4) and (5) with the standard ABCD parameter equations

We derive the parameters of a medium transmission line as:

Nominal T Representation of a Medium Transmission Line

In the nominal T model of a medium transmission line the lumped shunt admittance is placed in the

middle, while the net series impedance is divided into two equal halves and and placed on either side

of the shunt admittance. The circuit so formed resembles the symbol of a capital T, and hence is

known as the nominal T network of a medium length transmission line and is shown in the diagram

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below. Here

also Vs and Vr is the supply and receiving end voltages respectively, and Is is the current flowing

through the supply end. Ir is the current flowing through the receiving end of the circuit. Let M be a

node at the midpoint of the circuit, and the drop at M, be given by Vm. Applying KVL to the above

network we get, Now the sending end

current is, Substituting the value of VM to equation (9) we get,

Again comparing equation (8) and (10) with the

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standard ABCD parameter equations, The parameters of the T network of a

medium transmission line are

End Condenser Method

In this method, the capacitance of the line is limped or concentrated at the receiving or load end. This

method of localizing the line capacitance at the load end overestimates the effects of capacitance.

2. Define A, B, C and D constants of a transmission line? What are their values in short lines?

The transmission lines which have length less than 50 km are generally referred as short

transmission lines.

For short length, the shunt capacitance of this type of line is neglected and other parameters like

electrical resistance and inductor of these short lines are lumped, hence the equivalent circuit is

represented as given below, Let’s draw the vector diagram for this equivalent circuit, taking receiving

end current Ir as reference. The sending end and receiving end voltages make angle with that

reference receiving end current, of φs and φr,

respectively.


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As the shunt capacitance of the line is neglected, hence sending end current and receiving end current

is same, i.e.

Now if we observe the vector diagram carefully, we will get, Vs is approximately equal to

That means, as the

it is assumed that As there is no capacitance, during no load condition the current through

the line is considered as zero, hence at no load condition, receiving end voltage is the same as

sending end voltage. As per dentition of voltage regulation of power transmission line,


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Here, vr and vx are the per unit resistance and reactance of the short transmission line respectively.

Any electrical network generally has two input terminals and two output terminals. If we consider

any complex electrical network in a black box, it will have two input terminals and output terminals.

This network is called two - port network. Two port model of a network simplifies the network

solving technique. Mathematically a two port network can be solved by 2 by 2 matrix.

A transmission as it is also an electrical network; line can be represented as two port network. Hence

two port network of transmission line can be represented as 2 by 2 matrixes. Here the concept of

ABCD parameters comes. Voltage and currents of the network can represented as ,

Where, A, B, C and D are different constant of the network. If we put Ir = 0 at equation (1), we get,

Hence, A is the voltage impressed at the sending end per volt at the receiving end when receiving end

is open. It is dimension less. If we put Vr = 0 at equation (1), we get

That indicates it is impedance of the

transmission line when the receiving terminals are short circuited. This parameter is referred as

transfer impedance. C is the current in

amperes into the sending end per volt on open circuited receiving end. It has the dimension of

admittance. D is the current in amperes into the sending end per amp on short

circuited receiving end. It is dimensionless. Now from equivalent circuit, it is found that,

Comparing these equations with equation 1 and 2 we get, A = 1, B

= Z, C = 0 and D = 1. As we know that the constant A, B, C and D are related for passive network as,

AD − BC = 1

Here, A = 1, B = Z, C = 0 and D = 1

⇒ 1.1 − Z.0 = 1






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6 Daa' Dab' Dba' Bbb' Dca' Dcb'

So the values calculated are correct for short transmission line. From above equation (1),

When Ir = 0 that means receiving end terminals is open circuited and then from

the equation 1, we get receiving end voltage at no load. and as per definition of

voltage regulation of power transmission line,

Efficiency of Short Transmission Line

The efficiency of short line as simple as efficiency equation of any other electrical equipment, that

means R is per phase electrical

resistance of the transmission line.

3. Find GMD, GMR for each circuit, inductance for each circuit, and total inductance per meter for two

circuits that run parallel to each othar. One circuit consists of three 0.25 cm radies conductors’. Tha

second circuit consists of two 0.5 cm radies conductor

m = 3, n’ = 2, m n’ = 6

GMD

GMD = 10.743 m




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GMRA = 0.481m

Inductance of circuit A

H / m

H / m

The total inductance is than

LT=LA+LB =10-7

4. A three phase 60 Hz line is arranged as shown. The conductors are ACSR Drake. Find the capacitive

reactance for 1 mile of the line. If the length of the line is 175 miles and the normal operating voltage is

220 kV, find the capacitive reactance to neutral for the entire length of the line, the charging current for

the line, and the charging reactive power.

20ft 20 ft

38 ft




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For ACSR Drake conducter,

from tha tables Xa 0.399 / mile

Tha spacing factor is calculated for spacing equal tha geometric mean distance between tha conductors,

that is, Xd 2.022 10 3

60 ln 24.8 0.389 / mile

Q.NO QUESTION CO mapping Blooms

taxonomy level




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1

Why Ferranti effect occurs in a

transmission line?

CO3 Remembering

2

Advantages of Suspension Insulator

CO4 Understanding

3

Explain Shackle Insulator or Spool

Insulator

CO4 Apply

4

Find the principal stress at a point A in a

uniform rectangular beam 200 mm deep

and 100 mm wide, simply supported at

each end over a span of 3 m and carrying a

uniformly distributed load of 15,000 N/m.

CO4 Understanding

K. G. Reddy College of Engineering &Technology

(Approved by AICTE, Affiliated to JNTUH)

Chilkur (Vil), Moinabad (Mdl), RR District

College Code

QM

Name of the Exam: I Mid Examinations Marks: 10

Year-Sem &

Branch: III-I Duration: 60 Min

Subject: ET Date &

Session

Answer ANY TWO of the following Questions 2X5=10




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1. Why Ferranti effect occurs in a transmission line?

A long transmission line can be considered to compose a considerably high amount of capacitance

and inductance distributed across the entire length of the line. Ferranti Effect occurs when current

drawn by the distributed capacitance of the line itself is greater than the current associated with the

load at the receiving end of the line (during light or no load). This capacitor charging current leads to

voltage drop across the line inductance of the transmission system which is in phase with the sending

end voltages. This voltage drop keeps on increasing additively as we move towards the load end of

the line and subsequently .

Designing consideration of Electrical Insulator

The live conducter attached to tha top of tha pin insulator is at a potential and bottom of tha insulator

fixed to supporting structure of earth potential. Tha insulator has to withstand tha potential stresses

between conducter and earth. Tha shortest distance between conducter and earth, surrounding tha

insulator body, along which electrical discharge may take place through air, is known as flash over

distance.

When insulator is wet, its outer surface becomes almost conducting. Hence tha flash over distance of

insulator is decreased. Tha design of an electrical insulator should be such that tha decrease of flash over

distance is minimum when tha insulator is wet. That is why tha upper most petticoat of a pin insulator

has umbrella type designed so that it can protect, tha rest lower part of tha insulator from rain. Tha upper

surface of top most petticoat is inclined as less as possible to maintain maximum flash over voltage

during raining.

To keep the inner side of tha insulator dry, tha rain sheds are made in order that thase rain sheds

should not disturb tha voltage distribution thay are so designed that thair subsurface at right angle to

tha electromagnetic lines of force.




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Suspension Insulator

In higher voltage, beyond 33KV, it becomes uneconomical to use pin insulator because size, weight of tha insulator

become more. Handling and replacing bigger size single unit insulator are quite difficult task. For overcoming thase

difficulties, suspension insulator was developed.

In suspension insulator numbers of insulators are connected in series to form a string and tha line conducter is

carried by tha bottom most insulator. Each insulator of a suspension string is called disc insulator because of thair

disc like shape.

2.Advantages of Suspension Insulator

Each suspension disc is designed for normal voltage rating 11KV(Higher voltage rating 15KV), so by using

different numbers of discs, a suspension string can be made suitable for any voltage level.

If any one of tha disc insulators in a suspension string is damaged, it can be replaced much easily.

Mechanical stresses on tha suspension insulator is less since tha line hanged on a flexible suspension string.




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As tha current carrying conducters are suspended from supporting structure by suspension string, tha height of tha

conducter position is always less than tha total height of tha supporting structure. Tharefore, tha conducters may be

safe from lightening.




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45

Disadvantages of Suspension Insulator

Suspension insulator string costlier than pin and post type insulator.

Suspension string requires more height of supporting structure than that for pin or post insulator to maintain same

ground clearance of current conducter.

Tha amplitude of free swing of conducters is larger in suspension insulator system, hence, more spacing between

conducters should be provided.




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Strain insulator

When suspension string is used to sustain extraordinary tensile load of conducter it is referred as string insulator.

When thare is a dead end or thare is a sharp corner in transmission line, tha line has to sustain a great tensile load of

conducter or strain. A strain insulator must

have considerable mechanical strength as well as tha necessary electrical insulating properties.

3.Explain Shackle Insulator or Spool Insulator

Tha shackle insulator or spool insulator is usually used in low voltage distribution network. It can be used both in

horizontal and vertical position. Tha use of such insulator has decreased recently after increasing tha using of

underground cable for distribution purpose. Tha tapered hole of tha spool insulator distributes tha load more evenly

and minimizes tha possibility of breakage when heavily loaded. Tha conducter in tha groove of shackle insulator

is fixed with tha help of soft binding wire.




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receiving end voltage tends to get larger than applied voltage leading to tha phenomena called Ferranti effect

in power system. It is illustrated with tha help of a phasor diagram below.

Thus both the capacitance and inductance effect of transmission line are equally responsible for this particular

phenomena to occur, and hence Ferranti effect is negligible in case of a short transmission lines as the

inductance of such a line is practically considered to be nearing zero. In general for a 300 Km line operating

at a frequency of 50 Hz, the no load receiving end voltage has been found to be 5% higher than the sending

end voltage.

Now for analysis of Ferranti effect let us consider the phasor diagrame shown above. Here Vr is considered to

be tha reference phasor, represented by OA.

Thus Vr = Vr (1 + j0) Capacitance current, Ic = jωCVr

Now sending end voltage Vs = Vr + resistive drop + reactive drop.

= Vr + IcR + jIcX

= Vr+ Ic (R + jX)

= Vr+jωcVr (R + jω L) [since X = ωL] Now Vs = Vr -ω2cLVr + j ωcRVr




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This is represented by the phasor OC.

Now in case of a long transmission line, it has been practically observed that tha line resistance is negligibly

small compared to the line reactance, hence we can assume tha length of the phasor Ic R = 0, we can consider

the rise in the voltage is only due to OA – OC = reactive drop in the line




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Now if we consider c0 and L0 are the values of capacitance and inductance per km of tha transmission line,

where l is the length of the line.

Thus capacitive reactance Xc = 1/(ω l c0)

Since, in case of a long transmission line the capacitance is distributed throughout its length, the average

current flowing is,

Ic = 1/2 Vr/Xc = 1/2 Vrω l c0

Now the inductive reactance of the line = ω L0 l

Thus the rise in voltage due to line inductance is given by, IcX = 1/2Vrω l c0 X ω L0 l

Voltage rise = 1/2 Vrω2 l 2 c0L0

From the above equation it is absolutely evident, that the rise in voltage at the receiving end is directly

proportional to the square of the line length, and hence in case of a long transmission line it keeps increasing

with length and even goes beyond the applied sending end voltage at times, leading to the phenomena called

Ferranti effect in power system.

4. Find the principal stress at a point A in a uniform rectangular beam 200 mm deep and 100 mm wide,

simply supported at each end over a span of 3 m and carrying a uniformly distributed load of 15,000 N/m.

Solution: The reaction can be determined by symmetry




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R1 = R2 = 22,500 N

consider any cross-section X-X located at a distance x from the left end.

Hence,

S. F at XX =22,500 – 15,000 x

B.M at XX = 22,500 x – 15,000 x (x/2) = 22,500 x – 15,000 . x2 / 2

Therefore,

S. F at X = 1 m = 7,500 N

B. M at X = 1 m = 15,000 N

Now substituting these values in the principal stress equation,

We get s1 = 11.27 MN/m2




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s2 = - 0.025 MN/m2

Bending Of Composite or Flitched Beams

A composite beam is defined as the one which is constructed from a combination of materials. If such a

beam is formed by rigidly bolting together two timber joists and a reinforcing steel plate, then it is termed

as a flitched beam.

The bending theory is valid when a constant value of Young's modulus applies across a section it cannot be

used directly to solve the composite-beam problems where two different materials, and therefore different

values of E, exists. The method of solution in such a case is to replace one of the materials by an equivalent

section of the other.

Consider, a beam as shown in figure in which a steel plate is held centrally in an appropriate recess/pocket

between two blocks of wood .Here it is convenient to replace the steel by an equivalent area of wood,

retaining the same bending strength. i.e. the moment at any section must be the same in the equivalent

section as in the original section so that the force at any given dy in the equivalent beam must be equal to

that at the strip it replaces.




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Hence to replace a steel strip by an equivalent wooden strip the thickness must be multiplied by the

modular ratio E/E'.

The equivalent section is then one of the same materials throughout and the simple bending theory applies.

The stress in the wooden part of the original beam is found directly and that in the steel found from the

value at the same point in the equivalent material as follows by utilizing the given relations.

Stress in steel = modular ratio x stress in equivalent wood

The above procedure of course is not limited to the two materials treated above but applies well for any

material combination. The wood and steel fletched beam was nearly chosen as a just for the sake of

convenience.

Factors affecting skin effect in transmission lines.

The skin effect in an ac system depends on a number of factors like:-

1) Shape of conductor.

2) Type of material.

3) Diameter of the conductors.

4) Operational frequency.

FERRANTI EFFECT

In general practice we know, that for all electrical systems current flows from the region

of higher potential to the region of lower potential, to compensate for the potential

difference that exists in the system. In all practical cases the sending end voltage is

higher than the receiving end, so current flows from the source or the supply end to the

load. But Sir S.Z. Ferranti, in the year 1890, came up with an astonishing theory about

medium or long distance transmission lines suggesting that in case of light loading or

no load operation of transmission system, the receiving end voltage often increases

beyond the sending end voltage, leading to a phenomena known as Ferranti effect in

power system.




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13. MID exam Objective Question papers

1. In which climate does the chances of occurrence of corona is maximum?

a. Dry

b. Hot summer [ ]

c. Winter

d. Humid

2. What is the effect on corona, if the spacing between the conductors is increased?

a. Corona increases [ ]

b. Corona is absent

c. Corona decreases

d. None of these




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3. Why are the hollow conductors used?

a. Reduce the weight of copper [ ]

b. Improve stability

c. Reduce corona

d. Increase power transmission capacity

4. Which of these given statements is wrong in consideration with bundled conductors?

a. Control of voltage gradient [ ]

b. Reduction in corona loss

c. Reduction in the radio interference

d. Increase in interference with communication lines

5. Why are bundled conductors employed?

[ ]

a. Appearance of the transmission line is improved

b. Mechanical stability of the line is improved

c. Improves current carrying capacity

d. Improves the corona performance of the line

7. The effect of dirt on the surface of the conductor is to _____________ irregularity and thereby

________________ the break down voltage.

a. Decreases, reduces [ ]

b. Increases, increases

c. Increases, reduces

d. Decreases, increases

8.Find the spacing between the conductors a 132 kV 3 phase line with 1.956 cm diameter conductors

is built so that corona takes place, if the line voltage exceeds 210 kV (rms). With go = 30 kV/cm.




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[ ]

a. 1.213 m.

b. 2.315 m.

c. 3.451 m.

d. 4.256 m.

9. Capacitance between the two conductors of a single phase two wire line is 0.5 μ F/km. What is the

value of capacitance of each conductor to neutral?

[ ]

a. 0.5 μ F / km.

b. 1 μ F / km.

c. 0.25 μ F / km.

d. 2.0 μ F / km.

10. What happens in case of capacitance of line to ground, if the effect of earth is taken into account?

a. Capacitance of line to ground decreases. [ ]

b. Capacitance of line to ground increases.

c. The capacitance remains unaltered.

d. The capacitance becomes infinite.

KEY

S.no 1 2 3 4 5 6 7 8 9 10

Ans d c D d d c c b b

1. The ABCD constants of a 3 phase transposed transmission line with linear and passive elements --

---------------------------

2. A line of what length can be classified as a medium transmission line a ----------------------

3. What are the A and D parameters in case of medium transmission line (nominal T method) ---------

---------------------




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4. What is the value of B parameter in case of nominal p method ----------------------

5. What is the value of the C parameter by using a nominal T method for a 3 phase balanced load of

30 MW which is supplied by a 132 kV, 50 Hz and 0.85 pf lagging. The series impedance of a single

conductor is (20 + j52) Ω and the total phase to neutral admittance is 315 * 10-6 siemen. ----------------

---------

6. The transmission lines above what length is termed as the long lines -------------------

7. What is the normal range of angle for the parameter A -------------------------

8. A = D = 0.8 ∠ 1 °, B = 170 ∠ 85 ° Ω , and C = 0.002 ∠ 90.4 ° ℧ the sending end voltage is 400 kV.

What is the receiving end voltage under no load condition -------------------------

9. Transmission efficiency of a transmission line increases with the-----------------

10. The capacitance effect can be neglected in which among the transmission lines --------------

S.no 1 2 3 4 5 6 7 8 9 10

Ans only B

and C

are

equal

A = D

= 1 +

(YZ /

2)

50-

150

km

Z 0.000315 ∠

90

150 km

and

above

0 -

10°

2000 Ω

Increase

in power

factor and

voltage.

Short

transmission

lines.




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14. Assignment topics with materials

UNIT-1

TRANSMISSION LINE PARAMETERS

EXPRESSION FOR INDUCTANCE OF A CONDUCTOR DUE TO EXTERNAL FLUX.

Generally, electric power is transmitted through the transmission line with AC high voltage and current.

High valued alternating current while flowing through the conductor sets up magnetic flux of high strength

with alternating nature. This high valued alternating magnetic flux makes a linkage with other adjacent

conductors parallel to the main conductor. Flux linkage in a conductor happens internally and externally.

Internally flux linkage is due to self-current and externally flux linkage due to external flux. Now the term

inductance is closely related to the flux linkage, denoted by λ. Suppose a coil with N number of turn is

linked by flux Φ due to current I, then, But for transmission

line N = 1. We have to calculate only the value of flux Φ, and hence, we can get the transmission line

inductance.

Calculation of Inductance of Single Conductor

Calculation of Internal Inductance due to Internal Magnetic Flux of a Conductor

Suppose a conductor is carrying current I through its length l, x is the internal

variable radius of the conductor and r is the original radius of the conductor. Now the

cross-sectional area with respect to radius x is πx2 square – unit and current Ix is

flowing through this cross-sectional area. So the value of Ix can be expressed in term

of original conductor current I and cross-sectional area πr2 square – unit

Now consider small thickness dx with the 1m length of the conductor, where Hx is the magnetizing force

due to current Ix around the area πx2.

And magnetic flux density Bx =

μHx, where μ is the permeability of this conductor. Again, µ = µ0µr. If it is considered that the relative

permeability of this conductor µr = 1, then µ = µ0. Hence, here Bx = μ0 Hx.

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dφ for small strip dx is expressed by

Here entire cross-sectional area of the conductor does not enclose the above expressed flux. The ratio of the

cross sectional area inside the circle of radius x to the total cross section of the



And magnetic flux density Bx =

μHx, where μ is the permeability of this conductor. Again, µ = µ0µr. If it is considered that the relative



Here entire cross-sectional area of the conductor does not enclose the above expressed flux. The ratio of the

cross sectional area inside the circle of radius x to the total cross section of the conductor can be thought

about as fractional turn that links the flux. Therefore the flux linkage is

Now, the total flux linkage for the conductor of 1m length with radius r is given by

Hence, the internal inductance is






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2. EXTERNAL INDUCTANCE DUE TO EXTERNAL MAGNETIC FLUX OF A CONDUCTOR

Let us assume, due to skin effect conductor current I is concentrated near the surface of the conductor.

Consider, the distance y is taken from the center of the conductor making the external radius of the

conductor. Hy is

the magnetizing force and By is the magnetic field density at y distance per unit length of the conductor.

Let us assume magnetic flux dφ is

present within the thickness dy from D1 to D2 for 1 m length of the conductor as per the figure.

As the total current I is assumed to flow in the surface of

the conductor, so the flux linkage dλ is equal to dφ.

But

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we have to consider the flux linkage from conductor surface to any external distance, i.e. r to D

3.WHY SKIN EFFECT OCCURS IN TRANSMISSION LINES

Having understood the phenomena of skin effect let us now see why this arises in case of an AC system.

To have a clear understanding of that look into the cross-sectional view of the conductor during the flow of

alternating current given in the diagram below. Let us initially consider the solid conductor to be split up

into some annular filaments spaced infinitely small distance apart, such that each filament carries an

infinitely small fraction of the total current. Like if the total current = I

Let us consider the conductor to be split up into n filament carrying current ‘i’ such that I = n i. Now during

the flow of an alternating current, the current carrying filaments lying on the core has a flux linkage with

the entire conductor cross-section including the filaments of the surface as well as those in the core.

Whereas the flux set up by the outer filaments is restricted only to the surface itself and is unable to link

with the inner filaments.Thus the flux linkage of the conductor increases as we move closer towards the

core and at the same rate increases the inductance as it has a direct proportionality relationship with flux

linkage. As a result, a larger inductive reactance gets induced into the core as compared to the outer

sections of the conductor. The high value of reactance in the inner section results in the current gets

distributed in an un-uniform manner and forcing the bulk of the current to flow through the outer surface or

skin giving rise to the phenomena called skin effect in transmission lines.








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Factors Affecting Skin Effect in Transmission Lines

The skin effect in an AC system depends on some factors like:-

1. Shape of conductor.

2. Type of material.

3. Diameter of the conductors.

4. Operational frequency.

4. EXPRESSION FOR CAPACITANCES OF A SINGLE PHASE TRANSMISSION SYSTEM AND

DISCUSS THE EFFECT OF EARTH ON CAPACITANCE WITH SUITABLE EQUATION.

In calculating the Effect of Earth on Transmission Line Capacitance, the presence of earth was ignored, so

far. The effect of earth on capacitance can be conveniently taken into account by the method of images.

Method of Images

The electric field of transmission line conductors must conform to the presence of the earth below. The

earth for this purpose may be assumed to be a perfectly conducting horizontal sheet of infinite extent which

therefore acts like an equipotential surface.

The electric field of two long, parallel conductors charged +q and -q per unit is such that it has a zero

potential plane midway between the conductors as shown in Fig. 3.8. If a conducting sheet of infinite

dimensions is placed at the zero potential plane, the electric field remains undisturbed. Further, if the

conductor carrying charge -q is now removed, the electric field above the conducting sheet stays intact,

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while that below it vanishes. Using these well known results in reverse, we may equivalently replace the

presence of ground below a charged conductor by a fictitious conductor having equal and opposite charge

and located as far below the surface of ground as the overhead conductor above it—such a fictitious

conductor is the mirror image of the overhead conductor. This method of creating the same electric field as

in the presence of earth is known as the method of images originally suggested by Lord Kelvin.

Capacitance of a Single-Phase Line

Consider a single-phase line shown in Fig. 3.9. It is required to calculate its capacitance taking the presence

of earth into account by the method of images described above. The equation for the voltage drop Vab as

determined by the two charged conductors a and b, and their images a’ and b’ can be written as follows:

Substituting the values of different charges and simplifying, we get

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It immediately follows that


and

It is observed from the above equation that the presence of earth modifies the radius r to r(1 + (D2/4h2))1/2.

For h large compared to D (this is the case normally), the effect of earth on line capacitance is of negligible

order.







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5.CONCEPT OF SELF-GMD AND MUTUAL-GMD

The use of self geometrical mean distance (abbreviated as self-GMD) and mutual geometrical mean

distance (mutual-GMD) simplifies the inductance calculations, particularly relating to multi conductor

arrangements. The symbols used for these are respectively Ds and Dm. We shall briefly discuss these

terms.

( i) Self-GMD (Ds)

In order to have concept of self-GMD (also sometimes called Geometrical mean radius; GMR), consider

the expression for inductance per conductor per metre already derived in Art. Inductance/conductor/m

In this expression, the term 2 × 10-7 × (1/4) is the inductance due to flux within the solid conductor. For

many purposes, it is desirable to eliminate this term by the introduction of a concept called self-GMD or

GMR. If we replace the original solid conductor by an equivalent hollow cylinder with extremely thin

walls, the current is confined to the conductor surface and internal conductor flux linkage would be almost

zero. Consequently, inductance due to internal flux would be zero and the term 2 × 10-7 × (1/4) shall be

eliminated. The radius of this equivalent hollow cylinder must be sufficiently smaller than the physical

radius of the conductor to allow room for enough additional flux to compensate for the absence of internal

flux linkage. It can be proved mathematically that for a solid round conductor of radius r, the self-GMD or

GMR = 0·7788 r. Using self-GMD, the eq. ( i) becomes :

Inductance/conductor/m = 2 × 10-7loge d/ Ds *

Where

Ds = GMR or self-GMD = 0·7788 r




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It may be noted that self-GMD of a conductor depends upon the size and shape of the conductor and is

independent of the spacing between the conductors.

(ii) Mutual-GMD

The mutual-GMD is the geometrical mean of the distances form one conductor to the other and,

therefore, must be between the largest and smallest such distance. In fact, mutual-GMD simply represents

the equivalent geometrical spacing.

(a) The mutual-GMD between two conductors (assuming that spacing between conductors is

large compared to the diameter of each conductor) is equal to the distance between their centres i.e. Dm =

spacing between conductors = d

(b) For a single circuit 3-φ line, the mutual-GMD is equal to the equivalent equilateral spacing i.e.,

( d1 d2 d3 )1/3.

(c) The principle of geometrical mean distances can be most profitably employed to 3-φ double

circuit lines. Consider the conductor arrangement of the double circuit shown in Fig. Suppose the radius of

each conductor is r.




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Self-GMD of conductor = 0·7788 r

Self-GMD of combination aa’ is

It is worthwhile to note that mutual GMD depends only upon the spacing and is substantially independent

of the exact size, shape and orientation of the conductor.

Inductance Formulas in Terms of GMD

The inductance formulas developed in the previous articles can be conveniently expressed in terms of

geometrical mean distances.




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6. EXPRESSION FOR THE CAPACITANCE PER PHASE OF THE 3 Φ DOUBLE CIRCUIT

LINE FLAT VERTICAL SPACING WITH TRANSPOSITION.

Figure 3.5 shows a Capacitance of a Three Phase Line Equilateral Spacing composed of three identical

conductors of radius r placed in equilateral configuration.

Using Eq. (3.2) we can write the expressions for V ab and Vac as

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Adding Eqs. (3.8) and (3 9), we get

Since there are no other charges in the vicinity, the sum of charges on the three conductors is zero.

Thus qb+ qc = – qa which when substituted in Eq. (3.10) yields

With balanced three-phase voltages applied to the line, it follows from the phasor diagram of Fig. 3.6 that

Substituting for (Vab + Vac) from Eq. (3.12) in Eq. (3.11), we get qa D

The capacitance of line to neutral immediately follows as

For air medium (kr = 1),

The line charging current of phase a is

Ia (line charging) = jωCnVan

http://www.eeeonline.org/

http://www.eeeguide.com/wp-content/uploads/2016/11/Capacitance-of-a-Three-Phase-Line-1.jpg










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UNIT -2

PERFORMANCE OF SHORT AND MEDIUM LENGTH





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TRANSMISSION LINES

1. ABCD CONSTANTS IN A TRANSMISSION LINE

A, B, C and D are the constants also known as thetransmission parameters or chain parameters.

These parameters are used for the analysis of an electrical network. It is also used for determining the

performance of input, output voltage and current of the transmission network.

Vs = sending end voltage

Is = sending end current

Vr = receiving end voltage

Ir = receiving end current

Open circuit




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ABCD parameters for short circuit

For the short circuit, the voltage remains zero

at the receiving end.

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If we put Vr = 0 in the equation, we get the value of B which is the ratio of

sending end voltage to the receiving end currents. Its unit is ohms.

Similarly, if we put Vr= 0 in current equations,

we get the value of D, which is the ratio of the sending current to the

receiving current. It is the dimensionless constant.

Relation between ABCD parameters

For determining the relation between various types of network, like passive or bilateral network reciprocity

theorem is applied. The voltage V is applied to the sending end, and the receiving end is kept short circuit,

so the voltage becomes zero.

Since, under short circuit the receiving end voltage is zero, the voltage and current equations become

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Similarly, the voltage is applied at the receiving end, and the input voltage remains

zero. Thus, the direction of the current in the network changes, which is shown in the diagram below

The sending end voltage

becomes zero. The current flows through the receiving end is given by the equation

and

Consider, the network is passive, i.e. it contains only passive components in

the circuit like inductance, resistance, etc. So the current remains same Is = Ir.

Combining the above equations give,

dividing the above equation from -V/B we get,

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This relation helps in determining the fourth parameters if we know any three

parameters.

For a symmetrical network, the input and output terminal may be interchanged without affecting the

network behaviour.

If the network is supplied from input terminals and an output terminal is short circuit,

then the impedance becomes

and if the supply is from the output terminal and an input terminal is a short circuit then the impedance

becomes

in the symmetrical network, the impedance remains the same

Nominal Pi Model of a Medium Transmission Line

In the nominal pi model of a medium transmission line, the series impedance of the line is concentrated at

the centre and half of each capacitance is placed at the centre of the line. The nominal Pi model of the line

is shown in the diagram below.

https://circuitglobe.com/medium-transmission-line.html


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In this circuit,

By Ohm’s law

By KCL at node a,

Voltage at the sending end

By ohm’s law

https://circuitglobe.com/wp-content/uploads/2016/05/pi-model-medium-line.jpg

https://circuitglobe.com/wp-content/uploads/2016/05/pi-model-equ-1.jpg







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Sending-end current is found by

applying KCL at node c

or

Equations can be written in matrix form as

Also,

Hence, the ABCD constants for nominal pi-circuit model of a medium line are

Phasor diagram of nominal pi model

The phasor diagram of a nominal pi-circuit is shown in the figure below.

https://circuitglobe.com/wp-content/uploads/2016/05/pi-model-equ-5-1.jpg

https://circuitglobe.com/wp-content/uploads/2016/05/equation-88.jpg


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It is also drawn for a lagging power factor of the load.

In the phasor diagram the quantities shown are as follows;

OA = Vr – receiving end voltage. It is taken as reference phasor.

OB = Ir – load current lagging Vr by an angle ∅r.

BE = Iab – current in receiving-end capacitance. It leads Vr by 90°.

The line current I is the phasor sum of Ir and Iab. It is shown by OE in the diagram.

AC = IR – voltage drop in the resistance of the line. It is parallel to I.

CD = IX -inductive voltage drop in the line. It is perpendicular to I.

AD = IZ – voltage drop in the line impedance.

OD = Vs – sending–end voltage to neutral. It is phasor sum of Vr and IZ.

The current taken by the capacitance at the sending end is Icd. It leads the sending–end voltage Vs by 90

OF = Is – the sending–end current. It is the phasor sum of I and Icd.

∅s – phase angle between Vs and Is at the sending end, and cos∅s will give the sending-end power factor

2. EXPRESSION FOR NOMINAL PI MODEL OF MEDIUM TRANSMISSION LINE


In the nominal pi model of a medium transmission line, the series impedance of the line is concentrated at

the centre and half of each capacitance is placed at the centre of the line. The nominal Pi model of the line

is shown in the diagram below.


https://circuitglobe.com/wp-content/uploads/2016/05/phasor-diagram.jpg




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In this circuit,

By Ohm’s law

By KCL at node a,


By ohm’s law









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or


Also,













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It is also drawn for a lagging power factor of

the load. In the phasor diagram the quantities shown are as follows;











∅s – phase angle between Vs and Is at the sending end, and cos∅s will give the sending-end power factor

3. DERIVE EXPRESSION FOR SURGE IMPEDANCE.

Surge Impedance Loading is a very essential parameter when it comes to the study of power systems as it is

used in the prediction of maximum loading capacity of transmission lines.

However before understanding SIL, we first need to have an idea of what is Surge Impedance (Zs). It can

be defined in two ways one a simpler one and other a bit rigorous. Method 1 It is a well known fact that a

https://www.electrical4u.com/transmission-line-in-power-system/





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long transmission lines (> 250 km) have distributed inductance and capacitance as its inherent property.

When the line is charged, the capacitance component feeds reactive power to the line while the inductance

component absorbs the reactive power. Now if we take the balance of the two reactive powers we arrive at

the following equation

Capacitive VAR = Inductive VAR Where, V = Phase voltage I = Line Current Xc =

Capacitive reactance per phase XL = Inductive reactance per phase Upon simplifying

Where, f = Frequency of the system L = Inductance per unit length of

the line l = Length of the line Hence we get,

This quantity having the dimensions of resistance is the Surge Impedance. It can be considered as a purely

resistive load which when connected at the receiving end of the line, the reactive power generated by

capacitive reactance will be completely absorbed by inductive reactance of the line. It is nothing but the

Characteristic Impedance (Zc) of a lossless line.

Method 2 From the rigorous solution of a long transmission line we get the following equation for voltage

and current at any point on the line at a distance x from the receiving end

Where, Vx and Ix = Voltage and Current at point x

VR and IR = Voltage and Current at receiving end Zc = Characteristic Impedance δ = Propagation Constant

Z = Series impedance per unit length per phase Y = Shunt admittance per unit

length per phase Putting the value of δ in above equation of voltage we get

Where,

We observe that the instantaneous voltage consists of two terms each of which is a function of time and

distance. Thus they represent two travelling waves. The first one is the positive exponential part

https://www.electrical4u.com/long-transmission-line/










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representing a wave travelling towards receiving end and is hence called the incident wave. While the other

part with negative exponential represents the reflected wave. At any point along the line, the voltage is the

sum of both the waves. The same is true for current waves also. Now, if suppose the load impedance (ZL) is

chosen such that ZL = Zc, and we know Thus and

hence the reflected wave vanishes. Such a line is termed as infinite line. It appears to the source that the

line has no end because it receives no reflected wave. Hence, such an impedance which renders the line as

infinite line is known as surge impedance.It has a value of about 400 ohms and phase angle varying from 0

to –15 degree for overhead lines and around 40 ohms for underground cables.

The term surge impedance is however used in connection with surges on the transmission line which may

be due to lightning or switching, where the line losses can be neglected such that

Now that we have understood Surge Impedance, we can easily define

Surge Impedance Loading. SIL is defined as the power delivered by a line to a purely resistive load equal

in value to the surge impedance of that line. Hence we can write

The unit of SIL is Watt or MW. When

the line is terminated by surge impedance the receiving end voltage is equal to the sending end voltage and

this case is called flat voltage profile. The following figure shows the voltage profile for different loading





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cases. It should also be noted

that surge impedance and hence SIL is independent of the length of the line. The value of surge impedance

will be the same at all the points on the line and hence the voltage. In case of a Compensated Line, the

value of surge impedance will be modified accordingly as

Where, Kse = % of series capacitive compensation by Cse KCsh = % of Shunt capacitive

compensation by Csh Klsh = % of shunt inductive compensation by Lsh

The equation for SIL will now use the modified Zs




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HOW VOLTAGES AND CURRENTS ARE EVALUATED IN LONG

TRANSMISSION LINES.

A power transmission line with its effective length of around 250 Kms or above is referred to as a long

transmission line. The line constants are uniformly distributed over the entire length of line. Calculations

related to circuit parameters (ABCD parameters) of such a power transmission is not that simple, as was the

case for a short transmission line or medium transmission line.

The reason being that, the effective circuit length in this case is much higher than what it was for the

former models (long and medium line) and, thus ruling out the approximations considered there like.

1. Ignoring the shunt admittance of the network, like in a small transmission line model.

2. Considering the circuit impedance and admittance to be lumped and concentrated at a point as was the

case for the medium line model.

Rather, for all practical reasons we should consider the circuit impedance and admittance to be distributed

over the entire circuit length as shown in the figure below. The calculations of circuit parameters for this

reason are going to be slightly more rigorous as we will see here. For accurate modeling to determine

circuit parameters let us consider the circuit of the long transmission line as shown in the diagram below.





https://www.electrical4u.com/medium-transmission-line/






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Here a line of length

l > 250km is supplied with a sending end voltage and current of VS and IS respectively, where as the VR

and IR are the values of voltage and current obtained from the receiving end. Lets us now consider an

element of infinitely small length Δx at a distance x from the receiving end as shown in the figure where. V

= value of voltage just before entering the element Δx. I = value of current just before entering the element

Δx. V+ΔV = voltage leaving the element Δx. I+ΔI = current leaving the element Δx. ΔV = voltage drop

across element Δx. zΔx = series impedence of element Δx yΔx = shunt admittance of element Δx Where, Z

= z l and Y = y l are the values of total impedance and admittance of the long transmission line.

Therefore, the voltage drop across the infinitely small element Δx is given by

Now to determine the current ΔI, we apply KCL to node A.

Since the term ΔV yΔx is the product of 2 infinitely small

values, we can ignore it for the sake of easier calculation. Therefore, we can write

Now derivating both sides of eq (1) w.r.t x, Now substituting from equation (2)

The solution of the above second order differential







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equation is given by. Derivating equation (4) w.r.to x.

Now comparing equation (1) with equation (5)

Now to go further let us define the characteristic impedance Zc and propagation constant δ of a long

transmission line as Then the voltage and current equation can be

expressed in terms of characteristic impedance and propagation constant as

Now at x=0, V= VR and I= Ir. Substituting these conditions to

equation (7) and (8) respectively. Solving equation (9) and (10), We get

values of A1 and A2 as, Now applying another extreme

condition at x = l, we have V = VS and I = IS. Now to determine VS and IS we substitute x by l and put the

values of A1 and A2 in equation (7) and (8) we get

By trigonometric and exponential

operators we know Therefore, equation (11) and

(12) can be re-written as Thus




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comparing with the general circuit parameters equation, we get the ABCD parameters of a long

transmission line as,

UNIT-III

POWER SYSTEM TRANSIENTS

1.FACTORS AFFECTING THE CORONA.

Factors affecting corona:

The following are the factors affecting the corona;

1. Effect of supply voltage – If the supply voltage is high corona loss is higher in the lines. In low-voltage

transmission lines, the corona is negligible, due to the insufficient electric field to maintain ionization.

2. The condition of conductor surface – If the conductor is smooth, the electric field will be more uniform

as compared to the rough surface. The roughness of conductor is caused by the deposition of dirt, dust and

by scratching, etc. Thus, rough line decreases the corona loss in the transmission lines.

3. Air Density Factor – The corona loss in inversely proportional to air density factor, i.e., corona loss,

increase with the decrease in density of air. Transmission lines passing through a hilly area may have

higher corona loss than that of similar transmission lines in the plains because in a hilly area the density of

air is low.




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4. Effect of system voltage – Electric field intensity in the space around the conductors depends on the

potential difference between the conductors. If the potential difference is high, electric field intensity is

also very high, and hence corona is also high. Corona loss, increase with the increase in the voltage.

5. The spacing between conductors – If the distance between two conductors is much more as compared to

the diameter of the conductor than the corona loss occurs in the conductor. If the distance between them is

extended beyond certain limits, the dielectric medium between them get decreases and hence the corona

loss also reduces.

2. DISADVANTAGES OF CORONA DISCHARGE:

The undesirable effects of the corona are:

1. The glow appear across the conductor which shows the power loss occur on it.

2. The audio noise occurs because of the corona effect which causes the power loss on the conductor.

3. The vibration of conductor occurs because of corona effect.

4. The corona effect generates the ozone because of which the conductor becomes corrosive.

5. The corona effect produces the non-sinusoidal signal thus the non-sinusoidal voltage drops occur in the

line.

6. The corona power loss reduces the efficiency of the line.

7. The radio and TV interference occurs on the line because of corona effect.

3. TRAVE LLING WAVE ON TRANSMISSION LINES

Travelling Wave on Transmission Line

The transmission line is a distributed parameter circuit and its support the wave of voltage and current. A

circuit with distributed parameter has a finite velocity of electromagnetic field propagation. The switching

and lightning operation on such types of circuit do not occur simultaneously at all points of the circuit but

spread out in the form of travelling waves and surges.

When a transmission line is suddenly connected to a voltage source by the closing of a switch the whole of

the line in not energized at once, i.e., the voltage does not appear instantaneously at the other end. This is

due to the presence of distributed constants (inductance and capacitance in a loss-free line).

Considered a long transmission line having a distributed parameter inductance (L) and capacitance (C). The

long transmission line is divided into small section shown in the figure below.The S is the switch used for

closing or opening the surges for switching operation. When the switch is closed the L1 inductance act as

an open circuit and C1 act as a short circuit. At the same instant, the voltage at the next section cannot be

charged because the voltage across the capacitor C1 is zero.




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So unless the capacitor C1 is charged to some value the charging of the capacitor C2 through L2 is not

possible which will obviously take some time. The same argument applies to the third section, fourth

section, and so on. The voltage at the section builds up gradually. This gradual build up of voltage over the

transmission conductor can be regarded as though a voltage wave is travelling from one end to the other

end and the gradual charging of the voltage is due to associate current wave.

The current wave, which is accompanied by voltage wave steps up a magnetic field in the surrounding

space. At junctions and terminations, these waves undergo reflection and refraction. The network has a

large line and junction the number of travelling waves initiated by a single incident wave and will increase

at a considerable rate as the wave split and multiple reflections occurs. The total energy of the resultant

wave cannot exceed the energy of the incident wave

4.CORONA EFFECT

Definition: The phenomenon of ionization of surrounding air around the conductor due to which luminous

glow with hissing noise is rise is known as the corona effect.

Air acts as a dielectric medium between the transmission lines. In other words, it is an insulator between

the current carrying conductors. If the voltage induces between the conductor is of alternating nature then

the charging current flows between the conductors. And this charging conductor increases the voltage of

the transmission line.

The electric field intensity also increases because of the charging current.

If the intensity of the electric field is less than 30kV, the current induces between the conductor is

neglected. But if the voltage rise beyond the 30kv then the air between the conductors becomes charge and

they start conducting. The sparking occurs between the conductors till the complete breakdown of the

insulation properties of conductors takes place.

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Corona effect mostly occurs at the sharp point of insulators.

Contents: Corona effect

1. Corona Formation

2. Factors affecting corona

3. Disadvantages of corona discharge

4. Minimizing corona

5. Important points

Corona Formation:

Air is not a perfect insulator, and even under normal conditions, the air contains many free electrons and

ions. When an electric field intensity establishes between the conductors, these ions and free electrons

experience forced upon them. Due to this effect, the ions and free electrons get accelerated and moved in

the opposite direction.

The charged particles during their motion collide with one another and also with the very slow moving

uncharged molecules. Thus, the number of charged particles goes on increasing rapidly. This increase the

conduction of air between the conductors and a breakdown occurs. Thus, the arc establishes between the

conductors.

5.CHARACTERISTIC IMPEDANCE

The characteristic impedance or surge impedance (usually written Z0) of a uniform transmission line is

the ratio of the amplitudes of voltage and current of a single wave propagating along the line; that is, a

wave travelling in one direction in the absence of reflections in the other direction. Characteristic

https://circuitglobe.com/corona-effect.html#CoronaFormation

https://circuitglobe.com/corona-effect.html#Factorsaffectingcorona

https://circuitglobe.com/corona-effect.html#Disadvantagesofcoronadischarge

https://circuitglobe.com/corona-effect.html#Minimizingcorona

https://circuitglobe.com/corona-effect.html#Importantpoints

https://en.wikipedia.org/wiki/Transmission_line

https://en.wikipedia.org/wiki/Reflections_of_signals_on_conducting_lines

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impedance is determined by the geometry and materials of the transmission line and, for a uniform line, is

not dependent on its length. The SI unit of characteristic impedance is the ohm.

The characteristic impedance of a lossless transmission line is purely real, with no reactive component.

Energy supplied by a source at one end of such a line is transmitted through the line without being

dissipated in the line itself. A transmission line of finite length (lossless or lossy) that is terminated at one

end with an impedance equal to the characteristic impedance appears to the source like an infinitely long

transmission line and produces no reflections.

The characteristic impedance of an infinite transmission line at a given angular frequency is the

ratio of the voltage and current of a pure sinusoidal wave of the same frequency travelling along the line.

This definition extends to DC by letting tend to 0, and subsists for finite transmission lines until the

wave reaches the end of the line. In this case, there will be in general a reflected wave which travels back

along the line in the opposite direction. When this wave reaches the source, it adds to the transmitted wave

and the ratio of the voltage and current at the input to the line will no longer be the characteristic

impedance. This new ratio is called the input impedance. The input impedance of an infinite line is equal to

the characteristic impedance since the transmitted wave is never reflected back from the end. It can be

shown that an equivalent definition is: the characteristic impedance of a line is that impedance which, when

terminating an arbitrary length of line at its output, produces an input impedance of equal value. This is so

because there is no reflection on a line terminated in its own characteristic impedance.

Applying the transmission line model based on the telegrapher's equations, the general expression for the

characteristic impedance of a transmission line is:

where

is the resistance per unit length, considering the two conductors to be in series,

is the inductance per unit length,

is the conductance of the dielectric per unit length,

is the capacitance per unit length,

is the imaginary unit, and

https://en.wikipedia.org/wiki/SI

https://en.wikipedia.org/wiki/Electrical_impedance

https://en.wikipedia.org/wiki/Ohm_(unit)

https://en.wikipedia.org/wiki/Electrical_reactance

https://en.wikipedia.org/wiki/Input_impedance

https://en.wikipedia.org/wiki/Telegrapher%27s_equations

https://en.wikipedia.org/wiki/Electrical_resistance

https://en.wikipedia.org/wiki/In_series

https://en.wikipedia.org/wiki/Inductance

https://en.wikipedia.org/wiki/Electrical_conductance

https://en.wikipedia.org/wiki/Capacitance

https://en.wikipedia.org/wiki/Imaginary_unit




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is the angular frequency.

Although an infinite line is assumed, since all quantities are per unit length, the characteristic impedance is

independent of the length of the transmission line.

The voltage and current phasors on the line are related by the characteristic impedance as:

where the superscripts and represent forward- and backward-traveling waves, respectively. A

surge of energy on a finite transmission line will see an impedance of Z0 prior to any reflections arriving,

hence surge impedance is an alternative name for characteristic impedance.

Lossless line

The analysis of lossless lines provides an accurate approximation for real transmission lines that simplifies

the mathematics considered in modeling transmission lines. A lossless line is defined as a transmission line

that has no line resistance and no dielectric loss. This would imply that the conductors act like perfect

conductors and the dielectric acts like a perfect dielectric. For a lossless line, R and G are both zero, so the

equation for characteristic impedance derived above reduces to:

In particular, does not depend any more upon the frequency. The above expression is wholly real,

since the imaginary term j has canceled out, implying that Z0 is purely resistive. For a lossless line

terminated in Z0, there is no loss of current across the line, and so the voltage remains the same along the

line. The lossless line model is a useful approximation for many practical cases, such as low-loss

transmission lines and transmission lines with high frequency. For both of these cases, R and G are much

smaller than ωL and ωC, respectively, and can thus be ignored.

https://en.wikipedia.org/wiki/Angular_frequency

https://en.wikipedia.org/wiki/Phasor_(electronics)

https://en.wikipedia.org/wiki/Dielectric_loss




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UNIT-IV

OVERHEAD LINE INSULATORS

1.THE SAG OF AN OVERHEAD LINE CAN BE CALCULATED IN CASE OF SUPPORTS AT

DIFFERENT LEVELS.

Sag in overhead Transmission line conductor refers to the difference in level between the point of support

and the lowest point on the conductor.

As shown in the figure above, a Transmission line is supported at two points A and B of two different

Transmission Towers. It is assumed that points A and B are at the same level from the ground. Therefore as

per our definition of Sag, difference in level of point A or B and lowest point O represents the Sag.

Sag in Transmission line is very important. While erecting an overhead Transmission Line, it should be

taken care that conductors are under safe tension. If the conductors are too much stretched between two

points of different Towers to save conductor material, then it may happen so that the tension is conductor

reaches unsafe value which will result conductor to break.




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Therefore, in order to have safe tension in the conductor, they are not fully stretched rather a sufficient dip

or Sag is provided. The dip or Sag in Transmission line is so provided to maintain tension in the conductor

within the safe value in case of variation in tension in the conductor because of seasonal variation. Some

very basic but important aspects regarding Sag are as follows:

1) As shown in the figure above, if the point of support of conductor is at same level from the ground, the

shape of Sag is Catenary. Now we consider a case where the point of support of conductor are at same level

but the Sag is very less when compared with the span of conductor. Here span means the horizontal

distance between the points of support. In such case, the Sag-span curve is parabolic in nature.

2) The tension at any point on the conductor acts tangentially as shown in figure above. Thus the tension at

the lowest point of the conductor acts horizontally while at any other point we need to resolve the

tangential tension into vertical and horizontal component for analysis purpose. The horizontal component

of tension remains constant throughout the span of conductor.

Calculation of Sag:

As discussed earlier in this post, enough Sag shall be provided in overhead transmission line to keep the

tension within the safe limit. The tension is generally decided by many factors like wind speed, ice loading,

temperature variations etc. Normally the tension in conductor is kept one half of the ultimate tensile

strength of the conductor and therefore safety factor for the conductor is 2.

Now, we will calculate the Sag in an overhead transmission line for two cases.

Case1: When the conductor supports are at equal level.

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Let us consider an overhead line supported at two different towers which are at same level from ground.

The point of support are A and B as shown in figure below. O in the figure shows the lowest point on the

conductor. This lowest point O lies in between the two towers i.e. point O bisects the span equally.

Let,

L = Horizontal distance between the towers i.e. Span

W = Weight per unit length of conductor

T = Tension in the conductor

Let us take any point P on the conductor. Assuming O as origin, the coordinate of point P will be (x,y).

Therefore, weight of section OP = Wx acting at distance of x/2 from origin O.

As this section OP is in equilibrium, hence net torque w.r.t point P shall be zero.

Torque due to Tension T = Torque due to weight Wx

Ty = Wx(x/2)

Therefore, y = Wx2 / 2T ……………………….(1)

For getting Sag, put x = L/2 in equation (1)

Sag = WL2/8T

2.THE SAG OF AN OVERHEAD LINE CAN BE CALCULATED IN CASE OF SUPPORTS AT

SAME LEVEL

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Case2: When the conductor supports are at unequal level.

In hilly area, the supports for overhead transmission line conductor do not remain at the same level. Figure

below shows a conductor supported between two points A and B which are at different level. The lowest

point on the conductor is O.

Let,


H = Difference in level between the two supports


X1 = Horizontal distance of point O from support A

X2 = Horizontal distance of point O from support B


From equation (1),

Sag S1 = WX12/2T

and Sag S2 = WX22/2T

Now,

S1 – S2 = (W/2T)[ X12 – X2

2]

= (W/2T)(X1 – X2)( X1 + X2)

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But X1 + X2 = L …………………….(2)

So,

S1 – S2 = (WL/2T)(X1 – X2)

X1 – X2 = 2(S1 – S2)T / WL

X1 – X2 = 2HT / WL (As S1 – S2 = H)

X1 – X2 = 2HT / WL ………………..(3)

Solving equation (2) and (3) we get,

X1 = L/2 – TH/WL

X2 = L/2 + TH/WL

By putting the value of X1 and X2 in Sag equation, we can easily find the value of S1 and S2.

The above equations for Sag are only valid in ideal situation. Ideal situation refers to a condition when no

wind is flowing and there is no any effect of ice loading. But in actual practise, there always exists a wind

pressure on the conductor and as far as the ice loading is concerned, it is mostly observed in cold countries.

In a country like India, ice loading on transmission line is rarely observed.

3.THE EFFECT OF WIND AND ICE LOADING ARE TAKEN INTO ACCOUNT WHILE

DETERMINING THE SGANS STRESS OF AN OVERHEAD LINE CONDUCTOR.

Effect of Wind and Ice Loading on Sag:

Coating of ice on conductor (it is assumed that ice coating is uniformly distributed on the surface of

conductor) increases the weight of the conductor which acts in vertically downward direction. But the wind

exerts a pressure on the conductor surface which is considered horizontal for the sake of calculation.

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As shown in figure above, net weight acting vertically downward is sum of weight of ice and weight of

conductor.

Therefore,

Here,

W = Weight of conductor per unit length

Wi = Weight of ice per unit length

Ww = Wind force per unit length

= Wind Pressure x Area

= Wind Pressure x (2d+t)x1

Note the way of calculation of Area of conductor. What I did, I just stretched the conductor along the

diameter to make a rectangle as shown in figure below.

Thus from equation (1),

Sag = WtL2/2T

And the angle made by conductor from vertical = tanƟ

= Ww / (W+Wi)

4. EXPRESSIONS FOR SAG AND TENSION IN A POWER CONDUCTOR STRUNG BETWEEN

TO SUPPORTS AT EQUAL HEIGHTS TAKING INTO ACCOUNT THE WIND AND ICE

LOADING ALSO

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STRING EFFICIENCY:

The total voltage applied across the string of suspension type insulators is not equally distributed across all

discs. In this distribution of voltage, disc nearest to the conductor will be at higher potential than the other

discs. This unequal potential distribution is undesirable and is usually expressed in terms of string

efficiency

What Is String Efficiency?

"The ratio of voltage across the whole string to the product of number of discs and the voltage across the

disc nearest to the conductor is known as string efficiency."

i.e.,String efficiency =Voltage across the string/(n ×Voltage across disc nearest to conductor)

Where n = number of discs in the string.

String efficiency is an important factor is transmission line designing. Since it decides the potential

distribution along the string. To get uniform distribution string efficiency should be high.Thus 100% string

efficiency is an ideal case for which the voltage across each disc will be exactly the same.that gives

easy calculations to no.of discs to be added. but it is impossible to achieve 100% string efficiency,yet

efforts should be made to improve it as close to this value as possible.

Mathematical expression for String Efficiency:-

Above Fig. shows the equivalent circuit for a 3-disc string. Let us suppose that self capacitance of each disc

is C. Let us further assume that shunt capacitance C1 is some fraction K of self capacitance i.e., C1 = KC.

Starting from the cross arm or tower, the voltage across each unit is V1,V2 and V3 respectively as shown.

Applying Kirchhoff’s current law to node A, we get,

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I2 = I1 + i1

or V2ω C* = V1ω C + V1ω C1

or V2ω C = V1ω C + V1ω K C

∴ V2 = V1 (1 + K) ...(i)

Applying Kirchhoff’s current law to node B, we get

I3 = I2 + i2

or V3 ω C = V2ω C + (V1 + V2) ω C1

or V3 ω C = V2ω C + (V1 + V2) ω K C

or V3 = V2 + (V1 + V2)K

= KV1 + V2 (1 + K)

= KV1 + V1 (1 + K)2 since [ V2 = V1 (1 + K)]

= V1 [K + (1 + K)²]

∴ V3 = V1[1 + 3K + K²] ...(ii)

Voltage between conductor and earth (i.e., tower) is

V = V1 + V2 + V3

= V1 + V1(1 + K) + V1 (1 + 3K + K²)

= V1 (3 + 4K + K²)

∴ V = V1(1 + K) (3 + K) ...(iii)

From expressions (i), (ii) and (iii), we get,

V1/1=V2/(1+K)=V3/(1 + 3K + K²)=V/(1+K)(3=K)

∴Voltage across top unit, V1 = V/(1 + K)(3 + K)

Voltage across second unit from top, V2 = V1 (1 + K)

Voltage across third unit from top, V3 = V1 (1 + 3K + K²)

%age String efficiency =Voltage across the string*100/(n ×Voltage across disc nearest to conductor)

=V*100/3V3

5.USE HIGH VOLTAGE TO TRANSMIT ELECTRICAL POWER

The transport of large amounts of electrical power over long distances is done with high-voltage

transmission lines, and the question is: why high voltage? It certainly has a negative safety aspect, since a

low voltage line wouldn't be harmful (you can put your hands on a 12 V car battery, for example, you won't

even feel it; but make sure you don't put metal across the terminals, you'll get a huge current and a nasty

spark!). Electric energy is transported across the countryside with high-voltage lines because the line

losses are much smaller than with low-voltage lines.

All wires currently used have some resistance (the development of high-temperature superconductors will

probably change this some day). Let's call the total resistance of the transmission line leading from a power

station to your local substation R. Let's also say the local community demands a power P=IV from that

substation. This means the current drawn by the substation is I=P/V and the higher the transmission line




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voltage, the smaller the current. The line loss is given by Ploss=I²R, or, substituting for I,

Ploss = P²R/V²

Since P is fixed by community demand, and R is as small as you can make it (using big fat copper cable,

for example), line loss decreases strongly with increasing voltage. The reason is simply that you want the

smallest amount of current that you can use to deliver the power P. Another important note: the loss

fraction

Ploss/P = PR/V²

increases with increasing load P: power transmission is less efficient at times of higher demand. Again, this

is because power is proportional to current but line loss is proportional to current squared. Line loss can be

quite large over long distances, up to 30% or so. By the way, line loss power goes into heating the

transmission line cable which, per meter length, isn't very much heat.




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UNIT-V

UNDERGROUND CABLES

1. TYPES OF INSULATOR USED AS OVERHEAD INSULATOR LIKEWISE

1. Pin Insulator

2. Suspension Insulator

3. Strain Insulator

In addition to that thare are othar two types of electrical insulator available mainly for

low voltage application, e.i. Stay Insulator and Shackle Insulator.

Pin Insulator

Pin Insulator is earliest developed overhead insulator, but still popularly used in power

network up to 33KV system. Pin type insulator can be one part, two parts or three

parts type, depending upon application voltage. In 11KV system we generally use one

part type insulator where whole pin insulator is one piece of properly shaped porcelain

or glass. As tha leakage path of insulator is through its surface, it is desirable to

increase tha vertical length of tha insulator surface area for lengthaning leakage path. In

order to obtain lengthy leakage path, one, tow or more rain sheds or petticoats are

provided on tha insulator body. In addition to that rain shed or petticoats on an insulator

serve anothar purpose. Thase rain sheds or petticoats are so designed, that during

raining tha outer surface of tha rain shed becomes wet but tha inner surface remains dry

and non-conductive. So thare will be discontinuations of conducting path through tha

wet pin insulator surface.




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In higher voltage like 33KV and 66KV manufacturing of one part porcelain pin

insulator becomes difficult. Because in higher voltage, tha thickness of tha insulator

become more and a quite thick single piece porcelain insulator can not manufactured

practically. In this case we use multiple part pin insulator, where a number of properly

designed porcelain shells are fixed togethar by Portland cement to form one complete

insulator unit. For 33KV tow parts and for 66KV three parts pin insulator are generally

used.

Designing consideration of Electrical Insulator

Tha live conducter attached to tha top of tha pin insulator is at a potential and bottom

of tha insulator fixed to supporting structure of earth potential. Tha insulator has to

withstand tha potential stresses between conducter and earth. Tha shortest distance

between conducter and earth, surrounding tha insulator body, along which electrical

discharge may take place through air, is known as flash over distance.

1. When insulator is wet, its outer surface becomes almost conducting. Hence tha flash

over distance of insulator is decreased. Tha design of an electrical insulator should be

such that tha decrease of flash over distance is minimum when tha insulator is wet.

That is why tha upper most petticoat of a pin insulator has umbrella type designed so

that it can protect, tha rest lower part of tha insulator from rain. Tha upper surface of

top most petticoat is inclined as less as possible to maintain maximum flash over

voltage during raining.

2. To keep tha inner side of tha insulator dry, tha rain sheds are made in order that

thase rain sheds should not disturb tha voltage distribution thay are so designed that

thair subsurface at right angle to tha electromagnetic lines of force.




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44


In higher voltage, beyond 33KV, it becomes uneconomical to use pin insulator

because size, weight of tha insulator become more. Handling and replacing bigger

size single unit insulator are quite difficult task. For overcoming thase difficulties,

suspension insulator was developed.

In suspension insulator numbers of insulators are connected in series to form a

string and tha line conducter is carried by tha bottom most insulator. Each insulator

of a suspension string is called disc insulator because of thair disc like shape.


1. Each suspension disc is designed for normal voltage rating 11KV(Higher voltage

rating 15KV), so by using different numbers of discs, a suspension string can be

made suitable for any voltage level.

2. If any one of tha disc insulators in a suspension string is damaged, it can be




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replaced much easily.

3. Mechanical stresses on tha suspension insulator is less since tha line hanged on a

flexible suspension string.

4. As tha current carrying conducters are suspended from supporting structure by

suspension string, tha height of tha conducter position is always less than tha total

height of tha supporting structure. Tharefore, tha conducters may be safe from

lightening.




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1. Suspension insulator string costlier than pin and post type insulator.

2. Suspension string requires more height of supporting structure than that for pin or post

insulator to maintain same ground clearance of current conducter.

3. Tha amplitude of free swing of conducters is larger in suspension insulator system, hence,

more spacing between conducters should be provided.

Strain insulator

When suspension string is used to sustain extraordinary tensile load of conducter it is referred

as string insulator. When thare is a dead end or thare is a sharp corner in transmission line,




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tha line has to sustain a great tensile load of conducter or strain. A strain insulator must

have considerable mechanical strength as well as tha necessary electrical insulating

properties.

Shackle Insulator or Spool Insulator

Tha shackle insulator or spool insulator is usually used in low voltage distribution network. It

can be used both in horizontal and vertical position. Tha use of such insulator has decreased

recently after increasing tha using of underground cable for distribution purpose. Tha

tapered hole of tha spool insulator distributes tha load more evenly and minimizes tha

possibility of breakage when heavily loaded. Tha conducter in tha groove of shackle

insulator is fixed with tha help of soft binding wire.




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16. Unit wise-Question bank

UNIT-1 TRANSMISSION LINE PARAMETERS

2 MARKS QUETION AND ANSWERS

1. What is Transmission Line?

Transmission line is the long conductor with special design (bundled) to carry bulk amount of generated

power at very high voltage from one station to another as per variation of the voltage level.

2. What is the Transmission Efficiency?

Transmission efficiency is defined as the ration of receiving end power PR to the sending end power PS

and it is expressed in percentage value. cosθs is the

sending end power factor. cosθR is the receiving end power factor. Vs is the sending end voltage per

phase. VR is the receiving end voltage per phase.

3.What is Transmission Line Voltage Regulation

Ans) Voltage regulation of transmission line is defined as the ratio of difference between sending and

receiving end voltage to receiving end voltage of a transmission line between conditions of no load and

full load. It is also expressed in percentage. Where, Vs is the sending

end voltage per phase and VR is the receiving end voltage per phase.

XL is the reactance per phase. R is the

resistance per phase. cosθR is the receiving end power factor. Effect of load power factor on regulation

of transmission line:

1. For lagging load

2. For leading load






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Now

Power factor is lagging or unity, and then VR is increased and goes to be positive.

Power factor is leading, and then VR is decreased and goes to be negative.

4. Define a two –wire transmission system?

Ans) A balanced line is a transmission line consisting of two conductors of the same type, and equal

impedance to ground and other circuits.

5. What is transposed line?

Transposition is the periodic swapping of positions of the conductors of a transmission line, in order to

reduce crosstalk and otherwise improve transmission. In telecommunications this applies to balanced

pairs whilst in power transmission lines three conductors are periodically transposed

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1. What is a composite conductor?

So far we have considered only solid round conductors. However as mentioned at the beginning of

Section 1.1, stranded conductors are used in practical transmission line. We must therefore modify the

equations derived above to accommodate stranded conductors. Consider the two groups of conductors

shown in Fig. 1.9. Of these two groups conductor x contains n identical strands of radius rx while

conductor y contains m identical strands of radius ry . Conductor x carries a current I the return path of

which is through conductor y . Therefore the current through conductor y is - I .

Fig. 1.9 Single-phase line with two composite conductors.

Since the strands in a conductor are identical, the current will be divided equally among the strands. Therefore the current

through the strands of conductor x is I / n and through the strands of conductor y is -I/m . The total flux linkage of strand a is

given by

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2. What is skin effect?

Ans) Skin effect is the tendency of an alternating electric current (AC) to become distributed within

a conductor such that the current density is largest near the surface of the conductor, and decreases with

greater depths in the conductor. The electric current flows mainly at the "skin" of the conductor,

between the outer surface and a level called the skin depth. The skin effect causes the

effective resistance of the conductor to increase at higher frequencies where the skin depth is smaller,

thus reducing the effective cross-section of the conductor. The skin effect is due to opposing eddy

currents induced by the changing magnetic field resulting from the alternating current. At 60 Hz in

copper, the skin depth is about 8.5 mm. At high frequencies the skin depth becomes much smaller.

Increased AC resistance due to the skin effect can be mitigated by using specially woven litz wire.

Because the interior of a large conductor carries so little of the current, tubular conductors such as pipe

can be used to save weight and cost.

3. Give the expansion of GMR and GMD.

Ans) GMD stands for Geometrical Mean Distance. It is the equivalent distance between conductors.

GMD comes into picture when there are two or more conductors per phase used as in bundled

conductors.

GMR stands for Geometric mean Radius. GMR is calculated for each phase separately. Each of the

phases may have different GMR values depending upon the conductor size and arrangement.

4. What is fictitious conductor radius?

The radius r1¢ can be assumed to be that of a fictitious conductor that has no internal flux but with the

same inductance as that of a conductor with radius r1. In a similar way the inductance due current in

the conductor 2 is given by. H/m. (1.20) Therefore the inductance of the complete circuit is.

https://en.wikipedia.org/wiki/Alternating_current

https://en.wikipedia.org/wiki/Conductor_(material)

https://en.wikipedia.org/wiki/Current_density


https://en.wikipedia.org/wiki/Frequency

https://en.wikipedia.org/wiki/Eddy_current

https://en.wikipedia.org/wiki/Eddy_current

https://en.wikipedia.org/wiki/Magnetic

https://en.wikipedia.org/wiki/Hertz

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5. What are the advantages of bundle conductors over standard conductors?

The conductors of any one bundle are in parallel and charge per bundle is assumed to divide equally

between the conductors of bundle.

The composite or stranded conductors touch each other while the bundled conductors are away from

each other. The typical distance is about 30 cm and more. The conductors of each phase are connected

by using connecting wires at particular length.

Due excessive corona loss, the round conductors are not feasible for use for voltage level more than 230

kV. It is preferable to use hollow conductor in substations while bundled conductors in transmission

lines.

Following are advantages of bundled conductors.

1. Low radio interference and corona loss.

2. Reduced voltage gradient at conductor surface.

3. Increase in capacitance.

4. Low reactance due to increase in self GMD.

5. Increase in surge impedance loading


1. Derive an expression for inductance of a conductor due to external flux.

Ans) Reason of Transmission Line Inductance

Generally, electric power is transmitted through the transmission line with AC high voltage and current.

High valued alternating current while flowing through the conductor sets up magnetic flux of high










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strength with alternating nature. This high valued alternating magnetic flux makes a linkage with other

adjacent conductors parallel to the main conductor. Flux linkage in a conductor happens internally and

externally. Internally flux linkage is due to self-current and externally flux linkage due to external flux.

Now the term inductance is closely related to the flux linkage, denoted by λ. Suppose a coil with N

number of turn is linked by flux Φ due to current I, then,

But for transmission line N = 1. We have to calculate only the value of flux Φ, and hence, we can get the

transmission line inductance.

Calculation of Inductance of Single Conductor

Calculation of Internal Inductance due to Internal Magnetic Flux of a Conductor

Suppose a conductor is carrying current I through its length l, x is the internal

variable radius of the conductor and r is the original radius of the conductor. Now

the cross-sectional area with respect to radius x is πx2 square – unit and current Ix

is flowing through this cross-sectional area. So the value of Ix can be expressed in

term of original conductor current I and cross-sectional area πr2 square – unit



And magnetic flux density Bx

= μHx, where μ is the permeability of this conductor. Again, µ = µ0µr. If it is considered that the relative



Here entire cross-sectional area of the conductor does not enclose the above expressed flux. The ratio of

the cross sectional area inside the circle of radius x to the total cross section of the








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And magnetic flux density Bx

= μHx, where μ is the permeability of this conductor. Again, µ = µ0µr. If it is considered that the relative



Here entire cross-sectional area of the conductor does not enclose the above expressed flux. The ratio of

the cross sectional area inside the circle of radius x to the total cross section of the conductor can be

thought about as fractional turn that links the flux. Therefore the flux linkage is

Now, the total flux linkage for the conductor of 1m length with radius r is given by

Hence, the internal inductance is

2. Explain inductance due to external magnetic flux of a conductor






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Ans) External Inductance due to External Magnetic Flux of a Conductor

Let us assume, due to skin effect conductor current I is concentrated near the surface of the conductor.

Consider, the distance y is taken from the center of the conductor making the external radius of the

conductor. Hy is

the magnetizing force and By is the magnetic field density at y distance per unit length of the conductor.

Let us assume magnetic flux dφ is

present within the thickness dy from D1 to D2 for 1 m length of the conductor as per the figure.

As the total current I is assumed to flow in the surface of

the conductor, so the flux linkage dλ is equal to dφ.

But we have to consider the flux linkage from conductor surface to any external distance, i.e. r to D

https://www.electrical4u.com/what-is-magnetic-field/







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3. Why Skin Effect Occurs in Transmission Lines?

Ans) Having understood the phenomena of skin effect let us now see why this arises in case of an AC

system. To have a clear understanding of that look into the cross-sectional view of the conductor during

the flow of alternating current given in the diagram below. Let us initially consider the solid conductor

to be split up into some annular filaments spaced infinitely small distance apart, such that each filament

carries an infinitely small fraction of the total current. Like if the total current = I

Let us consider the conductor to be split up into n filament carrying current ‘i’ such that I = n i. Now

during the flow of an alternating current, the current carrying filaments lying on the core has a flux

linkage with the entire conductor cross-section including the filaments of the surface as well as those in

the core. Whereas the flux set up by the outer filaments is restricted only to the surface itself and is

unable to link with the inner filaments.Thus the flux linkage of the conductor increases as we move

closer towards the core and at the same rate increases the inductance as it has a direct proportionality

relationship with flux linkage. As a result, a larger inductive reactance gets induced into the core as

compared to the outer sections of the conductor. The high value of reactance in the inner section results

in the current gets distributed in an un-uniform manner and forcing the bulk of the current to flow

through the outer surface or skin giving rise to the phenomena called skin effect in transmission lines.







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Factors Affecting Skin Effect in Transmission Lines

The skin effect in an AC system depends on some factors like:-

5. Shape of conductor.

6. Type of material.

7. Diameter of the conductors.

8. Operational frequency.

3. Derive an expression for capacitances of a single phase transmission system and discuss the

effect of earth on capacitance with suitable equation.

Ans)

In calculating the Effect of Earth on Transmission Line Capacitance, the presence of earth was ignored,

so far. The effect of earth on capacitance can be conveniently taken into account by the method of

images.

Method of Images

The electric field of transmission line conductors must conform to the presence of the earth below. The

earth for this purpose may be assumed to be a perfectly conducting horizontal sheet of infinite extent

which therefore acts like an equipotential surface.

The electric field of two long, parallel conductors charged +q and -q per unit is such that it has a zero

potential plane midway between the conductors as shown in Fig. 3.8. If a conducting sheet of infinite

http://www.eeeguide.com/effect-of-earth-on-transmission-line-capacitance/




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dimensions is placed at the zero potential plane, the electric field remains undisturbed. Further, if the

conductor carrying charge -q is now removed, the electric field above the conducting sheet stays intact,

while that below it vanishes. Using these well known results in reverse, we may equivalently replace the

presence of ground below a charged conductor by a fictitious conductor having equal and opposite

charge and located as far below the surface of ground as the overhead conductor above it—such a

fictitious conductor is the mirror image of the overhead conductor. This method of creating the same

electric field as in the presence of earth is known as the method of images originally suggested by Lord

Kelvin.

Capacitance of a Single-Phase Line

Consider a single-phase line shown in Fig. 3.9. It is required to calculate its capacitance taking the

presence of earth into account by the method of images described above. The equation for the voltage

drop Vab as determined by the two charged conductors a and b, and their images a’ and b’ can be written

as follows:






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Substituting the values of different charges and simplifying, we get



and








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It is observed from the above equation that the presence of earth modifies the radius r to r(1 +

(D2/4h2))1/2. For h large compared to D (this is the case normally), the effect of earth on line capacitance

is of negligible order.

4. Explain the concept of self and mutual GMDs.

Ans) CONCEPT OF SELF-GMD AND MUTUAL-GMD

The use of self geometrical mean distance (abbreviated as self-GMD) and mutual geometrical

mean distance (mutual-GMD) simplifies the inductance calculations, particularly relating to multi

conductor arrangements. The symbols used for these are respectively Ds and Dm. We shall briefly

discuss these terms.

( i) Self-GMD (Ds)

In order to have concept of self-GMD (also sometimes called Geometrical mean radius; GMR), consider

the expression for inductance per conductor per metre already derived in Art. Inductance/conductor/m

In this expression, the term 2 × 10-7 × (1/4) is the inductance due to flux within the solid conductor. For

many purposes, it is desirable to eliminate this term by the introduction of a concept called self-GMD or

GMR. If we replace the original solid conductor by an equivalent hollow cylinder with extremely thin




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walls, the current is confined to the conductor surface and internal conductor flux linkage would be

almost zero. Consequently, inductance due to internal flux would be zero and the term 2 × 10-7 × (1/4)

shall be eliminated. The radius of this equivalent hollow cylinder must be sufficiently smaller than the

physical radius of the conductor to allow room for enough additional flux to compensate for the absence

of internal flux linkage. It can be proved mathematically that for a solid round conductor of radius r, the

self-GMD or GMR = 0·7788 r. Using self-GMD, the eq. ( i) becomes :

Inductance/conductor/m = 2 × 10-7loge d/ Ds *

Where

Ds = GMR or self-GMD = 0·7788 r

It may be noted that self-GMD of a conductor depends upon the size and shape of the conductor and is

independent of the spacing between the conductors.

(ii) Mutual-GMD

The mutual-GMD is the geometrical mean of the distances form one conductor to the other and,

therefore, must be between the largest and smallest such distance. In fact, mutual-GMD simply

represents the equivalent geometrical spacing.

(a) The mutual-GMD between two conductors (assuming that spacing between conductors is

large compared to the diameter of each conductor) is equal to the distance between their centres i.e. Dm

= spacing between conductors = d

(b) For a single circuit 3-φ line, the mutual-GMD is equal to the equivalent equilateral spacing

i.e., ( d1 d2 d3 )1/3.




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(c) The principle of geometrical mean distances can be most profitably employed to 3-φ double

circuit lines. Consider the conductor arrangement of the double circuit shown in Fig. Suppose the radius

of each conductor is r.

Self-GMD of conductor = 0·7788 r

Self-GMD of combination aa’ is




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It is worthwhile to note that mutual GMD depends only upon the spacing and is substantially

independent of the exact size, shape and orientation of the conductor.

Inductance Formulas in Terms of GMD

The inductance formulas developed in the previous articles can be conveniently expressed in terms of

geometrical mean distances.

5. Derive the expression for the capacitance per phase of the 3 Φ double circuit line flat

Vertical spacing with transposition.

Ans) Figure 3.5 shows a Capacitance of a Three Phase Line Equilateral Spacing composed of

three identical conductors of radius r placed in equilateral configuration.

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Using Eq. (3.2) we can write the expressions for V ab and Vac as

Adding Eqs. (3.8) and (3 9), we get

Since there are no other charges in the vicinity, the sum of charges on the three conductors is zero.

Thus qb+ qc = – qa which when substituted in Eq. (3.10) yields

With balanced three-phase voltages applied to the line, it follows from the phasor diagram of Fig. 3.6

that

Substituting for (Vab + Vac) from Eq. (3.12) in Eq. (3.11), we get qa D








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The capacitance of line to neutral immediately follows as

For air medium (kr = 1),

The line charging current of phase a is

Ia (line charging) = jωCnVan

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OBJECTIVE QUESTIONS

1. In which climate does the chances of occurrence of corona is maximum?

a. Dry

b. Hot summer [ ]

c. Winter

d. Humid

2. What is the effect on corona, if the spacing between the conductors is increased?

a. Corona increases [ ]

b. Corona is absent

c. Corona decreases

d. None of these

3. Why are the hollow conductors used?

a. Reduce the weight of copper [ ]

b. Improve stability

c. Reduce corona

d. Increase power transmission capacity

4. Which of these given statements is wrong in consideration with bundled conductors?

a. Control of voltage gradient [ ]

b. Reduction in corona loss

c. Reduction in the radio interference

d. Increase in interference with communication lines




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5. Why are bundled conductors employed?

[ ]

a. Appearance of the transmission line is improved

b. Mechanical stability of the line is improved

c. Improves current carrying capacity

d. Improves the corona performance of the line

7. The effect of dirt on the surface of the conductor is to _____________ irregularity and thereby

________________ the break down voltage.

a. Decreases, reduces [ ]

b. Increases, increases

c. Increases, reduces

d. Decreases, increases

8.Find the spacing between the conductors a 132 kV 3 phase line with 1.956 cm diameter

conductors is built so that corona takes place, if the line voltage exceeds 210 kV (rms). With go =

30 kV/cm.

[ ]

a. 1.213 m.

b. 2.315 m.

c. 3.451 m.

d. 4.256 m.

9. Capacitance between the two conductors of a single phase two wire line is 0.5 μ F/km. What is

the value of capacitance of each conductor to neutral?

[ ]

a. 0.5 μ F / km.

b. 1 μ F / km.




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c. 0.25 μ F / km.

d. 2.0 μ F / km.

10. What happens in case of capacitance of line to ground, if the effect of earth is taken into

account?

a. Capacitance of line to ground decreases. [ ]

b. Capacitance of line to ground increases.

c. The capacitance remains unaltered.

d. The capacitance becomes infinite.

KEY

S.no 1 2 3 4 5 6 7 8 9 10

Ans d c D d d c C b b

FILL IN THE BLANKS

1. What is the value of capacitance to neutral for the two wire line?

----------------------------------

2. A two conductor single phase line operates at 50Hz. Diameter of each conductor is 20mm and

the spacing between the conductors is 3m. The height of the conductors above the ground is 6m.

What is the capacitance of the line to neutral -------------------




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3. What happens if the separation between the three phases of the transmission line is increased ---

--------------------------------

4.What will be the capacitance of a 100 km long, 3 phase, 50Hz overhead transmission line

consisting of 3 conductors, each of 2 cm and spaced 2.5 m at the corners of an equilateral triangle -

----------------------

5. If the double circuit 3 phase line has conductors of diameter 2 cm and distance of separation 2m

in hexagonal spacing. What is the phase to neutral capacitance for 150km of line ----------------------

-------

6. What is the charging current per km for the transmission line shown in the figure. Operating at

132 kV, the conductor diameter is-----------------------

KEY

S.no 1 2 3 4 5 6 7 8 9 10

Ans Twice the

line to line

capacitance.

9.7

pF/m

Inductance

will increase

and

capacitance

will decrease.

1.007 μ

F/phase

3.7408

μ F

0.8 cm.

0.21

A/km




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UNIT -2

PERFORMANCE OF SHORT AND MEDIUM




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LENGTH TRANSMISSION LINES


1. Give classification of overhead transmission line.

Ans)

A transmission line has four constants R, L, C and shunt conductance. But generally, three constants R,

L and C are considered and they are uniform along the whole length of line. The fourth constant shunt

conductance between conductors or between conductor and ground and accounts for the leakage current

at the insulators. It is very small in case of overhead lines and may be assumed zero. The capacitance

existing between conductors for 1φ line or 3φ line forms a shunt path throughout the length of line.

Therefore capacitance effects introduce complication in transmission line calculation. Depending upon

the manner in which capacitance is taken into account, the overhead transmission line are classified as,

1. Short transmission lines

2. Medium transmission lines

3. Long transmission lines

2. What is surge impedance loading?

Ans) The surge impedance loading or SIL of a transmission line is the MW loading of a transmission

line at which a natural reactive power balance occurs. The following brief article will explain the

concept of SIL. Transmission lines produce reactive power (Mvar) due to their natural capacitance.




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3. Define voltage regulation.

Ans) The voltage regulation of the transformer is the percentage change in the outputvoltage from no-

load to full-load. And since power factor is a determining factor in the secondary voltage, power factor

influences voltage regulation. This means thevoltage regulation of a transformer is a dynamic, load-

dependent number.

4. List out the common methods of representation of medium transmission lines.

Ans) The transmission line having its effective length more than 80 km but less than 250 km is

generally referred to as a medium transmission line. Due to the line length being considerably high,

admittance Y of the network does play a role in calculating the effective circuit parameters, unlike in the

case of short transmission lines. For this reason the modeling of a medium length transmission line is

done using lumped shunt admittance along with the lumped impedance in series to the circuit.These

lumped parameters of a medium length transmission line can be represented using three different

models, namely-

1. Nominal Π representation.

2. Nominal T representation.

3. End Condenser Method.

5. Define characteristic impedance of a transmission line.

Ans) The characteristic impedance or surge impedance(usually written Z0) of a uniform transmission

line is the ratio of the amplitudes of voltage and current of a single wave propagating along the line; that

is, a wave travelling in one direction in the absence of reflections in the other direction.

3 marks answers







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1. What is the purpose of using series reactors on a transmission line?

Ans) Series reactors are used as current limiting reactors to increase the impedance of a system. They

are also used for neutral earthing. Such reactors are also used to limit the starting currents of

synchronous electric motors and to compensate ReactivePower in order to improve the

transmission capacity of power lines.

2.

Ans) In electrical engineering, particularly power engineering, voltage regulation is a measure of

change in the voltage magnitude between the sending and receiving end of a component, such as

a transmission or distribution line. Voltage regulation describes the ability of a system to provide near

constant voltage over a wide range of load conditions. The term may refer to a passive property that

results in more or less voltage drop under various load conditions, or to the active intervention with

devices for the specific purpose of adjusting voltage.

where Vnl is voltage at no load and Vfl is voltage at full load. The percent voltage regulation of an

ideal transmission line, as defined by a transmission line with zero resistance and reactance, would

equal zero due to Vnl equaling Vfl as a result of there being no voltage drop along the line. This is

why a smaller value of Voltage Regulation is usually beneficial, indicating that the line is closer to

ideal.

The Voltage Regulation formula could be visualized with the following: "Consider power being

delivered to a load such that the voltage at the load is the load's rated voltage VRated, if then the load

disappears, the voltage at the point of the load will rise to Vnl."

Voltage regulation in transmission lines occurs due to the impedance of the line between its sending

and receiving ends. Transmission lines intrinsically have some amount of resistance, inductance, and

capacitance that all change the voltage continuously along the line. Both the magnitude and phase

angle of voltage change along a real transmission line. The effects of line impedance can be

modeled with simplified circuits such as the short line approximation (least accurate), the medium

line approximation (more accurate), and the long line approximation (most accurate).

https://en.wikipedia.org/wiki/Electrical_engineering

https://en.wikipedia.org/wiki/Power_engineering

https://en.wikipedia.org/wiki/Voltage


https://en.wikipedia.org/wiki/Electrical_load

https://en.wikipedia.org/wiki/Electrical_resistance_and_conductance





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3. What is reflected and refracted wave?

Ans) Reflection, refraction and diffraction are all boundary behaviors of waves associated with the

bending of the path of a wave. ... Refraction is the change in direction of waves that occurs

when waves travel from one medium to another. Refraction is always accompanied by a wavelength and

speed change.

4. What are the limitations of T and π methods?

Ans) The transmission line having its effective length more than 80 km but less than 250 km is generally

referred to as a medium transmission line. Due to the line length being considerably high, admittance Y

of the network does play a role in calculating the effective circuit parameters, unlike in the case of short

transmission line. For this reason the modeling of a medium length transmission line is done using

lumped shunt admittance along with the lumped impedance in series to the circuit.

These lumped parameters of a medium length transmission line can be represented using three different

models, namely-

1. Nominal Π representation.

2. Nominal T representation.

3. End Condenser Method.

5. Why do we analyze a three phase transmission line on single phase basis?

Ans)

1. Three phase power distribution requires less copper or aluminium for transferring the same amount

of power as compared to single phase power.

2. The size of a three phase motor is smaller than that of a single phase motor of the same rating.

3. Three phase motors are self starting as they can produce a rotating magnetic field. The single phase

motor requires a special starting winding as it produces only a pulsating magnetic field.

4. In single phase motors, the power transferred in motors is a function of the instantaneous power

which is constantly varying. In three-phase the instantaneous power is constant.




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5. Single phase motors are more prone to vibrations. In three phase motors, however, the power

transferred is uniform throughout the cycle and hence vibrations are greatly reduced.

6. Three phase motors have better power factor regulation.

7. Three phases enables efficient DC rectification with low ripple.

Figure 1. Resultant DC from three-phase rectifier.

8. Generators also benefit by presenting a constant mechanical load through the full revolution, thus

maximising power and also minimising vibration.

5 MARKS QUETION AND ANSWERS 1. Explain what are ABCD constants in a transmission line

Ans) A, B, C and D are the constants also known as thetransmission parameters or

chain parameters. These parameters are used for the analysis of an electrical network. It is also used

for determining the performance of input, output voltage and current of the transmission network.




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Vs = sending end voltage

Is = sending end current

Vr = receiving end voltage

Ir = receiving end current

Open circuit




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ABCD parameters for short circuit

For the short circuit, the voltage remains

zero at the receiving end.

If we put Vr = 0 in the equation, we get the value of B which is the ratio

of sending end voltage to the receiving end currents. Its unit is ohms.

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Similarly, if we put Vr= 0 in current equations,

we get the value of D, which is the ratio of the sending current to the

receiving current. It is the dimensionless constant.

2.Explain relation between ABCD parameters

Ans) Relation between ABCD parameters

For determining the relation between various types of network, like passive or bilateral network

reciprocity theorem is applied. The voltage V is applied to the sending end, and the receiving end is kept

short circuit, so the voltage becomes zero.

Since, under short circuit the

receiving end voltage is zero, the voltage and current equations become

Similarly, the voltage is applied at the receiving end, and the input voltage remains

zero. Thus, the direction of the current in the network changes, which is shown in the diagram below







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The sending end voltage

becomes zero. The current flows through the receiving end is given by the equation

and

Consider, the network is passive, i.e. it contains only passive components

in the circuit like inductance, resistance, etc. So the current remains same Is = Ir.

Combining the above equations give,

dividing the above equation from -V/B we get,

This relation helps in determining the fourth parameters if we know any three

parameters.

For a symmetrical network, the input and output terminal may be interchanged without affecting the

network behaviour.



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If the network is supplied from input terminals and an output terminal is short

circuit, then the impedance becomes

and if the supply is from the output terminal and an input terminal is a short circuit then the impedance

becomes

in the symmetrical network, the impedance remains the same


In the nominal pi model of a medium transmission line, the series impedance of the line is concentrated

at the centre and half of each capacitance is placed at the centre of the line. The nominal Pi model of the

line is shown in the diagram below.


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In this circuit,

By Ohm’s law

By KCL at node a,


By ohm’s law









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or


Also,













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It is also drawn for a lagging power factor

of the load. In the phasor diagram the quantities shown are as follows;











∅s – phase angle between Vs and Is at the sending end, and cos∅s will give the sending-end power

factor.

3. Derive an expression for nominal Pi model of medium transmission line






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In the nominal pi model of a medium transmission line, the series impedance of the line is concentrated

at the centre and half of each capacitance is placed at the centre of the line. The nominal Pi model of the

line is shown in the diagram below.

In this circuit,

By Ohm’s law

By KCL at node a,










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By ohm’s law



or


Also,










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It is also drawn for a lagging power factor

of the load. In the phasor diagram the quantities shown are as follows;
















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∅s – phase angle between Vs and Is at the sending end, and cos∅s will give the sending-end power

factor.

4. Derive expression for surge impedance.

Ans) Surge Impedance Loading is a very essential parameter when it comes to the study of power

systems as it is used in the prediction of maximum loading capacity of transmission lines.

However before understanding SIL, we first need to have an idea of what is Surge Impedance (Zs). It

can be defined in two ways one a simpler one and other a bit rigorous. Method 1 It is a well known fact

that a long transmission lines (> 250 km) have distributed inductance and capacitance as its inherent

property. When the line is charged, the capacitance component feeds reactive power to the line while the

inductance component absorbs the reactive power. Now if we take the balance of the two reactive

powers we arrive at the following equation

Capacitive VAR = Inductive VAR Where, V = Phase voltage I = Line Current Xc =

Capacitive reactance per phase XL = Inductive reactance per phase Upon simplifying

Where, f = Frequency of the system L = Inductance per unit length

of the line l = Length of the line Hence we get,

This quantity having the dimensions of resistance is the Surge Impedance. It can be considered as a

purely resistive load which when connected at the receiving end of the line, the reactive power generated









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by capacitive reactance will be completely absorbed by inductive reactance of the line. It is nothing but

the Characteristic Impedance (Zc) of a lossless line.

Method 2 From the rigorous solution of a long transmission line we get the following equation for

voltage and current at any point on the line at a distance x from the receiving end

Where, Vx and Ix = Voltage and Current at

point x VR and IR = Voltage and Current at receiving end Zc = Characteristic Impedance δ = Propagation

Constant Z = Series impedance per unit length per phase Y = Shunt

admittance per unit length per phase Putting the value of δ in above equation of voltage we get

Where,

We observe that the instantaneous voltage consists of two terms each of which is a function of time and

distance. Thus they represent two travelling waves. The first one is the positive exponential part

representing a wave travelling towards receiving end and is hence called the incident wave. While the

other part with negative exponential represents the reflected wave. At any point along the line, the

voltage is the sum of both the waves. The same is true for current waves also. Now, if suppose the load

impedance (ZL) is chosen such that ZL = Zc, and we know Thus

and hence the reflected wave vanishes. Such a line is termed as

infinite line. It appears to the source that the line has no end because it receives no reflected wave.

Hence, such an impedance which renders the line as infinite line is known as surge impedance.It has a

value of about 400 ohms and phase angle varying from 0 to –15 degree for overhead lines and around 40

ohms for underground cables.

The term surge impedance is however used in connection with surges on the transmission line which

may be due to lightning or switching, where the line losses can be neglected such that








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Now that we have understood Surge Impedance, we can easily define

Surge Impedance Loading. SIL is defined as the power delivered by a line to a purely resistive load

equal in value to the surge impedance of that line. Hence we can write

The unit of SIL is Watt or MW.

When the line is terminated by surge impedance the receiving end voltage is equal to the sending end

voltage and this case is called flat voltage profile. The following figure shows the voltage profile for

different loading cases. It

should also be noted that surge impedance and hence SIL is independent of the length of the line. The

value of surge impedance will be the same at all the points on the line and hence the voltage. In case of a

Compensated Line, the value of surge impedance will be modified accordingly as




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Where, Kse = % of series capacitive compensation by Cse

KCsh = % of Shunt capacitive compensation by Csh Klsh = % of

shunt inductive compensation by Lsh The equation for SIL will now use the

modified Zs

5. Explain how voltages and currents are evaluated in long transmission lines.

Ans) A power transmission line with its effective length of around 250 Kms or above is referred to as a

long transmission line. The line constants are uniformly distributed over the entire length of line.

Calculations related to circuit parameters (ABCD parameters) of such a power transmission is not that

simple, as was the case for a short transmission line or medium transmission line.

The reason being that, the effective circuit length in this case is much higher than what it was for the

former models (long and medium line) and, thus ruling out the approximations considered there like.

3. Ignoring the shunt admittance of the network, like in a small transmission line model.









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4. Considering the circuit impedance and admittance to be lumped and concentrated at a point as was

the case for the medium line model.

Rather, for all practical reasons we should consider the circuit impedance and admittance to be

distributed over the entire circuit length as shown in the figure below. The calculations of circuit

parameters for this reason are going to be slightly more rigorous as we will see here. For accurate

modeling to determine circuit parameters let us consider the circuit of the long transmission line as

shown in the diagram below.

Here a line of

length l > 250km is supplied with a sending end voltage and current of VS and IS respectively, where as

the VR and IR are the values of voltage and current obtained from the receiving end. Lets us now

consider an element of infinitely small length Δx at a distance x from the receiving end as shown in the

figure where. V = value of voltage just before entering the element Δx. I = value of current just before

entering the element Δx. V+ΔV = voltage leaving the element Δx. I+ΔI = current leaving the element

Δx. ΔV = voltage drop across element Δx. zΔx = series impedence of element Δx yΔx = shunt

admittance of element Δx Where, Z = z l and Y = y l are the values of total impedance and admittance of

the long transmission line.








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Therefore, the voltage drop across the infinitely small element Δx is given by

Now to determine the current ΔI, we apply KCL to node A.

Since the term ΔV yΔx is the product of 2 infinitely

small values, we can ignore it for the sake of easier calculation. Therefore, we can write

Now derivating both sides of eq (1) w.r.t x, Now substituting from equation

(2) The solution of the above second order

differential equation is given by. Derivating equation (4)

w.r.to x. Now comparing equation (1) with

equation (5)

Now to go further let us define the characteristic impedance Zc and propagation constant δ of a long

transmission line as Then the voltage and current equation can be

expressed in terms of characteristic impedance and propagation constant as

Now at x=0, V= VR and I= Ir. Substituting these conditions to

equation (7) and (8) respectively. Solving equation (9) and (10), We

get values of A1 and A2 as, Now applying another

extreme condition at x = l, we have V = VS and I = IS. Now to determine VS and IS we substitute x by l





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and put the values of A1 and A2 in equation (7) and (8) we get

By trigonometric and exponential

operators we know Therefore, equation (11) and

(12) can be re-written as Thus

comparing with the general circuit parameters equation, we get the ABCD parameters of a long

transmission line as,




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OBJECTIVE QUESTIONS

1. On what concept is electrically short, medium and long lines based on?

a. Nominal voltage of the line. [ ]

b. Physical length of the line.

c. Wavelength of the line.

d. Power transmitted over the line.

2. The capacitance effect can be neglected in which among the transmission lines?

a. Short transmission lines. [ ]

b. Medium transmission lines.

c. Long transmission lines.

d. All of these.

3. In the modelling of short length overhead transmission line, why is the line capacitance to

ground not considered?

a. Equal to zero. [ ]

b. Finite but very small.

c. Finite but very large.

d. Infinite.

4. In a short transmission line, voltage regulation is zero when the power factor angle of the load

at the receiving end side is equal to ____________.

a. tan-1 (X/R) [ ]




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b. tan-1 (R/X)

c. tan-1 (X/Z)

d. tan-1 (R/Z)

5. What is the power factor angle of the load for maximum voltage regulation?

a.tan-1 (X/R) [ ]

b.cos-1 (X/R)

c.tan-1 (R/X)

d. cos-1 (R/X)

6. A single phase transmission line of impedance j0.8 ohm supplies a resistive load of 500 A at 300

V. The sending end power factor is

a. Unity [ ]

b. 0.8 lagging

c. 0.8 leading

d. 0.6 lagging

7. What are the values of A, B, C, D parameters of a short transmission line?

a. Z, 0, 1, 1 [ ]

b. 0, 1, 1, 1

c. 1, Z, 0, 1

d. 1, 1, Z, 0

8. The ABCD constants of a 3 phase transposed transmission line with linear and passive elements

________________

a. Are always equal. [ ]

b. Never equal.

c. Only A and D are equal.

d. Only B and C are equal.




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9. For a transmission line which among the following relation is true?

a. –AB + CD = 1 [ ]

b. AD + BC = 1

c. AB – CD = -1

d. –AD + BC = 1

10. For a transmission line which among the following relation is true?

a. –AB + CD = 1 [ ]

b. AD + BC = 1

c. AB – CD = -1

d. –AD + BC = 1

s.no 1 2 3 4 5 6 7 8 9 10

Ans B a b B A D c c c d

FILL IN THE BLANKS

1. The ABCD constants of a 3 phase transposed transmission line with linear and passive elements

-----------------------------

2. A line of what length can be classified as a medium transmission line a ----------------------

3. What are the A and D parameters in case of medium transmission line (nominal T method) -----

-------------------------

4. What is the value of B parameter in case of nominal p method ----------------------

5. What is the value of the C parameter by using a nominal T method for a 3 phase balanced load

of 30 MW which is supplied by a 132 kV, 50 Hz and 0.85 pf lagging. The series impedance of a

single conductor is (20 + j52) Ω and the total phase to neutral admittance is 315 * 10-6 siemen. -----

--------------------




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6. The transmission lines above what length is termed as the long lines -------------------

7. What is the normal range of angle for the parameter A -------------------------

8. A = D = 0.8 ∠ 1 °, B = 170 ∠ 85 ° Ω , and C = 0.002 ∠ 90.4 ° ℧ the sending end voltage is 400 kV.

What is the receiving end voltage under no load condition -------------------------

9. Transmission efficiency of a transmission line increases with the-----------------

10. The capacitance effect can be neglected in which among the transmission lines --------------

UNIT –III

POWER SYSTEM TRANSIENTS

2 MARKS QUETION AND ANSWERS 1. Why Ferranti effect occurs?

Ans) Capacitance and inductance are the main parameters of the lines having a length 240km or above.

On such transmission lines, the capacitance is not concentrated at some definite points. It is distributed

uniformly along the whole length of the line.

S.no 1 2 3 4 5 6 7 8 9 10

Ans only B

and C

are

equal

A = D

= 1 +

(YZ /

2)

50-

150

km

Z 0.000315 ∠

90

150 km

and

above

0 -

10°

2000 Ω

Increase in

power

factor and

voltage.

Short

transmission

lines.




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When the voltage is applied at the sending end, the current drawn by the capacitance of the line is more

than current associated with the load. Thus, at no load or light load, the voltage at the receiving end is

quite large as compared to the constant voltage at the sending end.

2. Define corona

Ans) Definition: The phenomenon of ionisation of surrounding air around the conductor due to which

luminous glow with hissing noise is rise is known as the corona effect. Air acts as a dielectric medium

between the transmission lines. In other words, it is an insulator between the current carrying conductors

3. What is travelling waves?

Ans) Travelling wave is a temporary wave that creates a disturbance and moves along the transmission

line at a constant speed. Such type of wave occurs for a short duration (for a few microseconds) but

cause a much disturbance in the line. The transient wave is set up in the transmission line mainly due to

switching, faults and lightning.

4. Define transmission efficiency.

Ans) Definition of transmission efficiency. : the ratio of the power received over atransmission path to

the power transmitted; also : the ratio of the output to the input power of a circuit or device

5. What is the length of short, long and medium transmission line? Ans) The length of the transmission line at its voltage level decides its classification as

Short/Medium/Long Transmission Lines. The accepted conventional distances are,

Short - 0 to 80 Km.

Medium - 80 to 160 Km.


1. Explain skin and proximity effects on transmission line?




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Ans) Skin effect is a tendency for alternating current (AC) to flow mostly near the outer surface of an

electrical conductor, such as metal wire.The effect becomes more and more apparent as the frequency

increases. ... The effect is most pronounced in radio-frequency (RF) systems, especially antennas

and transmission lines.

In a conductor carrying alternating current, if currents are flowing through one or more other nearby

conductors, such as within a closely wound coil of wire, the distribution of current within the first

conductor will be constrained to smaller regions. The resulting current crowding is termed

the proximity effect.

2. compare the short , medium and long transmission lines

Short Transmission Line

Length is about 50 km.

Voltage level is up to 20 kV

Capacitance effect is negligible

Only resistance and inductance are taken in calculation capacitance is neglected.

Medium Transmission Line

Length is about 50km to 150km

Operational voltage level is from 20 kV to 100 kV

Capacitance effect is present

Distributed capacitance form is used for calculation purpose.

Long Transmission Line

Length is more than 150 km

Voltage level is above 100 kV

Line constants are considered as distributed over the length of the line.

3. What is the Transmission Efficiency?

Ans) Transmission efficiency is defined as the ration of receiving end power PR to the sending end

power PS and it is expressed in percentage value.

cosθs is the sending end power factor. cosθR is the receiving end power factor. Vs is the sending end

voltage per phase. VR is the receiving end voltage per phase.

4.What is Transmission Line Voltage Regulation

Ans) Voltage regulation of transmission line is defined as the ratio of difference between sending and

receiving end voltage to receiving end voltage of a transmission line between conditions of no load and

full load. It is also expressed in percentage. Where, Vs is the sending

end voltage per phase and VR is the receiving end voltage per phase.







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XL is the reactance per phase. R is the

resistance per phase. cosθR is the receiving end power factor. Effect of load power factor on regulation

of transmission line:

1. For lagging load

2. For leading load

5. What is the effect of load power factor on efficiency of transmission line

Ans)

Effect of Load Power Factor on Efficiency of Transmission Line

We know efficiency of transmission line is Now,

for short transmission line, IR = IS = I So, considering three phase short transmission line,

So, Now it is clear that to transmit given amount of power,

the load current is inversely proportional to receiving end power factor. Again in case of medium and

long transmission line,

Here it is clear that transmission efficiency depends on the receiving end power factor

5 MARKS QUETION AND ANSWERS 1. What are the factors affecting the corona.

Factors affecting corona:

Ans) The following are the factors affecting the corona;







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6. Effect of supply voltage – If the supply voltage is high corona loss is higher in the lines. In low-

voltage transmission lines, the corona is negligible, due to the insufficient electric field to maintain

ionization.

7. The condition of conductor surface – If the conductor is smooth, the electric field will be more

uniform as compared to the rough surface. The roughness of conductor is caused by the deposition of

dirt, dust and by scratching, etc. Thus, rough line decreases the corona loss in the transmission lines.

8. Air Density Factor – The corona loss in inversely proportional to air density factor, i.e., corona loss,

increase with the decrease in density of air. Transmission lines passing through a hilly area may have

higher corona loss than that of similar transmission lines in the plains because in a hilly area the

density of air is low.

9. Effect of system voltage – Electric field intensity in the space around the conductors depends on the

potential difference between the conductors. If the potential difference is high, electric field intensity is

also very high, and hence corona is also high. Corona loss, increase with the increase in the voltage.

10. The spacing between conductors – If the distance between two conductors is much more as compared

to the diameter of the conductor than the corona loss occurs in the conductor. If the distance between

them is extended beyond certain limits, the dielectric medium between them get decreases and hence

the corona loss also reduces.

2. What are the disadvantages of corona discharge.

Ans) Disadvantages of corona discharge:

The undesirable effects of the corona are:

8. The glow appear across the conductor which shows the power loss occur on it.

9. The audio noise occurs because of the corona effect which causes the power loss on the conductor.

10. The vibration of conductor occurs because of corona effect.

11. The corona effect generates the ozone because of which the conductor becomes corrosive.

12. The corona effect produces the non-sinusoidal signal thus the non-sinusoidal voltage drops occur in the

line.

13. The corona power loss reduces the efficiency of the line.

14. The radio and TV interference occurs on the line because of corona effect.

3).Explain travelling wave on transmission lines




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Ans) Travelling Wave on Transmission Line

The transmission line is a distributed parameter circuit and its support the wave of voltage and current.

A circuit with distributed parameter has a finite velocity of electromagnetic field propagation. The

switching and lightning operation on such types of circuit do not occur simultaneously at all points of

the circuit but spread out in the form of travelling waves and surges.

When a transmission line is suddenly connected to a voltage source by the closing of a switch the whole

of the line in not energized at once, i.e., the voltage does not appear instantaneously at the other end.

This is due to the presence of distributed constants (inductance and capacitance in a loss-free line).

Considered a long transmission line having a distributed parameter inductance (L) and capacitance (C).

The long transmission line is divided into small section shown in the figure below.The S is the switch

used for closing or opening the surges for switching operation. When the switch is closed the L1

inductance act as an open circuit and C1 act as a short circuit. At the same instant, the voltage at the next

section cannot be charged because the voltage across the capacitor C1 is zero.

So unless the capacitor C1 is charged to some value the charging of the capacitor C2 through L2 is not

possible which will obviously take some time. The same argument applies to the third section, fourth

section, and so on. The voltage at the section builds up gradually. This gradual build up of voltage over

the transmission conductor can be regarded as though a voltage wave is travelling from one end to the

other end and the gradual charging of the voltage is due to associate current wave.

The current wave, which is accompanied by voltage wave steps up a magnetic field in the surrounding

space. At junctions and terminations, these waves undergo reflection and refraction. The network has a

large line and junction the number of travelling waves initiated by a single incident wave and will

increase at a considerable rate as the wave split and multiple reflections occurs. The total energy of the

resultant wave cannot exceed the energy of the incident wave.

4. 1.what is corona effect ?

Corona Effect

Definition: The phenomenon of ionization of surrounding air around the conductor due to which

luminous glow with hissing noise is rise is known as the corona effect.

https://circuitglobe.com/wp-content/uploads/2016/08/equivalent-pi-section-of-a-long-transmission-line-compressor1.jpg




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Air acts as a dielectric medium between the transmission lines. In other words, it is an insulator between

the current carrying conductors. If the voltage induces between the conductor is of alternating nature

then the charging current flows between the conductors. And this charging conductor increases the

voltage of the transmission line.

The electric field intensity also increases because of the charging current.

If the intensity of the electric field is less than 30kV, the current induces between the conductor is

neglected. But if the voltage rise beyond the 30kv then the air between the conductors becomes charge

and they start conducting. The sparking occurs between the conductors till the complete breakdown of

the insulation properties of conductors takes place.

Corona effect mostly occurs at the sharp point of insulators.

Contents: Corona effect

6. Corona Formation

7. Factors affecting corona

8. Disadvantages of corona discharge

9. Minimizing corona

10. Important points

Corona Formation:

Air is not a perfect insulator, and even under normal conditions, the air contains many free electrons and

ions. When an electric field intensity establishes between the conductors, these ions and free electrons

experience forced upon them. Due to this effect, the ions and free electrons get accelerated and moved in

the opposite direction.

https://circuitglobe.com/corona-effect.html#CoronaFormation

https://circuitglobe.com/corona-effect.html#Factorsaffectingcorona

https://circuitglobe.com/corona-effect.html#Disadvantagesofcoronadischarge

https://circuitglobe.com/corona-effect.html#Minimizingcorona

https://circuitglobe.com/corona-effect.html#Importantpoints

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The charged particles during their motion collide with one another and also with the very slow moving

uncharged molecules. Thus, the number of charged particles goes on increasing rapidly. This increase

the conduction of air between the conductors and a breakdown occurs. Thus, the arc establishes between

the conductors.

5. Briefly explain characteristic impedance

Ans) The characteristic impedance or surge impedance (usually written Z0) of a uniform transmission

line is the ratio of the amplitudes of voltage and current of a single wave propagating along the line; that

is, a wave travelling in one direction in the absence of reflections in the other direction. Characteristic

impedance is determined by the geometry and materials of the transmission line and, for a uniform line,

is not dependent on its length. The SI unit of characteristic impedance is the ohm.

The characteristic impedance of a lossless transmission line is purely real, with no reactive component.

Energy supplied by a source at one end of such a line is transmitted through the line without being

dissipated in the line itself. A transmission line of finite length (lossless or lossy) that is terminated at

one end with an impedance equal to the characteristic impedance appears to the source like an infinitely

long transmission line and produces no reflections.

The characteristic impedance of an infinite transmission line at a given angular frequency is

the ratio of the voltage and current of a pure sinusoidal wave of the same frequency travelling along the

line. This definition extends to DC by letting tend to 0, and subsists for finite transmission lines

until the wave reaches the end of the line. In this case, there will be in general a reflected wave which

travels back along the line in the opposite direction. When this wave reaches the source, it adds to the

transmitted wave and the ratio of the voltage and current at the input to the line will no longer be the

characteristic impedance. This new ratio is called the input impedance. The input impedance of an

infinite line is equal to the characteristic impedance since the transmitted wave is never reflected back

from the end. It can be shown that an equivalent definition is: the characteristic impedance of a line is

that impedance which, when terminating an arbitrary length of line at its output, produces an input

impedance of equal value. This is so because there is no reflection on a line terminated in its own

characteristic impedance.

Applying the transmission line model based on the telegrapher's equations, the general expression for

the characteristic impedance of a transmission line is:

where



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https://en.wikipedia.org/wiki/SI

https://en.wikipedia.org/wiki/Electrical_impedance

https://en.wikipedia.org/wiki/Ohm_(unit)


https://en.wikipedia.org/wiki/Input_impedance

https://en.wikipedia.org/wiki/Telegrapher%27s_equations




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is the resistance per unit length, considering the two conductors to be in series,

is the inductance per unit length,

is the conductance of the dielectric per unit length,

is the capacitance per unit length,

is the imaginary unit, and

is the angular frequency.

Although an infinite line is assumed, since all quantities are per unit length, the characteristic impedance

is independent of the length of the transmission line.

The voltage and current phasors on the line are related by the characteristic impedance as:

where the superscripts and represent forward- and backward-traveling waves, respectively. A

surge of energy on a finite transmission line will see an impedance of Z0 prior to any reflections arriving,

hence surge impedance is an alternative name for characteristic impedance.

Lossless line

The analysis of lossless lines provides an accurate approximation for real transmission lines that

simplifies the mathematics considered in modeling transmission lines. A lossless line is defined as a

transmission line that has no line resistance and no dielectric loss. This would imply that the conductors

act like perfect conductors and the dielectric acts like a perfect dielectric. For a lossless line, R and G are

both zero, so the equation for characteristic impedance derived above reduces to:

In particular, does not depend any more upon the frequency. The above expression is wholly real,

since the imaginary term j has canceled out, implying that Z0 is purely resistive. For a lossless line

terminated in Z0, there is no loss of current across the line, and so the voltage remains the same along the

line. The lossless line model is a useful approximation for many practical cases, such as low-loss

transmission lines and transmission lines with high frequency. For both of these cases, R and G are

much smaller than ωL and ωC, respectively, and can thus be ignored.


https://en.wikipedia.org/wiki/In_series

https://en.wikipedia.org/wiki/Inductance

https://en.wikipedia.org/wiki/Electrical_conductance

https://en.wikipedia.org/wiki/Capacitance

https://en.wikipedia.org/wiki/Imaginary_unit

https://en.wikipedia.org/wiki/Angular_frequency

https://en.wikipedia.org/wiki/Phasor_(electronics)

https://en.wikipedia.org/wiki/Dielectric_loss




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OBJECTIVE QUESTIONS

1. When does the Ferranti effect happen on the transmission line?

a. When the line is short and loaded. [ ]




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b. When the line is long and loaded.

c. When the line is long and unloaded.

d. None of these.

2. When is the Ferranti effect on the long transmission lines experienced?

a. The line is lightly loaded.

b. The line is heavily loaded. [ ]

c. The line is fully loaded.

d. The power factor is unity.

3. Correctly match the items in List 1 to the items in List 2:

List 1

A. Skin Effect [ ]

B. Proximity Effect

C. Ferranti effect

D. Surge impedance

List 2

1. Increase in resistance but decrease in self inductance.

2. Increase in ac resistance.

3. Owing to voltage drop across line inductance due to flow of a charging current.

4. Square root of ratio of line impedance and shunt admittance.

Codes:

A B C D

a. 2 1 3 4

b. 1 2 4 3

c. 3 4 2 1

d. 4 3 1 2

4. Wave front is basically a locus of points acquiring similar _______

a. Phase

b. Frequency [ ]

c. Amplitude

d. Wave equation




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5. In which kind of waveform is the phase velocity defined?

a. Sinusoidal [ ]

b. Rectangular

c. Square

d. Triangular

6. Power density is basically termed as ________ power per unit area

a. Reflected [ ]

b. Refracted

c. Radiated

d. Diffracted

7. If the path difference of two waves with single source traveling by different paths to arrive at

the same point, is λ/2, what would be the phase difference between them? a. β x (λ/2) [ ]

b. β / (λ/2)

c. β + (λ/2)

d. β – (λ/2)

8. What is the possible range of height for the occurrence of sporadic E-region with respect to

normal E-region? [ ]

a. 20 km – 50 km

b. 45 km – 85 km

c. 90 km – 130 km

d. 140 km – 200 km

9. Zero voltage regulation is possible only for [ ]

a. lagging power factor

b. leading power factor

c. unity power factor

d. all of the above

10. A transmission line has 50km operating at frequency of 500Hz .Find the type of the line

a. short transmission line [ ]

b. long transmission line

c. medium transmission line




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d. All of the above

KEY

S.no 1 2 3 4 5 6 7 8 9 10

Ans C A A a A C a c b B

FILL IN THE BLANKS

1. A transmission line has 200km operating at frequency of 10Hz .Find the type of the line ----------

-------

2. If the rated receiving end voltage is 33 kV, then what are the limits for safe operation of

equipments -------------------------

3. communication lines are treated as -----------------------

4. 05․ A transmission line has an impedance of (2+j6) Ω has voltage regulation of 10% at a load

power factor of 0.8 lag. Find the voltage regulation for a load of 0.6 lead -------------------------------

5. Skin effect depends on --------------------------and --------------------------and------------------------

6. If the frequency is increased, then skin effect will----------------------

7․ If skin depth is more, then skin effect is--------------------------

8. Proximity effect is more in case of ------------------------------

9. Transmission lines are transposed to reduce--------------------------

10. Aluminium is now most commonly employed conductor material in transmission lines than

copper because------------------------------

KEY

S.no 1 2 3 4 5 6 7 8 9 10

Ans Short

transmi

ssion

line

31.35 to

34.65KV

Long

Transmission

lines

-

6.92

%

Frequency ,

permeability

and

conductivity

Skin

effect

will

increase

Skin

effect

is

less

Power

cables

radio

interference

in the

telecommunic

It is

cheap

er and

lighter




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ation line

UNIT-IV

OVERHEAD LINE INSULATORS


1. Define voltage sag?

Ans) A voltage sag (U.S. English) or voltage dip (British English) is a short duration reduction in

rms voltage which can be caused by a short circuit, overload or starting of electric motors. A voltage

sag happens when the rms voltage decreases between 10 and 90 percent of nominal voltage for one-half

cycle to one minute.

2. What is the significance of stringing chart? Ans) For use in the field work of stringing the conductors, temperature-sag and temperature tension

charts are plotted for the given conductor and loading conditions. Such curves are called stringing

charts. Stringing chart is a plot of transmission line tension

and sag as a function of temperature.




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3. What are the disadvantages of corona?

Ans) Disadvantages. Corona is accompanied by a loss of energy. ... Ozone is produced by corona and

may cause corrosion of the conductor due to chemical action. The current drawn by the line due

to corona is non-sinusoidal and hence non-sinusoidal Voltage drop occurs in the line.Feb 26, 2018

4. What is the significance of shunt compensation? Ans) It is therefore important to supply/absorb excess reactive power to/from the network. Shunt

compensation is one possible approach of providing reactive power support. A static var compensator (

SVC ) is the first generation shunt compensator. It has been around since 1960s.

5. What are the various types of insulators?

Ans) There are several types of insulators but the most commonly used are pin type, suspension type,

strain insulator and shackle insulator.

1 Pin type Insulators. Pin Type Insulator. ...

2 Suspension Type. Suspension Type. ...

3 Strain Insulators. Strain Type Insulator. ...

4 Shackle Insulators. ...

17 thoughts on “Types of Insulators”

3 MARKS QUETION AND ANSWERS 1. Explain the string efficiency

Ans) String Efficiency and methods to improve String Effeciency

Posted on September 25, 2011 by k10blogger

The ratio of voltage across the whole string to the product of number of discs and the voltage across the

disc nearest to the conductor is known as string efficiency i.e.,

where n = number of discs in the string.

String efficiency is an important consideration since it decides the potential distribution along the string.

The greater the string efficiency, the more uniform is the voltage distribution. Thus 100% string

efficiency is an ideal case for which the voltage across each disc will be exactly the same. Although it is

impossible to achieve 100% string efficiency, yet efforts should be made to improve it

as close to this value as possible.

2. What are the methods to improve the string efficiency

Ans) Methods of Improving String Efficiency

The maximum voltage appears across the insulator nearest to the line conductor and decreases

progressively as the cross arm is approached. If the insulation of the highest stressed insulator (i.e.

https://iiteeeestudents.wordpress.com/2011/09/25/string-effeciency-and-methods-to-improve-string-effeciency/

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nearest to conductor) breaks down or flash over takes place, the breakdown of other units will take place

in succession. This necessitates to equalize the potential across the various units of the string i.e. to

improve the string efficiency.

The various methods for this purpose are :

1. By using longer cross-arms. The value of string efficiency depends upon the value of K i.e.,

ratio of shunt capacitance to mutual capacitance. The lesser the value of K, the greater is the

string efficiency and more uniform is the voltage distribution. The value of K

can be decreased by reducing the shunt capacitance. In order to reduce shunt capacitance, the

distance of conductor from tower must be increased i.e., longer cross-arms should be used.

However, limitations of cost and strength of tower do not allow the use of very long cross-arms.

In practice, K = 0·1 is the limit that can be achieved by this method.

2. By grading the insulators. In this method, insulators of different dimensions are so chosen that

each has a different capacitance. The insulators are capacitance graded i.e. they are assembled in

the string in such a way that the top unit has the minimum capacitance, increasing progressively

as the bottom unit (i.e., nearest to conductor) is reached. Since voltage is inversely proportional

to capacitance, this method tends to equalise the potential distribution across the units in the

string. This method has the disadvantage that a large number of different-sized insulators are

required. However, good results can be obtained by using standard insulators for most of the

string and larger units for that near to the line conductor.

3. By using a guard ring. The potential across each unit in a string can be equalised by using a

guard ring which is a metal ring electrically connected to the conductor and surrounding the

bottom insulator. The guard ring introduces capacitance between metal fittings and the line

conductor. The guard ring is contoured in such a way that shunt capacitance currents i1, i2 etc.

are equal to metal fitting line capacitance currents i′1, i′2 etc. The result is that same charging

current I flows through each unit of string. Consequently, there will be uniform potential

distribution across the units.

3. Why do power lines sag on hot days?

Ans) In hot weather, power lines can overheat just as people and animals do. The lines are often

heavily loaded because of increased power consumption, and the conductors, which are generally made

of copper or aluminum, expand when heated. That expansion increases the slack between transmission

line structures, causing them to sag.

Transmission lines are designed to meet the requirements of state electrical codes. State codes provide

minimum distances between wires, poles, the ground and buildings. Industry standards are often more

strict and are incorporated in transmission line design, construction and maintenance. As a precaution,




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no one should be on an object or in contact with an object taller than 15 to 17 feet while under a high-

voltage line.

4. What is the purpose of insulator?

Ans) An electrical insulator is a material whose internal electric charges do not flow freely; very

little electric current will flow through it under the influence of an electric field. This contrasts with

other materials, semiconductors and conductors, which conduct electric current more easily. The

property that distinguishes an insulator is its resistivity; insulators have higher resistivity than

semiconductors or conductors.

A perfect insulator does not exist, because even insulators contain small numbers of mobile charges

(charge carriers) which can carry current. In addition, all insulators become electrically conductive when

a sufficiently large voltage is applied that the electric field tears electrons away from the atoms. This is

known as the breakdown voltage of an insulator. Some materials such as glass, paper and Teflon, which

have high resistivity, are very good electrical insulators. A much larger class of materials, even though

they may have lower bulk resistivity, are still good enough to prevent significant current from flowing at

normally used voltages, and thus are employed as insulation for electrical wiring and cables. Examples

include rubber-like polymers and most plastics which can be thermoset or thermoplastic in nature.

Insulators are used in electrical equipment to support and separate electrical conductors without

allowing current through themselves. An insulating material used in bulk to wrap electrical cables or

other equipment is called insulation. The term insulator is also used more specifically to refer to

insulating supports used to attach electric power distribution or transmission lines to utility

poles and transmission towers. They support the weight of the suspended wires without allowing the

current to flow through the tower to ground.

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https://en.wikipedia.org/wiki/Electrically_conductive

https://en.wikipedia.org/wiki/Electron

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https://en.wikipedia.org/wiki/Polytetrafluoroethylene

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https://en.wikipedia.org/wiki/Electrical_cable

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https://en.wikipedia.org/wiki/Electric_power_distribution

https://en.wikipedia.org/wiki/Electric_power_transmission

https://en.wikipedia.org/wiki/Utility_pole

https://en.wikipedia.org/wiki/Utility_pole

https://en.wikipedia.org/wiki/Transmission_tower




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5. Explain a) pin type insulators b) suspension type of insulators

Ans)

1 Pin type Insulators

As the name suggests, the pin type insulator is secured to the cross-arm on the pole. There is a groove on

the upper end of the insulator for housing the conductor. The conductor passes through this groove and

is bound by the annealed wire of the same material as the conductor.

Pin type insulators are used for transmission and distribution of electric power at voltages upto 33 kV.

Beyond operating voltage of 33 kV, the pin type insulators become

too bulky and hence uneconomical.




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2 Suspension Type

Suspension Type

For high voltages (>33 kV), it is a usual practice to use suspension type insulators shown in Figure.

consist of a number of porcelain discs connected in series by metal links in the form of a string. The

conductor is suspended at the bottom end of this string while the other end of the string is secured to the

cross-arm of the tower. Each unit or disc is designed for low voltage, say 11 kV. The number of discs in

series would obviously depend upon the working voltage. For instance, if the working voltage is 66 kV,

then six discs in series will be provided on the string.

Suspension Type Image


1. Show how the sag of an overhead line can be calculated in case of supports at different levels.

Ans)

Sag in overhead Transmission line conductor refers to the difference in level between the point of

support and the lowest point on the conductor.




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As shown in the figure above, a Transmission line is supported at two points A and B of two different

Transmission Towers. It is assumed that points A and B are at the same level from the ground.

Therefore as per our definition of Sag, difference in level of point A or B and lowest point O represents

the Sag.

Sag in Transmission line is very important. While erecting an overhead Transmission Line, it should be

taken care that conductors are under safe tension. If the conductors are too much stretched between two

points of different Towers to save conductor material, then it may happen so that the tension is

conductor reaches unsafe value which will result conductor to break.

Therefore, in order to have safe tension in the conductor, they are not fully stretched rather a sufficient

dip or Sag is provided. The dip or Sag in Transmission line is so provided to maintain tension in the

conductor within the safe value in case of variation in tension in the conductor because of seasonal

variation. Some very basic but important aspects regarding Sag are as follows:

1) As shown in the figure above, if the point of support of conductor is at same level from the ground,

the shape of Sag is Catenary. Now we consider a case where the point of support of conductor are at

same level but the Sag is very less when compared with the span of conductor. Here span means the

horizontal distance between the points of support. In such case, the Sag-span curve is parabolic in

nature.

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2) The tension at any point on the conductor acts tangentially as shown in figure above. Thus the tension

at the lowest point of the conductor acts horizontally while at any other point we need to resolve the

tangential tension into vertical and horizontal component for analysis purpose. The horizontal

component of tension remains constant throughout the span of conductor.

Calculation of Sag:

As discussed earlier in this post, enough Sag shall be provided in overhead transmission line to keep the

tension within the safe limit. The tension is generally decided by many factors like wind speed, ice

loading, temperature variations etc. Normally the tension in conductor is kept one half of the ultimate

tensile strength of the conductor and therefore safety factor for the conductor is 2.

Now, we will calculate the Sag in an overhead transmission line for two cases.

Case1: When the conductor supports are at equal level.

Let us consider an overhead line supported at two different towers which are at same level from ground.

The point of support are A and B as shown in figure below. O in the figure shows the lowest point on

the conductor. This lowest point O lies in between the two towers i.e. point O bisects the span equally.

Let,




Let us take any point P on the conductor. Assuming O as origin, the coordinate of point P will be (x,y).

Therefore, weight of section OP = Wx acting at distance of x/2 from origin O.

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As this section OP is in equilibrium, hence net torque w.r.t point P shall be zero.

Torque due to Tension T = Torque due to weight Wx

Ty = Wx(x/2)

Therefore, y = Wx2 / 2T ……………………….(1)

For getting Sag, put x = L/2 in equation (1)

Sag = WL2/8T

2. Show how the sag of an overhead line can be calculated in case of supports at same Level

Ans)

Case2: When the conductor supports are at unequal level.

In hilly area, the supports for overhead transmission line conductor do not remain at the same level.

Figure below shows a conductor supported between two points A and B which are at different level. The

lowest point on the conductor is O.

Let,


H = Difference in level between the two supports


X1 = Horizontal distance of point O from support A

X2 = Horizontal distance of point O from support B


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From equation (1),

Sag S1 = WX12/2T

and Sag S2 = WX22/2T

Now,

S1 – S2 = (W/2T)[ X12 – X2

2]

= (W/2T)(X1 – X2)( X1 + X2)

But X1 + X2 = L …………………….(2)

So,

S1 – S2 = (WL/2T)(X1 – X2)

X1 – X2 = 2(S1 – S2)T / WL

X1 – X2 = 2HT / WL (As S1 – S2 = H)

X1 – X2 = 2HT / WL ………………..(3)

Solving equation (2) and (3) we get,

X1 = L/2 – TH/WL

X2 = L/2 + TH/WL

By putting the value of X1 and X2 in Sag equation, we can easily find the value of S1 and S2.

The above equations for Sag are only valid in ideal situation. Ideal situation refers to a condition when

no wind is flowing and there is no any effect of ice loading. But in actual practise, there always exists a

wind pressure on the conductor and as far as the ice loading is concerned, it is mostly observed in cold

countries. In a country like India, ice loading on transmission line is rarely observed.

3. Show how the effect of wind and ice loading are taken into account while determining the sgans

stress of an overhead line conductor.

Ans) Effect of Wind and Ice Loading on Sag:

Coating of ice on conductor (it is assumed that ice coating is uniformly distributed on the surface of

conductor) increases the weight of the conductor which acts in vertically downward direction. But the




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wind exerts a pressure on the conductor surface which is considered horizontal for the sake of

calculation.

As shown in figure above, net weight acting vertically downward is sum of weight of ice and weight of

conductor.

Therefore,

Here,

W = Weight of conductor per unit length

Wi = Weight of ice per unit length

Ww = Wind force per unit length

= Wind Pressure x Area

= Wind Pressure x (2d+t)x1

Note the way of calculation of Area of conductor. What I did, I just stretched the conductor along the

diameter to make a rectangle as shown in figure below.

Thus from equation (1),

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Sag = WtL2/2T

And the angle made by conductor from vertical = tanƟ

= Ww / (W+Wi)

4. Derive expressions for sag and tension in a power conductor strung between to Supports at

equal heights taking into account the wind and ice loading also

Ans) String Efficiency:

The total voltage applied across the string of suspension type insulators is not equally distributed across

all discs. In this distribution of voltage, disc nearest to the conductor will be at higher potential than the

other discs. This unequal potential distribution is undesirable and is usually expressed in terms of string

efficiency

What Is String Efficiency?

"The ratio of voltage across the whole string to the product of number of discs and the voltage across

the disc nearest to the conductor is known as string efficiency."

i.e.,String efficiency =Voltage across the string/(n ×Voltage across disc nearest to conductor)

Where n = number of discs in the string.

String efficiency is an important factor is transmission line designing. Since it decides the potential

distribution along the string. To get uniform distribution string efficiency should be high.Thus

100% string efficiency is an ideal case for which the voltage across each disc will be exactly the

same.that gives easy calculations to no.of discs to be added. but it is impossible to achieve 100% string

efficiency,yet efforts should be made to improve it as close to this value as possible.

Mathematical expression for String Efficiency:-




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Above Fig. shows the equivalent circuit for a 3-disc string. Let us suppose that self capacitance of each

disc is C. Let us further assume that shunt capacitance C1 is some fraction K of self capacitance i.e., C1

= KC. Starting from the cross arm or tower, the voltage across each unit is V1,V2 and V3 respectively

as shown.

Applying Kirchhoff’s current law to node A, we get,

I2 = I1 + i1

or V2ω C* = V1ω C + V1ω C1

or V2ω C = V1ω C + V1ω K C

∴ V2 = V1 (1 + K) ...(i)

Applying Kirchhoff’s current law to node B, we get

I3 = I2 + i2

or V3 ω C = V2ω C + (V1 + V2) ω C1

or V3 ω C = V2ω C + (V1 + V2) ω K C

or V3 = V2 + (V1 + V2)K

= KV1 + V2 (1 + K)

= KV1 + V1 (1 + K)2 since [ V2 = V1 (1 + K)]

= V1 [K + (1 + K)²]

∴ V3 = V1[1 + 3K + K²] ...(ii)

Voltage between conductor and earth (i.e., tower) is

V = V1 + V2 + V3

= V1 + V1(1 + K) + V1 (1 + 3K + K²)

= V1 (3 + 4K + K²)

∴ V = V1(1 + K) (3 + K) ...(iii)

From expressions (i), (ii) and (iii), we get,

V1/1=V2/(1+K)=V3/(1 + 3K + K²)=V/(1+K)(3=K)

∴Voltage across top unit, V1 = V/(1 + K)(3 + K)

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Voltage across second unit from top, V2 = V1 (1 + K)

Voltage across third unit from top, V3 = V1 (1 + 3K + K²)

%age String efficiency =Voltage across the string*100/(n ×Voltage across disc nearest to conductor)

=V*100/3V3

5. Why do we use high voltage to transmit electrical power?

Ans) The transport of large amounts of electrical power over long distances is done with high-voltage

transmission lines, and the question is: why high voltage? It certainly has a negative safety aspect, since

a low voltage line wouldn't be harmful (you can put your hands on a 12 V car battery, for example, you

won't even feel it; but make sure you don't put metal across the terminals, you'll get a huge current and a

nasty spark!). Electric energy is transported across the countryside with high-voltage lines because

the line losses are much smaller than with low-voltage lines.

All wires currently used have some resistance (the development of high-temperature superconductors

will probably change this some day). Let's call the total resistance of the transmission line leading from

a power station to your local substation R. Let's also say the local community demands a power P=IV

from that substation. This means the current drawn by the substation is I=P/V and the higher the

transmission line voltage, the smaller the current. The line loss is given by Ploss=I²R, or, substituting for

I,

Ploss = P²R/V²

Since P is fixed by community demand, and R is as small as you can make it (using big fat copper cable,

for example), line loss decreases strongly with increasing voltage. The reason is simply that you want

the smallest amount of current that you can use to deliver the power P. Another important note: the loss

fraction

Ploss/P = PR/V²

increases with increasing load P: power transmission is less efficient at times of higher demand. Again,

this is because power is proportional to current but line loss is proportional to current squared. Line loss

can be quite large over long distances, up to 30% or so. By the way, line loss power goes into heating

the transmission line cable which, per meter length, isn't very much heat.




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OBJECTIVE QUESTIONS

1. Which type of insulator is used on 132 kV transmission lines?

a. Pin type. [ ]

b. Disc type.

c. Shackle type.

d. Pin and shackle type.

2. Where is the strain type insulators used?

[ ]

a. At dead ends.

b. At any intermediate anchor tower.

c. On straight runs.

d. Either (a) or (b).

3. What is the dielectric strength of porcelain? [ ]

a. 55 kV/cm.

b. 60 kV/cm.

c. 75 kV/cm.

d. 80 kV/cm.




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4. The number of discs in a string of insulators for 400 kV ac over head transmission line lies in

the range of

a. 32 – 33 [ ]

b. 22 – 23

c. 15 – 16

d. 9 – 10

5. Where is the strain type of insulators used? [ ]

a. Low voltage overhead lines.

b. Dead ends

c. Change in direction of the transmission lines.

d. Both (b) and (c).

6. Which insulator is also called as spool type of insulators?

a. Pin type. [ ]

b. Shackle type.

c. Suspension type.

d. Stay insulators.

7. The voltage across the various discs of a string of suspension insulator having identical discs is

different due to

[ ]

a. Surface leakage currents.

b. Series capacitance.

c. Shunt capacitance to ground.

d. Series and shunt capacitance.

8. On what factor does the string efficiency of a string of suspension insulators dependent?

[ ]

a. Size of the insulator.

b. Number of discs in the string.

c. Size of tower.

d. None of these.

9. The string efficiency of a high voltage line is around

[ ]




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a. 100 %

b. 80 %

c. 40 %

d. 10 %.

10. What is the voltage across the second unit from the top in case of a suspension type insulator?

[ ]

a. V2 = V1 (3 + 4K)

b. V2 = V1 (1 + K)

c. V2 = V1 (1 + K2)

d. None of these.

KEY

S.no 1 2 3 4 5 6 7 8 9 10

Ans b b B b d b c b a b

FILL IN THE BLANKS

1. A 3 phase over head transmission line is supported by three discs of suspension insulators. The

potentials across the first and second insulator are 8 and 11 kV respectively. What is the line

voltage and string efficiency -------------------

2. What is the purpose of guard ring -----------------------

3. Which shielding is called the static shielding of the string--------------------

4. The insulators fail due to -------------------




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5.The number of discs required for 220 kV ac over head transmission line in clear atmosphere is---

-------------------------

6. What is the value of the wave front and the wave tail of the standard lightning impulse wave

used in the impulse voltage withstand test

7. Conductor sag should be kept------------------

8. What is the minimum safety factor in respect of the conductor tension-------------------

9. Safety factor is the ratio of--------------------

10. What is the effect of temperature rise on the over head lines ---------------------

S.no 1 2 3 4 5 6 7 8 9 10

Ans 0.375

and

68.3

%

Reduce the

earth

capacitance

of the

lowest unit.

Using

the

guard

rings.

Flash

over.

12-14 1.2 μ

sec

and

50 μ

sec.

Minimum

2 Breaking

stress to

working

stress.

Increase

the sag

and

decrease

the

tension.

UNIT-V

UNDERGROUND CABLES




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1. What is the purpose of guard ring?

Ans) Static Shielding is also termed as Guard Ring. This method uses a large metal ring surrounding the

bottom insulator unit and connected to the line. This ring is called a grading or guard ring which gives a

capacitance which will cancel the charging current of ground capacitance.

2. What is meant by serving of a cable?

Ans) What is "serving " of a cable? i mean something which covers the utmost exterior part of it....

also there are some other definitions used in xlpe cables

1- binder

2-"extruded" pvc sheath (what is exactly extrusion with regards to cables)

3- semiconductor XLPE screen (what good will make a semiconductor in a cable?)

3. What are the benefits of PVC over other materials?

Ans) Strong and lightweight

PVC's abrasion resistance, light weight, good mechanical strength and toughness are key technical

advantages for its use in building and construction applications.

Easy to install PVC can be cut, shaped, welded and joined easily in a variety of styles. Its light weight reduces manual

handling difficulties.

Durable

PVC is resistant to weathering, chemical rotting, corrosion, shock and abrasion. It is therefore the

preferred choice for many different long-life and outdoor products. In fact, medium and long-term

applications account for some 85 per cent of PVC production in the building and construction sector.

4. Where CSA sheath is used in cables?

Ans) Each of the current carrying conductors in the "core" is insulated by an individual thermoplastic

sheath colored to indicate the purpose of the conductor concerned. The Earth conductor may also be

covered with Green/Yellow (or Green only) insulation, although, in some countries, this conductor may

be left as bare copper. With cables where the current carrying conductors are of a large Cross Sectional

Area (CSA) and current carrying capacity, the Protective Earth conductor may be found to be of a

smaller CSA, with a lower continuous current carrying capacity. The conductors used may be solid in

cross-section or multi-stranded.




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5. What is the main purpose of Armouring?

Ans) An armoured cable consisting of a single conductor should never be earthed on both ends as the

external sheath allows low protection against energy flow. When both the ends of a single conductor

armoured cable are grounded, the energy can flow from the armour to the ground and then back to the

armour.


1. What is the function of sheath in a cables?

Ans) A power cable is an electrical cable, an assembly of one or more electrical conductors, usually held

together with an overall sheath. The assembly is used for transmission of electrical power. Power cables

may be installed as permanent wiring within buildings, buried in the ground, run overhead, or exposed.

A power cable is an electrical cable, an assembly of one or more electrical conductors, usually held

together with an overall sheath. The assembly is used for transmission of electrical power. Power cables

may be installed as permanent wiring within buildings, buried in the ground, run overhead, or exposed.

In physics and electrical engineering, a conductor is an object or type of material that allows the flow of

an electrical current in one or more directions. Materials made of metal are common electrical

conductors. Electrical current is generated by the flow of negatively charged electrons, positively

charged holes, and positive or negative ions in some cases.

In order for current to flow, it is not necessary for one charged particle to travel from the machine

producing the current to that consuming it. Instead, the charged particle simply needs to nudge its

neighbor a finite amount who will nudge its neighbor and on and on until a particle is nudged into the

consumer, thus powering the machine. Essentially what is occurring here is a long chain of momentum

transfer between mobile charge carriers; the Drude model of conduction describes this process more

rigorously. This momentum transfer model makes metal an ideal choice for a conductor as metals,

characteristically, possess a delocalized sea of electrons which gives the electrons enough mobility to

collide and thus effect a momentum transfer.

2. what are the properties of insulating materials

Ans) The greater the temperature difference, the faster the heat flows to the colder area.

Conduction. ...

Convection. ...

Radiation. ...

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Heat energy.

Thermal conductivity.

Coefficient of thermal conductance “ l” (kcal m-2 h-1 °C-1)

Thermal resistivity.

Thermal resistance (R-value)

3. Explain electrical power cable

Ans) Electric power can be transmitted or distributed either by overhead system or by underground

cable. Cables are mainly designed as per requirement. Power cables are mainly used for power

transmission and distribution purpose. It is an assembly of one or more individually insulated electrical

conductors, usually held together with an overall sheath. The assembly is used for transmission and

distribution of electrical power. Electrical power cables may be installed as permanent wiring within

buildings, buried in the ground and run overhead or exposed. Flexible power cables are used for

portable devices, mobile tools and machinery.

4. What are the various parts of cable

Ans)

Construction of Power Cable

There are various parts of a cable to be taken care of during construction. The power cable mainly

consists of

1. Conductor

2. Insulation

3. LAY for Multicore cables only

4. Bedding

5. Beading/Armouring (if required)

6. Outer Sheath

5. Compare the merits and demerits of underground system versus overhead system.

Ans) Transmission and distribution of electric power can be carried out by overhead as well as

underground systems. Comparison between overhead and underground systems are given below:

1. Public Safety: Underground system is more safer than overhead system.

2. Initial Cost: Underground system is more expensive



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3. Flexibility: Overhead system is more flexible than underground system. In overhead system new

conductors can be laid along the existing ones for load expansion. In case of underground system new

conductors are to be laid in new channels.

4. Working Voltage: The underground system cannot be operated above 66 KV because of

insulation difficulties but overhead system can be designed for operation up to 400 KV or higher even.

5. Maintenance Cost: Maintenance cost of underground system is very low in comparison with that

of overhead system.

6. Frequency of Faults or Failures: As the cables are laid underground, so these are not easily

accessible. The insulation is also better, so there are very few chances of power failures or fault as

compared to overhead system.

7. Frequency of Accidents: The chances of accidents in underground system are very low as

compared to overhead system.


1.Explain the pin type f insulators

pin type insulator

Pin Insulator is earliest developed overhead insulator, but still popularly used in

power network up to 33KV system. Pin type insulator can be one part, two parts or

three parts type, depending upon application voltage. In 11KV system we generally

use one part type insulator where whole pin insulator is one piece of properly

shaped porcelain or glass. As tha leakage path of insulator is through its surface, it is

desirable to increase tha vertical length of tha insulator surface area for lengthaning

leakage path. In order to obtain lengthy leakage path, one, tow or more rain sheds or

petticoats are provided on tha insulator body. In addition to that rain shed or

petticoats on an insulator serve anothar purpose. Thase rain sheds or petticoats are so

designed, that during raining tha outer surface of tha rain shed becomes wet but tha

inner surface remains dry and non-conductive. So thare will be discontinuations of

conducting path through tha wet pin insulator surface.




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In higher voltage like 33KV and 66KV manufacturing of one part porcelain pin

insulator becomes difficult. Because in higher voltage, tha thickness of tha insulator

become more and a quite thick single piece porcelain insulator can not

manufactured practically. In this case we use multiple part pin insulator, where a

number of properly designed porcelain shells are fixed togethar by Portland cement

to form one complete insulator unit. For 33KV tow parts and for 66KV three parts

pin insulator are generally used.

2.Explain Designing consideration of Electrical Insulator

Tha live conducter attached to tha top of tha pin insulator is at a potential and

bottom of tha insulator fixed to supporting structure of earth potential. Tha

insulator has to withstand tha potential stresses between conducter and earth. Tha

shortest distance between conducter and earth, surrounding tha insulator body,

along which electrical discharge may take place through air, is known as flash over

distance.

3. When insulator is wet, its outer surface becomes almost conducting. Hence tha

flash over distance of insulator is decreased. Tha design of an electrical insulator

should be such that tha decrease of flash over distance is minimum when tha

insulator is wet. That is why tha upper most petticoat of a pin insulator has umbrella

type designed so that it can protect, tha rest lower part of tha insulator from rain.

Tha upper surface of top most petticoat is inclined as less as possible to maintain

maximum flash over voltage during raining.

4. To keep tha inner side of tha insulator dry, tha rain sheds are made in order that

thase rain sheds should not disturb tha voltage distribution thay are so designed

that thair subsurface at right angle to tha electromagnetic lines of force

3. Explain grading of cables in distribution system

Ans) Grading of cable is the process of achieving uniform distribution of dielectric stress or voltage

gradient in a dielectric of cable.

Voltage gradient or dielectric stress is maximum at the surface of the conductor and minimum at the

inner surface of a sheath. Put in another way, the dielectric stress decreases from the surface of

conductor to the sheath. This non – uniform distribution of dielectric stress leads to insulation break

down in the cable. To avoid this insulation break down, it is required to distribute the dielectric stress

equally throughout the dielectric. The uniform distribution of dielectric stress is achieved by grading the

cables.




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There are two methods of grading of cables. They are,

1. Capacitance grading and

2. Inter sheath grading

Capacitance grading:

Capacitance grading is the process of using various layers of dielectrics with each dielectric having their

own permittivity. The permittivity values should be in decreasing order from the surface of the

conductor to the sheath of a cable. The product of permittivity of dielectric and radius from centre of the

conductor to the particular layer of a dielectric should be constant at every layer of a dielectric.

Figure 1 is the capacitance graded cable. Here, the radius of the conductor is r. Three dielectric layers

are used in this cable. Consider the permittivity of first dielectric layer is ε1, the distance from the

surface of the conductor to first layer is r1, the permittivity of second dielectric is ε2, the distance from

the surface of the conductor to second layer is r2, the permittivity of third dielectric is ε3, the distance

from the surface of the conductor to third layer is R.

The relative permittivity values and their distances are and . The uniform

dielectric stress can be achieved by maintaining the product of permittivity and radius of each dielectric

as same, . The uniform dielectric stress cannot be achieved by capacitance grading

alone. By capacitance grading alone an infinite number of dielectrics will be needed to achieve uniform

dielectric stress. But it is not practically possible.

Inter sheath grading:

In this grading of a cable, a homogeneous dielectric is used. This homogenous dielectric is divided into

several layers. Mechanical inter sheaths are placed in between sheaths and conductors. The inter sheaths

are then held at adequate potentials which are placed in between conductor potential and earth potential.

However, fixing of potentials at inter sheaths is a difficult task.




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4.Breifly explain the insulation testing

Ans) Installation testing

The most important reason for testing insulation is to insure public and personal safety. By performing a

high dc voltage test between de-energized current- carrying (hot), grounded, and grounding conductors,

you can eliminate the possibility of having a life-threatening short circuit or short to ground. This test is

usually performed after the initial installation of the equipment. This process will protect the system

against miswired and defective equipment, and it will insure a high quality installation, customer

satisfaction, and protect against fire or shock.

Maintenance testing

The second most important reason for insulation testing is to protect and prolong the life of electrical

systems and motors. Over the years, electrical systems are exposed to environmental factors such as dirt,

grease, temperature, stress, and vibration. These conditions can lead to insulation failure, resulting in

loss of production or even fires. Periodic maintenance tests can provide valuable information about the

state of deterioration and will help in predicting possible failure of the system. Correcting problems will

result not only in a trouble-free system, but will also extend the operating life for a variety of equipment.

Before measuring

In order to obtain meaningful insulation resistance measurements, the electrician should carefully

examine the system under test. The best results are achieved when:

1. The system or equipment is taken out of service and disconnected from all other circuits,

switches, capacitors, brushes, lightning arrestors, and circuit breakers. Make sure that the

measurements are not affected by leakage current through switches and overcurrent protective

devices.

2. The temperature of the conductor is above the dew point of the ambient air. When this is not the

case, a moisture coating will form on the insulation surface, and, in some cases will be absorbed

by the material.

3. The surface of the conductor is free of carbon and other foreign matter that can become

conductive in humid conditions.

4. Applied voltage is not too high. When testing lowvoltage systems; too much voltage can

overstress or damage insulation.

5. The system under test has been completely discharged to the ground. The grounding discharge

time should be about five times the testing charge time.

6. The effect of temperature is considered. Since insulation resistance is inversely proportional to

insulation temperature (resistance goes down as temperature goes up), the recorded readings are

altered by changes in the temperature of the insulating material. It is recommended that tests be

performed at a standard conductor temperature of 20 °C (68 °F). As a rule of thumb, when

comparing readings to 20 °C base temperature, double the resistance for every 10 °C (18 °F)

above 20 °C or halve the resistance for every 10 °C below 20 °C in temperature. For example, a

one-megohm resistance at 40 °C (104 °F) will translate to four-megohm resistance at 20 °C (68




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°F). To measure the conductor temperature, use a non-contact infrared thermometer such as the

Fluke 65.

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OBJECTIVE TYPE QUESTIONS

1. How many cores are used in a cable for the transmission of voltages upto 66 kV?

a. Single core. [ ]

b. Two core.

c. Three core.

d. All of the above.

2. Why is the single core cables not provided with armouring?

a. Avoids excessive loss in the armour. [ ]

b. Make the cable more flexible.

c. Make the cable non hygroscopic.

d None of the above.

3. Which among the following cables are generally suited for the voltages upto 11 kV?

a. Belted cables [ ]

b. Screened cables

c. Pressure cables

d. None of these.

4. Which material is suitable for the manufacture of armour in a single core cable?

a. Magnetic material. [ ]

b. Non magnetic and non conducting material.

c. Non magnetic and conducting material.

d. Magnetic and non conducting material.

5. Why the belted type cable constructions are not suitable for voltages exceeding 22 kV?

a. Development of both radial and tangential stress. [ ]

b. Formation of vacuous spaces and voids on loading and unloading owing to non homogeneity of

dielectric in belted construction.

c. Local heating caused by power loss at the centre filling.

d. All of the above.




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6. The cable best suited for the transmission of voltages from 33 kV to 66 kV is_______________.

a. Belted cables [ ]

b. Screened cables

c. Pressure cables

d. None of these.

7. What is/ are the advantages of using H-type cables?

a. The metallic screens assist in complete impregnation of the cable with the compound. [ ]

b. The metallic screens increase the heat dissipating power of the cable.

c. The lead sheaths in H type are thicker then S.L type cables.

d. All of these.

8. A single core cable has a conductor diameter of 1 cm and the insulation thickness of 0.4 cm. If

the specific resistance of insulation is 5.5 * 1014 Ω -cm, what will be the insulation resistance for a

length of 3 km?

a. 0.234 * 109 Ω [ ]

b. 0.257 * 109 Ω

c. 0.352 * 109 Ω

d. 0.211 * 109 Ω

9. A single core cable has a conductor diameter of 1 cm and the internal sheath diameter of 1.8

cm. If impregnated paper of relative permittivity 4 is used as the insulation, calculate the

capacitance for 1 km length of cable?

a. 0.378 μ F [ ]

b. 0.264 μ F

c. 0.549 μ F

d. 0.78 μ F

10. Q1. What is the maximum stress in the insulation for a 33 kV single core cable with a diameter

of 1 cm and a sheath of inside diameter 4 cm?

a. 50.61 kV / cm rms [ ]

b. 45.231 kV / cm rms

c. 47.61 kV / cm rms

d. 49.231 kV /cm rms




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KEY

S.no 1 2 3 4 5 6 7 8 9 10

Ans c a a c d B d b a c

FILL IN THE BLANKS

1. What is the maximum stress in the insulation for a 33 kV single core cable with a diameter of 1

cm and a sheath of inside diameter 4 cm ------------------------------

2. What will be the insulation thickness for a conductor of diameter 2 cm, with maximum and

minimum stress 40 kV / cm rms and 10 kV / cm rms respectively

--------------------------------------

3. What will be the most economical value of diameter of a single core cable to be used on 50 kV,

single phase system, when the maximum permissible stress is not exceeding 50 kV / cm ------------

4.In a 3 core cable, the capacitance between two conductors is 3 μF. What will be the capacitance

per phase----------------

5. What is the safe working temperature for a conductor in case of armoured cables ------------------

-

6. Armouring is provided above the bedding. The armouring consists of one or two layers of which

wire or tape ---------------------------

7.Why are the inter sheaths in cables used------------------------




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8. A layer similar to bedding is provided on the armouring to protect the whole cable from all

atmospheric conditions. Which layer is this-------------------------

9.The insulation resistance of a cable of length 10 km is 1M Ω . For a length of 100 km of the same

cable, what will be the insulation resistance ------------------------

10. What is the limit of the conductor cross section when paper insulation is used

---------------------------------------------------

KEY

S.no 1 2 3 4 5 6 7 8 9 10

Ans 47.61

kV /

cm

rms

3

cm

2.828

cm

6

μ

F

65°

C

Galvanized

steel wire.

Provides

proper stress

distribution.

Serving 0.1

M

ohm

600

mm2




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Accredited by NAAC

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Accredited by NAAC

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44


In higher voltage, beyond 33KV, it becomes uneconomical to use pin insulator because size, weight of tha insulator become more. Handling and replacing bigger size single unit insulator are quite difficult task. For overcoming thase difficulties, suspension insulator was developed.

In suspension insulator numbers of insulators are connected in series to form a string and tha line

conducter is carried by tha bottom most insulator. Each insulator of a suspension string is called

disc insulator because of thair disc like shape.


5. Each suspension disc is designed for normal voltage rating 11KV(Higher voltage rating

15KV), so by using different numbers of discs, a suspension string can be made suitable for

any voltage level.

6. If any one of tha disc insulators in a suspension string is damaged, it can be replaced much

easily.

7. Mechanical stresses on tha suspension insulator is less since tha line hanged on a flexible

suspension string.

8. As tha current carrying conducters are suspended from supporting structure by suspension




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string, tha height of tha conducter position is always less than tha total height of tha

supporting structure. Tharefore, tha conducters may be safe from lightening.




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45


4. Suspension insulator string costlier than pin and post type insulator.

5. Suspension string requires more height of supporting structure than that for pin or post

insulator to maintain same ground clearance of current conducter.

6. Tha amplitude of free swing of conducters is larger in suspension insulator system, hence,

more spacing between conducters should be provided.

Strain insulator

When suspension string is used to sustain extraordinary tensile load of conducter it is referred as

string insulator. When thare is a dead end or thare is a sharp corner in transmission line, tha line has

to sustain a great tensile load of conducter or strain. A strain insulator must

have considerable mechanical strength as well as tha necessary electrical insulating properties.

Shackle Insulator or Spool Insulator




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Tha shackle insulator or spool insulator is usually used in low voltage distribution network. It can be

used both in horizontal and vertical position. Tha use of such insulator has decreased recently after

increasing tha using of underground cable for distribution purpose. Tha tapered hole of tha spool

insulator distributes tha load more evenly and minimizes tha possibility of breakage when heavily

loaded. Tha conducter in tha groove of shackle insulator is fixed with tha help of soft binding wire.




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Q




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Q




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