propagation of transmission level switching overvoltages.pdf

Propagation of transmission level switching overvoltages, related to the distribution level

a product of :Projectgroup - EPSH-810Institute of Energy TechnologyAalborg University, AAU31st may 2007

Institute of Energy TechnologyAalborg University, AAU

Title:

Propagation of transmission level switching overvoltages,

related to the distribution level

Theme:

Control and Surveillance

of Electrical Power Systems

Project period:

1st feb 2007 - 31st may 2007

Group:

EPSH-810

Members:

Víðir Már Atlason

Jakob Kessel

Supervisor:

Claus Leth Bak

Publications: 5

Number of pages: 193

Finished: May 31st 2007

Institute of Energy TechnologyPontoppidanstræde 1019220 Aalborg ØstTelephone +45 96 35 92 40http://www.iet.aau.dk

SYNOPSIS:This project focuses on the problem with over-voltages, generated by reconnection of a grid seg-ment in a transmission system. In order to eval-uate this problem, the 150 kV grid segment be-tween Ferslev and Tinghøj is examined which isa combination of two underground cable sections,one overhead line section and one shunt reactor.Measurements are performed on the grid seg-ment between Ferslev and Tinghøj, manifestingthe presence of switching overvoltages. An anal-ysis of these overvoltages is performed. First, thefrequencies experienced after switch-on are evalu-ated theoretically. Thereafter, a simplified modelof the 150 kV grid is analyzed mathematically,before a model of the system is implemented inthe simulation software PSCAD.Both the mathematical analysis and the simula-tion in PSCAD shows lack of damping comparedwith the measurements on the system. This dif-ference is considered to originate from an insuffi-cient simulation model for the reactor componentand frequency independence of the short circuitresistance.Two proposals for minimizing the switching over-voltages on the 150 kV level are given: the zerocrossing closure principle and the pre-insertionresistor principle. Simulation of both proposalsshows limitation of the experienced overvoltagesto a maximum of 6 kV, compared with a maxi-mum of 64 kV without transient limitation.From an evaluation of the switching overvoltagesin the system between Ferslev and Tinghøj, it wasconcluded that such solution is not necessary.

Institut for EnergiteknikAalborg Universitet, AAU

TITEL:

Udbredelse af koblingsoverspændinger på transmissionsniveau

relateret til distributionsniveau

TEMA:

Regulering og overvågning

af elektriske anlæg

Projekt periode:

1. feb 2007 - 31. maj 2007

Gruppe:

EPSH-810

Medlemmer:

Víðir Már Atlason

Jakob Kessel

Vejleder:

Claus Leth Bak

Oplag: 5

Antal sider: 193

Afsluttet: 31. maj 2007

Institut for EnergiteknikPontoppidanstræde 1019220 Aalborg ØstTelefon 96 35 92 40http://www.iet.aau.dk

SYNOPSIS:Dette projekt fokuserer på problemet omkringoverspændinger forårsaget af genindkobling afen transmissionslinje. Til behandling af prob-lemet, undersøges 150 kV linjen mellem Ferslevog Tinghøj, bestående af to kabelstykker, etlinjestykke samt en reaktor.Målinger på linjen mellem Ferslev og Tinghøj,viser tilstedeværelsen af koblingsoverspændinger.Der foretages derfor en analyse af disse over-spændinger. Først foretages en teoretisk evaluer-ing af de frekvenser der optræder efter indkoblin-gen. Herefter udføres en matematisk analyse af ensimplificeret model for 150 kV linjen, før en modelfor systemet implementeres i simuleringsprogram-met PSCAD.Både den matematiske analyse samt simuleringeni PSCAD viser mangel på dæmpning, sammen-lignet med målinger på systemet. Denne forskelantages at stamme fra en utilstrækkelig model forreaktoren, samt manglende frekvensafhængighedi systemets kortslutningsmostand.Der opstilles to forslag til minimering af in-dkoblingsoverspændingerne på 150 kV linjen:et omhandlende nulgennemgangen for spænd-ingskurverne samt et omhandlende en oplad-ningsmodstand. Simuleringer af de to forslagviser en begrænsning af overspændingerne tilmaksimalt 6 kV, sammenlignet med 64 kV udentransient begrænsning.Ud fra en evaluering af overspændingerne på sys-tem mellem Ferslev og Tinghøj konkluderes det,at en begrænsning af overspændingerne ikke ernødvendig her.

Preface

This report is prepared as an 8th semester study by project group EPSH-810 on Institute of

Energy Technology, Aalborg University. The report is produced in the period from the 1st of

February to the 31th of May 2007, under the project theme Control and Surveillance of ElectricalPower Systems.

The report contains eight chapters, and eight appendices, supporting the report. Each chapter

is structured by the IMRaD model. Thereby, each chapter starts with a Introduction where

the objectives are presented, then Methods and Results are stated, and Discussion of the most

significant findings given, leading over to next chapter.

Citations are given in square brackets, first with a unique number followed by name of the author

and the used page number(s), e.g. [1, S. Vørts, p. 33].

The attached CD, contains files used for calculations and simulations, articles referred to in the

report, data regarding the shunt reactor and transformers, and the report self in .pdf and .ps

form. Details of the CD content, are given on page H-65.

The authors wish to express appreciation to the following:

Kim Søgaard and Lars Rasmussen from Energinet.dk and NV Net respectively, for informations

regarding different aspects of the reactor, the line sections and the substations, David Eklundfrom ABB for quickly responding to request for informations regarding the shunt reactor, and

Stefan Sörensen from HEF for good response regarding data for transformers.

Víðir Már Atlason Jakob Kessel

Contents

1 Introduction 1

2 Problem analysis 5

2.1 Problem statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Description and modelling of the 150 kV grid 9

3.1 Main system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.2 150 kV source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.3 Circuit breaker . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.4 Underground cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.5 Overhead transmission lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.6 Shunt reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.7 Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.8 Three phase representation of the grid . . . . . . . . . . . . . . . . . . . . . . . 29

3.9 Summary & discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 Analysis of switching overvoltages on the 150 kV grid 31

4.1 Estimation of the transient behaviour . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2 Simplified model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.3 Mathematical analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.4 PSCAD simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.5 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46


5 Configuration and simulation in PSCAD 57

5.1 PSCAD model subparts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.2 Verification of reactor model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71


viii CONTENTS

6 Evaluation of the PSCAD simulation 83

6.1 Estimation of cable and line models . . . . . . . . . . . . . . . . . . . . . . . . . 83

6.2 Estimation of models for shunt reactor and short circuit impedance . . . . . . . 87

6.3 Correction of cable capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

6.4 Estimation of missing resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . 93


7 Proposal for minimizing switching overvoltages 97

7.1 Evaluation of switching overvoltages at the 150 kV level . . . . . . . . . . . . . . 97

7.2 Evaluation of propagation of switching overvoltages . . . . . . . . . . . . . . . . 97

7.3 Proposal for minimization of switching overvoltages . . . . . . . . . . . . . . . . 98


8 Conclusion 107

8.1 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

8.2 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Bibliography 110

Appendix A-1

A Parameter determination for the overhead line system A-1

A.1 Series impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-1

A.2 Shunt admittance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-14

A.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-24

B Parameter determination in the shunt reactor B-25

B.1 Theoretical background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B-25

B.2 Inductance determination for the shunt reactor . . . . . . . . . . . . . . . . . . . B-27

B.3 Calculation of the inductances in the shunt reactor . . . . . . . . . . . . . . . . B-32

B.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B-34

C Measurements of transmission line switching C-35

C.1 General test procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-35

C.2 Measurements on 150 kV and 20 kV level . . . . . . . . . . . . . . . . . . . . . . C-36

C.3 Measurements on 0,4 kV level . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-41

C.4 Conclusion & discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-44

CONTENTS ix

D Transient recording using Omicron CMC 256-6 D-47

D.1 Hardware . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D-47

D.2 Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D-47

D.3 Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D-48

D.4 Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D-48

E Single-line diagrams E-49

E.1 Ferslev substation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E-49

E.2 Tinghøj substation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E-50

E.3 Explanation of symbols in single-line diagrams . . . . . . . . . . . . . . . . . . . E-51

F Parameter matrices for PSCAD F-53

G M-files used for calculations G-57

G.1 M-files for determination of parameters in the overhead line system . . . . . . . G-57

G.2 M-files for determination of parameters in the shunt reactor . . . . . . . . . . . G-59

G.3 M-files for mathematical analysis . . . . . . . . . . . . . . . . . . . . . . . . . . G-60

G.4 M-files for miscellaneous purposes . . . . . . . . . . . . . . . . . . . . . . . . . . G-63

H CD content H-65

Chapter 1

Introduction

Electricity is one of the cornerstones of modern society. Over the past decades, a steady rise

of electric power consumption has been experienced, which is also forecasted for the foreseeable

future. This is shown in figure 1.1.

TWh

Wind Decentral Central Hp Central Consumption

1990 1992 1994 1996 1998 2000 2002 2004 2007 2009 2011 2013 2015

0

10

20

30

40

50

60

Figure 1.1: Development in electricity production and consumption in Denmark the last 15 years,

including forecast for the next 10 years [2, www.miljorapport2006.dk]. Hp Central (High price central),

means that central production is activated when export prices are high.

In order to keep up with the increasing power consumption, the electric power generation must

respond. As shown in figure 1.1 the increment of power generation consist mainly of wind energy

and decentralized production. This was decided politically, in order to reduce the emission of

CO2. This is in accordance with sharper focus on environmental issues, visual pollution amongst

other, where renewable power sources for electric generation, and grounding of transmission lines

are to be favoured.

In the early years of electric power systems, the power was consumed close to its generation.

Nowadays electric power systems are connected across national borders in large networks, to

ensure high reliability of electric power supply to the costumer. Generally, the electric power

system is divided into three subparts, namely: power generation, power transmission and power

distribution, shown in figure 1.2.

An important aspect in the design of electric power systems is the efficiency. Higher efficiency

indicates lower prices for the consumers. A key issue in order to obtain a high efficiency in the

transmission system, is to avoid transmission of reactive power. Beside reactive power caused

by the consumers consumption, the reactive power is caused by inductances and capacitances in

the transmission system. A transmission system, consisting of overhead lines, is predominantly

inductive in normal drift as the inductances are larger than the capacitances, and the system

consumes thereby reactive power. Underground cables are, on the other hand, mainly capacitive,

and are therefore seen as producing reactive power.

In order to minimize transmission of reactive power, compensation of the production or

consumption of reactive power is made.

2 Introduction

Generation

Step-up transformer Step-down transformer Consumer level transformer

Transmission ConsumerDistribution

Figure 1.2: Graphical overview of the electrical power system

By law, the generation and transmission of electric energy, is to be separated. As the payload

from the power plant is active power, the production is according to this, and compensation is

therefore mostly left to the transmission company. This compensation is generally approached

by distributing reactive components along the power grid. These reactive components are chosen

to cancel out the reactive power, e.g. if the grid is inductive a capacitance is added, and vice

versa if the grid is capacitive an inductance is added. These reactive components can either be

stationary connected to the grid, or connected by circuit breakers to allow disconnections.

The power consumption is dynamic in its nature, and to compensate for different kind of loads,

the reactive components are connected/disconnected as the situation demands. A connection of

a single shunt reactor e.g. would therefore be performed in a case of a capacitive grid, and again

disconnected with the grid becoming inductive.

As the transmission lines at and below the 150 kV level, are to be substituted with underground

cables, an increment of the capacitance in the grid, and thereby reactive power production will

occur. This reactive power production has to be compensated for, and the predominant method

is implementation of shunt reactors. This power factor correction, is not achieved without a cost,

as problems due to connection of these reactive components to the grid arise. These are mainly

in the form of charging/discharging currents. Connection of an uncharged reactive component,

causes large inrush current resulting in overvoltages at the moment of the switching. These

overvoltages can propagate from the transmission level to the distribution level, and thereby

affecting the consumers. As the voltage quality affects the consumers electronic devices, the

propagation of these overvoltages is an actual problem, e.g. sensitive devices could be harmed

by overvoltages, resulting in malfunction. This leads to the general problem:

Inrush currents caused by switching of grid components on the transmission level,

affect the power quality at the distribution level.

In the year 2004 it was decided by the Danish Parliament, to build an offshore windmill park at

Horns Rev, in the North sea, called Horns Rev 2. The park is placed 23 km north-west of an

existing park Horns Rev 1. The geographical configuration of the two offshore parks, and the

energy transfer systems are shown in figure 1.3. The unified energy production of the Horns Rev2, will be 215 MW.

3

Horns Rev 1

Horns Rev 2

Blåbjerg

Blåvand

Karlsgårde

Galtho

EndrupEsbjerg

Sea cable

Land cable

Overhead lines

North sea

Ringkøbing Fjord

10 km

150 kV stationCable station

Future overhead linesFuture underground cablesExisting overhead linesExisting underground cables

Figure 1.3: The geographical configuration of the two offshore parks, Horns Rev 1 and Horns Rev 2 [3,

hornsrev2.energinet.dk].

The electricity produced by the windmills is collected at a transformer platform, transforming

their voltages, from 33 kV to 150 kV. From the transformer platform the electricity of the windmill

park is transported to the mainland by a 42 km long 150 kV sea cable. In a cable station, placed

at Blåbjerg Klitplantage, the sea cable is connected to a 30 km long 150 kV underground cable,

establishing connection to Galtho in the Tistrup region. In Galtho the cable will be connected

to a 26 km long overhead line. The overhead line connects to an existing transformer station

in Endrup, north east of Esbjerg, where the 150 kV is transformed to 400 kV, enabling the

electricity from the windmill park to enter the 400 kV system [3, hornsrev2.energinet.dk].

The resulting transmission system, with two long cable sections, will consequently require a

large reactor to compensate the reactive power production. As the windmill park only produces

electricity when the wind blows, the cable and hence the shunt reactor are disconnected from the

400 kV transmission system when the windmills are shut down. When the grid is reconnected,

over-voltages can occur. This leads to the initiating problem:

Connection of a grid segment, can cause overvoltages at transmission level, that

propagate to the distribution level.

In order to clarify and improve this problem, a model of the system is needed.

As Horns Rev 2 is not constructed, it is not possible to verify a model of the system through

measurements. Instead a model is constructed, on the basis of a similar system. Simulation

results can then be compared with measurements to verify the model.

In association with Energinet.dk, it is chosen to focus on the 150 kV grid between Ferslev and

Tinghøj as shown in figure 1.4.

This system has similar characteristics as the system at Horns Rev 2, as it contains a shunt

reactor, underground cables and overhead lines.

4 Introduction

Ferslev

Haverslev

Tinghøj

Nørresundby

Aalborg

Hobro

Aars

Støvring

Hadsund

To Hornbæk

To ÅdalenTo Mosbæk

Limfjorden

Mariager Fjord

Aalborg Bugt

10 km

150 kV stationCable station

Overhead linesUnderground cables

Figure 1.4: Location of the 150 kV grid from Ferslev to Tinghøj.

The grid is subdivided into three different subsections, containing two underground cable sections

and one overhead line section. A simplified outline of this system is shown in figure 1.5.

FerslevHaverslevTinghøj

0,43 km18,93 km21 km

Reactor Section 3 Section 2 Section 1

Figure 1.5: Subsections of the system.

At the transformer station in Tinghøj, a 40 Mvar shunt reactor is mounted, to compensate the

reactive power production of the system. The system can be disconnected from the rest of the

150 kV system through circuit breakers, in each of the two transformer stations.

In the next chapter, the problem of the experienced switching overvoltages is analysed, and the

solution method presented.

Chapter 2

Problem analysis

As stated in the initial problem, the switch-on of a grid segment at transmission level can

cause overvoltages that propagate to lower voltage levels. In order to improve on this problem,

the nature of this phenomenon must be analyzed. To investigate the magnitude of the problem,

measurements were made at the moment of switch-on, of the 150 kV grid section betweenTinghøj-Ferslev1. The voltage waveforms shown in figure 2.1 were recorded in the test, where the Tinghøj-Ferslev section was isolated.

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065

−100

0

100

Vol

tage

[kV

]

Time [s]

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065

−20

−10

0

10

20

Vol

tage

[kV

]

Time [s]

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065−500

0

500

Vol

tage

[V]

Time [s]

Figure 2.1: Voltage waveforms recorded in Ferslev at the moment of switch-on. From top-to-bottom

150 kV, 20 kV, 0,4 kV.

In figure 2.1, it is apparent that transients arise after the switch-on of the section. These are most

significant at the 150 kV level, where a high frequency component is apparent for a brief period,

plus a lower frequency component. At the 20 kV and 0,4 kV levels, the high frequency component

is no longer apparent. The lower frequency component is still present on both the 20 kV and

the 0,4 kV level, though damped in comparison with the 150 kV level. After approximately one

period of the fundamental frequency, the distortion at the 0,4 kV level is no longer present.

1The measuring reports are in appendix C on page C-35.

6 Problem analysis

The magnitudes of the recorded voltages are summarized in table 2.1.

Level [kV] Peak voltage Peak voltages at Percent

before switch-on [kV] switch-on instant [kV]

150 ±134 +159 -162 20,8%

20 ±16,8 +17,3 -17,4 3,5%

0,4 ±0,316 +0,381 -0,332 18,5%

Table 2.1: Recorded voltages from the test in Ferslev, sees in relation to figure 2.1.

From figure 2.1 and table 2.1, it is apparent that the switching overvoltages propagate to the

distribution level.

2.1 Problem statement

The switching overvoltages experienced when grid sections are switched-on, are unwanted as they

affect the consumer, and can in worst case result in damaged equipment. It is therefore desirable

to investigate this phenomenon, and provide a proposal for minimization of the switching

overvoltages.

Solution Method

In order to examine the problem, a dynamic model is constructed, and an analysis of the 150

kV grid between Ferslev and Tinghøj is performed. First a simplified mathematical model is

arranged, where the most significant parameters are included, in order to give a theoretical

understanding of the system. Later, a detailed version is implemented in the simulation software

PSCAD. In relation to the switch-on for the section, a method that minimizes the switch-on

transients is to be implemented in the model.

The main objectives for this project are listed below:

• Arrange a three phase representation of the 150 kV transmission system between Ferslev-Tinghøj, valid for the switch-on in Ferslev.

• Explain the experienced transients from a theoretical analysis viewpoint.

• Arrange a simplified dynamic model for the 150 kV system between Ferslev-Tinghøj.

• Arrange a mathematical solution for the simplified model in MatLab.

• Verify the MatLab model through measurements and simulation in PSCAD.

• Arrange a simulation of the model in PSCAD.

• Verify the PSCAD model by comparing to measurements and calculated values.

• Arrange a proposal for an optimization of the reconnection.

2.2 Limitations 7

2.2 Limitations

To limit the scope of the project the following limitations are made:

• The 400 kV system present in the overhead line section of the 150 kV grid between Ferslevand Tinghøj is neglected.

• For the simplified model, other voltage levels than 150 kV are neglected.

Further limitations are described in the respective sections.

Chapter 3

Description and modelling of the 150

kV grid

The objective of this chapter is to give a description of the system subparts, and their respective

electric equivalents. This is done in order to generate a three phase electric representation of

the 150 kV grid segment, valid for switch-on in Ferslev. A simplified line diagram of the system

is given, where all subparts can be seen, where after the system subparts are described. The

chapter concludes with a three phase electric representation of the grid.

3.1 Main system

In figure 3.1 is the single-line diagram1 for the connection between the Ferslev (FER3) and

Tinghøj (THØ3) transformer stations, and the rest of the transmission system shown.

S1 S1 S2

FER3THØ3

400 kV

60 kV

60 kV

20 kV 0,4 kV

HNB ADL1

MOSABB

ABB

40 Mvar

40 Mvar

Siemens80 MVA

KT31

KT31

Strømberg

Strømberg

80 MVA

80 MVA16 MVA 0,630 MVA

KT51ABB400 MVA

123

Reactor

Figure 3.1: Simplified single-line system of the two stations and their interconnection.

The single line diagram, shown in figure 3.1, shows the main components of the system, namely:

underground cables, overhead line, shunt reactor, and transformers. These components are

described in the following sections, along with circuit breaker and the voltage source.

3.2 150 kV source

The 150 kV grid section, is energized from Ferslev substation. When producing a model for

the 150 kV system, the remainder of the transmission system is represented by a voltage source

behind short circuit impedance, as shown in figure 3.2.

1It should be noted that the single-line diagram is simplified, the non-simplified single-line diagrams for the

two transformer stations are shown and explained in appendix E.

10 Description and modelling of the 150 kV grid

3.2.1 Electrical representation

Generally, the energization of grid sections is made through substations, whose characteristics can

vary considerably. The two extreme cases are: Energization from the busbar, that is solely fed

from generators, and energization from a busbar solely fed from long transmission lines/cables.

The former appears as a source behind an inductance, while the latter is represented by a source

behind a resistance [4, Bickford, p.97].

In this project, the energization of the Ferslev-Tinghøj grid section, is through FER3 substation.

The source impedance, seen from the 150 kV grid in Ferslev, is somewhere in between the above

mentioned extremes, as a 400/150kV transformer placed at the station, feeds the busbar from the

Nordjytland power plant, but transmission lines on the 150 kV level from Ådalen and Mosbæk,

supply short circuit power to Ferslev, as is shown in figure 3.1. The short circuit apparent power

in Ferslev substation, is not known at the moment of testing. Under normal conditions, the

short circuit impedance is mostly reactive, with angle of 6 80 to 6 85 assumed.

ua

ub

uc

Zs

Zs

Zs

Figure 3.2: The remainder of the transmission system can be equated with a voltage source and a

source impedance.

3.2.2 Electrical parameters

The voltage was measured at the 150 kV grid in Ferslev to a line voltage of 165 kV, which

will be used in simulations and calculations, as this will give more comparable results to the

measurements.

The maximum and minimum short-circuit apparent power is given by Energinet.dk, and is listed

in table 3.1.

Maximum Minimum

S 5908 MVA 2366 MVA

U 165 kV 165 kV

Table 3.1: Short circuit information for the 150 kV grid in Ferslev, given by Energinet.dk.

The maximum apparent short-circuit power is given for the following conditions:

3.2 150 kV source 11

• Intact transmission network, including AC-connection to Germany

• All available central power plants are running

• Synchronous compensation is activated for both machines

• All decentralized power plants are running

• All grid-connected wind mills are running

The minimum apparent short-circuit power is given for the following conditions:

• Three central production units are on the grid

• Synchronous compensation is not activated

• One important grid component is out of drift

• The short-circuit apparent power in the German 400 kV net is low

• Two thirds of the total decentralized power production is out

• All installed windpower is out

The short circuit impedance for each phase is given by:

Zs =U2

S(3.1)

The angle is unknown, and the short circuit impedance is therefore calculated for four different

angles, both for the maximum and minimum short circuit apparent power. The results are listed

in table 3.2.

Maximum power Minimum power

Zs = 4, 61 6 70 Ω Zs = 11, 37 6 70 Ω

Rs Ls Rs Ls

1,58 Ω 13,78 mH 3,94 Ω 34,4 mH

Zs = 4, 61 6 75 Ω Zs = 11, 37 6 75 Ω

Rs Ls Rs Ls

1,19 Ω 14,2 mH 2,98 Ω 35,6 mH

Zs = 4, 61 6 80 Ω Zs = 11, 37 6 80 Ω

Rs Ls Rs Ls

0,800 Ω 14,5 mH 2,00 Ω 36,1 mH

Zs = 4, 61 6 85 Ω Zs = 11, 37 6 85 Ω

Rs Ls Rs Ls

0,402 Ω 14,6 mH 1,00 Ω 36,5 mH

Table 3.2: Values for short circuit impedance, calculated for different angles and different apparent

power.

For a first approach, the short circuit impedance will be used for the maximum short circuit

apparent power with an angle of 80.


3.3 Circuit breaker

The circuit breaker connects the 150 kV grid segment to the rest of the 150 kV grid when switched

on.

CB

Figure 3.3: Model diagram for the circuit breaker, shown for one phase.


The electrical equivalent for the circuit breaker is shown in figure 3.4. Generally, a circuit breaker

will contain a leakage resistance when switched off, and a series resistance when switched on.

Ron

Roff

Figure 3.4: Equivalent diagram for the circuit breaker, shown for one phase.

As focus is at the switch-on instant, electric arcs are considered to be of no influence, as the

breaker closes almost instantaneously, and the breaker can therefore be considered ideal in this

operation.


In table 3.3 are the equivalent series resistances for the circuit breaker given, both for the on-state

and the off-state. These values are proposed by Energinet.dk.

Ron 100 µΩ

Roff 1 MΩ

Table 3.3: Resistances in the circuit breaker in its two states.

3.4 Underground cable

There are two underground cable sections in the system, a short section of 0,43 km (Section

1) connecting the overhead lines with the transformer station in Ferslev, and a 21 km section

(Section 3) connecting the cable station in Haverslev with the transformer station in Tinghøj.In these two sections, the cables are of the same type: 1200 mm2 aluminium conductors, with

each phase screened with 95 mm2 copper conductors. The maximum operating voltage for the

cable is 170 kV and the current capacity is 750 A.

3.4 Underground cable 13

3.4.1 Physical layout

The cable is from ABB, of the type AXLJ 1x1200/95 170kV, its physical layout is shown in figure

3.5.

Conductor: Aluminium, round, compact, A=1200 mm2, D=41,5 mm

Conductor screen: Thickness 1,2 mm

Insulation: XLPE, thickness 16 mm, D=75,9 mm, ǫ= 2,3

Insulation screen: Thickness 1,0 mm

Bedding: Semi conductive tapes, thickness 0,1 mm

Metallic screen: Copper wires, A=95 mm2

Longitudinal water seal: Conductive swelling tapes, thickness 0,65 mm

Radial water seal: Aluminium laminate, thickness 0,2 mm

Outer sheath: PE, thickness 3,9 mm

Complete cable: D ≈ 92 mm

Figure 3.5: The physical layout for AXLJ 1x1200/95 mm2 170 kV cable.

As each phase is independently screened, mutual capacitances between phases can be excluded.

The capacitance consist therefore only of the capacitance between conductor and ground (C0),

as shown to the right in figure 3.6.

Grounding & Cross bonding

A triangular cable layout is used in the two cable sections. The cables are buried at a depth of

1200 mm as shown in figure 3.6.

1200 mm

ConductorIsolationScreen

C0

C0 C0

Figure 3.6: Cable duct and triangular cable layout.

In the cable section between Haverslev and Tinghøj cross bonding is applied. In cross bonding,

the metallic screen for each phase is crossed with the other phases, and grounded along the

section. By doing this, the total induced voltage in the screen will be minimized, thereby

reducing losses. The cross bonding layout for Haverslev-Tinghøj is shown in figure 3.7. At every

second joint, the cable is grounded. Joints with screen separation are crossed and grounded, but

straight-through joints are grounded without crossing.


Straight-through joint Joint & screen-seperation

Linkbox for earth-connection

of straight-through jointLinkbox for crossbonding

Haverslev Tinghøj

Figure 3.7: The cross bonding of the cable section between Haverslev and Tinghøj.

In the cable section leading into the transformer station in Ferslev, one end of the screen is

grounded. No cross bonding is used in this section, due to its short distance.


In order to obtain a three phase representation for the two underground cable sections, a general

representation for one conductor is needed. A conductor of an infinitesimal length is shown in

figure 3.8.

rdx ldx

gdx cdxu

i

x

Figure 3.8: An infinitesimal line section[1, Vørts, p. 158].

As shown in figure 3.8, the conductor contains four parameters, namely: the series resistance,

the series inductance, the shunt conductance and the shunt capacitance. These parameters

are distributed over the full length of the conductor. In order to make a less complicated

representation, the distributed parameters are lumped in one component for each parameter.

The conductor can now be represented in to ways, by the π equivalent or by the T equivalent. In

the π equivalent one half of the shunt admittance is placed at each end of the conductor, and the

full series impedance is placed in between as shown to the left in figure 3.9. In the T equivalent,

the full shunt admittance is placed in between one half of there series impedance, as shown to

the right in figure 3.9.

3.4 Underground cable 15PSfrag

u1u1 u2u2

Z

Y2

Y2

Z2

Z2

Y

Figure 3.9: Electric representation of one conductor. To the left π equivalent. To the right T equivalent.

Where:Z = R + jX is the series impedance [Ω]

Y = G + jB is the shunt admittance [S]

In the three phase representation of the cable sections, the π equivalent will be used.

Series impedance

The series impedance is determined by the series resistance and series reactance. These are

mainly dependent on conductor type, length, geometrical configuration and frequency.

Shunt admittance

The shunt admittance is determined by the conductance and the susceptance of the conductors.

The shunt conductance consists of the leakage conductance and losses due to partial discharges.

In a cable the partial discharges will consist of internal discharges in the isolation material.

The shunt susceptance is determined by the capacitance between the conductor and ground, and

the capacitance between the phase conductors. As mentioned previously, the screen in the cable

allows the capacitance between the conductors to be neglected.

Three phase representation

In order to represent a three phase system, the flux linkage between the phases the must be

taken into consideration. The flux linkage gives rise to mutual inductances between the phases,

which is included in the three phase representation for the cable section, as shown in figure 3.10.


Can

2Can

2

Cbn

2Cbn

2

Ccn

2Ccn

2

Ra

Rb

Rc

La

Lb

Lc

Gan

2Gan

2

Gbn

2Gbn

2

Gcn

2Gcn

2

Mac

Mab

Mbc

Figure 3.10: π-model for a three phase cable section with mutual inductances.

Where:Mab is the mutual inductance between conductor a and conductor b [H]

Mac is the mutual inductance between conductor a and conductor c [H]

Mbc is the mutual inductance between conductor b and conductor c [H]

3.4.3 Electrical paramters

In table 3.4 are the most important electrical characteristics from the datasheet2 listed.

a.c. resistance at 57 C 0,0330 Ω/km

d.c. resistance at 20 C 0,0247 Ω/km

Capacitance per phase 0,243 µF/km

Inductance between conductors,per phase 0,343 mH/km

Dielectric loss per phase 0,1144 kW/km

Positive sequence impedance Z1 0,0331 + j0,1077 Ω/km

Negative sequence impedance Z2 0,0331 + j0,1077 Ω/km

Zero sequence impedance Z0 † 0,1876 + j0,0706 Ω/km

Table 3.4: Electrical characteristics for the AXLJ 1x1200/95 mm2 170 kV cable.

† The conditions for which the zero sequence impedance is calculated by the producer, are

unknown. The mutual inductances for the cable are not provided.

The shunt conductance per phase can be found from the dielectric loss per phase and the phase

voltage, here calculated for section 3:

Pdiel = U2 · G (3.2)

2The datasheet can be found on the CD

3.5 Overhead transmission lines 17

G =Pdiel · l

U2(3.3)

G =0, 1144 · 21(

165·103√3

)2 = 265 nS (3.4)

The phase vales for the two cable sections are listed in table 3.5, due to the symmetric

configuration, the phase vales are equal for all phases.

Section 1 (0,43 km) Section 3 (21 km)

Rx 0,0142 Ω 0,693 Ω

Lx 0,147 mH 7,20 mH

Cxn 104,5 nF 5,10 µF

Gxn 5,42 nS 265 nS

Table 3.5: Electric parameters fro the two cable sections. x=a,b,c.

3.5 Overhead transmission lines

Section 2 is a 18,93 km long overhead line section. The conductor is of the type Martin simplex

772 mm2, with a current rating of 835 A. The ground conductor is of the type, Dorking simplex

153 mm2.


The section makes use of the Donau mast, which is shown in figure 3.11. As shown in figure

3.11, there can be two systems on the mast, which also is the case as a 400 kV system is present

at the opposite side of the tower.

41,6 m

33,5 m

24,7 m

7,2 m5,8 m

10,4 m

20,8 m

Figure 3.11: Donau mast, type D11, with different systems each side of the tower.


The simplex configuration is used in the section, meaning that there is only one conductor per

phase. In this section the conductor material is aluminium over a steel core, as shown in figure

3.12.

Aluminium

Steelr2

r1

Figure 3.12: Aluminium conductor with steel core.


The overhead line section can be represented with the π equivalent as described in section 3.4.2

on page 14. Beside the inductance between the phases, capacitance between the phases will also

be present in an overhead line system. This gives the representation shown in figure 3.10.

Can

2Can

2

Cbn

2Cbn

2

Ccn

2Ccn

2

Ra

Rb

Rc

La

Lb

Lc

Gan

2Gan

2

Gbn

2Gbn

2

Gcn

2Gcn

2

Cab

2Cab

2

Cac

2Cac

2Cbc

2Cbc

2

Mac

Mab

Mbc

Figure 3.13: π-model for a three phase line section with mutual inductances.


The provided information for the phase conductor and the ground wire used in the overhead line

section are listed in table 3.6.

3.5 Overhead transmission lines 19

Phase conductor Ground wire

Type simplex 772 mm2 153 mm2

Self GMD 0,0146 m 0,0070 m

Resistance 0,0423 Ω/km 0,2982 Ω/km

Length 18,93 km 18,93 km

Table 3.6: Parameters for the conductor and ground wire in the overhead line system.

The calculated DC resistances in table 3.6 will always have smaller value than the actual

resistance, due to the following:

• Temperature

• Skin effect

• Proximity effect

• Conductor sag

• Spiraling

These terms are considered to give a correction factor of 1,153, resulting in an AC resistance of

0,898 Ω for section 2.

The shunt conductance of the system has not been calculated, however an estimation of its value

is performed in the following. Generally, the shunt conductance in an overhead line system is

given by the leakage losses and dielectric losses.

The leakage conductance is given by the surface conductance of the isolators, from the conductors

to earth. The leakage losses on a 132 kV line have been found to vary between 0,2 and 0,6 kW/km

[5, Weedy, p. 130]. This voltage level is close to the 150 kV level, and the worst case loss of 0,6

kW/km is therefore used. This gives the following conductance for the leakage part:

Gleakage =Pleakage

U2=

0,6·103

3(

165·103√3

)2 = 22, 0 · 10−9 S/km (3.5)

The partial discharges in the overhead line are in the form of corona, which occurs as the air

near the high voltage conductor is ionized. A method to evaluate the corona losses is presented

by A.-Salam & A.-Aziz [6]. Here the three phase power losses for a 210 kV overhead line (the

closest value to the 150 kV of the system), is found to vary 1-3 W/m (1-3 kW/km). By taking

the highest value of 3 kW/km, and knowing that the losses increase with higher voltage, this

value is used as a worst case scenario. This gives the conductance per phase, per/km for the

partial discharges part:

Gpd =Ppd

U2=

3·103

3(

165·103√3

)2 = 110 · 10−9 S/km (3.6)

3The correction factor is found in appendix A.1.3 on page A-10.


The approximated worst case shunt conductance is found from the parallel configuration of the

leakage conductance and partial discharge conductance:

Gshunt = Gleakage + Gpd (3.7)

Gshunt = 22, 0 · 10−9 + 110 · 10−9 = 132 · 10−9 S/km (3.8)

The electrical parameters of the overhead line section are calculated in appendix A. In table 3.7

are the phase values for the parameters in figure 3.13 listed.

Ra 0,898 Ω La 36,2 mH Can 0,149 µF

Rb 0,898 Ω Lb 36,3 mH Cbn 0,149 µF

Rc 0,898 Ω Lc 34,4 mH Ccn 0,165 µF

Gan 2,5 µS Mab 13,6 mH Cab 27,5 nF

Gbn 2,5 µS Mac 10,9 mH Cac 11,4 nF

Gcn 2,5 µS Mbc 11,0 mH Cbc 12,7 nF

Table 3.7: Electrical parameters for the overhead line section.

3.6 Shunt reactor

The shunt reactor is mounted at the transformer station in Tinghøj, to compensate the reactive

power production in the cable. The shunt reactor is produced by ABB Transformers, and shown

in figure 3.14.

Figure 3.14: The shunt reactor placed at Tinghøj transformer station.


The shunt reactor has a five-limbed core, with 8 gaps distributed across the three middle limbs,

which are for the phase windings. These gaps are inserted to avoid saturation and have the

reactor operating in the linear region of its B/H curve. Furthermore, it helps to decrease the

3.6 Shunt reactor 21

mutual inductances between the phases. The geometry of the reactor core is shown in figure

3.15.

a

b

c

c

dd ee

f

g

h

i

j

k

l

mm

n oo

p

Figure 3.15: Geometry of the reactor core from the front.

The distances from figure 3.15 are listed in table 3.8.

a 2213 mm i 151,9 mm

b 4182 mm j 32,7 mm

c 210 mm k 10 mm

d 888 mm l 2,5 mm

e 1203 mm m 193 mm

f 600 mm n 109 mm

g 1548 mm o 76 mm

h 70 mm p 65 mm

Table 3.8: Distances from figure 3.15.

The geometry of the reactor core from above is shown in figure 3.16.

a

b

c

d

ef

Figure 3.16: Geometry of the reactor core from above.

The distances from figure 3.16 are listed in table 3.9.


a 794 mm d 1038 mm

b 4182 mm e 976 mm

c 210 mm f 600 mm

Table 3.9: Distances from figure 3.16.


In this section, the electric equivalent for the shunt reactor described. In order to arrange such

model, the following terms must be included:

• Self inductances

• Mutual inductances

• Core losses

• I2R losses in the windings

These terms are described in the following, resulting in a three phase electric representation for

the shunt reactor.

Self inductances

The reactor core has five limbs, which give the flux paths for the flux generated by phase a,

shown in figure 3.17, where it is assumed that all the flux passes through the core.

Φa Φab ΦacΦa1 Φa2

Figure 3.17: The flux paths in the reactor core for phase a.

The solid line in figure 3.17 represent the primary flux path for the flux generated by the phase

a winding and is determined by the reluctances in the core and the gaps. The self inductance

for phase a, can be found if the reluctance for the primary flux path and the number of turns in

the winding are known4.

4The inductances in the shunt reactor are determined in appendix B.


Mutual inductances

The dashed lines in figure 3.17 represents the alternative flux path for the flux generated by the

phase a winding. Here the flux passes through the limb for phase b and phase c inducing voltages

in those two phases. In the same manner phase a will be affected with a contribution from phase

b and phase c. These contributions are represented by the mutual inductances, and determined

by the reluctances and number of turns in the same manner as the self inductance.

Core losses

As a consequence of the fluctuating magnetic field in the core, losses due to eddy currents and

hysteresis will occur. The magnetic flux in the core induces circulating currents, which causes

heat. In order to reduce the loss caused by these induced currents, the core is laminated, thereby

reducing the current pathway.

The hysteresis losses occur as the magnetic field changes polarity. Due to the magnetic

characteristics of the core material, a hysteresis loop will be present, where the hysteresis losses

is determined by the area in the hysteresis loop. Hence, is a slender hysteresis loop desirable.

Usually, a compromise between high saturation and slenderness of the hysteresis loop is taken,

due to the characteristics of the core material.

Copper losses in the windings

The conductor in each winding contains a certain resistance. Hence, losses will arise due to the

current passing through the resistance of the conductor, as given by:

Pcu = I2 · Rcu (3.9)

Electric circuit model for the shunt reactor

In figure 3.18 is the electric circuit equivalent for phase a of the reactor shown.

ua(t)

ia(t)

Rcu,a

Rfe,a La

Mab Mac

+

−

Figure 3.18: Electric equivalent diagram for phase a in the shunt reactor.

Where:Rcu,a represents the I2R losses in the windings [Ω]

Rfe,a represents the losses in the core [Ω]

La is the self inductance for phase a [H]

Mab represents the mutual inductance between phase a and phase b [H]

Mac represents the mutual inductance between phase a and phase c [H]


The electric equivalent of the shunt reactor is arranged with the winding loss resistance in

series with the parallel connection of the core loss resistance and the self inductance. The

parallel connection represents that the current flowing through the core loss resistance, does not

contribute to the induction. The inductance is linking with the inductances in the two other

phases, represented by the mutual inductances Mab and Mac.

This model does not account for saturation of the reactor core, or the magnetic remanence, as

the system is assumed totally discharged at the moment of switch-on.

The representation for phase a of the shunt reactor, can be used for the two other phases as well.

In figure 3.19 is the electric circuit equivalent for the shunt reactor shown for all three phases.

ua(t)

ia(t)

ub(t)

ib(t)

uc(t)

ic(t)

Mac

Mab

Mbc

Rcu,a

Rcu,b

Rcu,c

Rfe,a

Rfe,b

Rfe,c

La

Lb

Lc

+

+

+

−

−

−

Figure 3.19: Equivalent diagram for the three phases in the shunt reactor, including mutual inductances.

3.6.3 Electrical characteristics

The rated values of the shunt reactor is listed in table 3.10.

Rated power 40 Mvar

Rated voltage 170 kV

Rated current 135,8 A

Table 3.10: Information of reactor [7, ABB Reactor].


The calculated parameters for the reactor are listed in table 3.11. Values for resistance are given

in the user manual, found on the CD. Mutual inductances are calculated from the geometry of

the reactor.

Calculated Designed Measured

Rcu,a — — 0,681 Ω

Rcu,b — — 0,680 Ω

Rcu,c — — 0,680 Ω

La 1,745 H 2,300 H 2,307 H

Lb 1,744 H 2,300 H 2,304 H

Lc 1,745 H 2,300 H 2,307 H

Mab -607 µH -32,0 mH —

Mac -434 µH -23,0 mH —

Mbc -607 µH -32,0 mH —

Table 3.11: Electrical parameters for the shunt reactor [7, ABB Reactor].

As the calculated values for the mutual inductances in the shunt reactor, are small compared

with the expected values, the expected values in the 1-2% region are used for simulations.

The losses in the reactor are given to Ploss=64,3 kW @ 170 kV. This includes both the copper

loss and the core loss for all three phases. In order to find the core loss, the copper losses must

be subtracted from the total loss:

Pcu,a = Rcu,a · I2 (3.10)

Pcu,a = 0, 6812 · 135, 82 = 12, 56 kW (3.11)

The resistance representing core loss is calculated from the loss in the core, and the voltage across

the resistance Rfe,a shown in figure 3.19:

URfe

Ufe

= Rfe (3.12)

Rfe,a =(Ua − I · Rcu,a)

2

Ploss,a − Pcu,a

(3.13)

Rfe,a =(170·103√

3− 135, 8 · 0, 681)2

64,3·103

3 − 12, 56 · 103= 1, 08 MΩ (3.14)

Rfe,a 1,08 MΩ

Rfe,b 1,08 MΩ

Rfe,c 1,08 MΩ

Table 3.12: Calculated loss resistance for the core.

With all values for the components in the reactor, shown in figure3.19 known, it is possible to

simplify this model, to the one shown in figure 3.20.


ua(t)

ia(t)

ub(t)

ib(t)

uc(t)

ic(t)

Mac

Mab

Mbc

Rreactor,a

Rreactor,b

Rreactor,c

Lreactor,a

Lreactor,b

Lreactor,c

+

+

+

−

−

−

Figure 3.20: Simplified equivalent diagram for the three phases in the shunt reactor, including mutual

inductances.

The simplification is done by calculating a replacement impedance:

Zreactor,a = Rcu,a + (Rfe,a)||(jωLa) (3.15)

Zreactor,a = 0, 681 +(

1, 08 · 106)

||(j · 2 · 50 · π · 2, 307) (3.16)

Zreactor,a = 1, 17 + j725 Ω (3.17)

This is done for all three phases resulting in the values for the components in figure 3.20 listed

in table 3.13

Rreactor,a 1,17 Ω Lreactor,a 2,307 H Mab -32,0 mH

Rreactor,b 1,17 Ω Lreactor,b 2,304 H Mac -23,0 mH

Rreactor,c 1,17 Ω Lreactor,c 2,307 H Mbc -32,0 mH

Table 3.13: Electrical parameters for the shunt reactor. For mutual inductances is estimate of 1% used.

3.7 Transformers

There are three transformers between the 150 kV level and the 0,4 kV level, namely: 150/60

kV, 60/20 kV and 20/0,4 kV. These transformers must be represented, in order to evaluate the

3.7 Transformers 27

voltages at the distributions level. The parameters needed to implement in PSCAD are presented

following.

These are:

• Rated power

• Winding voltages (line-to-line)

• Rated frequency

• Coupling type

• Leakage reactance

• No load losses

• Copper losses

3.7.1 150/60 kV transformer

The values given in table 3.14 and 3.15, are taken from the 150/60 kV transformer test report,

found on the CD.

Primary winding Secondary winding

Rated power 80 MVA 80 MVA

Rated voltage 165 kV 67 kV

Rated current 280 A 689 A

Table 3.14: Rated values for the 150/60 kV transformer.

Coupling YNd11

No-load losses 41 kW

er 0,35%

ex 12,76%

Table 3.15: Electrical parameters for the 150/60 kV transformer.

In order to find the copper losses, Pcu, the short circuit impedance of the transformer is found,

and the rated current is used to evaluate Pcu. Impedances are calculated from the high voltage

side of the transformer, as this is seen from the source.

Zk = Rk + jXk =er + jex

100· U2

S(3.18)

Zk =0, 35 + j12, 76

100· 1652

80= 0, 9843 + j35, 88 Ω (3.19)

Pcu = 3 · I2r · Rk (3.20)

Pcu = 3 · 2802 · 0, 9843 = 231, 5 kW (3.21)

Pcu =231, 5 kW

80 MVA= 0, 00289 pu (3.22)



The data in table 3.16 and 3.17, for the 60/20 kV transformer, is given by HEF net A/S. This is

to be found on the CD.


Rated power 16 MVA 16 MVA

Rated voltage 60 kV 20 kV

Rated current 154 A 462 A

Table 3.16: Rated values for the 60/20 kV transformer.

Coupling YNyn0

No load losses 7,92 kW

er 0,44%

ek 7,91%

Table 3.17: Electrical parameters for the 60/20 kV transformer.

The copper losses are calculated in the same manner as for the 150/60 kV transformer:

Pcu = 3 · 1542 ·(

0, 44

100· 602

16

)

= 70, 4 kW (3.23)

Pcu =70, 4 kW

16 MVA= 0, 0044 pu (3.24)

3.7.3 20/0,4 kV transformer

The data given for the 20/0,4 kV transformer, are limited as only the name plate was avalible.

This is to be found on the CD. Due to this, losses in the transformer are not given, but a short

circuit voltage drop of 4% is given. Hence, typical values for transformer of this size, given in [1,

Vørts, p. 207], are used. The data is listed in table 3.18 and 3.19, estimated values are annoted

with †.


Rated power 630 kVA 630 kVA

Rated voltage 24 kV 0,420 kV

Rated current 15,15 A 866 A

Table 3.18: Rated values for the 20/0,4 kV transformer.

Coupling ZNyn5

No load losses 0,0025 pu†

ek 4%

er 1%†

ex 3,87%†

Table 3.19: Electrical parameters for the 20/0,4 kV transformer.

3.8 Three phase representation of the grid 29

The copper losses are calculated in the same manner as above:

Pcu = 3 · 15, 152 ·(

1

100· 242

0, 630

)

= 6, 3 kW (3.25)

Pcu =6, 3 kW

0, 630MV A= 0, 01 pu (3.26)

3.8 Three phase representation of the grid

The electric representations, obtained in the previous sections, are gathered as shown in figure

3.21.

ua

ub

uc

Shortcircuitimpedance

Circuitbreaker

Section 1 Section 2 Section 3 Reactor

Figure 3.21: Cumulated model for the system. Arrows indicate mutual inductances between phases.

The diagram shown in figure 3.21, represents the three phase 150 kV grid section between Ferslev-

Tinghøj. Transformers have been omitted from this representation. Parameter values, for each

part of the electric representation shown in figure 3.21, are given in table 3.20 on the next page.


Voltage Supply

ua 165sin(ω(t)) kV ub 165sin(ω(t) − 2π/3) kV uc 165sin(ω(t) + 2π/3) kV

Short circuit impedance

Zs 4, 61 6 80 Ω Rs 0,800 Ω Ls 14,5 mH

Circuit breaker

Ron 100 µΩ Roff 1 MΩ

Section 1 - 0,43 km long cable

Can/2 52,2 nF Ra 0,0142 Ω La 0,147 mH

Cbn/2 52,2 nF Rb 0,0142 Ω Lb 0,147 mH

Ccn/2 52,2 nF Rc 0,0142 Ω Lc 0,147 mH

Gan/2 2,71 nS Gbn/2 2,71 nS Gcn/2 2,71 nS

Section 2 - 18,93 km long line

Can/2 74,5 nF Ra 0,898 Ω La 36,2 mH

Cbn/2 74,5 nF Rb 0,898 Ω Lb 36,3 mH

Ccn/2 82,5 nF Rc 0,898 Ω Lc 34,4 mH

Mab 13, 6 mH Mac 10,9 mH Mbc 11,0 mH

Gan/2 1,25 µS Gbn/2 1,25 µS Gcn/2 1,25 µS

Cab/2 13,75 nF Cac/2 5,7 nF Cbc/2 6,35 nF

Section 3 - 21 km long cable

Can/2 2,55 µF Ra 0,693 Ω La 7,20 mH

Cbn/2 2,55 µF Rb 0,693 Ω Lb 7,20 mH

Ccn/2 2,55 µF Rc 0,693 Ω Lc 7,20 mH

Gan/2 132,5 nS Gbn/2 132,5 nS Gcn/2 132,5 nS

Reactor

Ra 1,17 Ω La 2,307 H Mab -32,0 mH

Rb 1,17 Ω Lb 2,304 H Mac -23,0 mH

Rc 1,17 Ω Lc 2,307 H Mbc -32,0 mH

Table 3.20: Parameters for the three phase representation, shown in figure 3.21.

3.9 Summary & discussion

In this chapter, the system subparts and their electric equivalents have been described. The last

section of the chapter summarizes the parameters and gives the three phase representation of

the grid segment that is to be analyzed.

The short circuit impedance was not known at the moment of measurements, but here a

value found from the maximum apparent power, and a typical angle was used. The reactor

representation does not account for magnetic saturation, as this is not expected to be a factor of

influence. With these limitations, the three phase model of the 150 kV grid section between

Ferslev and Tinghøj is assumed to represent the system adequately for analysis, which is

conducted in the next chapter.

Chapter 4

Analysis of switching overvoltages on

the 150 kV grid

The objective of this chapter is to analyze the transients that are experienced at the switch-on

moment. In the first section of the chapter, the measured transients are analyzed, and estimated

from a theoretical point of view. In order to conduct a mathematical analysis of the system, a

simplified version is produced and two different analysis approaches are taken. The results from

the mathematical analysis are compared with a simulation in PSCAD of the simplified circuit.

The chapter ends with a discussion of the validity of the simplified model, and the factors that

have the greatest influence on the transient behaviour of the model. In order to give a clearer

analysis, focus is only on the 150 kV, and other voltage levels are therefore not considered here.

4.1 Estimation of the transient behaviour

Switching transients occur when the switching circuit changes from one state to another. In

the following section, the frequencies experienced after switch-on is analyzed from a theoretical

viewpoint.

In the transition between states, transient currents and voltages are superimposed on the power

frequency of the current and voltage. The frequency of these superimposed currents and voltages,

is determined by the natural frequencies of the power grid [8, Laubst, p.31]. For a series element,

the natural frequency is determined by the inductance and the capacitance:

f0 =1

2π√

L · C(4.1)

The voltage waveform, during switch-on of the grid section, is measured1 and shown in figure

4.1.

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065

−100

0

100

Voltage [kV]

Time [s]

Figure 4.1: The recorded voltage waveforms for 150 kV.

A Fourier analysis of the frequencies of the measured voltage waveform was made, the results of

this can be seen in figure 4.2. Results for a Fourier-analysis for all phases, and voltage levels can

be seen in appendix C.

1The measurements are described in appendix C on page C-35.

32 Analysis of switching overvoltages on the 150 kV grid

0 1000 2000 3000 4000 5000 60000

5

10

0 1000 2000 3000 4000 5000 60000

5

10

Per

cent

age

of th

e fu

ndam

enta

l [%

]15

0 kV

leve

l

0 1000 2000 3000 4000 5000 60000

5

10

Frequency [Hz]

Figure 4.2: Fourier analysis of the 150 kV waveform, of harmonic frequencies up to 6,5 kHz.

The Fourier analysis is performed over one period of the fundamental frequency, after the

switching instant. The Fourier analysis is conducted in MatLab, with the m-file shown in appendix

G.10. As shown in figure 4.2, two frequency band with large magnitude are present for all three

phases, one around 350 Hz and one around 3800 Hz. These are the frequencies that can be seen

in the time domain in figure 4.1.

The grid section and its equivalent model, described in the previous chapter, can be analyzed for

one phase to estimate the frequencies that will occur at switch-on. This is shown in figure 4.3.

Section 1 Section 2 Section 3 ReactorCircuit

breaker

Short

circuit

impedance

Section 1

430 m cable

Seen as a capacitance

Section 2

18,93 km line

Seen as an inductance

Section 3

21 km cable

Seen as a capacitance

ReactorShort circuit

impedance

Seen as an inductance

Ls L2

LrC1 C2

ia

ib

Figure 4.3: Electric circuit for estimation of the transient behaviour. Top: non simplified representation.

Bottom: simplified representation for frequency analysis.

In order to obtain the circuit shown at the bottom of figure 4.3, the following simplifications

4.1 Estimation of the transient behaviour 33

are made. All resistances are stripped from the circuit, as these do not affect the frequencies,

only the damping. The short cable section is seen as purely capacitive, the overhead line is seen

as purely inductive, and the 21 km cable section is seen as purely capacitive. Furthermore, the

circuit breaker is eliminated, along with the mutual inductances and capacitances in the system.

From the current flowing into the circuit, two currents are interesting: Current a that charges

the capacitor in section 1, and current b that charges the capacitor in section 3. The parameters

for the circuit are:

Ls C1 L2 C2 Lr

14,4 mH 0,105 µF 35,6 mH 5,10 µF 2,306 H

Table 4.1: The parameters used to evaluate the frequencies of the Ferslev-Tinghøj section. These are

calculated average phase values.

For the current ia the natural frequency is found:

f0,ia =1

2π√

Ls · C1(4.2)

f0,ia =1

2π√

0, 0144 · 0, 105 · 10−6= 4, 1 kHz (4.3)

For the current ib the natural frequency is found:

f0,ib =1

2π√

(Ls + L2) · C2(4.4)

f0,ib =1

2π√

(0, 0144 + 0, 0356) · 5, 1 · 10−6= 315 Hz (4.5)

The two calculated frequencies f0,ia and f0,ib, can be recognized in the measured frequencies

of ~3800 Hz and ~350 Hz. As the short circuit impedance at the time of the measurements is

unknown, the calculated frequencies deviate from the measurements. Furthermore, inductances

in the cable sections are neglected. These would also give slight changes in the calculated vales.

The natural frequency in the parallel coupling between reactor and the capacitor of section 3 is

found:

f0 =1

2π√

2, 306 · 5, 10 · 10−6= 46, 4 Hz (4.6)

This frequency is expected, as the size of the reactor is chosen to compensate the capacitance of

the cable. If the reactance of the cable equals the reactance of the reactor, the frequency would

be 50 Hz.

The Fourier analysis of the measurement shown in figure 4.2, shows other frequencies than

the two frequencies caused by the capacitances in the cables. These frequencies arise due to

different capacitances in the system. E.g. the capacitances in the overhead line will cause

natural frequencies in the same way as the two cable sections did.


4.2 Simplified model

In this section, a simplified model is produced on the basis of the system, shown in figure 3.21 on

page 29. For this simplification, different parts of the system are compared, and parameters

deemed with little significance are neglected. The objective of this simplification is a less

complicated model, which can be analyzed mathematically with more ease.

A circuit breaker in a closing operation behaves near ideally, and as the project deals with the

switch-on of a grid segment, the circuit breaker is assumed to operate as an ideal switch. The

series resistance in the circuit breaker is assumed of no influence compared to the other resistances

in the circuit, and is therefore neglected.

As the length of section 1 is only 0,43 km, this section is negligible by comparison to the length

of the other sections2. This leaves two sections, one 18,93 km long overhead line section and one

21 km long underground cable section. As the phase values for the parameters in the system

do not differ much between the three phases, average values are used in the simplified model,

resulting in equal parameters for the three phases. The simplification is therefore explained for

only one phase. A comparison of the parameters in the two sections is shown in table 4.2.

Section 2 Section 3

Overhead line Underground cable Reactor

C 154 nF 5,10 µF —

L 35,6 mH 7,20 mH 2,306 H

R 0,898 Ω 0,693 Ω 1,17 Ω

G 132 nS 265 nS —

Table 4.2: Comparison of parameters between overhead lines and underground cable. Average phase

values are used.

From table 4.2 it can be seen, that the capacitance in the cable is approximately 30 times the

capacitance of the overhead lines. It is therefore reasonable for the overhead line section, to

neglect the capacitance, leaving only the series impedance. The shunt conductance in both

sections are considered to be of no influence, and are therefore neglected.

The mutual inductances in the shunt reactor, the overhead line section and the underground

cable section are compared in table 4.3.

Shunt reactor Section 2

Overhead line

Mab -32,0 mH 13,6 mH

Mac -23,0 mH 10,9 mH

Mbc -32,0 mH 11,0 mH

Table 4.3: Comparison of mutual inductances between the reactor, overhead line and underground

cable.

2With this simplification, the high frequency transient caused by this section will not be present in the analysis.

The simplification is taken to obtain a simple circuit.

4.2 Simplified model 35

From the table it can be seen that the mutual inductances in the shunt reactor are greater than the

mutual inductances in the overhead line section with a factor of ca. 2. The mutual inductances

in the cable is not provided but is assumed to be smaller than the mutual inductances in the

overhead line, due to the small distance between the conductors. As the mutual inductances in

the reactor are the greatest, these are the only mutual coupling included in the simplification. The

issue concerning mutual couplings will therefore be included, without complicating the model too

much. All mutual inductances and capacitances in the transmission lines are therefore neglected.

With the above-mentioned considerations, phase a from figure 3.21 can be simplified as shown

in figure 4.4.

ua(t)

MacMab

Rsection2 Lsection2

Rreactor1

2Csection3

1

2Csection3

Rsection3 Lsection3

Lreactor

Section 2 Section 3 Reactor

Rs Ls

Figure 4.4: Simplified representation for phase a.

The simplified circuit for phase a in figure 4.4 has two capacitances. The analysis is considered

to be less complicated if only one capacitance is present, the cable section is therefore expressed

with a T -equivalent as shown in figure 4.5.

ua(t)

MacMab

Rsection2 Lsection2

Rreactor

Csection3

1

2Rsection3

1

2Rsection3

1

2Lsection3

1

2Lsection3

Lreactor

Section 2 Section 3 Reactor

Rs Ls

Figure 4.5: Simplified model of phase a with cable represented by a T-equivalent.

The circuit in figure 4.5 is simplified further by unifying the series impedances resulting in the

circuit shown in figure 4.6


ua(t)

MacMab

R1

R2

L1

L2

C

Rs Ls

Figure 4.6: Simplified representation of phase a.

Where the following applies:

R1 = Rsection2 + 12Rsection3

R2 = 12Rsection3 + Rreactor

L1 = Lsection2 + 12Lsection3

L2 = 12Lsection3 + Lreactor

C = Csection3

With these simplifications, the three phase representation for the system, can be expressed as

shown in figure 4.7.

u1,a(t)

u1,b(t)

u1,c(t)

ua(t)

ub(t)

uc(t)

u2,a(t)

u2,b(t)

u2,c(t)

ia(t)

ib(t)

ic(t)

i2,a(t)

i2,b(t)

i2,c(t)

i1,a(t)

i1,b(t)

i1,c(t)

Mac

Mab

Mbc

R1

R1

R1

R2

R2

R2

C

C

C

L1

L1

L1

L2

L2

L2

Rs

Rs

Rs

Ls

Ls

Ls

++

++

++

−−

−−

−−

Figure 4.7: The simplified model of the system.

4.3 Mathematical analysis 37

The voltages u1,a(t), u1,b(t) and u1,c(t) and currents ia(t), ib(t) and ic(t) shown in figure 4.7

correspond to the voltages and currents measured in Ferslev.

The parameter values for the circuit model, shown in figure 4.7, are listed in table 4.4.

R1 1,245 Ω L1 39,2 mH Mab -32,0 mH C 5,10 µF

R2 1,517 Ω L2 2,310 H Mac -23,0 mH — —

Rs 0,800 Ω Ls 14,4 mH Mbc -32,0 mH — —

Table 4.4: Parameters for the simplified circuit shown in figure 4.7.

The equations describing the circuit can be found by analyzing the circuit. Only one phase is

shown in figure 4.8, as this will give a clearer view of the analysis. The principle used for phase

a is used for the other two phases as well.

+

-

+

-

u1,a

Rs Ls

ua

R1 L1ia

i1,a

i2,a

C

R2

L2

u2,a

Mab Mac

Figure 4.8: Phase model for mathematical analysis

Under the general notation of the voltage drop across an inductor, ul = L · dildt

, and the current

into a capacitor, ic = C · dul

dtthe equations describing the system can be found. The circuit in

figure 4.8, can be expressed by four equations:

ua = Rs · ia + Ls ·diadt

+ R1 · ia + L1 ·diadt

+ u2,a (4.7)

u2,a = R2 · i2,a + L2di2,a

dt− Mab

di2,b

dt− Mac

di2,c

dt(4.8)

i1,a = C · du2,a

dt(4.9)

ia = i2,a + i1,a (4.10)

In the next section, a solution for the equations describing the system, is found.

4.3 Mathematical analysis

In the following section, the simplified model of the system is analyzed. The analysis is conducted

with two different approaches, namely an analytical approach and a numerical approach. An

analytical solution of the equations describing the system will have a general appearance. Such

solution can be obtained without knowing the values for the components, allowing for a direct


analysis of the significance of each component in the system. A numerical solution will only give

results for a specific system, and will thereby not provide any information about the significance

of the particular parameters.

An analytical solution of the circuit would be preferable, as this will give a general expression for

the voltages and currents in the system. An analytical approach is attempted in the following.

4.3.1 Analytical approach

The principle for the analytical solution is to find the current into the circuit. The desired voltage

u1,a after the short circuit impedance, shown in figure 4.8, is then found from the current. This

approach lets the short circuit resistance and inductance being unified with R1 and L1 from

figure 4.8. The mutual inductances are neglected in this approach, as this allows analysis of one

phase at a time. Furthermore, it is assumed that the mutual inductances will only have a small

influence on the voltages and currents in the circuit.

The simplified circuit is shown in figure 4.9, where the circuit is expressed in the Laplace domain.

+

-

Ua(s)

R1 sL1Ia(s)

I1,a(s)

I2,a(s)

1sC

R2

ZinsL2

U2(s)

Figure 4.9: Phase model for mathematical analysis

Where the components in the circuit shown in figure 4.9 are:

R1 = Rsection2 + 12Rsection3 + Rs

L1 = Lsection2 + 12Lsection3 + Ls



C = Csection3

The initial values for the currents and voltages in the circuit are assumed to be zero, as the

system is assumed totally discharged at the switch-on of the system. The current into the circuit

Ia(s) can be found from the voltage Ua(s) and the input impedance Zin. The input impedance

is found to be:

Zin = R1 + sL1 +1

sC· (R2 + sL2)

1sC

+ (R2 + sL2)(4.11)

Zin =s3CL1L2 + s2C(L1R2 + L2R1) + s(CR1R2 + L1 + L2) + R1 + R2

s2CL2 + sCR2 + 1(4.12)

The current Ia(s) can now be found:

Ia(s) = YinUa(s) = Z−1in Ua(s) (4.13)

Ia(s) =s2CL2 + sCR2 + 1

s3CL1L2 + s2C(L1R2 + L2R1) + s(CR1R2 + L1 + L2) + R1 + R2Ua(s) (4.14)


The voltage Ua(s) is known, and can be expressed in the Laplace domain:

Ua(s) = L [Up · sin(ωt − ϕ)] (4.15)

Ua(s) = Up ·(sin(ϕ)s + cos(ϕ)ω)

s2 + ω2(4.16)

Where:Up is the peak voltage of the voltage source

ω is the power frequency of the voltage source

ϕ is the phase angle of the voltage source

The expression for the input current is then:

Ia(s) =(s2CL2 + sCR2 + 1) · Up · (sin(ϕ)s + cos(ϕ)ω)

(s3CL1L2 + s2C(L1R2 + L2R1) + s(CR1R2 + L1 + L2) + (R1 + R2))(s2 + ω2)(4.17)

With the current obtained, the voltage U1,a(s), represented with u1,a(t) in figure 4.8, can befound by subtracting the voltage drop across short circuit impedance from the source voltage:

U1,a(s) = Ua(s) − Ia(s) · (Rs + sLs) (4.18)

U1,a(s) = Up · (sin(ϕ)s + cos(ϕ)ω)

s2 + ω2− ...

...(s2CL2 + sCR2 + 1) · Up · (sin(ϕ)s + cos(ϕ)ω) · (Rs + sLs)

(s3CL1L2 + s2C(L1R2 + L2R1) + s(CR1R2 + L1 + L2) + (R1 + R2))(s2 + ω2)(4.19)

In order to obtain an expression in the time domain, the inverse Laplace transform must be

applied. To use this, an often used method is partial fraction. In partial fraction, the polynomial

is fractioned into an expression from which the inverse Laplace transform can easily be found.

In general, partial fraction is performed by evaluating the poles of a polynomial. The poles are

then used to expand the denominator into fractions, containing one pole each:

F (s) =num(s)

(s − p1)(s − p2) . . . (s − pn)(4.20)

By use of partial fraction the coefficients (r1, r2, . . . , r3) can be found from the following relations:

num(s)

(s − p1)(s − p2) . . . (s − pn)=

r1

(s − p1)+

r2

(s − p2)+ . . . +

rn

(s − pn)(4.21)

The coefficients are found by replacing s with each of the poles (p1, p2, . . . , pn), one at a time:

r1 =num(s)

(s − p2)(s − p3) . . . (s − pn)

∣∣∣∣s=p1

(4.22)

r2 =num(s)

(s − p1)(s − p3) . . . (s − pn)

∣∣∣∣s=p2

(4.23)

...

rn =num(s)

(s − p1)(s − p2) . . . (s − pn−1)

∣∣∣∣s=pn

(4.24)


When the coefficients are found, the inverse Laplace transform can be applied on the right hand

side of equation 4.21, which gives:

f(t) = r1ep1t + r2e

p2t + . . . + rnepnt (4.25)

As shown in equation 4.17 and 4.19, the denominator of the expressions for Ia(s) and U1,a(s)

will be of 5th order. The algebraic expression for each pole will then become large, and thereby

difficult to analyze and comprehend. It is therefore chosen to insert the values for each of the

components, resulting in an numerical solution, and not the desired analytical representation.

The expressions are therefore solved numerically in MatLab3. The partial fraction and inverse

Laplace transform of the current in phase gives:

ia(t) = (0, 233 − j652, 073)e(−18,651+j1934,601)t + (0, 233 + j652, 073)e(−18,651−j1934,601)t . . .

+ (0, 202 + j15, 163)ej314,159t + (0, 202 − j15, 163)e−j314,159t − 0, 870e−1,507t (4.26)

In the same manner, the partial fraction and inverse Laplace transform of the voltage is found:

u1,a(t) = (−18166 + j340)e(−18,651+j1934,601)t + (−18166 − j340)e(−18,651−j1934,601)t . . .

+ (67429 − j13)ej314,159t + (67429 + j13)e−j314,159t + 0, 677e−1,507t (4.27)

The solution of u1,a is explained in the following. From the Euler identities, cos(θ) is expressed

as:

cos(θ) =ejθ + e−jθ

2(4.28)

By taking the absolute value of the complex magnitude (−18166 + j340) and (−18166 − j340)

a magnitude of -18169 is found. The first term of equation 4.27 is be simplified:

− 18169 · (e(−18,651+j1934,601)t + e(−18,651−j1934,601)t) (4.29)

− 18169 · (e−18,651tej1934,601t + e−18,651te−j1934,601t) (4.30)

− 18169 · (ej1934,601t + e−j1934,601t) · e−18,651t (4.31)

By use of equation 4.28, the term is further simplified:

− 18169 · 2 · cos(1934, 601t) · e−18,651t (4.32)

− 36338 · cos(1934, 601t) · e−18,651t (4.33)

With this procedure a new expression of u1,a(t) is found:

u1,a(t) = 134858cos(314, 159t︸︷︷︸

50Hz

) − 36338cos(1934, 601t︸︷︷︸

308Hz

) · e−18,651t︸︷︷︸

τ=53,6 ms

+ 0, 677e−1,507t

︸︷︷︸

DC− τ=664 ms

(4.34)

It can now be seen that the voltage consist of a 50 Hz cosine with 134,858 kV magnitude.

Superimposed is a 308 Hz cosine with a magnitude of -36,338 kV, further damped by a 53,6 ms

time constant. Furthermore, the voltage has a DC component of 0,677 V damped by a time

constant of 664 ms.

It is now possible to obtain expressions in the time domain for the currents and voltages in the

two other phases, by using the same method. The results for the three phases are plotted and

compared in the last section of this chapter.

As shown, the solution to a second order system can be difficult to conduct, with an analytical

method, in this case even with further simplification in the form of mutual inductances. In the

following section, the simplified circuit is analyzed through a numerical approach.

3The m-file is shown in appendix G.8


4.3.2 Numerical approach

The numerical solution is conducted with the same principle as in the analytical approach, by

first finding the current into the circuit. The voltage u1 after the short circuit impedance, is then

found from this current. The circuit for this analysis is shown in figure 4.10.

+

-

ua

R1 L1ia

i1,a

i2,a

C

R2

L2

u2

Mab Mac

Figure 4.10: Phase model for mathematical analysis.

Where the components in the circuit shown in figure 4.10 are:R1 = Rsection2 + 1

2Rsection3 + Rs

L1 = Lsection2 + 12Lsection3 + Ls



C = Csection3

These equations describing the voltages an currents in the simplified system are:

ua = R1 · ia + L1 ·diadt

+ u2,a (4.35)

u2,a = R2 · i2,a + L2di2,a

dt− Mab

di2,b

dt− Mac

di2,c

dt(4.36)

i1,a = Ca ·du2,a

dt(4.37)

ia = i2,a + i1,a (4.38)

The wanted values from these equation are besides ia and u2,a, the current i2,a flowing through

inductance L2, as this is needed for the mutual couplings. These three variables are isolated on

the left side:

L1 ·diadt

= ua − R1 · ia − u2,a (4.39)

L2 ·di2,a

dt= u2,a − i2,a · R2,a − Mab ·

di2,b

dt− Mac ·

di2,c

dt(4.40)

Ca ·du2,a

dt= ia − i2,a (4.41)

By integrating on both sides of equation 4.39, 4.40 and 4.41, expressions for the currents and

voltages are found. As this analysis focuses on the reconnection of the grid, the system is assumed


totally discharged at the moment of switch-on of the system, and initial values of currents and

voltages therefore zero. The integration constants can therefore be neglected.

ia =

(∫

uadt − R1 ·∫

iadt −∫

u2,adt

)

· 1

L1(4.42)

i2,a =1

L2·(∫

u2,adt − R2 ·∫

i2,adt − Mab · i2,b − Mac · i2,c

)

(4.43)

u2,a =1

C·∫

(ia − i2,a) dt (4.44)

The expressions above are solved by constructing a numerical solver in Matlab-Simulink. The

numerical solver, consists of the blocksets shown in figure 4.11.

[U_a] [I_1_a]

1/L_1

1

sdu/dtdu/dtdu/dt

Signal block, for input voltage

From block, takes input from goto blocks

Integral block

Summation block

Gain block

Goto block, sends signals to from block

Differentiation block

Subsystem, used to write data to workspace

Ramp, used to trigger switch Switch block

Figure 4.11: The blocks used in the numerical solver.

The solver is constructed from equation 4.42, 4.43 and 4.44 and shown in figure 4.12. The values

for the parameters are given in table 4.4 on page 37, where R1 and Rs is unified along with

L1 and Ls. The values for the mutual inductances are inserted with opposite sign, as they are

defined to be negative in the numerical solver.

1

s

1

s

1

s

1

s

1

s

[U_2_a]

[I_2_a]

[I_a]

[I_2_c][I_2_a][U_2_a][I_a]

[U_a]

[I_2_b]

1/C

1/L_2

Mab Mac

R_2

1/L_1

R_1

Figure 4.12: Numerical solver for ia, i2,a and u2,a, corresponding to equations 4.42, 4.43 and 4.44.


By use of the solver, shown in figure 4.12, numerical solutions for the signals ia, i2,a and u2,a are

achieved.

The next step is to find the voltage u1,a measured in Ferslev, shown in figure 4.8. This voltage

can be found by subtracting the voltage drop across the short circuit impedance from the grid

voltage ua. This operation is shown in figure 4.13 and is described by:

u1 = ua − us (4.45)

u1 = ua − (Rs · ia + Ls ·diadt

) (4.46)

The subsystem shown in figure 4.13, serves the purpose of writing results to the Matlab

workspace.

U_Ferslev

voltage

writeout

[I_c]

[I_b]

[I_a]

[U_c]

[I_c]

[U_b]

[I_b]

[U_a]

[I_a] du/dt

du/dt

du/dt

Rs

Rs

Rs

Ls

Ls

Ls

Figure 4.13: Solver to find ua, by subtracting us from ua, corresponding to equations 4.45 and 4.46.

By using the above described method for all phases, a Simulink model is constructed that can

solve the simplified model, shown in figure 4.7. This solver is shown in figure 4.14.


U_Ferslev

voltage

writeout

U_2

voltage

writeout

1

s

1

s

1

s

1

s

1

s

1

s

1

s

1

s

1

s

1

s

1

s

1

s

1

s

1

s

1

s

Input

current

writeout

[U_2_c]

[I_2_c]

[I_c]

[U_2_a]

[I_2_a]

[I_a]

[U_2_b]

[I_2_b]

[U_c]

[U_b]

[U_a]

[I_b]

[I_2_c][I_2_a][U_2_a][I_a]

[I_2_c][I_2_b]

[I_c]

[I_b]

[I_a]

[U_2_a]

[U_2_b]

[U_2_b]

[U_2_c]

[I_a]

[I_b]

[I_c]

[U_c]

[I_c]

[U_b]

[I_b]

[U_a]

[I_b]

[I_a]

[U_c]

[U_b]

[U_a]

[I_2_b][I_2_c][U_2_c][I_c] [I_2_a]

[I_2_b]

[I_2_a]

du/dt

du/dt

du/dt

R_2

1/L_1

R_1

1/L_2

Mab Mbc

R_2

Rs

Rs

Rs

Ls

Ls

Ls

1/C

1/C

1/L_1

1/C

1/L_2

Mac Mbc

1/L_2

Mab Mac

R_2

1/L_1

R_1

R_1

Figure 4.14: The constructed simulink solver, which is used for analysis of the simple model.

The results from the numerical approach are compared with the analytical approach in the last

section of this chapter. In order to verify the results from the two approaches, a simulation of

the simplified circuit is performed in PSCAD, this is described in the next section.

4.4 PSCAD simulation

In order to verify the results obtained through the numerical solution in Simulink, a simulation

in PSCAD is performed. The setup for the PSCAD simulation is shown in figure 4.15.

4.4 PSCAD simulation 45

C

B

A

BRK

TimedBreaker

LogicOpen@t0

BRK

U_2b

I_b

I_a

I_c

U_2a

U_2c

U_1a

U_1b

U_1c

0.0392 [H] 1.245 [ohm]

5.1

0 [u

F]

0.0392 [H] 1.245 [ohm]

0.0392 [H] 1.245 [ohm]

5.1

0 [u

F]

5.1

0 [u

F]

A

B

C

R=0

ic ib ia

oc ob oa

Ra

La

Rb

Lb

Rc

Lc

0.0144 [H]0.800 [ohm]

0.0144 [H]0.800 [ohm]

0.0144 [H]0.800 [ohm]

Figure 4.15: Graphical configuration in PSCAD for simulation.

The component values given in table 4.4 are entered. As the source impedance is present in the

circuit model, an ideal source is used in PSCAD.

4.4.1 The reactor component in PSCAD

The reactor component, is constructed with background in the reactor described in section 3.6 on

page 20. PSCAD does not contain a component for the shunt reactor, a new component is therefore

created, in order to include the mutual inductances in the reactor. The graphical layout of the

created component is shown in figure 4.16.

ic ib ia

oc ob oa

Ra

La

Rb

Lb

Rc

Lc

Figure 4.16: Graphical representation of the shunt reactor in PSCAD.

In PSCAD, mutually coupled windings are constructed through the use of the # TRANSFORMERS

script directive. A basic Transformers Segment has the general format [9, PSCAD User manual]:

1 #TRANSFORMERS <Number_of_Transformers >2 <Prefix >< Number_of_Windings > /3 <Node_1 > <Node_2 > <R_11 > <L_11 > /4 <Node_2 > <Node_3 > <R_12 > <L_12 > <R_22 > <L_22 > /5 ...

Here <Node_1> <Node_2> represents the two nodes, which the following expression takes place

between. The expression <R_12> <L_12> <R_22> <L_22> can then be conceived as the equation

unode1 − unode2 = R12 · i1 + L12di1dt

+ R22 · i2 + L22di2dt

.

Using this, the computation script can be written from the inductance matrix for the shunt

reactor and the resistance in each winding:

1 #TRANSFORMERS 12 3 /3 $ia $oa $Ra $La 0 $Mab 0 $Mac /4 $ib $ob 0 $Mab $Rb $Lb 0 $Mbc /5 $ic $oc 0 $Mac 0 $Mbc $Rc $Lc /


Where:$ia, $ib and $ic are the input nodes for each phase.

$oa, $ob and $oc are the output nodes for each phase.

$Ra, $Rb and $Rc are the resistances in the windings.

$La, $Lb and $Lc are the inductances in the windings.

$Mab, $Mac and $Mbc are the mutual inductances between the windings.

Note that the mutual resistances are set to zero, in order not to contribute with voltage between

the two nodes.

The values for the reactor is entered in the configuration as shown in figure 4.17.

Figure 4.17: Configuration of the shunt reactor in PSCAD.

The resistance R2 and inductance L2 from figure 4.7 are entered in the reactor interface, resulting

in: R2 = Ra = Rb = Rc and L2 = La = Lb = Lc. The values are found in table 4.4 on page 37.

With the parameters entered, the system is simulated. The results from this simulation can now

be compared with the two mathematical solutions and the actual measured values.

4.5 Results

In the following section, the waveforms from the analytical approach (section 4.3.1), the numerical

approach in Simulink (section 4.3.2) and the simulation in PSCAD (section 4.4) are compared with

the measured waveforms4. The comparison will be performed for u1 and i, which represent

the voltage and current measured in Ferslev. The switch-on instant, for simulations and

measurement, is at t=0.

4The measurements in Ferslev are described in appendix C.

4.5 Results 47

4.5.1 Comparison of u1

The voltage waveforms for u1 on phase a are shown in figure 4.18. The waveforms are taken

from the analytical approach, numerical solution in Simulink, the simulation in PSCAD and the

measurements in Ferslev.

0 0.05 0.1 0.15

−200

−100

0

100

200

Ana

lytic

app

roac

h [k

V]

0 0.05 0.1 0.15

−200

−100

0

100

200

Num

eric

al a

ppro

ach

[kV

]

0 0.05 0.1 0.15

−200

−100

0

100

200

PS

CA

D s

imul

atio

n [k

V]

0 0.05 0.1 0.15

−200

−100

0

100

200

Mea

sure

men

ts [k

V]

Time [s]

Figure 4.18: Comparison of the voltage u1 on phase a.

The voltage u1 on phase b is shown, for the analytical approach, numerical solution in Simulink,

the simulation in PSCAD and the measurements in Ferslev, in figure 4.19.


0 0.05 0.1 0.15

−200

−100

0

100

200

Ana

lytic

app

roac

h [k

V]

0 0.05 0.1 0.15

−200

−100

0

100

200

Num

eric

al a

ppro

ach

[kV

]

0 0.05 0.1 0.15

−200

−100

0

100

200

PS

CA

D s

imul

atio

n [k

V]

0 0.05 0.1 0.15

−200

−100

0

100

200

Mea

sure

men

ts [k

V]

Time [s]

Figure 4.19: Comparison of the voltage u1 on phase b.

The voltage u1 on phase c is shown for the analytical approach, numerical solution in Simulink,

the simulation in PSCAD and the measurements in Ferslev, in figure 4.20.

0 0.05 0.1 0.15

−200

−100

0

100

200

Ana

lytic

app

roac

h [k

V]

0 0.05 0.1 0.15

−200

−100

0

100

200

Num

eric

al a

ppro

ach

[kV

]

0 0.05 0.1 0.15

−200

−100

0

100

200

PS

CA

D s

imul

atio

n [k

V]

0 0.05 0.1 0.15

−200

−100

0

100

200

Mea

sure

men

ts [k

V]

Time [s]

Figure 4.20: Comparison of the voltage u1 on phase c.

The three figures above, indicate that the voltage u1 is identical for the numerical solution in

Simulink and the simulation in PSCAD. The analytical approach gives a slightly different result,

which is assumed to stem from the neglected mutual inductance in the reactor. Compared to

4.5 Results 49

measurements, there is much lesser damping in the simulation results. The simulation results

also indicate, that switch-on at the peak gives the greatest oscillation.

4.5.2 Comparison of i for a duration of 0,15 s

The current i in phase a is shown for the analytical approach, numerical solution in Simulink,

the simulation in PSCAD and the measurements in Ferslev, in figure 4.21, for a duration of 0,15 s.

0 0.05 0.1 0.15

−1

0

1

Ana

lytic

app

roac

h [k

A]

0 0.05 0.1 0.15

−1

0

1

Num

eric

al a

ppro

ach

[kA

]

0 0.05 0.1 0.15

−1

0

1

PS

CA

D s

imul

atio

n [k

A]

0 0.05 0.1 0.15

−1

0

1

Mea

sure

men

ts [k

A]

Time [s]

Figure 4.21: Comparison of the current i in phase a for a duration of 0,15 s.

The current i in phase b is shown for the analytical approach, numerical solution in Simulink,



0 0.05 0.1 0.15

−1

0

1

Ana

lytic

app

roac

h [k

A]

0 0.05 0.1 0.15

−1

0

1

Num

eric

al a

ppro

ach

[kA

]

0 0.05 0.1 0.15

−1

0

1

PS

CA

D s

imul

atio

n [k

A]

0 0.05 0.1 0.15

−1

0

1

Mea

sure

men

ts [k

A]

Time [s]

Figure 4.22: Comparison of the current i in phase b for a duration of 0,15 s.



0 0.05 0.1 0.15

−1

0

1

Ana

lytic

app

roac

h [k

A]

0 0.05 0.1 0.15

−1

0

1

Num

eric

al a

ppro

ach

[kA

]

0 0.05 0.1 0.15

−1

0

1

PS

CA

D s

imul

atio

n [k

A]

0 0.05 0.1 0.15

−1

0

1

Mea

sure

men

ts [k

A]

Time [s]

Figure 4.23: Comparison of the current i in phase c for a duration of 0,15 s.

The simulation current waveforms are all comparable, while the measured deviates from the

others. Lack of damping is clearly seen compared with measurements, as the damping time of

the current is larger than the damping time of the measurements.

4.5 Results 51

4.5.3 Comparison of i for a duration of 2,8 s

The current i in phase a is shown for the analytical approach, numerical solution in Simulink,


0 0.5 1 1.5 2 2.5−0.5

0

0.5

Ana

lytic

app

roac

h [k

A]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

Num

eric

al a

ppro

ach

[kA

]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

PS

CA

D s

imul

atio

n [k

A]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

Mea

sure

men

ts [k

A]

Time [s]

Figure 4.24: Comparison of the current i in phase a for a duration of 2,8 s.



0 0.5 1 1.5 2 2.5−0.5

0

0.5

Ana

lytic

app

roac

h [k

A]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

Num

eric

al a

ppro

ach

[kA

]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

PS

CA

D s

imul

atio

n [k

A]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

Mea

sure

men

ts [k

A]

Time [s]

Figure 4.25: Comparison of the current i in phase b for a duration of 2,8 s.


The current i in phase c is shown for the analytical approach, numerical solution in Simulink,


0 0.5 1 1.5 2 2.5−0.5

0

0.5

Ana

lytic

app

roac

h [k

A]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

Num

eric

al a

ppro

ach

[kA

]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

PS

CA

D s

imul

atio

n [k

A]

0 0.5 1 1.5 2 2.5−0.5

0

0.5

Mea

sure

men

ts [k

A]

Time [s]

Figure 4.26: Comparison of the current i in phase c for a duration of 2,8 s.

Measured Simulated

110 A 60 A

Table 4.5: A comparison of steady state magnitude at 150 kV level, for the simplified system.

Once again, all the current waveforms from simulations are all comparable, while the measured

deviates from the others. The time constant for the DC component is though of the same size in

both simulations and measurements. As it can be seen from the figures and table 4.5.3, the steady

state current from the measurement has considerably higher magnitude than for the simulated

currents.

4.5.4 Harmonic analysis of u1

The harmonic analysis for u1 on phase a is shown in figure 4.27.

4.5 Results 53

0 1000 2000 3000 4000 5000 60000

10

20

Analytical

0 1000 2000 3000 4000 5000 60000

10

20

Simulink

0 1000 2000 3000 4000 5000 60000

10

20

Per

cent

age

of th

e fu

ndam

enta

l [%

]15

0 kV

leve

l

Pscad

0 1000 2000 3000 4000 5000 60000

10

20

Frequency [Hz]

Measurement

Figure 4.27: Harmonic analysis of the voltage u1 for phase a.

The harmonic analysis for u1 on phase b is shown in figure 4.28.

0 1000 2000 3000 4000 5000 60000

10

20

Analytical

0 1000 2000 3000 4000 5000 60000

10

20

Simulink

0 1000 2000 3000 4000 5000 60000

10

20

Per

cent

age

of th

e fu

ndam

enta

l [%

]15

0 kV

leve

l

Pscad

0 1000 2000 3000 4000 5000 60000

10

20

Frequency [Hz]

Measurement

Figure 4.28: Harmonic analysis of the voltage u1 for phase b.

The harmonic analysis for u1 on phase c is shown in figure 4.29.


0 1000 2000 3000 4000 5000 60000

10

20

Analytical

0 1000 2000 3000 4000 5000 60000

10

20

Simulink

0 1000 2000 3000 4000 5000 60000

10

20

Per

cent

age

of th

e fu

ndam

enta

l [%

]15

0 kV

leve

l

Pscad

0 1000 2000 3000 4000 5000 60000

10

20

Frequency [Hz]

Measurement

Figure 4.29: Harmonic analysis of the voltage u1 for phase c.

The Fourier analysis shows the same frequencies for the analytical approach, the numerical

approach and the simulation in PSCAD. Here, the largest frequency component is the 6th harmonic.

For the measurements the largest frequency component is the 7th harmonic. As expected the

frequencies around 3,8 kHz experienced in the measurements is not present in the simulations,

due to the neglected short cable section.


The solutions from the numerical approach in Simulink and the simulation in PSCAD give

identical results, both for the voltages u1,a,b,c, and the input currents ia,b,c. Small difference

is apparent between the analytical approach conducted in MatLab and the results from Simulink

and PSCAD. This difference is assumed to stem from the neglected mutual inductance in the

analytical approach, as that is the only factor separating the solutions. The small magnitude

of this difference indicates, the mutual inductances have only small influence on the generated

frequencies.

As expected, the high frequency voltage component appears in none of the simulations. This is

due to the fact that the short cable section is neglected in the simulation.

Common for all simulated voltage waveforms, is the appearance of the 6th harmonic, which is the

dominant one, and in accordance to the frequency predicted in section 4.1. For the measurements,

the 7th harmonic has the highest magnitude. This difference can be explained by the fact, that the

source inductance is estimated, which will influence the frequencies. Furthermore, the difference

can originate from the fact, that simplifications are made using average values, and distributed

parameters are in the measured system, whereas the cable section cannot be seen purely as a

capacitance.

Similarities between voltage waveforms and the Fourier analysis of the first period after switching,

in all simulations, indicate correct solution of the simplified model.

4.6 Summary & discussion 55

Similarities between simulation and test results are present, however a noticeable deviation is in

the steady state current magnitude, and the damping rate. The steady state current magnitude,

has a value of ca. 110 A for measurements, whereas in simulation the magnitude is ca. 60 A.

This difference can be explained by the neglected shunt capacitances and conductances in the

overhead lines and the capacitances in the short cable section.

Damping in the measurements is much larger than in the simplified model. In an oscillating

system, the damping is given by resistances, which are energy consuming elements. When

simplifying the three phase representation of the system, shown in figure 3.21, care was taken

not to neglect elements adding to the damping ratio of the system. The AC-resistance was

therefore estimated for the overhead line. There are though a few factors that can influence the

transient behaviour of the system, these are discussed below:

• The simplified model does not use distributed parametersIt is therefore difficult to make valid assumptions about frequency dependent components,

as the parameters are distributed and not lumped as assumed. The use of T equivalent,

for the long cable section, results in only one frequency component and thereby a high

magnitude in the Fourier analysis. The measured frequencies will, due to the distributed

parameters, contain several of frequency components, this can also be seen in the Fourier

analysis of the measurements. This is assumed to be the reason for difference between the

Fourier analysis for measurement and simulations.

• Mutual inductances in overhead lines and underground cable are not included in the sim-plified modelAs shown, only very small differences between the analytical (without mutual inductances)

and the numerical approach (with mutual inductances) were present. The mutual induc-

tances are therefore considered to have only little effect.

• Neglected capacitances in the overhead line sectionThese capacitances will give rise to high frequency components, though with small

magnitude. If the transmission line is considered as a π-equivalent, the current running into

one-half of the line capacitance, will give rise to frequency around 6 kHz. This frequency

band can be seen in the measurements, where frequencies around 5700 Hz is present.

• Shunt conductance neglected in simplified modelThis is one of the simplifications that decrease the damping ratio. There is a reason to

believe that this will not add significantly to the damping, as simulations were briefly

conducted with this component present, without noticeable increase of damping ratio.

• Neglected short cable sectionThe 430 m cable section in Ferslev, together with the source inductance, gives rise to a

frequency component of 4 kHz which is therefore not experienced in simulations. This is

one of the greater deviations between measurements and simulations.

• Skin- and proximity effect for higher frequenciesThe resistance in the transmission line is frequency dependent as a function of the skin

depth. This will lead to greater resistance for frequencies higher than 50 Hz, and thereby

greater damping. This, along with proximity effect in the cables, is assumed to be of

great significance for the validity of simulations, as the overvoltages are caused by higher

frequencies.


This chapter has dealt with analysis of the transient behaviour of the Ferslev-Tinghøj grid

section, with focus solely on the 150 kV level. First, the system was simplified to find its

natural frequency response, in order to explain the measured frequencies. In order to ease

the mathematical analysis, the three phase representation of the system, shown in figure 3.21,

was simplified. When attempting to solve the differential equations describing the simplified

system, it appeared that the analytical solution was too complicated to comprehend. These

equations were therefore solved numerically, and a simulation of the circuit model conducted, in

order to evaluate the validity of the mathematical solutions. Simulations bear resemblance with

measurements, large difference in the damping is though present.

In order to perform a more precise analyze of the system, a more detailed system simulation is

conducted with the simulation software PSCAD. Here models for overhead lines and underground

cables are included, which eliminates the simplifications. This is done in the next chapter.

Chapter 5

Configuration and simulation in

PSCAD

The objective of this chapter is to obtain a more precise simulation of the system, in order to

obtain a more valid estimation of the switching overvoltages. The configuration of the subparts,

shown in figure 5.1, are first described, whereafter the simulation is performed and the results

are presented.

5.1 PSCAD model subparts

The simulation model for the Ferslev-Tinghøj grid section, is shown in figure 5.1.

C

B

A

BRK1

TimedBreaker

LogicOpen@t0

BRK1

U_b

I150k_b

I150k_a

I150k_c

U_a

U_c

C1Section3

S1

C2

S2

C3

S3

C1Section3

S1

C2

S2

C3

S3

U150k_a

U150k_b

U150k_c

Section1

C

Section2

1

Section2

1

Section2

T

A

B

C

A

B

C60 [kV]

#2#1

150 [kV]

80.0 [MVA]A

B

C

A

B

C20.0 [kV]

#2#1

60.0 [kV]

16.0 [MVA]A

B

C

A

B

C0.42 [kV]

#2#1

24 [kV]

0.630 [MVA]

U20k_c

U20k_b

U20k_aU60k_a

U60k_b

U60k_c U04k_c

U04k_b

U04k_a 1 [ohm]

1[ohm]

1 [ohm]

3 [o

hm

]

3 [o

hm

]

C1Section1

S1

C2

S2

C3

S3

C1Section1

S1

C2

S2

C3

S3

A

B

C

R=0 14.4 [mH]0.8 [ohm]

14.4 [mH]0.8 [ohm]

14.4 [mH]0.8 [ohm]

1 [ohm] 1 [ohm]

1 [ohm] 1 [ohm]

1 [ohm] 1 [ohm]

1 [ohm]

1 [ohm]

1 [ohm]

I

ic ib ia

oc ob oa

Ra

La

Rb

Lb

Rc

Lc

Section3

C

Figure 5.1: PSCAD model of the system.

The model is a combination of a voltage source, source impedances, circuit breakers, two cable

sections, one overhead line section, a shunt reactor, and three transformers. The configuration

of these subparts is described in the following.

5.1.1 Voltage source

The voltage source is represented by an ideal Three Phase Voltage Source, as shown in figure 5.2,

with resistance and inductance in series, to represent the short circuit impedance. The values

for the short circuit impedance is found in section 3.2 on page 9.

A

B

C

R=0 0.0144 [H]0.8 [ohm]

0.0144 [H]0.8 [ohm]

0.0144 [H]0.8 [ohm]

Figure 5.2: Representation of the source in PSCAD.

58 Configuration and simulation in PSCAD

The configuration of the source is shown in fig 5.3. The voltage level is configured to 165 kV, as

this was the voltage measured in Ferslev, and a phase shift of -90, in order to match a switch-on

moment at peak for phase a.

Figure 5.3: Configuration of the source in PSCAD.

5.1.2 Circuit breaker

To simulate the reconnection of the grid, the Three Phase Breaker is used as shown in figure 5.4.

C

B

A

BRK1

TimedBreaker

LogicOpen@t0

BRK1

Figure 5.4: Representation of the breaker in PSCAD.

The breaker is controlled by the Timed Breaker Logic, shown at the bottom in figure 5.4. The

breaker is configured to switch-on after 0,01 s, as shown in figure 5.5, in order to have phase a in

peak at the moment of switch-on. It has been checked, that the current is in steady state at this

moment, thereby will the switching transients not be influenced by transformer inrush current.

5.1 PSCAD model subparts 59

Figure 5.5: Configuration of the circuit breaker and timed breaker logic.

The values for the resistances in figure 5.5 is given in section 3.3, where the earth resistance is

assumed 1 Ω as the cable is grounded in the station, and will hence have low resistance value.

5.1.3 Section 1

Section 1, which is a 430 m long underground cable section, is represented by the CableComponent and Cable Interface as shown in figure 5.6.

Section1

C

C1Section1

S1

C2

S2

C3

S3

C1Section1

S1

C2

S2

C3

S3

1 [ohm]

1 [ohm]

1 [ohm]

Figure 5.6: Representation of section 1 in PSCAD.

The physical layout configuration of the cable component is shown in figure 5.7.


100.0 [ohm*m]

Relative Ground Permeability:

Ground Resistivity:

1.0

Earth Return Formula:Analytical Approximation

20.75 [mm]

Cable # 1

39.05 [mm]39.435 [mm]

46 [mm]

6.97 [mm]

1.134[m]

0 [m]

Conductor

Insulator 1

Sheath

Insulator 2

20.75 [mm]

Cable # 2

39.05 [mm]39.435 [mm]

46 [mm]

6.97 [mm]

1.21367 [m]

-0.046 [m]

Conductor

Insulator 1

Sheath

Insulator 2

20.75 [mm]

Cable # 3

39.05 [mm]39.435 [mm]

46 [mm]

6.97 [mm]

1.21367 [m]

0.046 [m]

Conductor

Insulator 1

Sheath

Insulator 2

Figure 5.7: Configuration of section 1 in PSCAD.

The analytical approximation for solving the ground impedance integral is chosen in preference

to the numerical integration, as this is more time effective and recommended in the PSCAD help

file. The solution is normally accurate to within 5% of exact solution [9].

Note should be taken, that the conductor is configured to contain a hollow centre. This is done

in order to keep the outer diameter in the right dimension, without increasing the conductor

cross section. In the datasheet for the cable, the conductor cross section is defined as 1200 mm2,

and the diameter of the conductor is defined to be 41,5 mm, as shown in figure 3.5 on page 13.

The area difference1 can then be calculated:

A = r2π (5.1)(

41, 5

2

)2

π = 1353 mm2 (5.2)

∆A = 1353 − 1200 = 153 mm2 (5.3)

The area difference of 153 mm2c is considered to consist of the air between the threads in the

conductor. As the outer radius of the conductor is important in determination of the capacitance

this is kept as described in the datasheet. In order to obtain the correct resistance in the

1The difference between the area calculated from the cross section of 1200 mm2 and the area calculated from

the given radius 41,5

2.


conductor, the area difference is implemented, by defining the conductor as it is hollow. The

radius of the hollow part is found from the above calculated area:

r =

√

153

π= 6, 97 mm (5.4)

The configuration of conductor resistivity and isolation permittivity, of the cable component is

shown in figure 5.8 and 5.9.

Figure 5.8: Electrical configuration of the cable component, section 1 in PSCAD.

Figure 5.9: Electrical configuration of the cable component, section 1 in PSCAD.


The 1st Conducting Layer Data represents the phase conductor, which is aluminium for the

cable.

The Insulator 1 Data represents the isolation material between conductor and screen.

The 2nd Conducting Layer Data represents the metallic screen in the cable, which is made up

of copper wires. The resistivities and permeabilities for copper and aluminium is found in [10,

Serway, p. 959]. The insulation data is given in the user guide for the cable [11, ABB, cable user

guide].

In PSCAD, there are three different types of distributed transmission line models, namely the

Bergeron model, the Frequency dependent (mode) model and the Frequency dependent (phase)model. These are described in the PSCAD help file [12, PSCAD, p.342], direct quote.

• Bergeron model

“The Bergeron model represents the L and C elements of a PI section in a

distributed manner (not using lumped parameters like PI sections). It is accurate

at the specified frequency and is suitable for studies where the specified frequency

load-flow is most important (e.g. relay studies).”[12, PSCAD, p.342]

• Frequency dependent (mode) model

“The Frequency-Dependent (Mode) model represents the frequency dependence of

all parameters (not just at the specified frequency as in the Bergeron model). This

model uses modal techniques to solve the line constants and assumes a constant

transformation. It is therefore only accurate for systems of ideally transposed

conductors (or two conductor horizontal configurations) or single conductors.”[12,

PSCAD, p.342]

• Frequency dependent (phase) model

“The Frequency-Dependent (Phase) model also represents the frequency depen-

dence of all parameters as in the ’Mode’ model above. However, the Frequency

Dependent (Phase) model circumvents the constant transformation problem by di-

rect formulation in the phase domain. It is therefore accurate for all transmission

configurations, including unbalanced line geometry. The Frequency-Dependent

(Phase) model should always be the model of choice, unless another model is cho-

sen for a specific reason. This model is the most advanced and accurate time

domain line model in the world! ”[12, PSCAD, p.342]

As the Frequency dependent (phase) model is recommended in the user guide, and fulfills the

simulation demand it is the model of choice for all cable and line sections.

5.1.4 Section 2

Section 2 is represented with the Transmission Line Component and Transmission Line Interfaceas show in figure 5.10. The earth resistance is chosen to 3 Ω, as good grounding conditions are

assumed [13, Vørts, p. 50].


Section2

1

Section2

1

Section2

T

3 [o

hm

]

3 [o

hm

]


The line component is configured as shown in figure 5.11. Instead of using a graphical tower, x-y

coordinates are used. The graphical configuration of the Donau mast, is shown in figure 3.11 on

page 17.

100.0 [ohm*m]

Relative Ground Permeability:

Ground Resistivity:

1.0

Earth Return Formula:Analytical Approximation

4

Cond. #Phasing # Phasing #

41.6[m]8.35 [m]11

2

3

24.7 [m]

13 [m] 24.7 [m]

7.2 [m]

10.4 [m] 33.5 [m]1

2

3

Connection X (from X (fromtower centre)

GW. #Connection

Tower: Donau

Conductors: Martin Ground_Wires: Dorking

Tower Centre 0 [m]

Ytower centre) (at tower)

Y(at tower)

Figure 5.11: Configuration of section 2 in PSCAD.

The electrical configuration for the tower, the phase conductors and the ground conductors are

shown in figure 5.12 and 5.13.

Figure 5.12: Electrical configuration of the overhead line model, section 2 in PSCAD.


Figure 5.13: Electrical configuration of the overhead line model, section 2 in PSCAD.

The parameters entered in figure 5.12 and 5.13 are found in section 3.5 on page 17.

5.1.5 Section 3

Section 1 is represented with the Cable Component and Cable Interface as shown in figure 5.14.

C1Section3

S1

C2

S2

C3

S3

C1Section3

S1

C2

S2

C3

S3

Section3

C

1 [ohm] 1 [ohm]

1 [ohm] 1 [ohm]

1 [ohm] 1 [ohm]


Section 3 is configured in the same way as section 1. Only the length of the section differs, for

section 3 the length is 21 km. For this cable section both ends are grounded as described in

section 3.4 on page 12.


5.1.6 Shunt reactor

The shunt reactor model, is not included in the Master Library in PSCAD. A new component

was therefore constructed, as described in section 4.4.1 on page 45. The graphical layout of the

created reactor component is shown in figure 5.15.

ic ib ia

oc ob oa

Ra

La

Rb

Lb

Rc

Lc

Figure 5.15: The layout of the reactor component.

The configuration of the shunt reactor is shown in figure 5.16.

Figure 5.16: The electrical configuration of the reactor model.

The parameter values entered in figure 5.16 are found in section 3.6 on page 20.


The modelling of the transformers in the system, is made with 3 Phase 2 Winding Transformercomponent. The 150/60 kV transformer is coupled in YNd11, meaning Y-∆ transformer with

30 phase shift. The Y-∆ transformer is configured, with the ∆ winding leading the Y winding,

to represent the phase shift. Magnetic saturation is not included in this component. The 150/60

kV transformer representation is shown in figure 5.17.


A

B

C

A

B

C60 [kV]

#2#1

150 [kV]

100.0 [MVA]

Figure 5.17: Representation of 150/60 kV transformer in PSCAD.

The parameters, shown in figure 5.18, for the 150/60 kV transformer model, and entered in the

transformer parameter interface, are calculated in section 3.7.1 on page 27.

Figure 5.18: Configuration of the 150/60 kV transformer in PSCAD.


The 60/20 kV transformer is coupled in YNyn0, resulting in no phase shift between the primary

and secondary side. The transformer is represented with the 3 Phase 2 Winding Transformer as

shown in figure 5.19: Magnetic saturation is not included in the model.


A

B

C

A

B

C20.0 [kV]

#2#1

60.0 [kV]

16.0 [MVA]

Figure 5.19: Representation of 60/20 kV transformer in PSCAD.

The configuration of the 60/20 kV transformer in PSCAD, is shown in figure 5.20, the values are

calculated in section 3.7.2 on page 28.

Figure 5.20: Configuration of the 60/20 kV transformer in PSCAD.

5.1.9 20/0,4 kV transformer

The 20/0,4 kV transformer is coupled in ZNyn5, meaning ZigZag-Y transformer with 210 phase

shift. A model for this exact transformer type is not given in the Master Library in PSCAD,

which is why the ∆-Y transformer is used. This will though not give the same characteristics,

such as the earth reference of the Zig-Zag. To represent the phase shift, the ∆-Y transformer is

configured, so that the ∆ winding leads the Y. The voltage will then obtain a wave form which

can be compared with the measurement. This procedure will give 0 displacement, and hence

a difference phase sequence after switch-on, compared with the measurements2. It is though

2The sequence in the measurement would be: phase b, phase c, phase a due to the 240 (30 + 210). For the

simulation the resulting phase shift from 150 kV to 0,4 kV will be 0, hence the same sequence as for the 150 kV

level: phase a, phase b, phase c.


chosen not to proceed to obtain the correct model for this transformer. Magnetic saturation is

not included in this component. The 20/0,4 kV transformer is represented with the 3 Phase 2Winding Transformer as shown in figure 5.21.

A

B

C

A

B

C0.42 [kV]

#2#1

24 [kV]

0.630 [MVA]

Figure 5.21: Representation of 20/0,4 kV transformer in PSCAD.

The configuration of the 20/0,4 kV transformer in PSCAD, is shown in figure 5.22, the entered

values were calculated in section 3.7.3 on page 28.

Figure 5.22: Configuration of the 20/0,4 kV transformer in PSCAD.

The components in the PSCAD simulation and their configuration has now been described. The

validity of the created reactor model, is examined in the next section, where the steady state

behaviour is controlled against calculated parameters.

5.2 Verification of reactor model

The reactor component is created in PSCAD in order to include the mutual inductances. The

reactor component is therefore tested for mutual inductances by applying voltage on phase a,

and measure the induced voltages on phases b and c. The equivalent circuit is shown in figure

5.23, where the resistances have been eliminated.

5.2 Verification of reactor model 69

+

-

+

-

ua(t)

ia(t)

ub(t)

uc(t) Mac

Mab

La

Lb

Lc

Figure 5.23: Equivalent circuit for the reactor. The phases b and c are not connected to the voltage

source.

As phases b and c are not connected to the voltage source, the voltages in these two phases are:

ub(t) = Mab ·dia(t)

dt(5.5)

uc(t) = Mac ·dia(t)

dt(5.6)

Here dia(t)dt

can be found from the voltage equation for phase a:

ua(t) = La ·dia(t)

dt(5.7)

dia(t)

dt=

1

La

· ua(t) (5.8)

The expressions for the voltages at the two non-connected phases can then be found:

ub(t) = Mab ·ua(t)

La(5.9)

uc(t) = Mac ·ua(t)

La

(5.10)

The peak voltage for the two non-connected phases can then be found, by inserting the peak

voltage for phase a and the values for Mab, Mac and La:

ub,peak = −0, 032 ·165

√2√

3

2, 307= −1, 34 kV (5.11)

uc,peak = −0, 023 ·165

√2√

3

2, 307= −1, 87 kV (5.12)

The test setup in PSCAD is shown in figure 5.24 to the left, and the parameters for the reactor

component is shown to the right.


A

B

C

R=0

ic ib ia

oc ob oa

Ra

La

Rb

Lb

Rc

Lc

Ua

Ub

Uc

Ia

Figure 5.24: Steady state verification of the reactor component in PSCAD.

The results from the simulation are shown in figure 5.25 for the three phases.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−100

0

100

Pha

se a

[kV

]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−4

−2

0

2

4

Pha

se b

[kV

]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−4

−2

0

2

4

Pha

se c

[kV

]

Time [s]

Figure 5.25: Voltages in the three phases. Note the different scale on the y-axis.

The values for phase b and c are measured at the peak voltage for phase a, to -1,34 kV and -1,87

kV respectively. These are the same as the calculated value, and the mutual inductances in the

reactor component are therefore verified.

A new simulation is performed with the resistance Ra, Rb and Rc set to 1,17 kΩ as stated in

table 3.13 on page 26. The current into the reactor is simulated, as shown to the left in figure

5.24. The current in phase a is shown in figure 5.26.

5.3 Results 71

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Cur

rent

[kA

]

Time [s]

Figure 5.26: Current flowing into phase a.

The peak current is 188 A, resulting in a rms current of 133 A. This is in accordance with the

rated current for the reactor of 135,8 A. The difference can be explained by the fact that the

rated current is provided for 170 kV, whereas the simulation is performed with 165 kV.

5.3 Results

As the validity of the reactor model has been confirmed against steady state, simulation of the

system is conducted. The general configuration of the simulation project is first described.

5.3.1 General simulation configuration

The general simulation settings are shown in figure 5.27.

Figure 5.27: General configuration of the simulation.


Here, the time step is set to 1 µs. This is done in order to keep to time step below the travel

time for section 1 of 430 m. The travel time for section is:

τ =l

c=

0, 43

300000= 1, 43 µs (5.13)

Where:τ is the travel time for the cable [s]

l is the length of the cable [km]

c is the propagation time (assumed speed of light) [km/s]

The PSCAD simulation is performed and the results are compared with measurements in the

following. Each figure shows all phases for each voltage level both for simulation and measure-

ments. Comparison is made for the voltage at 150 kV, 20 kV and 0,4 kV. Current waveforms

are compared for the 150 kV level. No comparison is for the 60 kV level, as these measurements

were corrupt.

5.3.2 Comparison of the voltages at 150 kV level

Figure 5.28 shows the results for the voltages at 150 kV level.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−200

−100

0

100

200

Phase A

PS

CA

D [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−200

−100

0

100

200

Mea

sure

men

ts [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−200

0

200

Phase B

PS

CA

D [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−200

−100

0

100

200

Mea

sure

men

ts [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−200

0

200

Phase C

PS

CA

D [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−200

−100

0

100

200

Mea

sure

men

ts [k

V]

Time [s]

Figure 5.28: The simulation and measurement results for the voltage at 150 kV level.

5.3 Results 73

The voltage behaviour on the 150 kV is shown in figure 5.29 for phase a two periods after

switch-on in t = 0.01 s, in order to show the appearance of the high frequencies.

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−200

−100

0

100

200

Phase A

PS

CA

D [k

V]

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−200

−100

0

100

200

Mea

sure

men

ts [k

V]

Time [s]

Figure 5.29: Voltage behaviour for phase a, shown for two periods after switch-on in t = 0.01 s.

For the voltages at 150 kV level, a comparison of peak voltages are made:

Phases [kV]

a b c

PSCAD +199/-159 +141/-140 +146/-154

Measurement +159/-134 +154/-133 +146/-162

Table 5.1: A comparison for the peak voltages at 150 kV level.

From figure 5.28 and table 5.3.2, it can be seen that overvoltages reach a value of 199 kV,

which is greatly above the measured value of 159 kV. It is also apparent how the lower

frequency component is less damped in the simulation than in measurements. The high frequency

component seems to be damped at almost the same rate in simulation and measurement.

5.3.3 Comparison of the currents at 150 kV level

Figure 5.30 shows the results for the currents at 150 kV level.


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1000

0

1000Phase A

PS

CA

D [A

]

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1000

0

1000

Mea

sure

men

ts [A

]

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1000

0

1000Phase B

PS

CA

D [A

]

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1000

0

1000

Mea

sure

men

ts [A

]

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1000

0

1000Phase C

PS

CA

D [A

]

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1000

0

1000

Mea

sure

men

ts [A

]

Time [s]

Figure 5.30: The simulation and measurement results for the currents at 150 kV level.

For the current at 150 kV level, a comparison is made of the highest peaks, and steady state

magnitude.

Phases [A]

a b c

PSCAD peaks +1474/-1377 +798/-589 +493/-896

Steady state magnitude, peak-to-peak 14 16 14

Measurement peaks +1103/-1423 +1134/-1773 +666/-1195

Steady state magnitude, peak-to-peak 110 110 110

Table 5.2: A comparison of peak currents, and steady state magnitude at 150 kV level.

Comparison of the currents show that, apart from phase a, the initial inrush current reaches

higher peak values in measurements than in simulation, as shown in table 5.3.3. This inrush

period is also much shorter in the measurements. As shown in table 5.3.3, the steady state

magnitude obtain a much higher magnitude than in simulations.

5.3 Results 75

5.3.4 Comparison of voltage at 20 kV level

Figure 5.31 shows the results for voltage at 20 kV level.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−20

0

20

Phase A

PS

CA

D [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−20

0

20

Mea

sure

men

ts [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−20

0

20

Phase B

PS

CA

D [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−20

0

20

Mea

sure

men

ts [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−20

0

20

Phase C

PS

CA

D [k

V]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−20

0

20

Mea

sure

men

ts [k

V]

Time [s]

Figure 5.31: The simulation and measurement results for the voltage at 20 kV level.

For the voltage at 20 kV level, a comparison of peak voltages is made:

Phases [kV]

a b c

PSCAD +18,4/-21,1 +18,3/-17,7 +19,8/-26,5

Measurement +16,6/-17,4 +17,3/-16,7 +17,3/-17,2

Table 5.3: A comparison for peak voltages at 20 kV level.

The noticeable deviation between simulation and measurements at the 20 kV level, is the damping

rate of the lower frequency component. That applies especially for phases a and c. Furthermore,

the overvoltages reach considerably higher values in the simulation than in measurement. The

higher frequency component experienced in the simulation is not present in the measurements

as these are damped/filtered.


5.3.5 Comparison of voltage at 0,4 kV level

Figure 5.32 shows the results for voltage at 0,4 kV level.

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−400

−200

0

200

400Phase A

PS

CA

D [V

]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−400

−200

0

200

400

Mea

sure

men

ts [V

]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−400

−200

0

200

400Phase B

PS

CA

D [V

]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−400

−200

0

200

400

Mea

sure

men

ts [V

]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−400

−200

0

200

400Phase C

PS

CA

D [V

]

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−400

−200

0

200

400

Mea

sure

men

ts [V

]

Time [s]

Figure 5.32: The simulation and measurement results for the voltage at 0,4 kV level.

For the voltage at 0,4 kV level, a comparison of peak voltages is made:

Phases [V]

a b c

PSCAD +440/-368 +328/-326 +339/-357

Measurement +369/-311 +339/-319 +318/-332

Table 5.4: A comparison for peak voltages at 0,4 kV level.

As the case for the 20 kV level, the damping at the 0,4 kV level is too low for the lower

frequency component. The peak voltages reach considerably higher values in the simulation

than in measurement. The higher frequency component is not damped through the 20/0,4 kV

transformer as experienced in measurements.

5.3 Results 77

5.3.6 Harmonic analysis at 150 kV level

A frequency analysis is made for the voltage at the 150 kV level, which represents the voltage

measured in Ferslev. This is compared with a Fourier analysis of the measured voltage.

The harmonic analysis for phase a is shown in figure 5.33.

0 1000 2000 3000 4000 5000 60000

10

20Phase A

Pscad

0 1000 2000 3000 4000 5000 60000

10

20

Measurement

0 1000 2000 3000 4000 5000 60000

10

20

Phase B

Per

cent

age

of th

e fu

ndam

enta

l [%

]15

0 kV

leve

l

Pscad

0 1000 2000 3000 4000 5000 60000

10

20

Measurement

0 1000 2000 3000 4000 5000 60000

10

20

Phase C

Pscad

0 1000 2000 3000 4000 5000 60000

10

20

Frequency [Hz]

Measurement

Figure 5.33: Results for the Fourier analysis of the voltage vaveform at 150 kV level, 20 ms after

switching.

The most significant results from the frequency analysis of the voltage at 150 kV are discussed.

The two frequency bands are apparent for both simulation and measurement, though with some

deviation. For the simulation, magnitudes of all frequencies above the fundamental are much

higher than for the measurement, indicating yet again lack of damping. In the simulation, the

appearance of the 8th harmonic is the most dominant for the lower frequency band, whereas the

7th harmonic is the dominant in the measurements. The simulation gives both higher magnitude,

and higher frequency for the high frequency band. The measurements give a center frequency of

ca. 3,8 kHz, whereas the PSCAD simulation gives a frequency of ca. 4,2 kHz.


A frequency analysis of the voltage at the other voltage levels is not conducted, as non significant

damping is experienced through the transformers, and hence will this yield the same result as

for the 150 kV level.

As seen from the results, both by comparison of voltage waveforms, and from Fourier analysis,

is a lack of damping apparent in the simulation. There is a suspicion that the skin effect may

not be accounted for in the simulation software. Before further conclusions can be made about

the validity of the simulations, an investigation for this is performed.

5.3.7 Test of frequency dependency of the transmission line model in PSCAD

In order to investigate the frequency dependency of the resistance in the transmission line model,

a test is conducted on section 2. An estimation of the expected PSCAD results is needed, in order

to evaluate the results from PSCAD. This is done by calculating the input impedance of the

transmission line for the case of a frequency dependent series resistance and for the case of a

frequency independent series resistance. The equivalent circuit for the transmission line is shown

in figure 5.34, when a π equivalent is used. To simplify the analysis, the cable end is short

circuited.

ua(t) C

1

2R1

2R

1

2Rg

1

2Rg

1

2L1

2L

Zin

Figure 5.34: A T equivalent circuit for the transmission line, where the resistance in the ground is

included. The end of the transmission line is short circuited.

The input impedance to the circuit shown in figure 5.34:

Zin = (1

2R +

1

2Rg + jω

1

2L) +

1jωC

· (12R + 1

2Rg + jω 12L)

1jωC

+ (12R + 1

2Rg + jω 12L)

(5.14)

This equation is used to calculate the input impedance for both the frequency dependent

resistances and the frequency independent resistances. The resistance in the ground Rg can

be found as a function for the frequency [1, Vørts, p. 116]:

Rg = 0, 00099 · f [Ω/km] (5.15)

The AC resistance is calculated by using a method presented by Hurley [14, Hurley, p. 6]. Here,

an approximated method for calculating the frequency dependent resistance for a round wire is

presented. First the skin depth is calculated for the given frequency:

δ =

√ρ

fπµ(5.16)

5.3 Results 79

Where:δ is the skin depth

ρ is the resistivity of the conductor

µ is the permeability of the conductor

The frequency dependent resistance Rac can now be calculated:

Rac = Rdc · ks (5.17)

ks = 1 +(r/δ)4

48 + 0, 8(r/δ)4. . . r/δ < 1, 7 (5.18)

ks = 0, 25 + 0, 5 ·(

r

δ

)

+3

32·(

δ

r

)

. . . r/δ > 1, 7 (5.19)

Where:ks is the correction factor between the DC and the frequency dependent

resistancer is the radius of the conductor

In the case of a radius of 18,1 mm, equation 5.18 is used for frequencies below 65 Hz and equation

5.19 is used for frequencies above 65 Hz.

The input impedances are now calculated, both with and without the frequency dependent

resistance, for frequencies between 1 Hz and 2000 Hz, by using equation 5.143. The results are


In order to evaluate the frequency dependency of the series resistance in the transmission line

model, simulations are performed as shown in figure 5.35. Here only one phase is used in order

to eliminate induced voltage. The end of the line is grounded through a small resistance, acting

as short circuit. The transmission line is configured as described in section 5.1.4 on page 62.

IaA

B

C

R=0 U_an

Section2

1

Section2

1

3 [o

hm

]

Section2

T

3 [o

hm

]

0.0

01

[oh

m]

Figure 5.35: Setup for evaluation of frequency dependency of the resistance of the transmission line

model in PSCAD.

The simulation is conducted for 15 frequencies in the interval between 50 Hz and 2000 Hz. The

phase angle is measured along with the peak voltage and the peak current. The input impedance

can then be found:

Zin =Ua,peak

Ia,peak

6 φ (5.20)

(5.21)

3This procedure is performed in MatLab as shown in appendix G.11.


The real part of the input impedance is now plotted as a function of the frequency. This is done

in figure 5.36, where the simulation results are compared with the calculated results for both the

frequency independent and the frequency dependent series resistance.

0 200 400 600 800 1000 1200 1400 1600 1800 20000

5

10

15

20

25

30

35

40

45

50

Rea

l par

t of i

nput

impe

danc

e (Ω

)

Frequency [Hz]

Calculated with frequency independent series and ground resistanceCalculated with frequency dependent series and ground resistanceCalculated from PSCAD simulation

Figure 5.36: Comparison of real part of the input impedance, between measurements in PSCAD

and calculated values with both frequency dependence and independence of the series resistance of the

transmission line.

In figure 5.36 the real part of the input impedance for the transmission line is found for different

frequencies.

The calculated real part of the input impedance increases with the frequency, as shown by the

dottet line in figure 5.36.

As shown by the slight rise of the solid line in figure 5.36, the real part of the input impedance,

calculated with frequency independence resistances, does rise with the frequency. This rise

is caused by the change in the reactances of the capacitance and inducatance in the parallel

connection.

As shown by the “–o–” line in figure 5.36, does the real part of the simulated input impedance

increase with frequency.

The real part of the input impedance measured in PSCAD varies from the expected values, as

shown in the figure. This difference can be caused by different calculation methods, and the fact

that the PSCAD model s more detailed than the simplified model which calculation is conducted

from. From the above-mentioned it can be concluded that the transmission line model do account

for frequency dependence of the resistances. This is also assumed to be the case for the cable

model.


As shown, there are some differences between the simulation and the measurements. The

difference is reviewed in the following.


• Lack of dampingAs shown in section 5.3.7, the skin effect is accounted for in the frequency dependent (phase)

model4. This rules out, that the lack of damping is originated from frequency independency

of the transmission line and cable models. Therefore, the only frequency independent,

resistive component in the PSCAD model, is the real part of the short circuit impedance.

Implementation of frequency dependency in the short circuit resistance would increase the

damping factor of the system. The short circuit power seen from Ferslev is given for

maximum and minimum apparent power, resulting in a maximum and minimum short

circuit impedance. The angle of this impedance is unknown, which can cause difference

between the measurements and the simulation. The angle is set to a typical value of 80

in the simulation, here a decrease in the angle would give a larger short circuit resistance,

and thereby more damping.

• FrequenciesThe frequency components from the Fourier analysis did indicate difference between the

simulation and the measurements. This difference could stem from the short circuit

impedance, as the short circuit inductance can obtain values varying between 14,4 mH

and 35,6 mH, depending on whether the maximum or minimum is used. Another factor

could be that the capacitance of the cable is too small.

• Steady state currentThe steady state current in the simulation did show a smaller magnitude than the measured.

This indicates that the simulated system has too high impedance. A possible reason for

this would be a too small capacitance in the cable sections.

• Damping through transformersIn the measurements, the higher frequencies are filtered through the transformers. This is

not the case for the simulation, where the high frequencies from the 150 kV level propagate

to the lower voltage levels, only with a little damping.

Due to the difference between the simulation and measurements, an evaluation of the simulation

models is conducted in the next chapter.

4This was also confirmed by mail from Dr. Dharshana Muthumuni from PSCAD.

Chapter 6

Evaluation of the PSCAD simulation

As shown in the previous chapter, large differences between the simulation and measurements

were present. In this chapter, the simulation is examined in order to minimize simulation errors.

6.1 Estimation of cable and line models

In order to investigate the correctness of the configuration, the parameters given in the cable

datasheet are compared with the parameters calculated by PSCAD. A line constant subprogram

is included in the PSCAD simulation software. This allows an output file to be written, containing

the parameters for the given system. These parameters are given in a matrix form, both in a

sequence and phase form. The output files are shown in appendix F.

6.1.1 Cable shunt admittance

The phase data for the shunt admittance, is a 6×6 matrix, as the cable has three phases each

containing one phase conductor and one screen conductor.

The shunt admittance for conductor 1 in section 3, is given for the full length of the cable (21

km), in the matrix shown in figure F.2 on page F-55:

Yshunt = 0, 656 · 10−6 + j1, 34 · 10−3 S (6.1)

The conductance and capacitance can now be found per km, and compared with values given in

the datasheet. This is done in table 6.1.

Datasheet [per km] PSCAD [per km]

Can 243 nF 202 nF

Gan 12,6 nS 31,2 nS

Table 6.1: A comparison of the shunt conductance and shunt capacitance of the cable, from values

calculated by PSCAD and calculated from datasheet.

6.1.2 Cable sequence impedance

The sequence impedance for the cable is compared with the values from the datasheet, given in

[Ω/km]:

Datasheet [Ω/km] PSCAD [Ω/km]

Z0 0,188+j0,0706 0,226+j1,87

Z1 0,0331+j0,108 0,0787 +j0,0711

Z2 0,0331+j0,108 0,0787 +j0,0711

Table 6.2: A comparison of the sequence impedance of the cable, from values calculated by PSCAD

and given in datasheet.

84 Evaluation of the PSCAD simulation

From the comparison of the shunt admittance in table 6.1 and sequence impedance in table 6.2

the following can be concluded:

• The shunt conductance calculated by PSCAD is 2,47 times the value given in the datasheet.

• The shunt capacitance in PSCAD is only 83% of the value given in the datasheet.

• The zero sequence resistance in PSCAD is 1,2 times the datasheet value†.

• The zero sequence reactance in PSCAD is 26,5 times the datasheet value†.

• The positive sequence resistance in PSCAD is 2,37 times the calculated value.

• The positive sequence inductance in the datasheet is 1,51 times the PSCAD value.

† A comparison of the zero sequences is difficult to make, as the circumstances it is calculated

for are unknown.

The above-mentioned comparison, verifies that the cable section is not configured to give the

same results, as supplied in the datasheet.

The importance of this is significant, as the capacitance of the simulated cable section, will give

rise to higher frequency components than were measured. Furthermore, the amplitude of the

steady state current will obtain a too low value, as the input impedance will obtain too high

value. A re-evaluation of the cable configuration is needed. In figure 6.1 the configuration of the

cable is shown.

20.75 [mm]39.05 [mm]

39.435 [mm]46 [mm]

6.97 [mm]

Conductor

Insul tor 1

She th

Insul tor 2

a

aa

a

b

Figure 6.1: The physical configuration of one of the cable conductors. The representation to the right,

is shown in connection to equation 6.2

From the distances shown in the figure above, the capacitance is calculated. The capacitance of

a cylindrical conductor is found from [10, Serway, p.808]:

C =l

2 · ke · ln(

ba

) [F] (6.2)

ke =1

4πǫ0= 8.987551787 · 109 [N · m2/C2] (6.3)

C =1000 · 2, 3

2keln(

39,0520,75

) = 202, 3 nF/km (6.4)

6.1 Estimation of cable and line models 85

Where:Where ke is the Coulomb constant.

a is the radius of the first conductor

b is the radius to the second conductor

This equation

is given for isolation material that is air. In the case of other dielectric material, the relative

permittivity, in this case 2,3, is multiplied by the capacitance. The length is set to 1000 m, as a

capacitance per/km is desired. This gives identical result to the one found by PSCAD.

A new cable configuration is obviously needed, and this is calculated before inserting in PSCAD The

1,2 mm thick conductor screen, is defined as a part of the conductor, and the dielectric material as

only the isolation material, but not insulation screen and bedding for screen. Thereby is the new

radius a, defined to be 21,95 mm, and radius b, is found by the thickness of the isolation material,

defined in figure 3.5 on page 13 to be 37,95 mm. Thereby is a new capacitance calculated:

C =1000 · 2, 3

2keln(

37,9521,95

) = 233, 7 nF (6.5)

As the conductor cross section has been increased, the hollow center is increased accordingly to

keep the conductor areal 1200 mm2.

Aconductor = 21, 952π − Acenter = 1200 mm2 (6.6)

Acenter = 21, 952π − 1200 = 313.6 mm2 (6.7)

rcenter =

√

Acenter

π= 9, 99 mm (6.8)

The configuration of the corrected cable is shown in figure 6.2

21.35 [mm]37.95 [mm]38.89 [mm]

46 [mm]

9.99 [mm]

Conductor

Insulator 1

Sheath

Insulator 2

Figure 6.2: The configuration of the cable, with corrected distances.

After this configuration, the parameter values are solved for again in PSCAD, the values are listed

in table 6.3 and table 6.4.

Datasheet [per km] PSCAD [per km]

Can 243 nF 222 nF

Gan 12,6 nS 21,4 nS

Table 6.3: A comparison of the shunt conductance and shunt capacitance for the corrected cable, from

values calculated by PSCAD and calculated from datasheet.


The new sequence impedance for the cable is compared with the values from the datasheet, given

in [Ω/km]:

Datasheet [Ω/km] PSCAD [Ω/km]

Z0 0,188+j0,0706 0,191 + j1,86

Z1 0,0331+j0,108 0,0434 + j0,0682

Z2 0,0331+j0,108 0,0434 + j0,0682

Table 6.4: A comparison of the newly found sequence impedance of the cable, from values calculated

by PSCAD and given in datasheet.

As the capacitance has yet not obtained completely the same value as given in the datasheet,

the permittivity of the isolation material is increased. By increasing the permittivity of the

isolation material, the correct value for the capacitance is obtained. In section 6.3, a simulation

is performed with this configuration. With this correction of the cable, the positive sequence

resistance has obtained a value more comparable to the datasheet value. Other parameters, given

in table 6.4 are accepted as adequately accurate. The matrices describing the parameters for the

corrected cable are given in appendix F.3 on page F-56.

6.1.3 Line parameters

The parameters for the overhead line section, calculated in PSCAD are compared with the

calculated parameters from appendix A.

Line shunt admittance

The phase data for the shunt admittance, is a 4×4 matrix, as the overhead line section consist

of three phase conductors and one ground conductor. The shunt admittance for conductor 1 is

given for the full length of section 2, 18,93 km, shown in the matrix in figure F.1 on page F-54:

Yshunt = 2, 5 + j46, 1 µS (6.9)

The conductance and capacitance can now be compared with the values calculated in appendix

A:

Calculated [per/km] PSCAD [per/km]

Can 7,87 nF 7,74 nF

Gan 132 nS 132 nS

Table 6.5: A comparison of the shunt admittance of the overhead line, from values calculated by PSCAD

and values calculated in appendix A.

6.2 Estimation of models for shunt reactor and short circuit impedance 87

Line series impedance

The sequence impedance for the line in PSCAD is compared with the calculated values in table

6.1.3.

Calculated [Ω/km] PSCAD [Ω/km]

Z0 0,175 + j0,983 0,186 + j1,32

Z1 0,0477 + j0,395 0,0439 +j0,397

Z2 0,0477 + j0,395 0,0439 +j0,397

Table 6.6: A comparison of the sequence impedance of the overhead lines, from values calculated by

PSCAD and values calculated in appendix A.

From the comparison of the shunt admittance and series impedance the following can be

concluded:

• The shunt capacitance and shunt conductance are approximately the same for PSCAD and

calculations.

• The positive sequence impedance is approximately the same for PSCAD and calculations.

An 8,6% difference in the resistive part is though present.

• The zero sequence reactance calculated by PSCAD is 34% to high compared to calculated

value.

The values from PSCAD are in good accordance with the calculated values. The PSCAD model is

therefore considered to be correct configured.

In the next section, a discussion of the validity of the short circuit impedance and reactor model

is performed.

6.2 Estimation of models for shunt reactor and short circuit

impedance

An estimation of the factors which can affect the simulation, are described in the following, for

the short circuit impedance and the reactor model.

6.2.1 Short circuit impedance

As described in the section 3.2 on page 9, the short circuit impedance was estimated to 4, 66 80 Ω.

As this is an estimate, the actual short circuit impedance can vary from the estimate due to the

following reasons:


• The short circuit impedance is calculated for maximum and minimum apparent powerIn section 3.2 on page 9, the short circuit impedance was calculated from both the maximum

and minimum short circuit power. At the time of the measurement the short circuit

impedance is unknown. As the data for the maximum short circuit power are used in the

simulation, this could give rise to results that are different to those measured. If the short

circuit impedance is calculated with a value closer to the minimum short circuit power,

the inductance will obtain a larger value. If this is the case, the frequencies obtained in

simulation will be affected.

• The angle for the short circuit impedance is unknownIn continuation of the previous item, the angle of the short circuit impedance is unknown.

This will affect the short circuit impedance in such way that a larger angle will decrease

the resistance/increase the inductance, and vice versa, a smaller angle will increase the

resistance/decrease the inductance. The angle of the short circuit impedance is an

important factor, as the short circuit resistance will affect the damping for the whole

system, and the inductance has influences on the generated frequencies.

• The real part of the short circuit impedance is frequency independentThe short circuit impedance used in the simulation consists of a series connection of a

resistor and an inductor. The resistance represents the replacement resistance of the current

path between Ferslev and the power plants. As this path will consist of a combination

of transmission lines and bus bars, the resistance will be frequency dependent. The

frequency dependency of the short circuit resistance can be difficult to determine. In a

single transmission line consisting of round wires, the skin dept at different frequencies can

be calculated, and thereby is it possible to find the frequency dependent resistance. This

is not the case for the short circuit impedance, as the path from the power plants consist

of multiple transmission lines and bus bars of different sizes.

6.2.2 Reactor model

The reactor model used in the simulation is a simple model which only includes the losses in the

core and windings, and the self and mutual inductances. The possible impacts caused by the

simplified model are described in the following.

• Saturation of core is not includedWhen the grid is switched-on, large inrush currents will be experienced in the system.

Though these currents will mostly charge the cable capacitances, a side effect of these

transient is that the currents obtain a slowly decreasing DC-component. As this DC

component will cause a larger current in the reactor and hence high flux density, saturation

can occur. In order to check the inrush current, a simulation is performed, resulting in the

current, shown in figure 6.3. This simulation is performed precisely in the same manner

as the simulation described in chapter 5. The only difference is the measurement of the

current that flows into the reactor at the switch-on moment measured.

6.2 Estimation of models for shunt reactor and short circuit impedance 89

0 0.05 0.1 0.15−0.5

0

0.5

Cur

rent

[kA

]

Time [s]

Figure 6.3: The initial current sequence in the reactor.

The magnetic flux density can be calculated from F = NI = ΦR. Knowing Φ = BA and

R = N2

L, an expression for B is given:

B =L · iN · A (6.10)

The reactor is described in appendix B, where the number of windings is 1194, the

inductance for phase a is 2,307 H and the cross section of the air gap in the limb is

0,2827 m2. The maximum current in the reactor has a magnitude of 341 A, from which

the peak magnetic flux density is calculated:

Bpeak =L · Ipeak

N · A (6.11)

Bpeak =2, 307 H · 341 A

1194 · 0, 2827m2= 2, 330 T (6.12)

The saturation limit for the reactor is defined to 2,03 T [15]. It is therefore apparent, that

the simple reactor model is not sufficient in the start sequence, where the reactor will enter

saturation. When the core goes into saturation, the permeability will approach the one of

air, resulting in a high reluctance, and thereby a small inductance, thereby causing larger

currents to be present. From Amperes law (∮

~H · d~s = N · I), can it be seen that when

the core enters saturation, non-linearity occurs due to hysteresis, and hence will harmonic

currents be present.

• RemanenceRemanence occurs when the magnetization is removed, as some flux density will remain.

As the focus is on the switch-on, and the system assumed totally discharged, it is assumed

that there will be no remanence present. Furthermore, the air gaps in the phase limbs of

the core will decrease the appearance of remanence due to its large reluctances.

As the short circuit impedance can not be estimated correctly, the value used earlier is unchanged.

The estimations above for short circuit impedance and reactor are deemed to give minimal impact

on the frequencies in simulations, compared to the significance of the size of cable capacitance.

In the next section is the system therefore simulated with the cable sections configured to give

the capacitance given in the datasheet.


6.3 Correction of cable capacitance

In this section, the system is simulated with the corrected cable section. The configuration of the

cable has been corrected as to obtain the right capacitance, as described earlier, given the value

of 0,243 µF/km in the datasheet. The simulation is conducted in precisely the same manner

as the simulation described in chapter 5. Only voltages and currents at 150 kV level are shown

here, as is the Fourier analysis.

The shunt admittance and sequence impedance of the corrected cable component, is found by

reading the line constant program output file, the matrices for the corrected cable are given in

appendix F.3:

Corrected Cable Parameters [per km]

Yshunt 21,4·10−9 + j74, 9 · 10−6 S/km

Z1 43,4·10−3 + j68, 3 · 10−3 Ω/km

Table 6.7: The parameter values for the corrected cable component.

6.3.1 The voltage waveform

The simulated voltage waveforms, that represent the voltage measured in Ferslev, are shown in

figure 6.4, for the corrected cable component.

0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055−200

−100

0

100

200

Time [s]

Vol

tage

[kV

]

Figure 6.4: The simulated voltage waveform, with corrected cable component

The peak values for the recorded voltages are given in table 6.8

Peak voltages [kV]

Phases a b c

Corrected cable +193/-155 +144/-138 +141/-150

Earlier simulation +199/-159 +141/-140 +146/-154

Measurement +159/-134 +154/-133 +146/-162

Table 6.8: A comparison for peak voltages at 150 kV level, simulated with the corrected cable component.

Results of this simulation show that the overvoltages for phase a are still out of proportion,

while the other phases are more like the measured overvoltages. Lack of damping of the lower

harmonic is still present.

6.3 Correction of cable capacitance 91

6.3.2 The current waveform

The initial current sequence, that represents the current that was measured in Ferslev at 150 kV

level is shown in figure 6.5, simulated with the corrected cable component.

0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055−1.5

−1

−0.5

0

0.5

1

1.5

Time [s]

Cur

rent

[kA

]

Figure 6.5: The initial current sequence, with the corrected cable component.

The current waveform is recorded over 3 seconds, to show how the change in capacitance

influences the steady state current magnitude. The peak and steady state currents are compared

in table 6.6, with the measurements and the earlier simulations where the cable capacitance was

deemed too small.

0.5 1 1.5 2 2.5 3−1.5

−1

−0.5

0

0.5

1

1.5

Time [s]

Cur

rent

[kA

]

Figure 6.6: The current waveform simulated over 3 s. The steady state magnitude is here 90 A, peak

to peak.

Peak currents [A]

Phase a Phase b Phase c

Corrected cable +1564/-1502 +893/-615 +558/-992

Earlier simulation +1474/-1377 +798/-589 +493/-896

Measurement +1103/-1423 +1134/-1773 +666/-1195

Steady state magnitude, peak-to-peak [A]

Corrected cable 90 90 90

Earlier simulation 14 16 14

Measurement 110 110 110

Table 6.9: A comparison of peak currents, and steady state magnitude at 150 kV level.

Here is the steady state current obviously more comparable to the measured. Initial peaks have

risen considerably.


6.3.3 Fourier analysis

A Fourier analysis of the frequency components was conducted for the simulation. This is shown

in figure 6.7.

0 1000 2000 3000 4000 5000 60000

10

20

30

Frequency [hz]

Per

cent

of t

he fu

ndam

enta

l [%

]

Figure 6.7: A Fourier analysis for one period after switching.

The Fourier analysis shows that the corrected cable component has lowered the center frequency

of the lower frequency band. The center frequency of the lower frequency band is now 350 Hz,

which is identical to the measured. The magnitude of the this harmonic component, of 26 % of

the fundamental frequency, is though still significantly higher than in measurements, where the

magnitude of the 7th harmonic was ca. 5% of the fundmental. For the high frequency component

is the magnitude ca. 8% of the fundamental, while in measurement it was ca. 3%.

6.3.4 Summary of simulation with corrected cable capacitance

The most significant results, obtained by correcting the capacitance of the cable, is summarized.

• The initial peak currents have risen considerably in for all phases in simulation with the

newly configured cable.

• The steady state currents are now comparable to the measured steady state current.

• The Fourier analysis indicates the 7th harmonic as the dominant one, as in the measure-

ments, though with larger magnitude. In measurements this component had ca. 5% of the

fundamental frequency.

• Lack of damping is still apparent.

The damping of the lower harmonic component was not increased by correcting the cable. The

resistance needed to obtain the desired damping rate, is found in the next section.

6.4 Estimation of missing resistance 93

6.4 Estimation of missing resistance

This estimation is made with the same simulation configuration as before. By increasing the

source resistance, until the voltage waveforms for the simulation and measurements, are close to

identical, the missing damping is found.

By tuning the source resistance to 20 Ω, the desired damping rate is obtained. This is shown in

figure 6.8.

0.005 0.01 0.015 0.02 0.025 0.03−200

−100

0

100

200

Vol

tage

[kV

]

Simulated

0.005 0.01 0.015 0.02 0.025 0.03−200

−100

0

100

200

Time [s]

Vol

tage

[kV

]

Measured

Figure 6.8: The voltage waveforms, with a 20 Ω resistance as a source resistance.

Estimation of the voltage waveforms shows how the desired damping rate is obtained for

two phases, whereas phase b is to much damped. A slight difference is apparent between

measurements and simulation, as the phases appear to experience different damping.


0.5 1 1.5 2 2.5

−1000

0

1000

Cur

rent

[kV

]

Simulated

0.5 1 1.5 2 2.5

−1000

0

1000

Cur

rent

[kV

]

Measured

2.72 2.73 2.74 2.75 2.76 2.77 2.78

−100

0

100

Time [s]

Cur

rent

[kV

]

Measured

Figure 6.9: The current waveforms over 100 ms, with the 20 Ω source resistance

The high amplitude current spikes are present for approximately the same amount of time. There

is though a significant difference in that the simulated system is seemingly free of transients, 20

ms after the switch-on instant. An interesting phenomena is seen, when closer look is taken at

the measured steady state current, shown in the bottom diagram in figure 6.9. A significant

DC-component is still apparent, which might stem from error in measurement, and the current

waveforms are highly distorted, indicating non-linearities of some kind in the system.

0 1000 2000 3000 4000 5000 60000

5

10

15

20

25

30

Frequency [hz]

Per

cent

of t

he fu

ndam

enta

l [%

]

Figure 6.10: A Fourier analysis of the frequency components when the system contains the corrected

cable, and a 20 Ω source resistance.

The Fourier analysis of the harmonic currents is now comparable to the Fourier analysis of the


measurements, shown in appendix C.2.5 on page C-39. In the simulation has the 7th harmonic a

magnitude of 8%, whereas in the measurements this was ca. 5% of the fundamental frequency.

The 77th harmonic in simulation has a magnitude of 3,5% and the 76th in measurements has a

magnitude of approximately 3%.


The cable section was evaluated by comparing the calculated values from PSCAD and the provided

data for the cable. Here, differences in the values were found, both for the capacitance and the

series impedance. As the capacitance in the cable is an important component, this was changed

to its correct value, first by configuring the cable component differently, and then by giving

new value to the permittivity of the isolation material. A new simulation with the corrected

capacitance was conducted, indicating more comparable results in relation to steady state current

magnitude and harmonic frequencies.

Both the magnitude of the short circuit impedance and its angle are estimated values, as the

actual value at the time of measurement was not provided. Ideally, a simulation of the system

would incorporate the correct short circuit impedance, and a more advanced model of the shunt

reactor, where saturation is accounted for.

A new simulation where other parameters are varied has been decided against. It is though

acknowledged, that an increased resistance, could be achieved by changing the short circuit

impedance, and by implementing a frequency dependent short circuit resistance. The increased

damping, obtained with such simulation is assumed too low. By replacing the source resistance

with a 20 Ω resistance, the sufficient damping is obtained, and waveforms are obtained that bear

resemblance to the measured waveforms. As this resistance is out of proportion for this type of

system, and that difference between phases is apparent, it is assumed that the damping is not

only to be found in the form of resistivity, but also in the simplified model for the reactor.

Closer look at the measured current waveform in assumed steady state, show how much the

current is distorted. This indicates non-linearity in the system, which is not accounted for in

the constructed model. As for this, a simulation that has identical waveforms compared with

measurements will not be possible to obtain.

In the next chapter, proposals for minimizing the switching overvoltages are made. These

proposals are supported by simulations of the system, where the above described cable component

is used.

Chapter 7

Proposal for minimizing switching

overvoltages

As shown both through measurements and in simulation, switching overvoltages do occur. In

the following chapter, the problem with switching overvoltages is reviewed, and a proposal for

minimization of such overvoltages is given.

7.1 Evaluation of switching overvoltages at the 150 kV level

Through measurements and simulations, a generation of overvoltages on the 150 kV level has

been shown. The voltages from simulations and measurements are compared with the insulation

level for 170 kV in table 7.1.

Phase Measured [% of prior voltage] Simulated [% of prior voltage] BIL

a 159 kV [18,7%] 199 kV [23,0 %] 550-750 kV

b 154 kV [14,9%] 141 kV [0,8 %] 550-750 kV

c -162 kV [20,9%] -154 kV [4,2 %] 550-750 kV

Table 7.1: A comparison of simulated and measured magnitude of the overvoltages, with the insulation

level for 170 kV given in [16, Kuffel, p. 493].

As shown in table 7.1, the overvoltages stay below the insulation level for 170 kV by a large

margin. It can therefore be concluded that the switching overvoltages are not likely to cause

faults in the system.

Even though the switching overvoltages stay below the isolation level by a large margin, such

overvoltages are still unwanted, as the equipments are subjected to stress load. This stress load

can cause a decrease in the lifetime of the insulation material, and thereby lower the lifetime for

the equipments. Furthermore, the switching overvoltages will propagated to lower voltage levels.

This issue is evaluated in the following section.

7.2 Evaluation of propagation of switching overvoltages

The voltages, on the 0,4 kV level, from measurements and simulations are listed in table 7.2.

Phase Measured [% of prior voltage] Simulated [% of prior voltage]

a 382 V [18,4%] 440 V [23,0 %]

b 339 V [7,2%] 328 V [0,8 %]

c -332 V [4,5%] -357 V [8,8 %]

Table 7.2: Maximum overvoltages at the 0,4 kV level shown in measurements and simulations. It should

be noted that the measured prior voltage of phase b and c, is not 230√

2 V, for more details see appendix

C.3.

98 Proposal for minimizing switching overvoltages

From the table it can be seen that the overvoltages from the 150 kV level do propagated to lower

voltage levels. It can though be concluded that the magnitude of these overvoltages are not very

high.

The variation in the supply voltage is defined in the standards [17, DS/EN 50160] and quoted

below:

2.3 Supply voltage variations

Under normal operating conditions, excluding situations arising from faults or voltage

interruptions,

• during each period of one week 95 % of the 10 min mean rms values of the supply voltage

shall be within the range of Un ± 10%.

NOTE 1: Until the year 2008 the range of voltage may differ from these standard values, in accordance with HD 472 S1.

• all 10 minutes mean rms values of the supply voltage shall be within the range of Un

+10%/-15%.

NOTE 2: In cases of electricity supplies in remote areas with long lines the voltage could be outside the range of Un

+10%/-15%. Costumers should be informed.

As this is defined for mean values for a period of 10 minutes, it is considered that the propogated

overvoltages will have minimal influence on the power quality, as they are present for only a few

hundreds of milliseconds or less.

From the above-mentioned, the propagation of switching overvoltages is deemed of little

significance. Though there are no problems with switching overvoltages in the system FER3-

THØ3, problems can occur in other systems. The transmission system from Horns Rev 2 contain

a series connection of two cables with a total length of 72 km. In such system, the size and the

propagation of switching overvoltages can become an issue. A proposal for minimization of

switching overvoltages is therefore examined in the following section.

7.3 Proposal for minimization of switching overvoltages

Switching overvoltages can be reduced in different manners. Two different principles are described

in the following, and afterwards simulated in order to evaluate the effect of such implementations.

7.3.1 Zero crossing closure

In this principle, each phase of the circuit breaker closes at the zero crossing of the voltage, as


7.3 Proposal for minimization of switching overvoltages 99

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04−400

−300

−200

−100

0

100

200

300

400

Phase a [kV,kA]

Time [s]

Switch-on instant

a b c

Figure 7.1: Principle of zero crossing closure.

This principle will limit the transient voltages, as none of the phases experience instantaneous

change in voltage. Instead, the transition follow the 50 Hz waveform, and thereby a more slowly

charging of the capacitances in the system. The downside of this, is that a large asymmetry in

the current will occur, as two phases will obtain the same DC component, and the last phase

will obtain a opposite DC component. This will cause a considerable zero sequence current at

the moment of switch-on.

7.3.2 Pre-insertion resistors in circuit breakers

In this principle, a resistance is inserted in series with the circuit breaker in order to increase the

initial resistance of the system and thereby limit the inrush current. The resistor is only present

for a short duration, typically for 5-15 ms [18, Beanland], or until the system is fully charged,

determined by the system time constant. The size of the resistor is typically 300 - 500 Ω, but can

vary for different cases [19, Ryan, p. 52]. At the time the resistor is short circuited in the setup,

transients will occur as the system is not totally charged at this time, due to the voltage drop

over the resistor. These transients are though small compared with those experienced without

pre-inserting resistors.

7.3.3 Simulation of zero crossing closure

The main simulation for the zero crossing closure is configured as described in chapter 5. As

described in section 6.3 on page 90 the capacitance is corrected to the value provided from the

datasheet. This correction is included in the following simulation. The setup for the simulation

is shown in figure 7.2.


U_b

I150k_b

I150k_a

I150k_c

U_a

U_c

C1Section3

S1

C2

S2

C3

S3

C1Section3

S1

C2

S2

C3

S3

U150k_a

U150k_b

U150k_c

Section1

C

Section2

1

Section2

1

Section2

T

A

B

C

A

B

C60 [kV]

#2#1

150 [kV]

80.0 [MVA]A

B

C

A

B

C20.0 [kV]

#2#1

60.0 [kV]

16.0 [MVA]A

B

C

A

B

C0.42 [kV]

#2#1

24 [kV]

0.630 [MVA]

U20k_c

U20k_b

U20k_aU60k_a

U60k_b

U60k_c U04k_c

U04k_b

U04k_a 1 [ohm]

1[ohm]

1 [ohm]3

[oh

m]

3 [o

hm

]

C1Section1

S1

C2

S2

C3

S3

C1Section1

S1

C2

S2

C3

S3

A

B

C

R=0 14.4 [mH]

14.4 [mH]

14.4 [mH]

1 [ohm] 1 [ohm]

1 [ohm] 1 [ohm]

1 [ohm] ____1 [ohm]

1 [ohm]

1 [ohm]

1 [ohm]

I

ic ib ia

oc ob oa

Ra

La

Rb

Lb

Rc

Lc

Section3

C

BRK_A

BRK_B

BRK_C

0.8 [ohm]

0.8 [ohm]

0.8 [ohm]

Figure 7.2: Simulation setup for the zero crossing closure.

To represent the circuit breakers, three individual breakers are used (BRK_A, BRK_B and BRK_C

in figure 7.2), each with the same configuration of resistances as described in 5.1.2 on page 58.

Each breaker is configured to close at the zero crossing of each phase resulting in the following

times, when phase a crosses zero at t = 0, 015 s, as shown in figure 7.1:

ta,zero = 0, 015 s (7.1)

tb,zero = ta,zero + 2 · Tperiod

6= 0, 015 + 2 · 0, 020

6= 0, 02167 s (7.2)

tc,zero = ta,zero +Tperiod

6= 0, 015 +

0, 020

6= 0, 01833 s (7.3)

As shown, the sequence of the connection is: phase a, phase c, phase b. This sequence is chosen

to lower the asymmetry of the currents after switch on. As the chosen sequence will force the

DC component for phase c to be opposite of the other phases.

The three breakers are configured as shown in figure 7.3.

TimedBreaker

LogicOpen@t0

BRK_A

TimedBreaker

LogicOpen@t0

BRK_B

TimedBreaker

LogicOpen@t0

BRK_C

Figure 7.3: Configuration of the timer for the three breakers.


Simulation results

The voltage behaviour is shown in figure 7.4 for zero crossing closure, where the switch-on time

for the three phases is given in equation 7.1, 7.2 and 7.3.

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−200

−100

0

100

200

Pha

se a

[kV

]

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−200

−100

0

100

200

Pha

se b

[kV

]

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−200

−100

0

100

200

Pha

se c

[kV

]

Time [s]

Figure 7.4: Voltage on the three phases with the zero crossing closure principle.

The simulation of the zero crossing closure shows significant decrease of switching overvoltages

on the 150 kV level. The peak voltages for the simulation with and without the zero crossing

closure are shown in table 7.3.

Phase Without zero crossing With zero crossing

a 199 kV -140 kV

b 141 kV -137 kV

c -154 kV 140 kV

Table 7.3: A comparison of simulated overvoltages, with and without zero crossing switching.

The current behaviour is shown in figure 7.5 for the zero crossing closure.


0 0.5 1 1.5−0.5

0

0.5

Pha

se a

[kV

]

0 0.5 1 1.5−0.5

0

0.5

Pha

se b

[kV

]

0 0.5 1 1.5−0.5

0

0.5

Pha

se c

[kV

]

Time [s]

Figure 7.5: Current in the three phases with the zero crossing closure principle.

All the currents in figure 7.5 have a DC component. Here two of the DC components are negative

and one is positive. This will result in a negative DC component, flowing as zero sequence current.

The peak currents from simulation with and without zero crossing closure are listed in table 7.4.


a 1474 A -405 A

b 798 A -405 A

c -896 A 402 A

Table 7.4: A comparison of simulated peak currents, with and without zero crossing switching.

As shown are the inrush currents greatly reduced by use of this principle.

7.3.4 Simulation of pre-insertion resistor

The main simulation is configured as described in chapter 5. Like in the previous simulation,

the corrected capacitance1 is used in the simulations. The setup of the simulation is shown in

figure 7.6.

1As described in section 6.3 on page 90.


C

B

A

BRK1

TimedBreaker

LogicOpen@t0

BRK1

U_b

I150k_b

I150k_a

I150k_c

U_a

U_c

C1Section3

S1

C2

S2

C3

S3

C1Section3

S1

C2

S2

C3

S3

U150k_a

U150k_b

U150k_c

Section1

C

Section2

1

Section2

1

Section2

T

A

B

C

A

B

C60 [kV]

#2#1

150 [kV]

80.0 [MVA]A

B

C

A

B

C20.0 [kV]

#2#1

60.0 [kV]

16.0 [MVA]A

B

C

A

B

C0.42 [kV]

#2#1

24 [kV]

0.630 [MVA]

U20k_c

U20k_b

U20k_aU60k_a

U60k_b

U60k_c U04k_c

U04k_b

U04k_a 1 [ohm]

1[ohm]

1 [ohm]

3 [o

hm

]

3 [o

hm

]

C1Section1

S1

C2

S2

C3

S3

C1Section1

S1

C2

S2

C3

S3

A

B

C

R=0 14.4 [mH]0.8 [ohm]

14.4 [mH]0.8 [ohm]

14.4 [mH]0.8 [ohm]

1 [ohm] 1 [ohm]

1 [ohm] 1 [ohm]

1 [ohm] 1 [ohm]

1 [ohm]

1 [ohm]

1 [ohm]

I

ic ib ia

oc ob oa

Ra

La

Rb

Lb

Rc

Lc

Section3

C

400.0 [ohm]

BRK

400.0 [ohm]

BRK

400.0 [ohm]

BRKTimed

BreakerLogic

Open@t0BRK

Figure 7.6: Simulation setup for the pre-insertion resistor.

The size of the pre-insertion resistor can be determined from demands for the time constant or

the maximum inrush current. The time constant for the system FER3-THØ3 can be calculated

for the typically values of the pre-insertion resistor 300-500 Ω, when neglecting other capacitances

than the one of section 3 with a vaule of 5,1 µF:

τ300 = R · C = 300 · 5, 1 · 10−6 = 1, 53 ms (7.4)

τ500 = R · C = 500 · 5, 1 · 10−6 = 2, 55 ms (7.5)

As shown, both time constants are below the 10 ms which the resistor is configured to stay in the

circuit. The choice of the resistor is now a compromise between limitation of the inrush current

and the voltage drop across the pre-insertion resistor. Choosing the largest resistor will give a

better limitation of the inrush current. However, the voltage drop across the pre-insertion resistor

will increase, resulting in larger transients when the resistor is short circuited. This will occur as

the voltage across the capacitance will change when the pre-insertion resistor is short circuited.

From these two considerations a compromise has been made, selecting the pre-insertion resistor

to 400 Ω, resulting in a time constant of 2,04 ms.

For each resistance one breaker is implemented in parallel, as shown in figure 7.7.

C

B

A

BRK1

400.0 [ohm]

BRK

400.0 [ohm]

BRK

400.0 [ohm]

BRK

Figure 7.7: Implementation of pre-insertion resistor.


Three single phase breakers are used to short circuit the charging resistance 10 ms after switch-

on. This is done with the same timer module for all three phases (BRK). The two timer modules

for the breakers (BRK1 and BRK) are configured as shown in figure 7.8.

TimedBreaker

LogicOpen@t0

BRK1

TimedBreaker

LogicOpen@t0

BRK

Figure 7.8: Configuration of the two different timer modules.

Simulation results

The voltage behaviour is shown in figure 7.9

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−200

−100

0

100

200

Pha

se a

[kV

]

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−200

−100

0

100

200

Pha

se b

[kV

]

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−200

−100

0

100

200

Pha

se c

[kV

]

Time [s]

Figure 7.9: Voltage on the three phases with the pre-insertion resistor principle.

The transient overvoltages are limited significantly, compared to earlier simulations, as shown in

figure 7.9. As shown, transient voltages occur again when the resistor is short circuited a t = 0.2.

The peak voltages for simulation with and without the pre-insertion resistor are listed in table

7.5



a 199 kV -141 kV

b 141 kV 135 kV†c -154 kV 135 kV†

Table 7.5: A comparison of simulated peak currents, with and without pre insertion of resistors. †Shows

no overvoltages.

The switching overvoltages are limited drastically by use of the pre-insertion resistor. For phases

b and c, no overvoltages are present.

The current behaviour is shown in figure 7.10

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2−0.5

0

0.5

Pha

se a

[kV

]

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2−0.5

0

0.5

Pha

se b

[kV

]

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2−0.5

0

0.5

Pha

se c

[kV

]

Time [s]

Figure 7.10: Current in the three phases with the pre-insertion resistor principle.

The peak currents for simulation with and without the pre-insertion resistor are listed in table

7.6


a 1474 A -336 A

b 798 A 146 A

c -896 A -185 A

Table 7.6: A comparison of simulated overvoltages, with and without pre-insertion of resistors.

From table 7.6, it can be seen that resistor pre-insertion, limits both the transient peak currents,

and the DC component of the current.


As shown in the simulations, both the principle with zero crossing closure and the principle with

pre-insertion resistor, shows large improvements of the switching overvoltages. With the results


from the simulation it has been concluded that the pre-insertion principle is the most favorable.

In comparison with the zero crossing closure, only slight differences in overvoltages are present.

The currents experienced in the resistor pre-insertion principle, are symmetrical with only a

small DC component. Whereas, the zero crossing closure has larger DC components, resulting

in a large zero sequence current flowing in the initial state.

Though the pre-insertion resistor principle is the most favorable when focus is on the electrical

behaviour of the system, the zero crossing closure is generally the one of choice. This is based

on the implementation of the two principles. The zero crossing closure is considered to be

relatively easy to implement compared with the pre-insertion resistor. The zero crossing closure

can be implemented with a basic circuit breaker, and an implementation of a control algorithm

to operate the breakers correctly. This principle will therefore not add mechanical parts to the

system. The pre-insertion resistor requires new implementation of equipment, which will increase

the cost and complexity of this solution.

In the system FER3-THØ3, it was concluded that improvements were not needed. The above-

mentioned proposals are therefore only to be implemented in a system where it is deemed

necessary.

Chapter 8

Conclusion

In the following chapter, a summary of the report is conducted, and the most essential conclusions

extracted. The chapter ends with a discussion of the experienced problems and proposals for

future works.

8.1 Conclusion

This project deals with the generated overvoltages in the 150 kV transmission system between

Ferslev-Tinghøj, which is a combination of underground cable and overhead line sections. The

main objectives for this project are listed in the problem statement, section 2.1 on page 6, and

discussed in the following.

• Arrange a three phase representation of the 150 kV transmission system between Ferslev-Tinghøj, valid for the switch-on in Ferslev.In chapter three, each component in the system was described. From these descriptions,

electrical models and their respective parameter values were generated for the respective

subsystems, ending up with a three phase representation of the transmission system

between Ferslev-Tinghøj.

• Explain the experienced transients from a theoretical analysis viewpoint.From the three phase representation, an examination of the switching transients was

conducted. Here the frequencies seen in the measurements were explained from a theoretical

viewpoint. The frequencies calculated from the theory were slightly different compared with

the measured frequencies. This difference was explained with simplifications made in the

theoretical calculations.

• Arrange a simplified dynamic model for the 150 kV system between Ferslev-TinghøjIn order to ease the analysis of the system, a simplified version of the three phase model

was produced, by neglecting components deemed of little significance. This simplification

resulted in a simple electric circuit for analysis. As a part of the simplification, the short

cable section in the grid was neglected.

• Arrange a mathematical solution for the simplified model in MatLab.The differential equations describing the simplified model, were solved by two different

methods. First, an attempt was made to solve these analytically, where it appeared that

the solution was to extensive to comprehend, resulting in a numerical solution. Another

solution method was implemented by arranging a numeric solver in Simulink.

• Verify the MatLab model through measurements and simulation in PSCAD.

The two solutions obtained, were verified with simulations of the simplified model in PSCAD.

This gave identical results to the numeric solver. A slight difference was seen, compared to

the analytical approach, explained by the fact that the mutual inductances in the reactor

model were neglected there. The two solution methods were therefore considered correct.

However, a comparison with the measurements showed lack of damping in all solutions.

108 Conclusion

• Arrange a simulation of the model in PSCAD.In order to evaluate the lack of damping in the simplified model, a simulation of the

transmission system between Ferslev-Tinghøj, was made in PSCAD. Here each model, for

the overhead transmission lines and underground cables, was constructed from physical

parameters. A simple reactor component was created in PSCAD to include mutual induc-

tances. This component does not account for magnetic saturation, which can occur at the

switch-on moment, as shown in section 6.2.2 on page 88.

• Verify the PSCAD model by comparing to measurements and calculated values.Simulations of the system indicated an error in the model, as significant deviation was

apparent in the damping of transients and the frequency components in the system. The

resistances in the line and cable models in PSCAD were therefore examined for frequency de-

pendency. Frequency dependency of the models was verified. For further evaluation of the

simulation, the parameters calculated by PSCAD were controlled against calculated values

and values given in datasheet. From this comparison, it was appeared that the overhead

line parameters, calculated in appendix A did match the same parameters calculated by

the PSCAD line constant program. However, the parameters for the cable sections deviated

significantly from the datasheet values. Adjustment of the cable configuration was made,

which contributed significantly to a solution closer to measurements. A lack of damping

was though still present. In order to evaluate the lack of damping, the value of the source

resistance was varied until the voltage waveform was comparable to the measurement. This

evaluation gave a series resistance of 20 Ω which is an excessive size in such system.

• Arrange a proposal for an optimization of the reconnection.From simulations and measurements, the problem concerning overvoltages was estimated

for the specific case of the transmission system between Ferslev-Tinghøj. Two well known

methods, for minimizing the switching overvoltages were given, and simulation of these were

conducted. The first incorporated the idea of switch-on at zero crossing of the voltages.

This principle showed great improvement in reduction of switching overvoltages, though

a large asymmetry in the current was present in the initial states. The second proposal

was to charge the system through a resistor that is later short circuited. Simulation of

this proposal also showed great improvement in minimizing generated overvoltages. Both

simulations showed limitations of the overvoltages to a maximum of 6 kV, compared with

a maximum of 64 kV without transient limitation.

In the measurements on the 150 kV level in Ferslev two frequency bands was present. These

frequencies are concluded, to stem from the two cable sections, in series with the inductances in

the system.

The two different solutions of the differential equations, describing the simplified model, are close

to identical. Furthermore, a simulation of the simplified model showed good accordance with the

solutions. It is therefore concluded that the solutions are correctly solved.

The overhead line parameters calculated by the project group, are in accordance with the

parameters calculated by the line constant program in PSCAD. It is therefore concluded that

these are correctly calculated.

From simulation with different frequencies is concluded, that the skin effect is accounted for in

the Frequency dependent (phase) model, in PSCAD.

8.2 Discussion 109

From simulation and measurements, it is concluded that the propagated overvoltages to lower

voltage levels, are not of a magnitude considered harmful. The proposals for minimizing the

switching overvoltages focuses therefore solely on the 150 kV transmission level.

From simulation of the two different proposals, it is concluded that the switching overvoltages

are limited significantly. Of the two proposals, the zero crossing principle is preferable, as this

does not add to the number of mechanical parts.

The constructed PSCAD model, does not represent the transmission system between Ferslev-Tinghøj adequately, as the switching transients in the simulations, are not damped as the

measured transients. It is therefore concluded that a more detailed simulation model of the

reactor is needed, along with implementation of frequency dependency in the short circuit

impedance of the system.

8.2 Discussion

Throughout the simulation, a lack of damping was present. The size of the needed series

resistance, from which the desired damping was achieved, is unreasonable in a transmission

system like between Ferslev-Tinghøj. It is therefore concluded, that a part of the reason is to

be found elsewhere. As the included models in PSCAD are very detailed, the created reactor

component is incomparable in precision, as nonlinearities like saturation, are not included. As

indicated in section 6.2.2, the reactor operates in saturation in the initial states. As saturation

will decrease the inductance in the reactor, the impedance of the reactor will consequently be

lowered, and thereby change the amplitude, and harmonic components of the currents in the

system.

In order to obtain a comparable simulation with the measurements on the system, a more detailed

reactor component in PSCAD is needed. This component should include saturation, in order to

obtain a sufficiently detailed model. The exact value of the short circuit impedance is also

imperative, as is a realistic frequency dependency of this, if more exact simulation results are to

be obtained.

Proposition for further work, is obviously a more detailed model of the reactor and transformers,which do not include magnetic saturation. Also of great interest, is why the PSCAD model doesnot give identical results to measurements. Is this only caused by a insufficient model of theaforementioned components, or are there other factors adding to the described inaccuracy indamping ratio?

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forlag, 2th edition, 1965.

[14] W. G. Hurley, W. H. Wölfle, J. G. Breslin. Optimized Transformer Design: Inclusive of High-

Frequency Effects. IEEE Transactions on Power Electronics, vol. 13 no. 4 page 651-659, 1998.

[15] ABB Power Technology Products AB. David Eklund and Pär Carlsson. 2007. Information provided

by ABB are attached on the CD.

[16] E. Kuffel, W.S.Zaengl and J. Kuffel. High Voltage Engineering: Fundamentals. Newnes, 2nd edition,

2000. ISBN: 0-7506-3634-3.

[17] DS/EN. Voltage characteristics of electricity supplied by public distribution systems. CENELEC,

1999. Corrected: 2004-04-16 and 2004-10-01.

BIBLIOGRAPHY 111

[18] M. Beanland, T. Speas, R. Rostron. Pre-insertion Resistors in High Voltage Capacitor Bank

Switching. Western Protective Relay Conference, October 2004.

[19] H. M. Ryan. High voltage engineering and testing. Peter Peregrinus Ltd. on behalf of the IEE., 1st

edition, 1994. ISBN: 0-86341-293-9.

[20] P. M. Anderson. Analysis of Faulted Power Systems. IEEE Press, 1st edition, 1995. ISBN: 0-7803-

1145-0.

[21] C. A. Gross. Power system analysis. John Wiley & Sons, Inc., 2nd edition, 1986. ISBN: 0-471-86206-

1.

[22] L. M. Faulkenberry and W. Coffer. Electrical Power Distribution and Transmission. Prentice-Hall,

1st edition, 1996. ISBN: 0-13-249947-9.

[23] V. Del Toro. Electric Power Systems. Prentice Hall, 1992. ISBN: 0-13-678228-0.

[24] D. E. Johnson, J. R. Johnson, J. L. Hilburn, P.D. Scott. Electric Circuit Analysis. John Wiley &

Sons, Inc., 3th edition, 1999. ISBN: 0-471-36571-8.

[25] E. Kreyszig. Advanced Engineering Mathematics. John Wiley & Sons. Inc., 8nd edition, 1999. ISBN:

0-471-15496-2.

[26] Omicron electronics. CMC 256 Operations Manual. 1999. Version: C256.AE.1.

[27] Omicron electronics. Omicron test universe 1.3 EnerLyser. 1999. Version: ENLY.AE.1.

Appendix

Propagation of transmission level switching overvoltages,

related to the distribution level

a product of :

Projectgroup - EPSH-810

Institute of Energy Technology

Aalborg University, AAU

31st may 2007

Appendix A

Parameter determination for the

overhead line system

In this appendix, the calculations of the parameters for the overhead transmission lines are

explained. The appendix is divided into two, one section for the series impedance, and one

section for the shunt admittance. Each section starts by giving theoretical background for the

calculations, before calculations of the parameters are performed.

A.1 Series impedance

In this section the series impedance for the transmission lines is calculated. The MatLab file used

for these calculations is to be found in appendix G.3, and on the attached CD.

A.1.1 Theoretical background

The theoretical background for calculating the inductance in the overhead lines is subsequently

presented.

Magnetic field around a conductor

When a conductor carries current, there will, as a result of that current, be generated a magnetic

field around it. This situation is shown in figure A.1.

B

xI

r

Figure A.1: The figure shows the variation of the magnetic flux density with distance from the conductor

[1, Vørts p.102].

The relation between the magnetic field strength and the current, is described by Ampere’s law:

∮

~H · d~s = I · N (A.1)

Where:~H is the field strength [A/m]

N is the number of current carrying conductors [-]

A-2 Parameter determination for the overhead line system

The field strength ~Ho at a distance x (see figure A.1) from the center of the conductor is:

~Ho · 2 · π · x = I

~Ho =I

2 · π · x [A/m] (A.2)

The relation between the magnetic flux density ~B and the field strength ~H is given by:

~B = µ · ~H [T] (A.3)

Where:

µ is the permeability constant for the material [H/m]

For the case shown in figure A.1 the permeability of air applies, which approaches permeability

of free space:

µ = µrµ0 ≃ 4 · π · 10−7 [H/m] (A.4)

By combining equation A.2, equation A.3 and equation A.4, the following is valid for the magnetic

flux density at a distance x outside the conductor:

~Bo =4 · π · 10−7 · I

2 · π · x= 2 · 10−7 · I

x[T] (A.5)

A magnetic flux density generated by the current, will also exist inside the conductor, as shown

in figure A.2.

~Bi xI

r

Figure A.2: The figure shows a cross section of a current carrying cylindrical conductor.

The field strength, at any point x inside the conductor, is found to be:

~Hi · 2 · π · x = I · x2

r2(A.6)

~Hi =I · x

2 · π · r2[A/m] (A.7)

If the current in a conductor, at 50 Hz is assumed equally spread, over its cross section. The

magnetic flux density at the point x, shown in figure A.2, is only generated by the fraction x2

r2 ,

and the current running in the conductor as equation A.6 states:

~Bi = µ · ~Hi (A.8)

=µ · I · x2 · π · r2

[T] (A.9)

A.1 Series impedance A-3

Inductance between two conductors

In figure A.3, two long parallel, identical conductors are shown.

d

1 2Field1 Field2

rx

B

Figure A.3: Two parallel conductors, where the current I generates flux linkage Ψ between them.

The current I flows forward in conductor 1, and backward in conductor 2. This system shows

some similarity with a loop of a coil. Between the two conductors, there will, as in a coil, exist

mutual inductance. This inductance can be found as [1, p. 102]:

L =Φ · N

I=

Ψ

I[H] (A.10)

Where:Φ is the magnetic flux [Wb]

N is the number of turns in the coil [-]

Ψ is the magnetic flux linkage. [Wb · turn]

The flux linkage is given as:

Ψ = Φ · N = B · A · N[Wb · turns

m

]

(A.11)

To calculate the flux linkage of the system, first the flux linkage Ψ1 from conductor 1 is calculated.

This is done by adding the two components from the inner and outer flux densities as shown by:

Ψ1 = Ψ1i + Ψ1o (A.12)

= Bi · A + Bo · A

Equations for the flux densities of the inner field ~Bi and for the outer field ~Bo are known, from

equation A.5 and equation A.8. The flux linkage for the inner field of conductor 1 is calculated

as:

Ψ1i =

∫ r

0

2 · 10−7 · I · xr2

· x2

r2dx

=2 · 10−7 · I

r4

∫ r

0x3dx

= 2 · 10−7 · I

4(A.13)

For the outer field, the flux linkage is calculated in the same way, to be:

Ψ1o =

∫ d

r

2 · 10−7 · Ix

dx

= 2 · 10−7 · I · (ln(d) − ln(r))

= 2 · 10−7 · I · ln(

d

r

)

(A.14)


The flux linkage in conductor 1, is found by combining equations A.12, A.13 and A.14:

Ψ1 = Ψ1i + Ψ1o

=2 · 10−7 · I

4+ 2 · 10−7 · I · ln

(d

r

)

= 2 · 10−7 · I(

1

4+ ln

(d

r

))

(A.15)

By combining equation A.10, with equation A.15 and assuming, because of symmetry of the

conductors, that Ψ1 = Ψ2, the total inductance of the system is found as:

L =Ψ1 + Ψ2

I

=2 ·(

10−7·I2 + ·10−7 · I · ln

(dr

))

I

= 4 · 10−7(

1

4+ ln

(d

r

))

[H/m] (A.16)

The inductance of each conductor per meter is:

L = 2 · 10−7(

1

4+ ln

(d

r

))

(A.17)

Flux linkage in overhead line transmission system

In order to calculate the parameters for the overhead lines, equations describing the flux linkage

for multiple conductors are needed.

In the following, a generalized equation will be derived, for a system consisting of n-number of

conductors, as shown in figure A.4.

1

2

3

n

4

xx

y

d1x

d2xd12

r1

Figure A.4: The figure shows a cross section of, n conductor’s.

The sum of the currents in the system is presumed to be zero. The flux linkage for conductor 1

is calculated as in equation A.15:

Ψ11 = 2 · 10−7(

I1

4+ I1 · ln

(d1x

r1

)) [Wb · turns

m

]

(A.18)

Where:d1x is the distance between the center of conductor 1 to the point x [mm]

r1 is the radius of conductor 1 [mm]


And the effect of the current in conductor 2, is found by:

Ψ12 = Ψ2x − Ψ21

= 2 · 10−7 · I2 · ln(

d2x

d12

)

(A.19)

The total flux linkage contribution in conductor 1, from the currents in all n conductors is:

Ψ1 = Ψ11 + Ψ12 + ... + Ψ1n (A.20)

Ψ1 = 2 · 10−7(

I1

4+ I1 · ln

(d1x

r1

)

+ I2 · ln(

d2x

d12

)

+ ... + In · ln(

dnx

d1n

))

As stated the sum of the currents in the system, is equal to zero:

In = −I1 − I2 − I3... − In−1 (A.21)

And then equation A.20 is re written as:

Ψ1 = 2 · 10−7

(

I1

4+ I1 · ln

(d1n · d1x

r1 · dnx

)

+ I2 · ln(

d1n · d2x

d12 · dnx

)

+ ... + In−1 · ln(

d1nd(n−1)x

d1(n−1)dnx

))

(A.22)

If the point x is moved to infinity along the x-axis (x → ∞), then equation A.22 is simplified to:

Ψ1 = 2 · 10−7

(

I1

4+ I1 · ln

(d1n

r1

)

+ I2 · ln(

d1n

d12

)

+ ... + In−1 · ln(

d1n

d1(n−1)

))

(A.23)

This equation can be rewritten from the fact that∑

I = 0, simplifying equation A.23 to:

Ψ1 = 2 · 10−7(

I1

4+ I1 · ln

(1

r1

)

+ I2 · ln(

1

d12

)

+ ... + In · ln(

1

d1n

))

(A.24)

Simplified again gives:

Ψ1 = 2 · 10−7(

I1 · ln(

1

r′1

)

+ I2 · ln(

1

d12

)

+ ... + In · ln(

1

d1n

))

(A.25)

Here is r′1 the self GMD of the conductor. This equation gives the flux linkage for conductor 1,

per meter. The same is derived for the other n-conductors.

A.1.2 Determination of series impedance for overhead line transmission

system

The objective of this section, is to produce equations for the transmission system, which can be

used directly to find the impedances of the overhead line section.

The overhead lines in the system, are arranged in a triangle, with a ground conductor above the

phase wire, which apart from giving return path for the currents, serves as a lightning protection

for the phases. The geometry of the overhead lines, is shown in figure A.5, where the distances

between phases are annotated.


dag

dab

dac

dbg

dbc

dcg

a b

c

g

De

ha,b

hc

hg

Figure A.5: The geometry of the conductors at the Donau mast.

The generalized equation A.25 can be used to describe the inductances, in the case shown in

figure A.5, a more convenient way, is to write the equation in its matrix form.

Ψ = LI

[Wb · turns

m

]

Ψa

Ψb

Ψc

Ψg

Ψd

= 2 · 10−7 ·

ln( 1r′

a

) ln( 1dab

) ln( 1dac

) ln( 1dag

) ln( 1dad

)

ln( 1dba

) ln( 1r′

b

) ln( 1dbc

) ln( 1dbg

) ln( 1dbd

)

ln( 1dca

) ln( 1dcb

) ln( 1r′

c

) ln( 1dcg

) ln( 1dcd

)

ln( 1dga

) ln( 1dgb

) ln( 1dgc

) ln( 1r′

g

) ln( 1dgd

)

ln( 1dda

) ln( 1ddb

) ln( 1ddc

) ln( 1ddg

) ln( 1r′

d

)

·

Ia

Ib

Ic

Ig

Id

(A.26)

The resistance matrix for a system with 5 current paths, is given by the line resistance on

the diagonal, and zero on non-diagonal, where the earth resistance can be approximated to 50

mΩ/km [1, Vørts p. 118].


Rabcgd =

Ra 0 0 0 0

0 Rb 0 0 0

0 0 Rc 0 0

0 0 0 Rg 0

0 0 0 0 Rd

(A.27)

The series-impedance matrix for the system is then found by adding the resistance and inductive

reactance.

X = L/ω (A.28)

Zabcgd = Rabcgd + jXabcgd (A.29)

For the case shown in figure A.5, there are five current paths. One for each phase, one in the

ground conductor, and one in the earth. The schematic of this is shown in figure A.6.

a

b

c

d

g

a′

b′

c′

d′

g′

Zaa

Zbb

Zcc

Zdd

Zgg

Zab

Zac

Zad

ZagZbc

Zbd

Zbg

Zcd

Zcg

Zgd

Va Vb Vc

Vd=0

Vg = 0

Ia

Ib

Ic

Id

Ig

Lenght

De

h

Figure A.6: The three phase oh-transmission lines with ground wire, and earth return

The notation in figure A.6 will subsequently be used, where a-b-c indicate values for each phase,

g indicates value for the ground conductor, and d indicates value for the earth return. For the

system shown in figure A.6, the voltage equation is written:

Vaa′

Vbb′

Vcc′

Vgg′

Vdd′

=

Va − Va′

Vb − Vb′

Vc − Vc′

0 − Vg′

0 − Vd′

=

Zaa Zab Zac Zag Zad

Zba Zbb Zbc Zbg Zbd

Zca Zcb Zcc Zcg Zcd

Zga Zgb Zgc Zgg Zgd

Zda Zdb Zdc Zdg Zdd

Ia

Ib

Ic

Ig

Id

(A.30)

A reduction of the 5×5 impedance matrix to a 3×3 matrix is desired, in order to transform

the impedances to a sequence form. An inspiration for this reduction is found in [20, Anderson

p.113]. In the system shown in figure A.6, the sum of the currents, is zero, and the earth return

is shunted with the ground conductor, giving the following current relation:


−Id − Ig = Ia + Ib + Ic

Id = −Ia − Ib − Ic − Ig (A.31)

The matrix A.30, can thereby be reduced by its 5th column, giving 5 equations and four

unknowns:

Vaa′

Vbb′

Vcc′

Vgg′

Vdd′

=

Zaa − Zad Zab − Zad Zac − Zad Zag − Zad

Zba − Zbd Zbb − Zbd Zbc − Zbd Zbg − Zbd

Zca − Zcd Zcb − Zcd Zcc − Zcd Zcg − Zcd

Zga − Zgd Zgb − Zgd Zgc − Zgd Zgg − Zgd

Zda − Zdd Zdb − Zdd Zdc − Zdd Zdg − Zdd

Ia

Ib

Ic

Ig

(A.32)

Now are the impedances for the earth return (row 5) subtracted from the other impedances (rows

1 to 5):

Va = Vaa′ − Vdd′

Vb = Vbb′ − Vdd′

Vc = Vcc′ − Vdd′

Vg = 0 = Vgg′ − Vdd′

(A.33)

Giving the following impedance matrix, where the 5th row has been removed, as all entries areequal to zero:

(Zaa − Zad) − (Zda − Zdd) (Zab − Zad) − (Zdb − Zdd) (Zac − Zad) − (Zdc − Zdd) (Zag − Zad) − (Zdg − Zdd)

(Zba − Zbd) − (Zdb − Zdd) (Zbb − Zbd) − (Zdb − Zdd) (Zbc − Zbd) − (Zdc − Zdd) (Zbg − Zbd) − (Zdg − Zdd)

(Zca − Zcd) − (Zdb − Zdd) (Zcb − Zcd) − (Zdb − Zdd) (Zcc − Zcd) − (Zdc − Zdd) (Zcg − Zcd) − (Zdg − Zdd)

(Zga − Zgd) − (Zdb − Zdd) (Zgb − Zgd) − (Zdb − Zdd) (Zgc − Zgd) − (Zdc − Zdd) (Zgg − Zgd) − (Zdg − Zdd)

(A.34)

Reduction of this matrix gives a 4×4 matrix Zabcg:

Zaa + Zdd − Zad − Zda Zab + Zdd − Zad − Zdb Zac + Zdd − Zad − Zdc Zag + Zdd − Zad − Zdg

Zba + Zdd − Zbd − Zda Zbb + Zdd − Zbd − Zdb Zbc + Zdd − Zbd − Zdc Zbg + Zdd − Zbd − Zdg

Zca + Zdd − Zcd − Zda Zcb + Zdd − Zcd − Zcb Zcc + Zdd − Zcd − Zdc Zbg + Zdd − Zcd − Zdg

Zga + Zdd − Zgd − Zda Zgb + Zdd − Zgd − Zcb Zgc + Zdd − Zgd − Zdc Zgg + Zdd − Zgd − Zdg

(A.35)

Each element in equation A.35 is in the following form:

Zij = Zij − Zid − Zdj + Zdd (A.36)

Using this fact, the matrix can be simplified:

Va

Vb

Vc

·Vg = 0

=

Zaa Zab Zac · Zag

Zba Zbb Zbc · Zbg

Zca Zcb Zcc · Zcg

· · · · ·Zga Zgb Zgc · Zgg

Ia

Ib

Ic

·Ig

(A.37)


For the diagonal in this matrix, the impedance is given by:

Zii = Ri + Rd + jω · 2 · 10−7 · ln(

De

ri′

)

(A.38)

And for the non diagonal terms:

Zij = Rd + jω · 2 · 10−7 · ln(

De

dij

)

(A.39)

Where:

De is the distance to the earth return [m]

With the knowledge that De >> h, the distance between the earth return and conductors is

approximated to De.

De = 660

√ρ

f[m] (A.40)

Where:ρ is the specific resistance of the earth [Ω m]

f is the power frequency [Hz]

In order to reduce the impedance matrix from a 4 × 4 matrix to a 3 × 3 matrix, the following

operation is performed [20, Anderson p.114]:

Zabc =

Zaa Zab Zac

Zba Zbb Zbc

Zca Zcb Zcc

−

Zag

Zbg

Zcg

[1

Zgg

] [

Zga Zgb Zgc

]

(A.41)

This leaves a 3 × 3 matrix for the impedance:

Zabc =

Zaa Zab Zac

Zba Zbb Zbc

Zca Zcb Zcc

(A.42)

Where each element is given by the general expression:

Zij = Zij −Zig Zgj

Zgg

(A.43)

With a general expression for the series impedance for each phase known, the next step is to find

the sequence matrix. In order to do this, the transformation matrix T, is introduced [21, Gross,

p. 61]:

T =

1 1 1

1 a2 a

1 a a2

(A.44)

a = ej2π

3 (A.45)

The transformation matrix explains the relationship between the phase values and the sequence

values for both voltage and current. This relationship can now be used to transform the phase

values of the impedance into the sequence values. This is done with the following equation [21,

Gross, p. 62]:

Z012 = T−1ZabcT (A.46)

The sequence matrix for the series impedance can now be calculated by use of the explained

method.


A.1.3 Calculation of the series impedance for the overhead line section

To calculate the impedances in the overhead line transmission system, physical values and

dimension for each section is needed. Those information is listed in table A.1 and A.2.

Section 2

Conductor name Martin

Length 18,93 km

Conductor resistivity 0,028264 Ωm

Cross section total 772 mm2

Cross section aluminum 685 mm2

Self GMD 0,0146 m

Number of threads in conductor 54

Ground conductor Dorking

Cross section of ground conductor 153 mm2

Table A.1: Physical values for parameter calculation.

dab 5,8 m dbc 9,176 m

dac 9,364 m dbg 17,528 m

dag 16,939 m dcg 8,355 m

De 660 m

Table A.2: Distances between conductors.

Series resistance

The DC resistance for conductor is given by:

Rdc = ρl

A(A.47)

Where:ρ=0,028264 is the specific resistivity for aluminum [Ωmm2/m]

l is the length of the conductor [m]

A is the cross section of the aluminum conductor [mm2]

The series resistance is calculated for DC at 20C. The actual resistance in the overhead line will

be affected by the terms listed below:

• Temperature

• Skin effect

• Proximity effect

• Conductor sag

• Spiraling

In order to include the above-mentioned terms, a correction factor is estimated in the following.


Temperature The operation temperature will have an effect on the series resistance according

to the following general equation [10, NKT, Kabelteknisk håndbog p. C3]:

Rt = 1 + R20(1 + α(t − 20)) = R20(1 + ktemp) (A.48)

Where:R20 is the reference resistance at a temperature of 20C [Ω]

α is the temperature coefficient for the conductor material [C−1]

t is the actual temperature [C]

For the overhead line system the conductor material is aluminium with a temperature coefficient

α = 0,00403C−1. The reference resistance R20 is the calculated DC resistance. The operational

temperature is unknown as the focus is at the switch-on of the system. The actual temperature

is therefore dependent on the time between the disconnection and reconnection of the grid. In

order to find a correction factor for the resistance, the operational temperature at the switch on

is assumed to 40C, which gives the following temperature correction factor:

ktemp = 0, 00403(40 − 20) = 0, 0806 (A.49)

Skin effect The skin effect influences the resistivity of a conductor when an AC source is

used. Due to the alternating current, will the current density increase near the surface of the

conductor, and vice versa decrease in the middle of the conductor. This phenomena will cause

a smaller effective conducting area, hence a larger resistance. An approximation for calculating

the AC resistance is presented in [14, Hurley, p. 6]. First is the skin depth calculated:

δ =

√ρ

fπµ(A.50)

δ = 11, 96 mm (A.51)

Where:δ is the skin depth

ρ is the resistivity of the conductor

µ is the permeability of the conductor

For a non-magnetic conductor, as the case is with the aluminum conductor, the total permeability

is the permeability of free space[10, Serway, p.959]. A factor describing the ratio between RAC

and Rdc is given.

Rac

Rdc

= 1 + kskin (A.52)

kskin =(r/δ)4

48 + 0, 8(r/δ)4for

roδ

> 1, 7 (A.53)

kskin =

(14,7711,96

)4

48 + 0, 8 ·(

14,7711,96

)4 = 0, 0465 (A.54)

Proximity effect When conductors are placed near each other, the current will be affected,

as the magnetic field presses the electrons, causing higher density. Due to this, the current

density will not be equally distributed in the conductor. This will result in an increased effective

resistance. However, the proximity effect of overhead lines is normally small, due to the large

distances between the conductors. It is therefore neglected here.


Conductor sag When the conductors are mounted between two towers, the length of the

conductor between the two towers will be larger than the actual length between the two towers.

The sag of the conductor will approximately add 0,2% of the total length. This is found from

the maximum sag of 7,6 m and the average distance between the towers of 292 m.

Spiraling The conductors in an overhead line system, consists of a bundle of small conductors.

These small conductors are twisted around the axis of the conductor, resulting in an increase of

the length of the current path in the conductor. The spiraling of the conductor usually adds 2%

to the length of the conductor, and there by an increase of the total resistance [22, Faulkenberry,

p. 22].

The total correction factor can be calculated from each of the correction factors:

ktot = (1 + ktemp) · (1 + kskin) · (1 + kprox) · (1 + ksag) · (1 + kspir) (A.55)

ktot = (1 + 0, 0806) · (1 + 0, 0465) · (1 + 0) · (1 + 0, 002) · (1 + 0, 02) = 1, 15 (A.56)

The actual resistance can now be calculated, by using the correction factor:

Rtot = Rdc · ktot (A.57)

The resistances for the overhead line section and the ground conductor are calculated in MatLab1.

The series resistances are shown in table A.3.

Total resistances for section 2

DC resistance AC resistance

Phase a-b-c (Rdc) 0,781 Ω 0,898 Ω

Ground wire (Rdc,g) 3,497 Ω

Earth return (Rd) Ω 0,947

Table A.3: Series resistances for the overhead line.

Series inductance

The calculations for the series inductance are carried out in MatLab. The m-files2 used for this

are impedance.m, constants.m and geometry.m.

The calculated inductance matrix for section 2 is:

L =

Laa Mab Mac

Mba Lbb Mbc

Mca Mcb Lcc

= 10−3 ·

36, 188 13, 570 10, 890

13, 570 36, 267 11, 014

10, 890 11, 014 34, 384

[H] (A.58)

The calculated sequence impedance matrix for section 2 is:

Z012 =

3, 311 + j18, 618 0, 399 + j0, 246 −0, 429 + j0, 189

−0, 429 + j0, 189 0, 903 + j7, 474 −0, 319 − j0, 160

0, 399 + j0, 246 0, 312 − j0, 169 0, 903 + j7, 474

[Ω] (A.59)

1The m-file is shown in appendix G.3 on page G-58.2The m-files are shown in appendix G.1 on page G-57.


In table A.4 the calculated sequence inductances for the overhead line section are listed, including

values ”per km”.

Inductance Section 2 Per km

L0 59,3 mH 3,13 mH/km

L1 23,8 mH 1,26 mH/km

L2 23,8 mH 1,26 mH/km

Table A.4: Positive, negative and zero sequence inductance for the overhead line section.

For comparison, typical values for the inductances in an overhead line system is found in

Elektriske fordelingsanlæg [1, Vørts, p. 133-134]. Here a typical value for the positive sequence

inductance is stated to be 1,27 mH/km, which is in accordance with the calculated value of

1,26 mH/km. For the zero sequence inductance, typical values is given in the range of 3,82

mH/km to 4,77 mH/km for overhead lines without earth conductor. For overhead lines with

earth conductor, the values are given to be 10-20% lower, which gives a range from 3,06 mH to

3,82 for 20% lower values. The calculated value for the zero sequence inductance is therefore

within this interval, and the calculated values for the inductances are assumed valid.

In table A.5 are the calculated sequence resistances for the overhead line section listed, including

values ”per km”.

Resistance Section 2 Per km

R0 3,311 Ω 0,175 Ω/km

R1 0,903 Ω 0,0477 Ω/km

R2 0,903 Ω 0,0477 Ω/km

Table A.5: Positive, negative and zero sequence resistance for the overhead line section.

The typical value for the positive sequence resistance is very dependent on the cross sectional

area of the conductor. The largest cross section given in Elektriske fordelingsanlæg [1, Vørts,

p. 133-134], for a typical value is a 402 mm2 + 52 mm2 Steel-Aluminium conductor. Here the

typical resistance is specified to 0,07 Ω/km. The conductors used in the overhead line system

have a cross section of 772 mm2, and the resistance for these conductors is therefore expected

to be lower than the typical value of 402 mm2. This is also the case with a positive sequence

resistance of 0,0477 Ω/km. For the zero sequence resistance, are the typical values usually slightly

above 0,15 Ω plus the positive sequence resistance (0,15 + R1). This is in accordance with the

calculated value of 0,175 Ω/km. These calculated values are therefore assumed valid.


A.2 Shunt admittance

The shunt admittance of a transmission line system is determined by the conductance and

capacitances between the conductors. Here is the conductance neglected, and focus only on

finding the capacitance of the overhead lines. The principle for determination of the shunt

capacitance will be described in this section.

A.2.1 Theoretical background

In order to obtain equations for calculations of the shunt admittance, the theoretical background

for those equations is presented.

Electric field created by conductors

The cross section of a single conductor, is shown in figure A.7. The conductor has the radius r

and is carrying a charge q.

q r

d1

d2

Figure A.7: Cross section of a single conductor.

If the conductor shown in figure A.7 is considered to be infinitely long, the electric flux out of the

conductor will spread radially. According to Gauss’s law, the total electric flux out of a closed

surface will be equal to the total charge in the conductor:

∮

S

~D · d ~A = qtotal [C] (A.60)

Where:~D is the electric flux density [C/m2]

d ~A is the area over which the surface integral is performed [m2]

qtotal is the total charge the conductor is carrying [C]

The electric flux density can now be found by differentiating both sides of equation A.60, resulting

in:

~D =dq

d ~A[C/m2] (A.61)

As the flux density is determined by the charge on a given surface area, the flux density can be

found by using the total charge and the total surface area of the conductor:

~D =q

A(A.62)

~D =q

2πrl(A.63)

A.2 Shunt admittance A-15

Where:r is the radius of the conductor [m]

l is the unit length of the conductor [m]

Here the unit length is chosen to 1 m, in order to obtain the results per meter. The electric flux

density is then:

~D =q

2πr(A.64)

The relationship between the electric flux density and the electric field strength, is determined

by the permittivity of the material the electric field passes:

~D = ε ~E (A.65)

Using this, the electric field around the conductor can be determined:

~E =~D

ε(A.66)

~E =q

2πεr[V/m] (A.67)

With the electric field known, it is possible to find the potential difference between two points.

Potential difference

The potential difference ∆U , between two points, a and b, is determined by the integral of the

electric field between these two points: [10, Serway, p. 768-774]:

Uab = ∆U = Ua − Ub =

∫ b

a

~E · d~l [V] (A.68)

Where:Uab is the potential difference between a and b [V]

Ua is the potential at point a [V]

Ub is the potential at point b [V]

The potential difference between the two points d1 and d2 in figure A.7, can then be calculated:

U12 =

∫ d2

d1

q

2πεr· d~r (A.69)

U12 =q

2πε

∫ d2

d1

1

r· d~r (A.70)

U12 =q

2πε· lnd2

d1(A.71)

With the ability determine the potential difference between two points, it is possible to determine

the capacitance between two conductors.


Capacitance between two conductors

Figure A.8 shows two parallel conductors, with opposite charge. The considered conductors are

located in air.

Conductor 1 Conductor 2

r1 r2

+q −qd

Figure A.8: Two parallel conductors.

For a conductor located in air, the relative permittivity εr is approximately unity, hence the

permittivity approaches the permittivity of free space [23, Del Toro, p. 192]:

ε = εrε0∼= ε0 =

1

36π· 10−9 [F/m] (A.72)

The electric field around the conductor can then be determined by replacing ε in equation A.67

with the permittivity in air:

~E =q

2π 136π

· 10−9 · r (A.73)

~E = 18 · 109 · q

r(A.74)

The potential difference between two conductors can be found through superposition, by applying

a sample charge on each conductor, one at a time, and note the potential created by each

conductor. In a overhead transmission line system, the distance d will be much greater than the

radius of the conductor. This allows the radius rb to be neglected when integrating the electric

field between the conductors, hence simplifying the calculations. Likewise, the electric field can

be assumed uninfluenced by the other conductors and therefore radially shaped [1, Vørts, p. 136].

The potential difference between the two conductors, stems from the charges on each conductor.

The potential U ′12 is the contribution created by the charge +q, and can be expressed as:

U ′12 = 18 · 109 · q

∫ d

r1

1

r· d~r (A.75)

U ′12 = 18 · 109 · q · ln

(d

r1

)

(A.76)

The potential U ′′12 is the contribution created by the charge −q, and can be expressed as:

U ′′12 = 18 · 109 · (−q) ·

∫ r2

d

1

r· d~r (A.77)

U ′′12 = 18 · 109 · (−q) · ln

(r2

d

)

(A.78)

U ′′12 = 18 · 109 · q · ln

(d

r2

)

(A.79)


The potential difference between the two conductors is now calculated by adding the two

contributions:

U12 = U ′12 + U ′′

12 (A.80)

U12 = 18 · 109 · q · ln(

d

r1

)

+ 18 · 109 · q · ln(

d

r2

)

(A.81)

U12 = 18 · 109 · q · ln(

d2

r1 · r2

)

(A.82)

If the radius r = r1 = r2, the equation is simplified:

U12 = 18 · 109 · q · ln(

d2

r2

)

(A.83)

U12 = 36 · 109 · q · ln(

d

r

)

(A.84)

The relationship between the potential difference and the charge is [1, Vørts, p. 137]:

U = p · q (A.85)

U =1

C· q (A.86)

Where:p is the potential coefficient [m/F]

C is the capacitance [F/m]

1 2C12

Cp Cp

Figure A.9: Schematic of the capacitance in a system of two parallel conductors.

Figure A.9 shows the three capacitances in a system with two conductors. The capacitance C12

is now found from the relation between the potential difference and the charge in equation A.84.

p12 =U12

q= 36 · 109 · ln

(d

r

)

(A.87)

C12 =q

U12=

1

36 · 109 · ln(

dr

) (A.88)

The capacitance per conductor Cp is representing the capacitance between the conductor and

neutral as shown in figure A.9. When r = r1 = r2 the capacitances per conductor will have the

same value, namely two-times the value of C12 according to the schematic on figure A.9:

pp = 18 · 109 · ln(

d

r

)

(A.89)

Cp =1

18 · 109 · ln(

dr

) (A.90)


Capacitance to ground

A conductor placed above ground, will have a capacitance to ground. The electric field around

the conductor will, because of the neutral potential of the ground, have the same appearance

as if there, instead of ground, were a conductor carrying the opposite sign of the charge, in its

mirror image as shown in figure A.10 [1, Vørts, p. 137].

q

r

r

h

h

G

−q

Figure A.10: A single conductor with its mirror image under ground.

The capacitance and potential coefficient to ground will then be the same as the capacitance and

potential coefficient per conductor in equation A.89 and equation A.90 according to figure A.9.

In figure A.10 the distance between the conductors is 2h, this gives the potential coefficient and

capacitance:

pa0 = 18 · 109 · ln(

2h

r

)

(A.91)

Ca0 =1

18 · 109 · ln(

2hr

) (A.92)

This capacitance is annotated Caa and can be conceived as a self capacitance. The self potential

coefficient is then given as:

paa = 18 · 109 · ln(

2h

r

)

(A.93)

If two conductors are placed above ground, the capacitance between the conductors will be

affected by the capacitances to ground.

In equation A.93 the potential coefficient to ground was found for a single conductor. If another

conductor b is present, the conductor a will be affected (shown in figure A.11). The potential

created between a and G is determined by contributions from conductor b and its mirror image:

Ua − UG = Uab+ − UG + Uab− − UG (A.94)

The potential at ground UG is defined to be neutral, which gives:

Ua = Uab+ + Uab− (A.95)


q

dab

Dab

h

h

G

−q

ab+

b−

Figure A.11: Two conductors above ground.

The potential at conductor a can now be expressed with the charge on conductor b and its

potential coefficients to conductor a:

Ua = Uab+ + Uab− (A.96)

Ua = 18 · 109 · q · ln(

h

dab

)

+ 18 · 109 · (−q) · ln(

h

Dab

)

(A.97)

Ua = 18 · 109 · q · ln(

Dab

dab

)

(A.98)

The mutual potential coefficient and capacitance between conductor a and conductor b are now

found:

pab = 18 · 109 · ln(

Dab

dab

)

(A.99)

Cab =1

18 · 109 · ln(

Dab

dab

) (A.100)

With the ability to determine both the mutual capacitance between two conductors (equation

A.100) and the capacitance between a conductor and ground (equation A.92), it is possible to

determine the capacitances in a system consisting of multiple conductors.

Capacitance between multiple conductors

In a system of multiple conductors, capacitances will exist between all of the conductors.

The potential on conductor a in such system, is determined by the contributions from all the

conductors, including contribution from its mirror image.

Ua =1

Caa· qa +

1

Cab

· qb +1

Cac· qc + ...

1

Can· qn (A.101)

Ua =n∑

j=a

1

Caj· qj (A.102)

Ua =n∑

j=a

paj · qj (A.103)


The potential in the other conductors is found by the same principle and gives:

Ub =n∑

j=a

pbj · qj (A.104)

Uc =n∑

j=a

pcj · qj (A.105)

...

Un =n∑

j=a

pnj · qj (A.106)

These equations can be arranged in a general matrix equation:

Ua

Ub

Uc

...

Un

=

paa pab pac . . . pan

pba pbb pbc . . . pbn

pca pcb pcc . . . pcn

......

.... . .

...

pna pnb pnc . . . pnn

qa

qb

qc

...

qn

(A.107)

With a potential matrix produced for a system with multiple conductors, it is now possible to

determine the appropriate equations for the overhead transmission line system.

A.2.2 Determination of shunt admittance for overhead line transmission

system

To determine the admittances for the overhead line system shown in figure A.12, a potential

matrix will be constructed.

1 2

3

g

D11

d12

D12

Figure A.12: Distances between conductor 1, 2 and its mirror images.


The general matrix equation, for a system of multiple conductors is shown in equation A.107.

For the system shown in figure A.12 the matrix equation is found:

U = P · q (A.108)

U1

U2

U3

Ug

=

p11 p12 p13 p1g

p21 p22 p23 p2g

p31 p32 p33 p3g

pg1 pg2 pg3 pgg

q1

q2

q3

qg

(A.109)

The conductors self potential coefficient is placed on the diagonal of the potential matrix and

represent the potential coefficient to ground. This coefficient was found in equation A.93, and

can be written for a general case:

pii = 18 · 109 · ln(

Dii

r

)

[m/F] (A.110)

Where:

Dii is the distance between the ith conductor and its own mirror image [m]

The mutual potential coefficient represent the potential coefficient between two conductors. This

coefficient was found in equation A.99, and can be written for a general case:

pij = pji = 18 · 109 · ln(

Dij

dij

)

[m/F] (A.111)

Where:

Dij is the distance between the ith conductor and the jth mirror image [m]

Inserting the potential coefficients for the system in figure A.12 into equation A.109 gives:

U1

U2

U3

Ug

= 18 · 109 ·

ln(D11

r) ln(D12

d12) ln(D13

d13) ln(

D1g

d1g)

ln(D21

d21) ln(D22

r) ln(D23

d23) ln(

D2g

d2g)

ln(D31

d31) ln(D32

d32) ln(D33

r) ln(

D3g

d3g)

ln(Dg1

dg1) ln(

Dg2

dg2) ln(

Dg3

dg3) ln(

Dgg

rg)

q1

q2

q3

qg

(A.112)

In order to use the transformation matrix T to find the sequence matrix for the shunt admittance,

the constructed potential matrix needs to be reduced to a 3× 3 matrix. To do this the potential

matrix is subdivided into four matrices:

Pabcg =

P1... P2

· · · · · · · · ·P3

... P4

(A.113)

Where:P1 is a 3 × 3 matrix

P2 is a 1 × 3 matrix




As the potential at the ground wire is neutral (Ug = 0) equation A.112 can be simplified:

U1

U2

U3

0

= 18 · 109 ·

ln(D11

r) ln(D12

d12) ln(D13

d13)

... ln(D1g

d1g)

ln(D21

d21) ln(D22

r) ln(D23

d23)

... ln(D2g

d2g)

ln(D31

d31) ln(D32

d32) ln(D33

r)

... ln(D3g

d3g)

· · · · · · · · · · · · · · ·ln(

Dg1

dg1) ln(

Dg2

dg2) ln(

Dg3

dg3)

... ln(Dgg

rg)

q1

q2

q3

qg

(A.114)

With the above-mentioned matrix equation defined, a 3 × 3 matrix can be obtained through

reduction [20, Anderson, p. 169]:

Pabc = P1 − P2P−14 P3 (A.115)

The capacitance matrix is now found by taking the inverse of the potential matrix:

Cabc = P−1abc (A.116)

The susceptance is found from the capacitance through their relationship:

Babc =1

Xabc

=11

ωCabc

(A.117)

Babc = ωCabc (A.118)

As the conductance in overhead line systems is very small, it is neglected here. The admittance

matrix can therefore be seen as only imaginary, determined by the susceptance:

Yabc = 0 + jBabc (A.119)

The sequence matrix can then be found using the transformation matrix:

Y012 = T−1YabcT (A.120)

A.2.3 Calculation of sequence matrix for the shunt admittance

In order to make the calculations described previously. The distances in equation A.114 must

be known. In figure A.13 those distances are shown.

d12

d13

d1g

D11 D12

D13

D1g

1 2

3

g

d21

d23

d2g

D21 D22

D23

D2g

1 2

3

g

d31 d32

d3g

D31 D32

D33

D3g

1 2

3

g

dg1 dg2

dg3

Dg1 Dg2

Dg3

Dgg

1 2

3

g

Figure A.13: Distances for the three phase conductors and the ground wire.


The distances in figure A.13 are calculated3 and listed in table A.6.

Distance Length

d12 5,80 m

d13 9,36 m

d1g 16,94 m

D11 49,40 m

D12 49,74 m

D13 58,29 m

D1g 66,31 m

Distance Length

d21 5,80 m

d23 9,18 m

d2g 17,53 m

D21 49,74 m

D22 49,40 m

D23 58,26 m

D2g 66,46 m

Distance Length

d31 9,36 m

d32 9,18 m

d3g 8,36 m

D31 58,29 m

D32 58,26 m

D33 67,00 m

D3g 75,13 m

Distance Length

dg1 16,94 m

dg2 17,53 m

dg3 8,36 m

Dg1 66,31 m

Dg2 66,46 m

Dg3 75,13 m

Dgg 83,20 m

Table A.6: Distances for the three phase conductors and the ground wire shown in figure A.13.

The capacitance matrix for section 2, can now be calculated with the method presented

previously4.

Cabc =

Caa Cab Cac

Cba Cbb Cbc

Cca Ccb Ccc

= 10−6

0, 149 −0, 0275 −0, 0114

−0, 0275 0, 149 −0, 0127

−0, 0114 −0, 0127 0, 165

[F] (A.121)

The sequence admittance sequence matrix for section 2 is calculated to:

Y012 = 10−6

−0, 000 + j37, 624 −2, 935 − j1, 509 2, 935 − j1, 509

2, 935 − j1, 509 −0, 000 + j53, 829 1, 433 + j0, 618

−2, 935 − j1, 509 −1, 433 + j0, 618 −0, 000 + j53, 829

(A.122)

The sequence susceptances for the section are shown in table A.7.

Susceptance Section 2 Per km

B0 37,62 µS 1,988 µS/km

B1 53,83 µS 2,844 µS/km

B2 53,83 µS 2,844 µS/km

Table A.7: Positive, negative and zero sequence susceptance for the overhead line section.

Table A.7 can be recalculated in order to find the capacitance using the relation in equation

A.118. The results for the capacitance are listed in table A.8.

Capacitance Section 2 Per km

C0 0,1198 µF 6,328 nF/km

C1 0,1713 µF 9,049 nF/km

C2 0,1713 µF 9,049 nF/km

Table A.8: Positive, negative and zero sequence capacitance for the overhead line section.

3The distances are calculated in appendix G.24The calculations are performed in MatLab as shown in appendix G.4.


To verify the calculated values of the line parameters, the ”per km” values are compared with

normal values found in literature. In ”Elektriske Fordelingsanlæg” [1, S. Vørts, p. 157] the

admittance for overhead transmission lines is stated to have a typical magnitude of 3 µS/km.

This is in accordance with the calculated values of 2,844 µS, shown in table A.7. For the zero

sequence susceptance the typical value is given to 1,8 µS, which is also in accordance with the

calculated values of 1,988 µS. On the basis of these comparison, the calculated values are assumed

to be valid.

A.3 Summary

In this chapter has the determination of the series impedance and the shunt admittance for the

overhead line section been presented. Each section started with a theoretical review in order to

achieve general equations for the calculations. On the basis of these equations, parameter values

for the overhead transmission system were determined. The calculated phase values for the series

inductance, series resistance and the shunt capacitance are shown in table A.9.

Inductance Section 2 Resistance Section 2 Capacitance Section 2

La 36,2 mH Ra 0,898 Ω Can 0,149 µF

Lb 36,3 mH Rb 0,898 Ω Cbn 0,149 µF

Lc 34,4 mH Rc 0,898 Ω Ccn 0,165 µF

Table A.9: Phase values for self inductance and resistance for the overhead line section.

The calculated mutual inductance and mutual capacitances are shown in table A.10.

Mutual inductance Section 2 Mutual capacitance Section 2

Mab 13,6 mH Cab 0,0275 µF

Mac 10,9 mH Cac 0,0114 µF

Mbc 11,0 mH Cbc 0,0127 µF

Table A.10: Phase values for mutual inductance and mutual capacitance in the overhead line section.

The calculated phase values listed in table A.9 and A.10 will be used for the electric representation

of the overhead line.

Appendix B

Parameter determination in the shunt

reactor

In this appendix an analysis will be performed on the magnetic relations in the shunt reactor,

resulting in a determination of the inductances in the shunt reactor.

B.1 Theoretical background

Before the inductances in the shunt reactor are determined, a brief theoretical review of magnetic

field theory will be performed to find the essential equation for determination of the inductances.

In figure B.1 is a simple element with the permeability µ shown.

+ -

Φ

l

H

F

Figure B.1: An element with a flux passing though it, generating a magneto-motive force.

If a flux is passing through the element, a magneto-motive force (mmf ) will occur between two

points of the element. The mmf is determined by the magnetic field strength and the distance

between the two points:

F = Hl [A · turns] (B.1)

The relationship between the magnetic field strength and the magnetic flux density is determined

by the permeability µ the field passes through:

H =B

µ[A/m] (B.2)

The flux density can be expressed as the flux on a given surface, in this case the cross sectional

area of the element in figure B.1:

B =Φ

A[T] (B.3)

By combining equation B.2 and B.3 and inserting in equation B.1, a new expression for the mmfcan be found:

F =l

µAΦ (B.4)

B-26 Parameter determination in the shunt reactor

The fraction in equation B.4 is termed the reluctance and is expressed as:

R =l

µA[A · turns

Wb] (B.5)

With this stated equation B.4 can be reduced, stating the mmf as:

F = RΦ (B.6)

This equation can be seen as an equivalent to Ohm’s law, allowing a comparison to a electric

circuit where F can be seen as a voltage source with magnitude Ni according to Ampere’s law.

Φ can be seen as an equivalent to the current, and the reluctance R can be seen as an equivalent

to a resistance. A core with N turns and a current i in the winding is shown in figure B.2.

+

-

Φ(t)

u(t)

i(t)

N

Figure B.2: Core with N turns and permeability µ.

By using the previously mentioned relation with electric circuit, a magnetic equivalent of figure

B.2 is shown in figure B.3.

+

-

Φ(t)

F = Ni(t)

R1

R2

R3

R4

Figure B.3: Magnetic equivalent circuit of figure B.2.

In the magnetic circuit shown in figure B.3 each side of the core has its own reluctance. By

unifying the four reluctances in one (Rc = R1 +R2 +R3 +R4), the circuit is greatly simplified,

and the following relation can be expressed:

Ni(t) = RcΦ (B.7)

Faraday’s law states that the electro-motive force (emf ) is given as:

u(t) = NdΦ(t)

dt(B.8)

B.2 Inductance determination for the shunt reactor B-27

Solving equation B.7 for Φ and inserting in equation B.8 gives:

u(t) =N2

Rc

di(t)

dt(B.9)

u(t) = Ldi(t)

dt(B.10)

The inductance can now be found from the number of turns and the reluctance:

L =N2

Rc

(B.11)

By use of this general theory, the inductances in the shunt reactor can be found.

B.2 Inductance determination for the shunt reactor

As described, a magnetic flux will flow in the core as a consequence of the current in the windings.

The mean flux path in the five limbed reactor core is shown in figure B.41.

+

-

+

-

+

-

Φa(t) Φb(t) Φc(t)

ua(t) ub(t) uc(t)

Figure B.4: Mean flux path in the shunt reactor.

In order to ease the determination of the inductances, the following limitations are taken:

• The cross sectional area of the flux path in the core limb air gap, is assumed equal to thecross sectional area of the coreIn order to reduce the cross sectional area of the flux path in the gap, air gaps are distributed

over the length of the limb, as shown in figure B.5. Furthermore, the distance elements,

consisting of a type of porcelain, are used between the limb segments. A cross section of a

limb segment with distance elements is shown in figure B.5 [15, ABB].

1The reactor core is briefly described in section 3.6 on page 20.


Core limb Limb segment

Distance element

Figure B.5: Mean flux path in the shunt reactor.

It is assumed that this distribution of the air gap and the distance elements minimizes the

flux path area in the gap to the area of the limb. The cross sectional area of the air gap is

therefore assumed to be equal to the cross sectional area of the core limb.

• The permeability of the gap equals airDue to the use of distance elements between the core elements the permeability in the gap

will obtain a different value than the one of air. This value has not been provided by ABB,

but it is assumed that it stays in the region just above air, as a lower value would increase

the cross sectional area flux path, and much higher value would diminish the purpose of

the air gap, as the reluctance would then approach the reluctance of the core.

• The flux passes only through the mean flux pathAs shown in figure B.4 the flux is represented by a mean path. As a consequence of this

representation, it is assumed that the flux only passes through the core, and not through

different paths in the air surrounding the core. This is reasonable as the reluctance in the

air is much greater than the one of the core.

The geometry of the core is symmetrical, this allows the core being represented by five different

flux paths (s1, s2, s3, s4 and sg) as shown in figure B.6.

s1

s2s2 s3s3

s4

sg

A1

A2A3

A4

Ag

Φa Φb Φc

Figure B.6: Flux path in the shunt reactor.

The lengths and areas in figure B.4 is calculated in MatLab2 from the geometry described in the

2The m-file is shown in appendix G.5.


system description, section 3.6 on page 20. The calculated values are listed in table B.1.

s1 1,716 m A1 0,2827 m2

s2 1,203 m A2 0,1667 m2

s3 0,783 m A3 0,1667 m2

s4 2,003 m A4 0,1667 m2

sg 0,2866 m Ag 0,2827 m2

Table B.1: Flux paths and areas.

The permeability in the core can be found by use of the DC magnetization for the core shown

in figure B.7.

DC MAGNETIZATION &

DC PERMEABILITY

M=5 0,30mm

Tested by Epstein test apparatus. Samples were sheared

in the rolling direction and subjected to Stress-relief

annealing at 800°C (1472 F). Assumed density 7,65 kg/m

Frequency 50 Hz.

3

B-H

Magnetizing Force H (A/m)

Ind

uct

ion

B (

Tesl

a),

(x 1

0; K

ilog

au

ss)

Figure B.7: DC magnetization curve for the core.

This is done by calculating the peak value of B and read off the corresponding value for H. The

permeability can then be found through the relationship between B and H.

The magnetic flux density can be calculated by using the expression for the mmf :

F = Ni = ΦR (B.12)

Knowing Φ = BA and R = N2

L, an expression for B is given:

B =L · iN · A (B.13)


The peak magnetic flux density can now be found by use of the rated peak current in the winding:

Bpeak =L · Ipeak

N · A (B.14)

Bpeak =2, 3 · 135, 8

√2

1194 · 0, 2827 = 1, 3087 T (B.15)

With this value for Bpeak, the magnetic field strength H is read off the B-H curve to 20 A/m.

The relative permeability can now be found from the relation between B and H.

B = µH = µ0µrH (B.16)

µr =B

µ0H=

1, 3087

4 · π · 10−7 · 20 = 52074 (B.17)

The permeability in the iron core and the gap are listed in table B.2.

Permeability µr

Iron core 52074

Gap 1

Table B.2: Relative permeability.

With flux paths, cross sectional areas and permeability known, it is possible to make an equivalent

diagram for the reactor core, shown in figure B.4. This is done by transforming the flux path

into reluctances by use of equation B.5. The equivalent diagram is shown in figure B.8.

+

-

+

-

+

-

R1R1R1

R2

R2

R2

R2

R3

R3

R3

R3

R4 R4

Rg RgRg

Fa Fb Fc

Φa Φb Φc

Figure B.8: Equivalent diagram for the reluctances in the reactor.

In order to simplify the analysis of the reactor core, the equivalent diagram is simplified by

unifying reluctances. The simplified equivalent diagram for shunt reactor is show in figure B.9.


+

-

+

-

+

-

R1gR1gR1g

R2

R2

R2

R2

R343 R343

Fa Fb Fc

Φ1 Φ2 Φ3 Φ4

Φa Φb Φc

Figure B.9: Simplified equivalent diagram for the reactor core.

Where:

R343 = R3 + R4 + R3 (B.18)

R1g = R1 + Rg (B.19)

To analyze the equivalent circuit, general methods from electric circuit analysis can be used. In

this case the ”mesh analysis” is the most convenient, as the equivalent contains both multiple

sources and meshes [24, Johnson, p. 138 and p. 372]. The four mesh equations from figure B.9

are stated below:

0 = Fa + Φ1 · (R343 + R1g) − Φ2 · R1g (B.20)

0 = Fb −Fa + Φ2 · (R2 + R1g + R2 + R1g) − Φ3 · R1g − Φ1 · R1g (B.21)

0 = Fc −Fb + Φ3 · (R2 + R1g + R2 + R1g) − Φ4 · R1g − Φ2 · R1g (B.22)

0 = −Fc + Φ4 · (R343 + R1g) − Φ3 · R1g (B.23)

These equations can be simplified:

−Fa = +Φ1 · (R343 + R1g) − Φ2 · R1g (B.24)

Fa −Fb = −Φ1 · R1g + Φ2 · (2 · R2 + 2 · R1g) − Φ3 · R1g (B.25)

Fb −Fc = −Φ2 · R1g + Φ3 · (2 · R2 + 2 · R1g) − Φ4 · R1g (B.26)

Fc = −Φ3 · R1g + Φ4 · (R343 + R1g) (B.27)

These four equations can be arranged in a matrix equation:

−Fa

Fa −Fb

Fb −Fc

Fc

=

R343 + R1g −R1g 0 0

−R1g 2R2 + 2R1g −R1g 0

0 −R1g 2R2 + 2R1g −R1g

0 0 −R1g R343 + R1g

Φ1

Φ2

Φ3

Φ4

(B.28)

To determine the inductances in and between each phase, the flux in each limb is required.

According to the equivalent diagram in figure B.9, the flux in each limb can be expressed by the

four mesh fluxes:

Φa = Φ2 − Φ1 (B.29)

Φb = Φ3 − Φ2 (B.30)

Φc = Φ4 − Φ3 (B.31)


In order to use this fact, the equation for each mesh flux must be found from equation B.28.

This is done by taking the inverse of the reluctance matrix, creating a permeance matrix:

Φ1234 = R−1F (B.32)

Φ1234 = PF (B.33)

The inversion of the reluctance matrix shown in equation B.28, will give a long algebraic equation

and is therefore performed through a numerical Gauss-Jordan Elimination [25, Kreyszig, p. 350]

in MatLab:

AA−1 = I (B.34)

The inversion of the reluctance matrix will give a 4 × 4 matrix, and the flux equation shown

below:

Φ1

Φ2

Φ3

Φ4

=

P11 P12 P13 P14

P21 P22 P23 P24

P31 P32 P33 P34

P41 P42 P43 P44

−Fa

Fa −Fb

Fb −Fc

Fc

(B.35)

This matrix can be reduced to a 4× 3 matrix by multiplying the 4× 1 F-matrix, and afterward

isolate a new F-matrix:

Φ1

Φ2

Φ3

Φ4

=

P21 − P11 P31 − P21 P41 − P31

P22 − P12 P32 − P22 P42 − P32

P23 − P13 P33 − P23 P43 − P33

P24 − P14 P34 − P24 P44 − P34

Fa

Fb

Fc

(B.36)

Equation B.29, B.30 and B.31 can now be used to set up an expression for each limb flux:

[Φa

Φb

Φc

]

=

[(P22 − P12) − (P21 −P11) (P32 −P22) − (P31 −P21) (P42 −P32) − (P41 −P31)

(P23 − P13) − (P22 −P12) (P33 −P23) − (P32 −P22) (P43 −P33) − (P42 −P32)

(P23 − P13) − (P22 −P12) (P33 −P23) − (P32 −P22) (P43 −P33) − (P42 −P32)

][Fa

Fb

Fc

]

(B.37)

The inductance matrix can now be found by use of equation B.11, as the permeance is the

reciprocal of the reluctance:

L =N2

R (B.38)

L = N2P (B.39)

B.3 Calculation of the inductances in the shunt reactor

Calculations of the inductance matrix are performed by the above-mentioned method in MatLab3.

The results from the calculations are shown below:

L =

Laa Mab Mac

Mba Lbb Mbc

Mca Mcb Lcc

=

1, 767 −358 · 10−6 −256 · 10−6

−358 · 10−6 1, 767 −358 · 10−6

−256 · 10−6 −358 · 10−6 1, 767

(B.40)

3The m-files are shown in appendix G.5, G.6 and G.7.

B.3 Calculation of the inductances in the shunt reactor B-33

Due to the symmetry of the core: Mab = Mba, Mac = Mca and Mbc = Mcb. Furthermore,

Mba = Mbc as phase b is placed on the symmetrical axis.

The calculated self inductances are compared with the designed and the measured values provided

by ABB, in table B.3.


La 1,767 H 2,300 H 2,307 H

Lb 1,767 H 2,300 H 2,304 H

Lc 1,767 H 2,300 H 2,307 H

Table B.3: Self inductances in the shunt reactor[7, ABB Reactor].

The calculated self inductance in table B.3 does vary from the values stated by ABB. This

variation is assumed to be inherited from the following assumptions:

• The relative permeability in the gap equals air.

• The cross sectional area of the gap equals the cross section of the limb.

As both the permeability and the cross sectional area of the gap will have larger values than

assumed, the reluctance in the gap will, according to equation B.5, obtain a lower value. In order

to find a suitable estimate of the mutual inductances, the reluctances in the gap are adjusted to

give a inductance per phase of 2,3 H. When this is the case, the reluctance in the gap can be

assumed to be correct. The reluctance can be adjusted, by changing both the cross section in

the gap and/or the relative permeability in the gap. The results will be the same whether the

cross sectional area or the permeability is changed, as long as the reluctance obtains the same

value. In this case the relative permeability in the gap is changed to a value of 1,3021, while the

cross sectional area of the gap is kept equal to the limb cross sectional area. The new inductance

matrix is:

L =

Laa Mab Mac

Mba Lbb Mbc

Mca Mcb Lcc

=

2, 300 −607 · 10−6 −434 · 10−6

−607 · 10−6 2, 300 −607 · 10−6

−434 · 10−6 −607 · 10−6 2, 300

(B.41)

The new results are listed in table B.4.


La 2,300 H 2,300 H 2,307 H

Lb 2,300 H 2,300 H 2,304 H

Lc 2,300 H 2,300 H 2,307 H

Table B.4: Recalculated values of the self inductances in the shunt reactor[7, ABB Reactor].

Now the calculated self inductance matches with the values stated by ABB. The reason this

correction is made is to obtain a more credible estimate of the mutual inductances, as the

measured values for the self inductances will be used in simulations.


The mutual inductances in the reactor have not been calculated nor tested by ABB. However,

ABB states that the value usually stays in the region of 1-2 % of the phase values [15, ABB]. The

caluclated values for the mutual inductances are listed along with the 1 %-value of the designed

inductance in table B.5.

Calculated [% of phase value] Designed [% of phase value]

Mab -607 µH [0,03 %] -32,0 mH [1,39 %]

Mac -434 µH [0,02 %] -23,0 mH [1,00 %]

Mbc -607 µH [0,03 %] -32,0 mH [1,39 %]

Table B.5: Mutual inductances in the shunt reactor[15, ABB].

The calculated values of the mutual inductance do not stay in the expected region of 1-2 % as

ABB stated. Instead the percentage value stay in the 0,02-0,03 % region. The calculated value

have though the internally expected relationship, as Mab equals Mbc which are larger than Mac.

The factor affecting the mutual inductances the most is the permeability of the core, which is

found to have a very high value, and thereby causing low mutual inductances. Whether, the

information about the mutual inductances of 1-2% is wrong, or miscalculations are present is

unexplained. As this is the case, mutual inductances in the region of 1-2% will be used, for worst

case. For Mac values of 1% will be used, and for Mbc and Mab values of 1,39% will be used.

The difference in the values of the mutual inductances is due to the symmetrical structure of the

reactor.

B.4 Summary

In the previous appendix, the determination of the inductances in the shunt reactor was described.

The inductances were determined and are listed in table B.4 and B.5. For simulations the

inductances in table B.6 will be used.

Inductances

La 2,307 H

Lb 2,304 H

Lc 2,307 H

Mab -32,0 mH

Mac -23,0 mH

Mbc -32,0 mH

Table B.6: Self and mutual inductances in the shunt reactor.

Appendix C

Measurements of transmission line

switching

In order to determine the scope of the problem concerning switching overvoltages, measurements

are made at the switching instant, on the 150 kV, 20 kV and 0,4 kV level. Measurements on the

150 kV and 20 kV level are performed by staff from Energinet.dk and measurements on the 0,4

kV level are performed by the project group.

C.1 General test procedure

A simplified single-line diagram of the THØ3-FER3 section is shown in figure C.1.

THØ3 FER3

150 kV line

400 kV

60 kV 20 kV 0,4 kV

40 Mvar Sec 1Sec 2Sec 3

Figure C.1: A simplified single-line diagram for the Tinghøj-Ferslev section and stations, showing the

most important components.

In order to record the switching transients, the following four steps are taken:

1. Disconnection of THØ3

• The line is switched-off in Tinghøj.

2. Disconnection of FER3

• The line is switched-off in Ferslev.

• Measurements during switch-off are performed.

3. Reconnection of FER3

• The line is switched-on in Ferslev.

• Measurements during switch-on are performed.

4. Reconnection of THØ3

• The line is switched-on in Tinghøj.

C-36 Measurements of transmission line switching

C.2 Measurements on 150 kV and 20 kV level

C.2.1 Purpose

This experiment is performed to investigate the switching transients of the 150 kV grid segment

FER3-THØ3.

C.2.2 Instruments

The instruments used in the experiment are listed in table C.1.

Equipment Type Information

2× Transient recorder Omicron CMC 256-6 See appendix D

2× PC IBM With EnerLyzer installed

3× 150 kV voltage transformer ASEA EMFC 145 Transformation ratio: 160√3/ 0,1√

3kV

3× 150 kV current transformer ASEA IMBE 145 A3 Transformation ratio: 1000/1 A

3× 20 kV voltage transformer Transformation ratio: 20√3/ 0,1√

3kV

Table C.1: Instruments used for transient measurements.

C.2.3 Procedure

This test is performed by engineers from Energinet.dk. The principle of the setup is shown in

figure C.2.

+

- CMC 256-6

OMICRON

OMICRON

CMC 256-6

N

N

150 kV current

150 kV voltage

20 kV voltage

measurement

measurement

measurement

THØ3 FER3150 kV section

60 kV 20 kV

RRS

S TT

Sec 1Sec 2Sec 3

ch1

ch2

ch3

ch4

ch5

ch6

ch7

ch8

ch9

40 Mvar

0 V

Trigger (400 Vdc)

Figure C.2: Test setup for transient measurements.

C.2 Measurements on 150 kV and 20 kV level C-37

C.2.4 Data

The voltage behaviour on the 150 kV line after a disconnection is shown in figure C.3.

0 0.5 1 1.5 2 2.5 3−200

−100

0

100

200

Pha

se a

[kV

]

0 0.5 1 1.5 2 2.5 3−200

−100

0

100

200

Pha

se b

[kV

]

0 0.5 1 1.5 2 2.5 3−200

−100

0

100

200

Pha

se c

[kV

]

Time [s]

Figure C.3: Measured voltages on 150 kV level after disconnection.

The voltage behaviour on the 150 kV line after a reconnection (t = 0, 034 s) is shown in figure

C.4.

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065

−150

−100

−50

0

50

100

150

Vol

tage

[kV

]

Time [s]

Figure C.4: Measured voltages on 150 kV level after reconnection.

The current behaviour on the 150 kV line after a reconnection (t = 0, 034 s) is shown in figure

C.5 for a short time period.


0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2−2000

−1000

0

1000

2000

Pha

se a

[A]

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2−2000

−1000

0

1000

2000

Pha

se b

[A]

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2−2000

−1000

0

1000

Pha

se c

[A]

Time [s]

Figure C.5: Measured current on 150 kV level after reconnection.

The current behaviour on the 150 kV line after a reconnection (t = 0, 034 s) is shown in figure

C.6 for a longer time period.

0 0.5 1 1.5 2 2.5−500

0

500

Pha

se a

[A]

0 0.5 1 1.5 2 2.5−500

0

500

Pha

se b

[A]

0 0.5 1 1.5 2 2.5−500

0

500

Pha

se c

[A]

Time [s]

Figure C.6: Measured current on 150 kV level after reconnection.

The voltage behaviour on the 20 kV line after a reconnection (t = 0, 034 s) is shown in figure

C.7.

C.2 Measurements on 150 kV and 20 kV level C-39

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065−25

−20

−15

−10

−5

0

5

10

15

20

25

Vol

tage

[kV

]

Time [s]

Figure C.7: Measured voltages on 20 kV level after reconnection.

C.2.5 Results

150 kV

The voltage behaviour on the 150 kV line after disconnection is shown in figure C.3, and shows

how the voltages slowly decreases.

The peak overvoltages are evaluated from the voltage waveforms in TransView1:

Phase Prior peak voltage Switching peak voltage Percentage

R ±134 kV 159 kV 18,7%

S ±134 kV 154 kV 14,9%

T ±134 kV -162 kV 20,9%

Table C.2: The peak voltages at the 150 kV level, before and after switch-on.

In figure C.8 are the results from the Fourier analysis2 of the first period after switch-on shown.

1TransView is a part of the Enerlyzer software package.2The Fourier analysis is performed in MatLab with the m-file shown in appendix G.10.


0 1000 2000 3000 4000 5000 60000

5

10

0 1000 2000 3000 4000 5000 60000

5

10

Per

cent

age

of th

e fu

ndam

enta

l [%

]15

0 kV

leve

l

0 1000 2000 3000 4000 5000 60000

5

10

Frequency [Hz]

Figure C.8: Fourier analysis of the 150 kV waveform, of harmonic frequencies up to 6,5 kHz.

Beside the fundamental, the Fourier analysis for the 150 kV level shows two large frequency

bands, one around 350 Hz and one around 3,8 kHz.

The peak inrush currents, during reconnection of the 150 kV line, are shown in figure C.5 and

the values are listed in table C.3.

Phase Peak current

R 1423 A

S 1195 A

T 1773 A

Table C.3: Peak magnitude of the inrush current on the 150 kV level.

In figure C.6 are the currents in each phase, shown over a longer time period. At the reconnection

moment, can it be seen, that the currents obtain a slowly decreasing DC-offset.

20 kV

Evaluation of overvoltages in TransView yields:


R ±16,8 kV -17,4 kV 3,6%

S ±16,8 kV 17,3 kV 3,0%

T ±16,9 kV 17,3 kV 3,0%

Table C.4: The peak voltages at the 20 kV level, before and after switch-on.

In figure C.9 are the results from the Fourier analysis3 of the first period after switch-on shown.

3The Fourier analysis is performed in MatLab with the m-file shown in appendix G.10.

C.3 Measurements on 0,4 kV level C-41

0 1000 2000 3000 4000 5000 60000

5

10

0 1000 2000 3000 4000 5000 60000

5

10

Per

cent

age

of th

e fu

ndam

enta

l [%

]20

kV

leve

l

0 1000 2000 3000 4000 5000 60000

5

10

Frequency [Hz]

Figure C.9: fft analysis of the 20 kV waveform, of harmonic frequencies up to 6,5 kHz

The Fourier analysis of the 20 kV level shows harmonics around 350 Hz. On the 20 kV level is

higher frequencies not present.

C.2.6 Summary

Voltage and current waveforms were measured, during switching on/off at the 150 kV level.

Measurements on the 20 kV level, show that the overvoltages caused by reconnection on the

150 kV grid, propagate to the 20 kV level, though considerably damped. Furthermore, the high

frequencies experienced at the 150 kV level did not propagate to the 20 kV level.

C.3 Measurements on 0,4 kV level

C.3.1 Purpose

This experiment is performed to investigate the propagation of overvoltages caused by switching

of 150 kV grid segment to 0,4 kV level.

C.3.2 Instruments

The instruments used in the experiment are listed in table C.5.

Equipment Type Information

Transient recorder Omicron CMC 256-6 See appendix D

PC Dell Inspiron 8200 AAU #: 55657

With EnerLyzer installed

Table C.5: Instruments used for transient measurements.


C.3.3 Procedure

The setup is connected as shown in figure C.10. As only three of the ten inputs are necessary,

input #1, #3 and #5 are used, as those are galvanic isolated from each other.

+

- CMC 256-6

OMICRON

Parallel

Three phase outlet

PC

abcn

Figure C.10: Test setup for transient measurements.

The hardware is configured4 to a measurement range of 600 V with a sampling rate of 28,4 kHz.

The transient recording is started manually, which gives a maximum recording length of 11,72 s

with the chosen sampling rate and three channels used. The switching of the 150 kV grid is then

performed manually within this time.

C.3.4 Data

The voltage behaviour on the 0,4 kV level after a reconnection (t = 0, 0378 s) is shown in figure

C.11.

4The configuration of CMC 256-6 is described in appendix D.

C.3 Measurements on 0,4 kV level C-43

0.04 0.045 0.05 0.055 0.06 0.065 0.07−500

−400

−300

−200

−100

0

100

200

300

400

500

Vol

tage

[V]

Time [s]

Figure C.11: Measured voltages on the 0,4 kV level after reconnection of the 150 kV section.

C.3.5 Results

As shown in figure C.11, do the switching overvoltages on the 150 kV propagate to the 400 V

level. In table C.6 are the peak values of the overvoltages on the 0,4 kV level listed.


a 322,1 V 381,5 V 18,4%

b 316,1 V 339,0 V 7,2%

c 316,8 V 331,2 V 4,5%

Table C.6: The peak voltages on the 0,4 kV level, after switch-on.

At the reconnection instant, switching harmonics are visible, shown and evaluated in figure C.12

with a Fourier analysis5.

5The Fourier analysis is performed in MatLab with the m-file shown in appendix G.10.


0 1000 2000 3000 4000 5000 60000

5

10

0 1000 2000 3000 4000 5000 60000

5

10

Per

cent

age

of th

e fu

ndam

enta

l [%

]0,

4 kV

leve

l

0 1000 2000 3000 4000 5000 60000

5

10

Frequency [Hz]

Figure C.12: fft analysis of the 0,4 kV waveform, of harmonic frequencies up to 6,5 kHz

C.3.6 Summary

Voltage and current waveforms were measured, during switching on at the 150 kV level. The test

indicates that switching transients on the 150 kV grid, propagate to the 0,4 kV level. Fourier

analysis shows that the high frequency experienced at the 150 kV, is filtered.

C.4 Conclusion & discussion

The measurements on the different voltage levels, show how switching transients on the 150 kV

level propagates to lower voltage levels. The magnitude of the propagated switching overvoltages

is though damped on the lower levels, from 150 kV level 20,9%, to a maximum of 3,6% for 20

kV and a maximum of 18,4% for 0,4 kV.

A collection of the voltage waveforms is given in figure C.13.

C.4 Conclusion & discussion C-45

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065

−100

0

100

Vol

tage

[kV

]

Time [s]

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065

−20

−10

0

10

20

Vol

tage

[kV

]

Time [s]

0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065−500

0

500

Vol

tage

[V]

Time [s]

Figure C.13: Voltage waveforms recorded in Ferslev at the moment of switch-on. From top-to-bottom

150 kV, 20 kV, 0,4 kV.

A high-frequency voltage component, is measured for a short period on the 150 kV level, to stays

around 3,8 kHz. This frequency is filtered through the transformers and is not present at the

lower voltage levels.

From the Fourier analysis voltage components around 350 Hz was also measured, which takes

slightly longer time to be damped than the high frequencies.

A single spike in measurements at 0,4 kV, gives 18,4% overshoot. A reason for this high overshoot

could be that the ground obtains a potential for a short period after the switch-on instant, caused

by a zero sequence current. The proposed current path of the zero sequence current is shown in

figure C.14.

150 kV 60 kV 20 kV 0,4 kVGrid

Figure C.14: The proposed zero sequence current path.

The 150 kV level is grounded in the neutral-point, as is the 0,4 kV neutral-point. It is assumed

that the 20 kV level is grounded through a Peterson coil. This could explain why the spike is not

present at the 20 kV level, and why the overvoltages have a lower values than the overvoltages

at the 0,4 kV level.

Throughout the three voltage levels, is the 7th harmonic the dominant one, with 7% of the

fundamental at 150 kV, 4,4% at 20 kV and 5% at 0,4 kV.

Appendix D

Transient recording using Omicron

CMC 256-6

In this appendix is the device for transient recording described. This appendix is written by use

of [26, CMC 256 user manual] and [27, EnerLyser user manual].

D.1 Hardware

The front panel of the CMC 256-6 is shown in figure D.1.

1 2 3 N 4 N

1 2 3 N

1 2 3 N

1 2 3 4

1 2 3 4 5 6 7 8 9 10

0..±20mA 0..±10V

+

-

!

!

!

!

CMC 256-6

OMICRON

Voltage ouput

Current ouput A

Current ouput B

aux dc Analog dc input

Binary/analog input

Binary output

Figure D.1: Front panel of Omicron CMC 256-6.

The ”Binary/analog input” is galvanically isolated in pairs from each other.

D.2 Configuration

In order to perform transient voltage measurements, the ”Binary/analog input” shown in figure

D.1 is used. The hardware configuration is set in the software “EnerLyzer”. For analog

measurements five different measurement ranges and three different sampling rates can be chosen

as listed in table D.1 and table D.2.

Measurement ranges

100 mV 1 V 10 V 100 V 600 V

Table D.1: Measurement ranges for analog measurements using Omicron CMC 256.

Sampling rates

3,16 kHz 9,48 kHz 28,4 kHz

Table D.2: Sampling rates for analog measurements using Omicron CMC 256.

For the 0,4 kV measurements, the range is chosen to 600 V and the sampling rate is chosen to

28,4 kHz.

D-48 Transient recording using Omicron CMC 256-6

D.3 Accuracy

The accuracy of the sampled values depends on the chosen measurement range as shown in table

D.3. As the measurement is on the 0,4 kV level, an accuracy of 0,5% is guaranteed.

Accuracy of the sampling value

typical 0.2 %

guaranteed 0.5 %

Table D.3: Accuracy of the sampling value for measurement range of 600 V, 100 V, 10 V, and 1 V.

The accuracy is relative to the full scale measurement.

D.4 Harmonics

The EnerLyzer software, has an inbuilt harmonic analyzer, described in the user manual:

Harmonics are calculated by means of a full-cycle DFT (Discrete Fourier Transformation) and

are always r.m.s. values. The measuring window is placed to the left of the reference instant

(e.g. cursor position) and its length corresponds to one period of the nominal frequency TN .

Appendix E

Single-line diagrams

The single line diagrams for the two substations are shown in this appendix. In the last section,

the symbols used in the single-line diagrams are breifly explained.

E.1 Ferslev substation

S1 S2

Disp. KT32

1000

11.8

2500 16001600 1600 1000 2500

11.8

1600

1000

11.8

2500 16001600 1600 1000 2500

11.8

1600

2000

11.8

3150 25002500 2500 2000 3150

11.8

2500

1600

11.8

2500 16001600 1600 1600 2500

11.8

1600

1600

11.8

2500 16001600 1600 1600 2500

11.8

1600

1000

11.8

2500 16001600 1600 1000 2500

11.8

1600

1000

11.8

2500 16001600 1600 1000 2500

11.8

1600

1600 600

Disp. KT52

300

300

s s

Disp.

400 kV

60 kV

THØ

1 x 1200 AI

1 x 772

1 x 1200 AI

ADL1

1 x 454

MOS

1 x 772

ZL1

Siemens

80 MVA

KT31

Strømberg

80 MVA

KT51

ABB

400 MVA

Figure E.1: Single-line diagram of transformer station in Ferslev (FER3).

E-50 Single-line diagrams

E.2 Tinghøj substation

S1

1250

1250

1250

Disp.

Disp.

Disp.

2000

7.5

800 1250

2000

7.5

800 1250

2000

7.5

30060 kV

HNB

1 x 1200 AI

1 x 772

FER

1 x 1200 AI

1 x 772

1 x 1200 IAI

ABB

ABB

40 Mvar

40 Mvar

KT31

Strømberg

80 MVA

Figure E.2: Single-line diagram of transformer station in Tinghøj (THØ3).

E.3 Explanation of symbols in single-line diagrams E-51

E.3 Explanation of symbols in single-line diagrams

1000

11.8

2500

1600

1600

1250

Busbar.

Busbar connection with disconnector.

Earthing switch of busbar.

Disconnector.

Circuit breaker.

Capacitive voltage transformer.

Inductive voltage transformer.

Current transformer.

Earthing switch.

Surge arrester.

Cable connection to station.

Overhead line connection to station.

Shunt reactor.

Autotransformer.

Transformer.

Figure E.3: Explanations of symbols used in single-line diagrams.

Appendix F

Parameter matrices for PSCAD

The matrices written by the PSCAD line constant program are shown in this appendix.

• Figure F.1 shows the output file for the overhead line (section 2) configured as described

in section 5.1.4 on page 62.

• Figure F.2 shows the output file for the underground cable (section 3) configured as

described in section 5.1.3 on page 59.

• Figure F.2 shows the output file for the corrected underground cable ( section 3) corrected

as described in section 6.3 on page 90.

F-5

4Para

mete

rm

atric

es

for

PSC

AD

----------------------------------------------------- SECTION 2

PHASE DOMAIN DATA @ 50.00 Hz: Corresponds to PSCAD simulation in chapter 5

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

SERIES IMPEDANCE MATRIX (Z) [ohms/m]:

0.908318063E-04,0.704128248E-03 0.472961756E-04,0.296014235E-03 0.472962763E-04,0.297286539E-03 0.466072857E-04,0.303917061E-03

0.472961756E-04,0.296014235E-03 0.915579651E-04,0.703353087E-03 0.476608051E-04,0.325722308E-03 0.469641348E-04,0.259125595E-03

0.472962763E-04,0.297286539E-03 0.476608051E-04,0.325722308E-03 0.915579651E-04,0.703353087E-03 0.469635548E-04,0.256978115E-03

0.466072857E-04,0.303917061E-03 0.469641348E-04,0.259125595E-03 0.469635548E-04,0.256978115E-03 0.344722914E-03,0.756378016E-03

SHUNT ADMITTANCE MATRIX (Y) [mhos/m]:

0.132230000E-09,0.243401255E-08 0.000000000E+00,-.356286202E-09 0.000000000E+00,-.366967625E-09 0.000000000E+00,-.453693548E-09

0.000000000E+00,-.356286202E-09 0.132230000E-09,0.251071841E-08 0.000000000E+00,-.523915305E-09 0.000000000E+00,-.182825993E-09

0.000000000E+00,-.366967625E-09 0.000000000E+00,-.523915305E-09 0.132230000E-09,0.251126867E-08 0.000000000E+00,-.170187185E-09

0.000000000E+00,-.453693548E-09 0.000000000E+00,-.182825993E-09 0.000000000E+00,-.170187185E-09 0.000000000E+00,0.206434327E-08

LONG-LINE CORRECTED SERIES IMPEDANCE MATRIX [ohms]:

0.171898642E+01,0.133278577E+02 0.894985334E+00,0.560274895E+01 0.894987022E+00,0.562683206E+01 0.881920983E+00,0.575239048E+01

0.894985334E+00,0.560274895E+01 0.173271281E+01,0.133131305E+02 0.901874291E+00,0.616505762E+01 0.888665793E+00,0.490452437E+01

0.894987022E+00,0.562683206E+01 0.901874291E+00,0.616505762E+01 0.173271268E+01,0.133131306E+02 0.888654969E+00,0.486387573E+01

0.881920983E+00,0.575239048E+01 0.888665793E+00,0.490452437E+01 0.888654969E+00,0.486387573E+01 0.652445415E+01,0.143171389E+02

LONG-LINE CORRECTED SHUNT ADMITTANCE MATRIX [mhos]:

0.250354656E-05,0.460775500E-04 0.617432785E-10,-.674454238E-05 0.594313587E-10,-.694674825E-05 -.125759253E-09,-.858857234E-05

0.617432785E-10,-.674454238E-05 0.250355759E-05,0.475297408E-04 0.440640851E-10,-.991781750E-05 -.445111646E-11,-.346083512E-05

0.594313587E-10,-.694674825E-05 0.440640851E-10,-.991781750E-05 0.250355734E-05,0.475401582E-04 0.108056924E-11,-.322157456E-05

-.125759253E-09,-.858857234E-05 -.445111647E-11,-.346083512E-05 0.108056924E-11,-.322157456E-05 0.766497849E-09,0.390794778E-04

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

SEQUENCE COMPONENT DATA @ 50.00 Hz:

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

SEQUENCE TRANSFORM MATRIX:

0.577350269E+00,0.000000000E+00 0.577350269E+00,0.000000000E+00 0.577350269E+00,0.000000000E+00

0.577350269E+00,0.000000000E+00 -.288675135E+00,-.500000000E+00 -.288675135E+00,0.500000000E+00

0.577350269E+00,0.000000000E+00 -.288675135E+00,0.500000000E+00 -.288675135E+00,-.500000000E+00

SEQUENCE IMPEDANCE MATRIX (Zsq) [ohms/m]:

0.186151417E-03,0.131629353E-02 -.730861826E-06,-.943222438E-05 0.370319963E-08,-.943228249E-05 0.000000000E+00,0.000000000E+00

0.370319963E-08,-.943228249E-05 0.438981598E-04,0.397270446E-03 0.735564908E-06,0.196396095E-04 0.000000000E+00,0.000000000E+00

-.730861826E-06,-.943222438E-05 -.733565143E-06,0.196397257E-04 0.438981598E-04,0.397270446E-03 0.000000000E+00,0.000000000E+00

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

SEQUENCE ADMITTANCE MATRIX (Ysq) [mhos/m]:

0.132230000E-09,0.165388712E-08 0.292461573E-11,0.284357996E-10 -.292461573E-11,0.284357996E-10 0.000000000E+00,0.000000000E+00

-.292461573E-11,0.284357996E-10 0.132230000E-09,0.290105625E-08 -.632576761E-11,-.133852592E-09 0.000000000E+00,0.000000000E+00

0.292461573E-11,0.284357996E-10 0.632576761E-11,-.133852592E-09 0.132230000E-09,0.290105625E-08 0.000000000E+00,0.000000000E+00

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

Phase A Phase B Phase C

Zero Positive Negative

Fig

ure

F.1

:T

he

figure

show

sth

eim

ped

ance

and

adm

ittance

matrix

for

the

overh

ead

line

section,

simula

tedin

chapter

5.

F-5

5

----------------------------------------------------- SECTION 3

PHASE DOMAIN DATA @ 50.00 Hz: Corresponds to PSCAD simulation in Chapter 5 -----------------------------------------------------


0.751203844E-04,0.686881613E-03 0.492649635E-04,0.633515432E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03

0.492649635E-04,0.633515432E-03 0.233613922E-03,0.633412694E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03

0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.751239380E-04,0.686892684E-03 0.492685171E-04,0.633526503E-03 0.492139199E-04,0.580160462E-03 0.492139199E-04,0.580160462E-03

0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492685171E-04,0.633526503E-03 0.233617476E-03,0.633423765E-03 0.492139199E-04,0.580160462E-03 0.492139199E-04,0.580160462E-03

0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492139199E-04,0.580160462E-03 0.492139199E-04,0.580160462E-03 0.751239380E-04,0.686892684E-03 0.492685171E-04,0.633526503E-03

0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492139199E-04,0.580160462E-03 0.492139199E-04,0.580160462E-03 0.492685171E-04,0.633526503E-03 0.233617476E-03,0.633423765E-03


0.000000000E+00,0.635748641E-07 0.000000000E+00,-.635748641E-07 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

0.000000000E+00,-.635748641E-07 0.000000000E+00,0.528921349E-06 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.635748641E-07 0.000000000E+00,-.635748641E-07 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,-.635748641E-07 0.000000000E+00,0.528921349E-06 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.635748641E-07 0.000000000E+00,-.635748641E-07

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,-.635748641E-07 0.000000000E+00,0.528921349E-06


0.145526307E+01,0.136712923E+02 0.831085123E+00,0.125565016E+02 0.911457828E+00,0.114323377E+02 0.836067738E+00,0.114384935E+02 0.911457828E+00,0.114323377E+02 0.836067738E+00,0.114384935E+02

0.831085123E+00,0.125565016E+02 0.461895548E+01,0.125885485E+02 0.836067738E+00,0.114384935E+02 0.760665677E+00,0.114445052E+02 0.836067738E+00,0.114384935E+02 0.760665677E+00,0.114445052E+02

0.911457828E+00,0.114323377E+02 0.836067738E+00,0.114384935E+02 0.145533208E+01,0.136715129E+02 0.831152685E+00,0.125567226E+02 0.911493777E+00,0.114323941E+02 0.836103306E+00,0.114385501E+02

0.836067738E+00,0.114384935E+02 0.760665677E+00,0.114445052E+02 0.831152685E+00,0.125567226E+02 0.461902160E+01,0.125887699E+02 0.836103306E+00,0.114385501E+02 0.760700863E+00,0.114445620E+02

0.911457828E+00,0.114323377E+02 0.836067738E+00,0.114384935E+02 0.911493777E+00,0.114323941E+02 0.836103306E+00,0.114385501E+02 0.145533208E+01,0.136715129E+02 0.831152685E+00,0.125567226E+02

0.836067738E+00,0.114384935E+02 0.760665677E+00,0.114445052E+02 0.836103306E+00,0.114385501E+02 0.760700863E+00,0.114445620E+02 0.831152685E+00,0.125567226E+02 0.461902160E+01,0.125887699E+02


0.655855576E-06,0.133523572E-02 -.492205859E-05,-.133520987E-02 -.267528143E-11,-.272280740E-10 -.520540718E-07,0.508068918E-08 -.267527889E-11,-.272280731E-10 -.520540718E-07,0.508068916E-08

-.492205859E-05,-.133520987E-02 0.497152012E-04,0.112169078E-01 -.520540718E-07,0.508068918E-08 0.974687681E-05,0.100660699E-03 -.520540718E-07,0.508068916E-08 0.974687681E-05,0.100660699E-03

-.267528142E-11,-.272280742E-10 -.520540718E-07,0.508068918E-08 0.655855576E-06,0.133523572E-02 -.492205957E-05,-.133520987E-02 -.267537233E-11,-.272282148E-10 -.520543411E-07,0.508085876E-08

-.520540718E-07,0.508068918E-08 0.974687681E-05,0.100660699E-03 -.492205957E-05,-.133520987E-02 0.497158405E-04,0.112169097E-01 -.520543411E-07,0.508085876E-08 0.974720432E-05,0.100661220E-03

-.267527888E-11,-.272280733E-10 -.520540718E-07,0.508068916E-08 -.267537234E-11,-.272282147E-10 -.520543411E-07,0.508085876E-08 0.655855576E-06,0.133523572E-02 -.492205957E-05,-.133520987E-02

-.520540718E-07,0.508068916E-08 0.974687681E-05,0.100660699E-03 -.520543411E-07,0.508085876E-08 0.974720432E-05,0.100661220E-03 -.492205957E-05,-.133520987E-02 0.497158405E-04,0.112169097E-01

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


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


0.577350269E+00,0.000000000E+00 0.577350269E+00,0.000000000E+00 0.577350269E+00,0.000000000E+00

0.577350269E+00,0.000000000E+00 -.288675135E+00,-.500000000E+00 -.288675135E+00,0.500000000E+00

0.577350269E+00,0.000000000E+00 -.288675135E+00,0.500000000E+00 -.288675135E+00,-.500000000E+00


0.226412156E-03,0.186494929E-02 -.264426031E-04,-.279647560E-04 -.263721310E-04,0.635700150E-04 0.147656267E-03,0.175826415E-02 0.181990735E-07,0.177886803E-04 0.181990735E-07,0.177886803E-04

-.263721310E-04,0.635700150E-04 0.787230445E-04,0.711188508E-04 -.726783698E-04,-.545979167E-04 0.153979893E-04,-.891317920E-05 0.153963495E-04,-.891010099E-05 0.153963495E-04,-.891010099E-05

-.264426031E-04,-.279647560E-04 0.198107463E-04,0.368452469E-04 0.787230445E-04,0.711188508E-04 -.154180342E-04,-.887846026E-05 -.154145485E-04,-.887857927E-05 -.154145485E-04,-.887857927E-05

0.147656267E-03,0.175826415E-02 -.154180342E-04,-.887846026E-05 0.153979893E-04,-.891317920E-05 0.279250535E-03,0.184714502E-02 0.418325381E-04,0.190529766E-04 0.109622431E-04,-.724533102E-04

0.181990735E-07,0.177886803E-04 -.154145485E-04,-.887857927E-05 0.153963495E-04,-.891010099E-05 0.109622431E-04,-.724533102E-04 0.131554178E-03,0.532975952E-04 0.418871353E-04,0.724190174E-04

0.181990735E-07,0.177886803E-04 -.154145485E-04,-.887857927E-05 0.153963495E-04,-.891010099E-05 0.418325381E-04,0.190529766E-04 0.110168404E-04,-.190872695E-04 0.131554178E-03,0.532975952E-04


0.000000000E+00,0.176307116E-06 0.115981477E-06,-.881535582E-07 -.115981477E-06,-.881535582E-07 0.000000000E+00,-.211916214E-07 0.000000000E+00,-.211916214E-07 0.000000000E+00,-.211916214E-07

-.115981477E-06,-.881535582E-07 0.000000000E+00,0.239881980E-06 0.171038924E-06,-.563661261E-07 -.183524824E-07,0.105958107E-07 -.183524824E-07,0.105958107E-07 -.183524824E-07,0.105958107E-07

0.115981477E-06,-.881535582E-07 -.171038924E-06,-.563661261E-07 0.000000000E+00,0.239881980E-06 0.183524824E-07,0.105958107E-07 0.183524824E-07,0.105958107E-07 0.183524824E-07,0.105958107E-07

0.000000000E+00,-.211916214E-07 0.183524824E-07,0.105958107E-07 -.183524824E-07,0.105958107E-07 0.000000000E+00,0.331422611E-06 -.134333959E-06,0.987493688E-07 0.134333959E-06,0.987493688E-07

0.000000000E+00,-.211916214E-07 0.183524824E-07,0.105958107E-07 -.183524824E-07,0.105958107E-07 0.134333959E-06,0.987493688E-07 0.000000000E+00,0.394997475E-06 -.134333959E-06,0.351745048E-07

0.000000000E+00,-.211916214E-07 0.183524824E-07,0.105958107E-07 -.183524824E-07,0.105958107E-07 -.134333959E-06,0.987493688E-07 0.134333959E-06,0.351745048E-07 0.000000000E+00,0.394997475E-06

Zero Positive Negative Zero Positive Negative


Fig

ure

F.2

:T

he

figure

show

sth

eim

ped

ance

and

adm

ittance

matrix

for

the

underg

round

cable

section,

simula

tedin

chapter

5.

F-5

6Para

mete

rm

atric

es

for

PSC

AD

----------------------------------------------------- SECTION 3

PHASE DOMAIN DATA @ 50.00 Hz: Corresponds to the corrected cable section simulated in chapter 6 -----------------------------------------------------


0.758331451E-04,0.682683339E-03 0.492661464E-04,0.634845066E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03

0.492661464E-04,0.634845066E-03 0.126386606E-03,0.634588855E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03

0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.758366987E-04,0.682694410E-03 0.492697000E-04,0.634856137E-03 0.492139199E-04,0.580160462E-03 0.492139199E-04,0.580160462E-03

0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492697000E-04,0.634856137E-03 0.126390160E-03,0.634599926E-03 0.492139199E-04,0.580160462E-03 0.492139199E-04,0.580160462E-03

0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492139199E-04,0.580160462E-03 0.492139199E-04,0.580160462E-03 0.758366987E-04,0.682694410E-03 0.492697000E-04,0.634856137E-03

0.492120739E-04,0.580157503E-03 0.492120739E-04,0.580157503E-03 0.492139199E-04,0.580160462E-03 0.492139199E-04,0.580160462E-03 0.492697000E-04,0.634856137E-03 0.126390160E-03,0.634599926E-03


0.000000000E+00,0.749271569E-07 0.000000000E+00,-.749271569E-07 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

0.000000000E+00,-.749271569E-07 0.000000000E+00,0.501703800E-06 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.749271569E-07 0.000000000E+00,-.749271569E-07 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,-.749271569E-07 0.000000000E+00,0.501703800E-06 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.749271569E-07 0.000000000E+00,-.749271569E-07

0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,0.000000000E+00 0.000000000E+00,-.749271569E-07 0.000000000E+00,0.501703800E-06


0.147917825E+01,0.136435749E+02 0.890255341E+00,0.126414123E+02 0.920430106E+00,0.114926792E+02 0.891470771E+00,0.114951657E+02 0.920430106E+00,0.114926792E+02 0.891470771E+00,0.114951657E+02

0.890255341E+00,0.126414123E+02 0.247764775E+01,0.126433483E+02 0.891470771E+00,0.114951657E+02 0.862509528E+00,0.114976311E+02 0.891470771E+00,0.114951657E+02 0.862509528E+00,0.114976311E+02

0.920430106E+00,0.114926792E+02 0.891470771E+00,0.114951657E+02 0.147924770E+01,0.136437964E+02 0.890324230E+00,0.126416340E+02 0.920466274E+00,0.114927360E+02 0.891506793E+00,0.114952226E+02

0.891470771E+00,0.114951657E+02 0.862509528E+00,0.114976311E+02 0.890324230E+00,0.126416340E+02 0.247771608E+01,0.126435701E+02 0.891506793E+00,0.114952226E+02 0.862545404E+00,0.114976881E+02

0.920430106E+00,0.114926792E+02 0.891470771E+00,0.114951657E+02 0.920466274E+00,0.114927360E+02 0.891506793E+00,0.114952226E+02 0.147924770E+01,0.136437964E+02 0.890324230E+00,0.126416340E+02

0.891470771E+00,0.114951657E+02 0.862509528E+00,0.114976311E+02 0.891506793E+00,0.114952226E+02 0.862545404E+00,0.114976881E+02 0.890324230E+00,0.126416340E+02 0.247771608E+01,0.126435701E+02


0.449379481E-06,0.157367584E-02 -.237634492E-05,-.157366405E-02 -.544875861E-12,-.554937560E-11 -.215233875E-07,0.203904846E-08 -.544871327E-12,-.554937773E-11 -.215233875E-07,0.203904846E-08

-.237634492E-05,-.157366405E-02 0.228333559E-04,0.106280469E-01 -.215233875E-07,0.203904846E-08 0.770283689E-05,0.844406496E-04 -.215233875E-07,0.203904846E-08 0.770283689E-05,0.844406496E-04

-.544875861E-12,-.554937552E-11 -.215233875E-07,0.203904846E-08 0.449379481E-06,0.157367584E-02 -.237634533E-05,-.157366405E-02 -.544893449E-12,-.554940585E-11 -.215234988E-07,0.203911822E-08

-.215233875E-07,0.203904846E-08 0.770283689E-05,0.844406496E-04 -.237634533E-05,-.157366405E-02 0.228338831E-04,0.106280485E-01 -.215234988E-07,0.203911822E-08 0.770310898E-05,0.844410870E-04

-.544871327E-12,-.554937765E-11 -.215233875E-07,0.203904846E-08 -.544893406E-12,-.554940584E-11 -.215234988E-07,0.203911822E-08 0.449379481E-06,0.157367584E-02 -.237634533E-05,-.157366405E-02

-.215233875E-07,0.203904846E-08 0.770283689E-05,0.844406496E-04 -.215234988E-07,0.203911822E-08 0.770310898E-05,0.844410870E-04 -.237634533E-05,-.157366405E-02 0.228338831E-04,0.106280485E-01

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


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


0.577350269E+00,0.000000000E+00 0.577350269E+00,0.000000000E+00 0.577350269E+00,0.000000000E+00

0.577350269E+00,0.000000000E+00 -.288675135E+00,-.500000000E+00 -.288675135E+00,0.500000000E+00

0.577350269E+00,0.000000000E+00 -.288675135E+00,0.500000000E+00 -.288675135E+00,-.500000000E+00


0.191145680E-03,0.186342892E-02 -.651709507E-05,0.252038515E-05 -.103172190E-04,0.317366066E-04 0.147656661E-03,0.175870736E-02 0.185933687E-07,0.182318915E-04 0.185933687E-07,0.182318915E-04

-.103172190E-04,0.317366066E-04 0.434553853E-04,0.682688441E-04 -.539049498E-04,-.247765680E-04 0.157816243E-04,-.913512628E-05 0.157799845E-04,-.913204806E-05 0.157799845E-04,-.913204806E-05

-.651709507E-05,0.252038515E-05 0.370165632E-04,0.434599724E-05 0.434553853E-04,0.682688441E-04 -.158020636E-04,-.909972440E-05 -.157985779E-04,-.909984340E-05 -.157985779E-04,-.909984340E-05

0.147656661E-03,0.175870736E-02 -.158020636E-04,-.909972440E-05 0.157816243E-04,-.913512628E-05 0.208004033E-03,0.184741613E-02 0.222906651E-04,-.116541116E-04 -.547669809E-05,-.408411660E-04

0.185933687E-07,0.182318915E-04 -.157985779E-04,-.909984340E-05 0.157799845E-04,-.913204806E-05 -.547669809E-05,-.408411660E-04 0.603064929E-04,0.522390670E-04 0.223464452E-04,0.430415628E-04

0.185933687E-07,0.182318915E-04 -.157985779E-04,-.909984340E-05 0.157799845E-04,-.913204806E-05 0.222906651E-04,-.116541116E-04 -.542091798E-05,0.138545084E-04 0.603064929E-04,0.522390670E-04


0.000000000E+00,0.167234600E-06 0.101570198E-06,-.836173000E-07 -.101570198E-06,-.836173000E-07 0.000000000E+00,-.249757190E-07 0.000000000E+00,-.249757190E-07 0.000000000E+00,-.249757190E-07

-.101570198E-06,-.836173000E-07 0.000000000E+00,0.242161757E-06 0.166459019E-06,-.461537215E-07 -.216296071E-07,0.124878595E-07 -.216296071E-07,0.124878595E-07 -.216296071E-07,0.124878595E-07

0.101570198E-06,-.836173000E-07 -.166459019E-06,-.461537215E-07 0.000000000E+00,0.242161757E-06 0.216296071E-07,0.124878595E-07 0.216296071E-07,0.124878595E-07 0.216296071E-07,0.124878595E-07

0.000000000E+00,-.249757190E-07 0.216296071E-07,0.124878595E-07 -.216296071E-07,0.124878595E-07 0.000000000E+00,0.309493481E-06 -.123199805E-06,0.961051595E-07 0.123199805E-06,0.961051595E-07

0.000000000E+00,-.249757190E-07 0.216296071E-07,0.124878595E-07 -.216296071E-07,0.124878595E-07 0.123199805E-06,0.961051595E-07 0.000000000E+00,0.384420638E-06 -.123199805E-06,0.211780026E-07

0.000000000E+00,-.249757190E-07 0.216296071E-07,0.124878595E-07 -.216296071E-07,0.124878595E-07 -.123199805E-06,0.961051595E-07 0.123199805E-06,0.211780026E-07 0.000000000E+00,0.384420638E-06


Zero Positive Negative Zero Positive Negative

Fig

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show

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ittance

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6.

Appendix G

M-files used for calculations

G.1 M-files for determination of parameters in the overhead line

system

1 % Predefined constants :2 p = 0.028264; % Resistivity of aluminium3 p_jord = 49.5; % Resistivity of earth return4 f = 50; % Frequency from power source5 w = 2*pi*f; % Angular frequency form power source67 A_dorkin = 153; % Cross section of dorkin8 A_martin = 685; % Cross section of martin9 l = 18930; % Distance for section 2

10 Ds = 0.0146; % Self GMD for Martin conductor11 Ds_g = 0.007; % Self GMD for Dorkin conductor1213 a = exp(j*((2* pi)/3)); % Displacement between phases (120 degrees )14 T = [[1 1 1 ] % Transformation matrix15 [1 a^2 a ]16 [1 a a^2]];1718 Rdc = p/A_martin ; % Resistivity of martin conductor19 Rdc_g = p/A_dorkin ; % Resistivity of ground conductor20 Rd = 49.5e-6; % Resistivity of earth return21 k_ac = 1.15; % Correction factor for resistance22 Rac = Rdc *k_ac;2324 r = sqrt(A_martin /pi)/1000; % Radius of conductor25 rg = sqrt(A_dorkin /pi); % Radius of ground conductor

MatLab script G.1: constants.m-M-file for constants for calculation of impedance and admittance.

1 %Geometry2 % G (8.35 41.6)3 % |4 % |5 % --------T (10.4 33.5)6 % / \7 % / \8 % / \9 % (7.2 24.7) S-------R (13.0 24.7)

1011 % Coordinates [x y]12 Sxy = [7.2 24.7];13 Rxy = [13.0 24.7];14 Txy = [10.4 33.5];15 Gxy = [8.35 41.6];1617 % Mirror images coordinates [x y]18 Sxym = [7.2 -24.7];19 Rxym = [13.0 -24.7];20 Txym = [10.4 -33.5];21 Gxym = [8.35 -41.6];22 Dj = 660* sqrt(p_jord /f); % distance to earth -return2324 %Distances between conductors25 d12 = Rxy (1) -Sxy (1); %phase s-to - r/t/g26 d13 = sqrt((Txy (1) -Sxy (1))^2 +(Txy (2) -Sxy (2))^2) ;27 d1g = sqrt((Gxy (1) -Sxy (1))^2 +(Gxy (2) -Sxy (2))^2) ;2829 d21 = d12 ; %phase r-to - s/t/g30 d23 = sqrt((Rxy (1) -Txy (1))^2 +(Rxy (2) -Txy (2))^2) ;31 d2g = sqrt((Gxy (1) -Rxy (1))^2 +(Gxy (2) -Rxy (2))^2) ;3233 d31 = d13 ; %phase t-to - s/r/g34 d32 = d23 ;35 d3g = sqrt((Gxy (1) -Txy (1))^2 +(Gxy (2) -Txy (2))^2) ;3637 dg1 = d1g ; %ground -to- s/r/t

G-58 M-files used for calculations

38 dg2 = d2g ;39 dg3 = d3g ;4041 % Distances to mirror images42 D11 = 2*Sxy (2); %phase S-to- S/R/T/G43 D12 = sqrt(( Sxym (1) -Rxy (1))^2 + (Sxym (2) -Rxy (2))^2);44 D13 = sqrt(( Txym (1) -Sxy (1))^2 + (Txym (2) -Sxy (2))^2);45 D1g = sqrt(( Gxym (1) -Sxy (1))^2 + (Gxym (2) -Sxy (2))^2);4647 D21 = D12 ; %phase R-to- S/R/T/G48 D22 = 2*Rxy (2);49 D23 = sqrt(( Rxym (1) -Txy (1))^2 + (Rxym (2) -Txy (2))^2);50 D2g = sqrt(( Gxym (1) -Rxy (1))^2 + (Gxym (2) -Rxy (2))^2);5152 D31 = D13 ; %phase T-to- S/R/T/G53 D32 = D23 ;54 D33 = 2*Txy (2);55 D3g = sqrt(( Gxym (1) -Txy (1))^2 +(Gxym (2) -Txy (2))^2);5657 Dg1 = D1g ; %ground -to - S/R/T/G58 Dg2 = D2g ;59 Dg3 = D3g ;60 Dgg = 2*Gxy (2);

MatLab script G.2: geometry.m-M-file for geometry of the overhead line system.

1 % overhead line series impedance calculations2 clc ; close all; clear all;34 % Load constants & geometry5 constants6 geometry78 % DC -resistance matrix9 R_abc = [[ Rac+Rd Rd Rd Rd ]

10 [ Rd Rac+Rd Rd Rd ]11 [ Rd Rd Rac+Rd Rd ]12 [ Rd Rd Rd Rdc_g +Rd ]]; %[ohm/m]1314 % Inductance matrix15 L_abc = 2e-7* [[log(Dj/Ds) log(Dj/d12) log(Dj/d13) log(Dj/d1g ) ]16 [log(Dj/d21) log(Dj/Ds) log(Dj/d23) log(Dj/d2g ) ]17 [log(Dj/d31) log(Dj/d32) log(Dj/Ds) log(Dj/d3g ) ]18 [log(Dj/dg1) log(Dj/dg2) log(Dj/dg3) log(Dj/Ds_g)]]; % [H/m

]1920 % Inductance phase values21 Labc = (L_abc (1:3 ,1:3) - L_abc (1:3 ,4) *(1/ L_abc (4,4))*L_abc (4 ,1:3) );22 % Impedance23 Z_abc = R_abc + j*w*L_abc; % [ohm/m]2425 % Reduction of 4x4 impedance matrix to 3x3 matrix26 Z_abc = Z_abc (1:3 ,1:3) - Z_abc (1:3 ,4) *(1/ Z_abc (4,4))*Z_abc (4 ,1:3) ;2728 %Sequence matrix29 Z_012 = T^-1 * Z_abc * T; % [ohm/m]3031 %Reactance32 X_012 = imag(Z_012); % [ohm/m]33 %Resistance34 R_012 = real(Z_012); % [ohm/m]3536 Z_abc_section2 = Z_abc*l37 R_abc_section2 = real( Z_abc_section2)38 L_abc_section2 = imag( Z_abc_section2)/w3940 Z_012_section2 = Z_012 % [ohm]41 L_012_section2 = X_012/w* l % [H]4243 R_ac_section2 = Rac*l % [ohm]44 R_dcg_section2 = Rdc_g*l % [ohm]

MatLab script G.3: impedance.m-M-file for calculations of the impedance sequence matrix.

1 % overhead line admittance calculations2 clc ; close all; clear all;34 % Load constants & geometry

G.2 M-files for determination of parameters in the shunt reactor G-59

5 constants6 geometry78 % Potential matrix9 P_abcg = 18e9*[[ log (D11/r) log(D12 /d12) log(D13/d13 ) log(D1g/d1g ) ]

10 [ log (D21/d21) log(D22 /r) log(D23/d23 ) log(D2g/d2g ) ]11 [ log (D31/d31) log(D32 /d32) log(D33/r) log(D3g/d3g ) ]12 [ log (Dg1/dg1) log(Dg2 /dg2) log(Dg3/dg3 ) log ((Dgg )/rg)]];1314 % Reduction to 3x3 matrix15 P_abc = P_abcg (1:3 ,1:3) - P_abcg (1:3 ,4) *(1/ P_abcg (4,4))*P_abcg (4 ,1:3) ;1617 C_abc = P_abc ^-1; % Capacitance18 X_abc = 1./(w*C_abc); % Reactance19 B_abc = 1./ X_abc ; % Susceptance20 Y_abc = 0 + j*B_abc; % Admittance matrix2122 % Sequence matrix23 Y_012 = T^-1 * Y_abc * T;2425 %Writeout of parameters26 C_abc_section2 = imag(Y_abc )/w * l * 1e62728 Y_012_section2 = Y_012 * l * 1e6 % [muS ]29 Y_012_section2km = Y_012 * 1000 * 1e6 % [muS/km]30 C_012_section2 = imag(Y_012 )/w * l * 1e6 % [muF]31 C_012_section2km = imag(Y_012)/w * 1000 * 1e6 % [muF/km]

MatLab script G.4: admittance.m-M-file for calculations of the admittance sequence matrix.

G.2 M-files for determination of parameters in the shunt reactor

1 % Predefined constants :2 mu_0 = 4*pi*10^ -7; % Permeability3 mu_r = 52073.5; % Relative permeability in the core4 mu_r_gap = 1.302; % Relative permeability in the gap5 N = 1194; % Number of windings

MatLab script G.5: constants.m-M-file for constants for calculation of the shunt reactor.

1 % l3 l2 l2 l32 % -------------------------3 % | | | | |4 % h | | | | |5 % | | | | |6 % -------------------------78 l = 4182e-3; % Total lenght9 l2 = 1203e-3; % Length l2 in figure above

10 l3 = 888e-3; % Length l3 in figure above11 h = 2213e-3; % Total height12 d_core = 210e-3; % Diameter of outer core13 w_core = 794e-3; % Width of outer core14 d_limb = 600e-3; % Diameter of limb1516 h_seg = 151.9e-3; % Height of one segment17 h_plate = 70e-3; % Cross flux plates18 N_seg = 9; % Number of segment in one limb19 h_limb = h_seg*N_seg +2* h_plate ; % Total limb height2021 h_g = 32.7e-3; % Height of one gap22 N_g = 8; % Number of gaps23 h_air = 2*10e -3+2*2.5 e-3; % Gap at flux plates24 h_gap = h_g*N_g + h_air; % Total height of gaps252627 % s3 s2 s2 s328 % ----------------------------------------29 % | | | | |30 % | | | sg | | sg | | sg |31 % s4 | | | | | s432 % | | s1 | s1 | s1 |33 % | | | | |34 % ----------------------------------------


35 % s3 s2 s2 s33637 % Flux mean paths:38 s2 = l2;39 s3 = l3 - d_core /2;40 s4 = h - d_core ;41 s1 = s4 - h_gap;42 sg = h_gap;4344 % Flux areas:45 A1 = pi*( d_limb /2) ^2;46 Ag = A1;47 A2 = d_core *w_core ;48 A3 = A2;49 A4 = A2;

MatLab script G.6: geometry.m-M-file for geometry of the shunt reactor.

1 clc ; close all; clear all;23 % Load constants & geometry4 constants5 geometry67 % Reluctance calculation :8 R1 = s1/( mu_0*mu_r*A1);9 R2 = s2/( mu_0*mu_r*A1);

10 R3 = s3/( mu_0*mu_r*A2);11 R4 = s4/( mu_0*mu_r*A3);12 Rg = sg/( mu_0*mu_r_gap *Ag);1314 % Uniting reluctances :15 R1g = R1 + Rg;16 R343 = R3 + R4 + R3;1718 % Reluctance matrix :19 R = [[ R343+R1g -R1g 0 0 ];20 [-R1g (2* R2 + 2* R1g) -R1g 0 ];21 [0 -R1g (2* R2 + 2*R1g) -R1g ];22 [0 0 -R1g R343+R1g ]];2324 % Inversion of reluctance matrix :25 R_inv = R^-12627 % Reduction premeance to 3x3:28 P_123 = [(( R_inv (2,2) -R_inv (1,2)) - (R_inv (2,1) -R_inv (1,1))) ...29 (( R_inv (3,2) -R_inv (2,2)) - (R_inv (3,1) -R_inv (2,1))) ...30 (( R_inv (4,2) -R_inv (3,2)) - (R_inv (4,1) -R_inv (3,1)));31 (( R_inv (2,3) -R_inv (1,3)) - (R_inv (2,2) -R_inv (1,2))) ...32 (( R_inv (3,3) -R_inv (2,3)) - (R_inv (3,2) -R_inv (2,2))) ...33 (( R_inv (4,3) -R_inv (3,3)) - (R_inv (4,2) -R_inv (3,2)));34 (( R_inv (2,4) -R_inv (1,4)) - (R_inv (2,3) -R_inv (1,3))) ...35 (( R_inv (3,4) -R_inv (2,4)) - (R_inv (3,3) -R_inv (2,3))) ...36 (( R_inv (4,4) -R_inv (3,4)) - (R_inv (4,3) -R_inv (3,3)));]3738 P_123 ^-139 1./( P_123 ^-1)4041 % Inductance matrix :42 L = N^2.* P_123

MatLab script G.7: reactor.m-M-file for calculations of inductances in the shunt reactor.

G.3 M-files for mathematical analysis

1 clc ; close all; clear all;23 Up = 165 e3*sqrt (2)/sqrt (3);45 % Parameters6 R1 = 1.245;7 L1 = 39.2e -3;8 R2 = 1.517;9 L2 = 2.310;

G.3 M-files for mathematical analysis G-61

10 Rs = 0.800;11 Ls = 14.4e-3;12 C = 5.10e-6;1314 % Unify short circuit impedance15 R1 = R1 + Rs;16 L1 = L1 + Ls;1718 omega = 2*pi *50;19 a = 2*pi/3;20 t=0:0.00005:2.8;2122 Zs = tf([Ls Rs],1);23 Yin = tf([C*L2 C*R2 1],[C*L1*L2 (R1*C*L2+R2*C*L1) (C*R2*R1+L1+L2) (R1+R2)]);2425 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%26 % Current %27 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%2829 % Phase a:30 alpha = pi/2;31 A = sin(alpha);32 B = cos(alpha);33 Ua = tf([A*Up B*Up*omega ],[1 0 omega ^2]);3435 Ia = Yin*Ua;3637 % Extract numerator and denumerator38 [Ia_num ,Ia_denum ] = tfdata (Ia,’v’);3940 % Partitial fraction41 [r,p,k] = residue (Ia_num ,Ia_denum );4243 ia = 0;44 for k = 2: length (r)+145 ia = ia + r(k-1)*exp (p(k-1)*t);46 end474849 % Phase b:50 alpha = pi/2 - a;51 A = sin(alpha)52 B = cos(alpha)53 Ub = tf([A*Up B*Up*omega ],[1 0 omega ^2]);5455 Ib = Yin*Ub;5657 % Extract numerator and denumerator58 [Ib_num ,Ib_denum ] = tfdata (Ib,’v’);5960 % Partitial fraction61 [r,p,k] = residue (Ib_num ,Ib_denum );6263 ib = 0;64 for k = 2: length (r)+165 ib = ib + r(k-1)*exp (p(k-1)*t);66 end6768 % Phase c:69 alpha = pi/2 + a;70 A = sin(alpha);71 B = cos(alpha);72 Uc = tf([A*Up B*Up*omega ],[1 0 omega ^2]);7374 Ic = Yin*Uc;7576 % Extract numerator and denumerator77 [Ic_num ,Ic_denum ] = tfdata (Ic,’v’);7879 % Partitial fraction80 [r,p,k] = residue (Ic_num ,Ic_denum );8182 ic = 0;83 for k = 2: length (r)+184 ic = ic + r(k-1)*exp (p(k-1)*t);85 end8687 figure88 plot(t,ia*1e-3,t,ib*1e-3,t,ic*1e-3);grid;89 title (’\ bfi(t) ’)


90 xlabel (’t [s]’)91 ylabel (’i [kA]’)9293 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%94 % Voltage %95 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%9697 % Phase a:98 U1a = Ua - Ia*Zs;99

100 % Extract numerator and denumerator101 [U1a_num ,U1a_denum ] = tfdata (U1a ,’v’);102103 % Partitial fraction104 [r,p,k] = residue (U1a_num ,U1a_denum )105106 u1a = 0;107 for k = 2: length (r)+1108 u1a = u1a + r(k -1)*exp(p(k-1)*t);109 end110111 % Phase b:112 U1b = Ub - Ib*Zs;113114 % Extract numerator and denumerator115 [U1b_num ,U1b_denum ] = tfdata (U1b ,’v’);116117 % Partitial fraction118 [r,p,k] = residue (U1b_num ,U1b_denum );119120 u1b = 0;121 for k = 2: length (r)+1122 u1b = u1b + r(k -1)*exp(p(k-1)*t);123 end124125 % Phase c:126 U1c = Uc - Ic*Zs;127128 % Extract numerator and denumerator129 [U1c_num ,U1c_denum ] = tfdata (U1c ,’v’);130131 % Partitial fraction132 [r,p,k] = residue (U1c_num ,U1c_denum );133134 u1c = 0;135 for k = 2: length (r)+1136 u1c = u1c + r(k -1)*exp(p(k-1)*t);137 end138139 figure140 plot(t,u1a *1e-3,t,u1b *1e-3,t,u1c *1e-3);grid;141 title (’\ bfu1(t) ’)142 xlabel (’t [s]’)143 ylabel (’u [kV]’)

MatLab script G.8: simpel.m-M-file for numerical approach in MatLab.

1 clc , clear all , close all2 t = 1:0.1:100;34 Ua = 165 e3*sqrt (2)/sqrt (3);5 alpha = -pi/2; % phase delay67 % Parameters8 R1 = 1.245;9 L1 = 39.2e -3;

10 R2 = 1.969;11 L2 = 2.310;12 Rs = 0.800;13 Ls = 14.4e -3;14 C = 5.10e-6;15 Mab = 32.0e-3; % [H]16 Mac = 23.0e-3; % [H]17 Mbc = 32.0e-3; % [H]1819 % Unify short circuit impedance20 R_1 = R1 + Rs;21 L_1 = L1 + Ls;22 R_2 = R2;

G.4 M-files for miscellaneous purposes G-63

23 L_2 = L2;2425 Fs = 20000;

MatLab script G.9: simplified_model_init.m-Initialization of calculated parameters for simulinkmodel.

G.4 M-files for miscellaneous purposes

1 clc;close all; clear all ;23 fs = 1/0.035 e-3; % sampling frequency (0.035e-3 s per sample )45 % Load file and voltage waveform6 ans = csvread (’ measurements_ferslev/Omicron /150kV.txt ’);7 U = ans (3,:) /1000; % in kV89 % Shorten data to one period after

10 N = 572; % period time = 0.02s => 572 samples /period11 U = U(286:286+571); % switch on t=0.01s => 286 samples1213 X = abs(fft(U,N))/(N/2);14 X = fftshift (X);15 F = fs * [-N/2 : N/2 - 1]/N; % Mirrored round x-axis1617 figure18 bar(F,X); % plot fft as bar ’s19 grid;20 axis ([0 6500 0 10]); % 0->6500 Hz21 xlabel (’Frequency [Hz]’);22 ylabel (’ Percentage of the fundamental [%] ’; ’150 kV level ’) ;

MatLab script G.10: fft_for_report.m-M-file for FFT in MatLab. This file shows the principle forFFT analysis of one phase.

1 clc;clear all;close all23 mu = 4*pi *10^ -7;4 rho = 2.826*10^ -8;5 sigma = 1/rho;6 r = 18.1e-3;7 Rdc = 0.0423*18.93;8 L = 36.2e-3;9 C = .149e-6;

10 Rg_dc = 0.00099*50*18.93;1112 Rc = 0.693;13 Lc = 7.20e-3;14 Cc = 5.10e-6;1516 for f = 1:200017 d(f) = sqrt(rho ./(f*pi*mu));18 omega (f) = 2*pi*f;1920 if f < 6521 ks(f) = 1 + (r/d(f))^4/(48 + 0.8*( r/d(f))^4);22 else23 ks(f) = 0.25 + 0.5*(r./d(f)) + 3/32*( d(f)/r);24 end2526 Rac(f) = Rdc*ks(f);27 Rg(f) = 0.00099* f*18.93;2829 Z_ac_in (f) = ((Rac(f)+Rg(f))/2+j*omega (f)*L/2) + (1./(j*omega(f)*C)) * ((Rac(f)+

Rg(f))/2+j*omega(f)*L/2) /((1./( j*omega (f)*C)) + ((Rac (f)+Rg(f))/2+j*omega(f)*L/2));

30 R_ac_in (f) = real(Z_ac_in (f));3132 Z_dc_in (f) = ((Rdc+Rg_dc)/2+j*omega (f)*L/2) + (1./(j*omega(f)*C)) * ((Rdc+Rg_dc)

/2+j*omega(f)*L/2) /((1./( j*omega (f)*C)) + ((Rdc +Rg_dc )/2+j*omega(f)*L/2));33 R_dc_in (f) = real(Z_dc_in (f));34 end35


36 f_ps = [50 100 150 200 250 300 350 400 500 750 1000 1250 1500 1750 2000 2500 3000] ;3738 U = [165* sqrt (2)/sqrt (3) ];39 I = [11.4477 5.93649 4.03 3.25 2.433 2.036 1.75 1.53 1.228 .8128 .5958 .4642 .37075

.30045 .243 .1572 .08842];4041 tu = [.180];42 ti = [.184484 .18233593 .18157438 .18118904 .18095537 .18079782 .180684628 .180599897

.180480785 .1803208678 .18024049 .180192066 .18015975 .180136694 .180118719.1800942975 .1800759090];

4344 for k = 1: length (f_ps)45 theta (k) = (ti(k)-tu(1))/(1/ f_ps(k)) * 360;46 Z(k) = U(1)/I(k);47 R(k) = cos(theta(k)*2* pi /360)*Z(k);48 end4950 f = 1:f(end);5152 plot(f,R_dc_in ,f,R_ac_in ,’-.’,f_ps ,R,’o:’);grid;53 axis ([0 f(end) 0 50]);54 ylabel (’Real part of input impedance (\ Omega) ’);55 xlabel (’Frequency [Hz]’)56 legend (’Calculated with frequency independent series and ground resistance ’,’

Calculated with frequency dependent series and ground resistance ’,’ Calculatedfrom PSCAD simulation ’,2);

57 set(gcf ,’PaperUnits ’,’ centimeters ’)58 set(gcf ,’PaperPosition ’,[1 0 32 16])59 %saveas (gca ,’../ report /billeder /pscad /verification_frequency .eps ’);

MatLab script G.11: pscad_frekvens_verification.m-M-file for calculating and plotting the frequencydependence of the transmission line in PSCAD.

Appendix H

CD content

In this appendix, the content of the CD is briefly described. Besides the three main folders described in

the following, an electronic copy of the report is placed in the root of the CD.

Data & information

Articles

• A_charge_simulation_based_method_for_calculating_corona_loss.pdf

This article concerns calculation of corona losses on AC transmission lines.

• Optimized_Transformer_Design_Inclusive_of_high_frequency_effects.pdf

This article concerns the frequency aspect in transformer design. Among others frequency

dependence of a round wire.

• Pre-insertion_resistor_in_HV_capacitor_bank_switching.pdf

This article concerns limitations of switching overvoltages, with different approaches.

Miscellaneous

• Description_of_the_system_Horns_Rev.pdf

Contains information about the two windmill parks at Horns Rev.

• Miljorapport2006.pdf

Contains information about energy consumption, production, transmission etc. in Denmark.

Overhead line

• Donau.bmp

Contains information about the physical layout of the overhead line section.

PSCAD

• PSCAD help file.chm

Contains the help file for PSCAD v4.2.

• PSCAD Users Guide.pdf

Contains the user’s guide for PSCAD v4.2.

Shunt reactor

• Core_and_winding_geometries.pdf

Contains information about the physical layout of the shunt reactor.

• Core_design.pdf

Contains information about the principle structure of the shunt reactor.

• DRAFT_IEC_60076-6_Reactors_Chapter_7_Shunt_reactors.pdf

Contain a draft of the standard IEC60076-6.

• Instruction_Manual.pdf

Contains information about the shunt reactor, including test report and rating plate.

H-66 CD content

• Magnetising_curve.pdf

Contains the DC magnetization curve for the reactor core (M-5 0.30 mm).

• The_ABB_Shunt_Reactor.pdf

Contains information about the shunt reactor, including test report and rating plate.

• Winding_and_core_data.pdf

Contains information about design of the shunt reactor.

Transformers

• 60-20kV_&_20-0,4kV.pdf

Contains information about the 60-20 kV and the 20/0,4 kV transformer.

• 150-60kV.pdf

Contains information about the 150-60 kV transformer.

Underground cable & stations

• one-line_diagram_and_cable_information.PDF

Contains the one-line diagrams for FER3 and THØ3. Furthermore, information about parameter,

layout etc. for the cable are included in this file.

• XLPE_Cable_Systems_Users_Guide.pdf

Contains the user’s guide for the cable.

M-files

FFT analysis

• principle_of_fft.m

M-file containing the principle of the Fourier analysis.

Mathematical analysis

• simpel.m

M-file for solving the simplified model through the analytical approach.

Parameter determination of the overhead line system

• admittance.m

M-file for calculations of the admittance sequence matrix.

• constants.m

M-file for constants for calculation of impedance and admittance.

• geometry.m

M-file for geometry of the overhead line system.

• impedance.m

M-file for calculations of the impedance sequence matrix.

Parameter determination of the shunt reactor

• constants.m

M-file for constants for calculation of the shunt reactor.

H-67

• geometry.m

M-file for geometry of the shunt reactor.

• reactor.m

M-file for calculations of inductances in the shunt reactor.

PSCAD frequency dependent resistance

• pscad_frequency_dependent_resistance.m

M-file for calculating and plotting the frequency dependence of the transmission line in PSCAD.

PSCAD

• Ferslev-Tinghoj_1.psc

Simulation of the system FER3-THØ3, described in chapter 5 on page 57.

• Ferslev-Tinghoj_2.psc

Simulation of the system FER3-THØ3, with the corrected cable as described in chapter 6 on

page 83.

• Pre-insertion_resistor.psc

Simulation of the pre-insertion resistor principle as described in chapter 7 on page 97.

• Reactor_verification.psc

Simulation of the reactor verification as described in section 5.2 on page 68.

• Simple_model.psc

Simulation of the simplified model as described in section 4.4 on page 44.

• Zero_crossing_closure.psc

Simulation of the zero crossing closure principle as described in chapter 7 on page 97.

propagation of transmission level switching overvoltages.pdf

Documents

bestende af

kv level

kv grid segment

transmission system

switching overvoltages

af ensimplificeret model

af disse overspndinger

switching overvoltagesin