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Polytechnic of Namibia

School of Engineering

Department of Electrical Engineering

Power systems Engineering(PSE610S) semester project

report

Title: Simulation of transmission line parameters

Compiled by: Hiskiel Stephanus

Student number: 201082616

Supervisor: Dr Al-mas Sendegeya

Abstract

Transmission line parameters are the most important aspects that can ever be considered when dealing with

transmission lines. Transmission lines are often categorised in three major categories which are, the short

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lines, the medium lines and the long lines. Upon obtaining the transmission line parameters, it is often much

easier to predict the performance of a transmission line through efficiency measurement and per unit

analysis.

The students were thus given this project in order to model the transmission lines with their parameters by

using a very important mathematical tool called matlab. The students were thus expected to also evaluate the

performance of these lines y mainly using the values allocated to them and obtaining the appropriate results.

Acknowledgments

The students would like to very much acknowledge and appreciate the course lecturer for giving them this

window of opportunity to enhance their analytical skills using this software. They would like to thank him

for also being there for them during the course of the semester especially when they had problems with the

project . He took the students in for guidance and was able to ensure they always go with the correct path intheir problem solving skills.

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The students would thus like to assure the course lecturer that the knowledge they have obtained from doing

this project will not go to waste and will stick with them as a very important tool when they enter the

electrical industry.

Table of ContentsIntroduction ............................................................................................................................................ 4

Methodology ........................................................................................................................................... 6

Derivation of the ABCD parameters ....................................................................................................... 7

Program outline ................................................................................................................................... 12

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Codes used for the simulation of the transmission lines ...................................................................... 13

Results obtained from the simulation .................................................................................................. 22

Discussion of results .............................................................................................................................. 28

Conclusion ............................................................................................................................................. 30

References ............................................................................................................................................ 32

IntroductionThe steady state operation of a transmission line is a very important state when it comes to transmission line

modelling. Results obtained from the steady state analysis are often ideally used in the analysis of

transmission line and apply when the line conditions are balanced and also during unbalance.

The ABCD parameters of a transmission line have a handful of importance and knowing them enables us to

predict and manipulate the performance of a transmission line. During design one often tends to look for the

most efficient power system ad yet also considers the costs to erect the power system. This all start with the

analysis and acquisition of the power systems ABCD parameters and hence continues with the further

manipulation of these parameters to obtain the required system

The are basically three types of transmission lines. These are the short transmission line model, the medium

transmission, the model and the ling transmission line model. The medium transmission line model is further

subdivided into the T lines and the PI lines. These are all explained in details below and are further analysed

in the course of the report.

Short transmission lines

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Figure1: the circuit equivalent for a short transmission line (ref1)

These are transmission lines that are usually less than 80KM long. In the analysis of these lines, the capacitance and the leakage resistance to the earth are neglected

and hence only the wire resistance and the inductive reactance are considered.

This is a simple series circuit and hence the current flowing through the circuit is simply the same. This is the easily analysed transmission line and it is not commonly used in power systems due to

the fact that transmission lines are usually aimed to go for a long distance as to reduce the

transmission costs.

Medium transmission lines (pi model)

Figure2: the circuit equivalent for a pi modelled transmission line (ref1)

Medium lines usually have a length that varies between 80KM to 250km In the model for this lines parameters, the total shunt capacitance is usually lumped together and is

located half at each end.

This model in analysed by assumed a delta connected circuit that has been discussed in the subjectelectrical engineering (EEG120S).

Medium transmission lines (T model)

Figure3: the circuit equivalent for T modelled transmission line (ref1)

This is another form of representing a medium transmission line.

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structure their program accordingly. The structure of the programs used is discussed in the sections

that follow.

7. The students than wrote their programs and simulated to obtain the results.8. The students did not just wait for the course lecturer to approve to approve whether their readings

were correct or not, they went as far as troubleshooting the program for possible mistakes. They

themselves did the calculations by hand and were hence able to compare the two results if they were

going together or if there was an error.

9. The format of the code used for short transmission lines was used as a standard for other codes andhence this enabled the students to come up with the codes for long and medium transmission lines

within a short time.10. Upon completion of the simulations, the student s than answered the questions that followed and

were hence able to write their project report on the findings of simulation and were able to

comment on the results

Derivation of the ABCD parametersShort lines

From the circuit in figure1:

By Kirchhoffs voltage law

VS=(R+JXL)IL+VL..1

Also for series circuit analysis

IS=IL+0VL..2

In vector matrix form, this are represented as

1

0 1

s L

S L

V VZ

I L

Hence these equations we can obtain the ABCD parameters as follows

s L

S L

V VA B

I LC D

From there we derive that

A=1.3

B=R+JXL4

C=05

D=1.6

For T-model (medium transmission lines)

From the circuit diagram in figure3:

By Kirchhoffs voltage law

1

2S S

ZV V I .7

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1

2L L

ZV V I 8

Also By Kirchhoffs current lawS C LI I I 9

But1C C

I V Y

2

LL CZIV Y

..........................10

Where by Ycis the shunt admittance

1S C L

I V Y I .11

Substituting in equation 8

2

S L L C L

ZI V I Y I

.13

2

CL C L L

ZYV Y I I

.14

12

CS L C L

ZYI V Y I

.15

Substituting the value of v1in equation 7,

2 2

S L L S

Z ZV V I I

..16

Then again substituting in Isof equation 9 in the above equation,

12 2 2

CS L L L C L

ZYZ ZV V I V Y I

17

2

12 2 4 2

C CL L

Y Z Z Y Z ZV I

18

2

12 4

C CS L L

Y Z Z Y V V Z I

..19

The matrix form for the equation of Vsand IS ,

2

12 4

1

2

C C

S L

S LCC

ZY Z YZ

V V

I IZYY

.20

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s L

S L

V VA B

I LC D

This is a standard matrix for transmission lines

Comparing the standard matrix and the T-method matrix of equation 20, we obtain that

A=12

CZY 21

B=

2

4

CZ YZ 22

C=C

Y and .23

D=12

CZY.24

For the conventional PI representation

From the circuit diagram in figure2 we obtain that

By Kirchhoffs current law

IS=IZ+IC2..25

But IC2 = VS2

CY

26

And IZ = IL+IC127

2

CL L

YI V

28

By Kirchhoffs voltage law; S L ZV V ZI 29

Substituting in forZ

I ,

2

CS L L L

YV V I V Z

..30

2

CS L L L

ZYV V I Z V

.31

12

CL L

ZYV I Z

32

Since IS=IZ+IC2,we can write the equation in terms of the line parameters Z and YC,therefore;

Substituting in the values of IZ and IC2in the above equations yields;

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2 2

C CS L L S

Y YI I V V

33

12 2 2

C C CL L L L

Y ZY Y I V V ZI

34

2

1

2 2 4 2

C C C C L L

Y Y ZY ZY V I

35

2

14 2

C CS C L L

ZY ZYI Y V I

...36

1 14 2

C CC L L

ZY ZYY V I

37

The matrix form for the equation of Vsand IS

2

12 4

12

C C

S L

S LCC

ZY Z YZ

V V

I IZYY

..38

Comparing the standard matrix used in both the T and the short line model and the pi matrix of equation 38

A=1

2

CZY..39

B=Z..40

C= 14

CC

ZYY

..41

D=12

CZY42

Long transmission lines

The long transmission line is a special type of transmission line. This transmission line is analysed below.

Inaccurate analysis of this transmission line requires that the line parameters be not limped as in other

analysis but should be uniformly distributed throughout the length of the line.

The analysis starts with the figure below.

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Figure 5: the one phase neutral connection of a three phase transmission line

The following are to be noted about the diagram above.

The incremental section dx is at a distance x from the receiving end. It has series impedance Zdx It has shunt admittance Ydx

Where Y and Z are the impedance and admittance per unit length.

Therefore

dVx=(Vx+dVx)-Vx.43

= ( Ix+dIx)Zdx..44

Where Ix>>dIx

dVxIxZdx..45In a similar manner, the incremental charging current is

dIxVxYdx46and hence

dVx/dx= IxZ47

also

dIx/dx= VxY..48

by differentiating equations 47 and 48, we get

d2Vx/dx2=z( dIx/dx)49

d2Ix/dx2=Y( dVx/dx)..50

Hence substituting 47 and 48 into 49 and 50 we obtain that

d2

Vx/dx2

= ZVxY..51

d2Ix/dx2 = YIxZ52

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At x=0, Vx=VLand Ix=IL

Hence the solutions to the differential equations 51 and 52 are

V(x) = ()VL+( )IL53Where V(x)=Vs54

Also

I(x) = ( ) VL+ ( )IL55Where x=l( the length of the line)

The new solutions to the two differential equations hence are

Where =..56

cZ zy.57

S Ccosh sinhL LV V l Z I l .58

S

sinh cosh

L

L

C

V lI I l

Z

.59

In vector matrix form

cosh sinh

1sinh cosh

C

S L

S L

C

l Z lV V

l lI IZ

..60

And hence comparing thi vector matrix representation with the standard matrix we obtain that

C

A cosh cosh

B Z sinh sinh

1C sinh =Y sinh

D cosh cosh

C

C

C

l ZY

l Z ZY

l ZYZ

l ZY

70

4. Program outline

The program will be outlined as below.

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1. The program must allow for its user to enter the values that are provided and will be the baselinefor the whole simulation.

2. The program will then be able to evaluate the impedance and admittance of the line.3. Since the ABCD parameters are represented using the impedances and admittances, they are than

evaluated.

4. By direct substitution, the paramters are than substituted into the two standard equations as shownbelow.

Vs=AVL+ BIL

IS= CVL+ DIL

5. Since the values of Vs and Is are now obtained, one is now able to evaluate the sending end power,the voltage regulation, the efficiency, the sending end power factor.

6. This part of the program should than be the part where the bases of for the system are chosen inorder to give all the circuit quantities in per unit.

7. The per unit quantities are thus evaluated and displayed.8. The program should then go on and allow the user to enter the power factor that they want their

system to be corrected to.

9. It will than calculate the value of the reactive power that needs to be supplied by the capacitor bankand also the per phase capacitance of the capacitor bank.

10.The program for the short line is than modified so that it accommodates a looping structure andalso be able to obtain the different values of the capacitance bank when the length of line is being

varied.

That would hence sum up the project and hopefully attain the main objectives of the experiment.

Codes used for the simulation of the transmission lines

Short transmission lines

%program for short linesclear;clc;

%continue with the programvl = input('enter line load voltage: ');f = input('enter local frequency: ');l = input('enter line constant in H/KM: ');length = input('enter length of line: ');r = input('enter the resistive constant: ');power = input('enter the load power: ');pf = input('enter load power factor: ');

%evaluate the parametersxl=l*length*pi*2*f;p=power/3;

res=r*length;z=complex(res,xl);A=1;B=z;C=0;D=1;vlo=vl/sqrt(3);%evaluate the load currentilmag=p/(vlo*pf);

%assuming a lagging power factoril=complex(ilmag*cos(acos(pf)),-ilmag*sin(acos(pf))); vload=complex(vlo,0);

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%evaluate the value of Vsending end and Isending end

vs=A*vload + B*il;is=C*vload + D*il;%calculte sending end powers=conj(is)*vs;%calculate voltage regulationvr=((abs(vs)-abs(vlo))/abs(vlo))*100;

%calculate efficiencypin=real(s);pout=p;n=(pout/pin)*100;%evaluating the power factortheta=atan(imag(s)/ real(s));pfactor=cos(theta);

fprintf ('VS=');disp(vs);

fprintf('IS=');disp(is);

fprintf('power factor at sending end=');disp(pfactor);

fprintf('sending end complex power=');disp(s);

fprintf('voltage regulation=');disp(vr);

fprintf('effiency=');disp(n);

%evaluate the per unit values of the line parameters%choosing the base values as followsvbase=vlo;

sbase=p/pf;

zbase=mpower(vbase,2)/sbase;

ibase=sbase/vbase;

zpu=z/zbase;

vreceivingpu=vload/vbase;

ireceivingpu=il/ibase;

vsendingpu=vs/vbase;

isendingpu=is/ibase;

ssendingpu=s/sbase;

precevingpu=p/sbase;

fprintf('the per unit value of the impedance=');disp(zpu);fprintf('the per unit value of the receiving end voltage=');

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disp(vreceivingpu);fprintf('the per unit value of the receiving end current=');disp(ireceivingpu);fprintf('the per unit value of the sending end voltage=');disp(vsendingpu);fprintf('the per unit value of the sending end current=');disp(isendingpu);fprintf('the per unit value of the sending end apparent power=');disp(ssendingpu);fprintf('the per unit value of the receiving end real power=');disp(precevingpu);

%for power factor correction

thetanew=acos(pfdesired);

qold=imag(s);

qnew=real(s)*tan(thetanew);%evaluating the value of the capacitor bank

qc=qold-qnew;pfdesired = input('enter the desired power factor: ');c=qc/(mpower(abs(vs),2)*2*pi*f);

fprintf('value of the capacitor bank in farads=');disp(c);fprintf('the reactive power to be supplied by the capacitane bank in

var=');disp(qc);

Medium transmission lines (pie)

%program for medium pie lines

clear;clc;

%start with the system inputsvl = input('enter load voltage: ');f = input('enter local frequency: ');l = input('enter line constant in H/KM: ');r = input('enter the resistive constant in OHM/KM: ');power = input('enter the load power in WATTS: ');pf = input('enter load power factor: ');m = input('enter your month of birth as a number: ');d = input('enter your date of birth: ');

%evaluate the capacitive constant

%specifically for the vaues assigned to myselfx=d;ca = x*(mpower(10,-1.05*m));

%evaluate the lengthlength=7*d;

%evaluate the parametersxl=l*length*pi*2*f;

p=power/3;

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xc=ca*length*2*pi*f;

res=r*length;vlo=vl/sqrt(3);z=complex(res,xl);

yc=complex(0,xc);squ=mpower(yc,2);A=(0.5*(z*yc))+1;B=z;

C=yc+(0.25*(squ*z));D=1+(0.5*(z*yc));

%evaluate the load current and load voltageilmag=p/(vlo*pf);il=complex(ilmag*cos(acos(pf)),-ilmag*sin(acos(pf))); vload=complex(vlo,0);


vs=A*vload + B*il;is=C*vload + D*il;%calculte sending end power

s=conj(is)*vs;%evaluating the power factortheta=atan(imag(s)/ real(s));pfactor=cos(theta);

%calculate voltage regulationvr=((abs(vs)-vlo)/vlo)*100;

%calculate efficiencypin=3*abs(vs)*abs(is)*pfactor;pout=3*abs(vload)*abs(il)*pf;n=(pout/pin)*100;

fprintf ('voltage at the sending end=');disp(vs);

fprintf('current at the sending end=');disp(is);




fprintf('effiency=');disp(n);%evaluate the per unit values of the line parameters%choosing the base values as followsvbase=vlo;

sbase=p/pf;


ibase=sbase/vbase;

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zpu=z/zbase;ycpu=yc/zbase;





ssendingpu=s/sbase;


fprintf('the per unit value of the impedance=');disp(zpu);fprintf('the per unit value of the receiving end voltage=');disp(vreceivingpu);fprintf('the per unit value of the receiving end current=');disp(ireceivingpu);fprintf('the per unit value of the sending end voltage=');disp(vsendingpu);fprintf('the per unit value of the sending end current=');disp(isendingpu);fprintf('the per unit value of the sending end apparent power=');disp(ssendingpu);fprintf('the per unit value of the receiving end real power=');disp(precevingpu);fprintf('the per unit value of the shunt admittance=');disp(ycpu);

%for power factor correctionpfdesired = input('enter the desired power factor: ');


qold=imag(s);


qc=qold-qnew;

capa=qc/(mpower(abs(vs),2)*2*pi*f);

fprintf('value of the capacitor bank in farads=');disp(capa);fprintf('the reactive power to be supplied by the capacitane bank in

var=');disp(qc);

Medium transmission (T model)%program for medium T linesclear;clc;

%start with the system inputsvl = input('enter load voltage: ');f = input('enter local frequency: ');l = input('enter line constant in H/KM: ');

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r = input('enter the resistive constant in OHM/KM: ');power = input('enter the load power in WATTS: ');pf = input('enter load power factor: ');m = input('enter your month of birth as a number: ');d = input('enter your date of birth: ');


%specifically for the vaues assigned to myselfx=d;ca = x*(mpower(10,-1.05*m));

%evaluate the lengthlength=7*d;

%evaluate the parametersxl=l*length*3.141592654*2*f;

p=power/3;


res=r*length;vlo=vl/sqrt(3);z=complex(res,xl);

yc=complex(0,xc);squ=mpower(z,2);A=(0.5*(z*yc))+1;B=z+(0.25*(squ*yc));C=yc;D=1+(0.5*(z*yc));

%evaluate the load current and load voltageilmag=p/(vlo*pf);

il=complex(ilmag*cos(acos(pf)),-ilmag*sin(acos(pf))); vload=complex(vlo,0);


vs=A*vload + B*il;is=C*vload + D*il;%calculte sending end powers=conj(is)*vs;%evaluating the power factortheta=atan(imag(s)/ real(s));pfactor=cos(theta);




fprintf('current at the sending end=');

disp(is);

fprintf('power factor at sending end=');

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disp(pfactor);



fprintf('effiency=');disp(n);

%evaluate the per unit values of the line parameters%choosing the base values as followsvbase=vlo;

sbase=p/pf;


ibase=sbase/vbase;

zpu=z/zbase;

ycpu=yc/zbase;





ssendingpu=s/sbase;


fprintf('the per unit value of the impedance=');disp(zpu);fprintf('the per unit value of the receiving end voltage=');disp(vreceivingpu);fprintf('the per unit value of the receiving end current=');disp(ireceivingpu);fprintf('the per unit value of the sending end voltage=');disp(vsendingpu);fprintf('the per unit value of the sending end current=');disp(isendingpu);fprintf('the per unit value of the sending end apparent power=');

disp(ssendingpu);fprintf('the per unit value of the receiving end real power=');disp(precevingpu);frintf('the per unit value of the shunt admittance=');disp(ycpu);

%for power factor correctionpfdesired = input('enter the desired power factor: ');


qold=imag(s);

qnew=real(s)*tan(thetanew);

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%evaluating the value of the capacitor bank

qc=qold-qnew;


fprintf('value of the capacitor bank in farads=');disp(capa);fprintf('the reactive power to be supplied by the capacitane bank in

var=');

disp(qc);

Long transmission lines

%program for long linesclear;clc;

%start with the system inputsvl = input('enter load line voltage: ');f = input('enter local frequency: ');l = input('enter line constant in H/KM: ');

r = input('enter the resistive constant in OHM/KM: ');power = input('enter the load power in WATTS: ');pf = input('enter load power factor: ');m = input('enter your month of birth as a number: ');d = input('enter your date of birth: ');


%specifically for the values assigned to myselfx=d;ca = x*(mpower(10,-0.93*m));

%evaluate the length of the line

length=18*d;

%evaluation of the parametersxl=l*length*pi*2*f;

p=power/3;


res=r*length;vlo=vl/sqrt(3);

z=complex(res,xl);y=complex(0,xc);

zc=sqrt(z/y);yc=sqrt(y/z);

Y=sqrt(y*z);

%the ABCD parameters hence are

A=cosh(Y);

B=zc*sinh(Y);C=yc*sinh(Y);D=cosh(Y);

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%evaluate the load current and load voltageilmag=p/(vlo*pf);il=complex(ilmag*cos(acos(pf)),-ilmag*sin(acos(pf))); vload=complex(vlo,0);


vs=A*vload + B*il;is=C*vload + D*il;

%calculte sending end powers=conj(is)*vs;%evaluating the power factortheta=atan(imag(s)/ real(s));pfactor=cos(theta);




fprintf('current at the sending end=');disp(is);





sbase=p/pf;


ibase=sbase/vbase;

zpu=z/zbase;ypu=y/zbase;





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ssendingpu=s/sbase;


fprintf('the per unit value of the impedance=');disp(zpu);fprintf('the per unit value of the receiving end voltage=');disp(vreceivingpu);fprintf('the per unit value of the receiving end current=');disp(ireceivingpu);

fprintf('the per unit value of the sending end voltage=');disp(vsendingpu);fprintf('the per unit value of the sending end current=');disp(isendingpu);fprintf('the per unit value of the sending end apparent power=');disp(ssendingpu);fprintf('the per unit value of the receiving end real power=');disp(precevingpu);fprintf('the per unit value of the shunt admittance=');disp(ypu);


pfdesired = input('enter the desired power factor: ');


qold=abs(imag(s));

qnew=abs(real(s))*tan(thetanew);%evaluating the value of the capacitor bank

qc=qold-qnew;


fprintf('value of the capacitor bank in farads=');disp(capa);fprintf('the reactive power to be supplied by the capacitance bank in

var=');disp(qc);

Results obtained from the simulation

Short lines

enter line load voltage: 38000

enter local frequency: 50

enter line constant in H/KM: 0.00113

enter length of line: 40

enter the resistive constant: 0.4

enter the load power: 20000

enter load power factor: 0.88

VS= 2.1947e+004 +1.6908e+000i

IS= 0.3039 - 0.1640i

power factor at sending end= 0.8800

sending end complex power= 6.6686e+003 +3.6000e+003i

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voltage regulation= 0.0328

effiency= 99.9714

the per unit value of the impedance= 2.5183e-004 +2.2350e-004i

the per unit value of the receiving end voltage= 1

the per unit value of the receiving end current= 0.8800 - 0.4750i

the per unit value of the sending end voltage= 1.0003 + 0.0001i

the per unit value of the sending end current= 0.8800 - 0.4750i

the per unit value of the sending end apparent power= 0.8803 + 0.4752i

the per unit value of the receiving end real power= 0.8800

enter the desired power factor: 1

value of the capacitor bank= 2.3791e-008

the reactive power to be supplied by the capacitane banki in var=

3.6000e+003

Medium (pie) transmission line

In this line, the values given by the course lecturer for the constants were inaccurate and do not fall in thenormal range for this type of transmission lines.The student thus used the following problem to test their simulation for both the pi and the wye connectedlines. (ref2)

From this problem, the students manipulated the values and obtained the values of the month of birth, thedate of birth, the capacitive constant and the inductive constant as well as the resistive constant.

The results from the simulation of this problem are thus given as follows

enter load voltage: 138000


enter line constant in H/KM: 0.0016902

enter the resistive constant in OHM/KM: 0.11287

enter the load power in VAs: 49000000


enter your month of birth as a number: 8.703

enter your date of birth: 25

voltage at the sending end= 8.9449e+004 +1.4846e+004i

current at the sending end= 1.6683e+002 -2.3433e+001i




effiency= 95.2555

the per unit value of the impedance= 0.0508 + 0.2391i



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the per unit value of the shunt admittance= 0 +2.5729e-006i

enter the desired power factor: 1value of the capacitor bank= 1.7704e-006

the reactive power to be supplied by the capacitance bank in var=

4.5728e+006

Medium (T) transmission linesThe same problem used for pi model was used for the T transmission lines so that the results can be comparable.

enter load voltage: 138000


enter line constant in H/KM: 0.0016902enter the resistive constant in OHM/KM: 0.11287




enter your date of birth: 25






effiency= 95.2273






the per unit value of the sending end apparent power= 0.8926 +

0.2670i





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the reactive power to be supplied by the capacitance bank in var=

4.3614e+006

Long transmission linesFor this transmission line, the values of the constants given were also very much inaccurate and did not

represent a real life situation of this type of transmission line.The students however took a more practical example and where able to evaluate and simulate for this

transmission line.

Further calculations to determine further constants led to the calculation of the birth month and the birthdate to be as followsd=8.33m=9.507enter load line voltage: 138000


enter line constant in H/KM: 0.0126enter the resistive constant in OHM/KM: 0.1858




enter your date of birth: 8.33






effiency= 93.8680











the reactive power to be supplied by the capacitane banki in var=

2.0353e+007

the code for short transmission lines was edited in order to obtain the value of the capacitance bank withvariable length of the transmission line. The code is as shown below.

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%program for short linesclear;clc;%continue with the programvl = input('enter line load voltage: ');f = input('enter local frequency: ');l = input('enter line constant in H/KM: ');length = input('enter length of line: ');r = input('enter the resistive constant: ');power = input('enter the load power: ');pf = input('enter load power factor: ');

pfdesired = input('enter the desired power factor: ');forlength=10:2:20;

%evaluate the parametersxl=l*length*pi*2*f;p=power/3;

res=r*length;z=complex(res,xl);A=1;B=z;

C=0;D=1;vlo=vl/sqrt(3);%evaluate the load currentilmag=p/(vlo*pf);%assuming a lagging power factoril=complex(ilmag*cos(acos(pf)),-ilmag*sin(acos(pf))); vload=complex(vlo,0);


vs=A*vload + B*il;is=C*vload + D*il;

%calculte sending end powers=conj(is)*vs;%calculate voltage regulationvr=((abs(vs)-abs(vlo))/abs(vlo))*100;

%calculate efficiencypin=real(s);pout=p;n=(pout/pin)*100;%evaluating the power factortheta=atan(imag(s)/ real(s));pfactor=cos(theta);

fprintf ('VS=');disp(vs);

fprintf('IS=');disp(is);




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sbase=p/pf;


ibase=sbase/vbase;

zpu=z/zbase;





ssendingpu=s/sbase;


fprintf('the per unit value of the impedance=');disp(zpu);fprintf('the per unit value of the receiving end voltage=');disp(vreceivingpu);fprintf('the per unit value of the receiving end current=');disp(ireceivingpu);fprintf('the per unit value of the sending end voltage=');disp(vsendingpu);fprintf('the per unit value of the sending end current=');disp(isendingpu);fprintf('the per unit value of the sending end apparent power=');disp(ssendingpu);fprintf('the per unit value of the receiving end real power=');disp(precevingpu);


pfdesired = input('enter the desired power factor: ');thetanew=acos(pfdesired);

qold=imag(s);


qc=qold-qnew;c=qc/(mpower(abs(vs),2)*2*pi*f);

fprintf('value of the capacitor bank=');disp(c);fprintf('the reactive power to be supplied by the capacitane banki in

var=');disp(qc);

end

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The plot of the variable length of the transmission line is plotted against the value of the capacitor bank asfollows.

In tabular formLength (KM) Capacitance bank value(VAR)10 2648712 2648814 2648816 2648918 2649020 26491

Discussion of results

In the simulations carried out above, the student has the following to clarify on.

Short transmission lines are very accurate for lengths that are within 80km. As the length and the voltage of the line increases, the short method becomes inaccurate as it doesnt

accommodate some of the parameters that come in the picture.

The results for short transmission lines were able to come out with the values allocated to thestudents.

In the results obtained in the short line analysis showed that For the lagging power factor of the load, the voltage at the receiving end is always greater

than the voltage at the sending end and hence ensuring a positive voltage regulation.

The power factor at the sending end is almost and very close to the power factor at thereceiving end.

The efficiency of these lines is very high and was obtained to be above 80%. The transmission line losses are very minimum and do not hence reduce the power by thatmuch.

When the per unit values are obtained chossing the base values to be those at the receivingend, the values at the sending end are usually greater than one because they are greaterthan the chosen base values.

The capacitor bank value increases almost with a straight line characteristic equation withthe increase in the length of the line.

Medium lines used the same values in the simulation for comparison purposes. These comparisonsare made below

By a small margin, the pi configuration has a higher efficiency than a T configuration. The voltage regulation for the T configuration is greater than that of the pi configuration by

small margin as well.

This simply implies that there is big gap between the sending end and receiving endvoltages of the T transmission line.

26486.5

26487

26487.5

26488

26488.5

2648926489.5

26490

26490.5

26491

26491.5

0 5 10 15 20 25

capacitor bank

value(VAR)

Line length(km)

line length vs capacitor bank value

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Large differences in these two voltages could mean that there are greater voltage drops inthe transmission line leading to less terminal voltage.

The power factor for the pi is smaller than that of the T model. This would imply thathowever, the pi model would need a much greater and bigger capacitor bank to correct thispower factor.

This is shown in the values of the capacitor values of the two models as they differed by alarge amount.

The sending end current of both the two models were moderately low as to reduce thetransmission line losses.

From the above comparisons, we can observe that each of these two models has its ownpros and cons

The pi model has less voltage drops in the line and have higher terminal voltage deliveringcapabilities.

The T model however is much more efficient and has a higher sending end power factormaking it more ideal in power system applications.

For long transmission lines, The efficiency was high enough and was found to be at least 93% This could imply that the system is efficient enough and comparing its length with its efficiency one

can say that it is best to use this line.

A longer distance will hence be covered and yet the power losses are at their very minimum. The current in this line is also very little and could mean that the losses are also kept at their

minimum. The power factor was less than the receiving end power factor. This could imply that more reactivepower is supplied and doesnt necessarily reach the load.

The line itself is having capacitive components and one can say that this line has possibilities ofcorrecting its own power factor.

The power factor at the sending end is also lagging just as that of the receiving end. This line model was also found to have a very high voltage regulation than all the other two lines.

This could imply that there is a big difference between the terminal voltage and the sending endvoltage

This is mostly due to the voltage drops in the line causing the terminal voltage to be low.

Comparison between calculated and simulated values

Short modelQuantity Simulated values Calculated valuesEfficiency 99.9714 99.9Sending end voltage 2.1947e+004 +1.6908e+000i 21946.5+j1.687Sending end current 0.3039 - 0.1640i 0.3426-j0.0433

Voltage regulation 0.0328 0.0328%Sending end power 6.6686e+003 +3.6000e+003i -7.5141e+003-j -982.8812Power factor at the sending end 0.8800 0.879Capacitance values 1.7704e-006F 23.84F

Pi model

quantity Simulated CalculatedSending end voltage 8.9449e+004 +1.4846e+004i 91940.2+j17352.8Sending end current 1.6683e+002 -2.3433e+001i 196.53-j39.51Sending end power factor 0.9541 0.927Efficiency 95.2555 94.16

T modelquantity Simulated CalculatedSending end voltage 8.9067e+004 +1.4586e+004i 91486+ j170486

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Sending end current 1.6722e+002 -2.1583e+001i 196.95-j39.5Sending end power factor 0.9580 0.929Efficiency 95.2273 94.27

Long model

quantity Simulated CalculatedSending end voltage 1.3770e+005 +1.1480e+005i 99470

Sending end current 1.3736e+002 -3.3293e+001i Sending end power factor 0.8556 0.8653Efficiency 93.8680 93.1

Voltage regulation 25.0150 24.1

From the above comparisons of results, it can well be obtained that there is a slight differencebetween the calculated and the simulated values.

This could be an indication that the simulated results are of much greater accuracy than thecalculated values.

With matlab being a very vital rule, it is hence proven that it is a platform that can give veryaccurate results without errors.

The causes of the errors during manual calculations could be human errors, technical errors such ascalibration errors,parallax errors or even errors due to electromagnetic inteferences.

Conclusion

The simulation enhanced not only the students ability to work with a matlab environment but alsohelped them to be able to visualise the performance of different types of transmission lines.

Critical thinking was alos one of the aspects being tested in this project and the students were ableto bring up a successful simulation within a limited amount of time.

The results of the project further emancipated the fact that, computer simulations are much moreaccurate than manual means of calculations as they do not present any errors unless there are input

errors in the first place.

Furthermore, the students understood the fact the short transmission line is the fundamental of allthe other transmission lines and is only variable and tends to increase in complexity when the length

increases.

The long transmission lines are the most preffered and the short are more of an ideal type of line asthey rarely occur in the electrical industry.

Per unit analysis was used in the lines and proved to use very small values of the quantities. Thiswould imply its goodness in transmission line modelling and hence would prevent the program from

crashing due to large values then the memory can accommodate.

The capacitance bank value was found to increase as the length of the line increases and this meansthat the power factor becomes poorer as the length of the line increases.

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Overally, analysis of all the types of transmission lines is equally important as all the tpes of lines occur in

the electrical industry. A thorough understanding of both the lines would do an electrical engineer a lot

of good.

To put it in a nutshell, this lab simulation was an ultimate success with both the lecturer and students

learning a great deal of information. The Namibian industry would feel proud to have very well equipped

engineers upon our graduation.

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References1. Lab sheet, power system engineering simulation lab, revised 20132. Introduction to power engineering research project; Dr Houssem Rafik El-Hana Bouchekara;

department of electrical engineering; umm Al-cura university; Saudi Arabia

3. Class notes, mr Sendegeya,Polytechnic of Namibia,revised 20134. http://www.ece.ualberta.ca/knight/electrical_machines/fundamentals/f_ac_3p.html5. http://www.epuniversity.org/w3.html6. www.insightmaker.com/7. http://openbookproject.net/electricCircuits/AC/AC_10.html

http://en.wikipedia.org/wiki/Symmetrical_components8. http://en.wikipedia.org/wiki/Transmission_line9. http://en.wikipedia.org/wiki/Transmission_line10. http://www.ece.uci.edu/docs/hspice/hspice_2001_2-26.html11. http://www.ece.uci.edu/docs/hspice/hspice_2001_2-26.html12. http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Characteristic_impedance.html13. http://www.mathworks.com/help/rfblks/transmissionline.html14.http://na.tm.agilent.com/plts/help/WebHelp/Analyzing/Analyzing_Transmission_Line_Parameters.html
http://www.insightmaker.com/http://openbookproject.net/electricCircuits/AC/AC_10.htmlhttp://en.wikipedia.org/wiki/Symmetrical_componentshttp://en.wikipedia.org/wiki/Transmission_linehttp://en.wikipedia.org/wiki/Transmission_linehttp://en.wikipedia.org/wiki/Transmission_linehttp://en.wikipedia.org/wiki/Transmission_linehttp://www.ece.uci.edu/docs/hspice/hspice_2001_2-26.htmlhttp://www.ece.uci.edu/docs/hspice/hspice_2001_2-26.htmlhttp://www.ece.uci.edu/docs/hspice/hspice_2001_2-26.htmlhttp://www.ece.uci.edu/docs/hspice/hspice_2001_2-26.htmlhttp://www.princeton.edu/~achaney/tmve/wiki100k/docs/Characteristic_impedance.htmlhttp://www.princeton.edu/~achaney/tmve/wiki100k/docs/Characteristic_impedance.htmlhttp://www.mathworks.com/help/rfblks/transmissionline.htmlhttp://www.mathworks.com/help/rfblks/transmissionline.htmlhttp://na.tm.agilent.com/plts/help/WebHelp/Analyzing/Analyzing_Transmission_Line_Parameters.htmlhttp://na.tm.agilent.com/plts/help/WebHelp/Analyzing/Analyzing_Transmission_Line_Parameters.htmlhttp://na.tm.agilent.com/plts/help/WebHelp/Analyzing/Analyzing_Transmission_Line_Parameters.htmlhttp://na.tm.agilent.com/plts/help/WebHelp/Analyzing/Analyzing_Transmission_Line_Parameters.htmlhttp://www.mathworks.com/help/rfblks/transmissionline.htmlhttp://www.princeton.edu/~achaney/tmve/wiki100k/docs/Characteristic_impedance.htmlhttp://www.ece.uci.edu/docs/hspice/hspice_2001_2-26.htmlhttp://www.ece.uci.edu/docs/hspice/hspice_2001_2-26.htmlhttp://en.wikipedia.org/wiki/Transmission_linehttp://en.wikipedia.org/wiki/Transmission_linehttp://en.wikipedia.org/wiki/Symmetrical_componentshttp://openbookproject.net/electricCircuits/AC/AC_10.htmlhttp://www.insightmaker.com/

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