transmission line repaired)

8/8/2019 Transmission Line Repaired)

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Transmission Line-2

OBSERVATION SHEET

Name : N.H.M. PrabhathIndex No : 080368FDepartment : Electrical EngineeringSignature :

Sub Date : 28th

October 2010


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CALCULATON

Properties of the given Transmission Line

Name : N.H.M. PrabhathIndex No : 080368FDepartment : Electrical EngineeringPer: Date : 8th oct 2010Instructed by:


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Transmission line model

R = 0.316 /kmL = 2.0 mH/kmC = 0.26 F/kmG = 0 (negligible)

Total resistance of the transmission line R

TR

= 75 km 0.316/km

=

23.7

Total induction of the transmission line L

LT = 75 km 2.0mH/km= 150 mH

Total capacitance of the transmission line C

CT = 75km 0.256F/km= 19.2 F

Theoretical calculations of transmission line parameters

Lets assume that the transmission line has an equivalent

model.

Ya Yb

Yc

V1 V2

I1 I2


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Ya & Yb consists with Capacitor

Yc consists with resistor as wellas inductor

So using well known method,

Zc = (23.7 + (250)0.15 j) where = 1S=( 23.7 + 47.12 j)

Yc = 1/ ZC= 1/ (23.7 + j 47.12)= (0.00852-0.017j) S

Za = Zb = 1/ ((250)(19.210-6)(1/2) j)

= (-331.572 j)

Ya = Yb = 1/ (-j 331.572)= (0.003016 j) S

c

cb

Y

YYA

+=

= 0.003016 j+(0.00852-0.017j)0.00852-0.017j

= 0.86 4.733

cY

1B =

B= 10.00852-0.017j

B=52.588 63.381

c

accbba

Y

YYYYYYC

++=

Since Ya=Yb


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C=Ya2+(2YaYc)YcType equation here.

C=(0.003016 j)2+(2(0.003016 j)(0.00852-0.017j))(0.00852-0.017j)

C=0.00561 92.19

And also due to the symmetry of the system,A = D

D=0.86 4.733

To check the circuit is a passive one.

A.D-B.C=(0.86 4.733)(0.86 4.733)-(52.588 63.381)(0.00561 92.19 ) =0.99981

Therefore the given circuit is a passive circuit.

Theoretical values of the parameters,A = 0.86 4.733B = 52.588 63.381C = 0.00561 92.19D = 0.86 4.733

Transmission line parameters calculated practically

From open circuit test

VICosP =

21 =601.9CosCos =21601.9

=79.38

0IS

SO/C

RI

VZ

=

=

= 601000 -79.38 1.9200

= 157.89 -79.38

From short circuit test

VICosP =

8=151Cos

Cos =8151


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Cos =0.5333333

=57.77

I

VZ

0VS

S

S/C

R=

=

= 151000 -57.771200

= 75 -57.77

Substituting values for

O/CZ

and

S/CZ

==

=

735.6849.0

47.137213.0

77.577538.7989.157

38.7989.157

A

A

A

Also A = D due to the symmetry of the system,

== 735.6849.0DA

From short circuit test

D

B

I

VZ

0VS

SS/C

R

===

=

=

515.64675.63

78.5775735.6849.0

B

B

From open circuit test

C

A

I

VZ

0IS

SO/C

R

===

( )

cZscZo

cZoA

cZscZo

cZoAA

DA

cZocZscZo

DcZoAcZscZo

DCCBDAcZscZo

D

B

C

AcZscZo

//

/

//

/

///

/

1//

//

//

=

=

=

=

=

=

S/CZDB =


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=

=

=

115.860054.0

38.7989.157

735.6849.0

/

C

C

cZo

AC

To check the circuit is a passive one.

[ ]

1

0006.1

)115.860054.0()515.64675.63()735.6849.0()735.6849.0(

=

=

CBDA

CBDA

CBDA

Therefore the given circuit is passive one.

Practical values of the parameters,A =

)735.6849.0(

B = 515.64675.63

C =

115.860054.0

D =

)735.6849.0(

Compare Practical A B C D parameters with theoretical values

Parameter Theoretically Practically

A 0.86 4.733 735.6849.0

B 52.588 63.381 515.64675.63

C 0.00561 92.19 115.860054.0

D 0.86 4.733

735.6849.0

Receiving end circle diagram of a Resistive Load


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( )

( ) ;1

1

+=

B

VrEquarion

BIrAVrVs

( ) ( ) ( )

( ) ( ) ( )

+=

+=

IrVrB

VrVrA

B

VsVr

VrIrB

AVrVr

B

VrVs

Now,

==

==

VsVs

Vr

B

A

075

515.64675.63

735.6849.0

Vr Vs IrB

VsVr

IrVrB

VrVrA

Pr

=

IrVrCos

Pr1

75V 60V 0.15A 70.67 11.25 75.00 11W 12.10

75V 67V 0.24A 78.92 18.00 75.00 18W 0.00

75V 68V 0.28A 80.09 21.00 75.00 21W 0.00

75V 69V 0.38A 81.27 28.5 75.00 28W 10.74

Re ceiving endcircle diagram forR esistive load using

AutoCAD

Pr

Vs

from the

practical

Vs

from the graph

11W 60V 69.75V

18W 67V 72.01V

21W 68V 73.7V28W 69V 80.67V


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Receiving end circle diagram of a Inductive Load

Vr Vs IrB

VsVr

IrVrB

VrVrA

Pr

=

IrVrCos

Pr1

75V 72V 0.6A 84.80 45 75 15W 70.5

75V 73V 0.7A 85.98 52.5 75 16.7W 71.5

75V 74V 0.8A 87.16 60 75 17W 73.5

75V 75V 0.8A 88.34 60 75 17W 73.5

75V 76V 0.9A 89.52 67.5 75 19W 73.6

75V 76.5V 1.0A 90.11 75 75 19W 75.3

75V 78V 1.2A 91.87 90 75 21W 76.5

75V 79V 1.2A 93.05 90 75 20W 77.2

75V 80V 1.3A 94.23 97.5 75 24W 75.7

Pr

Vs

from the

Vs

from the graph


10/13

practical

15W 72V -

16.7W 73V -

17W 74V -

17W 75V -

19W 76V -

19W 76.5V -

21W 78V -

20W 79V -24W 80V -

Receiving end circle diagram for Inductive load using AutoCAD


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Variation of Vs with Pr for the resistive load

Vs with Qr for inductive resistive loads

DISCUSSION:

1) Reasons for variations of theoretical values with practical values.

There is a difference between observed values and the theoretical values. Some of thereasons for those deviations as follow,

1) Human errors can be caused during the practical or during the calculations. That is either

due to the invalid calculations or due to the invalid readings. Therefore we must takereadings during the practical with high accuracy and must do the calculations correctly.

2) For easier calculations we change the appearance of the transmission line in to a

-

model. So there will be some significant errors due to this lumping of the transmission line.3) We did the practical using ammeters, voltmeters and wattmeter. But sometimes these

instruments are not ideal or not sensitive enough to take measurements. We did open circuittest and short circuit test to calculate the transmission line parameters practically. Due tothe errors in the instruments, the calculated values of

S/CZ

and

O/CZ

can have some errors.

And also we used those two terms to calculate A, B, C, D parameters. Calculating theparameters with values which are having errors may increase the value of the error.

4) There were small resistances in the wires which are used for this practical. Sometimes inthe open circuit test the receiving end may not be open circuited well by voltmeter. And inthe short circuit test the receiving end may not be short circuited by the ammeter due tosmall amount of voltage drop across the ammeter. Due to those reasons there is a different

between the transmission line parameters calculated theoretically and practically.5) We must do these kind of practical as much as quickly. Because if we consumed

considerable time to take readings, resistance values might be changed due to the increasein the temperature as the current passing through it.

6) And also we assumed the frequency is 50 Hz. But it can be vary between the (50

1%) Hz.

so this also introduced some errors.

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Reasons for not giving identical diagram for the theoretical and observed.For the receiving end,

1) Due to the variations of the values of A, B, C, D parameters. Because we plot the circlediagrams using these A, B, C, D parameters.

2) Resistances at the connections and wires are neglected and also we use a 1.5 resistor asthe average resistance. For the high accuracy we must consider the resistances of wires andconnecting points.

3) Some of the values we calculate approximately to the first decimal point. There foreaccuracy is reduced when we are plotting the diagrams.

Importance of Circle Diagrams

There are three types of power circle diagrams. Their names are Receiving end circlediagram, Sending end circle diagram and Universal circle diagram. Importance of these diagramsas follow,

1) Power circle diagrams are used to analyze what happen to the power delivered withvariation of various parameters of the transmission line.

2) To determine active and reactive power at the receiving end for any load angle from thereceiving end circle diagram.

3) We can get the power at either receiving end or sending end for any given values oftransmission line parameters and voltages and currents at the sending and the receiving end.

4) The power flow at any point along transmission line can always be found if the voltage,current and power factor are known or can be calculated.

4) Because of the variations of A B C D parameter values we plot the circle diagrams.5) Resistances at the connections and wires are neglected and also we use a 1.5 resistor as

the average resistance. For the high accuracy we must consider the resistances of wires andconnecting points.

6) When we calculate the values, we got the approximately decimal point around to first orsecond. The exactness of the plotting diagrams are reduced due to differ in small decimal

point.

Importance of the usage of Circle Diagrams.

To obtain the system performance directly we could draw the power circle diagrams withthe aid of A B C D parameters. So there are three forms of the power circle diagrams.

Receiving end circle diagram Sending end circle diagram Universal circle diagram

3)

Although power flow at any point along a transmission line can always be found if the voltage, current

and power factor are known or can be calculated, very interesting equation for power can be derived in

terms of ABCD constants. The equation applies to any network of two ports or two terminal pairs. As

shown in below we can draw a power circle diagram

by using the equation actually this type of circle diagram

is quite important since we can derive how much power

12


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is delivered by transmission line. And also we can see

the behavior when the sending end voltages varying

and how the power factor varying according to that. and

also we can derive how we can increase the power factor

and the maximum power that can be carried by the line.

We can keep some variables constant as we wish and

vary others, then we can see how the all components are

behaving.

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transmission line repaired)

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