TRANSMISSION LINE-2
Name
: D.K.Pathirana
Index No : 080332P
Field
: EE
Group
: 09
Date of sub: 2010.12.06
OBSERVATION SHEET
Name :
Index no :
Group :
Date of per :
Instructed By :
D.K.Pathirana
080332P
EE09
06-12-2010
1) Open circuit test
Vs = 63V
Is = 2.55A
Ps = 33W
2) Short circuit test
Vs = 15V
Is = 0.9A
Ps = 8W
3) Load tests
Vr = 76V
4) Resistive load
Vs (V)
Ir (A)
Pr (W)
64
66
67
68
0.15
0.32
0.4
0.52
12
24
31
40
69
0.64
49
Vs (V)
Ir (A)
Pr (W)
70
71
71
72
73
75
0.64
0.67
0.7
0.76
0.86
1.0
15
16
16
17
17.5
19
5) Inductive load
CALCULATON
Properties of the given Transmission Line
Transmission line model
R = 0.316 Ω/km
L = 2.0 mH/km
C = 0.26 µF/km
G = 0 (negligible)
Total resistance of the transmission line R
R T = 0.316  75
= 23.7 
Total induction of the transmission line L
LT = 2.0×75 mH
= 150 mH
Total capacitance of the transmission line C
CT = 0.26×75 µF
= 19.5 µF
Theoretical calculations of transmission line parameters
Let’s assume that the transmission line has an equivalent
 model.
I2
I1
Yc
V1
Ya
Yb
V2
Zc = 23.7 + j (2π×50)×0.15
= 23.7 + j 47.124 Ω
= 52.75∠63.30 Ω
Yc = 1/ ZC
= 1/ (23.7 + j 47.124)
= (0.00852 - j 0.0169) S
Za = Zb = 1/ j (2π×50)×(19.5×10-6)×(1/2)
= (-j 331.57) Ω
=331.57∠ − 90Ω
Ya = Yb = 1/ (-j 331.565)
= j 0.003016 S
A
Yb  Yc
Yc
A   j 0.003016  0.00852  j 0.0169  0.00852  j 0.0169
A  0.85787  j 0.07148
A  0.8604.76
1
Yc
B  1  0.00852  j 0.0169
B
B  52.7563.30
C
Ya Yb  Yb Yc  Yc Ya
Yc
C  Ya  Ya   Ya  Ya   Yc   Yc
C  Ya.  Ya  Yc   2  Ya 
But , Ya  Yb  j 0.003016,
C  0.003016 j   0.003016 j   0.00852  j 0.0169  2  0.003016 j 
C  0.0056  0.0002156 j
C  0.005692.20
And also due to the symmetry of the system,
A=D
 D  0.8604.76
To check the circuit is a passive one.
A  D  B  C  0.8614.763  0.8614.763  52.74863.3  0.0056192.20
A  D  B  C  1.00037
 A D  B C 1
There fore the given circuit is a passive circuit.
Theoretical values of the parameters,
A = 0.8604.76
B = 52.7563.30
C = 0.005692.20
D = 0.8604.76
Transmission line parameters calculated practically
From open circuit test
Z O/C 
VS
IS
I R 0
63  1000
  90
2.55  200
 123.53  90

From short circuit test
P  VICos
8  15  0.9  Cos
  53.658
ZS/C 
VS
IS
VR 0
15  1000
53.658
0.9  200
 83.3353.658

A B

C D
A D  B C
Zo / c  Zs / c 
CD
1
Zo / c  Zs / c 
 A  Zo / c   D
Zo / c
Zo / c  Zs / c 
A D
Zo / c
A A 
Zo / c  Zs / c
Zo / c
A
Zo / c  Zs / c
Zo / c  Zs / c 
Substituting values for ZO/C and ZS/C
A
123.53  90
123.53  90  83.3353.658
A  0.627214.52
A  0.7937.26
Also A = D due to the symmetry of the system,
A  D  0.7937.26
From short circuit test
V
B
ZS/C  S

I S V 0 D
R
B  D  ZS/C
B  0.7937.26  83.3353.658
B  66.0860.92
From open circuit test
V
A
Z O/C  S

I S I 0 C
R
A
Zo / c
0.7937.26
C
123.53  90
C  0.006497.26
C
Practical values of the parameters,
A = 0.7937.26
B = 66.0860.92
C = 0.006497.26
D = 0.7937.26
Compare Practical A B C D parameters with theoretical values
Parameter
Theoretically
Practically
A
0.8604.76
0.7937.26
B
52.7563.30
66.0860.92
C
0.005692.20
0.006497.26
D
0.8604.76
0.7937.26
|Vs ||Vr |∠(θ − b) |A||Vr2 |∠(a − b)
=
+ |Vr ||Ir |∠ − φ
|B|
|B|
|A||Vr2 |∠(a − b) 0.793(762 )∠(7.26 − 60.92)
=
= 69.315∠−53.66𝑜
|B|
66.08
Pr = Vr Ir cos φ
For resistive load φ can be obtain only for second set of value
φ = cos−1 [
Pr
24
] = cos −1 [
] = 9.30o
Vr Ir
76 × 0.32
|Vr ||Ir | = 76 × 0.32 = 24.32
From the receiving end circle diagram,
|Vs ||Vr |
= 88
|B|
|Vs | =
88 × 66.08
= 76.51V
76
For inductive load
Vr =76V
Pr (W)
Ir (V)
Φ
|Vr ||Ir |
15
0.64
72.03
48.64
16
0.67
71.67
50.92
16
0.7
72.49
53.2
17
0.76
72.88
57.76
17.5
0.86
74.46
65.36
19
1.0
75.52
76
Pr (W)
|Vs ||Vr |
|B|
Vs (V) from
graph
Vs from
practical
15
116
100.78
70
16
118
102.52
71
16
120
104.25
71
17
129
112.07
72
17.5
132
114.68
73
19
142
123.37
75
130
Values from graph
Values from practical
120
110
Vs (V)
100
90
80
70
60
13
14
15
16
17
Pr (W)
18
19
20
DISCUSSION:
Reasons for variations of theoretical values with practical values.
During the practical, human errors can be happened. That is either due to the invalid
calculations or due to the invalid readings. There fore we must take readings during the practical with
high accuracy and must do the calculations correctly. 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. This practical is done by using ammeters, voltmeters and wattmeter. But
sometimes these instruments are not ideal or not sensitive enough to take measurements. Open circuit
test and short circuit test are also done to calculate the transmission line parameters practically. Due to
the errors in the instruments, the calculated values of ZS/C and ZO/C can have some errors. And also A,
B, C, D parameters were calculated using these two. Calculating the parameters with values which are
having errors may increase the value of the error. There were small resistances in the wires which are
used for this practical. Sometimes in the open circuit test the receiving end may not be open circuited
well by voltmeter. And in the short circuit test the receiving end may not be short circuited by the
ammeter due to small amount of voltage drop across the ammeter. Due to those reasons there is a
different between the transmission line parameters calculated theoretically and practically. 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 increase in the temperature as the current passing
through it. 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.

Reasons for not giving identical diagram for the theoretical and observed.
For the receiving end, due to the variations of the values of A, B, C, D parameters,
it is not given identical diagram for the theoretical and observed. Because we plot the circle diagrams
using these A, B, C, D parameters. Resistances at the connections and wires are neglected and also we
use a 1.5 Ω resistor as the average resistance. For the high accuracy we must consider the resistances
of wires and connecting points. Some of the values we calculate approximately to the first decimal
point. There fore accuracy is reduced when we are plotting the diagrams.
Importance of Circle Diagrams
Circle diagrams can be categorized into three types. They are,
1) Receiving end circle diagram
2) Sending end circle diagram
3) Universal circle diagram
Power circle diagrams are used to analyze what happen to the power delivered with variation of
various parameters of the transmission line. To determine active and reactive power at the receiving
end for any load angle from the receiving end circle diagram. We can get the power at either receiving
end or sending end for any given values of transmission line parameters and voltages and currents at
the sending and the receiving end. 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.
REFERENCE: Principles of power system by V.K Mehta & Rohit Mehta.