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Electrical Power Systems
6
Characteristics and Performance of
Transmission Lines
6.1 INTRODUCTION
This chapter deals with the characteristics and performance of transmission lines. It is convenient
to represent a transmission line by the two-port network, wherein the sending-end voltage VS
and current IS are related to the receiving-end voltage VR and current IR through A, B, C and
D parameters as
...(6.1)
VS = A VR + B IR Volts
IS = CVR + DIR Amp
or, in matrix form,
LMVI OP = LMCA DBOP LMVI OP
N Q N QN Q
S
R
S
R
...(6.2)
...(6.3)
A, B, C and D are the parameters that depend on the transmission-line constants R, L, C
and G. The ABCD parameters are, in general, complex numbers. A and D are dimensionless. B
has units of ohms and C has units of siemens.
Also the following identify holds for ABCD constants.
AD – BC = 1
...(6.4)
To avoid confusion between total series impedance and series impedance per unit length,
the following notation is used.
z = g + j wL W/m, series impedance per unit length
y = G + j wC S/m, shunt admittance per unit length
Z = zl W, total series impedance
Y = yl S, total shunt admittance
l = line length, m.
Note that the shunt conductance G is usually neglected for overhead transmission system.
6.2 SHORT TRANSMISSION LINE
Capacitance may be ignored without much error if the lines are less than 80 km long or if the
voltage is not over 66 kV. The short line model on a per-phase basis is shown in Fig. 6.1.
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This is a simple series circuit. The relationship
between sending-end, receiving-end voltages and
currents can be written as:
LMVI OP = LM01 Z1 OP LMVI OP
N Q N QN Q
S
R
S
R
...(6.4)
The phasor diagram for the short-line is shown in
Fig. 6.2 for lagging load current.
From Fig. 6.2, we can write
Fig. 6.1: Short line model.
|VS| cos (dS – dR) = |I|R cos dR + |I|
X sin dR + |VR|
(dS – dR) is very small,
\
...(6.5)
\ cos (dS – dR) » 1.0
|VS| = |VR| + |I| (R cos dR
+ X sin dR)
...(6.6)
Equation (6.6) is quite accurate for the normal range
of load.
Fig. 6.2: Phasor diagram.
6.3 VOLTAGE REGULATION
Voltage regulation of the transmission line may be defined as the percentage change in voltage
at the receiving end of the line (expressed as percentage of full-load voltage) in going from noload to full-load.
Percent voltage regulation =
where
|VRNL|-|VRFL|
|VRFL|
´ 100
...(6.7)
|VRNL| = magnitude of no-load receiving end voltage
|VRFL| = magnitude of full-load receiving end voltage
At no load, IR = 0, VR = VRNL and from eqn. (6.3),
VRNL =
VS
A
...(6.8)
Using eqns. (6.7) and (6.8), we get,
Percentage Voltage regulation =
|VS| -|A||VRFL|
|A||VRFL|
´ 100
...(6.9)
For a short line, |A| = 1.0, |VRFL| = |VR|
\ Percent voltage regulation =
|VS|-|VR|
´ 100
|VR|
...(6.10)
Using eqns. (6.10) and (6.6), we get,
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Percent voltage regulation =
|I |( R cos d R + X sin d R )
´ 100
|V R|
...(6.11)
In the above derivation, dR has been considered positive for a lagging load. dR will be
negative, for leading load. Therefore, for leading power factor load,
Percent voltage regulation =
|I|( R cos d R - X sin d R )
|VR|
...(6.12)
From the above equations, it is clear that the voltage regulation is a measure of line voltage
drop and depends on the load power factor.
6.4 MEDIUM TRANSMISSION LINE
For the lines more than 80 km long and below 250 km in length are treated as medium length
lines, and the line charging current becomes appreciable and the shunt capacitance must be
considered. For medium length lines, half of the shunt capacitance may be considered to be
lumped at each end of the line. This is referred to as the nominal p model as shown in Fig. 6.3.
The sending end voltage and current for the nominal p model are obtained as follows:
Fig. 6.3: Medium length line, nominal p representation.
From KCL, the current in the series impedance designated by IL is
Y
VR
2
From KVL, the sending end voltage is
IL = IR +
...(6.13)
VS = VR + ZIL
...(6.14)
From eqns. (6.14) and (6.13), we get,
FG
H
VS = 1 +
IJ
K
ZY
V R + ZIR
2
...(6.15)
The sending end current is,
Y
VS
2
From eqns. (6.16), (6.15) and (6.13), we get,
IS = IL +
FG
H
IS = Y 1 +
IJ FG
K H
...(6.16)
IJ
K
ZY
ZY
VR + 1 +
IR
4
2
...(6.17)
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Eqns (6.15) and (6.17) can be written in matrix form.
LMFG1 + ZY IJ Z OP LV O
LMVI OP = MMH 2ZYK ZY PP MM PP
N Q MY FGH1 + IJK FGH1 + IJK P MMN I PPQ
2 PQ
MN 4
R
S
S
...(6.18)
R
Therefore, the ABCD constants for the nominal p model are given by
FG
H
A= 1+
FG
H
IJ
K
ZY
, B = Z,
2
C=Y 1+
IJ
K
FG
H
ZY
ZY
, D= 1+
2
4
IJ
K
6.5 LONG TRANSMISSION LINE
For short and medium length lines, accurate models were obtained by assuming the line
parameters to be lumped. In case the lines are more than 250 km long, for accurate solutions
the parameters must be taken as distributed uniformly along the length as a result of which the
voltages and currents will vary from point to point on the line. In this section, expressions for
voltage and current at any point on the line are derived. Then, based on these equations, an
equivalent p model is obtained for long transmission line. Figure 6.4 shows one phase of a
distributed line of length l km.
Fig. 6.4: Schematic diagram of a long transmission line with
distributed parameters.
From KVL, we can write,
V(x + Dx) = z.Dx.I(x) + V(x)
\
As
V ( x + Dx) - V ( x)
= z.I(x)
Dx
...(6.19)
Dx ® 0
dV ( x)
= z.I(x)
dx
...(6.20)
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From KCL, we can write,
I(x + Dx) = I(x) + y. Dx .V(x + Dx)
\
As
I ( x + Dx) - I ( x)
= y.V(x + Dx)
Dx
...(6.21)
Dx ® 0
dI ( x)
= y.V(x)
dx
...(6.22)
Differentiating eqn. (6.20) and substituting from eqn. (6.22), we get,
d 2 V ( x)
dx
\
2
d 2 V ( x)
dx 2
dI ( x)
= z. yV(x)
dx
– zy V(x) = 0
g2 = zy
Let
Therefore,
= z×
d 2 V ( x)
dx 2
...(6.23)
...(6.24)
– g2V(x) = 0
...(6.25)
The solution of the above equation is
V(x) = C1 eg x + C2 e–g x
...(6.26)
where, g, known as the propagation constant and is given by,
g = a + jb =
zy
...(6.27)
The real part a is known as the attenuation constant, and the imaginary part b is known as
the phase constant. b is measured in radian per unit length.
From eqn. (6.20), the current is,
I(x) =
1 dV ( x)
×
Z
dx
\
I(x) =
g
(C1 egx – C2 e–gx)
z
\
I(x) =
\
I(x) =
y
(C1 egx – C2 e–gx)
z
1
(C egx – C2 e–gx)
ZC 1
...(6.28)
where, ZC is known as the characteristic impedance, given by
ZC =
z
y
...(6.29)
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Now note that, when x = 0, V(x) = VR and from eqn. (6.26), we get
VR = C1 + C2
...(6.30)
also when x = 0, I(x) = IR and from eqn. (6.28), we get,
IR =
1
(C1 – C2)
ZC
...(6.31)
Solving eqns. (6.31) and (6.32), we obtain,
C1 =
VR + ZC IR
2
...(6.32)
C2 =
(VR - ZC IR )
2
...(6.33)
Substituting the values of C1 and C2 from eqns. (6.32) and (6.33) into eqns. (6.26) and (6.28),
we get
V(x) =
(VR + Z C I R ) gx (V R - Z C I R ) - gx
e +
e
2
2
...(6.34)
I(x) =
(VR + Z C I R ) g x (V R - ZC I R ) - g x
e e
2 ZC
2 ZC
...(6.35)
The equations for voltage and currents can be rearranged as follows:
( e gx + e - gx )
(e gx - e - gx )
+
V
Z
IR
V(x) =
R
C
2
2
I(x) =
or
(e gx + e - g x )
( e gx - e - gx )
VR +
IR
2 ZC
2
...(6.36)
...(6.37)
V(x) = cosh(gx)VR + ZC sinh(gx) IR
...(6.38)
1
sinh(gx)VR + cosh(gx)IR
ZC
...(6.39)
I(x) =
Our interest is in the relation between the sending end and the receiving end of the line.
Therefore, when x = l, V(l) = VS and I(l) = IS. The result is
VS = cosh (g l )VR + ZC sinh(g l )IR
IS =
Therefore, ABCD constants are:
1
sinh (gl ) VR + cosh(g l )IR
ZC
A = cosh(g l ) ; B = ZC sinh(g l )
...(6.40)
...(6.41)
...(6.42)
1
sinh(g l) ; D = cosh(g l )
...(6.43)
ZC
It is now possible to find an accurate equivalent p model for long transmission line as shown
in Fig. 6.5.
C=
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Similar to the expressions of eqns. (6.15) and (6.17) obtained for the nominal p, for equivalent
p model we have,
FG
H
VS = 1 +
FG
H
IJ
K
Z¢Y ¢
VR + Z ¢ I R
2
IS = Y ¢ 1 +
IJ
K
FG
H
...(6.44)
IJ
K
Z¢Y ¢
Z ¢Y ¢
VR + 1 +
IR
4
2
...(6.45)
Now comparing eqns. (6.44) and (6.45) with eqns. (6.40) and (6.41), respectively and making
use of the identity
tanh
FG gl IJ = cosh(g l) - 1
H 2 K sinh (gl)
...(6.46)
the parameters of equivalent p model are obtained as:
Z¢ = ZC sinh(g l) =
FG IJ
H K
Z × sinh(g l)
gl
1
Y tanh (g l 2)
gl
Y¢
tanh
=
=
ZC
2
2
gl 2
2
...(6.47)
...(6.48)
Fig. 6.5: Equivalent p model for long transmission line.
Example 6.1: A single phase 60 Hz generator supplies an inductive load of 4500 kW at a power
factor of 0.80 lagging by means of an 20 km long overhead transmission line. The line resistance
and inductance are 0.0195 W and 0.60 mH per km. The voltage at the receiving end is required
to be kept constant at 10.2 kV.
Find (a) the sending end voltage and voltage regulation of the line; (b) the value of the
capacitors to be placed in parallel with the load such that the regulation is reduced to 60% of that
obtained in part (a); and (c) compare the transmission efficiencies in parts (a) and (b).
Solution: The line constants are:
R = 0.0195 × 20 = 0.39 W
X = 0.60 × 10–3 × 2p × 60 × 20 = 4.52 W
(a) This is a short line with I = IR = IS given by
|I| =
4500
Amp = 551.47 Amp.
10.2 ´ 0.80
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From eqn. (6.6),
|VS| » |VR| + |I|(R cos dR + X sin dR)
Here
|VR| = 10.2 kV = 10200 Volt
cosdR = 0.8, sindR = 0.6
\
|VS| = 10200 + 551.47 (0.39 × 0.8 + 4.52 × 0.6)
\
|VS| = 11.867 kV
\
Voltage regulation =
(11.867 - 10.2)
´ 100 = 16.34%
10.2
(b) Voltage regulation desired = 0.60 × 16.34 = 9.804%
Therefore, under this condition we can write
|VS| - 10.2
= 0.09804
10.2
\
|VS| = 11.2 kV
Figure 6.6 shows the equivalent circuit of the
line with a capacitor in parallel with the load.
Assuming combined power factor of the load and
capacitor = cos dR¢
Fig. 6.6
By using eqn. (6.6), we can write,
(11.2 – 10.2) × 10–3 = |IR| (R cos dR¢ + X sin dR¢)
...(i)
Since the capacitance does not draw any real power, we have,
|IR| =
4500
10.2 cosd ¢R
...(ii)
From eqns. (i) and (ii), we get
4.52 tan dR¢ = 1.876
\
tan dR¢ = 0.415
\
dR¢ = 22.5°
\
\
Now
cos dR¢ = 0.9238
|IR| = 477.56 Amp.
IC = IR – I,
IR = 477.56 -22.5° = 441.2 – j182.75
I = 551.47 -36.87° = 441.2 – j330.88
\
IC = 441.2 – j182.75 – 441.2 + j330.88
\
IC = j148.13 Amp.
Now
XC =
V
10.2 ´ 1000
1
= R =
148.13
2p ´ 60 ´ C
IC
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\
C = 38.5 mF Ans.
(c) Efficiency of transmission
Output
4500
=
Output + losses 4500 + (55147
. )2 ´ 0.39 ´ 10 -3
= 97.43%
Case (a)
h=
Case (b)
h=
4500
= 98.06%
4500 + (477.56) 2 ´ 0.39 ´ 10 -3
It is to be noted that by placing a capacitor in parallel with the load, the receiving end power
factor improves from 0.80 to 0.9238.
Example 6.2: A 220 kV, three phase transmission line is 60 km long. The resistance is 0.15 W/
Km and the inductance 1.4 mH/Km. Use the short line model to find the voltage and power at
the sending end and the voltage regulation and efficiency when the line is supplying a three
phase load of
(a) 300 MVA at 0.8 pf lagging at 220 kV
(b) 300 MVA at 0.8 pf leading at 220 kV
Solution:
R = 0.15 × 60 = 9 W
Assuming
\
f = 50 Hz.
X = 2p × 50 × 1.4 × 10–3 × 60 = 26.39 W.
(a) Receiving end voltage per phase is
VR =
220 0°
3
= 127 0° kV
The three phase apparent power is 300 MVA at 0.8 pf lagging
\
f = 36.87°
\
S = 300 36.87° = (240 + j180) MVA
The current per phase is given by
IR =
\
S*
3 VR*
=
300 -36.87°
3 ´ 127 0°
´ 10 3 Amp
IR = 787.4 -36.87° Amp
From eqn. (6.6), the sending end voltage magnitude is
|VS| = |VR| + |I| (R cos dR + X sin dR)
|VR| = 127 kV, |I| = 787.4 Amp = 0.7874 kA,
R = 9 W, X = 26.39 W
cosdR = 0.8, sindR = 0.6
\
|VS| = 127 + 0.7874 (9 × 0.8 + 26.39 × 0.6)
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\
|VS| = 145.13 kV.
\ Sending end line-to-line voltage
|VS|L–L =
Voltage regulation =
3 × 145.13 = 251.37 kV
. - 220
25137
= 14.26%
220
Per phase real power loss in the line
PLoss = |I|2 R = (787.4)2 × 9 × 10–6 MW = 5.58 MW.
Per phase receiving end power
PR =
300
´ 0.8 = 80 MW
3
\ Per phase sending end power
PS = (80 + 5.58) = 85.58 MW.
Transmission line efficiency is
h=
(b)
as:
PR
80
=
= 93.47%.
PS 85.58
3 |VR||IR| = 300
\
|VR| = 220 kV
\
|IR| = 787.4 Amp
Load is at 0.8 power factor leading. For leading power factor load, eqn. (6.6) can be written
|VS| = |VR|+ |I| (R cos dR – X sin dR)
\
\
|VS| = 127 + 787.4 (9 × 0.8 – 26.39 × 0.6) = 120.2 kV
|VS|L–L =
Voltage regulation =
3 × 120.2 = 208.2 kV
208.2 - 220
= –5.36%
220
Per phase real power loss = (787.4)2 × 9 × 10–6 = 5.58 MW
Per phase receiving end power,
PR =
300
´ 0.8 = 80 MW
3
Per phase sending end power,
PS = (80 + 5.58) = 85.58 MW
Transmission line efficiency
h=
PR
80
=
= 93.47%
PS 85.58
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Example 6.3: Determine the efficiency and regulation of a 3-phase, 150 km long, 50 Hz
transmission line delivering 20 MW at a power factor of 0.8 lagging and 66 kV to a balanced
load. Resistance of the line is 0.075 W/km, 1.5 cm outside dia, spaced equilaterally 2 meters
between centres. Use nominal p method.
Solution:
R = 0.075 × 150 = 11.25 W
diameter of the conductor = 1.5 cm
1.5
\
radius r =
= 0.75 cm
2
d = 2 mt = 200 cm
FGH 200 IJK Henry
0.75
\
L = 2 × 10–7 × (150 × 1000) ln
\
\
L = 0.1675 Henry
X = 2 × p × 50 × 0.1675 = 52.62 W.
The capacitance per phase =
\
\
\
2 ´ p ´ 8.854 ´ 10 -12
´ (150 ´ 1000) = 1.49 mF.
200
ln
0.75
FGH IJK
Y = jwC = j 2p × 50 × 1.49 × 10–6 mho
Y = j 468.1 × 10–6 mho
Y
= j 234 × 10–6 mho
2
Z = (11.25 + j52.62) = 53.809 77.9° W.
Now
3 × |IR|× 66 × 0.8 = 20 × 1000
\
|IR| = 218.7 Amp. at 0.8 pf lagging
Receiving end phase voltage
|VR| =
From eqn. (6.15), we have
66
3
FGH
= 38.104 kV
IJK
VS = 1 + ZY VR + Z IR
2
IR = 218.7 -36.87°
VR = 38.104 0°
ZY
= 53.809 × 234 × 10–6 77.9° + 90°
2
= 0.01259 167.9° = (–0.0123 + j0.00264)
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VS = (1 – 0.0123 + j0.00264) × 38.104 0° +
\
(53.809 77.9° ´ 218.7 -36.87°)
1000
\
. °
VS = (0.9877 + j0.00264) × 38.104 0° + 11.76 4103
\
VS = 37.63 + j0.1 + 8.87 + j7.72
\
VS = (46.5 + j7.82) = 47.15 9.54° kV
\
.
9.54° kV
3 × 47.15 9.54° kV = 8166
VS(L – L) =
FGH
IJK
81.66
|VS|
- 66
-|VR|
0.9877
Voltage regulation VR = |A|
= 25.26%
=
|VR|
66
Power loss per phase = |I|2 R = (218.7)2 × 11.25 × 10–6 MW = 0.538 MW
Per phase receiving end power
PR =
20
MW
3
Per phase sending end power
PS =
20
+ 0.538 = 7.204 MW
3
Transmission efficiency
h=
20 3
= 92.54%.
7.204
Example 6.4: Determine the voltage, current and power factor at the sending end of a 3 phase,
50 Hz, overhead transmission line 160 km long delivering a load of 100 MVA at 0.8 pf lagging
and 132 kV to a balanced load. Resistance per km is 0.16 W, inductance per km is 1.2 mH and
capacitance per km per conductor is 0.0082 mF. Use nominal p method.
Solution:
R = 0.16 × 160 = 25.6 W
X = 1.2 × 10–3 × 2p × 50 × 160 = 60.3 W.
Y = j2p × 50 × 0.0082 × 10–6 × 160 = j4.12 × 10–4 mho
Z = R + jX = 25.6 + j60.3 = 65.51 67° W.
From eqn. (6.15),
FGH
VS = 1 +
IJK
ZY
VR + ZIR
2
Phase voltage at the receiving end,
VR =
132
3
0° kV = 76.21 0° kV
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Receiving end current
100 ´ 10 6
IR =
3 ´ 132 ´ 10 3
Amp = 437.38 Amp
Load has lagging power factor of 0.80, i.e., dR = 36.87°.
IR = 437.38 -36.87° Amp
\
ZY
= 65.51 67° ´ 4.12 ´ 10 -4 90°
2
= (–0.0124 + j0.0053)
\
VS = (1 – 0.0124 + j0.0053) × 76.21 0° +
\
VS = 101.07 8.18° kV
65.51 67° ´ 437.38 -36.87°
1000
Sending end line to line voltage
VS, L–L =
. 8.18° kV
3 ´ 10107
= 175.05 8.18° kV
From eqn. (6.17)
FG
H
IS = Y 1 +
IJ
K
FG
H
IJ
K
ZY
ZY
VR + 1 +
IR
4
2
ZY
= 0.00675 157°
4
1+
FG
H
Y 1+
ZY
= 0.9938 0.15°
4
IJ
K
ZY
= 4.12 × 10–4 × 0.9938 90.15°
4
= 4.094 × 10–4 90.15°
\
IS = 4.094 × 10–4 90.15° × 76.21 0°
+
(1 - 0.0124 + j0.0053) ´ 437.38 -36.87°
1000
\
IS = 0.0311 90.15° + 0.432 -36.56°
\
IS = 0.414 -33.06° k Amp
\
IS = 414 -33.06° Amp
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Sending end power factor angle = 8.18° – (–33.06°) = 41.24°
Sending end power factor = cos (41.24° ) = 0.752
Example 6.5: A long transmission line delivers a load of 60 MVA at 124 kV, 50 Hz, at 0.8 power
factor lagging. Resistance of the line is 25.3 W, reactance is 66.5 W and admittance due to
charging capacitance is 0.442 × 10 3 mho. Find (a) A, B, C, D constants (b) Sending end voltage,
current and power factor (c) regulation (d) efficiency of the line.
Solution:
(a)
R = 25.3 ohm, X = 66.5 ohm, Z = (25.3 + j66.5) ohm
Y = j0.442 × 10–3 mho.
gl =
ZY =
zy l =
zl . yl =
ZY
(25.3 + j66.5) ( j0.442 ´ 10 –3 )
= (0.0327 + j0.174)
\
A = D = cosh (g l) = cosh
e ZY j
= cosh (0.03217 + j 0.174)
= 0.986 0.32°
B = ZC sinh (g l) =
FG Z IJ sinh e ZY j
H YK
Z
= (393 – j72.3)
Y
\
B = 70.3 69.2°
C=
1
sinh (g l) =
ZC
FG Y IJ sinh e ZY j
H ZK
= 4.44 × 10–4 90° = j4.44 × 10–4
(b) Load at 60 MVA at 124 kV (line-to-line)
\
Load current,
IR =
60 ´ 1000
3 ´ 124
Power factor is 0.80 (lagging)
\
IR = 279.36 -36.87° Amp
VR =
Now
Amp = 279.36 Amp
124
3
= 71.6 kV (phase voltage)
VS = AVR + BIR
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\
VS = 0.986 0.32° ´ 71.6 0° +
\
VS = 87.84 7.12° kV
\
70.3 69.2° ´ 279.36 -36.87°
1000
VS, L-L = 3 ´ 87.84 7.12° = 152.14 7.12° kV
IS = CVR + DIR
\
IS = j4.44 ´ 10 -4 ´ 716
. 0° ´ 1000 + 0.986 0.32° ´ 279.36 -36.87°
\
IS = 221.28 – j132.24
\
IS = 257.78 -30.86° Amp
Power factor angle at the sending end
= 7.12° – (–30.86°)
= 37.98°
Sending end power factor = cos (37.98°)
= 0.788
(c) Sending end power
PS =
3 ´ 152.14 ´ 257.78 ´ cos (37.98° ) kW
\
PS = 53520 kW = 53.52 MW
Receiving end power
PR = 60 × 0.80 = 48 MW
\ efficiency
(d) Voltage regulation
h=
=
48
= 89.68%
.
5352
VS
- VR
A
VR
´ 100
FG 152.14 - 124IJ
H 0.986 K ´ 100
=
124
= 24.43%
Example 6.6: A 60 Hz, 250 km long transmission line has an impedance of (33 + j104) ohm and
a total shunt admittance of 10 3 mho. The receiving end load is 50 MW at 208 kV with 0.80 power
factor lagging. Find the sending-end voltage, current, power and power factor using
(a) short line approximation (b) nominal P method (c) exact transmission line equations.
Solution:
Z = (33 + j104) = 109.11 72.4° ohm
Y = j10–3 mho
Receiving end load is 50 MW at 208 kV, 0.80 lagging power factor.
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\
IR =
VR =
50
3 ´ 208 ´ 0.80
208
3
-36.87° = 0.173 -36.87° kA
0° = 120.08 0° kV
(a) Short line approximation
VS = VR + IZ
\
VS = 120.08 0° + 0.173 -36.87° ´ 109.11 72.4°
\
VS = 135.87 4.62° k V
\
VS(L-L) =
3 ´ 135.87 4.62° k V
= 235.33 4.62° kV
IS = IR = 0.173 -36.87° k A
Sending end power factor = cos (36.87° + 4.62°) = 0.75
\
PS =
(b) Nominal-P method
3 ´ 235.33 ´ 0.173 ´ 0.75 MW = 52.88 MW
A=D= 1+
YZ
j ´ 10 -3 ´ 109.11 72.4°
=1+
2
2
= 0.9481 1°
B = Z = 109.11 72.4°
FG
H
C=Y 1+
\
YZ
4
VS = AVR + BIR
IJ = j ´ 10
K
-3
= 0.9481 1° ´ 120.08 0° + 109.11 72.4° ´ 0.173 -36.87°
\
\
VS = 129.817 5.72° kV
VS (L-L) =
3 ´ 129.817 5.72° k V = 224.85 5.72° k V
IS = CVR + DIR
-3
= j ´ 10 ´ 120.08 0° + 0.9481 1° ´ 0.173 -36.87°
= 0.135 10.23° k A
Sending end power factor = cos (10.23° – 5.72°)
= 0.997 (leading)
PS = 3 ´ 224.85 ´ 0.135 ´ 0.997 MW
= 52.4 MW
(c) Exact transmission line equation
gl =
\
gl =
e zy j l =
ZY
j ´ 10 -3 ´ 109.11 72.4°
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g l = 0.33 81.2°
\
ZC =
Z
=
Y
109.11 72.4°
j ´ 10 -3
FG
H
A = D = cosh (g l) = 1 +
sinh (g l) »
FG
H
YZ 1 +
YZ
6
= 330.31 -8.8°
YZ
2
IJ = 0.9481 1°
K
IJ = 0.33 81.2°
K
. °
B = ZC sinh (g l) = 330.31 -8.8° ´ 0.33 812
= 109 72.4°
C=
sinh ( g l)
0.33 81.2°
=
» j ´ 10 -3
ZC
330.31 -8.8°
VS = AVR + BIR
= 0.9481 1° ´ 120.08 0° + 109 72.4° ´ 0.173 -36.87°
= 113.84 1° + 18.85 35.53°
= 129.806 5.72° k V
\
VS (L-L) =
3 ´ 129.806 5.72° k V
= 224.83 5.72° k V
IS = CVR + DIR
IS = j ´ 10 -3 ´ 120.08 0° + 0.948 1° ´ 0.173 -36.87°
\
= 0.135 10.23° kA.
Sending end power factor = cos (10.23° – 5.72°)
= 0.997 (leading)
PS =
3 ´ 224.83 ´ 0.135 ´ 0.997 = 52.4 MW
Results are tabulated below:
Short line approx.
8S (L-L)
|1S|
FB
FI
235.33 kV
0.173 kA
0.75
52.88 MW
Nominal-p
224.85 kV
0.135 kA
0.997
52.40 MW
Exact
224.83 kV
0.135 kA
0.997
52.40 MW
From the above table, we can see that the results obtained by the nominal-p and exact
methods are practically the same. On the other hand, the results obtained by the short line
approximation are in considerable error.
Therefore, for a long line, it is sufficiently accurate to use the nominal-p method.
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6.6 VOLTAGE WAVES
By using eqns. (6.26) and (6.27), we obtain,
V(x) = C1.eax.e jbx + C2.e ax.e jbx
...(6.49)
Transforming eqn. (6.49) to time domain, the instantaneous voltage as a function of t and
x becomes
V(t, x) = 2 Real {C1eaxe j(wt + bx)} + 2 Real {C2.e ax.e j(wt – bx)}
...(6.50)
Note that V(x) in eqn. (6.49) is the rms phasor value of voltage at any point along the line.
As x increases (moving from receiving end to sending end), the first term becomes larger
because of eax and is called the incident wave. The second term e ax becomes smaller and is
called the reflected wave. At any point along the line, voltage is the sum of two components.
...(6.51)
V(t, x) = V1(t, x) + V2(t, x)
where
V1(t, x) =
2 C1eax cos(w t + bx)
V2(t, x) =
2 C2e
ax
cos(wt – bx)
...(6.52)
...(6.53)
As we move along the line, eqns. (6.52) and (6.53) behave like travelling waves. Now
consider the reflected wave V2(t, x) and imagine that we are riding along with the wave. For
observing instantaneous value, peak amplitude requires that
wt – bx = 2 k p
\
x=
w
2k p
tt
b
b
...(6.54)
The speed can be given as
\
dx
w
=
dt
b
Thus, the velocity of propagation is given by v =
...(6.55)
w
2p f
=
b
b
...(6.56)
A complete voltage cycle along the line corresponds to a change of 2p radian in the angular
argument bx. The corresponding line length is defined as the wavelength. If b is expressed in
rad/mt.
B l = 2p
\
l=
2p
b
...(6.57)
When line losses are neglected, i.e., when g = 0 and g = 0, then the real part of the
propagation constant a = 0. From eqn. (6.27)
g = a + jb = zy = (r + jwL ) ( g + jwc)
\
b = w LC
...(6.58)
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From eqn. (6.29), the characteristic impedance
Zc =
z
L
=
y
C
...(6.59)
Which is commonly referred to as the surge impedance. Its value varies between 250 ohm
and 400 ohm in case of overhead transmission lines and between 40 ohm and 60 ohm in case of
underground cables.
From eqns. (6.56) and (6.57), we get,
v=
1
...(6.60)
LC
and
l=
1
Now for a single phase line
L=
\
Approximating
\
FG IJ
H K
2 p Î0
m0
D
ln
; C=
D
2p
r¢
ln
r
LC = m0Î0 ln
ln
...(6.61)
f LC
FG D IJ » ln (D /r)
H r¢ K
FG D IJ
H r¢ K
FG IJ
H K
ln ( D /r )
LC » m0Î0
...(6.62)
Substituting the expression LC into eqn. (6.60) and (6.61), we get
v»
l»
1
m 0 Î0
1
f m 0 Î0
...(6.63)
...(6.64)
6.7 SURGE IMPEDANCE
When the transmission line is loaded by being terminated with an impedance equal to its
characteristic impedance, the receiving end current is
IR =
VR
ZC
...(6.65)
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For a loss less line ZC is purely resistive. Surge impedance loading (SIL) of a transmission
line is defined as the power delivered by a line to purely resistive load equal in value to the
surge impedance of the line. SIL is given by
SIL =
Since
3|V R|2
ZC
...(6.66)
|VR| = |VL (rated)|
SIL =
3 , SIL becomes
|V L (rated)|2
...(6.67)
ZC
Example 6.7: A three phase, 50 Hz, 400 kV transmission line is 300 km long. The line inductance
is 0.97 mH/km per phase and capacitance is 0.0115 mF/km per phase. Assume a loss less line.
Determine the line phase constant b, ZC, v and l.
Solution:
b = w L C = 2 p ´ 50 0.97 ´ 0.0115 ´ 10 -9
\
Surge impedance
b = 0.00105 rad/km
ZC =
0.97 ´ 10 -3
L
=
C
0.0115 ´ 10 -6
= 290.43 ohm
Velocity of propagation is
v=
\
Line wavelength is
LC
=
1
0.97 ´ 0.0115 ´ 10 -9
v = 2.994 × 105 km/sec.
l=
\
1
v
1
=
´ 2.994 ´ 105
f
50
l = 4990 km.
6.8 POWER FLOW THROUGH TRANSMISSION LINE
Consider a sample power system as shown in Fig. 6.7.
Fig. 6.7: Two bus sample power system.
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Let
A = |A| d A , B =|B| d B , VS =|VS| d S , V R =|VR| 0° , D = A =|A| d A
From eqn. (6.1), we obtain
\
IR =
VS - AVR |VS| d S -| A| d A .|V R| 0°
=
|B| d B
B
IR =
|VS|
|A||V R|
d S - dB d S - dB
|B|
|B|
...(6.68)
The receiving end complex power
SR(3 f) = PR(3 f) + jQR(3 f) = 3VR IR*
...(6.69)
Using eqns. (6.69) and (6.68), we get
|V S||V R|
| A||V R|2
dB - dS - 3
dB - dA
SR(3 f) = 3 .
|B|
|B|
or, in terms of line-to-line voltage
SR(3 f) =
|VS(L-L)||V R(L-L)|
|B|
d B - dS -
|A||VR(L-L)|2
dB - d A
...(6.70)
b
g
...(6.71)
b
g
...(6.72)
|B|
Separating real and imaginary parts of eqn. (6.70),
PR(3 f) =
QR(3 f) =
|VS(L-L)||V R(L-L)|
|B|
|VS(L-L)||V R(L-L)|
|B|
b
g
| A||VR(L-L)|2
b
g
| A||V R(L-L)|2
cos d B - d S -
sin d B - d S -
|B|
|B|
cos d B - d A
sin d B - d A
Similarly we can obtain
PS(3 f) =
QS(3 f) =
|A||VS(L-L)|2
|B|
|A||VS(L-L)|2
|B|
b
g
|VS(L-L)||VR(L-L)|
b
g
|VS(L-L)||VR(L-L)|
sin d B + d S
|B|
cos d B - d A -
sin d B - d A -
|B|
b
g
...(6.73)
b
g
...(6.74)
cos d B + d S
The real and reactive power losses are
PLoss(3 f) = PS(3 f) – PR(3 f)
...(6.75)
QLoss(3 f) = QS(3f) – QR(3 f)
...(6.76)
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6.9 FERRANTI EFFECT
During light-load or no-load condition, receiving end voltage is greater than sending end
voltage in long transmission line or cables. This happens due to very high line charging current.
This phenomenon is known as ferranti effect. A charged open circuit line draws significant
amount of current due to capacitive effect of the line. This is more in high-voltage long
transmission line. Under no load condition, IR = 0, therefore, eqn. (6.40) can be written as:
VS = VR cosh (g l)
\
VR =
VS
cosh ( g l)
...(6.77)
Now cosh (g l) £ 1, therefore VR is always greater or equal to VS.
EXERCISE
6.1. Find the characteristics of the load at the sending end and the efficiency of a three phase transmission
line 160 km long delivering 15 MVA load at 110 kv, 50 Hz and 0.9 power factor (lagging) having
inductance 1.356 mH/km per phase, capacitance 0.0085 mF/km per phase and resistance 40 ohms.
Use nominal P method
Ans. IS
70.3 20.8° Amp, VS(L-L)
power factor
117.6 29.2° kV,
0.9893, h
95.3%
6.2. A long three phase transmission line has resistance 63.5 ohms, reactance per phase 167 ohms,
capacitive susceptance to neutral is 1.1 × 10–3 mho. Determine ABCD constants.
Ans. A
D
0.91 2.13° , B
C
173.3 69.9°
1.067 × 10–3 90.7°
6.3. A 50 Hz, three phase, 275 kV, 400 km long transmission line has resistance 0.035 W/km, inductance
1 mH/km and capacitance 0.01 mF/km. If the line is supplied at 275 kV, find out the MVA rating of
a shunt reactor that would be required to maintain 275 kV at the receiving end, when the line is
delivering no load. Use nominal P method.
Ans. 47.56 MVAr (lagging)
6.4. The line constants of a transmission line are A 0.9301 0.98° and B 14124
. 8187
. ° ohm. The load
at the receiving end is 60 MVA, 50 Hz, 0.8 pf lagging. The supply voltage is 220 kV. Calculate the
load voltage.
Ans. 202.2 kV
6.5. A three phase 50 Hz transmission line has impedance of (25.3 + j66.5) ohms and a shunt admittance
of 4.42 × 10–4 mho per phase. If it delivers a load of 50 MW at 220 kV at 0.8 power factor lagging,
determine the sending end voltage (a) by short line approximation (b) nominal P method (c) exact
transmission line equations.
Ans. (a) 233.8 2.2° kV, (b) 232.2 2.33° , (c) 230.52 2.50° kV,
6.6. The line constants of a transmission line are A 0.986 0.320° and B 70.3 69.2° ohms. Determine
the capacity of a reactor to be installed at the receiving end so that when a load of 50 MVA is
delivered at 132 kV and power factor 0.707 lagging the sending end voltage can also be 132 kV.
Ans. 48.47 MV Ar.
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. ° , B 75.5 80° and C 4 × 10–4 91° . If an
6.7. A three phase transmission line has A D 0.98 15
impedance (2.64 + j42.3) ohms is connected in series with the line at the sending end, determine
the new values of ABCD constants.
Ans. Ao
0.963 153
. ° ; Bo
Co
4 ´ 10
–4
116.8 82.8° ohms
90° ; Do
0.98 15
. °
6.8. A three phase, 50 Hz, 100 km transmission line has resistance 0.1 ohms/km, inductance
111.7 mH/km and capacitance 0.9954 × 10–2 mF/ph per km. The line is delivering 20 MW at a 0.8
power factor lagging and 66 kV to a balanced load. Determine the efficiency and regulation. Use
nominal p method.
Ans. h 93.5%, VR 17.47%
6.9. A 132 kV, 50 Hz transmission line has the following generalised constants, A
D
0.9696 0.49° ,
-4
B 52.88 74.79° ohms and C 11.77 ´ 10
90.15° mho. It is supplying a load of 125 MVA, 0.9
power factor lagging at 132 kV. Calculate the sending end voltage and current.
Ans. VS
166.1 kV, IS
554 Amp
6.10. A three phase transmission line delivers a load of 5 MW at 0.8 power factor lagging. Resistance of
each conductor is 0.5 W/km. Receiving end voltage is 33 kV. If the line loss is not to exceed 10%,
determine the length of the line
Ans. 27.9 km.
6.11. The ABCD parameters of two transmission lines are A1, B1, C1, D1 and A2, B2, C2, D2. Find the
overall ABCD parameters when (i) the two networks are connected in series (ii) in parallel.
6.12. Two overhead lines are connected in parallel to supply a load of 10 MW at 0.8 power factor lagging
at 30 kV. The resistance and reactance of one line (A) are 5.5 W and 13.5 W respectively and for line
(B) are 6 ohm and 11 ohm respectively, calculate (a) the KVA supplied by each line and (b) power
supplied by each line
Ans. (a) 5790, 6730 (b) 4415 kW, 5585 kW.
6.13. A long transmission line has the following generalised constants: A
D
0.92 5.3° ; B
65.3 81°
ohm. Two identical transformers, each of series impedance 100 70° ohms and admittance
-4
2 ´ 10
-75° mho are connected at the two ends of the line. Determine new values of generalised
constants.
Ans. Ao
Do
0.843 26.38° , Bo
Co
249.3 76.85° ohm
34.38 ´ 10
–4
50.9 ° mho
23