EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
Example 1
Find the inductive reactance per km, capacitance to the neutral
per km, line to line capacitance per km of a single phase line
operating at 50Hz. The conductors are 6 m apart and the
diameter of the conductor is 25 mm.
Dia=25mm
D=6m
Solution1:
Given D=6 m and r=12.5x10-3 m
Inductance per conductor per m
4 10
6
0.7788 12.5 10
4 10
0.7788
.
/
The inductive reactance per km:
2
2
50 25.7 10
/
.
/
Capacitance to the neutral per km:
2
2
8.85 10
6
12.5 10
0.0556
6.1737
/
Line to line capacitance per km:
/
2
Example 2
A stranded conductor consists of seven identical strands each having a
radius r as shown in figure 1.Determine the GMR of the conductor in
terms of r.
1
2
r
7
6
Solution2:
5
3
4
The distance from strand 1 to all other strands is:
2
4
2√3
GMR of conductor:
· 2 · 2√3 · 4 · 2√3 · 2 · 2
·
2
GMR of conductor in terms of r:
2
3
2
.
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 1
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
Example 3
One circuit of a single phase transmission line composed of three
solid wires, each 5mm in diameter. The return circuit composed of
two wires, each 10mm in diameter. The arrangement of the
conductor is shown in figure 3. Find the inductance due to current
in each side of the line and the inductance of the complete line.
10m
a
a’
b
b’
5m
5m
c
Solution 3:
10
10
10
√10
14.14
5
11.18
Geometric Mean Distance (GMD):
10 11.18 11.18 10 14.14 11.18
.
Geometric Mean Radius (GMR):
For three solid wires (x):
5 10
0.7788 2.5 10
62500
5 5
5 10
.
For two solid wires (y):
5
5
0.7788 5 10
5 5
.
The inductance of conductor x is,
2 10
2 10
11.2
0.4259
.
/
2 10
11.2
0.1395
.
/
The inductance of conductor y is,
2 10
The total inductance will be,
6.54 10
8.77 10
/
.
/
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 2
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
Example 4
A three phase transmission line has its conductors at the corner of an
equilateral triangle with side 4.5 m. The diameter of each conductor is
20 mm. Find the inductance per km and capacitance to the neutral per
km of the line.
D
D
D
Solution 4:
D=4.5 m
and
r =10x10-3 m
r’=0.7788x10x10-3 =7.788x10-3 m
For three phase transmission line,
2 10
2 10
4.5
7.788 10
Inductance per km:
.
/
8.85 10
4.5
10 10
0.0556
6.10924
Capacitance to neutral per km:
2
2
.
/
Example 5
Find the inductive reactance per km at 50 Hz of three phase bundled conductor line with two
conductors per phase as shown in figure 5. Diameter of each conductor is 30 mm.
0.4m
a
0.4m
a’
b
0.4m
b’
9m
c
c’
9m
Solution 5:
GMR, for the two sub-conductor bundle:
0.7788 15 10
0.4
.
Distance from conductor to another,
Dab=Dbc=9m,
Dab’=Dbc’=9.4m, Da’b=Db’c=8.6m, Da’b’=Db’c’=9m, Dac=18m, Dac’=18.4m,
Da’c=17.6m, Da’c’=18m
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 3
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
GMD between each phase group:
9 9.4 8.6 9
8.9956
9 9.4 8.6 9
8.9956
17.9978
18 18.4 17.6 18
The equivalent GMD,
8.9956 8.9956 17.9978
Inductance is given by
2 10
11.3351
0.068
2 10
.
.
/
Inductance reactance per km is given by
2
2
.
50 10.23 10
/
Example 6
A 132kV 3 phase single circuit bundle conductor line with two sub-conductors per phase is shown
in figure 6. The radius of each conductor is 10mm. The centre to centre distance between adjacent
phases is 12m and the spacing between the sub-conductors is 0.5m. Find the capacitance per km.
Solution 6:
Distance between a bundles of conductors to another bundle:
12 ,
12 ,
24
Geometric Mean Distance (GMD):
12 12 24
15.119
GMR, rb for two sub-conductor bundle:
√
10 10
0.5
0.0707
Line capacitance to neutral per km:
0.0556
0.0556
15.119
0.0707
.
/
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 4
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
Example 7
A 345 kV double circuit three phase transposed
line is composed of two ACSR 1431000-cmil,
45/7 Bobolink conductor per phase with
vertical conductor configuration as shown in
figure 7. The conductors have a diameter of
1.427 in and a GMR of 0.564 in. The bundle
spacing in 18 in. Find the inductance and
capacitance per phase per km of the line.
a
S11=11m
c’
H12=7m
b
b’
S22=16.5m
H23=6.5m
S33=12.5m
c
a’
Solution 7:
Given that,
GMR=0.564 inch= 0.01433 m
Conductor diameter=1.427 inch=0.03625 m
Bundle spacing=18 inch=0.4572 m
Distance of conductors:
Dab’=15.43m,
Dbc’=15.43m,
Dca’=12.50m,
Dab=7.52m,
Dbc=6.80m,
Dca=13.52m,
Daa’=17.90m,
Da’b=15.21m,
Db’c=15.21m,
Dc’a=11.00m,
Dbb’=16.50m,
Da’b’=6.80 m
Db’c’=7.52m
Dc’a’=13.52m
Dcc’=17.90m
GMD between each phase group:
15.43 7.52 15.21 6.80
10.4666
15.43 6.80 15.21 7.52
10.4666
12.5 13.52 11.00 13.52
12.5911
The equivalent Geometric Mean Distance (GMD):
10.4666 10.4666 12.5911
11.132
Then,
GMR for each subconductor and spacing d:
0.01432 0.4572
0.03625
2
√
0.4572
0.08091
0.091
GMR of each phase group for inductance,
0.08091 17.9
0.08091 16.5
0.08091 17.9
1.2034
1.1554
1.2034
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 5
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
GMR of each phase group for capacitance,
0.091 17.9
0.091 16.5
0.091 17.9
1.276
1.225
1.276
The equivalent GMR,
1.2034 1.1554 1.2034
1.276 1.225 1.276
Inductance per phase:
0.2
0.2
Capacitance per phase:
0.0556
11.132
1.187
0.0556
11.132
1.2587
1.187
1.2587
.
/
.
/
Example 8
A 220 kV, 150 MVA, 60 Hz, three phase transmission line, 140 km long, characteristic parameters
of the transmission line are,
r = 0.12 Ω/km
x = 0.88 Ω/km
y = 4.1 x 10 -6 S/km
The voltage at the receiving end of the transmission line is 210 kV. If considered transmission line
is a short length.
Find,
a) What is series impedance and shunt admittance of this transmission line?
b) What is the sending end voltage if the line is supplying rated voltage and apparent power at
0.85 pf lagging? At unity pf? At 0.85 leading pf?
c) What is the voltage regulation of the transmission line for each of the cases (b)?
d) What is the efficiency of the transmission line when it is supplying rated apparent power at
0.85 pf lagging?
Solution 8:
a) The series resistance, and series reactance of this transmission line are :
R = r.d = (0.12 Ω/km) (140 km) = 16.8 Ω
XL = x.d = (0.88 Ω/km) (140 km) = 123.2 Ω
b) The current out of this transmission line will be given by
S
√3
√3VLLR IR
150
√3 210
412
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 6
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
The line to neutral voltage out of the line will be,
210
√3
121
Since this transmission line as short.
The line to neutral voltage VS at the sending end of the line when the power factor is 0.85 lagging
will be,
121
0°
412
31.8° 16.18
√3 158
123.3
158 14.4°
The line to neutral voltage VS at the sending end of the line when the power factor is 1.0 will be,
121
0°
412 0° 16.18
√3 137.4
123.3
137.4 21.6°
The line to neutral voltage VS at the sending end of the line when the power factor is 0.85 leading
will be,
121 0°
412 31.8° 16.18
VLLS √3 110.3k V
123.3
110.3 25.0°
c) The voltage regulation of a transmission line is given by
100%
The no load received voltage will be the same as the source voltage, since there will be no current
flowing in the line at no load conditions. The full load received voltage will be 210 kV.
The resulting voltage regulation at 0.85 power factor lagging is
.
274 210
100%
210
.
%
The resulting voltage regulation at unity power factor is
238 210
100%
210
.
%
The resulting voltage regulation at 0.85 power factor leading is
.
191 210
100%
210
.
%
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 7
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
d) The output power from the transmission line at a power factor of 0.85 lagging is
Pout = 3 VR IR Cos θR = 3 (121kV) (412A) (0.85) = 127 kW
The input power from the transmission line is
Pin = 3 VS IS cos θS= 3(158kV) (412A) Cos [14.1-(-31.8)] = 135.7 kW
The resulting transmission line efficiency at full load and 0.85 power factor lagging is
127
135.7
100%
100%
. %
Example 9
A three phase 132 kV transmission line is connected to a 50 MW load a 0.85 power factor lagging.
The line constants (resistance and susceptance) of the 85 km long line are
Z=95∠78° Ω
Y=1x10-3∠90° mho
Using nominal T circuit representation calculate
a) The ABCD of the line
b) Sending end voltage
c) Sending end current
d) Sending end power factor
e) Efficiency of transmission
Solution 9:
a) The ABCD of the line
The receiving end voltage is
132
76.21
√3
The receiving end current is
50 10
√3 132 10 0.85
257.29
For T circuit representation (Medium Line)
1
1
4
1
2
1
1 10
90° 95 78°
.
2
1
1 10
95 78°
90° 95 78°
4
C=Y=0.001∠90°
D=A=0.9536∠0.6°
1
. °
.
. °
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 8
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
b) Sending end voltage
VS=AVR+BIR
VS= (0.9536∠0.6°) (76.21∠0°kV) + (92.79∠78.3°) (257.29∠-31.79°) =90.92∠11.5°kV
Sending end line to line voltage,
VLLS=90.92x√3=15.46kV
c) Sending end current is
IS=CVR+DIR
IS=0.001∠90(76.21∠0°kV) + (0.9536∠0.6°) (257.29∠-31.79°) = 215.18∠-14.27°
d) Sending end power factor
PF = Cosθ = Cos (11.5°+14.27°) =0.9
e) Efficiency of transmission line
√3
√3
100%
√3 132 257.29 0.85
√3 157.46 215.18 0.9
.
%
Example 10
Find the voltage at the generator end of the following three phase line:
Length of the line, km
320
Resistance, ohm/km
0.055
Reactance, ohm/km
0.505
Susceptance, mho/km
3.24x10-6
Load measured at the load end of the line is 60 MW at 0.80 power factor lagging at 132 kV, using
long line equation.
a) Find the ABCD of the line
b) Find the sending end voltage, current and power factor
Solution 10:
a) The ABCD constants,
Given,
z = r +jx = 0.055 + j0.505 = 0.508 ∠83.78°
y = j3.24x10-6 = 3.24x10-6∠90°
Also, the propagation constant;
3.24 10
90° 0.508 83.78°
1.283 10
86.89°
γl = (1.283x10-3) (320km) = 0.411∠86.89 = 0.022+ j0.4104
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 9
EXAMPLES
TOPIC II
LINE PARAMETERS & MODELS
Therefore,
Attenuation constant (α) and phase constant (β):
αl =0.022 and βl=0.4104
Cosh γl=cosh (αl + jβl) = (coshαl cosβl) + j (sinhαl sinβl)
=cosh (0.022) cos (0.4104) +j (sinh (0.022) sin (0.4104))
=0.917∠0.55°
Sinh γl=sinh (αl + jβl) = (sinhαl cosβl) + j (coshαl sinβl)
= (sinh (0.022) cos (0.4104)) + j (cosh (0.022) sin (0.4104))
=0.3995∠87.13°
Characteristic impedance:
0.508 83.8°
90°
3.24 10
Now we have,
.
.
°
A=cosh γl=0.917∠0.55°
B=ZC sinh γl = (395.7∠-3.11°) (0.3995∠87.13°) =158.19∠84.02°
1
0.3995 87.13°
395.97
3.11°
sinh
.
.
°
D=A
b) Sending end voltage, current and PF
132
√3
And
60
√3 132
0.8
76210 0°
36.86°
328.04
36.86°
VS=AVR+BIR=112067.92∠20.21° V =112.06792∠20.21°kV
Sending end line voltage:
√3 112.06972
.
Also,
Sending end line current:
IS=CVR+DIR=262.90∠-22.61°A
Sending end power factor PF:
PF=cos θ=cos (20.21°+22.61°) =0.73 lagging
EET 305 Power System Fundamental ~ Mohd Rafi (SoESE) 10