EEL303: Power Engineering I - Tutorial 3
1. Find the self GMD for each conductor configuration shown in Figure 1.
1.7036r; (b)GMR = 1.6921r]
[(a)GMR =
Figure 1:
Solution:
(a)
GMR =
p
3
GMR1 × GMR2 × GMR3
√
GMR1 = GMR3 = 3 0.7788r × 2r × 4r = 1.84r
√
GMR2 = 3 0.7788r × 2r × 2r = 1.4604r
√
GMR = 3 1.84r × 1.4604r × 1.84r = 1.7036r
(b)
√
GMR1 = GMR3 = 4 0.7788r × 2r × 2r × 2r = 1.5798r
q
√
4
GMR2 = GMR4 = 0.7788r × 2r × 2r × 12r = 1.8124r
√
GMR = 4 1.5798r × 1.5798r × 1.8124r × 1.8124r = 1.6921r
2. Determine the inductance of a 3 phase line operating at 50 Hz and conductors arranged
as given in figure 2. The conductor diameter is 0.8 cm.
[L = 1.294mH/km]
Figure 2:
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EEL303: Power Engineering I - Tutorial 3
Solution:
GMR = 0.7788 × 0.004 = 0.0031152m
p
GMD = 6 (1.6 × 1.6)(1.6 × 3.2)(1.6 × 3.2) = 2.0158m
L = 2 × 10−7 ln
2.0158
= 1.294mH/km
0.0031152
3. Determine the inductance of a single phase transmission line consisting of three conductors of 2.5 mm radii in the ‘go’ conductor and 5 mm radii in the return conductor. The
configuration of line is as shown in figure 3.[(a)L = 1.42mH/km; (b)L = 1.485mH/km]
Figure 3:
Solution:
(a)
GMRA =
p
3
GMRa × GMRb × GMRc
√
GMRa = GMRc = 3 0.7788 × 0.0025 × 6 × 12 = 0.51947m
√
GMRb = 3 0.7788 × 0.0025 × 6 × 6 = 0.4123m
√
GMRA = 3 0.51947 × 0.4123 × 0.51947 = 0.4809m
GMRB =
p
4
(0.7788 × 5 × 6)(0.7788 × 5 × 6) = 0.1528m
GMDA = GMDB =
√
6
9 × 10.81 × 9 × 10.81 × 10.81 × 15 = 10.74m
10.74
LA = 2 × 10−7 ln
= 0.62mH/km
0.4809
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EEL303: Power Engineering I - Tutorial 3
LB = 2 × 10−7 ln
10.74
= 0.8mH/km
0.1528
L = LA + LB = 1.42mH/km
(b)
GMRA =
p
3
GMRa = GMRc =
√
3
GMRb =
GMRa × GMRb × GMRc
0.7788 × 0.0025 × 4 × 8 = 0.3964m
√
3
0.7788 × 0.0025 × 4 × 4 = 0.3146m
√
GMRA = 3 0.3964 × 0.3146 × 0.3964 = 0.367m
GMRB =
GMDA = GMDB =
q
6
√
√
0.7788 × 5 × 4 = 0.1248m
68 × 10 ×
LA = 2 × 10−7 ln
√
68 ×
√
68 ×
√
68 × 10 = 8.7936m
8.7936
= 0.635mH/km
0.367
LB = 2 × 10−7 ln
10.74
= 0.85mH/km
0.1248
L = LA + LB = 1.485mH/km
4. Determine the inductance per km of a transposed double circuit 3 phase line shown in
figure 4. Each circuit of the line remains on its own side. The dia of the conductor is
2.532 cm.
[L = 0.606mH/km]
Solution:
GMRA =
p
3
GMRa × GMRb × GMRc
p
′
4
ra′ Daa′ × ra′ Da′ a
GMRa =
√
= 4 0.7788 × 0.01266 × 10.965 × 0.7788 × 0.01266 × 10.965 = 0.3288m
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 3
Figure 4:
p
′
4
GMRc =
rc′ Dcc′ × rc′ Dc′ c
√
= 4 0.7788 × 0.01266 × 10.965 × 0.7788 × 0.01266 × 10.965 = 0.3288m
p
′
′
4
rb Dbb′ × rb′ Db′ b
GMRb =
√
= 4 0.7788 × 0.01266 × 9 × 0.7788 × 0.01266 × 9 = 0.3288m
√
GMRA = 3 0.3288 × 0.2978 × 0.3288 = 0.3181m
p
GMD = 3 GMDab × GMDbc × GMDca
p
√
GMDab = 4 Dab Dab′ Da′ b Da′ b′ = 4 4.069 × 9.168 × 9.168 × 4.069 = 6.1077m
p
√
GMDbc = 4 Dbc Dbc′ Db′ c Db′ c′ = 4 4.069 × 9.168 × 9.168 × 4.069 = 6.1077m
p
√
GMDca = 4 Dca Dca′ Dc′ a Dc′ a′ = 4 8 × 7.5 × 7.5 × 8 = 7.7459m
GMD = 6.610m
L = 0.606mH/km
5. Determine the inductance per km per phase of a double circuit 3 phase line arranged as
shown in figure 5. The radius of each conductor is 15 mm. [L = 0.51035mH/km/ph]
Figure 5:
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EEL303: Power Engineering I - Tutorial 3
Solution:
GMRA =
=
GMRa × GMRb × GMRc
p
4
GMRa =
p
4
p
3
′
ra′ Daa′ × ra′ Da′ a
0.7788 × 0.015 × (1.75 × 3) × 0.7788 × 0.015 × (1.75 × 3) = 0.2476m
Since GMRa = GMRb = GMRc , GMR = GMRa = 0.2476m.
p
√
4
Dab Dab′ Da′ b Da′ b′ = 4 1.75 × 7 × 3.5 × 1.75 = 2.9431m
p
√
GMDbc = 4 Dbc Dbc′ Db′ c Db′ c′ = 4 1.75 × 7 × 3.5 × 1.75 = 2.9431m
p
√
GMDca = 4 Dca Dca′ Dc′ a Dc′ a′ = 4 3.5 × 1.75 × 8.75 × 3.5 = 3.7m
√
GMD = 3 2.9431 × 2.9431 × 3.7 = 3.1766m
GMDab =
L = 0.51035mH/km/ph
6. Find the inductive reactance per phase in Ω/m and the capacitive reactance to neutral
in Ω − m for a three phase line that has equilaterally spaced conductors of ACSR ‘Dove.
The distance between lines of various phases is 3.048m and the operating frequency is
60 Hz. From standard table, Dove GMR = 0.00954024 m, Dia of Conductor = 0.02354
m.
[XL = 4347.86 × 10−7 Ω/m/ph, XC = 0.000265266 × 1012 Ω − m/ph]
Solution: Since conductors are spaced equilaterally,
GMD = 3.048m
GMR = 0.00954m
3.048
= 11.5334 × 10−7 H/m/ph
L = 2 × 10−7 ln
0.00954
XL = 2 × 3.1415 × 60 × 11.5334 × 10−7 = 4347.86 × 10−7 Ω/m/ph
For calculation of C, we have to use radius of conductor as GMR.
C=
XC =
2 × π × ǫ0
= 10 × 10−12 F/m
3.048
ln
0.01177
1
= 0.000265266 × 1012 Ω − m/ph
−12
2 × 3.1415 × 60 × 10 × 10
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 3
7. A 3 phase double circuit line is composed of 300,000-cmil 26/7 Ostrich conductors arranged as shown in Figure 6. Find the 60 Hz inductive reactance and capacitive reactance
in Ω/km/ph and Ω − km/ph. From standard table, Ostrich GMR = 0.00697 m, Dia of
Conductor = 0.017272m.
[XL = 2311 × 10−7 Ω/m/ph ,
XC = 0.000141117 × 1012 Ω − m/ph]
Figure 6:
Solution:
p
√
4
Dab Dab′ Da′ b Da′ b′ = 4 3.082 × 6.6792 × 6.6792 × 3.082 = 4.5371m
p
√
GMDbc = 4 Dbc Dbc′ Db′ c Db′ c′ = 4 3.082 × 6.6792 × 6.6792 × 3.082 = 4.5371m
p
√
GMDca = 4 Dca Dca′ Dc′a Dc′ a′ = 4 6.096 × 5.4864 × 5.4864 × 6.096 = 5.7831m
GMDab =
GMD =
GMRa =
q
4
√
3
4.5371 × 4.5371 × 5.7831 = 4.9192m
′
ra′ Daa′ × ra′ Da′ a =
√
4
0.00697 × 8.201 × 0.00697 × 8.201 = 0.239m
Similarly, GMRc = 0.239m.
q
√
′
GMRb = 4 rc′ Dcc′ × rc′ Dc′ c = 4 0.00697 × 6.4 × 0.00697 × 6.4 = 0.2112m
√
3
0.239 × 0.2112 × 0.239 = 0.2293m
4.9192
L = 2 × 10−7ln
= 6.131 × 10−7 H/m/ph
0.2293
GMR =
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 3
XL = 2 × 3.1415 × 60 × 6.131 × 10−7 = 2311 × 10−7 Ω/m/ph
For calculation of capacitance of the line, r=0.008636 m.
√
GMRa = GMRc = 0.008636 × 8.2 = 0.266m
√
GMRb = 0.008636 × 6.4 = 0.2350m
√
GMR = 3 0.266 × 0.2350 × 0.266 = 0.255m
2 × π × ǫ0
= 18.797 × 10−12 F/m
C=
4.9192
ln
0.255
1
XC =
= 0.000141117 × 1012 Ω − m/ph
2 × 3.1415 × 60 × 18.797 × 10−12
8. Each conductor of the bundled conductor line shown in figure. is ACSR, ’pheasant’.
Find (a) the inductive reactance in ohms per km per phase for d=45 cm. (b) the per
unit series reactance of the line if its length is 160 km and the base is 100 MVA, 345
kV. (c) the capacitive reactance to neutral of the line in ohm-km/phase. From standard
table, GMR=0.0466 ft (convert this into meters), outside diameter of each conductor is
1.382 inches (convert this into meters) and system frequency=60 Hz.
[(a)XL =
6
0.365Ω/km/phase; (b)Xp.u = 0.049p.u; (c)Xcn = 0.22589 × 10 Ω − km/ph to neutral]
Figure 7:
Solution:
(a)
GMR of each conductor = 0.0466 × 0.3048 = 0.014204m
√
GMRa = GMRb = GMRc = 0.014204 × 0.45 = 0.08m
p
√
GMD = 3 Dab × Dbc × Dca = 3 8 × 8 × 16 = 10.08m
XL = 2 × π × 60 × 2 × 10−7 × ln
10.08
= 0.365Ω/km/phase
0.08
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 3
(b)
(basekV )2
(345 × 103 )2
=
= 1190.25Ω
baseMV A
100 × 106
Xactual
0.365 × 160
=
=
= 0.049p.u
Zbase
1190
Baseimpedance, Z =
Xp.u
(c)
1.382 × 2.54
= 0.01755m
2 × 100
√
GMRa = GMRb = GMRc = 0.01755 × 0.45 = 0.0889m
p
√
GMD = 3 Dab × Dbc × Dca = 3 8 × 8 × 16 = 10.08m
r=
2πǫo
2 × π × 8.842 × 10−12
= 11.743 × 10−12 F/m
=
10.08
GMD
ln
ln
0.0889
GMR
12
−3
1
10 × 10
=
=
= 0.22589 × 106 Ω − km/ph to neutral
2πf C
2 × π × 60 × 11.743
Cn =
Xcn
9. A 3 phase 60 Hz transmission line has its conductors arranged in a triangular formation
so that two of the distances between conductors are 25 ft and the third is 42 ft. The
conductors are ACSR Osprey. Determine the capacitance to neutral in microfarads per
mile and the capacitive reactance to neutral in Ω-miles. If the line is 150 miles long, find
the capacitance to neutral and capacitive reactance of the line. From standard table,
GMR=0.0284 ft (convert this into meters), outside diameter of each conductor is 0.879
inches (convert this into meters).
[Cn = 13.34 × 10−9F/mi; Xc =
6
0.1988 × 10 Ω − mi; Cn,T otal = 2.001µF ; Xc,T otal = 1325Ω]
Solution:
GMD =
√
3
25 × 42 × 25 = 29.72f t
r = 0.879/2 = 0.4395inches
2πǫo
2 × π × 8.842 × 10−12
=
= 8.293 × 10−9 F/km
29.72 × 12
GMD
ln
ln
0.4395
r
−9
= 8.293 × 10 × 1.609F/mi = 13.34 × 10−9 F/mi
109
1
=
= 0.1988 × 106 Ω − mi
Xc =
2πf C
2 × π × 60 × 13.34
Cn,T otal = 150 × 13.34 × 10−9 = 2.001µF
Cn =
Xc,T otal =
0.1988 × 106
= 1325Ω
150
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 3
10. Determine the capacitance and charging current per km when transmission line conductors are
(a) placed as given in figure 2 and operated at 132 kV. Radius of conductor is 0.4 cm.
(b) spaced horizontally with 3 m spacing. Dia of conductor is 20 mm.
[(a)C =
8.928 × 10−9 F/km, Ic = 0.2136A/km; (b)C = 9.356 × 10−9 F/km, Ic = 0.224A/km]
Take frequency as 50 Hz.
Solution:
(a)
GMD = 2.015m
2 × π × ǫ0
= 8.928 × 10−9 F/km
C=
2.015
ln
0.004
Ic = V × w × C = 0.2136A/km
(b)
√
GMD = 3 3 × 3 × 6 = 3.7797m
2 × π × ǫ0
C=
= 9.356 × 10−9 F/km
3.7797
ln
0.01
Ic = V × w × C = 0.224A/km
11. Determine the capacitance and charging current per km when the transmission line given
in figure 4 operates at 220 kV - 50 Hz, diameter of conductor is 2.5 cm.
[C =
0.019056µF/km, Ic = 0.76A/km]
Solution:
GMD = 6.61m
√
GMRa = GMRc = 0.0125 × 10.965 = 0.3702m
√
GMRb = 0.0125 × 9 = 0.3354m
√
GMR = 3 0.3702 × 0.3354 × 0.3702 = 0.3582m
2 × π × ǫ0
C=
= 0.019056µF/km
6.61
ln
0.3582
Ic = V × w × C = 0.76A/km
Electrical Engineering Dept - IIT Delhi