determination of inductance, capacitance, and resistance of the transmission line
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8/6/2019 Determination of Inductance, Capacitance, And Resistance of the Transmission Line
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EE-372 FUNDAMENTALS OF POWER SYSTEMS
Project-2 Determination of Inductance, Capacitance, and Resistance of the
Transmission Line
Submitted by
Mehmet Serdar Teke
ID: 260701050
Submitted to:Prof. Canbolat Uçak
Submission Date: 02.04.2010
Department of Electrical and Electronics Engineering
Engineering and Architecture Faculty
Yeditepe University
2010
8/6/2019 Determination of Inductance, Capacitance, And Resistance of the Transmission Line
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1. Current location
I am residing in the dormitory which is located in the campus of Yeditepe University.
2. The place where the photo of transmission line pole was taken
This two bundled transmission pole line was found on the road between Sinop and
Samsun. When my father said that he would go to Samsun for business, I wanted him to take
a picture of a transmission line pole on the road between Sinop and Samsun.
3. The photo of the transmission line pole
4. Determination of the distances between phases and between conductors of one
phase
The distances were determined approximately. The distance between the conductors of
one phase is 60 cm. The distance between phases is 7 m.
5. Simple drawing of the transmission line
8/6/2019 Determination of Inductance, Capacitance, And Resistance of the Transmission Line
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6. Determination of inductance, capacitance, and resistance per 1 meter
It is assumed that ACSR is used in the transmission line. Thus,
r = 1.75cm = 0.0175m
GMR = Ds = 1.42cm = 0.0142m
• R = 0.1208 Ω/km for AC with the frequency of 60 Hz at 20 0 C
• R = 0.1321 Ω/km for AC with the frequency of 60 Hz at 50 0 C
• R = 0.0217 Ω/km for DC at 20 0 C
Deq = (DAB x DBC x DAC)1/3 = (7m x 7m x 14m)1/3 = 8.819447349m
Ds b
= (Ds x d)1/2 = (0.0142m x 0.6m)1/2 = 0.092303846m
Dsc b
= (r x d)1/2 = (0.0175m x 0.6m)1/2 = 0.102469507m
Calculation of inductance
L = 2 x 10-7 x ln(Deq / Ds b) = 9.11925 x 10-7 H/m
Calculation of capacitance
C = (2 x π x Ɛ0) / ln(Deq / Dsc b) = 1.24813 x 10-11 F/m
Calculation of resistance
For AC with the frequency of 60 Hz at 200C:
R t = (0.1208Ω/km) x (0.001km/1m) = 0.0001208Ω/m
Since there are two bundles and they are creating parallel resistances:
R eq = R t / 2 = 0.0000604Ω/m
For AC with the frequency of 60 Hz at 500C:
R t = (0.1321Ω/km) x (0.001km/1m) = 0.0001321Ω/m
Since there are two bundles and they are creating parallel resistances:
R eq = R t / 2 = 0.00006605Ω/m
For DC at 200C:
R t = (0.0217Ω/km) x (0.001km/1m) = 0.0000217Ω/m
Since there are two bundles and they are creating parallel resistances:
R eq = R t / 2 = 0.00001085Ω/m
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Conclusion
If the distance between the conductors in one phase increases, Ds b and Dsc
b values
increase. The increase in Ds
b
causes a decrease in inductance of transmission line while theincrease in Dsc
b causes an increase in capacitance of transmission line. Therefore, the distance
between the conductors in one phase should be chosen to optimize the inductance and the
capacitance of the transmission line.
If the distance between the phases increases, Deq increases. The increase in Deq causes
the increase in the inductance of transmission line while the increase in Deq causes a decrease
in the capacitance of the transmission line. Therefore, this distance should also be chosen to
minimize both inductance and capacitance. Thus, an optimization in the length between
phases is also needed.