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

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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.