Transmission Line Inductance

Demonstrative Video

Problem-1

A single-phase, two-wire transmission line, 15 km long, is made up of round conductors, each 0.8 cm in diameter, separated from each other by 40 cm. Calculate the equivalent diameter of a fictitious hollow, thin-walled conductor having the same inductance as the original line. What is the value of this inductance?

Solution-1

\[2r^{'}=re^{-1/4}=0.8\times0.7788=0.623~\text{cm}\]

and whose diameter is therefore The fictitious conductor is one whose radius is

\[\begin{aligned} L & =4\times10^{-7}\times l\times ln\dfrac{GMD}{GMR}\\ & =4\times10^{-7}\times15\times10^{3}\times ln\dfrac{40}{0.5\times0.623}\\ & =29.13~\text{mH} \end{aligned}\]

Then the inductance of 15 km of such a conductor is

Problem-2

A seven identical copper strands conductor, each having a radius \(r\), is shown in the fig. Find the GMR of the conductor.

Solution-2

\[\begin{aligned} D_{s} & =\sqrt[49]{\left(r^{'}\right)^{7}\left(D_{12}^{2}D_{26}^{2}D_{14}D_{17}\right)^{6}\left(2r\right)^{6}}\\ & =\sqrt[49]{\left(0.7788r\right)^{7}\left(2^{2}r^{2}\times3\times2^{2}r^{2}\times2^{2}r\times2r\times2r\right)^{6}}\\ & =\dfrac{2r\times\sqrt[7]{3\times0.7788}}{\sqrt[49]{6}}=2.177r \end{aligned}\]

The GMR of the 7 strand conductor

Problem-3

The outside diameter of the single layer of aluminum strands of an ACSR conductor shown in fig. is 5.04 cm. The diameter of each strand is 1.68 cm. Determine the 50 Hz reactance at 1 m spacing; neglect the effect of the central strand of steel and advance reasons for the same.

Solution-3

\[\begin{aligned} D_{12} & =D_{16}=d\\ D_{13} & =D_{15}=\sqrt{3}d\\ D_{14} & =2d\\ D_{s} & =\sqrt[36]{\left[\left(\dfrac{d^{'}}{2}\right)d^{2}\left(\sqrt{3}d\right)^{2}\left(2d\right)\right]^{6}} \end{aligned}\]

. For the arrangement of the strands Thus all strands are of diameter 1.68 = 1.68 cmDiameter of steel stand = 5.04 - 2

Substituting \(d^{'}=0.7788d\) and simplifying

\(D_{s} = 1.155d = 1.155 \times 1.68 = 1.93~\text{cm}\) \(D_m \simeq D\) since \(D \gg d\)

\[L = 0.461 log \dfrac{100}{1.93} = 0.789 mH/km\]

\(1.578 \times 314 \times 10^{-3} = 0.495\)
\(2 \times 0.789 = 1.578\)Now, the inductance of each conductor is

Problem-4

One circuit of a single-phase transmission line is composed of three solid 0.25 cm radius wires. The return circuit is composed of two 0.5 cm radius wires. The arrangement of conductors is shown in the Fig. Find the inductance due to the current in each side of the line and the inductance of the complete line?

Solution-4

\[\begin{aligned} GMD=D_{m} & =\sqrt[6]{D_{ad}D_{ae}D_{bd}D_{be}D_{cd}D_{ce}}\\ D_{ad} & =D_{be}=9m\\ D_{ac} & =D_{bd}=D_{ce}=\sqrt{6^{2}+9^{2}}=\sqrt{117}\\ D_{cd} & =\sqrt{9^{2}+12^{2}}=15m\\ D_{m} & =\sqrt[6]{9^{2}\times15\times117^{3/2}}=10.743m \end{aligned}\]

and Find the GMD between

\[\begin{aligned} GMR & =D_{s}=\sqrt[9]{D_{aa}D_{ab}D_{ac}D_{ba}D_{bb}D_{bc}D_{ca}D_{cb}D_{cc}}\\ & =\sqrt[9]{\left(0.25\times0.7788\times10^{-2}\right)^{3}\times6^{4}\times12^{2}}=0.481m \end{aligned}\]

Find the GMR for side

and GMR for side \(Y\)

\[\begin{aligned} D_{s} & =\sqrt[4]{\left(0.5\times0.7788\times10^{-2}\right)^{2}\times6^{2}}=0.153~\text{m}\\ L_{x} & =2\times10^{-7}\times ln\dfrac{10.743}{0.481}=6.212\times10^{-7}H/m\\ L_{y} & =2\times10^{-7}\times ln\dfrac{10.743}{0.153}=8.503\times10^{-7}H/m\\ L & =L_{x}+L_{y}=14.715\times10^{-7}H/m \end{aligned}\]

Problem-5

Determine the inductance of a 3-phase line operating at 50 Hz and conductors arranged as follows. The conductor diameter is 0.8 cm.

Solution-5

\[=\dfrac{0.7788\times0.8}{2\times100}=0.003115~m\]

\[\begin{aligned} & =2\times10^{-4}ln\dfrac{2.015}{0.003115}\\ & =1.294~\text{mH/km} \end{aligned}\]

\(\therefore\)

\[=\sqrt[3]{1.6\times3.2\times1.6}=2.015~m\]

The self GMD or GMR of the conductor

Problem-6

Determine the inductance per km per phase of a single circuit 460 kV line using two bundle conductors per phase as shown in the fig. The diameter of each conductor is 5 cm.

Solution-6

\[\begin{aligned} D_{s} & =\sqrt{0.025\times0.4\times0.7788}=0.08825m\\ D_{m} & =\sqrt[3]{6.5\times13\times6.5}=8.19m \end{aligned}\]

Assuming the effect of transposition to be negligibly small,

Inductance per km/phase = \(2\times10^{-4}ln\dfrac{8.19}{0.08825}=0.906~\text{mH/km/phase}\)