Question

A completely transposed three-phase transmission line of ${}^{\mathbf{200}\mathbf{}\mathbf{k}\mathbf{m}}$in length has the following symmetrical sequence impedances and sequence admittances:

$$\begin{gathered} {\mathit{Z}}_{\mathbf{1}}\mathbf{=}{\mathit{Z}}_{\mathbf{2}}\mathbf{=}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{5}\frac{\mathbf{\Omega }}{\mathbf{k}\mathbf{m}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{Z}}_{\mathbf{0}}\mathbf{=}\mathit{j}\mathbf{2}\frac{\mathbf{\Omega }\mathbf{}}{\mathbf{k}\mathbf{m}} \\ {\mathit{Y}}_{\mathbf{1}}\mathbf{=}{\mathit{Y}}_{\mathbf{2}}\mathbf{=}\mathit{j}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\frac{\mathbf{s}}{\mathbf{m}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{Y}}_{\mathbf{0}}\mathbf{=}\mathit{j}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\frac{\mathbf{s}}{\mathbf{m}} \\ \mathit{S}\mathit{e}\mathit{t}\mathbf{}\mathit{u}\mathit{p}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{n}\mathit{o}\mathit{m}\mathit{i}\mathit{n}\mathit{a}\mathit{l}\mathbf{}\mathbf{\pi }\mathbf{}\mathbf{sequence}\mathbf{}\mathbf{circuits}\mathbf{}\mathbf{of}\mathbf{}\mathbf{this}\mathbf{}\mathbf{medium}\mathbf{-}\mathbf{length}\mathbf{}\mathbf{line}\mathbf{.} \end{gathered}$$

Short answer

The $\mathrm{\pi }$sequence network for the medium line

Step-by-step solution

Write the given data from the question:

The length of the transmission line, $/=200k{m}^{}$

Write the symmetrical impedances.

$$\begin{gathered} Z_1=Z_2 \\ {Z}_1=j0.5\frac{\Omega }{km} \\ {Z}_0=j2\frac{\Omega }{km} \\ Writethesymmetricaladmit\tan ces. \\ {Y}_1=Y_{2a} \\ {Y}_1=j_3\times {10}^{-9}\frac{s}{m} \\ {Y}_0=j_1\times {10}^{-9}\frac{s}{m} \end{gathered}$$

Determine the formulas the calculate the actual impedance and admittance for transmission line.

The equation to calculate the actual zero sequence impedance is given as follows.

${\mathit{Z}}_{\mathbf{0}\mathbf{a}}\mathbf{=}{\mathit{Z}}_{\mathbf{0}\mathbf{I}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(1\right)}$

The equation to calculate the actual positive and negative sequence impedance is given as follows.

$$\begin{gathered} {\mathit{Z}}_{\mathbf{1}\mathbf{A}}\mathbf{=}{\mathit{Z}}_{\mathbf{2}\mathbf{A}} \\ {\mathit{Z}}_{\mathbf{1}\mathbf{A}}\mathbf{=}{\mathit{Z}}_{\mathbf{1}\mathbf{I}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}······\left(2\right) \end{gathered}$$

The equation to calculate the actual zero sequence admittance is given as follows.

${\mathit{Y}}_{\mathbf{0}\mathbf{A}}\mathbf{=}{\mathit{Y}}_{\mathbf{0}\mathbf{I}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(3\right)}$

The equation to calculate the actual positive and negative sequence admittance is given as follows.

$$\begin{gathered} {\mathit{Y}}_{\mathbf{1}\mathbf{A}}\mathbf{=}{\mathit{Y}}_{\mathbf{2}\mathbf{A}} \\ {\mathit{Y}}_{\mathbf{1}\mathbf{A}}\mathbf{=}{\mathit{Y}}_{\mathbf{1}\mathbf{I}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}·······\left(4\right) \end{gathered}$$

The equation to calculate the zero-sequence admittance of nominal $\mathrm{\pi }$network is given as follow.

${\mathit{Y}}_{\mathbf{\pi }}\mathbf{=}\frac{{\mathbf{Y}}_{\mathbf{0}\mathbf{A}}}{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}······\left(5\right)$

The equation to calculate the positive and negative sequence admittance of the parallel limbs of nominal $\mathrm{\pi }$network is given as follow.

${\mathit{Y}}_{\mathbf{\pi }}\mathbf{=}\frac{{\mathbf{Y}}_{\mathbf{1}\mathbf{A}}}{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}......\left(6\right)$

Set up the nominal π sequence circuits of this medium-length line.

Calculate the actual zero sequence impedance.

Substitute for and for into equation (1).

$$\begin{gathered} Z_{0A=}j2\times 200 \\ {Z}_{0A}=j400\Omega \end{gathered}$$

Calculate the actual positive and negative sequence impedance.

Substitute $j0.5^{}$$\frac{\Omega }{km}$for $Z_1$and 200 km for $/$into equation (2).

$$\begin{gathered} Z_{1A}=j0.5\times 200 \\ {Z}_{1A}=j100\Omega \end{gathered}$$

Calculate the actual zero sequence admittance.

Substitute j1$\times {10}^{-9}\frac{s}{m}for{Y}_{0}and200kmfor/intoequation\left(3\right).$

$$\begin{gathered} Y_{0A}=j1\times {10}^{-9}\times 200\times {10}^3 \\ {Y}_{0A}=j2\times {10}^{-4}s \\ Calculatetheactualpositiveandnegativesequenceadmit\tan ce. \\ Substitutej3\times {10}^{-9\frac{s}{m}for{Y}_{1}and200kmfor/intuequation\left(4\right).} \end{gathered}$$

$$\begin{gathered} Y_{1A}=j3\times {10}^{-9}\times 200\times {10}^3 \\ {Y}_{1A}=j6\times {10}^{-4}s \end{gathered}$$

In the nominal $\mathrm{\pi }$sequence network, the admittance is represented on parallel limbs and impedance is represented between the two parallel limbs.

Calculate the zero-sequence admittance of nominal $\mathrm{\pi }$network.

Substitute localid="1654499255284" $j2\times {10}^{-4}s$ for localid="1654499267695" $Y_{0A}$into equation localid="1654499281860" $\left(5\right)$.

localid="1654499405997" $$\begin{gathered} Y_{0\mathrm{\pi }}=\frac{j2\times {10}^{-4}}{2} \\ {Y}_{0\mathrm{\pi }}=j1\times {10}^{-4}s \end{gathered}$$

Calculate the positive and negative sequence admittance of nominal $\mathrm{\pi }$network.

Substitute localid="1654499445009" $j6\times {10}^{-4}sfor{Y}_{1A}$for into equation (6).

localid="1654499454972" $$\begin{gathered} Y_{1\mathrm{\pi }}=\frac{j6\times {10}^{-4}}{2} \\ {Y}_{1\mathrm{\pi }}=j3\times {10}^{-4}s \end{gathered}$$

Therefore the $\mathrm{\pi }$sequence network for the medium line