Question

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

$$\begin{gathered} {\mathit{Z}}_{\mathbf{1}}\mathbf{=}{\mathit{Z}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{5}\frac{\mathbf{\Omega }}{\mathbf{k}\mathbf{m}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{Z}}_{\mathbf{0}}\mathbf{=}\mathit{j}\mathbf{2}\frac{\mathbf{\Omega }}{\mathbf{k}\mathbf{m}} \\ \mathit{Y}\mathbf{}\mathbf{=}{\mathit{Y}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathit{j}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{}\frac{\mathbf{s}}{\mathbf{m}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{Y}}_{\mathbf{0}}\mathbf{=}\mathit{j}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{}\frac{\mathbf{s}}{\mathbf{m}} \end{gathered}$$

Set up the nominal $\mathbf{\pi }$sequence circuits of this medium-length line.

Short answer

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

Step-by-step solution

Write the given data from the question:

The length of the transmission line,$\mathit{I}\mathbf{=}\mathbf{200}\mathit{k}\mathit{m}$

Write the symmetrical impedances.

$$\begin{gathered} {\mathit{Z}}_{\mathbf{1}\mathbf{}}\mathbf{=}{\mathit{Z}}_{\mathbf{2}} \\ {\mathit{Z}}_{\mathbf{1}\mathbf{}}\mathbf{=}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{5}\frac{\mathbf{\Omega }}{\mathbf{k}\mathbf{m}} \\ {\mathit{Z}}_{\mathbf{0}\mathbf{}}\mathbf{=}\mathit{j}\mathbf{2}\frac{\mathbf{\Omega }}{\mathbf{k}\mathbf{m}} \end{gathered}$$

Write the symmetrical admittances.

$$\begin{gathered} {\mathit{Y}}_{\mathbf{1}}\mathbf{=}{\mathit{Y}}_{{\mathbf{2}}^{\mathbf{a}}} \\ {\mathit{Y}}_{\mathbf{1}}\mathbf{=}\mathit{j}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\frac{\mathbf{s}}{\mathbf{m}} \\ {\mathit{Y}}_{\mathbf{0}}\mathbf{=}\mathit{j}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\frac{\mathbf{s}}{\mathbf{m}} \end{gathered}$$

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.

${\mathit{Z}}_{\mathbf{0}\mathbf{A}}\mathbf{=}{\mathit{Z}}_{\mathbf{0}}\mathbf{I}$ …… (1)

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

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

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

${\mathit{Y}}_{\mathbf{0}\mathbf{A}}\mathbf{=}{\mathit{Y}}_{\mathbf{0}}\mathbf{I}$ …… (3)

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

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

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

${\mathbf{Y}}_{\mathbf{\pi }}\mathbf{=}\frac{{\mathbf{Y}}_{\mathbf{0}\mathbf{A}}}{\mathbf{2}}$ …… (5)

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

${\mathbf{Y}}_{\mathbf{\pi }}\mathbf{=}\frac{{\mathbf{Y}}_{\mathbf{1}\mathbf{A}}}{\mathbf{2}}$ …… (6)

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

Calculate the actual zero sequence impedance.

Substitute $\mathit{j}\mathbf{2}\frac{\mathbf{\Omega }}{\mathbf{km}}$for ${\mathit{Z}}_{\mathbf{0}}$and $\mathbf{200}\mathit{k}\mathit{m}$for $\mathit{I}$into equation (1).

$$\begin{gathered} {\mathbf{Z}}_{\mathbf{0}\mathbf{A}}\mathbf{=}\mathbf{j}\mathbf{2}\mathbf{\times }\mathbf{200} \\ {\mathbf{Z}}_{\mathbf{0}\mathbf{A}}\mathbf{=}\mathit{j}\mathbf{400}\mathbf{\Omega } \end{gathered}$$

Calculate the actual positive and negative sequence impedance.

Substitute$\mathit{j}\mathbf{0}\mathbf{.}\mathbf{5}\frac{\mathbf{\Omega }}{\mathbf{km}}$ for ${\mathit{Z}}_{\mathbf{1}}$and $\mathbf{200}\mathit{k}\mathit{m}$for $\mathit{I}$into equation (2).

localid="1656099633238" $$\begin{gathered} {\mathbf{Z}}_{\mathbf{1}\mathbf{A}}\mathbf{=}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\times }\mathbf{200} \\ {\mathbf{Z}}_{\mathbf{1}\mathbf{A}}\mathbf{=}\mathit{j}\mathbf{100}\mathbf{\Omega } \end{gathered}$$

Calculate the actual zero sequence admittance.

Substitutelocalid="1656099645854" $\mathit{j}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\frac{\mathbf{s}}{\mathbf{m}}$ for localid="1656099654097" ${\mathit{Y}}_{\mathbf{0}}$and $\mathbf{200}\mathit{k}\mathit{m}$for localid="1656099954448" $\mathit{I}$into equation (3).

localid="1656099661151" $$\begin{gathered} {\mathit{Y}}_{\mathbf{0}\mathbf{A}}\mathbf{=}\mathit{j}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{\times }\mathbf{200}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}} \\ {\mathit{Y}}_{\mathbf{0}\mathbf{A}}\mathbf{=}\mathit{j}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{s} \end{gathered}$$

Calculate the actual positive and negative sequence admittance.

Substitute localid="1656099668834" $\mathit{j}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\frac{\mathbf{s}}{\mathbf{m}}$for localid="1656099676164" ${\mathit{Y}}_{\mathbf{1}}$and $\mathbf{200}\mathit{k}\mathit{m}$for $\mathit{I}$into equation (4).

localid="1656099684620" $$\begin{gathered} {\mathit{Y}}_{\mathbf{1}\mathbf{A}}\mathbf{=}\mathit{j}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{\times }\mathbf{200}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}} \\ {\mathit{Y}}_{\mathbf{1}\mathbf{A}}\mathbf{=}\mathit{j}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{s} \end{gathered}$$

In the nominal localid="1656099692966" $\mathbf{\pi }$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 localid="1656099710990" $\mathbf{\pi }$network.

Substitutelocalid="1656099701227" $\mathit{j}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{s}$for localid="1656099719163" ${\mathit{Y}}_{\mathbf{0}\mathbf{A}}$into equation (5).

localid="1656099727195" $$\begin{gathered} {\mathit{Y}}_{\mathbf{0}\mathbf{\pi }}\mathbf{=}\frac{\mathbf{j}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}{\mathbf{2}} \\ {\mathit{Y}}_{\mathbf{0}\mathbf{\pi }}\mathbf{=}\mathit{j}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{s} \end{gathered}$$

Calculate the positive and negative sequence admittance of nominal localid="1656099745007" $\mathbf{\pi }$network.

Substitutelocalid="1656099735896" $\mathit{j}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{s}$for localid="1656099752791" ${\mathit{Y}}_{\mathbf{1}\mathbf{A}}$into equation (6).

localid="1656099761323" $$\begin{gathered} {\mathit{Y}}_{\mathbf{1}\mathbf{\pi }}\mathbf{=}\frac{\mathbf{j}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}{\mathbf{2}} \\ {\mathit{Y}}_{\mathbf{1}\mathbf{\pi }}\mathbf{=}\mathit{j}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{s} \end{gathered}$$

Therefore the localid="1656099770111" $\mathbf{\pi }$sequence network for the medium line