Question

The conductor configuration of a bundled single-phase overhead transmission line is shown in Figure 4.31. Line x has its three conductors situated at the corners of an equilateral triangle with spacing. Line$\mathbf{Y}$has its three conductors arranged in a horizontal configuration with$\mathbf{10}\mathbf{}\mathbf{cm}$spacing. All conductors are identical, solid-cylindrical conductors each with a radius of$\mathbf{2}\mathbf{cm}$. Find the equivalent representation in terms of the geometric mean radius of each bundle and a separation that is the geometric mean distance.

Short answer

The geometric mean distance, geometric mean radius for line$X$and $Y$are for line $6.149\mathrm{m}$,$0.0537\mathrm{m}$ and$0.0626\mathrm{m}$ respectively.

Step-by-step solution

Write the given data from the question.

The number of sub-conductors in conductor$X,N=3$

The number of sub-conductors in conductor$Y,m=3$

The radius of the sub-conductors of conductor$X,r_x=0.02\mathrm{cm}$

The radius of the sub-conductors of conductor$Y,r_y=0.02\mathrm{cm}$

The spacing between the conductors is$0.1\mathrm{m}$

Determine the formula geometric mean distance and geometric mean radius of each bundle.

The equation to calculate the geometric mean radius of the conductor $\mathrm{X}$as follows,

$$\begin{gathered} {\mathbf{D}}_{\mathbf{xx}}\mathbf{=}\sqrt[{\mathbf{N}}^{\mathbf{2}}]{\mathbf{\prod }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{N}}\mathbf{\prod }_{\mathbf{m}\mathbf{=}\mathbf{1}}^{\mathbf{N}}{\mathbf{D}}_{\mathbf{km}}\mathbf{}\mathbf{}\mathbf{}} \\ {\mathbf{D}}_{\mathbf{xx}}\mathbf{=}\sqrt[{\mathbf{3}}^{\mathbf{2}}]{\mathbf{\prod }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{3}}\mathbf{\prod }_{\mathbf{m}\mathbf{=}\mathbf{1}}^{\mathbf{N}\mathbf{3}}{\mathbf{D}}_{\mathbf{km}}\mathbf{}\mathbf{}\mathbf{}} \\ {\mathbf{D}}_{\mathbf{xx}}\mathbf{=}\sqrt[\mathbf{9}]{\mathbf{(}{\mathbf{D}}_{\mathbf{11}}\mathbf{}{\mathbf{D}}_{\mathbf{12}}{\mathbf{D}}_{\mathbf{13}}\mathbf{)}\left(D_{21}{D}_{22}{D}_{23}\right)\left(D_{31}{D}_{32}\right)\mathbf{}} \end{gathered}$$

Here represents the distance between the conductors and number$\mathbf{1}\mathbf{,}\mathbf{2}$and$\mathbf{3}$represents the number of the sub-conductors$\mathbf{x}$.

The equation to calculate the geometric mean radius of the conductor as follows,

$$\begin{gathered} {\mathbf{D}}_{\mathit{Y}\mathit{Y}}\mathbf{=}\sqrt[{\mathit{M}}^{\mathbf{2}}]{\mathbf{\prod }_{\mathbf{k}\mathbf{=}\mathbf{1}}^M\mathbf{\prod }_{\mathbf{m}\mathbf{=}\mathbf{1}}^M{\mathbf{D}}_{\mathbf{km}}\mathbf{}\mathbf{}\mathbf{}} \\ {\mathbf{D}}_{\mathit{Y}\mathit{Y}}\mathbf{=}\sqrt[{\mathbf{3}}^{\mathbf{2}}]{\mathbf{\prod }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{3}}\mathbf{\prod }_{\mathbf{m}\mathbf{=}\mathbf{1}}^{\mathbf{N}\mathbf{3}}{\mathbf{D}}_{\mathbf{km}}\mathbf{}\mathbf{}\mathbf{}} \\ {\mathbf{D}}_{\mathbf{Y}\mathbf{Y}}\mathbf{=}\sqrt[\mathbf{9}]{\mathbf{(}{\mathbf{D}}_{\mathbf{1}\mathbf{\text{'}}\mathbf{1}\mathbf{\text{'}}}\mathbf{}{\mathbf{D}}_{\mathbf{1}\mathbf{\text{'}}\mathbf{2}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{1}\mathbf{\text{'}}\mathbf{3}\mathbf{\text{'}}}\mathbf{)}\mathbf{\left(}{\mathbf{D}}_{\mathbf{2}\mathbf{\text{'}}\mathbf{1}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{2}\mathbf{\text{'}}\mathbf{2}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{2}\mathbf{\text{'}}\mathbf{3}\mathbf{\text{'}}}\mathbf{\right)}\mathbf{\left(}{\mathbf{D}}_{\mathbf{3}\mathbf{\text{'}}\mathbf{1}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{3}\mathbf{\text{'}}\mathbf{2}\mathbf{\text{'}}}\mathbf{\right)}\mathbf{}} \end{gathered}$$

Here represents the distance between the sub-conductors and number $\mathbf{1}\mathbf{\text{'}}\mathbf{,}\mathbf{2}\mathbf{\text{'}}$and$\mathbf{3}\mathbf{\text{'}}$represents the number of the sub-conductors$\mathbf{Y}$.

The equation to calculate the geometric mean distance between the sub-conductors of $X$and $Y$.

$$\begin{gathered} {\mathbf{D}}_{\mathit{X}\mathit{Y}}\mathbf{=}\sqrt[\mathit{N}\mathit{M}]{\mathbf{\prod }_{\mathbf{k}\mathbf{=}\mathbf{1}}^N\mathbf{\prod }_{\mathbf{m}\mathbf{=}\mathbf{1}}^M{\mathbf{D}}_{\mathbf{km}}\mathbf{}\mathbf{}\mathbf{}} \\ {\mathbf{D}}_{\mathit{X}\mathit{Y}}\mathbf{=}\sqrt[\mathbf{3}\mathbf{\times }\mathbf{3}]{\mathbf{\prod }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{3}}\mathbf{\prod }_{\mathbf{m}\mathbf{=}\mathbf{1}}^{3\text{'}}{\mathbf{D}}_{\mathbf{km}}\mathbf{}\mathbf{}\mathbf{}} \\ {\mathbf{D}}_{\mathbf{XY}}\mathbf{=}\sqrt[\mathbf{9}]{\mathbf{(}{\mathbf{D}}_{\mathbf{11}\mathbf{\text{'}}}\mathbf{}{\mathbf{D}}_{\mathbf{12}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{13}\mathbf{\text{'}}}\mathbf{)}\mathbf{\left(}{\mathbf{D}}_{\mathbf{21}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{22}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{23}\mathbf{\text{'}}}\mathbf{\right)}\mathbf{\left(}{\mathbf{D}}_{\mathbf{31}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{32}\mathbf{\text{'}}}\mathbf{\right)}\mathbf{}} \end{gathered}$$

Calculate the geometric mean distance and geometric mean radius of each bundle.

The distance between the sub-conductors $1$of conductor $X$and sub conductors of $Y$.

$$\begin{gathered} D_{11\text{'}}=6.1\mathrm{m} \\ {\mathrm{D}}_{12\text{'}}=6.2\mathrm{m} \\ {\mathrm{D}}_{13\text{'}}=6.3\mathrm{m} \end{gathered}$$

The distance between the sub-conductors $2$of conductor $X$and sub conductors of $Y$.

$$\begin{gathered} D_{21\text{'}}=6\mathrm{m} \\ {\mathrm{D}}_{22\text{'}}=6.1\mathrm{m} \\ {\mathrm{D}}_{23\text{'}}=6.2\mathrm{m} \end{gathered}$$

The mean distance between the sub conductor $3$and $1\text{'}$.

$$\begin{gathered} D_{31\text{'}}=\sqrt{{\left(0.1^2-0.{05}^2\right)}^2+6.{05}^2} \\ {D}_{31\text{'}}=\sqrt{36.602} \\ {D}_{31\text{'}}=6.05\mathrm{m} \end{gathered}$$

The mean distance between the sub conductor $3$and $2\text{'}$.

$$\begin{gathered} D_{32\text{'}}=\sqrt{{\left(0.1^2-0.{05}^2\right)}^2+6.{15}^2} \\ {D}_{32\text{'}}=\sqrt{37.822} \\ {D}_{32\text{'}}=6.15\mathrm{m} \end{gathered}$$

The mean distance between the sub conductor $3$and $3$.

$$\begin{gathered} D_{31\text{'}}=\sqrt{{\left(0.1^2-0.{05}^2\right)}^2+6.{25}^2} \\ {D}_{31\text{'}}=\sqrt{39.062} \\ {D}_{31\text{'}}=6.25\mathrm{m} \end{gathered}$$

Calculate the geometric mean distance.

Substitute forrole="math" localid="1655199083941" $$\begin{gathered} 6.1\mathrm{m}\mathrm{for}{D}_{11\text{'}},6.2\mathrm{m}\mathrm{for}6.2\mathrm{m},6.3\mathrm{m}\mathrm{for}{D}_{13\text{'}},6\mathrm{m}\mathrm{for}{D}_{21\text{'}},for{D}_{22\text{'}}, \\ 6.2mfor{D}_{23\text{'},}6.05m\mathrm{for}{D}_{32\text{'},}\mathrm{and}6.25\mathrm{m}\mathrm{for}{D}_{31\text{'},} \end{gathered}$$

into equation (3).

$$\begin{gathered} D_{XY}=\sqrt[9]{\left(6.\times 6.2\times 6.3\right)\left(6\times 6.1\times 6.2\right)\left(6.05\times 6.15\times 6.25\right)} \\ {D}_{XY}=\sqrt[9]{12573186.473} \\ {D}_{XY}=6.149\mathrm{m} \end{gathered}$$

Hence the geometric mean distance$D_{XY}$is $6.149\mathrm{m}$

The distance for the calculation of the geometric mean radius for line $X$.

$$\begin{gathered} D_{11}=D_{22} \\ {D}_{11}=D33\text{'} \\ {D}_{11}=0.7788\times 0.02 \\ {D}_{11}=0.0155\mathrm{m} \end{gathered}$$

The distance between conductors 1 and 2.

$$\begin{gathered} D_{12}=D_{21} \\ {D}_{12}=0.1\mathrm{m} \end{gathered}$$

The distance between conductors 2 and 3.

$$\begin{gathered} D_{23}=D_{32} \\ {D}_{23}=0.1\mathrm{m} \end{gathered}$$

The distance between conductors 1 and 3.

Calculate the geometric mean distance for the line $X$.

Substitute$0.0155\mathrm{m}$for $D_{11\text{'}}{D}_{22}\&D_{33}$and $0.1\mathrm{m}$for$D_{12\text{'}}{D}_{21\text{'}}{D}_{13\text{'}}{D}_{31}{D}_{23}\&D_{32}$into equation (1).

$$\begin{gathered} D_{XX}=\sqrt[9]{\left(0.0155\times 0.1\times 0.1\right)\left(0.1\times 0.0155\times 0.1\right)\left(0.1\times 0.1\times 0.0155\right)} \\ {D}_{XX}=\sqrt[9]{3.72\times {10}^{-12}} \\ {D}_{XX}=0.0537 \end{gathered}$$

The distances for the calculation of the geometric mean radius for line$Y$.

$$\begin{gathered} D_{1\text{'}1\text{'}}=D_{2\text{'}2\text{'}} \\ {D}_{1\text{'}1\text{'}}=D_{3\text{'}3\text{'}} \\ {D}_{1\text{'}1\text{'}}=0.7788\times 0.02 \\ {D}_{1\text{'}1\text{'}}=0.0155\mathrm{m} \end{gathered}$$

The distance between conductors $1\text{'}$and $2\text{'}$.

$$\begin{gathered} D_{1\text{'}2\text{'}}=D_{2\text{'}1\text{'}} \\ {D}_{1\text{'}2\text{'}}=0.1\mathrm{m} \end{gathered}$$

The distance between conductorrole="math" localid="1655200668091" $2\text{'}$and $3\text{'}$.

$$\begin{gathered} D_{2\text{'}3\text{'}}=D_{2\text{'}3\text{'}} \\ {D}_{2\text{'}3\text{'}}=0.1\mathrm{m} \end{gathered}$$

The distance between conductors$1\text{'}$and $3\text{'}$.

$$\begin{gathered} D_{1\text{'}3\text{'}}=D_{3\text{'}1\text{'}} \\ {D}_{1\text{'}3\text{'}}=0.2\mathrm{m} \end{gathered}$$

Calculate the geometric mean distance for the line .

Substitute for$0.0155\mathrm{m}\mathrm{for}{D}_{1\text{'}1\text{'},}{D}_{2\text{'}2\text{'}}\&D_{3\text{'}3\text{'},}0.1\mathrm{m}$for $D_{1\text{'}2\text{'},}{D}_{2\text{'}3\text{'},}\&D_{3\text{'}2\text{'},}0.2\mathrm{m}\mathrm{for}{\mathrm{D}}_{1\text{'}3\text{'},}\&{\mathrm{D}}_{3\text{'}1\text{'}}$

$$\begin{gathered} D_{YY}=\sqrt[9]{\left(0.0155\times 0.1\times 0.2\right)\left(0.1\times 0.0155\times 0.1\right)\left(0.2\times 0.1\times 0.0155\right)} \\ {D}_{YY}=\sqrt[9]{1.489\times {10}^{-11}} \\ {D}_{YY}=0.0626\mathrm{m} \end{gathered}$$

Therefore, the geometric mean distance, geometric mean radius for line$X$ and $Y$are for line $6.149\mathrm{m},0.0537\mathrm{m}$, and $0.0606\mathrm{m}$respectively.