Question

The line of problem 4.23 as shown in figure 4.32 is operating at$\mathbf{60}\mathbf{Hz}$determine (a) the line-to-neutral capacitance in$\mathbf{nF}\mathbf{/}\mathbf{km}$per phase and in $\mathbf{nF}\mathbf{/}\mathbf{mi}$

Per phase; (b) the capacitive reactance in$\mathbf{\Omega }\mathbf{-}\mathbf{km}$per phase and in $\mathbf{\Omega }\mathbf{-}\mathbf{mi}$per phase; and (c) the capacitive reactance in $\mathbf{\Omega }$per phase for a line length of$\mathbf{100}\mathbf{mi}$.

Short answer

(a) The capacitance to neutral is $10.949\mathrm{nF}/\mathrm{km}\mathrm{and}17.616\mathrm{nF}/\mathrm{mi}.$

(b) The capacitive reactance per phase is $2.43\times {10}^{-5}\mathrm{\Omega }-\mathrm{km}\mathrm{and}1.505\times {10}^{-5}\mathrm{\Omega }-\mathrm{mi}$.

(c) The capacitive reactance in ohm’s per phase is$1.505\times {10}^{-3}\mathrm{\Omega }$.

Step-by-step solution

Write the given data of the question

Consider the given figure shown below.

The bundle spacing is $30\mathrm{cm}$.

The distances between the conductor is given by,

$D_{AB}=6\mathrm{m},D_{BC}=6\mathrm{m}\mathrm{and}{D}_{\mathit{A}\mathit{C}}=12\mathrm{m}.$

Write the formulae for capacitance to neutral and admittance to neutral;

$$\begin{gathered} \mathbf{The}\mathbf{}\mathbf{capacitance}\mathbf{}\mathbf{to}\mathbf{}\mathbf{neutral}\mathbf{}\mathbf{is}\mathbf{}\mathbf{given}\mathbf{}\mathbf{by}\mathbf{,} \\ {\mathbf{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau }}{\mathbf{In}\left({\displaystyle \frac{D_{\mathrm{eq}}}{D_{\mathrm{sc}}}}\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathbf{Here}\mathbf{}{\mathbf{D}}_{\mathbf{eq}}\mathbf{}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{equivalent}\mathbf{}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{}\mathbf{the}\mathbf{}\mathbf{conductors}\mathbf{,}{\mathbf{D}}_{\mathbf{sc}}\mathbf{}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the} \\ \mathbf{equivalent}\mathbf{}\mathbf{radius}\mathbf{}\mathbf{of}\mathbf{}\mathbf{two}\mathbf{}\mathbf{conductor}\mathbf{}\mathbf{bundle}\mathbf{.} \\ \mathbf{The}\mathbf{}\mathbf{capacitive}\mathbf{}\mathbf{reactance}\mathbf{}\mathbf{is}\mathbf{}\mathbf{given}\mathbf{}\mathbf{by}\mathbf{,} \\ {\mathbf{X}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}} \\ \mathbf{Here}\mathbf{}{\mathbf{X}}_{\mathbf{C}}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{capacitive}\mathbf{}\mathbf{reactance}\mathbf{,}\mathbf{\omega }\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{frequency}\mathbf{}\mathbf{and}\mathbf{}\mathbf{C}\mathbf{}\mathbf{is} \\ \mathbf{the}\mathbf{}\mathbf{capacitance}\mathbf{.}\mathbf{} \end{gathered}$$

Calculate the capacitance to neutral.

(a)

Now consider the given diagram is completely transposed three-phase overhead transmission line with bundled phase conductors.

The bundle spacing between the conductors is given by,

$$\begin{gathered} d=30\mathrm{cm} \\ =0.30\mathrm{m} \\ \mathrm{The}\mathrm{radius}\mathrm{of}\mathrm{each}\mathrm{conductor}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{r}=0.74\mathrm{cm} \\ =0.0074\mathrm{m} \\ \mathrm{The}\mathrm{equivalent}\mathrm{radius}\mathrm{of}\mathrm{two}\mathrm{conductors}\mathrm{is}\mathrm{calculated}\mathrm{by},\mathrm{\epsilon } \\ {D}_{\mathit{s}\mathit{c}}\mathit{=}\sqrt{\mathit{r}\mathit{d}} \\ Here\mathit{}r\mathrm{is}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{each}\mathrm{conductor}\mathrm{and}\mathit{}d\mathit{}\mathrm{is}\mathrm{the}\mathrm{bundle}\mathrm{spacing}\mathrm{between}\mathrm{the}\mathrm{conductors}. \\ \mathrm{substitute}0.0074\mathrm{m}\mathrm{for}\mathit{}r\mathit{}\mathrm{and}0.3\mathrm{m}\mathrm{for}d\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {D}_{\mathrm{sc}}=\sqrt{\left(0.0074\right)(0.3)} \\ =0.0471\mathrm{m} \\ \mathrm{The}\mathrm{equivalent}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{conductors}\mathrm{is}\mathrm{calculated}\mathrm{by}, \\ {D}_{\mathit{e}\mathit{q}}\mathit{=}\sqrt[\mathit{3}]{\mathit{}{\mathit{D}}_{\mathit{A}\mathit{B}}{\mathit{D}}_{\mathit{B}\mathit{C}}{\mathit{D}}_{\mathit{C}\mathit{A}}\mathit{}\mathit{}} \\ \mathrm{Here},{\mathrm{D}}_{\mathit{A}\mathit{B}}{\mathrm{D}}_{\mathrm{BC}}{\mathrm{D}}_{\mathrm{CA}}\mathrm{are}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{conductos}. \\ \mathrm{Substitute}6\mathrm{m}{\mathrm{forD}}_{\mathrm{AB}},6\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{BC}}\mathrm{and}12\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{CA}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {D}_{\mathit{e}\mathit{q}}=\sqrt[3]{6\times 6\times 12} \\ =7.559\mathrm{m} \\ \mathrm{The}\mathrm{capacitance}\mathrm{to}\mathrm{neutral}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{C}}_{\mathrm{n}}=\frac{2\mathrm{\pi \epsilon }}{\mathrm{In}\left({\displaystyle \frac{{\mathrm{D}}_{\mathrm{eq}}}{{\mathrm{D}}_{\mathrm{sc}}}}\right)} \\ \mathrm{Here}{\mathrm{D}}_{\mathrm{eq}}\mathrm{is}\mathrm{the}\mathrm{equivalent}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{conductor}\mathrm{and}{\mathrm{D}}_{\mathrm{sc}}\mathrm{is}\mathrm{the} \\ \mathrm{equivalent}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{conductor}. \\ \mathrm{substitute}8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\mathrm{for}\mathrm{\epsilon },7.559\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{eq}}\mathrm{and}0.0471\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{sc}}\mathrm{in}\mathrm{the} \\ \mathrm{above}\mathrm{equation}. \\ {C}_{\mathrm{n}}\mathit{=}\frac{\mathit{2}\mathit{\pi }\mathit{(}\mathit{8}\mathit{.}\mathit{85}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{12}}\mathit{)}}{\mathit{In}\left({\displaystyle \frac{7.559}{0.0471}}\right)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}10.949\times {10}^{-12}\mathrm{F}/\mathrm{m} \\ \mathrm{Capacitance}\mathrm{to}\mathrm{neutral}\mathrm{in}\mathrm{F}/\mathrm{km}\mathrm{per}\mathrm{phase}\mathrm{is}, \\ {\mathrm{C}}_{\mathrm{n}}=10.949\times {10}^{-12}\mathrm{F}/\mathrm{m} \\ =0.949\times {10}^{-12}\times \frac{1}{\left({10}^{-9}\right)\left({10}^{-3}\right)}\mathrm{nF}/\mathrm{km} \\ =10.949\mathrm{nF}/\mathrm{km} \\ \mathrm{Capacitance}\mathrm{to}\mathrm{neutral}\mathrm{in}\mathrm{nF}/\mathrm{m}\mathrm{i}\mathrm{per}\mathrm{phase}\mathrm{is}, \\ {\mathrm{C}}_{\mathrm{n}}=10.949\mathrm{nF}/\mathrm{km} \\ =10.949\times 1.609\mathrm{nF}/\mathrm{mi} \\ =17.616\mathrm{nF}/\mathrm{m}\mathrm{i} \\ \mathrm{Therefore}\mathrm{capacitance}\mathrm{to}\mathrm{neutral}\mathrm{is}10.949\mathrm{nF}/\mathrm{km}\mathrm{and}17.616\mathrm{nF}/\mathrm{m}\mathrm{i}. \end{gathered}$$

Calculate the capacitive reactance per phase.

b)

The capacitive reactance per phase is given by,

$$\begin{gathered} {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\mathrm{\omega C}} \\ =\frac{1}{2\mathrm{\pi }f\mathrm{C}} \\ \mathrm{Here}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{frequency},\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{equivalent}\mathrm{capacitance}. \\ \mathrm{Substitute}60\mathrm{Hz}\mathrm{for}10.949\times {10}^{-12}F/\mathrm{m}\mathrm{for}\mathrm{c}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {\mathrm{X}}_{\mathrm{c}}=\frac{1}{2\mathrm{\pi }\left(60\right)\left(10.949\times {10}^{-12}\right)} \\ =2.423\times {10}^8\mathrm{\Omega }-\mathrm{m} \\ \mathrm{The}\mathrm{capacitive}\mathrm{reactance}\mathrm{in}\mathrm{\Omega }-\mathrm{km}\mathrm{per}\mathrm{phase}\mathrm{is}, \\ {\mathrm{X}}_{\mathrm{c}}=2.423\times {10}^8\mathrm{\Omega }-\mathrm{km} \\ =2.423\times {10}^8\frac{1}{{10}^3}\mathrm{\Omega }-\mathrm{km} \\ =2.423\times {10}^5\mathrm{\Omega }-\mathrm{km} \\ \mathrm{The}\mathrm{capacitive}\mathrm{reactance}\mathrm{in}\mathrm{\Omega }-\mathrm{mi}\mathrm{per}\mathrm{phase}\mathrm{is}, \\ {\mathrm{X}}_{\mathrm{c}}=2.423\times {10}^8\mathrm{\Omega }-\mathrm{mZ} \\ =2.423\times {10}^8\frac{1}{{10}^3\times (1.609)}\mathrm{\Omega }-\mathrm{km} \\ =1.505\times {10}^5\mathrm{\Omega }-\mathrm{m}\mathrm{i} \\ \mathrm{Therefore}\mathrm{capacitive}\mathrm{reactance}\mathrm{per}\mathrm{phase}\mathrm{is}2.423\times {10}^5\mathrm{\Omega }-\mathrm{km}\mathrm{and} \\ 1.505\times {10}^5\mathrm{\Omega }-\mathrm{m}\mathrm{i}. \end{gathered}$$

Calculate the capacitive reactance per phase for a line length.

(c)

The capacitive reactance in $\mathrm{\Omega }-\mathrm{mi}$per phase is calculated above as,

$$\begin{gathered} {\mathrm{X}}_{\mathrm{c}}=1.505\times {10}^5\mathrm{\Omega }-\mathrm{mi} \\ \mathrm{Now}\mathrm{capacitive}\mathrm{reactance}\mathrm{for}\mathrm{a}\mathrm{length}\mathrm{of}100\mathrm{m}\mathrm{i}. \\ {\mathrm{X}}_{\mathrm{c}}=1.505\times {10}^5\mathrm{\Omega }-\frac{1}{100} \\ =1.505\times {10}^5\mathrm{\Omega } \\ \mathrm{Therefore}\mathrm{capacitive}\mathrm{reactance}\mathrm{in}\mathrm{ohm}’\mathrm{s}\mathrm{per}\mathrm{phase}\mathrm{is}1.505\times {10}^3\mathrm{\Omega }. \end{gathered}$$