Question

(a) In practice, one deals with the capacitive reactance of the line in$\mathbf{o}\mathbf{}\mathbf{hm}\mathbf{}\mathbf{s}\mathbf{-}\mathbf{m}\mathbf{}\mathbf{i}$to neutral. Show that Eq. (4.9.15) of the text can be rewritten as

role="math" localid="1655189968906" $$\begin{gathered} {\mathit{X}}_{\mathbf{c}}\mathbf{=}{\mathbf{k}}^{\mathbf{,}}\mathbf{log}\frac{\mathbf{D}}{\mathbf{r}}\mathbf{ohms}\mathbf{-}\mathbf{mi} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{{\mathbf{x}}^{\mathbf{,}}}_{\mathbf{d}}\mathbf{+}{{\mathbf{x}}^{\mathbf{,}}}_{\mathbf{a}} \\ \mathbf{Where}\mathbf{}{\mathbf{x}}_{\mathbf{d}}\mathbf{=}{\mathbf{k}}^{\mathbf{,}}\mathbf{logD}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{capacitive}\mathbf{}\mathbf{reactance}\mathbf{}\mathbf{spacing}\mathbf{}\mathbf{factor} \\ {{\mathbf{X}}^{\mathbf{,}}}_{\mathbf{d}}\mathbf{=}{\mathbf{k}}^{\mathbf{,}}\mathbf{log}\frac{\mathbf{1}}{\mathbf{r}}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{capacitive}\mathbf{}\mathbf{reactance}\mathbf{}\mathbf{at}\mathbf{}\mathbf{1}\mathbf{-}\mathbf{ft}\mathbf{}\mathbf{spacing} \\ {\mathbf{k}}^{\mathbf{,}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{)}\mathbf{/}\mathbf{f}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{06833}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{f}\mathbf{=}\mathbf{60}\mathbf{}\mathbf{Hz} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{Determine}\mathbf{}\mathbf{the}\mathbf{}\mathbf{capacitive}\mathbf{}\mathbf{reactance}\mathbf{}\mathbf{in}\mathbf{}\mathbf{\Omega }\mathbf{}\mathbf{mi}\mathbf{}\mathbf{}\mathbf{for}\mathbf{}\mathbf{a}\mathbf{}\mathbf{single}\mathbf{-}\mathbf{phase}\mathbf{}\mathbf{line}\mathbf{}\mathbf{of} \\ \mathbf{problem}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{14}\mathbf{.}\mathbf{}\mathbf{if}\mathbf{}\mathbf{the}\mathbf{}\mathbf{spacing}\mathbf{}\mathbf{is}\mathbf{}\mathbf{doubled}\mathbf{,}\mathbf{}\mathbf{how}\mathbf{}\mathbf{does}\mathbf{}\mathbf{the}\mathbf{}\mathbf{reactance}\mathbf{}\mathbf{change}\mathbf{?} \end{gathered}$$

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{capacitive}\mathrm{reactance}{X}_{c}is0.06833\times {10}^6\log \frac{\mathrm{D}}{\mathrm{r}}. \\ \left(b\right)\mathrm{The}\mathrm{value}\mathrm{of}\mathrm{capacitive}\mathrm{reactance}\mathrm{is}14.869\times {10}^4\mathrm{\Omega mi}\mathrm{and}\mathrm{the}\mathrm{capacitive} \\ \mathrm{reactance}\mathrm{is}\mathrm{increased}\mathrm{by}13.78\%\mathrm{when}\mathrm{the}\mathrm{spacing}\mathrm{is}\mathrm{doubled}. \end{gathered}$$

Step-by-step solution

Write the given data of the question

The capacitive reactance at$1-\mathrm{ft}$spacing is given by the expression $x_d^{,}=k^{,}\log \frac{1}{r}.$

The capacitive reactance spacing factor is given by$x_d^{,}=k^{,}\log D.$

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

The capacitance to neutral is given by,

$$\begin{gathered} {\mathbf{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau }}{\mathbf{In}\left({\displaystyle \frac{D}{r}}\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathbf{Here}\mathbf{}\mathbf{D}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{conductor}\mathbf{}\mathbf{spacing}\mathbf{,}\mathbf{}\mathbf{r}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{radius}\mathbf{}\mathbf{of}\mathbf{}\mathbf{each}\mathbf{}\mathbf{solid}\mathbf{}\mathbf{cylinder}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{C}}_{\mathbf{n}}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{capacitance}\mathbf{}\mathbf{to}\mathbf{}\mathbf{neutral}\mathbf{.} \\ \mathbf{The}\mathbf{}\mathbf{capacitive}\mathbf{}\mathbf{reactance}\mathbf{}\mathbf{is}\mathbf{}\mathbf{given}\mathbf{}\mathbf{by}\mathbf{,} \\ {\mathbf{X}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}} \\ \mathbf{Here}\mathbf{}{\mathbf{X}}_{\mathbf{C}}\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{capacitive}\mathbf{}\mathbf{reactance}\mathbf{,}\mathbf{}\mathbf{\omega }\mathbf{}\mathbf{is}\mathbf{}\mathbf{the}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{frequency}\mathbf{}\mathbf{and}\mathbf{}\mathbf{}\mathbf{C}\mathbf{}\mathbf{is} \\ \mathbf{the}\mathbf{}\mathbf{capacitance}\mathbf{.} \end{gathered}$$

Calculate the capacitive reactance.

(a)

The capacitance to neutral is given by,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{n}}=\frac{2\mathrm{\tau \tau }}{\mathrm{In}\left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}\mathrm{F}/\mathrm{m} \\ \mathrm{Here}D\mathrm{is}\mathrm{the}\mathrm{conductor}\mathrm{spacing},\mathit{}r\mathrm{is}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{each}\mathrm{solid}\mathrm{cylinder}\mathrm{and}{\mathrm{C}}_{\mathrm{n}} \\ \mathrm{is}\mathrm{the}\mathrm{capacitance}\mathrm{to}\mathrm{neutral}. \\ \mathrm{Convert}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{in}\mathrm{F}/\mathrm{mi}\mathrm{units}. \\ \mathrm{substitute}8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\mathrm{for}\mathrm{\epsilon }\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {\mathrm{C}}_{\mathrm{n}}=\frac{2\mathrm{\pi }(8.85\times {10}^{-12})}{\mathrm{In}\left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}\times \frac{1}{1609}\mathrm{F}/\mathrm{mi} \\ =\frac{8.496\times {10}^{-8}}{\mathrm{In}\left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}\mathrm{F}/\mathrm{mi} \\ \mathrm{Now}\mathrm{convert}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{in}\mathrm{terms}\mathrm{of}\log \mathrm{to}\mathrm{base}10. \\ {\mathrm{C}}_{\mathrm{n}}=\frac{8.946\times {10}^{-8}}{2.302\log \left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}\mathrm{F}/\mathrm{mi} \\ =\frac{3.866\times {10}^{-8}}{\log \left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)} \\ \mathrm{The}\mathrm{capacitive}\mathrm{reactance}\mathrm{at}60\mathrm{Hz}\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mathrm{X}}_{\mathrm{c}}=\frac{1}{{\mathrm{\omega C}}_{\mathrm{n}}} \\ =\frac{1}{\mathit{2}\pi f{C}_{\mathit{n}}} \\ \mathrm{subsititute}\frac{3.866\times {10}^{-8}}{\log \left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}\mathrm{for}{\mathrm{C}}_{\mathrm{n}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {\mathrm{X}}_{\mathrm{c}}=\frac{1}{2\mathrm{\pi f}{\displaystyle \frac{3.886\times {10}^{-8}}{\log \left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}}} \\ =\frac{4.1\times {10}^6}{\mathrm{f}}\log \left(\frac{\mathrm{D}}{\mathrm{r}}\right) \\ =\frac{4.1\times {10}^6}{60\mathrm{Hz}}\log \left(\frac{\mathrm{D}}{\mathrm{r}}\right) \\ =0.06833\times {10}^6\log \left(\frac{\mathrm{D}}{\mathrm{r}}\right) \\ \mathrm{Therefore}\mathrm{the}\mathrm{capacitive}\mathrm{reactance}\mathrm{is}\frac{4.1\times {10}^6}{\mathrm{f}}\log \left(\frac{\mathrm{D}}{\mathrm{r}}\right) \end{gathered}$$

 Step 4: Calculate the increment in capacitive reactance.

(b)

The capacitive reactance in terms of log is given by,

$$\begin{gathered} X_{\mathit{c}}\mathit{=}{k}^{,}\log \frac{\mathrm{D}}{\mathrm{r}}\mathrm{\Omega }-\mathrm{mi}\mathrm{to}\mathrm{netural}. \\ {X}_{\mathit{c}}\mathit{=}{x}_{\mathit{d}}^{\mathit{,}}+{\mathrm{x}}_{\mathrm{a}}^{,} \\ \mathrm{Here} \\ {\mathrm{x}}_{\mathrm{d}}^{,}=k^{,}\log \mathit{}D\mathit{}\mathrm{is}\mathrm{the}\mathrm{capacitive}\mathrm{reactance}\mathrm{spacing}\mathrm{factor}. \\ {\mathrm{x}}_{\mathrm{d}}^{,}=k^{\mathit{,}}\log \frac{1}{\mathrm{r}}\mathrm{is}\mathrm{the}\mathrm{capacitive}\mathrm{reactance}\mathrm{at}1-\mathrm{ft}\mathrm{spacing}. \\ \mathrm{The}\mathrm{value}\mathrm{of}{k}^{\mathit{,}}\mathrm{is}, \\ {k}^{,}\mathit{=}\frac{4.1\times {10}^6}{f} \\ \mathrm{Substitute}60\mathrm{Hz}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {\mathrm{k}}^{,}\mathit{=}\frac{4.1\times {10}^6}{60} \\ =0.06833\times {10}^6 \\ \mathrm{The}\mathrm{phase}\mathrm{separation}\mathrm{between}\mathrm{the}\mathrm{conductors}\mathrm{is}, \\ D\mathit{=}10\mathrm{ft} \\ =10(0.3)\mathrm{m} \\ =3\mathrm{m} \\ \mathrm{The}\mathrm{conductor}\mathrm{radius}\mathrm{is}, \\ \mathrm{r}=0.06677\mathrm{ft} \\ =0.06677\times 0.3\mathrm{m} \\ =0.02\mathrm{m} \\ {{\mathrm{x}}^{,}}_{\mathrm{d}}=0.06833\times {10}^6{\log }^3 \\ =3.260\times {10}^4\mathrm{\Omega }-\mathrm{mi} \\ {\mathrm{x}}_{\mathrm{a}}^{,}=0.06833\times {10}^6\log \frac{1}{0.02} \\ =11.609\times {10}^4\mathrm{\Omega }-\mathrm{mi} \\ {X}_{\mathit{c}}\mathit{=}{{\mathit{x}}^{\mathit{,}}}_{\mathit{d}}\mathit{+}{\mathrm{x}}_{\mathit{a}}^{\mathit{,}} \\ =3.260\times {10}^4+11.609\times {10}^4 \\ =14.869\times {10}^4\mathrm{\Omega }-\mathrm{mi} \\ \mathrm{Therefore}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{capacitive}\mathrm{reactance}\mathrm{is}14.869\times {10}^4\mathrm{\Omega }-\mathrm{mi}. \\ \mathrm{When}\mathrm{the}\mathrm{spacing}\mathrm{is}\mathrm{getting}\mathrm{doubled},\mathrm{that}\mathrm{is}\mathrm{D}=6\mathrm{m} \\ {{\mathrm{x}}^{,}}_{\mathrm{d}}=0.06833\times {10}^6{\log }^6 \\ =5.31\times {10}^4\mathrm{\Omega }-\mathrm{mi} \\ {\mathrm{x}}_{\mathrm{a}}^{,}=0.06833\times {10}^6\log \frac{1}{0.02} \\ =11.609\times {10}^4\mathrm{\Omega }-\mathrm{mi} \\ {\mathrm{x}}_{\mathrm{c}}={{\mathrm{x}}^{,}}_{\mathrm{d}}+{\mathrm{x}}_{\mathrm{a}}^{,} \\ =5.31\times {10}^4+11.609\times {10}^4 \\ =16.919\times {10}^4\mathrm{\Omega }-\mathrm{mi} \\ \mathrm{The}\mathrm{change}\mathrm{in}\mathrm{reactance}\mathrm{is}\mathrm{calculated}\mathrm{below}, \\ \mathrm{change}\mathrm{in}\mathrm{reactance}=\frac{(16.919\times {10}^4)-(14.869\times {10}^4)}{14.869\times {10}^4} \\ =13.78\% \\ \mathrm{Therefore}\mathrm{the}\mathrm{capacitive}\mathrm{reactance}\mathrm{is}\mathrm{increased}\mathrm{by}13.78\%\mathrm{when}\mathrm{the}\mathrm{spacing}\mathrm{is}\mathrm{doubled}. \end{gathered}$$