Question

Rework Problem 4.39 if the phase spacing between adjacent conductors is (a) increased by 10% to 7.7 m or (b) decreased by 10% to 6.3m . Compare the results with those of Problem 4.39.

Short answer

(a) The capacitance to neutral, admittance to neutral and charging current are $8.61\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}},\mathrm{j}3.25\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{and}0.047\frac{\mathrm{kA}}{\mathrm{phase}}\mathrm{respectively}.$The percentage change in the capacitance, admittance and charging current due to change in spacing are 1.29%,1.2% and 2% respectively.

(b) The capacitance to neutral, admittance to neutral and charging current are$8.89\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}},\mathrm{j}3.25\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{and}0.048\frac{\mathrm{kA}}{\mathrm{phase}}\mathrm{respectively}.$The percentage change in the capacitance, admittance and charging current due to change in spacing are -1.72%,-1.8% and 0% respectively.

Step-by-step solution

Write the given data by the question.

The voltage of the line,V=230kV

The frequency of the system,$\mathrm{f}=60\mathrm{Hz}$

Spacing between the conductor is 7m .

The length of the line, $I=110\mathrm{m}$.

From appendix A4 the outer diameter of conductor, $d=1.196\mathrm{in}$

From problem 4.39

The capacitance to neutral, admittance to neutral and charging current are , and respectively.

$8.74\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}},\mathrm{j}3.29\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{and}0.048\frac{\mathrm{kA}}{\mathrm{phase}}\mathrm{respectively}.$

Determine the formulas to calculate the capacitance-to-neutral, admittance to neutral and line charging current.

The equation to calculate the radius of the conductor is given as follows.

$\mathbf{r}\mathbf{=}\frac{\mathbf{d}}{\mathbf{2}}$ …… (1)

The equation to calculate the geometric mean distance is given as follows.

${\mathbf{D}}_{\mathbf{eq}}\mathbf{=}\sqrt[\mathbf{3}]{{\mathbf{D}}_{\mathbf{AB}}{\mathbf{D}}_{\mathbf{BC}}{\mathbf{D}}_{\mathbf{CA}}}$ …… (2)

Here, ${\mathbf{D}}_{\mathbf{AB}}$is the distance between conductor A and B ,${\mathbf{D}}_{\mathbf{BC}}$is the distance

between conductor B and C, ${\mathbf{D}}_{\mathbf{CA}}$is distance between C and A.

The equation to calculate the capacitance is given as follows.

${\mathbf{C}}_{\mathbf{AN}}\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau \epsilon }}{\mathbf{In}\left({\displaystyle \frac{D_{\mathrm{eq}}}{r}}\right)}\frac{\mathbf{F}}{\mathbf{m}}$ …… (3)

The equation to calculate the admittance neutral is given as follows.

${\mathbf{Y}}_{\mathbf{AN}}\mathbf{=}\mathbf{j}\mathbf{2}{\mathbf{\tau \tau fC}}_{\mathbf{AN}}$ …… (4)

The equation to calculate the charging current is given as follows.

${\mathbf{I}}_{\mathbf{c}\mathbf{h}}\mathbf{=}\left|Y_{AN}I\frac{V_L}{\sqrt{3}}\right|$ …… (5)

Conversion from inch to meter.

1in=0.0254 m

Calculate the capacitance to neutral, admittance to neutral and charging current for spacing 7.7m .

(a)

Calculate the radius of the conductor.

Substitute 1.196 in for d into equation (1).

$$\begin{gathered} r=\frac{1.196}{2} \\ r=0.598\mathrm{in} \end{gathered}$$

Convert radius from inch to meter.

$$\begin{gathered} \mathrm{r}=0.598\times 0.0254 \\ \mathrm{r}=0.0152\mathrm{m} \end{gathered}$$

Calculate the geometric mean distance.

Substitute 7.7m for $D_{AB}$, 7.7m for $D_{BC}$and 15.4 for $D_{CA}$into equation (2).

$$\begin{gathered} D_{eq}=\sqrt[3]{7.7\times 7.7\times 15.4} \\ {D}_{eq}=9.7\mathrm{m} \\ \mathrm{Calculate}\mathrm{the}\mathrm{capacitance}\mathrm{per}\mathrm{phase}. \\ \mathrm{Substitute}9.7\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{eq}}\mathrm{and}0.0152\mathrm{m}\mathrm{for}\mathrm{r}\mathrm{into}\mathrm{equation}\left(3\right). \\ \mathrm{c}=\frac{2\mathrm{\pi }\times 8.85\times {10}^{-12}}{\mathrm{In}\left({\displaystyle \frac{9.7}{0.0152}}\right)} \\ \mathrm{C}=\frac{2\mathrm{\pi }8.85\times {10}^{-12}}{6.458} \\ \mathrm{C}=8.61\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \\ \mathrm{Calculate}\mathrm{the}\mathrm{percentage}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{capacitance}\mathrm{to}\mathrm{neutral}. \\ \%\mathrm{C}=\left(\frac{8.74\times {10}^{-12}-8.61\times {10}^{-12}}{8.74\times {10}^{-12}}\right)\times 100 \\ \%\mathrm{C}=0.0149\times 100 \\ \%\mathrm{C}=1.49\% \end{gathered}$$

$$\begin{gathered} \mathrm{Calculate}\mathrm{the}\mathrm{admittance}\mathrm{to}\mathrm{neutral}. \\ \mathrm{Substitute}60\mathrm{Hz}\mathrm{for}\mathrm{f}\mathrm{and}8.61\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\mathrm{for}{\mathrm{C}}_{\mathrm{AN}}\mathrm{into}\mathrm{equation}\left(4\right). \\ {\mathrm{Y}}_{\mathrm{AN}}=\mathrm{j}2\mathrm{\pi }\times 60\times 8.61\times {10}^{-12} \\ {\mathrm{Y}}_{\mathrm{AN}}=\mathrm{j}3.25\times {10}^{-9}\frac{\mathrm{S}}{\mathrm{m}} \\ {\mathrm{Y}}_{\mathrm{AN}}=\mathrm{j}3.25\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{m}} \end{gathered}$$

$$\begin{gathered} \mathrm{Calculate}\mathrm{the}\mathrm{percentage}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{admittance}\mathrm{to}\mathrm{neutral}. \\ \%\mathrm{Y}=\left(\frac{3.29\times {10}^{-6}-3.25\times {10}^{-6}}{3.29\times {10}^{-6}}\right)\times 100 \\ \%\mathrm{Y}=0.012\times 100 \\ \%\mathrm{Y}=1.2\% \\ \mathrm{Calculate}\mathrm{the}\mathrm{charging}\mathrm{current}. \\ \mathrm{Substitute}\mathrm{j}3.25\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{m}}\mathrm{for}{\mathrm{Y}}_{\mathrm{AN}},110\mathrm{km}\mathrm{for}I\mathrm{and}230\mathrm{V}\mathrm{for}{\mathrm{V}}_{\mathrm{L}}\mathrm{into}\mathrm{equation}\left(5\right). \\ {I}_{\mathit{c}\mathit{h}}\mathit{=}\left|\mathrm{j}3.25\times {10}^{-6}110\times \frac{230}{\sqrt{3}}\right| \\ {I}_{\mathit{c}\mathit{h}}=357.5\times {10}^{-6}\times 132.8\mathrm{kV} \\ {\mathrm{I}}_{\mathrm{ch}}=0.047\frac{\mathrm{kA}}{\mathrm{phase}} \\ \mathrm{Calculate}\mathrm{the}\mathrm{percentage}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{admittance}\mathrm{to}\mathrm{neutral}. \\ \%{\mathrm{I}}_{\mathrm{ch}}=\left(\frac{0.048-0.047}{0.048}\right)\times 100 \\ \%{\mathrm{I}}_{\mathrm{ch}}=0.02\times 100 \\ \%{\mathrm{I}}_{\mathrm{ch}}=2\% \end{gathered}$$

Hence the capacitance to neutral, admittance to neutral and charging current are role="math" localid="1655891792598" $8.61\times {10}^{-12}\frac{F}{m},j3.25\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{m}}\mathrm{and}0.047\frac{\mathrm{kA}}{\mathrm{phase}}\mathrm{respectively}$.The percentage change in the capacitance, admittance and charging current due to change in spacing are 1.29% ,1.2% and 2% respectively.

Calculate the capacitance to neutral, admittance to neutral and charging current for spacing 6.3m .

(b)

Calculate the geometric mean distance.

Substitute 6.3m for $D_{AB}$,6.3m for $D_{BC}$and 12.6m for $D_{CA}$into equation (1).

$$\begin{gathered} D_{eq}=\sqrt[3]{6.3\times 6.3\times 12.6} \\ {D}_{eq}=7.94m \\ \mathrm{Calculate}\mathrm{the}\mathrm{capacitance}\mathrm{per}\mathrm{phase}. \\ \mathrm{Substitute}7.94\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{eq}}\mathrm{and}0.0152\mathrm{m}\mathrm{for}\mathrm{r}\mathrm{into}\mathrm{equation}\left(3\right). \\ \mathrm{C}=\frac{2\mathrm{\pi }\times 8.85\times {10}^{-12}}{\mathrm{In}\left({\displaystyle \frac{7.94}{0.0152}}\right)} \\ \mathrm{C}=\frac{2\mathrm{\pi }\times 8.85\times {10}^{-12}}{6.258} \\ \mathrm{C}=8.89\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \\ \mathrm{Calculate}\mathrm{the}\mathrm{percentage}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{capacitance}\mathrm{to}\mathrm{neutral}. \\ \%\mathrm{C}=\left(\frac{8.74\times {10}^{-12}-8.89\times {10}^{-12}}{8.74\times {10}^{-12}}\right)\times 100 \\ \%\mathrm{C}=-0.0172\times 100 \\ \%\mathrm{C}=-1.72\% \\ \mathrm{Calculate}\mathrm{the}\mathrm{admittance}\mathrm{to}\mathrm{neutral}. \\ \mathrm{Substitute}60\mathrm{Hz}\mathrm{for}\mathrm{f}\mathrm{and}8.89\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\mathrm{for}{\mathrm{C}}_{\mathrm{AN}}\mathrm{into}\mathrm{equation}\left(4\right). \end{gathered}$$

$$\begin{gathered} Y_{AN}=j2\mathrm{\pi }\times 60\times 8.89\times {10}^{-12} \\ {\mathrm{Y}}_{\mathrm{AN}}=\mathrm{j}3.35\times {10}^{-9}\frac{\mathrm{S}}{\mathrm{m}} \\ {\mathrm{Y}}_{\mathrm{AN}}=\mathrm{j}3.35\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{Km}} \\ \mathrm{Calculate}\mathrm{the}\mathrm{percentage}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{admittance}\mathrm{to}\mathrm{neutral}. \\ \%\mathrm{Y}=\left(\frac{3.29\times {10}^{-6}-3.35\times {10}^{-6}}{3.29\times {10}^{-6}}\right)\times 100 \\ \%\mathrm{Y}=-0.018\times 100 \\ \%\mathrm{Y}=-1.8\% \end{gathered}$$

Calculate the charging current.

$$\begin{gathered} \mathrm{Substitute}\mathrm{j}3.35\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{m}}\mathrm{for}{\mathrm{Y}}_{\mathrm{AN}},110\mathrm{km}\mathrm{for}I\mathrm{and}230\mathrm{V}\mathrm{for}{\mathrm{V}}_{\mathrm{L}}\mathrm{into}\mathrm{equation}\left(5\right). \\ {I}_{\mathit{c}\mathit{h}}\mathit{=}\left|\mathrm{j}3.35\times {10}^{-6}\times 110\times \frac{230}{\sqrt{3}}\right| \\ {I}_{\mathit{c}\mathit{h}}\mathit{=}368.5\times {10}^{-6}\times 132.8\mathrm{kV} \\ {\mathrm{I}}_{\mathrm{ch}}=0.047\frac{\mathrm{kA}}{\mathrm{phase}} \\ \mathrm{Calculate}\mathrm{the}\mathrm{percentage}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{admittance}\mathrm{to}\mathrm{neutral}. \\ \%{\mathrm{I}}_{\mathrm{ch}}=\left(\frac{0.048-0.048}{0.048}\right)\times 100 \\ \%{\mathrm{I}}_{\mathrm{ch}}=0\times 100 \\ \%{\mathrm{I}}_{\mathrm{ch}}=0\% \end{gathered}$$

Hence the capacitance to neutral, admittance to neutral and charging current are $8.89\times {10}^{-12}\frac{F}{m},J3.35\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{m}}\mathrm{and}0.048\frac{\mathrm{kA}}{\mathrm{phase}}\mathrm{respectively}$. The percentage change in the capacitance, admittance and charging current due to change in spacing are -1.72%,-1.8% and 0% respectively.