Question

Calculate the capacitance-to-neutral in $\frac{\mathbf{F}}{\mathbf{m}}$and the admittance-to-neutral in$\frac{\mathbf{S}}{\mathbf{km}}$for the three-phase line in Problem 4.18. Also calculate the line charging current in kA/phase if the line is 110 m in length and is operated at 230kV . Neglect the effect of the earth plane.

Short answer

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

$8.74\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}},j3.29\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{and}0.048\frac{\mathrm{kA}}{\mathrm{phase}}\mathrm{respectively}$.

Step-by-step solution

Write the given data by the question.

The voltage of the line,V=230kV

The frequency of the system,f=60Hz

Spacing between the conductor is 7m.

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

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

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.

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

Conversion from inch to meter.

1in=0.0254m

Calculate the capacitance to neutral, admittance to neutral and charging current.

Calculate the radius of the conductor.

Substitute 1.196 in for into equation (1).

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

Convert radius from inch to meter.

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

Calculate the geometric mean distance.

Substitute 7m for $D_{AB}$,7m for$D_{BC}$and 14m for $D_{CA}$into equation (2).

$$\begin{gathered} D_{eq}=\sqrt[3]{7\times 7\times 14} \\ {D}_{eq}=8.819\mathrm{m} \\ \mathrm{Calculate}\mathrm{the}\mathrm{capacitance}\mathrm{per}\mathrm{phase}. \\ \mathrm{Substitute}8.819\mathrm{m}\mathrm{forb}{\mathrm{D}}_{\mathrm{eq}}\mathrm{and}0.0152\mathrm{for}\mathrm{r}\mathrm{into}\mathrm{equation}\left(3\right). \\ \mathrm{C}=\frac{2\mathrm{\pi }\times 8.85\times {10}^{-12}}{\mathrm{In}\left({\displaystyle \frac{8.819}{0.0152}}\right)} \\ \mathrm{C}=\frac{2\mathrm{\pi }\times 8.85\times {10}^{-12}}{6.363} \\ \mathrm{C}=8.74\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Calculate the admittance to neutral.

Substitute 60Hz for f and role="math" localid="1655883589394" $8.74\times {10}^{-12}\frac{F}{m}\mathrm{for}{\mathrm{C}}_{\mathrm{AN}}$for into equation (4).

$$\begin{gathered} Y_{AN}=j2\mathrm{\pi }\times 60\times 8.74\times {10}^{-12} \\ {\mathrm{Y}}_{\mathrm{AN}}=\mathrm{j}3.29\times {10}^{-9}\frac{\mathrm{S}}{\mathrm{m}} \\ {\mathrm{Y}}_{\mathrm{AN}}=\mathrm{j}3.29\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}} \end{gathered}$$

Calculate the charging current.

Substitute role="math" localid="1655883960875" $\mathrm{j}3.29\times {10}^{-6}\frac{S}{Km}$for $Y_{AN}$,110km for $I$and 230V for $V_L$into equation (5).

$$\begin{gathered} I_{ch}=\left|j3.29\times {10}^{-6}\times 110\times \frac{230}{\sqrt{3}}\right| \\ {I}_{ch}=361.9\times {10}^{-6}\times 132.8kV \\ {I}_{ch}=0.048\frac{kA}{\mathrm{phase}} \end{gathered}$$

Hence, the capacitance to neutral, admittance to neutral and charging current are , and respectively.

$8.74\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}},j3.29\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{and}0.048\frac{\mathrm{ka}}{\mathrm{phase}}\mathrm{respectively}$