Question

Calculate the capacitance-to neutral in $\mathbf{F}\mathbf{/}\mathbf{m}$and the admittanceto

neutral in$\mathbf{s}\mathbf{/}\mathbf{km}$for the three phase line in problem 4.10. neglect the effect of the

earth plane.

Short answer

The capacitance to neutral is $10.632\times {10}^{-12}\mathrm{F}/\mathrm{m}$and admittance to neutral is $j4.006\times {10}^{-6}\mathrm{S}/\mathrm{km}$.

Step-by-step solution

Write the given data of the question

The conductor spacing is 4ft.

The diameter of the solid cylinder is 0.5in.

The system frequency is given as 60Hz.

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

The capacitance to neutral is given by,

${\mathbf{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi \epsilon }}{\mathbf{In}\left({\displaystyle \frac{D}{r}}\right)}$

Here is D the conductor spacing, r is the radius of each solid cylinder and ${\mathrm{C}}_{\mathrm{n}}$is the capacitance to neutral.

The admittance to neutral is given by,

${\mathbf{Y}}_{\mathbf{n}}\mathbf{=}{\mathbf{j\omega C}}_{\mathbf{n}}$

Here ${\mathbf{Y}}_{\mathbf{n}}$is the admittance to neutral, $\mathbf{\omega }$is the angular frequency,${\mathrm{C}}_{\mathrm{n}}$ is the capacitance to neutral.

Calculate the capacitance to neutral and admittance to neutral.

Calculate the conductor spacing in meter.

$$\begin{gathered} D=4ft \\ =\frac{4}{3.28}m \\ =1.219m \end{gathered}$$

Calculate the radius of each solid cylindrical conductor.

$$\begin{gathered} \mathrm{r}=\frac{0.5}{2}\mathrm{in} \\ =\frac{0.5}{2}\times 0.0254\mathrm{m} \\ =0.00635\mathrm{m} \end{gathered}$$

The capacitance to neutral is given by,

${\mathrm{C}}_{\mathrm{n}}=\frac{2\mathrm{\pi \epsilon }}{\mathrm{In}\left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}$

Here D is the conductor spacing, r is the radius of each solid cylinder and ${\mathrm{C}}_{\mathrm{n}}$is the capacitance to neutral.

Substitute 1.219m for D,0.00635m for r and $8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}$for $\epsilon$in the above

equation.

$$\begin{gathered} {\mathrm{C}}_{\mathrm{n}}=\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{\mathrm{In}\left({\displaystyle \frac{1.219}{0.00635}}\right)} \\ =10.632\times {10}^{-12}\mathrm{F}/\mathrm{m} \end{gathered}$$

Therefore the capacitance to neutral is$10.632\times {10}^{-12}\mathrm{F}/\mathrm{m}$.

The admittance to neutral is given by,

$Y_n=j\omega C_n$

Here $Y_n$is the admittance to neutral, $\omega$is the angular frequency and $C_n$is the capacitance to neutral.

Substitute for in the above equation.

Therefore the admittance to neutral is $\mathrm{j}4.006\times {10}^{-6}\mathrm{S}/\mathrm{km}$.