Question

For an average line height of$\mathbf{10}\mathbf{}\mathbf{m}$, determine the effect of the earth on capacitance for the single-phase line in Problem 4.32. Assume a perfectlyconducting earth plane.

Short answer

Therefore, the capacitance is $1.3247\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}}$ and it is larger than the capacitance in problem 4.32.

Step-by-step solution

Write the given data.

From Problem 4.32 and problem 4.8, the radius is 0.0075 m and the distance is 0.5 m .

The distance $H_{xx}$ is 20 m .

The distance $H_{xy}$is 20.006 m .

Determine the formula of line-to-line capacitance.

Write the formula of line-to-line capacitance.

${\mathbf{C}}_{\mathbf{xn}}\mathbf{=}\frac{\mathbf{\tau \tau \epsilon }}{\mathbf{In}\left({\displaystyle \frac{D}{r}}\right)\mathbf{-}\mathbf{In}\left({\displaystyle \frac{H_{\mathrm{xy}}}{H_{\mathrm{xx}}}}\right)}\frac{\mathbf{F}}{\mathbf{m}}$ ……. (1)

Here, Dis the distance between two conductors, r is the radius of conductor, $H_{xy}$ is the hypotenuse distance between conductors x and y and $H_{xx}$is the vertical distance between two conductots.

Determine the single phase to line capacitance.

Determine the capacitance.

Substitute 0.5 m for D , 0.0075 m for r , $8.854\times {10}^{-12}$for $\epsilon$, 20 m for $H_{xx}$and 20.006 m for $H_{xy}$ in equation (1).

$$\begin{gathered} {\mathrm{C}}_{\mathrm{xn}}=\frac{2\mathrm{\pi }\times 8.854\times {10}^{-12}}{\mathrm{In}\left({\displaystyle \frac{0.5}{0.0075}}\right)-\mathrm{In}\left({\displaystyle \frac{20.006}{20}}\right)} \\ =1.3247\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

In problem 4.32, the capacitance is $1.3246\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}}$.

So, the capacitance in this problem is slightly larger than the previous.

Therefore, the capacitance is $1.3247\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}}$.