Question

Short answer

Therefore, the capacitance is $11.856\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and the capacitive reactance is$0.1392\times {10}^6\mathrm{\Omega }.\mathrm{m}\mathrm{i}$.

Step-by-step solution

Given data.

The outside diameter of conductor is ${\mathrm{d}}^{\text{'}}=1.16\mathrm{in}.$.

The sub-conductor spacing is $\mathrm{S}=\frac{18}{12}\mathrm{ft}.$.

The number of sub-conductors are N = 4.

From the diagram, the phase shifts are ${\mathrm{D}}_{\mathrm{AB}}={\mathrm{D}}_{\mathrm{BC}}=60\mathrm{ft}\mathrm{and}{\mathrm{D}}_{\mathrm{AC}}=120\mathrm{ft}.$

Determine the formulas of capacitance and capacitive reactance.

Write the formula of radius of conductors in ft. converting from inches.

$\mathbf{r}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{\text{'}}}}{\mathbf{2}\mathbf{\times }\mathbf{12}}\mathbf{ft}$ ……. (1)

Here, $d^{\text{'}}$is the outside diameter.

Write the formula of capacitance.

${\mathbf{C}}_{\mathbf{an}}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi \epsilon }}{\mathbf{In}\left(\begin{array}{c}{\displaystyle \frac{\mathrm{GMD}}{\mathrm{GMR}}}\end{array}\right)}\frac{\mathbf{F}}{\mathbf{m}}\mathbf{line}\mathbf{-}\mathbf{to}\mathbf{-}\mathbf{neutral}$ ……. (2)

Here, G M D is geometric mean diameter and G M R is geometric mean radius.

Write the formula of geometric mean diameter.

$\mathbf{G}\mathbf{\hspace{0.17em}}\mathbf{M}\mathbf{\hspace{0.17em}}\mathbf{D}\mathbf{}\mathbf{=}\sqrt[\mathbf{3}]{{\mathbf{D}}_{\mathbf{AB}}{\mathbf{D}}_{\mathbf{BC}}{\mathbf{D}}_{\mathbf{AC}}}$ ..........(3)

Here, $D_{AB}$is the distance between conductors A and B ,$D_{BC}$ is the distance between conductors B and C,$D_{AC}$ , is the distance between conductors A and B .

Write the formula of geometric mean radius.

$\mathbf{G}\mathbf{}\mathbf{M}\mathbf{}\mathbf{R}\mathbf{=}\sqrt[\mathbf{N}]{\mathbf{r}\mathbf{}\mathbf{N}{\mathbf{\left(}\mathbf{A}\mathbf{\right)}}^{\mathbf{N}\mathbf{-}\mathbf{1}}}$ ...........(4)

Here,

$\mathbf{A}\mathbf{=}\frac{\mathbf{S}}{\mathbf{2}\mathbf{}\mathbf{sin}\left({\displaystyle \frac{\pi }{N}}\right)}$ ....................(5)

Write the formula of capacitive reactance.

${\mathbf{X}}_{\mathbf{c}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{06883}\mathbf{}\mathbf{log}\left(\frac{\mathrm{GMD}}{\mathrm{GMR}}\right)$

Determine the capacitance.

Determine the capacitive reactance.

Substitute 0.63 ftfor GMR , 75.6ftfor GMD in equation (6).

$$\begin{gathered} X_C=0.068In\left(\frac{75.6\mathrm{ft}}{0.693\mathrm{ft}}\right) \\ =0.1392\times {10}^6\Omega .\mathrm{m}\mathrm{i} \end{gathered}$$

Therefore, thecapacitive reactance is$0.1392\times {10}^6\mathrm{\Omega }.\mathrm{m}\mathrm{i}$.

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