Question

Consider the line of Problem 4.25. Calculate the capacitive reactance per phase in$\mathbf{\Omega }\mathbf{.}\mathbf{mi}$.

Short answer

Therefore, the capacitance is $0.212\frac{{\displaystyle \frac{\mathrm{\mu F}}{\mathrm{mi}}}}{\mathrm{phase}}$and the reactance is$-\mathrm{j}125.122\times {10}^3\mathrm{\Omega }.\mathrm{mi}$ .

Step-by-step solution

Given data.

From Problem 4.25, the GMD of conductor is 51.87 ft .

Determine the formulas of capacitance per-phase per mile, equivalent radius and capacitive reactance.

Write the formula of equivalent radius for four conductor bundled.

${\mathbf{D}}_{\mathbf{SC}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{091}\sqrt[\mathbf{4}]{{\mathbf{rd}}^{\mathbf{3}}}$ ……. (1)

Here,ris the radius of conductor andis the distance between two conductors.

Write the formula of capacitance.

${\mathbf{C}}_{\mathbf{an}}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi \epsilon }}{\mathbf{In}\left({\displaystyle \frac{D_{\mathrm{eq}}}{D_{\mathrm{SC}}}}\right)}\frac{\mathbf{F}}{\mathbf{m}}$ ……. (2)

Here,$D_{eq}$is geometric mean distance and$D_{SC}$is geometric mean radius.

Write the formula of capacitive reactance.

${\mathbf{X}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{j}\mathbf{2}{\mathbf{\pi fC}}_{\mathbf{an}}}\mathbf{\Omega }$ ……. (3)

Here, fis frequency.

Determine the capacitance, and capacitive reactance per phase.

Determine the equivalent radius of conductors.

Substitute 0.0498 in for r and 1.667 infor d in equation (1).

$$\begin{gathered} {\mathrm{D}}_{\mathrm{SC}}=1.0.91\sqrt[4]{\left(0.0498\right){\left(1.667\right)}^3} \\ =0.7561\mathrm{ft} \end{gathered}$$

Determine the capacitance.

Substitute 0.7561 ftfor $D_{SC}$, 51.87 ft for r and $8.854\times {10}^{-12}$for $\epsilon$in equation (2).

$$\begin{gathered} {\mathrm{C}}_{\mathrm{an}}=\frac{2\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left({\displaystyle \frac{51.87\mathrm{ft}}{0.7561\mathrm{ft}}}\right)} \\ =0.212\frac{{\displaystyle \frac{\mathrm{\mu F}}{\mathrm{mi}}}}{\mathrm{phase}} \end{gathered}$$

Determine the capacitive reactance.

Substitute $0.212\frac{{\displaystyle \frac{\mathrm{\mu F}}{\mathrm{mi}}}}{\mathrm{phase}}$ for $C_{an}$ and 60 Hz for f in equation (3).

$$\begin{gathered} {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\mathrm{j}2\mathrm{\pi }\left(60\right)(0.212\times {10}^{-6})} \\ =-\mathrm{j}125.122\times {10}^3\mathrm{\Omega }.\mathrm{mi} \end{gathered}$$

Therefore, the capacitance is role="math" localid="1655881559689" $0.212\frac{{\displaystyle \frac{\mathrm{\mu F}}{\mathrm{mi}}}}{\mathrm{phase}}$and the reactance is$-\mathrm{j}125.122\times {10}^3\mathrm{\Omega }.\mathrm{mi}$ .