Question

The capacitance of a single-circuit, three-phase transposed line with the configuration shown in Figure 4.38, including ground effect, and with conductors not equilaterally spaced is given by

(a) Now consider Figure 4.39 in which the configuration of a three-phase, single circuit,line with conductors having an outside diameter of. is shown. Determine the capacitance to neutral in F/m, including the ground effect.

(b) Next, neglecting the effect of ground, see how the value changes.

Short answer

(a)Therefore, thecapacitance is$9.7695\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$.

(b) Therefore, the capacitance is $8.4746\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and error after neglecting earth effect is 13.25%.

Step-by-step solution

Given data.

The outside diameter of conductor is $d^{\text{'}}=1.065$in.

From the figure, 4.39 and 4.38, the distances , $H_1,H_2\mathrm{and}{H}_3$are $2\times 50\mathrm{ft}=100\mathrm{ft}.$.

localid="1655198385311" $Also,thedis\tan ces{H}_{12},H_{23}$,

$\sqrt{{\left(100\mathrm{ft}\right)}^2+{\left(25\mathrm{ft}\right)}^2}=103.08\mathrm{ft}.$

.

$$\begin{gathered} Thedistances{H}_{31}is, \\ \sqrt{{(100ft)}^2+{\left(50ft\right)}^2}=111.8ft. \end{gathered}$$

.

From the diagram, the phase shifts are$D_{12}=D_{23}=25\mathrm{ft}\mathrm{and}{D}_{13}=50\mathrm{ft}.$

Determine the formulas of line-to-line capacitance with earth effect, geometric mean distance and height between conductors.

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

$\mathbf{r}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{\text{'}}}}{\mathbf{2}\mathbf{\times }\mathbf{12}}\mathit{f}\mathit{t}$ ……. (1)

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

Write the formula of capacitance with earth effect.

localid="1655197885389" ${\mathbf{C}}_{\mathbf{an}}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi \epsilon }}{\mathbf{In}\mathbf{\left(}\begin{array}{c}{\displaystyle \frac{D_{\mathrm{eq}}}{r}}\end{array}\mathbf{\right)}\mathbf{-}\mathbf{In}\mathbf{\left(}\begin{array}{c}{\displaystyle \frac{H_m}{H_S}}\end{array}\mathbf{\right)}}\frac{\mathbf{F}}{\mathbf{m}}\mathbf{line}\mathbf{-}\mathbf{to}\mathbf{-}\mathbf{neutral}$ ……. (2)

Determine the capacitance with earth effect.

Therefore, the capacitance is$9.7695\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}.$

Determine the capacitance without earth effect.

$$\begin{gathered} \left(\mathrm{b}\right) \\ \mathrm{Substitute}31.5\mathrm{ft}\mathrm{for}{\mathrm{D}}_{\mathrm{eq}},0.0444\mathrm{ft}\mathrm{for}8.854\times {10}^{-12}\mathrm{for}\mathrm{\epsilon }\mathrm{in}\mathrm{equation} \\ \left(6\right) \\ {{\mathrm{C}}_{\mathrm{an}}}^{\text{'}}=\frac{2\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left(\frac{31.5}{0.0444}\right)} \\ =8.4746\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

localid="1655197957052" $$\begin{gathered} \mathrm{Substitute}9.7695\times {10}^{-12}\frac{F}{m}for{C}_{an}and8.4746\times {10}^{-12}\frac{F}{m}\mathrm{for}{C_{an}}^{\text{'}} \\ \mathrm{in}\mathrm{equation}\left(7\right) \\ percentageerror=\frac{9.7695\times {10}^{-12}\frac{F}{m}-8.4746\times {10}^{-12}\frac{F}{m}}{9.7695\times {10}^{-12}\frac{F}{m}} \\ =13.25\% \end{gathered}$$

Therefore, the capacitance is $8.4746\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and error after neglecting earth effect is 13.25% .