Question

For the single-phase line of Problem 4.14 (b), if the height of the conductor above ground is 80ft, determine the line-to-line capacitance in F/m. Neglecting earth effect, evaluate the relative error involved. If the phase separation is doubled, repeat the calculations.

Short answer

Therefore, the capacitance is $5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and error after neglecting earth effect is 0.036%.

As the phase separation is doubled, the capacitance is $4.885\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and error after neglecting earth effect is 0.143 %.

Step-by-step solution

Write the given data.

The phase separation is D = 10 ft .

The frequency is f = 60 Hz.

The conductors radius is r = 0.06677 ft.

The height of the conductor above ground is h = 80 ft.

Determine the formulas of line-to-line capacitance with earth effect and neglecting earth effect, and error percentage.

Write the formula of capacitance with earth effect.

${\mathrm{C}}_{\mathrm{xy}}=\frac{\mathrm{\pi }\mathrm{\epsilon }}{\mathrm{In}\left(\begin{array}{c}\frac{\mathrm{D}}{\mathrm{r}}\end{array}\right)-\mathrm{In}\left(\begin{array}{c}{\displaystyle \frac{{\mathrm{H}}_{\mathrm{xy}}}{{\mathrm{H}}_{\mathrm{xx}}}}\end{array}\right)}\frac{\mathrm{F}}{\mathrm{m}}$ ……. (1)

Here, Dis phase separation, ris radius of conductor,$H_{xx}$is the vertical distance$H_{xy}$and isthe hypotenuse distances between two conductors

Write the formula of capacitance neglecting earth effect.

${\mathrm{C}}_{\mathrm{xy}}=\frac{\mathrm{\pi }\mathrm{\epsilon }}{\mathrm{In}\left(\begin{array}{c}\frac{\mathrm{D}}{\mathrm{r}}\end{array}\right)}\frac{\mathrm{F}}{\mathrm{m}}$ ……. (2)

Write the formula of error in capacitance.

$\mathrm{percentage}\mathrm{error}=\frac{{\mathrm{C}}_{\mathrm{xy}}-{\mathrm{C}}_{\mathrm{xy}}}{{\mathrm{C}}_{\mathrm{xy}}}\times 100$ ……. (3)

Determine the capacitance with and without earth effect and error.

The vertical distances is,

$$\begin{gathered} H_{xx}=80\mathrm{ft}+80\mathrm{ft} \\ =160\mathrm{ft} \end{gathered}$$

The hypotenuse distances is,

$$\begin{gathered} H_{xy}=\sqrt{{160}^2+{10}^2} \\ =160.3\mathrm{ft} \end{gathered}$$

Determine the capacitance.$$\begin{gathered} \mathrm{Substitute}10\mathrm{ft}\mathrm{for}D,0.06677\mathrm{ft}\mathrm{for}\mathrm{r},160.3\mathrm{for}{\mathrm{H}}_{\mathrm{xy},}160\mathrm{ft}\mathrm{for}{\mathrm{H}}_{\mathrm{xx}}\mathrm{and} \\ 8.854\times {10}^{-12}\mathrm{for}\mathrm{\epsilon }\mathrm{in}\mathrm{equation}\left(1\right). \\ {C}_{xy}=\frac{\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left(\begin{array}{c}\frac{10}{0.06677}\end{array}\right)-\mathrm{In}\left(\begin{array}{c}\frac{160.3}{160}\end{array}\right)} \\ =5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}10\mathrm{ft}\mathrm{for}D,0.06677\mathrm{ft}\mathrm{for}\mathrm{r},8.854\times {10}^{-12}\mathrm{\epsilon }\mathrm{in}\mathrm{equation}\left(2\right). \\ {C_{xy}}^{\text{'}}=\frac{\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left(\begin{array}{c}\frac{10}{0.06677}\end{array}\right)} \\ =5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Determine the error.

$$\begin{gathered} \mathrm{Substitute}5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\mathrm{for}{\mathrm{C}}_{\mathrm{xy}}\mathrm{and}5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\mathrm{for}{\mathrm{C}}_{\mathrm{xy}} \\ \mathrm{in}\mathrm{equation}\left(3\right). \\ \mathrm{percentage}\mathrm{errer}=\frac{5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}-5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}}{5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}} \\ =0.036\% \end{gathered}$$

Therefore, the capacitance is$5.555\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and error after neglecting earth effect is 0.036%.

Determine the capacitance with and without earth effect and error.

Now, the phase separation is doubled, that is

The vertical distances is,

$$\begin{gathered} H_{xx}=80\mathrm{ft}+80\mathrm{ft} \\ =160\mathrm{ft} \end{gathered}$$

The hypotenuse distances is,

$$\begin{gathered} H_{xy}=\sqrt{{160}^2+{20}^2} \\ =161.245\mathrm{ft} \end{gathered}$$

Determine the capacitance.

$$\begin{gathered} \mathrm{Substitute}20\mathrm{ft}\mathrm{for}D,0.06677\mathrm{ft}\mathrm{for}\mathrm{r},161.245\mathrm{for}{\mathrm{H}}_{\mathrm{xy},}160\mathrm{ft}\mathrm{for}{\mathrm{H}}_{\mathrm{xx}}\mathrm{and} \\ 8.854\times {10}^{-12}\mathrm{for}\mathrm{\epsilon }\mathrm{in}\mathrm{equation}\left(1\right). \\ {C}_{xy}=\frac{\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left(\begin{array}{c}\frac{10}{0.06677}\end{array}\right)-\mathrm{In}\left(\begin{array}{c}\frac{160.3}{160}\end{array}\right)} \\ =4.885\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}10\mathrm{ft}\mathrm{for}D,0.06677\mathrm{ft}\mathrm{for}\mathrm{r},8.854\times {10}^{-12}\mathrm{\epsilon }\mathrm{in}\mathrm{equation}\left(2\right). \\ {C_{xy}}^{\text{'}}=\frac{\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left(\begin{array}{c}\frac{20}{0.06677}\end{array}\right)} \\ =4.878\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Determine the error.

$$\begin{gathered} \mathrm{Substitute}4.885\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\mathrm{for}{\mathrm{C}}_{\mathrm{xy}}\mathrm{and}4.878\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\mathrm{for}{\mathrm{C}}_{\mathrm{xy}} \\ \mathrm{in}\mathrm{equation}\left(3\right). \\ \mathrm{percentage}\mathrm{errer}=\frac{4.885\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}-4.878\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}}{4.885\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}} \\ =0.143\% \end{gathered}$$

Therefore, the capacitance is$4.885\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and error after neglecting earth effect is 0.143% .