Question

Calculate the capacitance-to-neutral in$\frac{\mathbf{F}}{\mathbf{m}}$ and the admittance-to-neutral in $\frac{\mathbf{S}}{\mathbf{Km}}$for the single-phase line in Problem 4.8. Neglect the effect of the earth plane.

Short answer

The for the single-phase line the capacitance to neutral is $13.245\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and admittance to neutral is $j4.993\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{Km}}$.

Step-by-step solution

Write the given data from the question.

The frequency of the system,f = 60 Hz

Diameter of Copper conductor, d = 1.5 cm

The spacing between the conductors, D = 0.5 m

Determine the formulas to calculate the capacitance and admittance for the single-phase line.

The expression to calculate the radius of the conductor is given by,

$\mathbf{r}\mathbf{=}\frac{\mathbf{d}}{\mathbf{2}}$ …… (1)

The equation to calculate capacitance to neutral is given as follows.

${\mathbf{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau \epsilon }}{\mathbf{In}\left({\displaystyle \frac{D}{r}}\right)}$ …… (2)

The equation to calculate the admittance to neutral is given as follows.

${\mathbf{Y}}_{\mathbf{n}}\mathbf{=}\mathbf{j}\mathbf{2}{\mathbf{\tau \tau fC}}_{\mathbf{n}}$ …… (3)

Calculate the capacitance and admittance of the for the single-phase line.

Calculate the radius of the solid cylinder.

Substitute 1.5 cm for D into equation (1).

$$\begin{gathered} \mathrm{r}=\frac{1.5\times {10}^{-2}}{2} \\ \mathrm{r}=0.0075\mathrm{m} \end{gathered}$$

Calculate the capacitance to neutral.

Substitute 1.5 cm for D, 0.0075 m for r into equation (2).

$$\begin{gathered} \mathrm{C}=\frac{2\mathrm{\pi }\times 8.85\times {10}^{-12}}{\mathrm{In}\left({\displaystyle \frac{1.5\times {10}^{-2}}{0.0075}}\right)} \\ \mathrm{C}=\frac{5.563\times {10}^{-11}}{4.2} \\ \mathrm{C}=13.245\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Calculate the admittance to neutral.

Substitute 60 Hz for f ,$13.245\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$ for C into equation (3).

$$\begin{gathered} {\mathrm{Y}}_{\mathrm{n}}=\mathrm{j}(2\mathrm{\pi }\times 60)\times 13.245\times 10\times {10}^{-12} \\ {\mathrm{Y}}_{\mathrm{n}}=\mathrm{j}4.993\times {10}^{-9} \\ {\mathrm{Y}}_{\mathrm{n}}=\mathrm{j}4.993\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{Km}} \end{gathered}$$

Therefore, the capacitance to neutral is $13.245\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$and admittance to neutral is $j4.993\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{Km}}$for the single-phase line.