Question

Rework problem 4.32 if the diameter of each conductor is (a) increased by 20% to 1.8 cmor is (b) decreased by 20% to 1.2 cm. Compare the results with those of Problem 4.32.

Short answer

(a) The capacitance to neutral is$13.841\times {10}^{-12}\mathrm{F}/\mathrm{m}$ and the capacitance to neutral is decreased by 4.57% when the diameter is increased by 20% . The admittance-to-neutral is$\mathrm{j}5.215\times {10}^{-6}\mathrm{S}/\mathrm{Km}$ and the admittance to neutral is increased by 4.46% when the diameter is increased by 20% .

(b) The capacitance to neutral is $12.57\times {10}^{-12}\mathrm{F}/\mathrm{m}$and the capacitance to neutral is decreased by 50.5% when the diameter is decreased by 20% . The admittance-to-neutral is$\mathrm{j}4.736\times {10}^{-6}\mathrm{S}/\mathrm{Km}$ and the admittance to neutral is decreased by$5.12\%$ when the diameter is decreased by 20% .

Step-by-step solution

Write the given data of the question.

Given configuration is a single-phase, two wire overhead line in which the conductors are arranged in horizontal configuration.

The conductor spacing is 0.5 m .

The diameter of each solid cylindrical copper conductor is 1.5 cm.

Write the formulae for capacitance to neutral and admittance to neutral.

The capacitance to neutral is given by,

${\mathbf{C}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau \epsilon }}{\mathbf{In}\left({\displaystyle \frac{D}{r}}\right)}$ …… (1)

Here D is the conductor spacing, r is the radius of each solid cylinder and${\mathit{C}}_{\mathbf{n}}$ is the capacitance to neutral.

The admittance to neutral is given by,

${\mathbf{Y}}_{\mathbf{n}}\mathbf{=}{\mathbf{j\omega C}}_{\mathbf{n}}$ …… (2)

Here ${\mathbf{Y}}_{\mathbf{n}}$ is the admittance to neutral, $\mathit{\omega }$is the angular frequency,${\mathit{C}}_{\mathbf{n}}$ is the capacitance to neutral.

 Step 3: Calculate the capacitance to neutral and admittance to neutral.

(a)

Calculate the radius of each solid cylindrical copper conductor.

$$\begin{gathered} \mathrm{r}=\frac{1.5}{2}\mathrm{cm} \\ =0.0075\mathrm{m} \end{gathered}$$

Substitute 0.5 m for D ,0.0075 m for r and $8.55\times {10}^{-12}\mathrm{F}/\mathrm{m}$ for$\epsilon$ in equation (1).

$$\begin{gathered} C_n=\frac{2\pi (8.85\times {10}^{-12})}{In\left({\displaystyle \frac{0.5}{0.0075}}\right)} \\ =13.239\times {10}^{-12}F/m \end{gathered}$$

The admittance to neutral is given by,

$Y_n=j\omega C_n$

Here $Y_n$ is the admittance to neutral,$\omega$ is the angular frequency and $C_n$ is the capacitance to neutral.

Substitute $13.239\times {10}^{-12}\mathrm{F}/\mathrm{m}$for$C_n$ in the above equation.

$$\begin{gathered} Y_n=j\omega C_n \\ =j(2\pi \times 60)(13.239\times {10}^{-12}) \\ =j4.992\times {10}^{-6}S/K\mathrm{m} \end{gathered}$$

The diameter of each cylindrical copper conductor is increased by 20% = 1.8 cm .

Calculate the radius of each solid cylindrical conductor.

$$\begin{gathered} \mathrm{r}=\frac{1.8}{2}\mathrm{cm} \\ =0.009\mathrm{m} \end{gathered}$$

Substitute 0.5 m for D , 0.009 m for r and $8.55\times {10}^{-12}\mathrm{F}/\mathrm{m}$ for$\epsilon$ in equation ().

$$\begin{gathered} {\mathrm{C}}_{\mathrm{n}}=\frac{2\mathrm{\pi }(8.85\times {10}^{-12})}{\mathrm{In}\left({\displaystyle \frac{0.5}{0.009}}\right)} \\ =13.841\times {10}^{-12}\mathrm{F}/\mathrm{m} \end{gathered}$$

Therefore the capacitance to neutral is$13.841\times {10}^{-12}\mathrm{F}/\mathrm{m}$ .

The admittance to neutral is given by,

$Y_n=j\omega C_n$

Here $Y_n$ is the admittance to neutral,$\omega$ is the angular frequency and$C_n$ is the capacitance to neutral.

Substitute$13.841\times {10}^{-12}\mathrm{F}/\mathrm{m}$ for$C_n$ in the above equation.

$$\begin{gathered} Y_n=j\omega C_n \\ =j(2\pi \times 60)(13.841\times {10}^{-12}) \\ =j5.215\times {10}^{-6}S/Km \end{gathered}$$

Therefore the admittance to neutral is role="math" localid="1655888462868" $j5.215\times {10}^{-6}S/Km$.

Compare the values of capacitance and admittance obtained by increasing the diameter of each conductor by with the actual value.

Determine the change in the value of capacitance to neutral is,

$$\begin{gathered} ∆{\mathrm{C}}_{\mathrm{n}}=\frac{13.841\times {10}^{-12}-13.239\times {10}^{-12}}{13.239\times {10}^{-12}} \\ =4.57\% \end{gathered}$$

Therefore the capacitance to the neutral is increased by 4.57% and diameter is increased by 20% .

The change in the value of admittance to neutral is,

$$\begin{gathered} ∆{\mathrm{Y}}_{\mathrm{n}}=\frac{5.215\times {10}^{-6}-4.992\times {10}^{-6}}{4.992\times {10}^{-6}} \\ =4.46\% \end{gathered}$$

Therefore the admittance to neutral is increased by 4.46% when the diameter is increased by 20% .

Calculate the capacitance to neutral and admittance to neutral.

(b)

The diameter of each cylindrical copper conductor is increased by 20% = 1.2 cm

Calculate the radius of each solid cylindrical conductor.

$$\begin{gathered} \mathrm{r}=\frac{1.2}{2}\mathrm{cm} \\ =0.006\mathrm{m} \end{gathered}$$

The capacitance to neutral is given by,

${\mathrm{C}}_{\mathrm{n}}=\frac{2\mathrm{\pi \epsilon }}{\mathrm{In}\left({\displaystyle \frac{\mathrm{D}}{\mathrm{r}}}\right)}$

Here D is the conductor spacing, r is the radius of each solid cylinder and$C_n$ is the capacitance to neutral.

Substitute 0.5 m for D , 0.006 m for r and$8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}$ for$\epsilon$ in the above equation.

$$\begin{gathered} C_n=\frac{2\pi (8.85\times {10}^{-12})}{In\left({\displaystyle \frac{0.5}{0.006}}\right)} \\ =12.57\times {10}^{-12}\mathrm{F}/\mathrm{m} \end{gathered}$$

Therefore the capacitance to neutral is role="math" localid="1655891433706" $12.57\times {10}^{-12}\mathrm{F}/\mathrm{m}$ .

The admittance to neutral is given by,

$Y_n=j\omega C_n$

Here $Y_n$ is the admittance to neutral,$\omega$ is the angular frequency and$C_n$ is the capacitance to neutral.

Substitute $12.57\times {10}^{-12}\mathrm{F}/\mathrm{m}$for$C_n$ in the above equation.

$$\begin{gathered} Y_n=j\omega C_n \\ =j(2\pi \times 60)(12.57\times {10}^{-12}) \\ =j4.736\times {10}^{-6}S/Km \end{gathered}$$

Therefore the admittance to neutral is $j5.215\times {10}^{-6}S/Km$ .

Compare the capacitance and admittance obtained by decreasing the diameter of each conductor by 20% .

Determine the change in the value of capacitance to neutral is,

$$\begin{gathered} ∆C_{\mathrm{n}}=\frac{13.239\times {10}^{-12}-12.57\times {10}^{-12}}{13.239\times {10}^{-12}} \\ =5.05\% \end{gathered}$$

Therefore the capacitance to neutral is decreased by 5.05 % and the diameter is decreased by 20% .

The change in the value of admittance to neutral is,

$$\begin{gathered} ∆{\mathrm{Y}}_{\mathrm{n}}=\frac{4.992\times {10}^{-6}-4.736\times {10}^{-6}}{4.992\times {10}^{-6}} \\ =5.12\% \end{gathered}$$

Therefore, the admittance to neutral is decreased by 5.12% and the diameter decrease by 20% .