Question

Rework problem 4.34 if the phase spacing is is (a) increased by 20%

to 4.8ftor is (b) decreased by 20%to 3.2ft.Compare the results with those of Problem 4.34.

Short answer

(a) The capacitance to neutral is $10.221\times {10}^{-12}\mathrm{F}/\mathrm{m}$and the capacitance to neutral is decreased by 3.865% when the phase spacing is increased by 20% .The admittance-to-neutral is $\mathrm{j}3.851\times {10}^{-6}\mathrm{S}/\mathrm{km}$and the admittance to neutral is increased by 3.869%

when phase spacing is increased by 20% .

(b) The capacitance to neutral is $11.045\times {10}^{-12}\mathrm{F}/\mathrm{m}$and the capacitance to neutral is

increased by 3.884% when the phase spacing is decreased by 20% . The admittance-to-

neutral is $\mathrm{j}4.161\times {10}^{-6}\mathrm{S}/\mathrm{km}$and the admittance to neutral is increased by 3.869% when

the phase spacing 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 0.5m .

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

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

The capacitance to neutral is given by,

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

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

Calculate the capacitance to neutral and admittance to neutral.

(a)

Calculate the radius of each solid cylindrical copper conductor.

$$\begin{gathered} r=\frac{0.5}{2}\mathrm{in} \\ =0.025\times 0.0254\mathrm{m} \\ =0.00635\mathrm{m} \end{gathered}$$

The capacitance to neutral is given by,

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

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

capacitance to neutral.

Substitute 1.219m for D,0.00635m for r and $8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}$for $\epsilon$in the above equation.

role="math" localid="1655881415708" $$\begin{gathered} C_n=\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{In\left({\displaystyle \frac{1.219}{0.00635}}\right)} \\ =10.632\times {10}^{-12}F/m \end{gathered}$$

The admittance to neutral is given by,

${\mathrm{Y}}_{\mathrm{n}}={\mathrm{j\omega C}}_{\mathrm{n}}$

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 in the above equation.

$$\begin{gathered} {\mathrm{Y}}_{\mathrm{n}}={\mathrm{j\omega C}}_{\mathrm{n}} \\ =\mathrm{j}\left(2\mathrm{\pi }\times 60\right)\left(10.632\times {10}^{-12}\right) \\ =\mathrm{j}4.006\times {10}^{-6}\mathrm{S}/\mathrm{km} \end{gathered}$$

The phase spacing between the conductor is getting increased by 20% = 4.8ft .

$$\begin{gathered} D=4.8ft \\ =\frac{4.8}{3.28}m \\ =1.463m \end{gathered}$$

The capacitance to neutral is given by,

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

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

capacitance to neutral.

Substitute $1.463\mathrm{m}$for D,0.00635m for r and $8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}$for in the above equation.

$$\begin{gathered} C_n=\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{In\left({\displaystyle \frac{1.463}{0.00635}}\right)} \\ =10.221\times {10}^{-12}F/m \end{gathered}$$

Therefore the capacitance to neutral is $10.221\times {10}^{-12}\mathrm{F}/\mathrm{m}$.

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 $10.221\times {10}^{-12}\mathrm{F}/\mathrm{m}$for $C_n$in the above equation.

role="math" localid="1655882672087" $$\begin{gathered} Y_n=j\omega C_n \\ =j\left(2\mathrm{\pi }\times 60\right)\left(10.221\times {10}^{-12}\right) \\ =j3.851\times {10}^{-6}S/km \end{gathered}$$

Therefore the admittance to neutral is $\mathrm{j}3.851\times {10}^{-6}\mathrm{S}/\mathrm{km}$.

Compare the values of capacitance and admittance obtained with the actual value when the phase spacing is increased by 20% .

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

$$\begin{gathered} ∆C_n=\frac{10.632\times {10}^{-12}-10.221\times {10}^{-12}}{10.632\times {10}^{-12}} \\ =3.865\% \end{gathered}$$

Therefore the capacitance to neutral is decreased by 3.865% if the phase spacing increased by 20% .

The change in the value of admittance to neutral is,

$$\begin{gathered} ∆Y_n=\frac{4.006\times {10}^{-6}-3.851\times {10}^{-6}}{4.006\times {10}^{-6}} \\ =3.869\% \end{gathered}$$

Therefore, the admittance to neutral is decreased by 3.869% when the phase spacing is increased by 20% .

Calculate the capacitance to neutral and admittance to neutral.

(b)

Calculate the radius of each solid cylindrical conductor.

$$\begin{gathered} r=\frac{0.5}{2}in \\ =0.006\times 0.0254 \\ =0.00635\mathrm{m} \end{gathered}$$

The phase spacing between the conductor is getting decreased by 20% = 3.2ft .

$$\begin{gathered} D=3.2ft \\ =\frac{3.2}{3.28}m \\ =0.975m \end{gathered}$$

The capacitance to neutral is given by,

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

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

Substitute 0.975m for D,0.00635 for r and $8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}$for $\epsilon$in the above equation.

$$\begin{gathered} C_n=\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{In\left({\displaystyle \frac{0.975}{0.00635}}\right)} \\ =11.045\times {10}^{-12}F/m \end{gathered}$$

Therefore the capacitance to neutral is$11.045\times {10}^{-12}\mathrm{F}/\mathrm{m}$.

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$11.045\times {10}^{-12}\mathrm{F}/\mathrm{m}$for $C_n$in the above equation.

$$\begin{gathered} {\mathrm{Y}}_{\mathrm{n}}={\mathrm{j\omega C}}_{\mathrm{n}} \\ =\mathrm{j}\left(2\mathrm{\pi }\times 60\right)\left(11.045\times {10}^{-12}\right) \\ =\mathrm{j}4.161\times {10}^{-6}\mathrm{S}/\mathrm{km} \end{gathered}$$

Therefore the admittance to neutral is $\mathrm{j}4.161\times {10}^{-6}\mathrm{S}/\mathrm{km}$.

Compare capacitance and admittance of the decreased phased shift and the actual value.

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

$$\begin{gathered} ∆C_n=\frac{11.045\times {10}^{-12}-10.632\times {10}^{-12}}{10.632\times {10}^{-12}} \\ =3.884\% \end{gathered}$$

Therefore the capacitance to neutral is increased by 3.884% for the phase spacing is decreased by 20% .

The change in the value of admittance to neutral is,

$$\begin{gathered} ∆Y_n=\frac{4.161\times {10}^{-6}-4.006\times {10}^{-6}}{4.006\times {10}^{-6}} \\ =3.869\% \end{gathered}$$

Therefore, the admittance to neutral is increased by for the phase spacing is

decreased by 20% .