Question

A$\mathbf{60}\mathbf{Hz}$, single-phase two-wire overhead line has solid cylindrical copper conductors with a$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{cm}$diameter. The conductors are arranged in a horizontal configuration with$\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{m}$spacing. Calculate in$\frac{\mathbf{mH}}{\mathbf{km}}$(a) the inductance of each conductor due to internal flux linkages only, (b) the inductance of each conductor due to both internal and external flux linkages, and (c) the total inductance of the line.

Short answer

(a) The inductance of each conductor due to internal flux linkage is$0.05\frac{\mathrm{mH}}{\mathrm{km}}$.

(b) The inductance for each conductor due to both internal and external flux linkage is$0.889\frac{\mathrm{mH}}{\mathrm{km}}$.

(c) The total inductance of the line is$1.778\frac{\mathrm{mH}}{\mathrm{km}}$.

Step-by-step solution

Write the given data by the question.

The frequency of the system$f=60\mathrm{Hz}$.

Diameter of Copper conductor$d=1.5\mathrm{cm}$.

The spacing between the conductors$D=0.5\mathrm{m}$.

Determine the formulas to calculate the inductance of each conductor due to internal flus linkage, due to internal and external flux linkage and total inductance.

The expression to calculate the inductance due to internal flux linkage in given by,

${\mathbf{L}}_{\mathbf{in}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\frac{\mathbf{H}}{\mathbf{m}}$ …… (1)

The expression to calculate the inductance of each conductor due to internal and external is given by,

$\mathit{L}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}\mathbf{}}\mathit{I}\mathit{n}\left(\frac{D}{0.7788r}\right)$ …… (2)

Here, $\mathbf{r}$is the radius of the conductor.

The expression for the total inductance of the line is given by,

$\mathbf{L}\mathbf{=}{\mathbf{L}}_{\mathbf{x}}\mathbf{+}{\mathbf{L}}_{\mathbf{y}}$ …… (3)

Here, ${\mathbf{L}}_{\mathbf{x}}\mathbf{\&}{\mathbf{L}}_{\mathbf{y}}$are the inductance of the two lines.

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

$\mathbf{r}\mathbf{=}\left(0.7788\right)\frac{\mathbf{d}}{\mathbf{2}}$ …… (4)

Calculate the inductance of each conductor due to internal flux linkage.

(a)

Convert the equation (1) from to for the calculation the inductance of each conductor due to internal flux linkage.

$$\begin{gathered} \mathrm{L}=\frac{1}{2}\times {10}^{-7}\left(\frac{\mathrm{H}}{\mathrm{m}}\right)\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1000\mathrm{mH}}{1\mathrm{H}} \\ \mathrm{L}=0.05\frac{\mathrm{mH}}{\mathrm{km}} \end{gathered}$$

Hence the inductance of each conductor due to internal flux linkage is$0.05\frac{\mathrm{mH}}{\mathrm{km}}$.

Calculate the inductance of each conductor due to internal and external flux linkage.

(b)

Calculate the radius of the conductor.

Substitute$1.5\mathrm{m}$for $d$into equation (4).

$$\begin{gathered} \mathrm{r}=\left(0.7788\right)\frac{1.5\times {10}^{-2}}{2} \\ \mathrm{r}=5.841\times {10}^{-3}\mathrm{m} \end{gathered}$$

The inductance of both the wire would be the same.

$L_x=L_y$

Calculate the inductance due internal and external flux linkage.

Substitute$5.841\times {10}^{-3}\mathrm{m}$for $r$, and$0.5\mathrm{m}$for $D$in the equation (2).

$$\begin{gathered} L_x=2\times {10}^{-7}In\left(\frac{0.5}{5.841\times {10}^{-3}}\right) \\ {L}_x=2\times {10}^{-7}In\left(85.60\right) \\ {L}_x=2\times {10}^{-7}\times 4.449 \\ {L}_x=0.889\frac{mH}{km} \end{gathered}$$

Hence, inductance for each conductor due to both internal and external flux linkage is$0.889\frac{mH}{km}$

Determine the total inductance of the line.

(c)

Calculate the total inductance.

Substitute $0.889\frac{\mathrm{mH}}{\mathrm{km}}$for $L_x$and $0.889\frac{\mathrm{mH}}{\mathrm{km}}$for$L_y$in equation (3).

$$\begin{gathered} \mathrm{L}=0.889+0.889 \\ \mathrm{L}=1.778\frac{\mathrm{mH}}{\mathrm{kmA}} \end{gathered}$$

Hence, the total inductance of the line is $1.778\frac{\mathrm{mH}}{\mathrm{kmA}}$.