Question

Rework Problem 4.8 if the diameter of each conductor is (a) increased by$\mathbf{20}\mathbf{\%}$to $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{cm}$or (b) decreased by$\mathbf{20}\mathbf{\%}$to$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{cm}$without changing the phase spacing. Compare the results with those Of Problem 4.8.

Short answer

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

The inductance for each conductor due to both internal and external flux linkage

is.$0.853\frac{\mathrm{mH}}{\mathrm{km}}$

The total inductance of the line is$1.706\frac{\mathrm{mH}}{\mathrm{km}}$.

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

The inductance for each conductor due to both internal and external flux linkage

is $0.0934\frac{\mathrm{mH}}{\mathrm{km}}$.

The total inductance of the line is $1.869\frac{\mathrm{mH}}{\mathrm{km}}$.

The inductance due to internal linkage is independent on the diameter of the conductor. The inductance due to both linkage and total inductance decreases when diameter increases and vice-versa.

Step-by-step solution

Write the given data by the question.

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

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

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

The increased diameter of the conductor$d_{in}=1.8\mathrm{cm}$.

The decreased diameter of the conductor$d_{de}=1.2\mathrm{cm}$.

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 radius of the conductor is given by,

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

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

$\mathbf{L}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{In}\left(\frac{D}{0.7788r}\right)$ …… (3)

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

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

$\mathbf{L}\mathbf{=}{\mathbf{L}}_{\mathbf{x}}\mathbf{+}{\mathbf{L}}_{\mathbf{y}}$ …… (4)

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

Calculate the inductance of each conductor due to internal flux linkage, due to internal and external linkage and total inductance when the conductor’s diameter is1.8 cm.

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 radius of the conductor.

Substitute $1.8\mathrm{m}$for$d$into equation (2).

$$\begin{gathered} r=\frac{1.8\times {10}^{-2}}{2} \\ r=0.9\times {10}^{-2}\mathrm{m} \end{gathered}$$

The inductance of both the wire would be the same.

$L_x=L_y$

Calculate the inductance due internal and external flux linkage.

Substitute $0.9\times {10}^{-2}$for $r$, $0.5\mathrm{m}$for $D$in the equation (3).

$$\begin{gathered} L_x=2\times {10}^{-7}In\left(\frac{0.5}{0.7788\times 0.9\times {10}^{-2}}\right) \\ {L}_x=2\times {10}^{-7}In\left(71.33\right) \\ {L}_x=2\times {10}^{-7}\times 4.267 \\ {L}_x=0.853\frac{\mathrm{mH}}{\mathrm{km}} \end{gathered}$$

Hence inductance for each conductor due to both internal and external flux linkage is $0.853\frac{\mathrm{mH}}{\mathrm{km}}$.

Calculate the total inductance.

Substitute $0.853\frac{\mathrm{mH}}{\mathrm{km}}$for $L_x$and $0.853\frac{\mathrm{mH}}{\mathrm{km}}$for $L_y$in equation (4).

$$\begin{gathered} L=0.853+0.853 \\ L=1.706\frac{\mathrm{mH}}{\mathrm{km}} \end{gathered}$$

Hence, the total inductance of the line is$1.706\frac{\mathrm{mH}}{\mathrm{km}}$ .

Calculate the inductance of each conductor due to internal flux linkage, due to internal and external linkage and total inductance when the conductor’s diameter is 1.2cm.

Convert the equation (1) from $\mathrm{H}/\mathrm{m}$to $\mathrm{mH}/\mathrm{km}$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 radius of the conductor.

Substitute $1.2\mathrm{m}$for$d$into equation (2).

$$\begin{gathered} r=\frac{1.2\times {10}^{-2}}{2} \\ r=0.6\times {10}^{-2}\mathrm{m} \end{gathered}$$

The inductance of both the wire would be the same.

$L_x=L_y$

Calculate the inductance due internal and external flux linkage.

Substitute $0.6\times {10}^{-2}$for $r$,$0.5\mathrm{m}$for $D$in the equation (3).

$$\begin{gathered} L_x=2\times {10}^{-7}\mathrm{In}\left(\frac{0.5}{0.7788\times 0.6\times {10}^{-2}}\right) \\ {L}_x=2\times {10}^{-7}\mathrm{In}\left(107\right) \\ {L}_x=2\times {10}^{-7}\times 4.672 \\ {L}_x=0.934\frac{\mathrm{mH}}{\mathrm{km}} \end{gathered}$$

Hence inductance for each conductor due to both internal and external flux linkage is $0.934\frac{\mathrm{mH}}{\mathrm{km}}$.

Calculate the total inductance.

Substitute$0.934\frac{\mathrm{mH}}{\mathrm{km}}$for $L_x$and $0.934\frac{\mathrm{mH}}{\mathrm{km}}$for $L_y$in equation (4).

$$\begin{gathered} L=0.934+0.934 \\ L=1.869\frac{\mathrm{mH}}{\mathrm{km}} \end{gathered}$$

Hence the total inductance of the line is $1.869\frac{\mathrm{mH}}{\mathrm{km}}$.

The inductance due to internal linkage is independent on the diameter of the conductor. The inductance due to both linkage and total inductance decreases when diameter increases and vice-versa.