Question

Figure 4.37 shows the conductor configuration of a three-phase transmission line and a telephone line supported on the same towers. The power line carries a balanced current of$\mathbf{250}\mathbf{A}\mathbf{/}\mathbf{phase}\mathbf{}$at$\mathbf{60}\mathbf{Hz}$,while the telephone line is directly located below phase$\mathbf{b}$. Assume balanced three-phase currents in the power line. Calculate the voltage per kilometre induced in the telephone line.

Short answer

The induced voltage in telephone line is$3.845\frac{\mathrm{V}}{\mathrm{km}}$.

Step-by-step solution

Write the given data from the question.

The current in the power,$/=250\mathrm{A}$per phase.

The spacing between the transmission line conductor is$4.6\mathrm{m}$.

The spacing between the telephone line is$1.2\mathrm{m}$.

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

Determine the formulas to calculate the voltage per kilometre induced in the telephone line.

The equation to calculate the flux linkage with conductor$1$and $2$due to current in the conductor is given as follows.

${\mathbf{\varphi }}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}{\mathbf{I}}_{\mathbf{a}}\mathbf{In}\left(\frac{D_{a2}}{D_{a1}}\right)$

Here,${\mathbf{I}}_{\mathbf{a}}$is the current in the conductor$\mathbf{a}$, ${\mathbf{D}}_{\mathbf{a}\mathbf{2}}$is the distance between the conductor$\mathbf{a}\mathbf{}\mathbf{,}\mathbf{2}$, and ${\mathbf{D}}_{\mathbf{a}\mathbf{1}}$is the distance between conductor$\mathbf{a}\mathbf{}\mathbf{,}\mathbf{1}$.

The equation to calculate the flux linkage with conductor and due to current in the conductor is given as follows.

${\mathbf{\varphi }}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}{\mathbf{I}}_{\mathbf{c}}\mathbf{In}\mathbf{\left(}\frac{{\mathbf{D}}_{\mathit{c}\mathbf{2}}}{{\mathbf{D}}_{\mathit{c}\mathbf{1}}}\mathbf{\right)}$

Here, ${\mathbf{I}}_{\mathbf{c}}$is the current in the conductor$\mathbf{c}\mathbf{}\mathbf{,}{\mathbf{D}}_{\mathbf{c}\mathbf{2}}$is the distance between the conductor$\mathbf{c}\mathbf{}\mathbf{,}\mathbf{}\mathbf{2}$and ${\mathbf{D}}_{\mathbf{c}\mathbf{1}}$is the distance between conductor$\mathbf{c}\mathbf{}\mathbf{,}\mathbf{}\mathbf{1}$.

The total flux linkage is the sum of the flux linkage due to current in conductor $a$and current in the conductor$b$.

$\mathbf{\varphi }\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}{\mathbf{I}}_{\mathbf{a}}\mathbf{In}\left(\frac{D_{a2}}{D_{a2}}\right)\mathbf{+}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}{\mathbf{I}}_{\mathbf{a}}\mathbf{In}\left(\frac{D_{c2}}{D_{c1}}\right)$

Since the conductor is at equal distance from$1$and $2$, So, the distances

$$\begin{gathered} {\mathbf{D}}_{\mathbf{a}\mathbf{1}}\mathbf{=}{\mathbf{D}}_{\mathbf{c}\mathbf{2}} \\ {\mathbf{D}}_{\mathbf{a}\mathbf{2}}\mathbf{=}{\mathbf{D}}_{\mathbf{c}\mathbf{1}} \end{gathered}$$

The modified the total flux linkage equation as,

$$\begin{gathered} \mathbf{\varphi }\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}{\mathbf{I}}_{\mathbf{a}}\mathbf{In}\mathbf{\left(}\frac{{\mathbf{D}}_{\mathbf{a}\mathbf{2}}}{{\mathbf{D}}_{\mathbf{a}\mathbf{1}}}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}{\mathbf{I}}_{\mathbf{C}}\mathbf{In}\mathbf{\left(}\frac{{\mathbf{D}}_{a2}}{{\mathbf{D}}_{\mathit{a}\mathbf{1}}}\mathbf{\right)} \\ \mathbf{\varphi }\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\left(I_a‐I_c\right)\mathbf{}\mathbf{In}\mathbf{\left(}\frac{{\mathbf{D}}_{\mathbf{a}\mathbf{2}}}{{\mathbf{D}}_{\mathbf{a}\mathbf{1}}}\mathbf{\right)} \end{gathered}$$

Since the current are balanced and${\mathbf{120}}^{\mathbf{\circ }}$shifted with each other.

${\mathbf{I}}_{\mathbf{a}}\mathbf{‐}{\mathbf{I}}_{\mathbf{c}}\mathbf{=}\sqrt{\mathbf{3}{\mathbf{I}}_{\mathbf{a}}}\mathbf{\angle }{\mathbf{30}}^{\mathbf{\circ }}$

Again, modified the total flux linkage equation as,

$\mathbf{\varphi }\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\sqrt{\mathbf{3}{\mathbf{I}}_{\mathbf{a}}}\mathbf{\angle }{\mathbf{30}}^{\mathbf{\circ }}\mathbf{In}\left(\frac{D_{a2}}{D_{a1}}\right)$

The equation to calculate the mutual inductance is given as follows.

$\mathbf{M}\mathbf{=}\frac{{\mathbf{\varphi }}_{\mathbf{12}}}{{\mathbf{I}}_{\mathbf{a}}}$

The modified equation of mutual inductance is given as follows.

$$\begin{gathered} \mathbf{M}\mathbf{=}\frac{\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\sqrt{\mathbf{3}}{\mathbf{I}}_{\mathbf{a}}\mathbf{In}\left({\displaystyle \frac{D_{a2}}{D_{a1}}}\right)}{{\mathbf{I}}_{\mathbf{a}}} \\ \mathbf{M}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\sqrt{\mathbf{3}}\mathbf{In}\mathbf{\left(}\frac{{\mathbf{D}}_{\mathbf{a}\mathbf{2}}}{{\mathbf{D}}_{\mathbf{a}\mathbf{1}}}\mathbf{\right)} \end{gathered}$$ …… (1)

The equation to calculate the voltage induced in the telephone line is given as follows.

$\mathbf{V}\mathbf{=}\mathbf{2}\mathbf{\tau \tau }\mathbf{}\mathbf{fMI}$ …… (2)

Calculate the induced voltage in the telephone line.   

Calculate the distance between the conductors.

Distance between conductor$1,\mathrm{a}$and $2,\mathrm{c}$.

$$\begin{gathered} D_{a1}=D_{c2} \\ {D}_{a1}=\sqrt{5^2+{\left(4.6-0.6\right)}^2} \\ {D}_{a1}=\sqrt{41} \\ {D}_{a1}=6.4\mathrm{m} \end{gathered}$$

Distance between conductor $2,a$and $1,c$.

$$\begin{gathered} D_{a2}=D_{c2} \\ {D}_{a2}=\sqrt{5^2+{\left(4.6-0.6\right)}^2} \\ {D}_{a2}=\sqrt{52.04} \\ {D}_{a2}=7.2\mathrm{m} \end{gathered}$$

Calculate the mutual inductance.

substitute$7.2\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{a}2}\mathrm{and}6.4\mathrm{for}{D}_{a1}$into equation (1).

$$\begin{gathered} M=2\times {10}^{-7}\times \sqrt{3}\times \mathrm{In}\left(\frac{7.2}{6.4}\right) \\ M=3.464\times {10}^{-7}\mathrm{In}\left(1.125\right) \\ M=3.464\times {10}^{-7}\times 0.1177 \\ M=4.08\times {10}^{-8}\frac{\mathrm{H}}{\mathrm{m}} \end{gathered}$$

Calculate the induced voltage in telephone line.

Substitute$60Hz\mathrm{for}f,250a\mathrm{A}\mathrm{for}/\mathrm{and}4.08\times {10}^{-8}\frac{\mathrm{H}}{\mathrm{m}}\mathrm{fo}M$into equation (2).

$$\begin{gathered} V=2\pi \times 60\times 250\times 4.08\times {10}^{-8} \\ V=384530km.94\times {10}^{-8} \\ V=3.845\times {10}^{-3} \\ V=3.845\frac{V}{\mathrm{km}} \end{gathered}$$

Hence the induced voltage in telephone line is$3.845\frac{V}{\mathrm{km}}$.