Question

A three-phase60 Hz, 125 Kmoverhead transmission line has flat horizontal spacing with three identical conductors. The conductors have an outside diameter of 3.28 cmwith12 mbetween adjacent conductors.

(a) Determine the capacitive reactance-to-neutral in V-m per phase and the capacitive reactance of the line in V per phase. Neglect the effect of the earth plane.

(b) Assuming that the conductors are horizontally placed 20 mabove ground, repeat part (a) while taking into account the effect of ground. Consider the earth plane to be a perfect conductor.

Short answer

(a) The capacitive reactance in $\Omega .m$is localid="1655897938346" $-j3.254\times {10}^8\Omega .m$and in $\Omega$ is localid="1655898387051" $-j2603\Omega$.

(b) The capacitive reactance in$\Omega .m$islocalid="1655898555362" $-j3.217\times {10}^8\Omega .m$ and in $\Omega$ islocalid="1655898526043" $-j2573\Omega$.

Step-by-step solution

Given data.

The ength of overhead line is 125 km .

The ourside diameter of conductor is d =3.28 cm .

The distance between conductors is D = 12 m

Determine the formulas of line-to-line capacitance.

Write the formula of radius of conductor.

$\mathbf{r}\mathbf{=}\frac{\mathbf{d}}{\mathbf{2}}$ ……. (1)

Here,dis the outside diameter of conductor.

Write the formula of equivalent distance between conductors.

${\mathbf{D}}_{\mathbf{eq}}\mathbf{=}\sqrt[\mathbf{3}]{{\mathbf{D}}_{\mathbf{ab}}{\mathbf{D}}_{\mathbf{bc}}{\mathbf{D}}_{\mathbf{ac}}}$ ……. (2)

Here, $D_{ab}$ is the distance between conductors a and b , $D_{bc}$ is the distance between conductors c and b and $D_{ac}$is the distance between conductors a and c .

Write the formula of capacitance.

${\mathbf{C}}_{\mathbf{an}}\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau \epsilon }}{\mathbf{In}\mathbf{\left(}{\displaystyle \frac{D_{\mathrm{eq}}}{r}}\mathbf{\right)}}\frac{\mathbf{F}}{\mathbf{m}}$ ……. (3)

Write the formula of capacitive reactance.

${\mathbf{X}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{j}\mathbf{2}{\mathbf{\tau \tau fC}}_{\mathbf{an}}}\mathbf{\Omega }$ ……. (4)

Here, fis frequency.

Write the formula of capacitance for horizontally placed conductors.

role="math" localid="1655896793083" ${\mathbf{C}}_{\mathbf{an}}\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau \epsilon }}{\mathbf{In}\mathbf{\left(}{\displaystyle \frac{D_{\mathrm{eq}}}{r}}\mathbf{\right)}\mathbf{-}{\displaystyle \frac{\mathbf{1}}{\mathbf{3}}}\mathbf{In}\mathbf{\left(}{\displaystyle \frac{H_{13}{H}_{23}{H}_{12}}{H_1{H}_2{H}_3}}\mathbf{\right)}}\frac{\mathbf{F}}{\mathbf{m}}$ ……. (5)

Here, $H_{1,}{H}_2$ and $H_3$ are the vertical distance between two conductors, are $H_{12}$, $H_{13}$and $H_{23}$ are the hypotenuse distances.

Determine the capacitance and capacitive reactance.

(a)

Determine the radius of conductors.

Substitute 3.28 cm for d in equation (1).

$$\begin{gathered} \mathrm{r}=\frac{3.28\mathrm{cm}}{2} \\ =1.64\mathrm{cm} \end{gathered}$$

Determine the equivalent distance.

Substitute localid="1655897357586" $12\mathrm{m}$for $D_{ab}$ ,$12\mathrm{m}$for $D_{bc}$and 24 m for $D_{ac}$in equation (2).

$$\begin{gathered} D_{eq}=\sqrt[3]{(12\mathrm{m})(12\mathrm{m})(24\mathrm{m})} \\ =15.12\mathrm{m} \end{gathered}$$

Determine the capacitance.

Substitute 15.12 mfor $D_{eq}$, 1.64 cm for r and $8.854\times {10}^{-12}$for $\epsilon$ in equation (3).

$$\begin{gathered} {\mathrm{C}}_{\mathrm{an}}=\frac{2\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left({\displaystyle \frac{15.12}{0.0164}}\right)} \\ =8.149\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Determine the capacitive reactance.

Substitute $8.149\times {10}^{-12}\frac{F}{m}$for $C_{an}$and 60 Hzfor f in equation (4).

$$\begin{gathered} {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\mathrm{j}2\mathrm{\pi }\left(60\right)\left(8.149\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\right)} \\ =-\mathrm{j}3.254\times {10}^8\mathrm{\Omega m} \end{gathered}$$

Determine the capacitove reactance per-kilometer.

$$\begin{gathered} X_c=\frac{-j3.254\times {10}^8\Omega m}{125\times {10}^{3}m} \\ =-j2603\Omega \end{gathered}$$

Therefore, the capacitive reactance in $\Omega .m$is $-j3.254\times {10}^8\Omega .m$and in $\Omega$ is $-j2603\Omega$ .

Determine the capacitance and capacitive reactance.

(b)

The vertical distances are $H_1=40\mathrm{m},H_2=40\mathrm{m}and{H}_3=40m$.

The hypotenuse distances are,

$$\begin{gathered} H_{13}=\sqrt{{40}^2+{12}^2} \\ =41.761m \end{gathered}$$

And,

$$\begin{gathered} H_{13}=\sqrt{{40}^2+{24}^2} \\ =46.648m \end{gathered}$$

Determine the capacitance.

Substitute 15.12 mfor $D_{eq}$, 1.64 cm for r , 41.761 m for $H_{12,}{H}_{23}$,46.648 m for role="math" localid="1655899397804" $H_{13,}40\mathrm{m}$ for $H_{1,}{H}_{2,}and{H}_3$ and $8.854\times {10}^{-12}$for $\epsilon$in equation (5).

$$\begin{gathered} {\mathrm{C}}_{\mathrm{an}}=\frac{2\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left({\displaystyle \frac{15.12}{0.0164}}\right)-{\displaystyle \frac{1}{3}}\mathrm{In}\left({\displaystyle \frac{41.761\times 41.761\times 46.648}{40\times 40\times 40}}\right)} \\ =8.245\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Substitute $8.245\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$ for $C_{an}$ and 60 Hzfor f in equation (4).

$$\begin{gathered} {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\mathrm{j}2\mathrm{\pi }\left(60\right)\left(8.245\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\right)} \\ =-\mathrm{j}3.217\times {10}^8\mathrm{\Omega m} \end{gathered}$$

Determine the capacitove reactance per-kilometer.

$$\begin{gathered} X_c=\frac{-j3.217\times {10}^8\Omega m}{125\times {10}^{3}m} \\ =-j2573\Omega \end{gathered}$$

Therefore, the capacitive reactance inrole="math" localid="1655898626944" $\Omega m$ is $-j3.217\times {10}^8\Omega .m$ and in $\Omega$ is $-j2573\Omega$ .