Question

Three ACSR Drakeconductors are used for a three-phase overhead transmission line operating at. The conductor configuration is in the form of an isosceles triangle with sides of 20, 20, and 38 ft.

(a) Find the capacitance-to-neutral and capacitive reactance-to-neutral for eachlength of line.

(b) For a line length of 175 mileand a normal operating voltage of220 kV, determine the capacitive reactance-to-neutral for the entire line length as well as the charging current per mile and total three-phase reactive power supplied by the line capacitance.

Short answer

(a)The capacitance to neutral is$8.8466\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$ and the capacitive reactance to neutral isrole="math" localid="1655877094040" $-\mathrm{j}0.1864\times {10}^6\mathrm{\Omega }.\mathrm{mi}$ .

(b) The capacitive reactance for 175 mile is $j1065\Omega$, the charging current is $119\mathrm{A}$and the three-phase reactive power is$43.5\mathrm{MVAR}$.

Step-by-step solution

Write the given data.

The number of ACSR conductors are three.

The line voltage given is$\mathrm{V}=220\mathrm{kV}$ .

From table A.4, for drake the outside diameter is 1.108 in .

The frequency is 60 Hz .

The distance between conductors are 20 ft , 20 ft and 38 ft .

Determine the formulas of radius of conductor, geometric mean diameter, capacitance, capacitive reactance, charging current and reactive power.

Write the formula of geometric mean radius.

$\mathbf{r}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{\text{'}}}}{\mathbf{2}\mathbf{\times }\mathbf{12}}\mathbf{ft}$ ……. (1)

Here,$d^{\text{'}}$is the diameter.

Write the formula of geometric mean diameter for three conductor bundle.

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

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

Write the formula of capacitive reactance.

${\mathbf{X}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{j}\mathbf{2}{\mathbf{\pi fC}}_{\mathbf{an}}\mathbf{\times }\mathbf{1069}}\mathbf{\Omega }\mathbf{.}\mathbf{mi}$ ……. (4)

Here, fis frequency.

Write the formula of charging current.

${\mathbf{I}}_{\mathbf{chg}}\mathbf{=}\frac{\mathbf{V}}{\sqrt{\mathbf{3}}\mathbf{\times }{\mathbf{X}}_{\mathbf{c}}}\frac{\mathbf{A}}{\mathbf{mi}}$ ……. (5)

Here,Vis the line voltage.

Write the formula of three-phase reactive power.

${\mathbf{Q}}_{\mathbf{c}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{VI}}_{\mathbf{chg}}$ ……. (6)

Here, $l_{chg}$is the charging current.

Determine the capacitance and capacitive reactance.

(a)

Determine the radius of conductors.

Substitute 1.108 in for$d^{\text{'}}$ in equation (1).

$$\begin{gathered} \mathrm{r}=\frac{1.108}{2\times 12}\mathrm{ft} \\ =0.0462\mathrm{ft} \end{gathered}$$

Determine the geometric mean diameter.

Substitute 20 ftfor$D_{ab}$ ,role="math" localid="1655879256994" $20\mathrm{ft}$for $D_{bc}$ and 38 ftfor $D_{ac}$ in equation (2).

$$\begin{gathered} {\mathrm{D}}_{\mathrm{eq}}=\sqrt[3]{(20\mathrm{ft})(20\mathrm{ft})(38\mathrm{ft})} \\ =24.8\mathrm{ft} \end{gathered}$$

Determine the capacitance.

Substitute 24.8 ftfor $D_{eq}$ , 0.0462 ft for r and $8.854\times {10}^{-12}$for $\epsilon$ in equation (3).

$$\begin{gathered} {\mathrm{C}}_{\mathrm{an}}=\frac{2\mathrm{\pi }(8.854\times {10}^{-12})}{\mathrm{In}\left({\displaystyle \frac{24.8}{0.0462}}\right)} \\ =8.8466\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}} \end{gathered}$$

Determine the capacitive reactance.

Substitute$8.8466\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$ for $C_{an}$ and 60 Hzfor f in equation (4).

$$\begin{gathered} {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\mathrm{j}2\mathrm{\pi }\left(60\right)\left(8.8466\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\right)\times 1069} \\ =-\mathrm{j}0.1864\times {10}^6\mathrm{\Omega }.\mathrm{mi} \end{gathered}$$

Therefore, the capacitance is role="math" localid="1655878700037" $8.8466\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}$ and the capacitive reactance is $-\mathrm{j}0.1864\times {10}^6\mathrm{\Omega }.\mathrm{mi}$.

Determine the capacitance, charging current and reactive power.

(b)

Determine the capacitive reactance for 175 mile .

$$\begin{gathered} {\mathrm{X}}_{\mathrm{c}}=\frac{-\mathrm{j}0.1864\times {10}^6\mathrm{\Omega }.\mathrm{mi}}{175\mathrm{mile}} \\ =-\mathrm{j}1065\mathrm{\Omega } \end{gathered}$$

Determine the charging current.

Substitute $0.1864\times {10}^6\mathrm{\Omega }.\mathrm{mi}$ for $X_c$ and 220 kV for V in equation (5).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{chg}}=\frac{220\mathrm{kV}}{0.1864\times {10}^6\mathrm{\Omega }.\mathrm{mi}} \\ =0.681\frac{\mathrm{A}}{\mathrm{mi}} \end{gathered}$$

Determine the charging current for 175 mile .

$$\begin{gathered} I_{chg}=0.681\frac{A}{mi}\times 175mile \\ =119A \end{gathered}$$

Determine the reactive power.

Substitute 119 A for $I_{chg}$ and 220 kVfor V in equation (5).

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{c}}=\sqrt{3(}119\mathrm{A}\left)\right(220\mathrm{kV}) \\ =43.5\mathrm{MVAR} \end{gathered}$$

Therefore, the capacitive reactance for 175 mile is $-j1065\Omega$ , the charging current is 119 Aand the three-phase reactive power is43.5 MVAR .