Question

Rework Problem 4.41 if the bundled line has (a) three ACSR,1351kcmilconductors per phase or (b) three ACSR, 900kcmilconductors per phase without changing the bundle spacing or the phase spacings between bundle centers.

Short answer

(a) The capacitance is $1.284\times {10}^{-11}\frac{\mathrm{f}}{\mathrm{m}}$, the admittance is $j4.842\times {10}^{-6}\frac{S}{m}$

and the reactive power is $1.21\frac{MVAR}{km}$ .

(b) The capacitance is $1.265\times {10}^{-11}\frac{F}{m}$, the admittance is $j4.842\times {10}^{-6}\frac{S}{km}$

and the reactive power is $1.192\frac{MVAR}{km}$.

Step-by-step solution

Write the given data.

The number of ACSR conductors are three.

The geometric mean radius from problem 4.41 is$D_{eq}=12.6m$ .

The line voltage given is V=500 kV .

From table A.4, for 1351kcmil conductor per phase, the outside diameter is

1.424 in and for 900kcmil conductor per phase, the outside diameter is 1.162 in .

The distance between two conductors given is d=0.5m .

Determine the formulas of radius of conductor, geometric mean radius, capacitance, admittance and reactive power.

Write the formula of radius of condutors in meters.

$\mathbf{r}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{,}}}{\mathbf{2}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{0254}\mathbf{m}$ ……. (1)

Here, is the diameter.

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

${\mathbf{D}}_{\mathbf{SC}}\mathbf{=}\sqrt[\mathbf{3}]{{\mathbf{rd}}^{\mathbf{2}}}$ ……. (2)

Here, ris the radius of conductor and dis the distance between two conductors.

Write the formula of capacitance.

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

Here, $D_{eq}$is geometric mean distance and$D_{SC}$is geometric mean radius.

Write the formula of admittance.

$\mathit{Y}\mathbf{=}\mathit{j}\mathbf{2}{\mathbf{\pi fC}}_{\mathbf{an}}$ ……. (4)

Here, fis frequency.

Write the formula of reactive power.

$\mathbf{Q}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{Y}}$ ……. (5)

Here, Vis line voltage.

Determine the capacitance, admittance and reactive power.

(a)

Determine the radius of conductors.

Substitute 1.424 for $d^{,}$in equation (1).

$$\begin{gathered} r=\frac{1.424}{2}\times 0.0254\mathrm{m} \\ =0.0181\mathrm{m} \end{gathered}$$

Determine the geometric mean radius.

Substitute 0.5m for d and 0.0181mfor r in equation (2).

$$\begin{gathered} D_{SC}=\sqrt[3]{90.0181m){(0.5m)}^2} \\ =0.1654m \end{gathered}$$

Determine the capacitance.

Substitute 0.1654mfor $D_{SC}$,12.6m for $D_{eq}$and $8.854\times {10}^{-12}$ for $\epsilon$in equation (3).

$$\begin{gathered} C_{an}=\frac{2\mathrm{\pi }\left(8.854\times {10}^{-12}\right)}{\mathrm{In}\left({\displaystyle \frac{12.6}{0.1654}}\right)} \\ =1.284\times {10}^{-11}\frac{F}{m} \end{gathered}$$

Determine the admittance.

Substitute $1.284\times {10}^{-11}\frac{F}{m}$ for $C_{an}$ and 60Hzfor f in equation (4).

$$\begin{gathered} Y=j2\mathrm{\pi }\left(60\right)\left(1.284\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}}\right) \\ =\mathrm{j}4.842\times {10}^{-9}\frac{\mathrm{S}}{\mathrm{m}} \\ =\mathrm{j}4.842\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}} \end{gathered}$$

Determine the reactive power.

Substitute 500kv for v and $4.842\times {10}^{-6}\frac{S}{m}$for Y in equation (5)

$$\begin{gathered} Q={\left(500kV\right)}^2\left(4.842\times \frac{S}{km}\right) \\ =1.21\frac{\mathrm{MVAR}}{\mathrm{km}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{capacitance}\mathrm{is}1.284\times \frac{\mathrm{F}}{\mathrm{m}},\mathrm{the}\mathrm{admittance}\mathrm{is} \\ \mathrm{j}4.842\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{m}}\mathrm{and}\mathrm{the}\mathrm{reactive}\mathrm{power}\mathrm{is}1.21\frac{\mathrm{MVAR}}{\mathrm{km}}. \end{gathered}$$

Determine the capacitance, admittance and reactive power.

(b)

Determine the radius of conductors.

Substitute 1.162 in for $d^{\text{'}}$in equation (1).

localid="1655908365727" $$\begin{gathered} r=\frac{1.162}{2}\times 0.0254\mathrm{m} \\ =0.01476\mathrm{m} \end{gathered}$$

Determine the geometric mean radius.

Substitute 0.5mfor d and $0.01476\mathrm{m}$for r in equation (2).

$$\begin{gathered} D_{sc}=\sqrt[3]{(0.01476\mathrm{m})(0.5)} \\ =0.1546m \\ \mathrm{Determine}\mathrm{the}\mathrm{capacitance}. \\ \mathrm{substitute}0.1546\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{SC}},12.6\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{eq}}\mathrm{and}8.854\times {10}^{-12}\mathrm{for}\mathrm{\epsilon }\mathrm{in} \\ \mathrm{equation}\left(3\right). \\ {\mathrm{C}}_{\mathrm{an}}=\frac{2\mathrm{\pi }\left(8.854\times {10}^{-12}\right)}{\mathrm{In}\left({\displaystyle \frac{12.6}{0.1546}}\right)} \\ =1.256\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}} \\ \mathrm{Determine}\mathrm{the}\mathrm{admittance}. \\ \mathrm{substitute}1.256\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}}\mathrm{for}{\mathrm{C}}_{\mathrm{an}}\mathrm{and}60\mathrm{Hz}\mathrm{for}\mathrm{f}\mathrm{in}\mathrm{equation}\left(4\right). \\ \mathrm{Y}=\mathrm{j}2\mathrm{\pi }\left(60\right)\left(1.265\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}}\right) \\ =\mathrm{j}4.77\times {10}^{-9}\frac{\mathrm{S}}{\mathrm{m}} \\ =\mathrm{j}4.77\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}} \end{gathered}$$

$$\begin{gathered} \mathrm{Determine}\mathrm{the}\mathrm{reactive}\mathrm{power}. \\ \mathrm{Substitute}500\mathrm{kV}\mathrm{for}\mathrm{V}\mathrm{and}4.77\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{m}}\mathrm{for}\mathrm{Y}\mathrm{in}\mathrm{equation}\left(5\right). \\ \mathrm{Q}={\left(500\mathrm{kV}\right)}^2\left(4.77\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\right) \\ =1.192\frac{\mathrm{MVAR}}{\mathrm{km}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{capacitance}\mathrm{is}1.265\times {10}^{-11}\frac{\mathrm{F}}{\mathrm{m}},\mathrm{the}\mathrm{admittance}\mathrm{is} \\ \mathrm{j}4.77\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{and}\mathrm{the}\mathrm{reactive}\mathrm{power}\mathrm{is}1.192\frac{\mathrm{MVAR}}{\mathrm{km}}. \end{gathered}$$