Question

From (4.6.22) and (4.10.4), the series inductance and shunt capacitance of a three-phase overhead line are

$$\begin{gathered} {\mathbf{L}}_{\mathbf{a}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{In}\mathbf{(}{\mathbf{D}}_{\mathbf{eq}}\mathbf{/}{\mathbf{D}}_{\mathbf{SL}}\mathbf{)}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{2}\mathbf{\pi }}\mathbf{In}\mathbf{(}{\mathbf{D}}_{\mathbf{eq}}\mathbf{/}{\mathbf{D}}_{\mathbf{SL}}\mathbf{)}\mathbf{H}\mathbf{/}\mathbf{m} \\ {\mathbf{C}}_{\mathbf{an}}\mathbf{=}\frac{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}{\mathbf{In}\mathbf{}\mathbf{(}{\mathbf{D}}_{\mathbf{eq}}\mathbf{/}{\mathbf{D}}_{\mathbf{SC}}\mathbf{)}}\mathbf{F}\mathbf{/}\mathbf{m} \end{gathered}$$

Using these equations, determine formulas for surge impedance and velocity of propagation of an overhead lossless line. Then determine the surge impedance and velocity of propagation for the three-phase line given in Example 4.5. Assume positive-sequence operation. Neglect line losses as well as the effects of the overhead neutral wires and the earth plane.

Short answer

(a)The value of surge impedance is ${\mathrm{Z}}_{\mathrm{C}}=267\mathrm{\Omega }.$

(b)The value of velocity of propagation for the three-phase line is$2.98\times {10}^8\frac{m}{s}$

Step-by-step solution

Given data.

The series inductance of three-phase overhead line is,

${\mathrm{L}}_{\mathrm{a}}=2\times {10}^{-7}\mathrm{In}({\mathrm{D}}_{\mathrm{eq}}/{\mathrm{D}}_{\mathrm{SL}})=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\mathrm{In}({\mathrm{D}}_{\mathrm{eq}}/{\mathrm{D}}_{\mathrm{SL}})\mathrm{H}/\mathrm{m}.$

The shunt capacitance of a three-phase overhead line is${\mathrm{C}}_{\mathrm{an}}=\frac{2{\mathrm{\pi \epsilon }}_0}{\mathrm{In}\left({\mathrm{D}}_{\mathrm{eq}}/{\mathrm{D}}_{\mathrm{SC}}\right)}\mathrm{F}/\mathrm{m}$

Determine the required formula.

Write the formula of characteristic impedance.

${\mathrm{Z}}_{\mathrm{c}}=\sqrt{\frac{\mathrm{z}}{\mathrm{y}}}$ …… (1)

Here, $\mathrm{Z}$is the impedance and$\mathrm{y}$is the admittance of the conductor.

Write the formula of velocity propagation factor.

$\mathrm{v}=\sqrt{\frac{1}{\mathrm{LC}}}$ …… (2)

Here, $\mathrm{L}$is inductance per phase of the conductor and $\mathrm{C}$is capacitance.

Determine surge impedance ZC.

Rework of problem (4.6.22) and (4.10.4).

Calculate the GMR of the conductor from appendix A.4.

$$\begin{gathered} {\mathrm{D}}_{\mathrm{S}}=0.037\mathrm{ft} \\ =\left(0.037\right)\left(\frac{1\mathrm{m}}{3.28\mathrm{ft}}\right) \\ =0.0114\mathrm{m} \end{gathered}$$

The arrangement consists of a two conductor bundle in which bundle spacing is $\mathrm{d}=0.4\mathrm{m}.$

Calculate the cube root of the product of the three-phase spacing (equivalent distance).

$D_{eq}=3\sqrt{D_{12}{D}_{23}{D}_{31}}$ …… (3)

Here, ${\mathrm{D}}_{12},{\mathrm{D}}_{23},{\mathrm{D}}_{31}$is the distance between positions, conductor positions are denoted by 1, 2 and 3.

Substitute $10$for ${\mathrm{D}}_{12},{\mathrm{D}}_{23}$and $20$for ${\mathrm{D}}_{31}$into equation (3).

$$\begin{gathered} {\mathrm{D}}_{\mathrm{eq}}=3\sqrt{\left(10\right)\left(10\right)\left(20\right)} \\ =3\sqrt{2000} \\ =12.6\mathrm{m} \end{gathered}$$

Determine the geometric mean radius (GMR) for inductor for a three conductor bundle.

${\mathrm{D}}_{\mathrm{SL}}=\sqrt{{\left({\mathrm{D}}_{\mathrm{S}}\times \mathrm{d}\times \mathrm{d}\right)}^3}\dots \dots \left(4\right)$

Here, ${\mathrm{D}}_{\mathrm{S}}$is geometric mean radius and $d$is distance between conductors.

$$\begin{gathered} \mathrm{Substitute}0.0114\mathrm{for}{\mathrm{D}}_{\mathrm{S}}\mathrm{and}0.4\mathrm{for}\mathrm{d}\mathrm{into}\mathrm{equation}\left(4\right) \\ {\mathrm{D}}_{\mathrm{SL}}=\sqrt{\left(0.0114\right)\left(0.4\right)} \\ =\sqrt{4.56\times {10}^{-3}} \\ =0.0676\mathrm{m} \\ \mathrm{Outside}\mathrm{diameter}\mathrm{of}795000\mathrm{cmil}26/2\mathrm{ACSR}\mathrm{conductor}\mathrm{is}1.108\mathrm{in}. \end{gathered}$$

Calculate the radius of the conductor.

$$\begin{gathered} \mathrm{d}=1.108\mathrm{in} \\ \mathrm{r}=\left(\frac{1.108}{2}\mathrm{in}\right)\left(\frac{0.0254\mathrm{m}}{\mathrm{in}}\right) \\ =0.0141\mathrm{m} \\ \mathrm{The}\mathrm{bundle}\mathrm{spacing}\mathrm{d}=0.4\mathrm{m}. \end{gathered}$$

Determine the radius of each conductor.

$D_{SC}=\sqrt{rd}$ …… (5)

$\mathrm{Here},\mathrm{r}\mathrm{is}\mathrm{radius}\mathrm{of}\mathrm{conductor}\mathrm{and}\mathrm{d}\mathrm{is}\mathrm{diameter}\mathrm{of}\mathrm{each}\mathrm{conductor}.$

$$\begin{gathered} \mathrm{Substitute}0.0141\mathrm{for}\mathrm{r}\mathrm{and}0.4\mathrm{for}\mathrm{d}\mathrm{into}\mathrm{equation}\left(5\right). \\ {\mathrm{D}}_{\mathrm{sc}}=\sqrt{\left(0.0141\right)\left(0.4\right)} \\ =\sqrt{5.64\times {10}^{-3}} \\ =0.075\mathrm{m} \end{gathered}$$

Determine the surge impedance.

$$\begin{gathered} \mathrm{Substitute}{\mathrm{L}}_{\mathrm{a}}\mathrm{for}\mathrm{z}\mathrm{and}{\mathrm{C}}_{\mathrm{an}}\mathrm{for}\mathrm{y}\mathrm{into}\mathrm{equation}\left(1\right). \\ {\mathrm{Z}}_{\mathrm{C}}=\sqrt{\frac{{\mathrm{L}}_{\mathrm{a}}}{{\mathrm{C}}_{\mathrm{an}}}} \\ =\sqrt{\frac{{\displaystyle \frac{{\mathrm{\mu }}_0}{2}}\mathrm{In}\left({\displaystyle \frac{{\mathrm{D}}_{\mathrm{eq}}}{{\mathrm{D}}_{\mathrm{SL}}}}\right)}{{\displaystyle \frac{2\mathrm{\pi \epsilon }}{\mathrm{In}\left({\displaystyle \frac{{\mathrm{D}}_{\mathrm{eq}}}{{\mathrm{D}}_{\mathrm{SC}}}}\right)}}}} \end{gathered}$$

Rewrite the equation as,

$Z_C=\frac{1}{2\pi }\sqrt{\frac{{\mu }_0}{\epsilon }}\left[\sqrt{In\left(\frac{D_{eq}}{D_{SL}}\right)In\left(\frac{D_{eq}}{D_{SC}}\right)}\right]$

$$\begin{gathered} \mathrm{Here},{\mathrm{Z}}_{\mathrm{c}}\mathrm{is}\mathrm{characteristics}\mathrm{impendence},\sqrt{\frac{{\mathrm{\mu }}_0}{\mathrm{\epsilon }}}\mathrm{is}\mathrm{characteristics}\mathrm{impendence}\mathrm{of}\mathrm{free}\mathrm{space}, \\ {\mathrm{D}}_{\mathrm{eq}}\mathrm{is}\mathrm{equivalent}\mathrm{distance},{\mathrm{D}}_{\mathrm{SL}}\mathrm{is}\mathrm{geometric},\mathrm{ean}\mathrm{radius}\left(\mathrm{GMR}\right)\mathrm{for}\mathrm{inductor}\mathrm{and}{\mathrm{D}}_{\mathrm{SC}}\mathrm{is} \\ \mathrm{radius}\mathrm{of}\mathrm{each}\mathrm{conductor}. \end{gathered}$$

Determine the value of characteristics impedance of free space:

$$\begin{gathered} \sqrt{\frac{{\mathrm{\mu }}_0}{\mathrm{\epsilon }}}=\sqrt{\frac{{\displaystyle \frac{4\mathrm{\pi }\times {10}^{-7}}{1\times {10}^{-9}}}}{36\mathrm{\pi }}} \\ =\sqrt{142122.3} \\ =377\mathrm{\Omega } \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}377\mathrm{for}\sqrt{\frac{{\mathrm{\mu }}_0}{\mathrm{\epsilon }}},12.6\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{eq}}\mathrm{and}0.075\mathrm{m}\mathrm{for}{\mathrm{D}}_{\mathrm{SC}}\mathrm{into}\mathrm{equation}\left(6\right) \\ {\mathrm{Z}}_{\mathrm{C}}=\frac{1}{2\mathrm{\pi }}\left(377\right)\left[\left(\frac{12.6}{0.0676}\right)\mathrm{In}\left(\frac{12.6}{0.075}\right)\right] \\ =\left(60\right)\left[\sqrt{\mathrm{In}\left(186.39\right)\mathrm{In}\left(168\right)}\right] \\ =\left(60\right)\left(5.17\right) \\ {\mathrm{Z}}_{\mathrm{C}}=310.54\mathrm{\Omega } \end{gathered}$$

Determine velocity of propagation for the three-phase line.

_{Determine the velocity propagation factor.}

_{$$\begin{gathered} v=\sqrt{\frac{1}{\mathrm{LC}}} \\ =\sqrt{\frac{1}{{\displaystyle \frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}}\mathrm{In}\left({\displaystyle \frac{{\mathrm{D}}_{\mathrm{eq}}}{{\mathrm{D}}_{\mathrm{SC}}}}\right)\mathrm{In}{\displaystyle \frac{2}{\left(\frac{{\mathrm{D}}_{\mathrm{eq}}}{{\mathrm{D}}_{\mathrm{SC}}}\right)}}}} \\ \mathrm{Rewrite}\mathrm{the}\mathrm{equation}\mathrm{as}, \\ \mathrm{v}=\frac{1}{\sqrt{{\mathrm{\mu }}_0\mathrm{\epsilon }}}\left[\sqrt{\frac{In\frac{{\mathrm{D}}_{eq}}{{\mathrm{D}}_{sc}}}{In\frac{{\mathrm{D}}_{eq}}{{\mathrm{D}}_{sc}}}}\right]\dots \dots \left(7\right) \end{gathered}$$}

_{Here, $\frac{1}{\sqrt{{\mathrm{\mu }}_0\mathrm{\epsilon }}}$is free space velocity of propagation, ${\mathrm{D}}_{\mathrm{eq}}$is equivalent distance, ${\mathrm{D}}_{\mathrm{SL}}$is geometric mean radius (GMR) for inductor and $D_{SC}$is radius of each conductor.}

_{Determine the value of free space velocity of propagation.}

_{$$\begin{gathered} \sqrt{\frac{1}{{\mathrm{\mu }}_0\mathrm{\epsilon }}}=\sqrt{\frac{1}{\left(4\mathrm{\pi }\times {10}^{-7}\right)\left({\displaystyle \frac{1\times {10}^{-9}}{36\mathrm{\pi }}}\right)}} \\ =\sqrt{\frac{36\mathrm{\pi }}{4\mathrm{\pi }\times {10}^{-16}}} \\ =\sqrt{9\times {10}^{16}} \\ =3\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$}

_{Substitute $3\times {10}^8\mathrm{m}/\mathrm{s}$for$\sqrt{\frac{1}{{\mu }_{0^{\in }}}},12.6\mathrm{m}$, for${\mathrm{D}}_{\mathrm{eq}}$,$0.0676\mathrm{m}$for ${\mathrm{D}}_{SL}$and $0.075\mathrm{m}$for ${\mathrm{D}}_{SC}$into equation (7).}

$$\begin{gathered} \mathrm{v}=\left(3\times {10}^8\right)\left[\sqrt{\frac{\mathrm{In}\left({\displaystyle \frac{12.6}{0.075}}\right)}{\mathrm{In}\left({\displaystyle \frac{12.6}{0.0676}}\right)}}\right] \\ =\left(3\times {10}^8\right)\left[\sqrt{\frac{\mathrm{In}\left(168\right)}{\mathrm{In}(186.39)}}\right] \\ =\left(3\times {10}^8\right)\left[\sqrt{\frac{5.124}{5.227}}\right] \\ =\left(3\times {10}^8\right)(0.99) \\ \mathrm{Solve}\mathrm{further}\mathrm{as}, \\ \mathrm{v}=2.97\times {10}^8\mathrm{m}/\mathrm{s} \\ \mathrm{Therefore},\mathrm{the}\mathrm{velocity}\mathrm{of}\mathrm{propagation}\mathrm{is}2.97\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$