Question

For the case of double-circuit, bundle-conductor lines, the same method indicated in Problem 4.27 applies with$\mathit{r}\mathbf{\text{'}}$replaced by the bundle's GMR in the calculation of the overall GMR.

Now consider a double-circuit configuration shown in Figure 4.36 that belongs to alocalid="1655292945286" $\mathbf{500}\mathbf{kV}$, three-phase line with bundle conductors of three sub conductors atlocalid="1655292951972" $\mathbf{21}\mathbf{in}$spacing. The GMR of each sub conductor is given to belocalid="1655292956229" $\mathbf{0}\mathbf{.}\mathbf{0485}\mathbf{ft}$.

Determine the inductive reactance of the line in ohms per mile phase. You may use
localid="1655292961375" ${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2794}\mathbf{log}\frac{\mathbf{GMD}}{\mathbf{GMR}}\frac{\mathbf{\Omega }}{\mathbf{mi}\mathbf{.}\mathbf{phase}}$

Short answer

The inductive reactance of the line in ohms per mile phase$0.25246\frac{\mathrm{\Omega }}{\mathrm{mi}.\mathrm{phase}}$

Step-by-step solution

Write the given data from question.

The voltage,$\mathrm{V}=500\mathrm{kV}$

The GMR of each sub conductor,$D_s=0.0485\mathrm{ft}$

Spacing between the conductor,

$$\begin{gathered} d=21in \\ d=21\times 0.0833 \\ d=1.75\mathrm{ft} \end{gathered}$$

Determine the formulas to calculate the inductive reactance of the line.

The equation to calculate the bungle conductor GMR,

$$\begin{gathered} {\mathbf{D}}_{\mathbf{SL}}\mathbf{=}\mathbf{r}\mathbf{\text{'}} \\ {\mathbf{D}}_{\mathbf{SL}}\mathbf{=}\sqrt[\mathbf{3}]{{\mathbf{D}}_{\mathbf{s}}{\mathbf{d}}^{\mathbf{2}}\mathbf{}\mathbf{}} \end{gathered}$$ …… (1)

The equation to calculate the GMR of conductor A is given as follows.

${\mathbf{\left(}\mathbf{GMR}\mathbf{\right)}}_{\mathbf{A}}\mathbf{}\mathbf{=}{\mathbf{[}\mathbf{r}\mathbf{\text{'}}{\mathbf{D}}_{\mathbf{11}\mathbf{\text{'}}}\mathbf{]}}^{\frac{\mathbf{1}}{\mathbf{2}}}$ …… (2)

The equation to calculate the GMR of conductor B is given as follows.

${\mathbf{\left(}\mathbf{GMR}\mathbf{\right)}}_{\mathbf{B}}\mathbf{}\mathbf{=}{\mathbf{[}\mathbf{r}\mathbf{\text{'}}{\mathbf{D}}_{\mathbf{22}\mathbf{\text{'}}}\mathbf{]}}^{\frac{\mathbf{1}}{\mathbf{2}}}$ …… (3)

The equation to calculate the GMR of conductor C is given as follows.

${\mathbf{\left(}\mathbf{GMR}\mathbf{\right)}}_{\mathbf{C}}\mathbf{}\mathbf{=}{\mathbf{[}\mathbf{r}\mathbf{\text{'}}{\mathbf{D}}_{\mathbf{33}\mathbf{\text{'}}}\mathbf{]}}^{\frac{\mathbf{1}}{\mathbf{2}}}$ …… (4)

The equation to calculate the total GMR of double circuit bundle conductor is given as follows.

$\mathbf{GMR}\mathbf{=}{\mathbf{\left[}{\left(\mathrm{GMR}\right)}_{\mathbf{A}}{\left(\mathrm{GMR}\right)}_{\mathbf{B}}{\left(\mathrm{GMR}\right)}_{\mathbf{C}}\mathbf{\right]}}^{\frac{\mathbf{1}}{\mathbf{3}}}$ …… (5)

The equation to calculate the equivalent GMD between conductors A and B is given as follows.

${\mathbf{D}}_{{\mathbf{AB}}_{\mathbf{eq}}}\mathbf{}\mathbf{=}{\mathbf{\left(}{\mathbf{D}}_{\mathbf{12}}{\mathbf{D}}_{\mathbf{1}\mathbf{\text{'}}\mathbf{2}}{\mathbf{D}}_{\mathbf{12}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{1}\mathbf{\text{'}}\mathbf{2}}\mathbf{\right)}}^{\frac{\mathbf{1}}{\mathbf{4}}}$ …… (6)

The equation to calculate the equivalent GMD between conductors B and C is given as follows.

${\mathbf{D}}_{{\mathbf{BC}}_{\mathbf{eq}}}\mathbf{}\mathbf{=}{\mathbf{\left(}{\mathbf{D}}_{\mathbf{23}}{\mathbf{D}}_{\mathbf{2}\mathbf{\text{'}}\mathbf{3}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{2}\mathbf{\text{'}}\mathbf{3}}{\mathbf{D}}_{\mathbf{23}\mathbf{\text{'}}}\mathbf{\right)}}^{\frac{\mathbf{1}}{\mathbf{4}}}$ …… (7)

The equation to calculate the equivalent GMD between conductors C and A is given as follows.

${\mathbf{D}}_{{\mathbf{CA}}_{\mathbf{eq}}}\mathbf{=}{\mathbf{\left(}{\mathbf{D}}_{\mathbf{13}}{\mathbf{D}}_{\mathbf{1}\mathbf{\text{'}}\mathbf{3}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{13}\mathbf{\text{'}}}{\mathbf{D}}_{\mathbf{1}\mathbf{\text{'}}\mathbf{3}}\mathbf{\right)}}^{\frac{\mathbf{1}}{\mathbf{4}}}$ …… (8)

The equation to calculate the total GMD of the bundle is given as follows.

$\mathbf{GMD}\mathbf{=}{\mathbf{\left(}{\mathbf{D}}_{{\mathbf{AB}}_{\mathbf{eq}}}{\mathbf{D}}_{{\mathbf{BC}}_{\mathbf{eq}}}{\mathbf{D}}_{{\mathbf{CA}}_{\mathbf{eq}}}\mathbf{\right)}}^{\frac{\mathbf{1}}{\mathbf{3}}}$ …… (9)

The equation to calculate the inductive reactance of the lineinohms per mile phase is given as follows.

${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2794}\mathbf{log}\frac{\mathbf{GMD}}{\mathbf{GMR}}$ …… (10)

Calculate the value of the inductive reactance of the line in ohms per mile phase.

Calculate the bundle GMR.

Substitute$0.0485\mathrm{ft}\mathrm{for}{\mathrm{D}}_{\mathrm{s}}\mathrm{and}1.75\mathrm{ft}\mathrm{for}d$into equation (1).

$$\begin{gathered} D_{SL}=\sqrt[3]{0.0485\times 1.{75}^2} \\ {D}_{SL}=0.5296\mathrm{ft} \end{gathered}$$

Determine the distance between the sub conductors.

Calculate the distance$D_{11\text{'}}$$D_{11\text{'}}$$D_{11\text{'}}$

localid="1655292515653" $$\begin{gathered} D_{11\text{'}}=\sqrt{{36}^2\times {32}^2} \\ {D}_{11\text{'}}=48.166\mathrm{ft} \end{gathered}$$

Calculate the distancelocalid="1655292522120" $D_{22\text{'}}$.

localid="1655292527323" $D_{22\text{'}}=96\mathrm{ft}$

Calculate the distancelocalid="1655292532727" $D_{33\text{'}}$.

localid="1655292536204" $$\begin{gathered} D_{33\text{'}}=D_{11\text{'}} \\ {D}_{33\text{'}}=48.166\mathrm{ft} \end{gathered}$$

Calculate the GMR of conductor A.

Substitutelocalid="1655292540647" $0.5296\mathrm{ft}\mathrm{for}r\text{'}\mathrm{and}48.166\mathrm{ft}\mathrm{for}{D}_{11\text{'}}$into equation (2).

localid="1655292544058" $$\begin{gathered} {\left(GMR\right)}_A={\left(0.5296\times 48.166\right)}^{\frac{1}{2}} \\ {\left(GMR\right)}_A=5.05\mathrm{ft} \end{gathered}$$

Calculate the GMR of conductor B.

Substitutelocalid="1655292547640" $0.529\mathrm{ft}\mathrm{for}r\text{'}\mathrm{and}96\mathrm{ft}\mathrm{for}{D}_{22\text{'}}$into equation (3).

localid="1655292552497" $$\begin{gathered} {\left(GMR\right)}_B={\left(0.5296\times 96\right)}^{\frac{1}{2}} \\ {\left(GMR\right)}_B=7.130\mathrm{ft} \end{gathered}$$

Calculate the GMR of conductor C.

Substitutelocalid="1655292562664" $0.5296\mathrm{ft}\mathrm{for}r\text{'}\mathrm{and}48.166\mathrm{ft}\mathrm{for}{D}_{33\text{'}}$into equation (4).

localid="1655292567155" $$\begin{gathered} {\left(GMR\right)}_c={\left(0.5296\times 48.166\right)}^{\frac{1}{2}} \\ {\left(GMR\right)}_c=5.05\mathrm{ft} \end{gathered}$$

Calculate the total GMR.

Substitutelocalid="1655292571798" $5.05\mathrm{ft}\mathrm{for}{\left(\mathrm{GMR}\right)}_{\mathrm{A}\text{'}}{\left(\mathrm{GMR}\right)}_{\mathrm{c}\text{'}}\mathrm{AND}7.130\mathrm{ft}\mathrm{for}{\left(\mathrm{GMR}\right)}_{\mathrm{B}}$into equation (5)

localid="1655292576883" $$\begin{gathered} \mathrm{GMR}={\left(5.05\times 7.130\times 5.05\right)}^{\frac{1}{3}} \\ \mathrm{GMR}=5.666\mathrm{ft} \end{gathered}$$

Calculate the distance between the sub conductor and double circuit configuration.

Calculate the distancelocalid="1655292586758" $D_{12\text{'}}$

localid="1655292594420" $$\begin{gathered} D_{12\text{'}}=\sqrt{{32}^2+{36}^2} \\ {D}_{12\text{'}}=48.166\mathrm{ft} \end{gathered}$$

Calculate the distancelocalid="1655292600876" $D_{12}$,'

localid="1655292620885" $$\begin{gathered} D_{12\text{'}}=\sqrt{{36}^2+{64}^2} \\ {D}_{12}\text{'}=73.43\mathrm{ft} \end{gathered}$$

Calculate the distancelocalid="1655292632488" $D_{1\text{'}2}$,

localid="1655292638572" $D_{1\text{'}2}=64\mathrm{ft}$

Calculate the distancelocalid="1655292643478" $D_{1\text{'}2\text{'}}$,

localid="1655292648894" $D_{1\text{'}2}=32\mathrm{ft}$

Calculate the distance localid="1655292656474" $D_{23\text{'}}$,

localid="1655292664305" $D_{23}=32\mathrm{ft}$

Calculate the distancelocalid="1655292682888" $D_{23\text{'}}$,

localid="1655292688841" $$\begin{gathered} D_{23\text{'}}=\sqrt{{36}^2+{64}^2} \\ {D}_{23\text{'}}=73.43\mathrm{ft} \end{gathered}$$

Calculate the distancelocalid="1655292699857" $D_{2\text{'}3\text{'}}$

localid="1655292707161" $D_{2\text{'}3}=64\mathrm{ft}$

Calculate the distance localid="1655292712774" $D_{2\text{'}3\text{'}}$,

localid="1655292718183" $$\begin{gathered} D_{2\text{'}3\text{'}}=\sqrt{{32}^2+{36}^2} \\ {D}_{2\text{'}3\text{'}}=48.166\mathrm{ft} \end{gathered}$$

Calculate the distance localid="1655292723447" $D_{13\text{'}}$

localid="1655292744544" $D_{13}=36\mathrm{ft}$

Calculate the distancelocalid="1655292752249" $D_{13\text{'}}$,

localid="1655292757691" $D_{13\text{'}}=32\mathrm{ft}$

Calculate the distancelocalid="1655292764746" $D_{1\text{'}3\text{'}}$,

localid="1655292771407" $D_{13\text{'}}=32\mathrm{ft}$

Calculate the distancelocalid="1655292776696" $D_{1\text{'}3\text{'}}$

localid="1655292781628" $D_{1\text{'}3\text{'}}=36\mathrm{ft}$

Calculate the equivalent GMD between conductors A and B.

Substitutelocalid="1655292803577" $48.166\mathrm{ft}\mathrm{for}{D}_{12\text{'}}73.43\mathrm{ft}\mathrm{for}{D}_{12\text{'}}\mathrm{and}32\mathrm{ft}\mathrm{for}{D}_{1\text{'}2}$into equation (6).

localid="1655292809720" $$\begin{gathered} D_{A{B}_{eq}}={\left(48.166\times 73.43\times 64\times 32\right)}^{\frac{1}{4}} \\ {D}_{A{B}_{eq}}={\left(7243426.5702\right)}^{\frac{1}{4}} \\ {D}_{A{B}_{eq}}=51.878\mathrm{ft} \end{gathered}$$

Calculate the equivalent GMD between conductors B and C.

Substitutelocalid="1655292822342" $32\mathrm{ft}\mathrm{for}{D}_{23\text{'}}73.43\mathrm{ft}\mathrm{for}{D}_{2\text{'}3}\mathrm{and}48.166\mathrm{ft}\mathrm{for}{D}_{23\text{'}}$into equation (7).

localid="1655292829030" $$\begin{gathered} D_{B{C}_{eq}}={\left(32\times 73.43\times \times 64\times 48.166\right)}^{\frac{1}{4}} \\ {D}_{B{C}_{eq}}={\left(7243426.5702\right)}^{\frac{1}{4}} \\ {D}_{B{C}_{eq}}=51.878\mathrm{ft} \end{gathered}$$

Calculate the equivalent GMD between conductors C and A.

Substitute localid="1655292835977" $36\mathrm{ft}\mathrm{for}{D}_{13\text{'}}32\mathrm{ft}\mathrm{for}{D}_{1\text{'}3\text{'}\text{'}}32\mathrm{ft}\mathrm{for}{D}_{13\text{'}}and36\mathrm{ft}\mathrm{for}{D}_{1\text{'}3}$into equation (8).

localid="1655292843747" $$\begin{gathered} D_{C{A}_{eq}}={\left(36\times 32\times 32\times 36\times \right)}^{\frac{1}{4}} \\ {D}_{C{A}_{eq}}={\left(1327104\right)}^{\frac{1}{4}} \\ {D}_{C{A}_{eq}}=33.941\mathrm{ft} \end{gathered}$$

Calculate the total GMD of the bundle.

Substitutelocalid="1655292869844" $51.878\mathrm{ft}\mathrm{for}{D}_{{\mathrm{AB}}_{\mathrm{eq}}},33.941\mathrm{ft}\mathrm{for}{D}_{{\mathrm{BC}}_{\mathrm{eq}}}\mathrm{and}51.878\mathrm{ft}\mathrm{for}{D}_{{\mathrm{CA}}_{\mathrm{eq}}}$into equation (9).

localid="1655292877242" $$\begin{gathered} GMD={\left(51.878\times 33.941\times 51.878\right)}^{\frac{1}{3}} \\ GMD={\left(91346.325\right)}^{\frac{1}{3}} \\ GMD=45.036\mathrm{ft} \end{gathered}$$

Calculate the inductive reactance of thelineinohms per mile phase.

Substitutelocalid="1655292889398" $45.036\mathrm{ft}\mathrm{for}\mathrm{GMD}\mathrm{and}5.666\mathrm{ft}\mathrm{for}\mathrm{GMR}$into equation (10).

localid="1655292902219" $$\begin{gathered} X_L=0.2794\log \left(\frac{45.036}{5.666}\right) \\ {X}_L=0.2794\times \log \left(7.9484\right) \\ {X}_L=0.2794\times 0.9 \\ {X}_L=0.25246\frac{\Omega }{\mathrm{miper}\mathrm{phase}} \end{gathered}$$

Therefore, the inductive reactance of the line in ohms per mile phaselocalid="1655292912821" $0.25246\frac{\Omega }{\mathrm{miper}\mathrm{phase}}$