Question

Figure 4.32 shows the conductor configuration of a completely transposed three-phasence per overhead transmission line with bundled phase conductors. All conductors have a radius of$\mathbf{0}\mathbf{.}\mathbf{74}\mathbf{}\mathbf{cm}$with a$\mathbf{30}\mathbf{}\mathbf{cm}$bundle spacing. (a) Determine the inductaphase in$\frac{\mathbf{mH}}{\mathbf{km}}$and in$\frac{\mathbf{mH}}{\mathbf{km}}$(b) Find the inductive line reactance per phase in$\frac{\mathbf{\Omega }}{\mathbf{mi}}$at$\mathbf{60}\mathbf{}\mathbf{Hz}$.

Short answer

(a) The inductance per phase in $\frac{\mathrm{mH}}{km}$and $\frac{\mathrm{mH}}{\mathrm{mi}}$are role="math" localid="1655202659118" $1.040\frac{\mathrm{mH}}{K\mathrm{m}}$and$1.673\frac{\mathrm{mH}}{\mathrm{mi}}$ respectively.

(b) The inductive reactance of transposed overhead line is $0.631\frac{\Omega }{\mathrm{mi}}$.

Step-by-step solution

Write the given data from the question.

The frequency of the system,$f=60\mathrm{Hz}$

Radius of the conductors,$r=0.0074\mathrm{m}$

Spacing between the conductors,$d=0.30\mathrm{m}$

Determine the equation to calculate the inductance per phase and inductive line reactance.

The equation to calculate the geometric mean distance of completely transposed overhead line is given as follows.

${\mathbf{D}}_{\mathbf{eq}}\mathbf{=}\sqrt[\mathbf{3}]{{\mathbf{D}}_{\mathbf{AB}}\mathbf{}{\mathbf{D}}_{\mathbf{BC}}\mathbf{}{\mathbf{D}}_{\mathbf{CA}}}$ …… (1)

Here, ${\mathbf{D}}_{\mathbf{AB}}$is the distance between conductor $\mathbf{A}$and $\mathbf{B}\mathbf{}\mathbf{,}{\mathbf{D}}_{\mathbf{BC}}$is the distance between conductor $\mathbf{B}$and$\mathbf{C}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{D}}_{\mathbf{CA}}$is distance between $\mathbf{C}$and $\mathbf{A}$.

The equation to calculate the bundle geometric mean radius of completely transposed overhead line is given as follows.

${\mathbf{D}}_{\mathbf{SL}}\mathbf{=}\sqrt{\mathbf{0}\mathbf{.}\mathbf{7788}\mathbf{rd}}$ …… (2)

The equation to calculate the inductance transposed overhead line inductance is given as follows.

$\mathbf{L}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{In}\left(\frac{D_{\mathrm{eq}}}{D_{\mathrm{SL}}}\right)\frac{\mathbf{H}}{\mathbf{m}}$ …… (3)

The equation to calculate inductive reactanceof transposed overhead lineis given as follows,

$\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{\tau \tau fL}$ …… (4)

Calculate the inductive reactance of the transposed overhead line.

(a)

Consider the schematic diagram for the calculation of inductance per phase.

Calculate the geometric mean distance.

Substitute $6\mathrm{m}$for $D_{\mathrm{AB}}$, $6\mathrm{m}$for $D_{\mathrm{BC}}$and$12\mathrm{m}$for $D_{CA}$into equation (1).

$$\begin{gathered} D_{eq}=\sqrt[3]{6\times 6\times 12} \\ {D}_{eq}={\left(432\right)}^{\frac{1}{3}} \\ {D}_{eq}=7.55\mathrm{m} \end{gathered}$$

Calculate the bundle GMR.

Substitute$0.0074\mathrm{m}$for $r$and $0.3\mathrm{m}$for$d$into equation (2).

$$\begin{gathered} D_{SL}=\sqrt{0.7788\times 0.0074\times 0.3} \\ {D}_{SL}=\sqrt{0.00172} \\ {D}_{SL}=0.0415\mathrm{m} \end{gathered}$$

Calculate the inductance of the transposed overhead conductors.

Substitute $7.55\mathrm{m}$for localid="1655204128703" $D_{eq}\mathrm{m},0.0415\mathrm{m}$for$D_{SL}$into equation (3).

$$\begin{gathered} L=2\times {10}^{-7}In\left(\frac{7.55}{0.0415}\right) \\ L=2\times {10}^{-7}In\left(181.927\right) \\ L=2\times {10}^{-7}\times 5.203 \\ L=1.040\frac{\mathrm{mH}}{\mathrm{km}} \end{gathered}$$

Calculate the inductance in $\frac{\mathrm{mH}}{\mathrm{mi}}$,

$$\begin{gathered} L=1.040\times 1.609 \\ L=1.673\frac{\mathrm{mH}}{\mathrm{mi}} \end{gathered}$$

Hence the inductance per phase in $\frac{\mathrm{mH}}{\mathrm{km}}$and $\frac{\mathrm{mH}}{\mathrm{mi}}$are $1.673\frac{\mathrm{mH}}{\mathrm{mi}}$and respectively.

Calculate the inductive line reactance of the transposed overhead line.

(b)

Calculate the inductive reactance of the transposed overhead conductors.

Substitute$1.040\times {10}^{-3}\frac{\Omega }{\mathrm{km}}$for $L$and $60\mathrm{Hz}$for $f$into equation (4).

$$\begin{gathered} X=2\pi \times 60\times 1.040\times {10}^{-3} \\ X=392.07\times {10}^{-3} \\ X=0.390\frac{\Omega }{\mathrm{km}} \end{gathered}$$

Calculate the inductive reactance in$\frac{\Omega }{\mathrm{km}}$.

$$\begin{gathered} X=0392\times 1.609 \\ X=0.631\frac{\Omega }{\mathrm{mi}} \end{gathered}$$

Therefore, the inductive reactance of transposed overhead line is $0.631\frac{\Omega }{\mathrm{mi}}$.