Question

Consider a three-phase overhead line made up of three phase conductors: Linnet$\mathbf{336}\mathbf{.}\mathbf{4}\mathbf{kcmil}\mathbf{,}$and ACSR $\mathbf{26}\mathbf{/}\mathbf{7}$.The line configuration is such that the horizontal separation between center of C and that of A is$\mathbf{40}\mathbf{\text{'}}\mathbf{\text{'}}$ , and between that of A and B is also $\mathbf{40}\mathbf{"}$in the same line; the vertical separation of A from the line of C-B is $\mathbf{16}\mathbf{"}$. If the line is at at a conductor temperature of , determine the inductive reactance per phase in role="math" localid="1655205883058" $\frac{\mathbf{\Omega }}{\mathbf{mi}}$,

(a) by using the formula given in Problem 4-14 (a), and

(b) by using from the text.

Short answer

(a) The inductive reactance by using the logarithmic formula is $6304\frac{\Omega }{\mathrm{mi}}$.

(b) The inductive reactance per phase of a three-phase overhead line is $0.6295\frac{\Omega }{\mathrm{mi}}$.

Step-by-step solution

Write the given data from the question.

The inductive reactance of the line per mile per phase$x=k\log \frac{D}{r\text{'}}$.

The frequency of the system.$60\mathrm{Hz}$

The horizontal separation between center of C and that of A is $40"$.

The separation between A and B is$40"$ .

The vertical separation of A from the line of C-B is$16"$ .

From appendix A4 the GMR of ACSR $\frac{26}{7}$conductor.

role="math" localid="1655206306641" $$\begin{gathered} D_s=0.0244\mathrm{ft} \\ {D}_s=0.0244\times 0.305 \\ {D}_s=0.00744\mathrm{m} \end{gathered}$$

Determine the formula for calculate the inductive reactance per phase.

The equation to calculate the geometric mean distance is given as follows.

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

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

The equation to calculate inductive reactance spacing factor in ohms per mile is given as follows,

${\mathbf{x}}_{\mathbf{d}}\mathbf{}\mathbf{=}{\mathbf{klogD}}_{\mathbf{eq}}$ …… (2)

The equation to calculate inductive reactance at spacing in Ohm’s mile per is given as follows.

${\mathbf{x}}_{\mathbf{a}}\mathbf{}\mathbf{=}\mathbf{klog}\frac{\mathbf{1}}{{\mathbf{D}}_{\mathbf{s}}}$ …… (3)

The equation to calculate total inductive reactance per phase is given as follows.

$\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{d}}\mathbf{+}{\mathbf{x}}_{\mathbf{a}}$ ……. (4)

The equation to calculate the inductance per phase of a three-phase overhead line is given as follows.

$\mathbf{L}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{In}\left(\frac{D_{\mathrm{eq}}}{D_s}\right)\frac{\mathbf{H}}{\mathbf{m}}$ …… (5)

The equation to calculate inductive reactanceper phase of a three-phase overhead line is given as follows.

$\mathbf{X}\mathbf{=}\mathbf{2}\mathbf{\tau \tau ft}$ …… (6)

Calculate the inductive reactance per phaseΩmi.

Draw the schematic diagram as,

Calculate the phase distance between $\mathbf{A}$and$\mathit{B}$ .

$$\begin{gathered} D_{AB}=\sqrt{{16}^2+{40}^2} \\ {D}_{AB}=\sqrt{1856} \\ {D}_{AB}=43.08\mathrm{in} \end{gathered}$$

Calculate the phase distance between$BandC$

$D_{BC}=80\mathrm{in}$

Calculate the phase distance between $A$and $C$.

$$\begin{gathered} D_{AC}=\sqrt{{16}^2+{40}^2} \\ {D}_{AC}=\sqrt{1856} \\ {D}_{AC}=43.08\mathrm{in} \end{gathered}$$

Calculate the geometric mean distance.

Substitute $43.08\mathrm{in}$forlocalid="1655208108036" $D_{AB}\&D_{AC}$and $80\mathrm{in}$for $D_{BC}$into equation (1).

$$\begin{gathered} D_{eq}=\sqrt[3]{43.08\times 80\times 43.08} \\ {D}_{eq}=\sqrt[3]{148470.911} \\ {D}_{eq}=52.952\mathrm{in} \end{gathered}$$

Calculate the geometric mean distance in meters.

$$\begin{gathered} D_{eq}=52.951\times 0.0254 \\ {D}_{eq}=1.344\mathrm{m} \end{gathered}$$

Calculate inductive reactance spacing factor in ohms per mile.

Substitute$0.2794\times \log \left(1.344\right)$for $k$and$1.344$for $D_{eq}$into equation (2).

$$\begin{gathered} x_d=0.2794\times \log \left(1.344\right) \\ {x}_d=0.0358\frac{\Omega }{\mathrm{mi}} \end{gathered}$$

Calculate inductive reactance at spacing in Ohm’s mile per.

Substitute $0.2794$for$k$and$0.007444\mathrm{m}$for$D_s$into equation (3).

$$\begin{gathered} X_a=0.2794\times \log \left(\frac{1}{0.00744}\right) \\ {X}_a=0.5946\frac{\Omega }{\mathrm{mi}} \end{gathered}$$

Calculate the total inductive reactance per phase.

Substitute $0.0358\frac{\Omega }{\mathrm{mi}}$for$X_d$and $0.5946\frac{\Omega }{\mathrm{mi}}$for$X_a$into equation (4).

$$\begin{gathered} X=0.0358+0.5946 \\ X=6304\frac{\Omega }{\mathrm{mi}} \end{gathered}$$

Hence, the inductive reactance by using the logarithmic formula is$6304\frac{\Omega }{\mathrm{mi}}$

Calculate the inductive reactance per phase.

Calculate the inductance per phase of three-phase overhead line.

Substitute $1.344\mathrm{m}$for $D_{eq}$and $0.007444\mathrm{m}$for$D_s$into equation (5).

$$\begin{gathered} L=2\times {10}^{-7}\mathrm{In}\left(\frac{1.344}{0.00744}\right) \\ L=2\times {10}^{-7}\mathrm{In}\left(180.64\right) \\ L=2\times {10}^{-7}\times 5.196 \\ L=1.039\times {10}^{-6}\frac{\mathrm{H}}{\mathrm{m}} \end{gathered}$$

Calculate the inductance in $\frac{\mathrm{H}}{\mathrm{mi}}$.

$$\begin{gathered} L=1.031\times {10}^{-6}\times 1609 \\ L=0.00167\frac{H}{\mathrm{mi}} \end{gathered}$$

Calculate inductive reactanceper phase of a three-phase overhead line.

Substitute $60\mathrm{Hz}$for $f$and $0.00167\frac{\mathrm{H}}{\mathrm{mi}}$for $L$into equation (6).

$$\begin{gathered} X=2\pi \times 60\times 0.00167 \\ X=0.6295\frac{\Omega }{\mathrm{mi}} \end{gathered}$$

Hence, the inductive reactance per phase of a three-phase overhead line is $0.6295\frac{\Omega }{\mathrm{mi}}$