Question

For the overhead line of configuration shown in Figure 4.33 operating at 60 Hz and a

conductor temperature of${\mathbf{70}}^{\mathbf{\circ }}\mathbf{C}$,determine the resistance per phase, inductive reactance

in ohms/mile/phase, and the current carrying capacity of the overhead line. Each conductor

is ACSR Cardinal of

Table A.4.

Short answer

The resistance per phase, inductive reactance and current carrying capacity are

$0.03\frac{\Omega }{mile}$,$2.08\frac{\Omega }{mile}$ and 4040 A respectively.

Step-by-step solution

Write the given data from the question.

The system frequency,$f=60Hz$

The conductor temperature,$T={70}^{\circ }C$

The bundle spacing,

$$\begin{gathered} \mathrm{d}=20\mathrm{in} \\ \mathrm{d}=20\times 0.0833 \\ \mathrm{d}=1.667\mathrm{ft} \end{gathered}$$

From Table A4, the geometrical mean distance for cardinal,${\mathrm{D}}_{\mathrm{s}}=0.0403\mathrm{ft}$.

Determine the formula to calculate the resistance per phase, inductive reactance and current carrying capacity of line.

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

${\mathbf{D}}_{\mathbf{SL}}\mathbf{=}\sqrt[\mathbf{4}]{{\mathbf{D}}_{\mathbf{s}}{\mathbf{d}}^{\mathbf{3}}}$ …… (1)

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

${\mathbf{D}}_{\mathbf{eq}}\mathbf{=}\sqrt[\mathbf{3}]{{\mathbf{D}}_{\mathbf{ab}}{\mathbf{D}}_{\mathbf{bc}}{\mathbf{D}}_{\mathbf{ca}}}$ …… (2)

Here, ${\mathbf{D}}_{\mathbf{ab}}$is the distance between conductor A and B, ${\mathbf{D}}_{\mathbf{bc}}$is the distance

between conductor B and C, ${\mathbf{D}}_{\mathbf{ca}}$is distance between C and A .

The equation to calculate the inductance per phase is given as follows.

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

The equation to calculate inductive reactanceper phaseis given as follows,

${\mathit{X}}_{\mathbf{a}}\mathbf{=}\mathbf{4}\mathbf{\times }\mathbf{2}{\mathbf{\pi fL}}_{\mathbf{a}}$ …… (4)

The equation to calculate the resistance is given as follows.

${\mathbf{R}}_{{\mathbf{70}}^{\mathbf{\circ }}\mathbf{C}}\mathbf{=}{\mathbf{R}}_{{\mathbf{50}}^{\mathbf{\circ }}\mathbf{C}}\left(\frac{70+T}{50+T}\right)$ …… (5)

Conversion from inches to feet.

1 ft = 0.0833 ft

Determine the resistance per phase, inductive reactance and current carrying capacity of line.

Consider the schematic diagram shown below.

Calculate the geometric mean distance of each phase.

Substitute 0.0403 ft for $D_s$and 1.667 ft for d into equation (1).

$$\begin{gathered} D_{SL}=1.091 \\ {D}_{SL}=1.091 \\ {D}_{SL}=0.717 \end{gathered}$$

Determine the distance $D_{ab}$.

$$\begin{gathered} D_{ab}=\sqrt{{40}^2+{12}^2} \\ {D}_{ab}=41.76ft \end{gathered}$$

Determine the distance $D_{ac}$.

$$\begin{gathered} {\mathrm{D}}_{\mathrm{ac}}=\sqrt{{40}^2+{12}^2} \\ {\mathrm{D}}^{\mathrm{ac}}=41.76\mathrm{ft} \end{gathered}$$

Determine the distance role="math" localid="1655892305844" $D_{bc}$.

$D_{bc}=80ft$

Calculate the geometric mean distance.

Substitute 41.76 ft for role="math" localid="1655892436077" $D_{ab}\&D_{ac}$and 80 ft for $D_{bc}$into equation (2).

$$\begin{gathered} D_{eq}=\sqrt[3]{41.76\times 41.76\times 80} \\ {D}_{eq}=51.86ft \end{gathered}$$

Calculate the per phase inductance per phase.

Substitute $51.86\mathrm{ft}$for $D_{eq}$and 0.717 ft for $D_{SL}$into equation (3).

$$\begin{gathered} L_a=2\times {10}^{-7}In\left(\frac{51.86}{0.717}\right) \\ {L}_a=2\times {10}^{-7}In\left(71.3229\right) \\ {L}_a=2\times {10}^{-7}\times 4.2812 \\ {L}_a=8.562\times {10}^{-7}\frac{\mathrm{H}}{\mathrm{m}} \end{gathered}$$

Calculate the inductive reactance per phase.

Substitute 60 Hz for and $8.562\times {10}^{-7}\frac{\mathrm{H}}{\mathrm{m}}$for $L_a$into equation (4).

$$\begin{gathered} X_a=4\times 2\mathrm{\pi }\times 60\times 8.562\times {10}^{-7} \\ {X}_a=1.29\times {10}^{-3}\frac{\Omega }{m} \end{gathered}$$

Convert the inductance from ohm per meter per phase to ohm per mile per phase.

$$\begin{gathered} X_a=\frac{1.29\times {10}^{-3}}{0.000621} \\ {X}_a=2.08\frac{\mathrm{\Omega }}{\mathrm{mile}} \end{gathered}$$

From the table A4, the current carrying capacity of each ACSR cardinal conductor is 1010 A.

Calculate the current carrying conductor capacity per phase.

$$\begin{gathered} I=4\times 1010 \\ I=4040\mathrm{A} \end{gathered}$$

The conductor resistance at ${50}^{\circ }C$is $0.1128\frac{\mathrm{\Omega }}{\mathrm{mile}}$per conductor.

The conductor resistance at ${25}^{\circ }C$is role="math" localid="1655893317948" $0.0982\frac{\mathrm{\Omega }}{\mathrm{mile}}$per conductor.

The temperature constant from Table 4.3,$T=228.1^{\circ }C$

Calculate the resistance at ${70}^{\circ }C$per phase.

$$\begin{gathered} R_{{50}^{\circ }\mathrm{C}}=0.1128\left(\frac{70+228.1}{50+228.1}\right) \\ {R}_{{50}^{\circ }\mathrm{C}}=0.1128\times \frac{298.1}{278.1} \\ {R}_{{50}^{\circ }\mathrm{C}}=0.1128\times 1.071 \\ {R}_{{50}^{\circ }\mathrm{C}}=0.1209\frac{\mathrm{\Omega }}{\mathrm{mile}} \end{gathered}$$

Calculate the resistance at ${70}^{\circ }C$per phase for four conductors.

$$\begin{gathered} R=\frac{0.1209}{4} \\ R=0.03\frac{\mathrm{\Omega }}{\mathrm{mile}} \end{gathered}$$

Hence, the resistance per phase, inductive reactance and current carrying capacity are $0.03\frac{\mathrm{\Omega }}{\mathrm{mile}}$,$2.08\frac{\mathrm{\Omega }}{\mathrm{mile}}$and 4040A respectively.