Question

One thousand circular mils or $\mathbf{1}\mathbf{}\mathit{k}\mathit{c}\mathit{m}\mathit{i}\mathit{l}$is sometimes designated by the abbreviation MCM. Data for commercial bare-aluminum electrical conductors lists a $\mathbf{60}\mathbf{}\mathit{H}\mathit{z}$resistance of $\mathbf{0}\mathbf{.}\mathbf{0080}\mathit{\Omega }$per kilometer at $\mathbf{75}\mathbf{°}\mathbf{}\mathit{C}$for a 793-MCM AAC conductor.

(a) Determine the cross-sectional conducting area of this conductor in square meters.

(b) Find the $\mathbf{60}\mathbf{}\mathit{H}\mathit{z}$resistance of this conductor in ohms per kilometer at $\mathbf{50}\mathbf{°}\mathbf{}\mathit{C}$.

Short answer

Answer

(a) The area of conductor is $4.0175\times {10}^{-4}{m}^2$.

(b) The resistance is $0.0807\Omega$per kilometer.

Step-by-step solution

Given data.

The given conductor is 793-MCM ACC conductor.

The area of the conductor is $A=1kcmil$.

The resistance of conductor at $75°C$is $R_{60Hz,75}=0.0080\Omega$per mile.

Determine the formulas of resistance and area of conductor.

Write the conversion formula of MCM.

$\mathit{x}\mathbf{}\mathit{M}\mathit{C}\mathit{M}\mathbf{=}\left(xcmil\right)\left(\frac{{\displaystyle \frac{\pi }{4}}sq.mil}{1cmil}\right){\left(\frac{1in}{1000mil}\right)}^{\mathbf{2}}{\left(\frac{0.0254m}{1in}\right)}^{\mathbf{2}}$ …… (1)

Here, $x$is MCM rating.

Write the formula of dc resistance at $50°C$.

${\mathit{R}}_{\mathbf{50}\mathbf{}\mathbf{Hz}\mathbf{,}\mathbf{50}}\mathbf{=}{\mathit{R}}_{\mathbf{60}\mathbf{}\mathbf{Hz}\mathbf{,}\mathbf{75}}\frac{\mathbf{50}\mathbf{+}\mathbf{T}}{\mathbf{75}\mathbf{+}\mathbf{T}}$ ……. (2)

Here, $T$is a temperature constant.

Determine the area of conductor.

(a)

Substitute 793 for x in equation (1).

$$\begin{gathered} 793MCM=\left(793cmil\right)\left(\frac{{\displaystyle \frac{\mathrm{\pi }}{4}}sq.mil}{1cmil}\right){\left(\frac{1in}{100mil}\right)}^2{\left(\frac{0.0254m}{1in}\right)}^2 \\ =4.0175\times {10}^{-4}{m}^2 \end{gathered}$$

Therefore, the area of conductor is $4.0175\times {10}^{-4}{m}^2$.

Determine the resistance.

(b)

Substitute $0.0080\Omega$for $R_{60Hz,75}$and $228.1$for $T$in equation (2).

$$\begin{gathered} R_{60Hz,50}=0.0080\left(\frac{50+228.1}{75+228.1}\right) \\ =0.0807\Omega \end{gathered}$$

Therefore, the resistance is $0.0807\Omega$per kilometer.