Question 1

The Aluminum Electrical Conductor Handbook lists a dc resistance of $0.01552 \mathrm{ohm}$ per $1000 \mathrm{ft}$ at $20^{\circ} \mathrm{C}$ and a $60-\mathrm{Hz}$ resistance of $0.0951 \mathrm{ohm}$ per mile at $50^{\circ} \mathrm{C}$ for the allaluminum Marigold conductor, which has 61 strands and whose size is $1113 \mathrm{kcmil}$. Assuming an increase in resistance of $1.6 \%$ for spiraling, calculate and verify the dc resistance. Then calculate the dc resistance at $50^{\circ} \mathrm{C}$, and determine the percentage increase due to skin effect.

Accepted Answer

$\begin{aligned}
& R_{d c, 20^{\circ} \mathrm{C}}=\frac{P_{20^{\circ} \mathrm{C}} l}{A}=\frac{(17.00)(1000 \times 1.016)}{1113 \times 10^{3}}=\underline{\underline{0.01552}} \frac{\Omega}{1000^{1}} \
& R_{d c, 50^{\circ} \mathrm{C}}=R_{d c, 20^{\circ} \mathrm{C}}\left(\frac{50+T}{20+T}\right)=0.01552\left(\frac{50+228.1}{20+228.1}\right) \
& R_{d c, 50^{\circ} \mathrm{C}}=(0.01552)(1.1209)=\underline{0.01739} \frac{\Omega}{1000^{1}} \
& \frac{R_{60 \mathrm{HƵ}, 50^{\circ} \mathrm{C}}}{R_{d c, 50^{\circ} \mathrm{C}}}=\frac{0.0951 \Omega / \mathrm{mi}}{\left(0.01739 \frac{\Omega}{1000^{1}}\right)\left(5.28 \frac{1000^{1}}{\mathrm{mi}}\right)}=\frac{0.0951}{0.0918}=\underline{\underline{1.035}}
\end{aligned}$
The $60 \mathrm{HƵ}$ resistance is $3.5 \%$ larger than the dc resistance, due to skin effect.

Question 2

One thousand circular mils or $1 \mathrm{kcmil}$ is sometimes designated by the abbreviation MCM. Data for commercial bare aluminum electrical conductors lists a $60-\mathrm{Hz}$ resistance of $0.0740 \mathrm{ohm}$ per kilometer at $75^{\circ} \mathrm{C}$ for a 954-MCM AAC conductor. (a) Determine the cross-sectional conducting area of this conductor in square meters. (b) Find the $60-\mathrm{Hz}$ resistance of this conductor in ohms per kilometer at $45^{\circ} \mathrm{C}$.

Accepted Answer

(a) $954 \mathrm{MCM}=\left(954 \times 10^{3} \mathrm{cmil}\right)\left(\frac{\frac{\pi}{4} \text { sqmil }}{1 \mathrm{cmil}}\right)\left(\frac{1 \mathrm{in}}{1000 \mathrm{mil}}\right)^{2}\left(\frac{.0254 \mathrm{~m}}{\text { in }}\right)^{2}$
$=\underline{\underline{4.834 \times 10^{-4}} \mathrm{~m}^{2}}$
(b) $R_{60 \mathrm{~HƵ}, 45^{\circ}}=R_{60 \mathrm{~Hz}, 75^{\circ}}\left(\frac{45+T}{75+T}\right)$
$\begin{aligned}
& =(0.0740)\left(\frac{45+228.1}{75+228.1}\right)=(0.0740)(0.9010) \
& =\underline{\underline{0.0667} \Omega / \mathrm{km}}
\end{aligned}$

Question 3

A $60-\mathrm{Hz}, 765-\mathrm{kV}$ three-phase overhead transmission line has four ACSR $1113 \mathrm{kcmil}$ $54 / 3$ conductors per phase. Determine the $60-\mathrm{Hz}$ resistance of this line in ohms per kilometer per phase at $50^{\circ} \mathrm{C}$.

Accepted Answer

From Table A. 4
$R_{60 \mathrm{~HƵ}, 50^{\circ} \mathrm{C}}=\left(0.0969 \frac{\Omega}{\mathrm{mi}}\right)\left(\frac{1 \mathrm{mi}}{1.609 \mathrm{~km}}\right)=0.0602 \frac{\Omega}{\mathrm{km}} \text { per conductor (at } 75 \% \text { current capacity) }$
For 4 conductors per phase:
$R_{60 \mathrm{~HƵ}, 50^{\circ} \mathrm{C}}=\frac{0.0602}{4}=\underline{\underline{0.0151}} \frac{\Omega}{\mathrm{km}} \text { per phase }$

Quiz 3: Transmission Line Parameters

A $60-\mathrm{Hz}, 765-\mathrm{kV}$ three-phase overhead transmission line has four ACSR $1113 \mathrm{kcmil}$ $54 / 3$ conductors per phase. Determine the $60-\mathrm{Hz}$ resistance of this line in ohms per kilometer per phase at $50^{\circ} \mathrm{C}$ .