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.

Question

Quizplus · Accepted Answer

The Answer is

\begin{aligned}& R_{d c, 20^{\circ} \mathrm{C}}=\frac{P_{20^{\circ} \mathrm{C}} l}{A}=\frac{(17.00)(1000 imes 1.016)}{1113 imes 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} ight)=0.01552\left(\frac{50+228.1}{20+228.1} ight) \& 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}} ight)\left(5.28 \frac{1000^{1}}{\mathrm{mi}} ight)}=\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.

The Aluminum Electrical Conductor Handbook Lists a Dc Resistance Of 0.01552