Question

The Aluminum Electrical Conductor Handbooklists a dc resistance of $\mathbf{0}\mathbf{.}\mathbf{01558}\mathit{\Omega }$ per$\mathbf{1000}\mathbf{ft}$ at $\mathbf{20}\mathbf{°}\mathbf{C}$ and a $\mathbf{60}\mathbf{Hz}$ resistance of role="math" localid="1655275109672" $\mathbf{0}\mathbf{.}\mathbf{0956}\mathit{\Omega }$ per mile at $\mathbf{50}\mathbf{°}\mathbf{C}$ for the all-aluminum Marigold conductor, which has 61 strands and whose size is$\mathbf{1113}\mathbf{kcmil}$ . Assuming an increase in resistance of 2% for spiraling, calculate and verify the dc resistance. Then calculate the dc resistance at $\mathbf{50}\mathbf{°}\mathbf{C}$, and determine the percentage increase due to skin effect.

Short answer

Therefore, the dc resistance at $20°\mathrm{C}$ is $0.01558\Omega$ and at$50°\mathrm{C}$ is $0.01746\Omega$. The increase in resistance is $3.68\%$.

Step-by-step solution

Given data.

Assume that, the resistivity of all-aluminium Marigold conductor at$20°\mathrm{C}$is$\mathrm{\rho }=17\mathrm{\Omega }-\mathrm{m}$ .

The area of the conductor is $\mathrm{A}=1113\mathrm{kcmil}$.

The length of the conductor is $\mathrm{I}=1000\mathrm{ft}$.

The resistance of conductor at $50°\mathrm{C}$ is ${\mathrm{R}}_{60,50}=0.0956\mathrm{\Omega }$ per mile.

Determine the formulas of dc resistance

Write the formula of dc resistance at $20°\mathrm{C}$.

${\mathrm{R}}_{\mathrm{dc},20}=\frac{\mathrm{\rho l}}{\mathrm{A}}$ ……. (1)

Here,$\rho$is resistivity, $l$is length of conductor and $A$ is it’s area.

Write the formula of dc resistance at $50°\mathrm{C}$.

${\mathbf{R}}_{\mathbf{dc}\mathbf{,}\mathbf{50}}\mathbf{=}{\mathbf{R}}_{\mathbf{dc}\mathbf{,}\mathbf{20}}\frac{\mathbf{50}\mathbf{+}\mathbf{T}}{\mathbf{20}\mathbf{+}\mathbf{T}}$ ……. (2)

Here,$T$ is a temperature constant.

Write the formula of increase in resistance due to skin effect at $50°\mathrm{C}$.

$\mathrm{x}=\frac{{\mathrm{R}}_{60,50}}{{\mathrm{R}}_{\mathrm{dc},50}}$ ……. (3)

Here,${\mathrm{R}}_{60,50}$ is an original dc resistance.

Determine the dc resistance with 2 % increase.

Determine the resistance at$20°\mathrm{C}$.

Substitute$17\mathrm{\Omega }-\mathrm{m}$for$\mathrm{\rho }$,$1113kcmil$for $A$ and $1000\mathrm{ft}$ for$l$in equation (1).

$\begin{array}{c}{\mathrm{R}}_{\mathrm{dc},20}=\frac{17\times 1000}{1113\times {10}^3}\\ =0.01527\frac{\mathrm{\Omega }}{1000\text{'}}\end{array}$

Now, determine the dc resistance with$2\%$ increase

$\begin{array}{c}{\mathrm{R}}_{\mathrm{dc},20}=0.01527\mathrm{\Omega }\times 1.02\\ =0.01558\frac{\mathrm{\Omega }}{1000\text{'}}\end{array}$

Determine the resistance at role="math" localid="1655275976926" $50°\mathrm{C}$.

Substitute $0.01558\Omega$for ${\mathrm{R}}_{\mathrm{dc},20}$ and $228.1$ for $\mathrm{T}$in equation (2).

$\begin{array}{c}{\mathrm{R}}_{\mathrm{dc},50}=0.01558\left(\frac{50+228.1}{20+228.1}\right)\\ =0.01746\frac{\mathrm{\Omega }}{1000\text{'}}\end{array}$

Therefore, the dc resistance at $20°\mathrm{C}$is $0.01558\Omega$ and at$50°\mathrm{C}$ is $0.01746\Omega$.

Determine the increase in resistance.

Substitute$0.01746\Omega /1000\text{'}$ for${\mathrm{R}}_{\mathrm{dc},50}$and$0.0956\Omega /1\text{'}$ for ${\mathrm{R}}_{60,50}$ in equation (3).

The increase in resistance is$3.68\%$.

Therefore, the increase in resistance is $3.68\%$.