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katovenus [111]

3 years ago

11

The Aluminum Electrical Conductor Handbook lists a dc resistance of 0.01558 ohm per 1000 ft at 208C and a 60-Hz resistance of 0.

0956 ohm per mile at 508C for the all-aluminum Marigold conductor, which has 61 strands and whose size is 1113 kcmil. Assuming an increase in resistance of 2% for spiraling, calculate and verify the dc resistance. Then calculate the dc resistance at 508C, and determine the percentage increase due to skin effect.

1 answer:

Bogdan [553]3 years ago

4 0

Answer:

a) 0.01558 Ω per 1000 feet

b) 0.0923 Ω per mile

c) 3.57%

Explanation:

<u>a) Calculate and verify the DC resistance </u>

Dc resistance = R = р $\frac{l}{A}$

for aluminum at 20°C

р = 17 Ωcmil/ft

hence R = 17 * 1000 / ( 113000 ) = 0.01527 Ω per 1000 feet

there when there is an increase in resistance of 2% spiraling

R = 1.02 * 0.01527 = 0.01558 Ω per 1000 feet

<u>b) Calculate the DC resistance at 50°C </u>

R2 = R1 ( $\frac{T+t2}{T+ t1}$ )

where ; R1 = 0.01558 , T = 228 , t2 = 50, t1 = 20 ( input values into equation above )

hence R2 = ( 0.01746 / 0.189 ) Ω per mile = 0.0923 Ω per mile

<u>c ) Determine the percentage increase due to skin effect </u>

AC resistance = 0.0956 ohm per mile

Hence; Increase in skin effect

= ( 0.0956 -0.0923 ) / 0.0923

= 0.0357 ≈ 3.57%

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