PE Integrated Concepts (a) An aluminum power transmission

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Problem 66PE Chapter 20

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 66PE

PE Integrated Concepts (a) An aluminum power transmission line has a resistance of 0.0580 ? / km . What is its mass per kilometer? (b) What is the mass per kilometer of a copper line having the same resistance? A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.

Step-by-Step Solution:

Step-by-step solution Given : 1) Resistance of (Al) wire = 0.0580 / km 2) Resistivity of the Al wire = 2.65 x 10 .m-8 3 3 3) Density of the Al wire= 2.70 x 10 kg /m 3 3 4) Density of copper wire = 8.90 x 10 kg /m 5) Resistivity of the Cu wire = 1.72 x 10 .m -8 Calculate; 1)Mass per km of the Al wire= 2) Mass per Km of Cu wire (with same resistance)= 3) Comment on practical limits of lowering the resistance and heat. Step 1 o f 6 We know that, The resistance R of a uniform cylindrical wire of length L, cross-section area A, and made of a material of material with resistivityis given by equation (1) as below. R = L/A R= L/r ……………………..(1) Here, r is the radius of the wire. Step 2 of 6 Calculating the resistance per unit length for 0.0580 / km, so we can get the resistance of 1.00 km wire is, R=( 0.0580 /km) (1 Km) = 0.0580 Assume, isRhe resistivity of the transmission line. Therefore, R = R/A 2 R = LR r 2 r = ( LRR ) ………………………..(2) Substitute 0.0580 for R, 1.00 km for L, 2.65 x 10 .m for . in equation (2) we R get ( Aluminium wire) r = ( L/R ) R = [ (2.65 x 10-8.m) ( , 1.00 km ) (1000 m/1 km) / (0.0580 )] = ( (2.65 x 10-8.m) (1000 m) / (0.0580 ) -4 2 = 1.45 x 10 m Step 3 of 6 , Let m and V are the mass and volume of the transmission line (Al) respectively and is the density of the transmission line is given by equation ( 3) as m below Thus, m= V m 2 m= (r mL……………………(3) -4 2 2 3 3 Substitute 1.45 x 10 m for r , 1.00 km for L and 2.70 x 10 kg /m for . in m equation (3) we get, m= ( 2.70 x 10 kg /m ) ( 1.45 x 10 m (1.00 km )( 10 m/1km)2) 3 3 3 -4 2) m= ( 2.70 x 10 kg /m ) ( 1.45 x 10 m (1000 m ) 3 = 1.23 x 10 kg 3 Hence, the mass is 1.23 x 10 kg.

Step 4 of 6

Chapter 20, Problem 66PE is Solved
Step 5 of 6

Textbook: Physics: Principles with Applications
Edition: 6th
Author: Douglas C. Giancoli
ISBN: 9780130606204

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PE Integrated Concepts (a) An aluminum power transmission

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