GATE 2016 ME (ME2) – Question 25

Heat Transfer · Modes of heat transfer, one-dimensional conduction, resistance concept · 1 mark · Multiple choice

A hollow cylinder has length L, inner radius r1, outer radius r2, and thermal conductivity k. The thermal resistance of the cylinder for radial conduction is

$\ln(r_2/r_1)/(2\pi kL)$
$\ln(r_1/r_2)/(2\pi kL)$
$2\pi kL/\ln(r_2/r_1)$
$2\pi kL/\ln(r_1/r_2)$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (A) $\ln(r_2/r_1)/(2\pi kL)$

Explanation

Steady cylindrical conduction has local area2πrL. Integrating dR=dr/(k·2πrL) from r1 to r2 gives Rth=ln(r2/r1)/(2πkL). This is **A**, positive for r2>r1. Reversing the logarithm would produce an unphysical negative resistance.