Question

2. Heat flows through the walls of an annular pipe whose inside and outside radii are R and Ro respectively. The thermal conductivity of the material is directly proportional to the square of temperature. k aT2 b (a, b constant) Derive the temperature profile of the wall and compute the heat loss through the walls. (20 points)

Answer 1

Given heat flows through wall of annual pipe inside radii R_1 outside radii R_0 The thermal conductivity of material is k = aT^2 + b (a, b = constant) To derive the temperature profile of wall and compute heat loss through walls now For steady heat conduction with no heat of generation 1/r d/dr (kr dT/dr) = 0 q_r = -kA dT/dr = -k (2 pi r L) dT/dr since q_r is constant because kr dT/dr is independent dr q_r/2 pi r^2 = -k dT/dr Given k = aT^2 + b q_r/2 pi l integral^R_0_R_i dr/r = -integral^T_0_T_i (aT^2 + b) dT \r\n q_r/2 pi L integral^R_0_R_i dr/r = -integral^T_0_T_i (aT^2 + b) dT q_r/2 pi L (2n R_0/R_i) = -a/3 (T_0^3 - T_^3) - b (T_0 - T_i) q_r/L = (T_i^3 - T_0^3) a/3 + (T_i - T_0) p/ln R_0/R_i times pi Therefore Temperature profile is a/3 T^3 + bT