Answer to Question #232803 in Molecular Physics | Thermodynamics for Unknown346307

Question

A thick walled tube of stainless steel with 20 mm inner diameter and

40 mm outer diameter is covered with a 30

mm layer of asbestos insulation (k = 0.2 W/

m°C). If the inside wall temperature of the pipe

is maintained at 600°C and the outside

insulation at 1000°C, calculate the heat loss

per metre of length.

Answer 1

Solution;

Given;

$r_1=\frac{20}{2}=10mm=0.01m$

$r_2=\frac{40}{2}=20mm=0.02m$

$r_3=20+30=50mm=0.05m$

$t_1=600°c$

$t_3=1000°c$

$k_B=0.2W/m°c$

Heat transfer per meter of length is;

From;

$Q=\frac{2πL(t_1-t_3)}{\frac{ln(\frac{r_2}{r_1})}{k_A}+\frac{ln(\frac{r_3}{r_2}}{k_B}}$

Since the thermal conductivity of satinless steel is not given, therefore, neglecting the resistance offered by stainless steel to heat transfer across the tube, we have;

$\frac{Q}{L}=\frac{2π(t_1-t_3)}{\frac{ln(\frac{r_3}{r_2})}{k_B}}$

$\frac{Q}{L}=\frac{2π(600-1000)}{\frac {ln(\frac{0.05}{0.02})}{0.2}}$

$\frac{Q}{L}=-548.57W/m$

Note:

Negative sign indicates that the heat transfer takes place radially inward.