Question #232803

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.


1
Expert's answer
2021-09-06T06:56:51-0400

Solution;

Given;

r1=202=10mm=0.01mr_1=\frac{20}{2}=10mm=0.01m

r2=402=20mm=0.02mr_2=\frac{40}{2}=20mm=0.02m

r3=20+30=50mm=0.05mr_3=20+30=50mm=0.05m

t1=600°ct_1=600°c

t3=1000°ct_3=1000°c

kB=0.2W/m°ck_B=0.2W/m°c

Heat transfer per meter of length is;

From;

Q=2πL(t1t3)ln(r2r1)kA+ln(r3r2kBQ=\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;

QL=2π(t1t3)ln(r3r2)kB\frac{Q}{L}=\frac{2π(t_1-t_3)}{\frac{ln(\frac{r_3}{r_2})}{k_B}}

QL=2π(6001000)ln(0.050.02)0.2\frac{Q}{L}=\frac{2π(600-1000)}{\frac {ln(\frac{0.05}{0.02})}{0.2}}

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

Note:

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


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