Solution;
Given;
r1=2120=60mm=0.06m
r2=60+60=120mm=0.12m
r3=60+60+40=160mm=0.16m
kA=0.24W/m°c
kB=0.4W/m°c
hhf=60W/m2°c
hcf=12W/m2°c
thf=65°c
tcf=20°c
Length of pipe,L=60m
The rate of heat loss is give by;
Q=hhf.r11+kAln(r1r2)+kBln(r2r3)+hcf.r312πL(thf−tcf)
Q=60×0.061+0.24ln(0.060.12)+0.4ln(0.120.16)+12×0.1612π×60(65−20)
Q=0.2777+2.8881+0.7192+0.520816964.6
Q=3850.5W
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