Answer to Question #272726 in Mechanical Engineering for Brian

Question

A steam pipe 100mm external diameter is to be lagged with two layers of different lagging material each 25mm each material A has a coefficient of thermal conductivity of 0.052W/mk while that of material B is 0.086W/mk Determine which of the two lagging materials must be on the inside to produce the best insulation arrangement if the internal surface temperature is 320°c and the external surface temperature is 20°c Determine the heat loss/h a 10m run of lagged pipe with the best lagging arrangement

Answer 1

Take $k_i=0.052W/mk, k_o=0.086W/mk$

$Q=\frac{T_i-T_o}{(\frac{ln\frac{\delta_2}{\delta_1}}{2πk_iL}+\frac{ln\frac{\delta_2}{\delta_1}}{2πk_iL})}$

$=\frac{320-20}{(\frac{ln\frac{(0.025)}{(0.050}}{2π*0.052*10}+\frac{ln\frac{(0.1)}{(0.075}}{2π*0.086*10})}$

=1691.674W

Now take $k_i=0.086W/mk, k_o=0.052W/mk$

$Q=\frac{320-20}{(\frac{ln\frac{(0.025)}{(0.050}}{2π*0.086*10}+\frac{ln\frac{(0.1)}{(0.075}}{2π*0.052*10})}$

=1839.509W

So for the best insulation arrangement the material with least thermal conductivity (k=0.052W/mk) should be placed inside.

Heat loss

Q=1691.674kJ/s

Q=6090.0264kJ/h