A large 35 mm thick slab of material has surface temperatures of 318.8 degrees K and...
A large 35 mm thick slab of material has surface temperatures of 318.8 degrees K and 303.2 degrees K, and heat flux through the material of 35.1 W/m^2. Calculate the thermal conductivity of the slab.
Homework Answers
Flux is given by:
Flux = k*dT/d where :
k = thermal conductivity
dT = temperature difference
d = thickness
Rearranging in terms of k:
k = flux*d/(dT)
= (35.1)(0.035m)/(318.8-303.2)
= 0.07875 W/(mK)
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