A flat surface is covered with insulation with a thermal conductivity of 0.08 W/m-K. The temperature at the interface between the surface and the insulation is 300oC. The outside of the insulation is exposed to air at 30oC and the convection heat transfer coefficient between the insulation and the air is 10 W/m2-K. Ignoring radiation heat transfer, determine the minimum thickness of the insulation, in m, such that the outside surface of the insulation is not hotter than 60oC at steady-state.
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ANSWERS:
Lins = 0.064 m
HINTS:
The key to this problem is that, at steady state (no change in time), all of the heat that arrives at the surface of the insulation by conduction through the insulation must leave the surface as heat transfer by convection. Otherwise, the temperature of the surface would rise or fall as energy accumulated or depleted at the surface. Therefore, q_ins = q_conv.
You can assume that the thermal conductivity of both the brick and the insulation is constant, and thus not a function of temperature, so you can approximate dT/dx as deltaT/delta_x. (delta_x is the thickness of insulation.)
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