The wall of a refrigerator of 4m^2 surface area consists of two metal sheets with insulation in between. The temperature of the inner wall surface is 5 C and that of the outer surface is 20 C. The thermal conductivity of the metal wall is 16 W/m C and that of the insulation is 0.017 W/m C. If the thickness of each metal sheet is 2mm, calculate the thickness of the insulation that is required so that the heat transferred to the refrigerator through the wall is 10 W/m^2.
By Fourier's law,
Q = -kA(dT/dx)
-ve sign indicates that heat is transferred from higher temperature to lower temperature.
For a slab,
Q = -kA T / x
A = surface area normal to the heat transfer
x = thickness of the slab
this can be written as
Q = T / (x/kA)
flow = driving force / resistance
For a composite slab, the resistances are in series
T1 = temperature of the outer surface
T2 = temperature of the inner surface
x1 = x3 = 2 mm = 0.002 m
1 --> outer metal sheet
2---> insulation
3 ---> inner metal sheet
heat flux,
.: x2 = 0.0255 m = 25.5 mm
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