Question

# The wall of a refrigerator of 4m^2 surface area consists of two metal sheets with insulation...

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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