A composite slab has two layers of different materials with thermal conductivities k₁ and k₂. If each layer has the same thickness, the equivalent thermal conductivity of the slab would be

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This question was previously asked in

BPSC AE Paper 5 (Mechanical) 25th March 2022 Official Paper

Answer 1

Concept:

Thermal resistance:

Thermal resistance is defined as the ratio of the temperature difference between the two faces of a material to the rate of flow per unit area.

\({\bf{i}}.{\bf{e}}.\;{{\bf{R}}_{{\bf{thermal}}}} = \frac{{{\bf{\Delta T}}}}{{\bf{Q}}}\)

For a solid plate, thermal resistance is given by:

\({{\bf{R}}_{{\bf{thermal}}}} = \frac{{\bf{L}}}{{{\bf{kA}}}} \)

cilmt001

Since the slabs are in contact with each other side by side. Therefore, the thermal resistances will be in series.

The equivalent thermal resistance to the flow of heat is given by,

\({R_{eq}} = {R_1} + {R_2}\)

\(\frac{{{L_{eq}}}}{{{k_{eq}}A}} = \frac{{{L_1}}}{{{k_1}A}} + \frac{{{L_2}}}{{{k_2}A}}\)

Since \({L_1} = {L_2} = L\)

\(\frac{{2L}}{{{k_{eq}}}} = \frac{L}{{{k_1}}} + \frac{L}{{{k_2}}}\)

\(\frac{{{k_{eq}}}}{2} = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}\\\)

\({k_{eq}} = \frac{{2{k_1}{k_2}}}{{{k_1} + {k_2}}}\)

Important Points

Thermal resistance for various shapes:

Heat conduction through the plane wall	\(Q = \frac{{{T_1} - {T_2}}}{{\frac{L}{{kA}}}}\)
Heat conduction through a hollow cylinder	\(Q = \frac{{{T_1} - {T_2}}}{{\frac{{\ln \left( {\frac{{{r_o}}}{{{r_i}}}} \right)}}{{2\pi kL}}}}\)
Heat conduction through the hollow sphere	\(Q = \frac{{{T_1} - {T_2}}}{{\frac{{{r_o} - {r_i}}}{{4\pi k{r_o}{r_i}}}}}\)