Consider a compound slab consisting of two different materials having equal lengths, thickness and thermal conductivities K and 2 K respectively. The equivalent thermal conductivity of the slab is

Step-by-step solution

Let the length of each slab be L and the cross-sectional area be A . The thermal conductivities are K₁ = K and K₂ = 2K . When the slabs are connected in series, the total thermal resistance R_ eq is the sum of individual resistances R₁ and R₂ . R₁ = L K₁ A = L KA R₂ = L K₂ A = L 2KA R_ eq = R₁ + R₂ = L KA + L 2KA = 2L + L 2KA = 3L 2KA The equivalent thermal conductivity K_ eq for a slab of total length 2L and area A is given by R_ eq = 2L K_ eq A . Equating the expressions for R_ eq : 2L K_ eq A = 3L 2KA 2 K_ eq =