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

2 k₁ k₂/ (k₁ + k₂)
k₁ k₂
(k₁ + k₂)/ k₁ k₂
(k₁ + k₂)

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Heat and Mass Transfer
Fourier's law of heat conduction is (where Q = Amount of heat flow through the body in unit time, A = Surface area of heat flow, taken at right angles to the direction of heat flow, dT = Temperature difference on the two faces of the body, dx = Thickness of the body, through which the heat flows, taken along the direction of heat flow, and k = Thermal conductivity of the body)

k. (dT/dx)
k. A. (dT/dx)
k. (dx/dT)
k. A. (dx/dT)

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Heat and Mass Transfer
The heat transfer by conduction through a thick cylinder (Q) is given by (where T₁ = Higher temperature, T₂ = Lower temperature, r₁ = Inside radius, r₂ = Outside radius, l = Length of cylinder, and k = Thermal conductivity)

Q = [2πlk (T₁ - T₂)]/2.3 log (r₂/r₁)
Q = 2.3 log (r₂/r₁)/[2πlk (T₁ - T₂)]
Q = [2π (T₁ - T₂)]/2.3 lk log (r₂/r₁)
Q = = 2πlk/2.3 (T₁ - T₂) log (r₂/r₁)

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