Heat flows from high to low temperature i.e., from left side to right side, as given in fig.

After some time, temperature of every section becomes constant with time. This is known as steady state.

If ΔQ is the amount of heat transferred through a cross-section in Δt second at steady state, then the rate of heat flow is shown by:        

K = coefficient of thermal conductivity of solids

Its units are:    J s^-1 m^-1 K^-1 (W s^-1 m^-1 K^-1)           or         cal  s^-1 m^-1

Heat Conduction through a Composite Slab (rod):

(A) Two rods connected in series             (B) Three rods connected in series

(A) Take a composite rod built up of two rods of lengths d1 and d2 and each of cross-section A, connected end to end. Let K1 and K2 be the coefficients of thermal conductivities of two nodes and θA and θAB be the temperatures of two points of the composite rod.

            Let us suppose θ as the temperature of the junction of two rods and θ_A > θ_AB. Heat flows from left to the right. In steady phase, heat flow per second is same through every rod. It is shown by:

                                        H = ΔQ/ΔT

            For lst rod:          H = K₁/d₁(θ_A-θ)                                     .... (i)

            For IInd rod:        H = K₂/d₂(θ-θ_B)                                          .... (ii)

            Eliminating q from (i) & (ii) ,             

            Equating (i)  & (ii),            

(B) Consider a composite rod build up of three rods of sizes d₁, d₂ and d₃ connected end to end. Let area of cross-section of every rod be A. The left side of the composite rod is maintained at θ_A and right side at θ_B. Let the coefficient of thermal conductivities of three rods be K₁, K₂­ and K₃ and the junction temperatures of rods 1 and 2 be θ₁ and that of 2 and 3 be θ₂.

            As θ_A > θ_B heat drowns from left to right side. At steady state let H be the heat going per second.

            First rod :                     H = K₁/d_1A(θ_A-θ₁)                               .... (i)

            Second rod :                 H = K₂/d₂A(θ₁-θ₂)                               .... (ii)

            Third rod :                     H = K₃/d₃A(θ₂-θ_B)                           .... (iii)

            From (i)  θ_A-θ₁ = H/A.d₁K₁          From (ii)θ₁-θ₂ = H/A.d₂K₂           From (iii)  θ₂-θ_B = H/A.d₃K₃    

            θ_A and θ_B are given, so θ₁ and θ₂ can be calculated.

            Adding the equations to remove θ₁ and θ₂, we get:

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