Question

Two walls of thickness d1 and d2 thermal conducticities k1 and k2 are incontact in the steady state if the temperature at the outer surfaces are t1 and t2 the temperture common wall

Answer 1

Let Δt be the small time in which heat flow from one from one end to the wall.

Now since, the two conductors are connected in series the heat current flow through them will remain same.

ΔH1/ Δt1 = ΔH2/ Δt2 ---1

Let x be the temperature of the wall

Heat current reaching the wall will be

ΔH1/ Δt1 = k1A(x – t1) /d1 ----2.

Thermal resistance R1 = k1A/d1

Now this amount of heat reaching the wall will travel to the other side

ΔH2/ Δt2 = k2A(t2 – x) /d2 ----3.

From 1, 2 and 3.

k1A(x – t1)/d1 = k2A(t2 – x)/d2

=> k1(x – t1)/d1 = k1(t2 – x)/d1

=> x = (k1t1d2 + k2t2d1)/( k1d2 + k2d1)

Answer 2

The answer is provided in the attachment.

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