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Two plates each of area A, thicknesses L1 and L2 and thermal conductivities K1 and K2 are joined to form a single plate of thickness (L1 + L2). If the temperatures of the free surfaces are T1 and T2,Find the equivalent thermal conductivity:

(A) L1 - L2 / [L1/K1 + L2/K2]

(B) L1+L2 / [L1/K1 - L2/K2]

(C) L1+L2 / [L1/K1 + L2/K2]

(D) L1+L2 / [L1/K2 + L2/K1]

1 Answer

If K is the equivalent conductivity of the composite slab, i.e, slab of thickness L1 + L2 and cross- sectional area A, then as in series
Rs = R1 + R2
So, (L1 + L2)/AKeq = R1 + R2
Therefore, Keq = (L1+L2)/A(R1+R2) = L1+L2 / [L1/K1 + L2/K2] (as R = L/KA)
answered Mar 10, 2014 by balaji.thirumalai

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