A composite slab consists of two plates of thicknesses L1 and L2 and thermal conductivities K1 and K2. The cross-sectional areas are equal. The equivalent thermal conductivity is:

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K1K2(L1+L2)K2L1+K1L2

K2L1+K1L2K1K2(L1+L2)

L1L2(K1+K2)(K2L1+K1L2)

K2L1+K1L2L1L2(K1+K2)

answer is A.

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Detailed Solution

In series connectionKeq=∑li∑liKi=L1+L2L1K1+L2K2=K1K2(L1+L2)K2L1+K1L2

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