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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a
K1K2(L1+L2)K2L1+K1L2
b
K2L1+K1L2K1K2(L1+L2)
c
L1L2(K1+K2)(K2L1+K1L2)
d
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