Question:-
A composite wall consists of two layer of different materials having conductivities k1 and k2. For equal thickness of the two layers the equivalent thermal conductivity of the slab will be
Option (A)
k1 + k2
Option (B)
k1k2
Option(C)
(2k1k2/(k1+k2))
Option(D)
((k1+k2)/k1k2)
Correct Option:
(C)
Question Solution:

Or T1-T2 = (2Qb/KA)
Or T1 - T3 = (Qb/K1A)
Or T3 -T2 = (Qb/K2A)
Or (2Qb/KA) = (Qb/K1A)+(Qb/K2A)
\ K = (2K1K2 / (K1+K2))
question-answer-faq-3080