Question

A composite slab is prepared by pasting two plates of thickness `L_(1)` and `L_(2)` and thermal conductivites `K_(1)` and `K_(2)` . The slab have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Answer 1

Correct Answer - A::B
Here the thermal conductivy are in series,
``((K_1A(theta_1-theta_2))/(l_2)xx(K_2A(theta_1-theta_2))/(l_1))/((K_1A(theta_1-theta_2))/(l_2)+(K_1A(theta_1-theta_2))/(l_2))`
`=(KA(theta_1-theta_2))/(l_1+l_2)`
`implies((K_1)/(L_1)xx(K_2)/(l_2))/((K_1)/(L_1)+(K_2)/(l_2))=K/(L_1+L_2)`
`implies (K_1 K_2)/((K_1)(L_2)+(K_2)(l_1))=K/(L_1+L_2)`.
`K=(K_1 K_2(L_1+L^2))/(K_1 l_2+K_2+l_1)`