There slabs of thickness `d_(1)`, `d_(2)`, `d_(3)` and thermal conductivites `lambda_(1)`, `lambda_(2)` and `lambda_(3)` are placed in contact. Show that in the steady the combination behaves as a single slab of material of conductivity `lambda` given by `(d_(1)+d_(2)+d_(3))/(lambda)=(d_(1))/(lambda_(1))+(d_(2))/(lambda_(2))+(d_(3))/(lambda_(3))`

Explanation

Suppose that the temperature of the faces are `theta_(1)`, `theta_(2)`, `theta_(3) theta_(4)`, respectively. Then in the steady state
`Q=(lambda_(1)A(theta_(1)-theta_(2)))/(d_(1))`
`=(lambda_(2)A(theta_(2)-theta_(3)))/(d_(2))=(lambda_(3)A(theta_(3)-theta_(4)))/(d_(3))`
`:. theta_(1)-theta_(2)=(Qd_(1))/(lambda_(1)A)`, `theta_(2)-theta_(3)=(qd_(2))/(lambda_(2)A)`
`theta_(3)-theta_(4)=(Qd_(3))/(lambda_(3)A)`
For the combiantion we have
`Q=(lambdaA(theta_(1)-theta_(4)))/(d_(1)+d_(2)+d_(3))` or `theta_(1)-theta_(4)(Q(d_(1)+d_(2)+d_(3)))/(lambdaA)`
Now `theta_(1)-theta_(4)=(theta_(1)-theta_(2))+(theta_(2)-theta_(3))+(theta_(3)-theta_(4))`
`:. (Q(d_(1)+d_(2)+d_(3)))/(lambdaA)=(Qd_(1))/(lambda_(1)A)+(Qd_(2))/(lambda_(2)A)+(Qd_(3))/(lambda_(3)A)`
or `(d_(1)+d_(2)+d_(3))/(lambda)=(d_(1))/(lambda_(1))+(d_(2))/(lambda_(2))+(d_(3))/(lambda_(3))`