a) 2k/3
b) 3k
c) 3k/2
d) 4k/3
Answer: d) 4k/3
Solution:
Let
A = Area
T1 is the temperature of the sink
T2 is the temperature of the source
T is the temperature of the junction
For the first conductor , the thermal rate is given by
(H1/t)=KA(T–T1)L ………….1
For the first conductor, the thermal rate is given by
(H2/t)=2KA(T2–T)L …………..2
The two slabs are connected in series so the heat current flowing through them will be the same
So, H1/t=H2/t
⇒KA(T–T1)/L=2KA(T2–T)/L
⇒T=(2T2+T1)/3 …………..3
Let the equivalent coefficient of thermal conductivity be K’
Adding equations (1) and (2)
[H1+H2/t]=(KA(T–T1)/L)+(2KA(T2–T)/L)\(\begin{array}{l}\Rightarrow \frac{H}{t}=\frac{[KA[{\frac{1}{3}{(2T_{2}+T_{1})-T_{1}} ]}+2KA[T_{2}-\frac{(2T_{2}+T_{1})}{3}]}{L}\end{array} \)
⇒(H/t)=KA(4/3)(T2−T1/L)
⇒(H/t)={(4/3)K}A(T2−T1)L
So, the equivalent thermal conductivity is (4/3)K
