# Find the equivalent thermal conductivity of the given rod

a) 2k/3

b) 3k

c) 3k/2

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’

[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