`A, B and C` are three identical rods made of different materials and placed end-to-end as shown in Fig. 11.16. The thermal conductivity of A is twice that of B and four times that of C. The free end of A is kept at 100° and that of C is kept at 0°C. Find the temperature of the AB junction and the BC junction.

Explanation

Thermal resistance `= R = (l)/( KA)`
The thermal resistance of the bar `AB, BC and BD` are `R_1` each and that of `CE` an `DE` are `R_2` each. When `R_1 and R_2` are in series the total thermal resistance is `R_1 + R_2 and R_1 and R_2` are in parallel the total thermal resistance is `(R_1 R_2)/( R_1 + R_2)`
The bars `BC and CE` and in series. Their total thermal resistance is `R_1 + R_2`. Total thermal resistance of `BD and DE` is also `R_1 + R_2`. These two are in parallel. So the total thermal resistance of the frame `BCED` is two `(R_1 + R_2)` in parallel `= (R_1 + R_2)/( 2)`
The bar AB of thermal resistance `R_1` is the series with `(R_1 + R_2)/( 2)`. So the net thermal resistance between A and E.
`= R_1 + (R_1 + R_2)/(2 ) = (3 R_1 + R_2)/( 2)`
The rate of heat flow through the system per second during steady state
`= ("Temperature difference between A and E")/("Net thermal resistance")`
`Q= (100-0)/( (3R_1 + R_2)/(2))" "...(i)`
where, `R_1 = (1)/( K_1 A), R_2 = (1)/( K_2 A) " "...(ii)`
Rate of heat flow through AB per second is `Q= (100- theta_B)/( R_1 )`, where `theta_B` is the temperature at B.
From equations (i) and (ii)
`( 2 xx 100)/( 3 R_1 + R_2 ) = ( 100 - theta_B)/( R_1 )`
`(200)/( (3)/( K_1) + (1)/( K_2) ) = ( 100 - theta_B)/( (1)/( K_1) )`. Given ` (K_1)/(K_2) = (2)/(1) , K_2 = (K_1)/( 2)`
`(200)/( (3)/(K_1) + (2)/( K_1) )= (100 - theta_B ) /( (1)/( K_1) ) , 40 = 100- theta_B , theta_B = 100 - 40 = 60^(@) C`