Three rods of material X and three rods of material Y are connected as shown in Fig. 1.62. All the rods are of identical lengths and cross-sectional areas, If the rod are of idnetical lengths and cross-sectional areas. If the rod end A is maintained at 60^(@)C` and the jucntion E at `10^(@)C`, calculate the temperature of jusnction B, C and D, the thermal conductivity of X is `9.2xx10^(-2)"kcal"//m//^(@)C` and that of Y is `4.6xx10^(-2) "kcal"//m//s//^(@)C`.

Explanation

Let `theta_(B), theta_(C)` and `theta_(D)` be the temperature of junction B,C and D respectively at staady state . Let `K_(x)` and `k_(y)`
be thermal conductivites of rods of material X and Y respectively .
The concept in solving the problem is that in steady state heat entering per second at any junction must be equal to that leaving the junction.

As `K_(x) = 9.2 xx 10^(-2)`
ans `K_(y) = 4.6 xx 10^(-2) kilocal m^(-1) (""^(@)C)^(-1) (sec)^(-1)`
i.e.., `(K_(x))/(K_(y)) = (9.2 xx 10^(-2))/(4.6 xx 10^(-2)) 2 `
Let `K_(x) = 2K, K_(y) = K`
(or) `(K_(y)A(60-theta_(B)))/(l) = (K_(x)A(theta_(B) - theta_(C)))/(l) + (K_(y) A(theta_(B) - theta_(C)))/(l)`
(or) `K (60 - theta_(B)) = 2K(theta_(B) - theta_(C)) + K(theta_(b) - theta_(D)) therefore (60-theta_(B)) = 2(theta_(B) - theta_(C)) + theta_(B) - theta_(D)`
`therefore 4 theta_(B) - 2 theta_(C) - 2 theta_(D) = 60...........(1)`
For junction `C_(1),H_(1) = H_(3) + H_(5) therefore (K_(y)A(theta_(B) - theta_(C)))/(l) = (K_(x)A(theta_(c) - 10))/(l) + (k_(y)A(theta_(D) - 10))/(l)`
This gives `therefore theta_(B) - 3 theta_(C) + theta_(D) = - 10..........(2)`
For junction D, `H_(2) + H_(3) = H_(5) therefore (k_(y) A(theta_(B) - theta_(D)))/(l) + (k_(y)A(theta_(C) - theta_(D)))/(l) = (k_(y) A(theta_(D) - 10))/(l)`
(or) ` K(theta_(B) - theta_(D)) + 2K(theta_(C) - theta_(D)) = k(theta_(D) - 10)`
This gives `theta_(B) + 2theta_(C) - 4 theta_(D) = - 10..............(3)` from (2) and (3)
`theta _(B)-3theta_(C) - theta_(D) = theta_(B) + 2 theta_(C) - 4 theta_(D)`
i.e., `5theta_(C) = 5 theta_(D) " " or " " theta_(C) = theta_(D) ............(4)` Substituting this in (1) and (2), we get .
` 4 theta_(B) - 3 theta_(C) = 60.............(5) , theta_(B) - 2theta_(C) = - 10............(6)`
Solving (5) and (6) , we get `theta_(B) = 30^(@)C, theta_(C) = 20^(@)C`
As `theta_(c) = theta_(D) " " therefore " " theta_(C) = theta_(D) = 20^(@)C`
`therefore` Thus the temperature of junction , B, C and D are `30^(@),20^(@)C and 20^(@)C` respectively.