Three rods of material X and three rods of material Y are connected as shown in the figure. All the rods are of identical length and cross-sectional area. If the end A is maintained at `60^@C` and the junction E at `10^@C`. Calculate the temperature of the junction B, C and D. The thermal conductivity of X is `0.92cal//sec-cm^@C` and that of Y is `0.46cal//sec-cm-^@C`.

Explanation

Let `K_(x)` and `K_(y)` be the thermal conductivites of X and Y respectively, and let `T_(B), T_(C)` and `T_(D)` be the temperature of junctions B,C and D, respectively Given `T_(A)=60^(@)C` and `T_(E)=10^(@)C`. In order to solve this problem, we will apply the principle that, in the steady state, the rate at which heat enters a junction is equal to the rate at which heat leaves that junction. In Fig. 1.63, the direction of flow of heat is given by arrows.
For junction B, we have

`(K_(y)A(T_(A)-T_(B)))/(d)=(K_(x)A(T_(B)-T_(C)))/(d)+(K_(y)A(T_(B)-T_(C)))/(d)`
or `K_(y)(T_(A)-T_(B))=Kx(T_(B)-T_(C))+K_(y)(T_(B)-T_(D))`
or `(T_(A)-T_(B))=(K_(x))/(K_(y))(T_(B)-T_(C))+K_(y)(T_(B)-T_(D))`
Given `(K_(x))/(K_(y))=(9.2xx10^(-2))/(4.6xx10^(-2))=2` and `T_(A)=60^(@)C`. Therefore
`(60-T_(B))=2(T_(B)-T_(C))+(T_(B)-T_(C))+(T_(B)-T_(D))`
or `4T_(B)-2T_(C)-T_(D)=60` (i)
For junction C, we have
`(K_(x)A(T_(B)-T_(C)))/(d)=(K_(x)A(T_(C)-T_(D)))/(d)+(K_(x)A(T_(C)-T_(B)))/(d)`
Solving we get
`-T_(B)+3T_(C)-T_(D)=10 ["At" T_(E)=10^(@)C]` (ii)
For junction D, we have
or `(K_(y)A(T_(B)-T_(D)))/(d)+(K_(x)A(T_(C)-T_(D)))/(d)=(K_(y)A(T_(D)-T_(E)))/(d)`
or `(T_(B)-T_(D))+(K_(x))/(K_(y))=2` and `T_(E)=10^(@)C`. In this equation, we have
`(T_(B)-T_(D))+2(T_(C)-T_(D))=(T_(D)-10)`
`T_(B)+2T_(C)-4T_(D)=-10` (iii)
Solving Eqs. (i),(ii) and (iii), we get
`T_(B)=30^(@)C` and `T_(C)=T_(D)=20^(@)C`