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`.

Answer

`[T_(B)=30^(@)C,T_(C)=T_(D)=20^(@)]`

Explanation

Let `T_(A), T_(B),T_(C),T_(D) and T_(E)` be the temperatures of junction A, B, C, D and E respectively. Then `T_(A)=60^(@)C, T_(E)=10^(@)C`
Let l be the length each rod and A be the area of cross-section of each rod. At steady state, heat entering a junction per second is equal to heat leaving that junction per second. For junction B, Heat entering the junction B per second `=(K_(y)A(T_(A)-T_(B)))/(l)` .....(i)
Heat leaving the junction B per second
`(K_(x)A(T_(B)-T_(C)))/(l)+(K_(Y)A(T_(B)-T_(D)))/(l)` ....(ii)
Equating (i) and (ii), we have
`(K_(Y)A(T_(A)-T_(B)))/(l)=(K_(X)A(T_(B)-T_(C)))/(l)+(K_(Y)A(T_(B)-T_(D)))/(l)`
or `K_(Y)(T_(A)-T_(B))=K_(X)(T_(B)-T_(C))+K_(Y)(T_(B)-T_(D))`
or `0.46(60-T_(B))=0.92(T_(B)-T_(C))+0.46(T_(B)-T_(D))`
or `(60-T_(B))=2 (T_(B)-T_(C))+(T_(B)-T_(D))`
or `4 t_(B)-2 T_(C)+T_(D)=-10` ...(iii)
For the junction C, Proseeding as above, we get
`K_(X)(T_(B)-T_(C))=K_(X)(T_(C)-T_(D))+K_(X)(T_(C)-T_(E))`
or `T_(B)-3 T_(C)+T_(D)= -T_(E)`
or `T_(B)-3 T_(C)+T_(D)= -10` .....(iv)
For the junction D, Prosseding as above, we obtain
`K_(Y) (T_(B)-T_(D))=K_(X)(T_(D)-T(C))+K_(Y)(T_(D)-T_(E))`
or `0.46 (T_(B)-T_(D))=0.92(T_(D)-T_(C))+0.46(T_(D)-T_(E))`
or `T_(B)-T_(D)=2(T_(D)-T_(C))+(T_(D)-T_(E)`
or `T_(B)+2 T_(C) -4T_(D)= -10` .....(v)
On solving the equation (iii), (iv) and (v), we get
`T_(B)=30^(@)C, T_(C)=20^(@)C=T_(D)`