Few rods of material X and Y are connected as shown in figure - 4.10. The cross sectional area of all the rods are same. If the end A is maintained at `80^(@)C` and the end F is maintained at `10^(@)C~ . Calculate the temperatures of junctions B and E in steady state . Given that thermal conductivty of material X is double that of Y.

Explanation

We first find the thermal resistances of the different rods shown in figure - 4.10 these are given as
`R_(AB) = (1)/(k_(x).(L)/(A)`
`R_(BCE) =(1)/(k_(y)).(pi^(1//2))/(A)`
`R_(BDE) =(1)/(k_(X)).(pi^(1//2))/(A)`
`R_(EF) =(1)/(k_(y)) .(L)/(A)`
Now in steady state the amount of heat flow end E to F remains constant as there is no absorption of heat . Then we must have that the amount of heat coming at junction B is equal to the amount of heat having B and same statement can be given for junction E. If temperature of junction B and E are taken as `T_(B)` and `T_(E)` then we have for junction B.
`(T_(A)-T_(B))/(R_(AB)) =(T_(B)-T_(E))/(R_(BCE)+(T_(B)-T_(E))/(R_(BDE)`
or `(80-T_(B))/(1//k_(X).L//A)=(T_(B)-T_(E))/(1//k_(Y).piL//2A)+(T_(B)-T_(E))/(1//k_(X).piL//2A)`
or ` 80-T_(B) = ((T_(B)-T_(E))/(pi)) +(T_(B)-T^(E)/(2pi))` `[As k_(X)=2k_(Y)]`
or `80-T_(B)=(3)/(2pi)(T_(B)-T_(E))`..(4.14)
Similarly for junction E, we can write
`(T_(B)-T_(E))/(R_(BCE))+(T_(B)-T_(E))/(R_(BDE)) = (T_(E)-10)/(R_(EF))`
or `(T_(B)-T_(E))/(1//k_(Y).piL//2A)+(T_(B)-T_(E))/(1//k_(X).piL//2A)+(T_(E)-10)/(1//k_(Y).L//A)`
or `(2(T_(B)-T_(E)))/(pi)+(4(T_(B)-T_(E)))/(pi)=T_(E)-10`
or `(3)/(4 pi) (T_(B)-T_(E))=T_(E)-10`
Solving equation -(4.14) and (4.15), we get
`T_(E)=19.74^(@)C`
and `T_(B)= 60.52^(@)C`