A few rods of materials `X` and `Y` are connected as shown in The cross-sectional areas of all the rods are same if the end A is maintaind at `80^(@)C` and the end `F` is maintained at `10^(2)C` If the temperature of junctions `B` and `E` in steady state are `(39.48^(@)C)/(n_(1)` and `(60.52^(@)C)/(n_(2)` Find `n_(1)` and `n_(2)` Given thermal conductivity of material `X` double that of `Y`
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Answer

2

Explanation

`(n_(1) =1` and `n_(2)` We first find the thermal resistances of the different rods shown in the These are given as `R_(AB) = (1)/(k_(x)).(L)/(A)`
`R_(BCE)=(1)/(k_(Y)).((piL//2))/(A),R_(BDE)=(1)/(k_X).((piL//2))/(A)`
`R_(EF)=(1)/(k_(Y))=(L)/(A)`
Also `k_(x) =2k_(Y)` Now in steady state the amount of heat flow from end `E` to `F` remains constant as there is no absorption of heat Then we must have that the amount of heat having `B` and same statement can be given for juction `E` If temperatures of junctions `B` and `E` arw taken as `T_(B)` and `T_(E)` respectively then we have for junction `beta`
`(T_(A)-T_(B))/(R_(AB))=(T_(B)-T_(E))/(R_(BCE))+(T_(B)-T_(E))/(R_(BDE))` or
`(80-TB)/(1//k_(x).L//A)=(T_(B)-T_(C))/(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))`...(i)
Similarly for juction `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)/(4pi) (T_(B) -T_(E)) = T_(E) - 10` ....(ii)
Solving equation (i) and (ii0 we get
`T_(E) = 19.74^(@)C` and `T_(B) = 60.52^(@)C` .