Two parallel plates `A` and `B` are joined together to form a compound plate . The thicknesses of the plate are `4.0cm` and `2.5cm` respectively and the area of cross section is `100cm^(2)` for each plate. The thermal conductivities are `K_(A)=200Wm^(-1).^(@)C^(-1)` for the plate `A` and `K_(B)=400Wm^(-1).^(@)C^(-1)` for the plate `B`. The outer surface of the plate `A` is maintained at `100^(@)C`and the outer surface of the plate A is maintained at `100^(@)C` . and the outer surface of the plate B is maintained at `0^(@)C`. Find
(a) the rate of heat flow through any cross section,
(b) the temperature at the interface and
(c) the equivalent thermal conductivity of the compound plate.

Explanation

`(a)` Let the temperature of the interface be `T`.
The area of cross-section of each pate is `A=100cm^(2)=0.01m^(2)`. The thickness are `x_(A)=0.04m` and `x_(B)=0.025m`
The thermal resistance of the plate `A` is
`R_(A)=(x_(A))/(K_(A)A)`
and that of the plate `B` is
`R_(B)=(x_(B))/(K_(B)A)`
The equivalent thermal resistance is
`R_(eq)=R_(A)+R_(B)=(1)/(A)((x_(A))/(K_(A))+(x_(B))/(K_(B)))`.........`(i)`
Thus, `(DeltaQ)/(Deltat)=(T_(1)-T_(2))/(R_(eq))=(A(T_(1)-T_(2)))/(x_(A)//K_(A)+x_(B)//K_(B))`
`=((0.01m^(2))(100^(@)C))/((0.04m)//(200W//m-^(@)C)+(0.025m)//(400W//m-^(@)C))=3810W`.
`(b)` We have `(DeltaQ)/(Deltat)=(A(T-T_(2)))/(x_(B)//K_(B))`
or, `3810W=((0.01m^(2))(T-0^(@)C))/((0.025m)//(400W//m-^(@)C))`
or , `T=25^(@)C`
`(c )` If `K` is the equivalent thermal conductivity of the compound plate, its thermal resistance is
`R_(eq)=(1)/(A)(x_(A)+x_(B))/(K_(eq))`
Comparing with `(i)`,
`(x_(A)+x_(B))/(K_(eq))=(x_(A))/(K_(A))+(x_(B))/(K_(B))`
or, `K_(eq)=(x_(A)+x_(B))/(x_(A)//K_(A)+x_(B)//K_(B))=248W//m-^(@)C`