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 palte B. The outer surface of the plate A is maintainned 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

Let the temperature of the interface be theta. The area of cross section of each plate is `A=100cm^(2)=0.01m^(2)` .The thicknesses are x_(A)=0.004m` and `x_(B)=0.025m.
The thermal resistance of the plate A is `R_(1)=1/K_(A)x_(A)/(A)` .and that of the plate B is `R_(2)=1/K_(B)x_(B)/(A)` . The equivalent thermal resistance is
`R=`R_(1)+R_(2)=(x_(A)/K_(A)+x_(B)/(K_(B)))` . thus, `(DeltaQ)/(Deltat)=(theta_(1)-theta_(2))/(R)` . `=(Atheta_(1)-theta_(2))/(x_(A)//K_(A)+x_(B)//K_(B))` . `((0.01m^(2))((100^(@)C)))/((0.04m)//(200Wm^(-1)`^(@)C^(-1))+(0.025m)//(400Wm^(-1)`^(@)C^(-1)))` . `=3810W` . (b) We have `(DeltaQ)/(Deltat)=(A(theta-theta_(2)))/(x_(B)//K_(B))` . or, `3810W=((0.01m^(2))(theta-0^(@)C))/((0.025m)(200Wm^(-1)`^(@)C^(-1))+(0.025m)//(400Wm^(-1)`^(@)C^(-1)))` . or, `theta=24^(@)C` . (c) If K is the equivalent thermal conductivity of the compound plate, its thermal resistance is
`R=(1)/(A)(x_(A)+x_(B))/(K)` . Comparing with (i) `(x_(A)+x_(A))/(K)=(x_(A))/(K_(A))+x_(B)/(K_(B))` . or, `K=(x_(A)+x_(B))/(x_(A)//K_(A)+x_(B)//K_(B))` . `=248Wm^(-1)`^(@)C^(-1)` .