Question

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.

Answer 1

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)` .