Answer:
[tex]\frac{\dot Q}{A} =20.129\ W.m^{-2}[/tex]
[tex]T_1=27.58\ ^{\circ}C[/tex] & [tex]T_2=2.41875\ ^{\circ}C[/tex]
Explanation:
Given:
From the Fourier's law of conduction:
[tex]\dot Q=\frac{dT}{(\frac{x}{kA}) }[/tex]
[tex]\dot Q=\frac{dT}{R_{th} }[/tex] ....................................(1)
Now calculating the equivalent thermal resistance for conductivity using electrical analogy:
[tex]R_{th}=R_p+R_s+R_p[/tex]
[tex]R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}[/tex]
[tex]R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})[/tex]
[tex]R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})[/tex]
[tex]R_{th}=\frac{1.4904}{A}[/tex] .....................(2)
Putting the value from (2) into (1):
[tex]\dot Q=\frac{30-0}{\frac{1.4904}{A} }[/tex]
[tex]\dot Q=\frac{30\ A}{1.4904}[/tex]
[tex]\frac{\dot Q}{A} =20.129\ W.m^{-2}[/tex] is the heat per unit area of the wall.
The heat flux remains constant because the area is constant.
For plywood-Styrofoam interface from inside:
[tex]\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}[/tex]
[tex] 20.129=0.104\times \frac{30-T_1}{0.0125}[/tex]
[tex]T_1=27.58\ ^{\circ}C[/tex]
&For Styrofoam-plywood interface from inside:
[tex]\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}[/tex]
[tex] 20.129=0.04\times \frac{27.58-T_2}{0.05}[/tex]
[tex]T_2=2.41875\ ^{\circ}C[/tex]
