Answer:
[tex]960*10^{-3}W/m\°c[/tex]
Explanation:
According to the problem we can define the different kind of energy on the wall.
The total conduction heat transfer is given by,
[tex]\dot{q}_{T} = \dot{q}_{conv}+\dot{q}_{rad}+\dot{q}_{solar}[/tex]
Where,
[tex]\dot{q}_{T} =[/tex]Total conduction heat transfer
[tex]\dot{q}_{conv}[/tex] = The convection heat transfer
[tex]\dot{q}_{rad}[/tex]= Radiation heat transfer
[tex]\dot{q}_{solar}[/tex]= Solar heat transfer
Each heat transfer equation hast its respective formula, so replacing in its respectively order,
[tex]k(\frac{T_2-T_1}{L})=h(T_0-T_2)+\epsilon\sigma(T^4_{surr}-T^4_2)+\alpha_s I[/tex]
Now we can compare of these variable with our values, and that is,
[tex]L=0.25m\\T_1=27\°c\\T_2=44\°c\\T_{surr}=40\°c \\h=8W/m^2 \\T_0=40\°c\\I=150W/m^3\\\sigma = 5.67*10^{-8}\\\epsilon=0.8\\\alpha_s = 0.8\\[/tex]
Replacing,[tex]k(\frac{44-27}{0.25}) = 8(40-44)+(0.8)(5.67*10^{-8})(313^4-317^4)+(0.8)(150)\\k=0.9605W/m.\°c[/tex]
Hence the required effective thermal conductivity of the wall is 960*10^{-3}W/m\°c
