Answer:
q=2313.04[tex]W/m^2[/tex]
T=690.86°C
Explanation:
Given that
Thickness t= 20 cm
Thermal conductivity of firebrick= 1.6 W/m.K
Thermal conductivity of structural brick= 0.7 W/m.K
Inner temperature of firebrick=980°C
Outer temperature of structural brick =30°C
We know that thermal resistance
[tex]R=\dfrac{t}{KA}[/tex]
These are connect in series
[tex]R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}[/tex]
[tex]R=\dfrac{0.2}{1.6A}+\dfrac{0.2}{0.7A}\ K/W[/tex]
[tex]R=\dfrac{23}{56A}\ K/W[/tex]
Heat transfer
[tex]Q=\dfrac{\Delta T}{R}[/tex]
[tex]Q=56A\times \dfrac{980-30}{23}\ W[/tex]
So heat flux
q=2313.04[tex]W/m^2[/tex]
Lets temperature between interface is T
Now by equating heat in both bricks
[tex]\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}[/tex]
So T=690.86°C