A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coefficients of 25 and 1000 W/m2 ⋅ K. The wall is composed of a 12-mm-thick layer of beryllium oxide on the gas side and a 24-mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is 0.05 m2 ⋅ K/W. What is the rate of heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

[tex]t_1[/tex] and [tex]t_2[/tex] are thicknesses of gas layer and steel slab in meters,
[tex]h_1[/tex], [tex]h_2[/tex] are convection coefficients gas and liquid in [tex]W/m^2 \cdot K[/tex],
[tex]R_c[/tex] is the contact resistance in [tex]m^2 \cdot K/W[/tex],

and [tex]k_1, k_2[/tex] are thermal conductivities of gas and steel in [tex]W/m \cdot K[/tex],

then: part(a):

[tex]\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}[/tex]

using known values:

[tex]\text {heat loss} = 2486.05 W/m^2[/tex]

part(b): Using the rate equation :

[tex]\text {heat loss} = h_1 (T_1 - T_{s1})[/tex]

the surface temperature [tex]T_{s1} = 1678.438 \ K[/tex]

and [tex]T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K[/tex]

Similarly

[tex]T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K[/tex]

[tex]T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K[/tex]

The temperature distribution is shown in the attached image