Answer:
The geometry is treated as a two surface enclosure because the two surfaces have the same properties.
Let's take the base surface to be surface 1, while the side surfaces are surface 2.
Let's take the heat transfer expression:
[tex] Q_1_2 = \frac{\sigma[(T_1)^4 - (T_2)^4]}{\frac{1 - E_1}{A_1 E_1} + \frac{1}{A_1 F_1_2} + \frac{1-E_2}{A_2 E_2}} [/tex]
Where,
[tex] \sigma = Boltmanz constant = 5.68*10^-^8[/tex]
[tex] T_1 [/tex] = base temperature
[tex] T_2 [/tex] = surface 2 temperature = 500K
[tex] E_1 [/tex] = emissivity of surface 1 = 0.8
[tex] E_2 [/tex] = emissivity of surface 2 = 0.5
[tex] A [/tex] = Area
[tex] F_1_2 [/tex] = shape factor
Substituting figures in the equation, we have:
[tex] 800= \frac{5.67*10^-^8[(T_1)^4 - (500)^4]}{\frac{1 - 0.8}{2*0.8} + \frac{1}{2*1} + \frac{1-0.5}{4*0.5}}[/tex]
[tex] [(T_1)^4 - (500)^4] = \frac{700}{5.67*10^-^8} [/tex]
[tex] (T_1)^4 = 1.234*10^1^0 + 6.25*10^1^0[/tex]
[tex] T_1 = (7.484*10^1^0)^0^.^2^5 [/tex]
[tex] T_1 = 523.038 K [/tex]
The base temperature is 523.038 k
