Question 12.60: KNOWN: Large furnace with diffuse, opaque walls (Tf, εf) and......

ANALYSIS: Consider first the furnace wall (A). Since the wall material is diffuse and gray, it follows that

$\varepsilon_{ A }=\varepsilon_{ f }=\alpha_{ A }=0.85$.

The emissive power is

$E _{ A }=\varepsilon_{ A } E _{ b }\left( T _{ f }\right)=\varepsilon_{ A } \sigma T _{ f }=0.85 \times 5.67 \times 10^{-8}  W / m ^2 \cdot K ^4(3000  K )^4=3.904 \times 10^6  W / m ^2$.

Since the furnace is an isothermal enclosure, blackbody conditions exist such that

$G _{ A }= J _{ A }= E _{ b }\left( T _{ f }\right)=\sigma T _{ f }^4=5.67 \times 10^{-8}  W / m ^2 \cdot K ^4(3000  K )^4=4.593 \times 10^6  W / m ^2$.

Considering now the semitransparent, diffuse, spectrally selective object at $\mathrm T_{\mathrm o}$ = 300 K. From the radiation balance requirement, find

$\alpha_\lambda=1-\rho_\lambda-\tau_\lambda \text { or } \alpha_1=1-0.6-0.3=0.1 \text { and } \alpha_2=1-0.7-0.0=0.3$

$\alpha_{ B }=\int_0^{\infty} \alpha_\lambda G _\lambda d \lambda / G = F _{0-\lambda T } \cdot \alpha_1+\left(1- F _{0-\lambda T }\right) \cdot \alpha_2=0.970 \times 0.1+(1-0.970) \times 0.3=0.106$

where $\mathrm F_{0 – \lambda T}$ = 0.970 at λT = 5 μm × 3000 K = 15,000 μm·K since $\mathrm G = \mathrm E_{\mathrm b}(\mathrm T_{\mathrm f})$. Since the object is diffuse, $ε_λ = α_λ$, hence

$\varepsilon_{ B }=\int_0^{\infty} \varepsilon_\lambda E _{\lambda, b }\left( T _{ o }\right) d \lambda / E _{ b , o }= F _{0-\lambda T } \alpha_1+\left(1- F _{0-\lambda T }\right) \cdot \alpha_2=0.0138 \times 0.1+(1-0.0138) \times 0.3=0.297$

where $\mathrm F_{0 – \lambda T}$ = 0.0138 at λT = 5 μm × 300 K = 1500 μmK. The emissive power is

$E _{ B }=\varepsilon_{ B } E _{ b , B }\left( T _{ o }\right)=0.297 \times 5.67 \times 10^{-8}  W / m ^2 \cdot K ^4(300  K )^4=136.5  W / m ^2$.

The irradiation is that due to the large furnace for which blackbody conditions exist,

$G _{ B }= G _{ A }=\sigma T _{ f }^4=4.593 \times 10^6  W / m ^2$.

The radiosity leaving point B is due to emission and reflected irradiation,

$J _{ B }= E _{ B }+\rho_{ B } G _{ B }=136.5  W / m ^2+0.3 \times 4.593 \times 10^6  W / m ^2=1.378 \times 10^6  W / m ^2$.

If we include transmitted irradiation, $\mathrm J_{\mathrm B} = \mathrm E_{\mathrm B} + (ρ_{\mathrm B} + τ_{\mathrm B}) \mathrm G_{\mathrm B} = \mathrm E_{\mathrm B} + (1  –  α_{\mathrm B}) \mathrm G_{\mathrm B} = 4.106 \times 10^6$ W/m².
In the first calculation, note how we set $ρ_{\mathrm B} ≈ ρ_λ (λ < 5  μ\mathrm m)$.