Question 12.4: Consider a large isothermal enclosure that is maintained at ...

Known: Large isothermal enclosure at uniform temperature.

Find:
1. Emissive power of a small aperture on the enclosure.
2. Wavelengths below which and above which 10% of the radiation is concentrated.
3. Spectral emissive power and wavelength associated with maximum emission.
4. Irradiation on a small object inside the enclosure.

Schematic:

Assumptions: Areas of aperture and object are very small relative to enclosure surface.

Analysis:
1. Emission from the aperture of any isothermal enclosure will have the characteristics of blackbody radiation. Hence, from Equation 12.32,

$E_{b} = σT^{4}$              (12.32)

$E =E_{b}(T) = σT^{4} = 5.670 × 10^{-8}  W/m^{2} · K^{4}(2000  K)^{4}\\ E = 9.07 × 10^{5}  W/m^{2}$

2. The wavelength $λ_{1}$ corresponds to the upper limit of the spectral band $(0 → λ_{1})$ containing 10% of the emitted radiation. With $F_{(0→λ_{1})} = 0.10$ it follows from Table 12.2 that $λ_{1}T = 2195  μm · K$. Hence

$λ_{1} = 1.1  μm$

The wavelength $λ_{2}$ corresponds to the lower limit of the spectral band $(λ_{2} → ∞)$ containing 10% of the emitted radiation. With

$F_{(λ_{2}→∞)} = 1  –  F_{(0→λ_{2})} = 0.1\\ F_{(0→λ_{2})} = 0.9$

it follows from Table 12.2 that $λ_{2}T = 9382  μm · K$. Hence

$λ_{2} = 4.69  μm$

3. From Wien’s displacement law, Equation 12.31, $λ_{\max}T = 2898  μm · K$. Hence

$λ_{\max}T = C_{3}$              (12.31)

$λ_{\max} = 1.45  μm$

The spectral emissive power associated with this wavelength may be computed from Equation 12.30 or from the third column of Table 12.2. For $λ_{\max}T = 2898  μm · K$ it follows from Table 12.2 that

$E_{λ,b}(λ, T) = \pi I_{λ,b}(λ, T) = \frac{C_{1}}{λ^{5}[\exp (C_{2}/λT)  –  1]}$              (12.30)

$I_{λ,b}(1.45  μm, T) = 0.722 × 10^{-4} σT^{5}$

Hence

$I_{λ,b}(1.45  μm, 2000  K) = 0.722 × 10^{-4}  (μm · K · sr)^{-1} × 5.67 × 10^{-8}  W/m^{2} · K^{4}  (2000  K)^{5}\\ I_{λ,b}(1.45  μm, 2000  K) = 1.31 × 10^{5}  W/m^{2} · sr · μm$

Since the emission is diffuse, it follows from Equation 12.16 that

$E_{λ}(λ) = \pi I_{λ,e}(λ)$              (12.16)

$E_{λ,b} = \pi I_{λ,b} = 4.12 × 10^{5}  W/m^{2} · μm$

4. Irradiation of any small object inside the enclosure may be approximated as being equal to emission from a blackbody at the enclosure surface temperature. Hence $G = E_{b}(T)$, in which case

$G = 9.07 × 10^{5}  W/m^{2}$