Question 12.5: A surface emits as a blackbody at 1500 K. What is the rate p...

Known: Temperature of a surface that emits as a blackbody.

Find: Rate of emission per unit area over all directions between θ = 0° and 60° and over all wavelengths between λ = 2 and 4 μm.

Schematic:

Assumptions: Surface emits as a blackbody.

Analysis: The desired emission may be inferred from Equation 12.15, with the limits of integration restricted as follows:

E = \int_{0}^{∞}\int_{0}^{2\pi}\int_{0}^{\pi/2}{ I_{λ,e} (λ, θ, \phi)  \cos θ \sin θ  dθ  d\phi  dλ}              (12.15)

ΔE = \int_{2}^{4}\int_{0}^{2\pi}\int_{0}^{\pi/3}{ I_{λ,b}  \cos θ \sin θ  dθ  d\phi  dλ}

or, since a blackbody emits diffusely,

ΔE = \int_{2}^{4}{I_{λ,b}}\left(\int_{0}^{2\pi}\int_{0}^{\pi/3}{\cos θ \sin θ  dθ  d\phi}\right)dλ

ΔE = \int_{2}^{4}{I_{λ,b}}\left(2\pi\frac{\sin^{2}θ}{2}\bigg|^{\pi/3}_{0}\right)dλ = 0.75 \int_{2}^{4}{\pi I_{λ,b}}dλ

Substituting from Equation 12.16 and multiplying and dividing by E_{b}, this result may be put in a form that allows for use of Table 12.2 in evaluating the spectral integration. In particular,

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

ΔE = 0.75E_{b} \int_{2}^{4}{\frac{E_{λ,b}}{E_{b}}}dλ = 0.75E_{b} [F_{(0→4)}  –  F_{(0→2)}]

where from Table 12.2

λ_{1}T = 2  μm × 1500  K = 3000  μm · K:\qquad F_{(0→2)} = 0.273

λ_{2}T = 4  μm × 1500  K = 6000  μm · K:\qquad F_{(0→4)} = 0.738

Hence

ΔE = 0.75(0.738  –  0.273)E_{b} = 0.75(0.465)E_{b}

From Equation 12.31, it then follows that

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

ΔE = 0.75(0.465)5.67 × 10^{-8}  W/m^{2} · K^{4}  (1500  K)^{4} = 10^{5}  W/m^{2}

Comments: The total, hemispherical emissive power is reduced by 25% and 53.5% due to the directional and spectral restrictions, respectively.