Question 12.3: Determine an expression for the net radiative heat flux at t...

Known: Surface temperature of a small blackbody, T_{s}, and the surroundings temperature, T_{sur}.

Find: Expression for the net radiative flux at the surface of the small object, q''_{rad}.

Assumptions: Small object experiences blackbody irradiation.

Schematic:

Analysis: Since none of the irradiation is reflected from the small object, Equation 12.28 may be written as

q''_{rad} = \int_{0}^{∞}\int_{0}^{2\pi}\int_{0}^{\pi/2}I_{λ,e+r}(λ, θ, \phi) \cos  θ  \sin  θ  dθ  d\phi  dλ  –  \int_{0}^{∞}\int_{0}^{2\pi}\int_{0}^{\pi/2}I_{λ,i}(λ, θ, \phi) \cos  θ  \sin  θ  dθ  d\phi  dλ              (12.28)

q''_{rad} = \int_{0}^{∞}\int_{0}^{2\pi}\int_{0}^{\pi/2}I_{λ,e}(λ, θ, \phi) \cos  θ  \sin  θ  dθ  d\phi  dλ  –  \int_{0}^{∞}\int_{0}^{2\pi}\int_{0}^{\pi/2}I_{λ,i}(λ, θ, \phi) \cos  θ  \sin  θ  dθ  d\phi  dλ              (1)

The intensity emitted by the small object corresponds to that of a blackbody. Hence

I_{λ,e}(λ, θ, \phi) = I_{λ,b}(λ, T_{s})              (2)

The intensity corresponding to the irradiation is also black. Therefore

I_{λ,i}(λ, θ, \phi) = I_{λ,b}(λ, T_{sur})              (3)

Since the blackbody intensity is diffuse, it is independent of angles θ and \phi. Therefore, substituting Equations 2 and 3 into Equation 1 yields

q''_{rad} = \int_{0}^{2\pi}\int_{0}^{\pi/2} \cos  θ  \sin  θ  dθ  d\phi × \int_{0}^{∞}I_{λ,b}(λ, T_{sur})  dλ  –  \int_{0}^{2\pi}\int_{0}^{\pi/2} \cos  θ  \sin  θ  dθ  d\phi × \int_{0}^{∞}I_{λ,b}(λ, T_{s})  dλ\\ = \pi \left[\int_{0}^{∞}I_{λ,b}(λ, T_{sur})  dλ  –  \int_{0}^{∞}I_{λ,b}(λ, T_{s})  dλ]\right]

Substituting from Equations 12.32 and 12.33 yields

E_{b} = σT^{4}              (12.32)

I_{b} = \frac{E_{b}}{\pi}              (12.33)

q''_{rad} = σ(T^{4}_{s}  –  T^{4}_{sur})

which is identical to Equation 1.7 with ε = 1.

q''_{rad} = \frac{q}{A} = \varepsilon E_{b}(T_{s})  –  \alpha G =\varepsilon \sigma (T_{s}^{4}  –  T_{sur}^{4})             (1.7)