Question 12.11: A small, solid metallic sphere has an opaque, diffuse coatin...

Known: Small metallic sphere with spectrally selective absorptivity, initially at T_{s} = 300 K, is inserted into a large furnace at T_{f} = 1200 K.

Find:
1. Total, hemispherical absorptivity and emissivity of sphere coating for the initial condition.
2. Values of α and ε after sphere has been in furnace a long time.

Schematic:

Assumptions:
1. Coating is opaque and diffuse.
2. Since furnace surface is much larger than that of sphere, irradiation approximates emission from a blackbody at T_{f}.

Analysis:
1. From Equation 12.52 the total, hemispherical absorptivity is

α = \frac{\int_{0}^{∞}{α_{λ}(λ)G_{λ}(λ)}  dλ}{\int_{0}^{∞}{G_{λ}(λ)}  dλ}              (12.52)

or, with G_{λ} = E_{λ,b}(T_{f}) = E_{λ,b}(λ, 1200  K),

α = \frac{\int_{0}^{∞}{α_{λ}(λ)E_{λ,b}(λ, 1200  K)}  dλ}{E_{b}(1200  K)}

Hence

α = α_{λ,1}\frac{\int_{0}^{λ_{1}}{E_{λ,b}(λ, 1200  K)}  dλ}{E_{b}(1200  K)} + α_{λ,2}\frac{\int_{λ_{1}}^{∞}{E_{λ,b}(λ, 1200  K)}  dλ}{E_{b}(1200  K)}

or

α = α_{λ,1}F_{(0→λ_{1})} + α_{λ,2}[1  –  F_{(0→λ_{1})}]

From Table 12.2,

λ_{1}T_{f} = 5  μm × 1200  K = 6000  μm · K:\qquad F_{(0→λ_{1})} = 0.738

Hence

α = 0.8 × 0.738 + 0.1 (1  –  0.738) = 0.62

The total, hemispherical emissivity follows from Equation 12.43.

ε(T) = \frac{\int_{0}^{∞}{ε_{λ}(λ, T)E_{λ,b}(λ, T)} dλ}{E_{b}(T)}              (12.43)

ε(T_{s}) = \frac{\int_{0}^{∞}{ε_{λ}E_{λ,b}(λ, T_{s})}  dλ}{E_{b}(T_{s})}

Since the surface is diffuse, ε_{λ} = α_{λ} and it follows that

ε = α_{λ,1}\frac{\int_{0}^{λ_{1}}{E_{λ,b}(λ, 300  K)}  dλ}{E_{b}(300  K)} + α_{λ,2}\frac{\int_{λ_{1}}^{∞}{E_{λ,b}(λ, 300  K)}  dλ}{E_{b}(300  K)}

or,

ε = α_{λ,1}  F_{(0→λ_{1})} + α_{λ,2}[1  –  F_{(0→λ_{1})}]

From Table 12.2,

λ_{1}T_{s} = 5  μm × 300  K = 1500  μm · K:\qquad F_{(0→λ_{1})} = 0.014

Hence

ε = 0.8 × 0.014 + 0.1(1  –  0.014) = 0.11

2. Because the spectral characteristics of the coating and the furnace temperature remain fixed, there is no change in the value of α with increasing time. However, as T_{s} increases with time, the value of ε will change. After a sufficiently long time, T_{s} = T_{f}, and ε = α (ε = 0.62).

Comments:
1. The equilibrium condition that eventually exists (T_{s} = T_{f}) corresponds precisely to the condition for which Kirchhoff’s law was derived. Hence α must equal ε.
2. Approximating the sphere as a lumped capacitance and neglecting convection, an energy balance for a control volume about the sphere yields

\dot{E}_{in}  –  \dot{E}_{out} = \dot{E}_{st}

(αG)A_{s}  –  (εσT_{s}^{4})A_{s} = Mc_{p}\frac{dT_{s}}{dt}

The differential equation could be solved to determine T(t) for t > 0, and the variation in ε that occurs with increasing time would have to be included in the solution.