A small, solid metallic sphere has an opaque, diffuse coating for which αλ = 0.8 for λ ≤ 5 μm and αλ = 0.1 for λ 5 μm. The sphere, which is initially at a uniform temperature of 300 K, is inserted into a large furnace whose walls are at 1200 K. Determine the total, hemispherical absorptivity and

Question

A small, solid metallic sphere has an opaque, diffuse coating for which [latex]α_{λ}[/latex] = 0.8 for λ ≤ 5 μm and [latex]α_{λ}[/latex] = 0.1 for λ > 5 μm. The sphere, which is initially at a uniform temperature of 300 K, is inserted into a large furnace whose walls are at 1200 K. Determine the total, hemispherical absorptivity and emissivity of the coating for the initial condition and for the final, steady-state condition.

Accepted Answer

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

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

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

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

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

Hence

[latex]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)}[/latex]

or

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

From Table 12.2,

TABLE 12.2 Blackbody Radiation Functions
[latex]\pmb{\frac{I_{λ, b}(λ, T)}{I_{λ,b}(λ_{\max}, T)}}[/latex]	[latex]\pmb{I_{λ, b}(λ, T)/σT^{5} (μm · K · sr)^{-1}}[/latex]	[latex]\pmb{F_{(0 → λ)}}[/latex]	λT (μm · K)
0.000000	[latex]0.375034 × 10^{-27}[/latex]	0.000000	200
0.000000	[latex]0.490335 × 10^{-13}[/latex]	0.000000	400
0.000014	[latex]0.104046 × 10^{-8}[/latex]	0.000000	600
0.001372	[latex]0.991126 × 10^{-7}[/latex]	0.000016	800
0.016406	[latex]0.118505 × 10^{-5}[/latex]	0.000321	1,000
0.072534	[latex]0.523927 × 10^{-5}[/latex]	0.002134	1,200
0.186082	[latex]0.134411 × 10^{-4}[/latex]	0.007790	1,400
0.344904	0.249130	0.019718	1,600
0.519949	0.375568	0.039341	1,800
0.683123	0.493432	0.066728	2,000
0.816329	[latex]0.589649 × 10^{-4}[/latex]	0.100888	2,200
0.912155	0.658866	0.140256	2,400
0.970891	0.701292	0.183120	2,600
0.997123	0.720239	0.227897	2,800
1.000000	[latex]0.722318 × 10^{-4}[/latex]	0.250108	2,898
0.997143	[latex]0.720254 × 10^{-4}[/latex]	0.273232	3,000
0.977373	0.705974	0.318102	3,200
0.943551	0.681544	0.361735	3,400
0.900429	0.650396	0.403607	3,600
0.851737	[latex]0.615225 × 10^{-4}[/latex]	0.443382	3,800
0.800291	0.578064	0.480877	4,000
0.748139	0.540394	0.516014	4,200
0.696720	0.503253	0.548796	4,400
0.647004	0.467343	0.579280	4,600
0.599610	0.433109	0.607559	4,800
0.554898	0.400813	0.633747	5,000
0.513043	[latex]0.370580 × 10^{-4}[/latex]	0.658970	5,200
0.474092	0.342445	0.680360	5,400
0.438002	0.316376	0.701046	5,600
0.404671	0.292301	0.720158	5,800
0.373965	0.270121	0.737818	6,000
0.345724	[latex]0.249723 × 10^{-4}[/latex]	0.754140	6,200
0.319783	0.230985	0.769234	6,400
0.295973	0.213786	0.783199	6,600
0.274128	0.198008	0.796129	6,800
0.254090	0.183534	0.808109	7,000
0.235708	[latex]0.170256 × 10^{-4}[/latex]	0.819217	7,200
0.218842	0.158073	0.829527	7,400
0.203360	0.146891	0.839102	7,600
0.189143	0.136621	0.848005	7,800
0.176079	0.127185	0.856288	8,000
0.147819	[latex]0.106772 × 10^{-4}[/latex]	0.874608	8,500
0.124801	[latex]0.901463 × 10^{-5}[/latex]	0.890029	9,000
0.105956	0.765338	0.903085	9,500
0.090442	[latex]0.653279 × 10^{-5}[/latex]	0.914199	10,000
0.077600	0.560522	0.923710	10,500
0.066913	0.483321	0.931890	11,000
0.057970	0.418725	0.939959	11,500
0.050448	[latex]0.364394 × 10^{-5}[/latex]	0.945098	12,000
0.038689	0.279457	0.955139	13,000
0.030131	0.217641	0.962898	14,000
0.023794	[latex]0.171866 × 10^{-5}[/latex]	0.969981	15,000
0.019026	0.137429	0.973814	16,000
0.012574	[latex]0.908240 × 10^{-6}[/latex]	0.980860	18,000
0.008629	0.623310	0.985602	20,000
0.003828	0.276474	0.992215	25,000
0.001945	[latex]0.140469 × 10^{-6}[/latex]	0.995340	30,000
0.000656	[latex]0.473891 × 10^{-7}[/latex]	0.997967	40,000
0.000279	0.201605	0.998953	50,000
0.000058	[latex]0.418597 × 10^{-8}[/latex]	0.999713	75,000
0.000019	0.135752	0.999905	100,000

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

Hence

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

The total, hemispherical emissivity follows from Equation 12.43.

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

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

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

[latex]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)}[/latex]

or,

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

From Table 12.2,

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

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 [latex]T_{s}[/latex] increases with time, the value of ε will change. After a sufficiently long time, [latex]T_{s} = T_{f}[/latex], and ε = α (ε = 0.62).

Comments:
1. The equilibrium condition that eventually exists [latex](T_{s} = T_{f})[/latex] 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

[latex]\dot{E}_{in} - \dot{E}_{out} = \dot{E}_{st}[/latex]

[latex](αG)A_{s} - (εσT_{s}^{4})A_{s} = Mc_{p}\frac{dT_{s}}{dt}[/latex]

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.

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

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