A diffuse, fire brick wall of temperature $T_{s}$ = 500 K has the spectral emissivity shown and is exposed to a bed of coals at 2000 K.

Determine the total, hemispherical emissivity and emissive power of the fire brick wall. What is the total absorptivity of the wall to irradiation resulting from emission by the coals?

Question

A diffuse, fire brick wall of temperature [latex]T_{s}[/latex] = 500 K has the spectral emissivity shown and is exposed to a bed of coals at 2000 K.

Determine the total, hemispherical emissivity and emissive power of the fire brick wall. What is the total absorptivity of the wall to irradiation resulting from emission by the coals?

Accepted Answer

Known: Brick wall of surface temperature [latex]T_{s}[/latex] = 500 K and prescribed [latex]ε_{λ}(λ)[/latex] is exposed to coals at [latex]T_{c}[/latex] = 2000 K.

Find:
1. Total, hemispherical emissivity of the fire brick wall.
2. Total emissive power of the brick wall.
3. Absorptivity of the wall to irradiation from the coals.

Schematic:

Assumptions:
1. Brick wall is opaque and diffuse.
2. Spectral distribution of irradiation at the brick wall approximates that due to emission from a blackbody at 2000 K.

Analysis:
1. 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}(λ)E_{λ,b}(λ, T_{s})} dλ}{E_{b}(T_{s})}[/latex]

Breaking the integral into parts,

[latex]ε(T_{s}) = ε_{λ,1} \frac{\int_{0}^{λ_{1}}E_{λ,b} dλ}{E_{b}} + ε_{λ,2} \frac{\int_{λ_{1}}^{λ_{2}}E_{λ,b} dλ}{E_{b}} + ε_{λ,3} \frac{\int_{λ_{2}}^{∞}E_{λ,b} dλ}{E_{b}}[/latex]

and introducing the blackbody functions, it follows that

[latex]ε(T_{s}) = ε_{λ,1} F_{(0→λ_{1})} + ε_{λ,2}[F_{(0→λ_{2})} - F_{(0→λ_{1})}] + ε_{λ,3}[1 - F_{(0→λ_{2})}][/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_{s} = 1.5 μm × 500 K = 750 μm · K:\qquad F_{(0→λ_{1})} = 0.000[/latex]

[latex]λ_{2}T_{s} = 10 μm × 500 K = 5000 μm · K:\qquad F_{(0→λ_{2})} = 0.634[/latex]

Hence

[latex]ε(T_{s})[/latex] = 0.1 × 0 + 0.5 × 0.634 + 0.8 (1 - 0.634) = 0.610

2. From Equations 12.32 and 12.36, the total emissive power is

[latex]E_{b} = σT^{4}[/latex] (12.32)

[latex]ε(T) \equiv \frac{E(T)}{E_{b}(T)}[/latex] (12.36)

[latex]E(T_{s}) = ε(T_{s})E_{b}(T_{s}) = ε(T_{s})σT^{4}_{s}[/latex]

[latex]E(T_{s}) = 0.61 × 5.67 × 10^{-8} W/m^{2} · K^{4}(500 K)^{4} = 2162 W/m^{2}[/latex]

3. From Equation 12.52, the total absorptivity of the wall to radiation from the coals is

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

Since the surface is diffuse, [latex]α_{λ}(λ) = ε_{λ}(λ)[/latex]. Moreover, since the spectral distribution of the irradiation approximates that due to emission from a blackbody at 2000 K, [latex]G_{λ}(λ) ∝ E_{λ,b}(λ, T_{c})[/latex]. It follows that

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

Breaking the integral into parts and introducing the blackbody functions, we then obtain

[latex]α = ε_{λ,1} F_{(0→λ_{1})} + ε_{λ,2}[F_{(0→λ_{2})} - F_{(0→λ_{1})}] + ε_{λ,3}[1 - F_{(0→λ_{2})}][/latex]

From Table 12.2

[latex]λ_{1}T_{c} = 1.5 μm × 2000 K = 3000 μm · K:\qquad F_{(0→λ_{1})} = 0.273[/latex]

[latex]λ_{2}T_{c} = 10 μm × 2000 K = 20,000 μm · K:\qquad F_{(0→λ_{2})} = 0.986[/latex]

Hence

α = 0.1 × 0.273 + 0.5(0.986 - 0.273) + 0.8(1 - 0.986) = 0.395

Comments:
1. The emissivity depends on the surface temperature [latex]T_{s}[/latex], while the absorptivity depends on the spectral distribution of the irradiation, which depends on the temperature of the source [latex]T_{c}[/latex].
2. The surface is not gray, α ≠ ε. This result is to be expected. Since emission is associated with [latex]T_{s}[/latex] = 500 K, its spectral maximum occurs at [latex]λ_{\max}[/latex] ≈ 6 μm. In contrast, since irradiation is associated with emission from a source at [latex]T_{c}[/latex] = 2000 K, its spectral maximum occurs at [latex]λ_{\max}[/latex] ≈ 1.5 μm. Even though [latex]ε_{λ}[/latex] and [latex]α_{λ}[/latex] are equal, because they are not constant over the spectral ranges of both emission and irradiation, α ≠ ε. For the prescribed spectral distribution of [latex]ε_{λ} = α_{λ}[/latex], ε and α decrease with increasing [latex]T_{s}[/latex] and [latex]T_{c}[/latex], respectively, and it is only for [latex]T_{s} = T_{c}[/latex] that ε = α. The foregoing expressions for ε and α may be used to determine their equivalent variation with [latex]T_{s}[/latex] and [latex]T_{c}[/latex], and the following result is obtained:

Question 12.10: A diffuse, fire brick wall of temperature Ts = 500 K has the...

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