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

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

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}(λ)E_{λ,b}(λ, T_{s})}  dλ}{E_{b}(T_{s})}

Breaking the integral into parts,

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}}

and introducing the blackbody functions, it follows that

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})}]

From Table 12.2

λ_{1}T_{s} = 1.5  μm × 500  K = 750  μm · K:\qquad F_{(0→λ_{1})} = 0.000

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

Hence

ε(T_{s}) = 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

E_{b} = σT^{4}              (12.32)

ε(T) \equiv \frac{E(T)}{E_{b}(T)}              (12.36)

E(T_{s}) = ε(T_{s})E_{b}(T_{s}) = ε(T_{s})σT^{4}_{s}

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

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

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

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

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

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

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})}]

From Table 12.2

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

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

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