Question 12.6: A diffuse surface at 1600 K has the spectral, hemispherical ...

Known: Spectral, hemispherical emissivity of a diffuse surface at 1600 K.

Find:
1. Total, hemispherical emissivity.
2. Total emissive power.
3. Wavelength at which spectral emissive power will be a maximum.

Assumptions: Surface is a diffuse emitter.

Analysis:
1. The total, hemispherical emissivity is given by Equation 12.43, where the integration may be performed in parts as follows:

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

ε = \frac{\int_{0}^{∞}{ε_{λ}E_{λ,b}}dλ}{E_{b}} = \frac{ε_{1}\int_{0}^{2} {E_{λ,b}}dλ}{E_{b}} + \frac{ε_{2}\int_{2}^{5} {E_{λ,b}}dλ}{E_{b}}

or

ε = ε_{1}F_{(0→2  μm)} + ε_{2}[F_{(0→5  μm)}  –  F_{(0→2  μm)}]

From Table 12.2 we obtain

λ_{1}T = 2  μm × 1600  K = 3200  μm · K:\qquad F_{(0→2  μm)} = 0.318

λ_{2}T = 5  μm × 1600  K = 8000  μm · K:\qquad F_{(0→5  μm)} = 0.856

Hence

ε = 0.4 × 0.318 + 0.8[0.856 – 0.318] = 0.558

2. From Equation 12.36 the total emissive power is

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

E = εE_{b} = εσT^{4}\\ E = 0.558(5.67 × 10^{-8}  W/m^{2} · K^{4})(1600  K)^{4} = 207  kW/m^{2}

3. If the surface emitted as a blackbody or if its emissivity were a constant, independent of λ, the wavelength corresponding to maximum spectral emission could be obtained from Wien’s displacement law. However, because ε_{λ} varies with λ, it is not immediately obvious where peak emission occurs. From Equation 12.31 we know that

λ_{\max}T = C_{3}              (12.31)

λ_{\max} = \frac{2898  μm · K}{1600  K} = 1.81  μm

The spectral emissive power at this wavelength may be obtained by using Equation 12.40 with Table 12.2. That is,

ε_{λ}(λ, T) \equiv \frac{E_{λ}(λ, T)}{E_{λ,b}(λ, T)}              (12.40)

E_{λ}(λ_{\max}, T) = ε_{λ}(λ_{\max})E_{λ,b}(λ_{\max}, T)

or, since the surface is a diffuse emitter,

E_{λ}(λ_{\max}, T) = \pi ε_{λ}(λ_{\max})I_{λ,b}(λ_{\max}, T)\\ = \pi ε_{λ}(λ_{\max})\frac{I_{λ,b}(λ_{\max}, T)}{σT^{5}} × σT^{5}

E_{λ}(1.81  μm, 1600  K) = \pi × 0.4 × 0.722 × 10^{-4}  (μm · K · sr)^{-1} × 5.67 × 10^{-8}  W/m^{2} · K^{4} × (1600  K)^{5} = 54  kW/m^{2} · μm

Since ε_{λ} = 0.4 from λ = 0 to λ = 2 μm, the foregoing result provides the maximum spectral emissive power for the region λ < 2 μm. However, with the change in ε_{λ} that occurs at λ = 2 μm, the value of E_{λ} at λ = 2 μm may be larger than that for λ = 1.81 μm. To determine whether this is, in fact, the case, we compute

E_{λ}(λ_{1}, T) = \pi ε_{λ}(λ_{1})\frac{I_{λ,b}(λ_{1}, T)}{σT^{5}} × σT^{5}

where, for λ_{1}T = 3200  μm · K, [I_{λ,b}(λ_{1}, T)/σT^{5}] = 0.706 × 10^{-4}  (μm · K · sr)^{-1}. Hence

E_{λ}(2  μm, 1600  K) = \pi × 0.80 × 0.706 × 10^{-4}  (μm · K · sr)^{-1} × 5.67 × 10^{-8}  W/m^{2} · K^{4}  (1600  K)^{5}

E_{λ}(2  μm, 1600  K) = 105.5  kW/m^{2} · μm > E_{λ}(1.81  μm, 1600  K)

and peak emission occurs at

λ = λ_{1} = 2  μm

Comments: For the prescribed spectral distribution of ε_{λ}, the spectral emissive power will vary with wavelength as shown.