Purdue ME 606: Beyond the Gray Approximation and Radiative Heat Flux (Lecture 10)

Beyond the Gray Approximation

Spectral enclosure energy balance:

$\frac{{q''_{\pi i} }}{{\varepsilon _i }} - \sum\limits_{j = 1}^i {\left[ {\frac{1}{{\varepsilon _{\lambda i} }} - 1} \right]} \hat q_{\lambda j} F_{ij} = E_{b\lambda i} - \sum\limits_{j = 1}^N {E_{b\lambda j} F_{ij} }$

Generally, divide spectrum into M bands $(Δλ s)$

$\to$ N surfaces , M bands $\to$ $M \cdot N$ equations

Integral coupling equation

${q''}_i = \int_0^\infty {{q''}_{\lambda i} } d\lambda = \sum\limits_{m = 1}^M {{q''}_{\lambda i}^{(m)} } \Delta \lambda _m$

$E_{bi} = \int_0^\infty {E_{b\lambda i} } d\lambda = \sum\limits_{m = 1}^M {E_{b\lambda i}^{(m)} } \Delta \lambda _m$

Simplest Approximation: Semi Gray

Two wavelength bands—most useful when source and wall temperature differ greatly.

M=2; # of equation =2N;

Unknowns: 4N

$q'' a n d T$ for each band on each surface

Going further, we could have more refinement of the spectrum

Non-gray directional radiation exchange

Need to use $I(\lambda ,\theta ,\bar r)$

$I_\lambda (\theta ,\bar r) = \varepsilon _\lambda (\theta )I_{b\lambda } + \int {\rho _{bd\lambda } (} \theta _i ,\theta ,\bar r)I_\lambda (\theta _i )\cos \theta _i d\omega _i$

Now, we must discreteize both $\bar r$ and $θ$

Let’s each has N increments

$\to M N 2$ equations (M is # of spectral segments)

Computer Graphics

Typically:

N=5,000 – 50,000 +

M=3 – 8 (ex. RGB, CMYK)

$ρ b d$ treated fairly simply

Often supplemented by a specular reflectivity

$ρ s = ρ s (θ i)$

Radiative Heat Flux

$\bar q_\eta ^{''} \cdot {\hat n} = \int_{4\pi } {I_\eta ({\hat s}} ){\hat s} \cdot \hat nd\omega$

$\bar q'' = \int_0^\infty {\bar q_\eta ^{''} } d\eta = \int_0^\infty {\int_{4\pi } {I_\eta ({\hat s}} )\hat sd\omega d\eta }$

Divergence:

$\nabla \cdot \bar q_\eta ^{''} = \int_{4\pi } {\frac{{dI_\eta }}{{ds}}d\omega }$

General Equation of Radiative Transfer

Derivation: $I η =$ net rate of radiant transfer

In direction of S per area per solid angle $\left[ {\frac{W}{{m^2 Sr}}} \right]$

$\eta = \frac{{2\pi }}{\lambda }$ is the wave number

Absorption/ Scattering

$- d I η = k η I η d s + σ η I η d s = β η I η d s$

$k η$ : linear, monochromatic absorption coefficient $\left[ {m^{ - 1} } \right]$

$σ η$ : linear, monochromatic scattering coefficient $\left[ {m^{ - 1} } \right]$

Outscattering

Total attenuation

$β η \equiv k η + σ η$ $\left[ {m^{ - 1} } \right]$ Extinction coefficient

For all incident beams in a volume:

$\int_{4\pi } {k_\eta I_\eta d\omega _i } + \int_{4\pi } {\sigma _\eta I_\eta d\omega _i } = \int {\beta _\eta I_\eta } d\omega _i$

If $I η$ is isotropic then:

$\int_{4\pi } {k_\eta d\omega _i } + \int_{4\pi } {\sigma _\eta d\omega _i } = \int {\beta _\eta } d\omega _i$

In general, we do not expect

$β η, k η,σ η$ to depend on direction

Last modified on 04 Nov, 2008

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