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Purdue ME 606: Radiative Heat Flux (Lecture 12)

Radiative Heat Flux

Consider a ray in direction \hat s through a plane within a participating medium with normal direction \hat n

\vec q''_\eta \cdot \hat n = \int_{4\pi}^{} I_\eta ({\hat s}) \left( {\hat n \cdot \hat s} \right) d\omega

Remember that \hat n \cdot \hat s = \cos \theta

Of typically more interest is the divergence of the heat flux vector:

\nabla \cdot \vec q''_\eta = \int_{4\pi} \frac{{dI_\eta}}{{ds}} d\omega = 4 \pi \kappa_\eta I_{b\eta} - \beta_\eta \int_{4\pi} I_\eta d\omega + \sigma_\eta \int_{4\pi} I_\eta d\omega_i = \kappa_\eta \left( {4 \pi I_{b\eta} - \int_{4\pi} I_\eta d\omega} \right) = \kappa_\eta \left( {4 \pi I_{b\eta} - G_\eta} \right)

where the incident radiation, G_\eta = \int_{4\pi}^{} {I_\eta d\omega}

Note: Gη = cuη G = cu, where u = \frac{{energy}}{{volume}} of radiation field.

FIGURE: FLUID ELEMENT IN SURROUNDINGS

Combined with the overall fluid energy equation, we have:

\rho c_p \frac{{DT}}{{Dt}} = \nabla \cdot \left({k \nabla T} \right) + \beta_t T \frac{{Dp}}{{Dt}} + \mu \Phi_t - \nabla \cdot \vec q''_{rad}

where:

\frac{{D}}{{Dt}} = \frac{{\partial}}{{\partial t}} + \vec v \cdot \nabla

\nabla \cdot \left({k \nabla T} \right) = - \nabla \cdot \vec q''_{rad}

\beta_t T \frac{{Dp}}{{Dt}} = (compressible) flow work

βt = thermal expansion coefficient

μΦt = viscous dissipation

1-D Transport

FIGURE: 1-D BOUNDED DOMAIN

Note: We assume that all quantities are spectral; therefore subscripts λ are implied in all terms.

The optical lengths in directions normal to the interface and along the wave are:

\tau_z = \int_0^z \beta dz
\tau_s = \int_0^s \beta ds

Multiply RTE by e^{\tau_s}:

\frac{{d \left( {I e^{\tau_s}} \right)}}{{d \tau_s}} = \left ( {source} \right) e^{\tau_s}

I \left( {\tau_s,\hat s} \right) = I_{wall} \left( \hat s \right) e^{-\tau_s} + \int_0^{\tau_s} S \left( {\tau'_s , \hat s} \right) e^{- \left( {\tau_s - \tau'_s} \right) } d\tau'_s

where S \left( {\tau'_s , \hat s} \right) = \left( {1-\omega} \right) I_b \left( \tau'_s \right) + \frac{{\omega}}{{4 \pi}} \int_{4\pi} I \left( {\tau'_s,\hat s} \right) \Phi \left( {\hat s, \hat s_i} \right) d\omega_i,

where \omega = \frac{{\sigma}}{{\kappa + \sigma}} = \frac{{\sigma}}{{\beta}} and \left({1-\omega} \right) = \frac{{\kappa}}{{\beta}}

Note: We normally have a similar equation for radiation originating from the top wall.

Observations

  • When τL > > 1, the medium is called optically thick, i.e., the photon mean free path is short relative to L.
  • When τL < < 1, the medium is called optically thin, i.e., the photon mean free path is large relative to L.
  • When τL = 0, the medium is non-participating.
  • If scattering is isotropic, Φ = 1.
  • For anisotropic scattering,

\Phi \left( {\hat s,\hat s_i } \right) = 1 + \sum\limits_{m = 1}^M {A_m P_m } \left( {\hat s,\hat s_i } \right)

  • If scattering is isotropic,

G \left( \tau'_s \right) = \int_{4\pi} I\left({\tau'_s , \hat s_i } \right) d\omega_i

  • Energy flux integrated along a direction in isotropically scattering medium,

\nabla \cdot \vec q''_{rad} = \kappa \left( {4 \pi I_b - G} \right) and \vec q''_{rad} = \int_{4\pi} I \hat s d\omega

From Modest’s text, Equation (9.53):

\frac{{dq''_{rad}}}{{d\tau_z}} = \frac{{\kappa}}{{\kappa + \sigma}} \left({4 \pi I_b - G}\right)

Summary

We have two simultaneous equations, both integro-differential:

  • 1 transport equation for I,
  • 1 energy balance,

which involve I_{b \eta} \left( \tau_z \right), G_\eta \left( \tau_\eta \right), q''_{rad,\eta} \left( \tau_z \right).

We often simplify the problem by assuming either:

  • T = constantIbη is known
  • \frac{{dq''_{rad}}}{{d\tau_z}} = 0 \to radiative equilibrium
Note: Many exponential integrals occur.

E_n \left( x \right) = \int_0^1 \mu^{n-z} e^{ - x / \mu} du

These integrals are tabulated. Hint: turn cosines into μ‘s.

Also, boundary conditions can be formulated as:

  • heat balances at the walls
  • specified Iω or radiosity

Last modified on 15 Oct, 2008