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Purdue ME 606: Solutions of the RTE

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Solutions of the RTE

(Prof. Fisher please include a figure of the differential section over which intensity is transmitted)

Assume that all quantities are spectral -> Subscript η is implied in this analysis

We define: \tau _s = \int_0^s {\beta ds'} and \tau _z = \int_0^z {\beta dz'}

Now multiply the RTE by e^{\tau_s}.

Now integrate the RTE:

I\left( {\tau _s ,\hat s} \right) = I_w \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) = \underbrace {\left( {1 - \omega } \right)}_{\kappa /\beta }I_b \left( {\tau _s '} \right) + {\omega \over {4\pi }}\int\limits_{4\pi } {I\left( {\tau _s ',\hat s} \right)\Phi \left( {\hat s,\hat s_i } \right)} d\omega _i

Notes: If scattering is isotropic then Φ = 1 otherwise:

\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)}

Energy Equation in direction integrated form:

\nabla \cdot \bar q_R'' = \kappa \left( {4\pi I_b - G} \right)

Express \bar q_R'' in terms of τ:

{{dq_R ''} \over {d\tau _z }} = {\kappa \over {\kappa + \sigma }}\left( {4\pi I_b - G} \right), G = \int\limits_{4\pi } {Id\omega _i }

Summary of equations:

  • 1 transport equation for I (RTE)
  • 1 energy balance equation
  • note that the equations are nonlinear in T

Note: often assume either 1.) T=const. (I b is known) or that 2.) radiative equilibrium prevails such that \frac{dq_R''}{dz} = 0.

Analytic solutions are typically complicated and therefore are rare.

Special Cases

(Prof. Fisher, please include figure of participating gas layer here)

Note: \tau _L = \int\limits_0^L {\beta dz} In this case, \tau _L << 1 ~; ~implies ~\beta very small -> optically thin

I + = const. = Iw1,I − = const. = Iw2

Invoke radiative equilibrium: \frac{dq_R''}{dz} = 0.

q_R''=\sigma \left( {T_1 ^4 - T_2 ^4 } \right)

\tau _L >> 1 ~; ~implies ~\beta very large -> optically thick

Evaluate q_R '' = - {{4\pi } \over 3}{{dS} \over {d\tau }} -> from Taylor Series expansion

q_R '' = - {{4\pi } \over {3 \beta}}{{dI_b} \over {dz }} -> applies to non-scattering media.

q_R '' = - {4 \over {3\underbrace {\bar \beta _R }_{Rosseland\,Mean1}}}{{dE_b } \over {dz}} = - {{4\sigma } \over {3\bar \beta _R }}{{d\left( {T^4 } \right)} \over {dz}} = - \underbrace {{{1 16 \sigma } \over {3\bar \beta _R }}T^3 }_{k_R - radiative\,heat\,conductivity}{{dT} \over {dz}}

At radiative equilibrium: q_R '' = {{4\sigma } \over {3\bar \beta _R }}{{T_1 ^4 - T_2 ^4 } \over L}

Intermediate τ regime:

(Prof. Fisher please put a graph of this correlation here – Thx)

semi-empirical correlation: q_R '' = \left( {{1 \over {1 + 3/4\bar \beta _R L}}} \right)\sigma \left( {T_1 ^4 - T_2 ^4 } \right)

General Formulation of RTE for plane-parallel medium

Dr Fisher, insert figure of this geometry

Start with the 1D transport equation for intensity

I\left( {\tau _s ,\hat s} \right) = I_w \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 '

We then convert the previous equation to the τ coordinate and consider intensity in both the positive (+) and negative (-) directions

I^+ (\tau,\theta) = I_1(\theta) e^{{-\tau}\over{cos\theta}}+\int_{0}^\tau S(\tau',\theta) e^{fract{-(\tau-\tau')}{cos\theta}}{{d\tau '}\over{cos\theta}}

I^- (\tau,\theta) = I_2(\theta) e^{{(\tau_L-\tau)}\over{cos\theta}}+\int_{\tau_L}^\tau S(\tau',\theta) e^{fract{(\tau'-\tau)}{cos\theta}}{{d\tau '}\over{cos\theta}}

These two equations can then be solved to find the full radiation field.

Last modified on 14 Dec, 2008