3.1 DUGKS with anisotropic scattering effects
In this section, the discrete unified gas kinetic scheme for gray radiative transfer equation involving the anisotropic scattering effects (Eq. (1)) is constructed in detail. Similar to the discretization approach in Ref [39], the solid angle space is discretized into M discrete angles using the discrete ordinates method based on certain spherical quadratures, and correspondingly we obtain M discrete directions sk. With these discrete directions, the RTE (1) can be expressed as
$$ \begin{aligned} \frac{1}{c} \frac{\partial I \left(\boldsymbol{x}, \boldsymbol{s}_{k}, t \right)}{\partial t} + \boldsymbol{s}_{k} \cdot \nabla I \left(\boldsymbol{x}, \boldsymbol{s}_{k}, t \right) = - \beta I \left(\boldsymbol{x}, \boldsymbol{s}_{k}, t \right) +\beta S \left(\boldsymbol{x}, \boldsymbol{s}_{k}, t \right), \end{aligned} $$
(10)
$$ \begin{aligned} S \left(\boldsymbol{x}, \boldsymbol{s}_{k}, t \right)=\left(1 - \omega \right) I_{b} \left(\boldsymbol{x},t \right) + \frac{ \omega}{4 \pi} \sum_{m=1}^{M} \left[ I \left(\boldsymbol{x}, \boldsymbol{s}_{m},t \right) \Phi \left(\boldsymbol{s}_{m}, \boldsymbol{s}_{k} \right) \omega_{m} \right], \end{aligned} $$
(11)
where k,m=1,2,...,M, and ωm is the weight assigned to the discretized direction sm. Following Refs. [1, 32], integrating Eq. (10) on a control volume Vj centered at xj from time tn to tn+1=tn+△t, we can obtain
$$ \begin{aligned} & I \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k}, t_{n+1} \right) - I \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k}, t_{n} \right) + \frac{c \triangle t}{\left| V_{j} \right|} F^{n+1/2} \hfill \\ & =\frac{c \beta \triangle t}{2} \left [ S \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k},t_{n+1} \right) - I \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k},t_{n+1} \right) \right] +\frac{c \beta \triangle t}{2} \left [ S \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k},t_{n} \right) - I \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k},t_{n} \right) \right] \hfill, \end{aligned} $$
(12)
where
$$ F^{n+1/2} = \sum_{f} \left(\boldsymbol{s}_{k} \cdot \boldsymbol{n}_{f} \right) I \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{k}, t_{n+1/2} \right) \triangle S_{f}, $$
(13)
is the flux across the cell interface, I(xj,sk,tn) denotes the cell averaged value for the diffuse intensity at time tn with control volume of Vj located at xj along photon propagation direction sk, nf is the outward unit normal vector at xf of an interface, and △Sf is the corresponding interface area. The midpoint rule for the integration of the second term on the left-hand of Eq. (12) and trapezoidal rule for the right-hand of Eq. (12) are used, respectively. Two new distribution functions are introduced to remove the implicitness in Eq. (12),
$$ \widetilde{I} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) =I\left(\boldsymbol{x}, \boldsymbol{s}, t \right)+\frac{\chi}{2} \left[ I\left(\boldsymbol{x}, \boldsymbol{s}, t \right) -S \left(\boldsymbol{x},\boldsymbol{s},t \right) \right], $$
(14)
$$ {\widetilde{I}}^{+} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) =I\left(\boldsymbol{x}, \boldsymbol{s}, t \right)-\frac{\chi}{2} \left[ I\left(\boldsymbol{x}, \boldsymbol{s}, t \right) -S \left(\boldsymbol{x},\boldsymbol{s},t \right) \right], $$
(15)
where χ=cβ△t. Then Eq. (12) can be rewritten as
$$ \widetilde{I} \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k}, t_{n+1} \right)={\widetilde{I}}^{+} \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k}, t_{n} \right) - \frac{c \triangle t}{\left| V_{j} \right|} F^{n+1/2}. $$
(16)
In order to evaluate the cell interface flux at the half time-step Fn+1/2, we integrate Eq. (10) along the characteristic line with a half time step,
$$ \begin{aligned} & I \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{k}, t_{n+1/2} \right) - I \left(\boldsymbol{x}_{f} - \boldsymbol{s}_{k} c h, \boldsymbol{s}_{k}, t_{n} \right) \hfill \\ & =\frac{c \beta h }{2} \left [ S \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{k},t_{n+1/2} \right) - I \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{k},t_{n+1/2} \right) \right] \hfill \\ & +\frac{c \beta h}{2} \left [ S \left(\boldsymbol{x}_{f} - \boldsymbol{s}_{k} c h, \boldsymbol{s}_{k},t_{n} \right) - I \left(\boldsymbol{x}_{f} - \boldsymbol{s}_{k} c h, \boldsymbol{s}_{k},t_{n} \right) \right] \hfill, \end{aligned} $$
(17)
where h=△t/2, and the trapezoidal rule is again used to evaluate the right-hand term of Eq. (10). Another two new distribution functions are also introduced to remove the implicitness in Eq. (17),
$$ \bar{I} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) = I \left(\boldsymbol{x},\boldsymbol{s},t \right) + \frac{\chi}{4} \left[ I \left(\boldsymbol{x},\boldsymbol{s},t \right) - S \left(\boldsymbol{x},\boldsymbol{s},t \right) \right], $$
(18)
$$ {\bar{I}}^{+} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) = I \left(\boldsymbol{x},\boldsymbol{s},t \right) - \frac{\chi}{4} \left[ I \left(\boldsymbol{x},\boldsymbol{s},t \right) -S \left(\boldsymbol{x},\boldsymbol{s},t \right) \right]. $$
(19)
Substituting Eqs. (18) and (19) into Eq. (17), we can obtain
$$ \bar{I} \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{k}, t_{n+1/2} \right) ={\bar{I}}^{+} \left(\boldsymbol{x}_{f} - \boldsymbol{s}_{k} c h, \boldsymbol{s}_{k}, t_{n} \right). $$
(20)
\(\bar {I}^{+} \left (\boldsymbol {x}_{f} - \boldsymbol {s}_{k} c h, \boldsymbol {s}_{k}, t_{n} \right)\) can be reconstructed by
$$ \begin{aligned} {\bar{I}}^{+} \left(\boldsymbol{x}_{f} - \boldsymbol{s}_{k} c h, \boldsymbol{s}_{k}, t_{n} \right)={\bar{I}}^{+} \left(\boldsymbol{x}_{j}, \boldsymbol{s}_{k}, t_{n} \right) + \left(\boldsymbol{x}_{f} - \boldsymbol{s}_{k} c h - \boldsymbol{x}_{j} \right) \cdot \mathbf{\sigma}_{j}, \\ \left(\boldsymbol{x}_{f} - \boldsymbol{s}_{k} c h \right) \in V_{j}, \end{aligned} $$
(21)
where σj is the slope of the distribution function \({\bar {I}}^{+} \) in cell j. In the present study, the van Leer limiter [40] is used to calculate the slope.
The new distribution functions \(\widetilde {I}, {\widetilde {I}}^{+}, \bar {I}, {\bar {I}}^{+}\) are all related to the original distribution function I and the scattering phase function Φ(cosψ). Their relations in the present work can be finally obtained as
$$ \begin{aligned} I \left(\boldsymbol{x},\boldsymbol{s},t \right)&=\frac{4}{4+\chi}\bar{I} \left(\boldsymbol{x}, \boldsymbol{s}, t \right)+ \frac{\chi}{4+\chi}\left(1- \omega \right) I_{b} \left(\boldsymbol{x}, t\right) \\ & +\frac{\chi\omega}{4 \pi \left(4 + \chi \right)}\int_{4 \pi} I\left(\boldsymbol{x}, \boldsymbol{s} ', t \right) {\Phi} (\boldsymbol{s} ',\boldsymbol{s}) d {{\Omega}'}, \end{aligned} $$
(22)
$$ \begin{aligned} {\bar{I}}^{+} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) & =\frac{4 - \chi}{4+2\chi}\widetilde{I} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) + \frac{3\chi}{4 + 2 \chi} \left(1- \omega \right) I_{b}\left(\boldsymbol{x}, t \right) \\ & +\frac{ \omega\chi}{8\pi \left(2+\chi \right)}\int_{4 \pi} \left[ \widetilde{I} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) + 2 {\bar{I}}^{+} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) \right] {\Phi} (\boldsymbol{s} ',\boldsymbol{s}) d {{\Omega}'}, \end{aligned} $$
(23)
$$ {\widetilde{I}}^{+} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) = \frac{4}{3}{\bar{I}}^{+} \left(\boldsymbol{x}, \boldsymbol{s}, t \right) - \frac{1}{3} \widetilde{I} \left(\boldsymbol{x}, \boldsymbol{s}, t \right). $$
(24)
In the isotropic scattering condition (Φ≡1), Eqs. (22) and (23) can be solved explicitly [1]. However, in anisotropic case, due to the complexity of the phase function (Φ(s′,s)), Eqs. (22) and (23) can not be solved explicitly. Here a simple iterative method is employed as follows. First, the iteration procedure for the calculation of the original distribution function I(xf,sk,tn+1/2) is
$$ \begin{aligned} I^{l+1} \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{k}, t_{n+1/2} \right) & = \frac{4}{4+\chi}{\bar{I}} \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{k}, t_{n+1/2} \right) + \frac{\chi}{4+\chi} (1-\omega) {I_{b}}^{l} \left(\boldsymbol{x}_{f}, t_{n+1/2} \right) \\ & + \frac{\chi\omega}{4 \pi \left(4 + \chi \right)} \sum_{m=1}^{M} \sum_{j=0}^{N} \omega_{m} \left[ I^{l} \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{m},t_{n+1/2} \right) C_{j} P_{j} (cos \psi) \right], \end{aligned} $$
(25)
where l is the iteration index, and
$$ {I_{b}}^{l} \left(\boldsymbol{x}_{f}, t_{n+1/2} \right)= \frac{1}{4 \pi} \sum_{k=1}^{M} \omega_{k} I^{l} \left(\boldsymbol{x}_{f}, \boldsymbol{s}_{k}, t_{n+1/2} \right). $$
(26)
The iteration stops as the intensity is converged, i.e., ∣Il+1−Il∣<ε, where ε is a small number which is set to be 10−4 in our simulations. Eq. (23) is also solved iteratively in a similar way. We found that the number of iterations is less than 10 in almost all the numerical tests in the present work. But for more complex problems, more advanced acceleration technique should be employed.
Finally, we note that with the use of the trapezoidal rule in Eqs. (12) and (17), the present DUGKS is a semi-implicit scheme and the time step Δt is not limited by the scattering, which is determined by the Courant-Friedrichs-Lewy (CFL) condition [41],
$$ \Delta t =\alpha \frac{\Delta x}{c}, $$
(27)
where 0<α<1 is the CFL number, and Δx is the minimal grid spacing.
3.2 Boundary conditions
In the present work, diffusely emitting and reflecting boundaries are considered. When the wall is black, a photon is absorbed as it hits the wall, and a new photon in thermal equilibrium with boundary temperature is emitted into the domain. When the wall is gray, some of the incident photons are absorbed and the rest are reflected diffusively back to the domain, depending on the reflectivity of the wall. The general boundary condition for Eq. (1) can be expressed as
$$ I \left (\boldsymbol{x}_{w},\boldsymbol{s}, t \right)= \varepsilon_{w} I_{b} \left (\boldsymbol{x}_{w} \right) +\frac{\rho_{w}}{\pi} \int_{\boldsymbol{n}_{w} \cdot {\boldsymbol{s}}' < 0} \left(\boldsymbol{n}_{w} \cdot {\boldsymbol{s}}' \right) I \left (\boldsymbol{x}_{w},\boldsymbol{s} ', t \right) d {\mathbf{\Omega}'}, $$
(28)
where εw is the diffuse emissivity, ρw is the diffuse reflectivity, and nw is the unit inner normal vector at the boundary. Ib(xw) is the blackbody radiation intensity at the boundary surface having a specified temperature. This boundary condition is implemented in the DUGKS straightforwardly by replacing the s with each discrete angle sk, and evaluating the integral with the numerical quadrature.
3.3 Algorithm
In summary, the main procedure of the DUGKS from time step tn to tn+1 can be summarized as follows:
- 1.
Calculate the microflux Fn+1/2 at cell interface xf and at time tn+1/2.
- (a)
Calculate \({\bar {I}}^{+}\) from \(\widetilde {I}\) at each cell center with the iterative method according to Eq. (23);
- (b)
Reconstruct the slope σj of \({\bar {I}}^{+}\) in each cell center;
- (c)
Reconstruct the distribution function \({\bar {I}}^{+}\) at xf−skch according to Eq. (21);
- (d)
Calculate the distribution function \(\bar {I}\) at cell interface at time tn+1/2 according to Eq. (20);
- (e)
Calculate the original distribution function I at cell interface and at time tn+1/2 with the iterative method according to Eq. (25);
- (f)
Calculate the microflux Fn+1/2 through each cell interface from I(xf,sk,tn+1/2) according to Eq. (13).
- 2.
Calculate \({\widetilde {I}}^{+} \) at cell center and at time tn according to Eq. (24).
- 3.
Update the cell averaged \(\widetilde {I}\) in each cell from tn to tn+1 according to Eq. (16).
When the transformed intensity distribution is known, the local incident radiation energy can be calculated based on Eq. (14)
$$ \begin{aligned} G \left(\boldsymbol{x}, t \right)= \frac{2}{2 + \chi \left(1- \omega \right)} \sum_{k=1}^{M} \omega_{k} \widetilde{I} \left(\boldsymbol{x}, \boldsymbol{s}_{k}, t \right) + \frac{\chi \left(1- \omega \right)}{2 + \chi \left(1- \omega \right)} 4 \pi I_{b} \left(\boldsymbol{x},t \right), \end{aligned} $$
(29)
and the net radiative heat flux can be calculated based on Eqs. (14) and (19),
$$ \begin{aligned} q \left(\boldsymbol{x}, t \right) = \frac{1}{3} \sum_{k=1}^{M} \left[ \omega_{k} \boldsymbol{s}_{k} \left(\widetilde{I} \left(\boldsymbol{x}, \boldsymbol{s}_{k}, t \right) + 2 \bar{I}^{+} \left(\boldsymbol{x}, \boldsymbol{s}_{k},t \right) \right) \right]. \end{aligned} $$
(30)