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5 Radiation

The radiative heating rate is calculated from the radiative flux by using the second order centered scheme.

$\displaystyle Q_{rad,i,j}^{n}$ $\textstyle =$ $\displaystyle Q_{rad,IR,i,j}^{n} + Q_{rad,NIR,i,j}^{n} +
Q_{rad,dust,SR,i,j}^{n} + Q_{rad,dust,IR,i,j}^{n},$ (51)
       
$\displaystyle Q_{rad,*,i,j}$ $\textstyle =$ $\displaystyle - \frac{g}{c_{p}}
\frac{F_{*,net,i,j+\frac{1}{2}} -
F_{*,net,i,j-\frac{1}{2}}}
{\Delta P _{0,j}},$ (52)


\begin{displaymath}
F_{*,net,i,j+\frac{1}{2}} =
F_{IR,i,j+\frac{1}{2}}^{\uparr...
...d
\Delta P_{0,j} = P_{0,j+\frac{1}{2}} - P_{0,j-\frac{1}{2}}.
\end{displaymath}

The radiative flux is evaluated on the half grid point The subscript $m$ shows grid point in the wave number. In the following sections, the superscript which shows time step is omitted.

5.1 Radiative transfer of atmospheric CO${}_{2}$

The finite difference form of the infrared radiative flux and the Plank function are represented as follows.

$\displaystyle F_{IR,i,j+\frac{1}{2}}^{\uparrow}$ $\textstyle =$ $\displaystyle \sum _{m}\Delta \nu _{m}\left\{
\pi B_{\nu _{i},T_{sfc,i}}{\cal T}_{i}(0,z_{j+\frac{1}{2}})
\right.$  
    $\displaystyle \left.
+ \sum _{j'=1}^{j} \pi B_{\nu _{m},T_{i,j'}}
\frac{ {\cal ...
...cal T }_{m}(z_{j+\frac{1}{2}},z_{j'-\frac{1}{2}}) }
{ \Delta z_{j'} } \right\},$ (53)
$\displaystyle F_{IR,i,j+\frac{1}{2}}^{\downarrow}$ $\textstyle =$ $\displaystyle \sum _{m}\Delta \nu _{m}\left\{
\sum _{j'=j}^{J}\pi B_{\nu _{m},T...
...al T }_{m}(z_{j+\frac{1}{2}},z_{j'+\frac{1}{2}}) }
{ \Delta z_{j'} }
\right\} ,$ (54)
$\displaystyle B_{\nu _{m},T_{i,j}}$ $\textstyle =$ $\displaystyle \frac{1.19\times 10^{-8}\nu _{m}^{3}}
{e^{1.4387\nu_{i}/T_{i,j}}-1},$ (55)
$\displaystyle T_{i,j}$ $\textstyle =$ $\displaystyle \Pi _{0,j}(\theta _{i,j}+\Theta _{0,j})$ (56)

where the averaged transmission function, the equivalent width, and the effective path length are represented as follows.


\begin{displaymath}
{\cal T}_{m}(z_{j+\frac{1}{2}},z_{j'+\frac{1}{2}}) =
\exp...
...{*}(z_{j+\frac{1}{2}},z_{j'+\frac{1}{2}})/
\alpha ^{*}_{m}}},
\end{displaymath}


\begin{displaymath}
u(z_{j+\frac{1}{2}},z_{j'+\frac{1}{2}}) = 1.67
g \vert P_...
...\alpha ^{*}_{m}=\alpha _{m} \frac{\overline{p}_{jj'}}{p_{0}},
\end{displaymath}


\begin{displaymath}
P_{0,j+\frac{1}{2}} = 0.5(P_{0,j}+P_{0,j+1}),
\quad
\over...
...\frac{1}{2}}\vert}
{u(z_{j+\frac{1}{2}},z_{j'+\frac{1}{2}})},
\end{displaymath}

The finite form of the near infrared radiative flux and the effective path length are represented as follows.

\begin{displaymath}
F_{NIR,i,j+\frac{1}{2}}^{\downarrow} = \sum _{m}\Delta \nu ...
...{i}(z_{j+\frac{1}{2}},z_{J+\frac{3}{2}})\cos \zeta
\right\}.
\end{displaymath} (57)


\begin{displaymath}
u(z_{j+\frac{1}{2}},z_{J+\frac{3}{2}}) =
\frac{1.67 g \ve...
...{0,J+\frac{3}{2}} \vert}
{\mbox{MAX}(\cos \zeta, \epsilon )}.
\end{displaymath}

where $\epsilon $ is a small parameter to ensure $u = 0$ when $\cos \zeta =0$.

5.2 Radiative transfer of dust

The finite difference form of the solar radiative transfer equation of dust are represented as follows.

$\displaystyle F_{dif,\nu_{m},i,j+\frac{1}{2}}^{\uparrow}$ $\textstyle =$ $\displaystyle F_{dif,\nu_{m},i,j-\frac{1}{2}}^{\uparrow} -
\Delta \tau _{\nu_{m...
...frac{1}{2}}^{\uparrow} +
F_{dif,\nu_{m},i,j-\frac{1}{2}}^{\uparrow}}{2}
\right.$  
    $\displaystyle - \left.
\gamma _{2,\nu_{m}}\frac{F_{dif,\nu_{m},i,j+\frac{1}{2}}...
...ga}_{\nu_{m}}^{*}
S_{0}e^{-\tau _{\nu_{m},j+\frac{1}{2}}^{*}
/\mu_{0}} \right],$ (58)
$\displaystyle F_{dif,\nu_{m},i,j-\frac{1}{2}}^{\downarrow}$ $\textstyle =$ $\displaystyle F_{dif,\nu_{m},i,j+\frac{1}{2}}^{\downarrow} +
\Delta \tau _{\nu_...
...frac{1}{2}}^{\uparrow} +
F_{dif,\nu_{m},i,j-\frac{1}{2}}^{\uparrow}}{2}
\right.$  
    $\displaystyle - \left.
\gamma _{1,\nu_{m}}\frac{F_{dif,\nu_{m},i,j+\frac{1}{2}}...
...ga}_{\nu_{m}}^{*}
S_{0}e^{-\tau _{\nu_{m},j+\frac{1}{2}} ^{*}/\mu_{0}}
\right],$ (59)


\begin{displaymath}
\Delta \tau _{\nu_{m}, j}= \frac{\overline{Q}_{e,\nu_{m}}}{...
...+\frac{1}{2}} = \sum _{j'=j+1}^{J+1}\Delta \tau _{\nu_{m},j'}.
\end{displaymath}

(58) and (59) can be represented in matrix form as follows.

$\displaystyle \Dvect{F}_{dif,\nu_{m}}^{\uparrow}$ $\textstyle =$ $\displaystyle \Dvect{A}\Dvect{F}_{dif,\nu_{m}}^{\uparrow} +
\Dvect{B}\Dvect{F}_{dif,\nu_{m}}^{\downarrow} + \Dvect{R},$ (60)
$\displaystyle \Dvect{F}_{dif,\nu_{m}}^{\downarrow}$ $\textstyle =$ $\displaystyle \Dvect{C}\Dvect{F}_{dif,\nu_{m}}^{\uparrow} +
\Dvect{D}\Dvect{F}_{dif,\nu_{m}}^{\downarrow} + \Dvect{S}.$ (61)

where $\Dvect{F}_{dif,\nu_{m}}^{\uparrow}=(
F_{dif,\nu_{m},i,\frac{1}{2}}^{\uparrow},
...
...,i,\frac{3}{2}}^{\uparrow},...,
F_{dif,\nu_{m},i,J+\frac{1}{2}}^{\uparrow})^{T}$ and so on. (60) and (61) are solved by iteration method. The elements of the matrixes in (60) and (61) are represented as follows.

\begin{eqnarray*}
A_{jk} &=& \left\{
\begin{array}{cl}
- \frac{1}{2}\Delta \t...
...}}^{*}
/\mu_{0}} &
j\leq J-1 \\
0 & j=J
\end{array} \right.
\end{eqnarray*}

The finite difference form of the solar radiative transfer equation of dust are represented as follows.

$\displaystyle F_{IR,\nu_{m},i,j+\frac{1}{2}}^{\uparrow}$ $\textstyle =$ $\displaystyle F_{IR,\nu_{m},i,j-\frac{1}{2}}^{\uparrow} -
\Delta \tau _{\nu_{m}...
...\frac{1}{2}}^{\uparrow} +
F_{IR,\nu_{m},i,j-\frac{1}{2}}^{\uparrow}}{2}
\right.$  
    $\displaystyle - \left.
\gamma _{2,\nu_{m}}\frac{F_{IR,\nu_{m},i,j+\frac{1}{2}}^...
...wnarrow}}{2}
-2\pi (1-\tilde{\omega}_{\nu_{m}}^{*})B_{\nu_{m},T_{i,j}}
\right],$ (62)
$\displaystyle F_{IR,\nu_{m},i,j-\frac{1}{2}}^{\downarrow}$ $\textstyle =$ $\displaystyle F_{IR,\nu_{m},i,j+\frac{1}{2}}^{\downarrow} +
\Delta \tau _{\nu_{...
...\frac{1}{2}}^{\uparrow} +
F_{IR,\nu_{m},i,j-\frac{1}{2}}^{\uparrow}}{2}
\right.$  
    $\displaystyle - \left.
\gamma _{1,\nu_{m}}\frac{F_{IR,\nu_{m},i,j+\frac{1}{2}}^...
...wnarrow}}{2}
+2\pi (1-\tilde{\omega}_{\nu_{m}}^{*})B_{\nu_{m},T_{i,j}}
\right].$ (63)

(62) and (63) can be represented in matrix form as follows.

$\displaystyle \Dvect{F}_{IR,\nu_{m}}^{\uparrow}$ $\textstyle =$ $\displaystyle \Dvect{A}\Dvect{F}_{IR,\nu_{m}}^{\uparrow} +
\Dvect{B}'\Dvect{F}_{IR,\nu_{m}}^{\downarrow} + \Dvect{R}',$ (64)
$\displaystyle \Dvect{F}_{IR,\nu_{m}}^{\downarrow}$ $\textstyle =$ $\displaystyle \Dvect{C}\Dvect{F}_{IR,\nu_{m}}^{\uparrow} +
\Dvect{D}\Dvect{F}_{IR,\nu_{m}}^{\downarrow} + \Dvect{S}'.$ (65)

where $\Dvect{F}_{IR,\nu_{m}}^{\uparrow}=(F_{IR,\nu_{m},i,\frac{1}{2}}^{\uparrow},
F_{...
...},i,\frac{3}{2}}^{\uparrow},...,
F_{IR,\nu_{m},i,J+\frac{1}{2}}^{\uparrow})^{T}$ and so on. (64) and 行列形式のダストの下向き赤外放射 are solved by iteration method. The elements of the matrixes in (64) and 行列形式のダストの下向き赤外放射 are represented as follows.

\begin{eqnarray*}
B'_{jk} &=& \left\{
\begin{array}{cl}
0 & j=k=1 \\
B_{jk}...
...\nu_{m}, T_{i,j}} & j\leq J-1 \\
0 & i=N
\end{array} \right.
\end{eqnarray*}


next up previous
: 6 Ground surface : Two dimensional anelastic model : 4 Dust transport
Odaka Masatsugu 平成19年4月26日