The narrow cavity limit

We seek an asymptotic solution of this problem for large vertical aspect ratio. In the limit $\mbox{$\mathcal A$}\rightarrow\infty$ , equations (4.7)-(4.10) and (4.12) become

$\displaystyle \frac{\partial u}{\partial X}$	$\textstyle =$	$\displaystyle 0,$	(4.13)
$\displaystyle \mbox{\textit{Gr}}(1+N)\mbox{\textit{Sc}}\; u\frac{\partial m}{\partial X}$	$\textstyle =$	$\displaystyle \frac{\partial^2 m}{\partial X^2},$	(4.14)
$\displaystyle 0$	$\textstyle =$	$\displaystyle \frac{\partial p}{\partial X},$	(4.15)
$\displaystyle \mbox{\textit{Gr}}(1+N)\;u\frac{\partial v}{\partial X}+\frac{\partial p}{\partial Y}-\frac{T+Nm}{1+N}$	$\textstyle =$	$\displaystyle \frac{\partial^2 v}{\partial X^2},$	(4.16)

$\displaystyle \mbox{\textit{Gr}}(1+N)\left[\mbox{\textit{Pr}}_r+\mbox{\textit{Pr}}_I\left(1-\mathrm{e}^{-\varPhi }\right)m\right] u\frac{\partial T}{\partial X}$	$\textstyle =$
$\displaystyle \frac{\mbox{\textit{Pr}}_I}{\mbox{\textit{Sc}}}\left(1-\mathrm{e}^{-\varPhi }\right) \frac{\partial m}{\partial X}\frac{\partial T}{\partial X}$	$\textstyle +$	$\displaystyle \frac{\partial^2 T}{\partial X^2},$	(4.17)

The vertical and spanwise derivatives of

and

have dropped out of the elliptic field equations (4.14)-(4.17). This means that the boundary conditions at the floor and ceiling and front and back walls must be abandoned. This is entirely analogous to the situation for the tangential velocity component in slightly viscous flow over a nonslip surface (Van Dyke 1964, ch. 7). It means that unless the solution fortuitously matches these conditions (without their being enforced) there will be regions of nonuniformity in the neighbourhoods of

and $Z=\pm1/2$ .

At the remaining boundaries; the vertical walls

; apply constant temperatures and vapour mass fractions and the transpiration (2.59) and no-slip boundary conditions:

$\displaystyle T=m$	$\textstyle =$	$\displaystyle X, \mbox{\hspace{46.65mm} on } X=0,1$	(4.18)
$\displaystyle u$	$\textstyle =$	$\displaystyle -\frac{1-\mathrm{e}^{-\varPhi }}{\mbox{\textit{Gr}}(1+N)\mbox{\textit{Sc}}}\frac{\partial m}{\partial X}, \mbox{\hspace{15mm} on } X=0$	(4.19)
$\displaystyle u$	$\textstyle =$	$\displaystyle -\frac{\mathrm{e}^{\varPhi }-1}{\mbox{\textit{Gr}}(1+N)\mbox{\textit{Sc}}}\frac{\partial m}{\partial X}, \mbox{\hspace{15mm} on } X=1$	(4.20)
$\displaystyle v$	$\textstyle =$	$\displaystyle 0, \mbox{\hspace{48mm} on } X=0,1.$	(4.21)

The equation set above, (4.13)-(4.17), subject to (4.18)-(4.21), will apply for the mid-section of the cavity, sufficiently far away from the floor and ceiling and front and back walls. The precise meaning of this statement must await the solution of the ceiling and floor problems in chapter 5, and the front and back wall problems in chapter 7.