Instability of fully developed mixed convection with viscous dissipation in a vertical porous channel

Abstract

Flow driven by an externally imposed pressure gradient in a vertical porous channel is analysed. The combined effects of viscous dissipation and thermal buoyancy are taken into account. These effects yield a basic mixed convection regime given by dual flow branches. Duality of flow emerges for a given vertical pressure gradient. In the case of downward pressure gradient, i.e. upward mean flow, dual solutions coincide when the intensity of the downward pressure gradient attains a maximum. Above this maximum no stationary and parallel flow solution exists. A nonlinear stability analysis of the dual solution branches is carried out limited to parallel flow perturbations. This analysis is sufficient to prove that one of the dual solution branches is unstable. The evolution in time of a solution in the unstable branch is also studied by a direct numerical solution of the governing equation.

Appendix

If we multiply Eq. (8a) by $T'$ and integrate, we obtain

$$\begin{aligned} \frac{1}{2}\left( \frac{\mathrm{d}T}{\mathrm{d}y} \right) ^2 + \frac{1}{3}\left( T-T_m-P \right) ^3= \frac{1}{3}\left[ T(0)-T_m-P \right] ^3, \end{aligned}$$

(17)

after noting that $T'(0)=0$. If we use Eq. (17) and scale with $T(0)-T_m-P$, we obtain two branches.

When $T(0)-T_m-P>0$, one can describe T parametrically as follows:

$$\begin{aligned} T(y)=6s^2[\varphi (s)^2-\varphi (s|y|)^2], \end{aligned}$$

(18)

where $0\le s<s_\mathrm{max}=\int _0^\infty (3-3x^2+x^4)^{-1/2}\mathrm{d}x=2.103$ and $\varphi$ is defined on $[0,s_\mathrm{max})$ by

$$\begin{aligned} \varphi '=\sqrt{3-3\varphi ^2+\varphi ^4},\quad \varphi (0)=0. \end{aligned}$$

(19)

In this case

$$\begin{aligned} T(0)= & {} 6s^2\varphi (s)^2,\quad P=-6s^2+6s\int _0^s \varphi (x)^2\,\mathrm{d}x,\nonumber \\ T_m= & {} 6s^2\varphi (s)^2-6s\int _0^s \varphi (x)^2\,\mathrm{d}x. \end{aligned}$$

(20)

These solutions make up the entire upper branch in Fig. 1 and the piece between C and A on the lower branch.

When $T(0)-T_m-P<0$ one can describe T parametrically as follows:

$$\begin{aligned} T(y)=6s^2[\psi (s)^2-\psi (s|y|)^2], \end{aligned}$$

(21)

where $0\le s<z_\mathrm{max}=\int _0^\infty (3+3x^2+x^4)^{-1/2}\mathrm{d}x=1.214$ and $\psi$ is defined on $[0,z_\mathrm{max})$ by

$$\begin{aligned} \psi '=\sqrt{3+3\psi ^2+\psi ^4},\quad \psi (0)=0. \end{aligned}$$

(22)

In this case

$$\begin{aligned} T(0)= & {} 6s^2\psi (s)^2,\nonumber \\ P= & {} 6s^2+6s\int _0^s \psi (x)^2\,\mathrm{d}x,\nonumber \\ T_m= & {} 6s^2\psi (s)^2-6s\int _0^s \psi (x)^2\,\mathrm{d}x. \end{aligned}$$

(23)

These solutions make up the lower branch in Fig. 1 without the piece between C and A.