Berman flow

Flow description

Consider a rectangular channel of width much longer than the height. Let the distance between the top and bottom wall be ${\displaystyle 2h}$ and choose the coordinates such that ${\displaystyle x=0,y=0}$ lies in the midway between the two walls, with ${\displaystyle y}$ points perpendicular to the planes. Let both walls be porous with equal velocity ${\displaystyle V}$. Then the continuity equation and Navier–Stokes equations for incompressible fluid become

${\displaystyle \begin{array}{cc}\frac{\mathrm{\partial }u}{\mathrm{\partial }x}+\frac{\mathrm{\partial }v}{\mathrm{\partial }y}& =0\\ u\frac{\mathrm{\partial }u}{\mathrm{\partial }x}+v\frac{\mathrm{\partial }u}{\mathrm{\partial }y}& =-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }x}+\nu \left(\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{x}^2}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{y}^2}\right),\\ u\frac{\mathrm{\partial }v}{\mathrm{\partial }x}+v\frac{\mathrm{\partial }v}{\mathrm{\partial }y}& =-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }y}+\nu \left(\frac{{\mathrm{\partial }}^{2}v}{\mathrm{\partial }{x}^2}+\frac{{\mathrm{\partial }}^{2}v}{\mathrm{\partial }{y}^2}\right)\end{array}}$

with boundary conditions

${\displaystyle u(x,\pm h)=0,{\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }y}\right)}_{y=0}=0,v(x,0)=0,v(x,\pm h)=V}$

The boundary conditions at the center is due to symmetry. Since the solution is symmetric above the plane ${\displaystyle y=0}$, it is enough to describe only half of the flow, say for ${\displaystyle y>0}$. If we look for ${\displaystyle v}$ a solution, that is independent of ${\displaystyle x}$, the continuity equation dictates that the horizontal velocity ${\displaystyle u}$ can at most be a linear function of ${\displaystyle x}$. Therefore, Berman introduced the following form,

${\displaystyle \eta =\frac{y}{h},\psi (x,\eta )=[h{\overline{u}}_o-xV]f(\eta ),u=\left({\overline{u}}_o-\frac{Vx}{h}\right)f^{\prime }(\eta ),v=Vf(\eta )}$

where ${\displaystyle {\overline{u}}_0}$ is the average value (averaged cross-sectionally) of ${\displaystyle u}$ at ${\displaystyle x=0}$, that is to say

${\displaystyle {\overline{u}}_0=\frac{1}{2}{\int }_{-1}^{1}u(0,\eta )d\eta =\frac{{\overline{u}}_0}{2}{\int }_{-1}^1{f}^{\prime }(\eta )d\eta =\frac{{\overline{u}}_0}{2}[f(1)-f(-1)]={\overline{u}}_0}$

This constant will be eliminated out of the problem and will have no influence on the solution. Substituting this into the momentum equation leads to

${\displaystyle \begin{array}{cc}-\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }x}& =\left({\overline{u}}_o-\frac{Vx}{h}\right)\left(-\frac{V}{h}[f^{\prime 2}-f{f}^{″}]-\frac{\nu }{h^2}{f}^{‴}\right),\\ -\frac{1}{\rho }\frac{\mathrm{\partial }p}{\mathrm{\partial }\eta }& =\nu \frac{dv}{d\eta }-\frac{\nu }{h}\frac{d^{2}v}{d{\eta }^2}\end{array}}$

Differentiating the second equation with respect to ${\displaystyle x}$ gives ${\displaystyle {\mathrm{\partial }}^{2}p/\mathrm{\partial }x\mathrm{\partial }\eta =0}$ this can substituted into the first equation after taking the derivative with respect to ${\displaystyle \eta }$ which leads to

${\displaystyle f^{iv}+\mathrm{Re}(f^{\prime 2}-f{f}^{″}{)}^{\prime }=0}$

where ${\displaystyle \mathrm{Re}=Vh/\nu }$ is the Reynolds number. Integrating once, we get

${\displaystyle f^{‴}+\mathrm{Re}(f^{\prime 2}-f{f}^{″})=C}$

with boundary conditions

${\displaystyle f(0)=f^{″}(0)=f(1)-1=f^{\prime }(1)=0}$

This third order nonlinear ordinary differential equation requires three boundary condition and the fourth boundary condition is to determine the constant ${\displaystyle C}$. and this equation is found to possess multiple solutions. The figure shows the numerical solution for low Reynolds number, solving the equation for large Reynolds number is not a trivial computation.

Limiting solutions

In the limit ${\displaystyle Re\to 0}$, the solution can be written as

${\displaystyle f(\eta )=\frac{1}{2}(3\eta -{\eta }^3)+\frac{Re}{280}(3{\eta }^3-2\eta -{\eta }^7)+O(R{e}^2).}$

In the limit ${\displaystyle Re\to -\mathrm{\infty }}$, the leading-order solution is given by

${\displaystyle f(\eta )=\sin \frac{\pi }{2}\eta .}$

The above solution satisfies all the necessary boundary conditions even though Reynolds number is infinite (see also Taylor–Culick flow)