# Berman flow

In fluid dynamics, Berman flow is a steady flow created inside a rectangular channel with two equally porous walls. The concept is named after a scientist Abraham S. Berman who formulated the problem in 1953.

## Flow description

Consider a rectangular channel of width much longer than the height. Let the distance between the top and bottom wall be $2h$ and choose the coordinates such that $x=0,\ y=0$ lies in the midway between the two walls, with $y$ points perpendicular to the planes. Let both walls be porous with equal velocity $V$. Then the continuity equation and Navier–Stokes equations for incompressible fluid become
    ${\begin{aligned}{\frac {\partial u}{\partial x}}+{\frac {\partial v}{\partial y}}&=0\\u{\frac {\partial u}{\partial x}}+v{\frac {\partial u}{\partial y}}&=-{\frac {1}{\rho }}{\frac {\partial p}{\partial x}}+\nu \left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}\right),\\u{\frac {\partial v}{\partial x}}+v{\frac {\partial v}{\partial y}}&=-{\frac {1}{\rho }}{\frac {\partial p}{\partial y}}+\nu \left({\frac {\partial ^{2}v}{\partial x^{2}}}+{\frac {\partial ^{2}v}{\partial y^{2}}}\right)\end{aligned}}$
with boundary conditions
    $u(x,\pm h)=0,\quad \left({\frac {\partial u}{\partial y}}\right)_{y=0}=0,\quad v(x,0)=0,\quad v(x,\pm h)=V$
The boundary conditions at the center is due to symmetry. Since the solution is symmetric above the plane $y=0$, it is enough to describe only half of the flow, say for $y>0$. If we look for $v$ a solution, that is independent of $x$, the continuity equation dictates that the horizontal velocity $u$ can at most be a linear function of $x$. Therefore, Berman introduced the following form,
    $\eta ={\frac {y}{h}},\quad \psi (x,\eta )=[h{\bar {u}}_{o}-xV]f(\eta ),\quad u=\left({\bar {u}}_{o}-{\frac {Vx}{h}}\right)f'(\eta ),\quad v=Vf(\eta )$
where ${\bar {u}}_{0}$ is the average value (averaged cross-sectionally) of $u$ at $x=0$, that is to say
    ${\bar {u}}_{0}={\frac {1}{2}}\int _{-1}^{1}u(0,\eta )d\eta ={\frac {{\bar {u}}_{0}}{2}}\int _{-1}^{1}f'(\eta )d\eta ={\frac {{\bar {u}}_{0}}{2}}[f(1)-f(-1)]={\bar {u}}_{0}$
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
    ${\begin{aligned}-{\frac {1}{\rho }}{\frac {\partial p}{\partial x}}&=\left({\bar {u}}_{o}-{\frac {Vx}{h}}\right)\left(-{\frac {V}{h}}[f'^{2}-ff'']-{\frac {\nu }{h^{2}}}f'''\right),\\-{\frac {1}{\rho }}{\frac {\partial p}{\partial \eta }}&=\nu {\frac {dv}{d\eta }}-{\frac {\nu }{h}}{\frac {d^{2}v}{d\eta ^{2}}}\end{aligned}}$
Differentiating the second equation with respect to $x$ gives $\partial ^{2}p/\partial x\partial \eta =0$ this can substituted into the first equation after taking the derivative with respect to $\eta$ which leads to
    $f^{iv}+\operatorname {Re} (f'^{2}-ff'')'=0$
where $\operatorname {Re} =Vh/\nu$ is the Reynolds number. Integrating once, we get
    $f'''+\operatorname {Re} (f'^{2}-ff'')=C$
with boundary conditions
    $f(0)=f''(0)=f(1)-1=f'(1)=0$
This third order nonlinear ordinary differential equation requires three boundary condition and the fourth boundary condition is to determine the constant $C$. 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 $Re\rightarrow 0$, the solution can be written as
    $f(\eta )={\frac {1}{2}}(3\eta -\eta ^{3})+{\frac {Re}{280}}(3\eta ^{3}-2\eta -\eta ^{7})+O(Re^{2}).$
In the limit $Re\rightarrow -\infty$, the leading-order solution is given by
    $f(\eta )=\sin {\frac {\pi }{2}}\eta .$
The above solution satisfies all the necessary boundary conditions even though Reynolds number is infinite (see also Taylor–Culick flow)

## Axisymmetric case

The corresponding problem in porous pipe flows was addressed by S. W. Yuan and A. Finkelstein in 1955.