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

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

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. Introduce the definition of stream function as

$u={\frac {\partial \psi }{\partial y}},\quad v=-{\frac {\partial \psi }{\partial x}}$

Since the solution is symmetric above the plane $y=0$, it is enough to describe only half of the flow, say for $y>0$. Berman introduced the following form for the stream function

$\eta ={\frac {y}{h}},\quad \psi (x,\eta )=[h{\bar {u}}_{o}-xV]f(\eta ),\quad u=\left({\bar {u}}_{o}-{\frac {Vx}{h}}\right)f'(\eta ),\quad v=Vf(\eta )$

where ${\bar {u}}_{o}$ is vertically averaged velocity at the starting point of the flow, which will eliminated out of the problem in due course. Substituting this into the momentum equation leads to

${\begin{aligned}-{\frac {1}{\rho }}{\frac {\partial p}{\partial x}}&=\left({\bar {u}}_{o}-{\frac {Vx}{h}}\right)\left(-{\frac {V}{h}}[f'^{2}-ff'']-{\frac {\nu }{h^{2}}}f'''\right),\\-{\frac {1}{\rho }}{\frac {\partial p}{\partial \eta }}&=\nu {\frac {dv}{d\eta }}-{\frac {\nu }{h}}{\frac {d^{2}v}{d\eta ^{2}}}\end{aligned}}$

Berman flow

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.