Berman flow

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[2]

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

with boundary conditions

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

The boundary conditions at the center is due to symmetry. Since the solution is symmetric above the plane ${\displaystyle y=0}$, 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 ${\displaystyle x}$.[3] Therefore, Berman introduced the following form,

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

where ${\displaystyle u_{o}(\eta )}$ is an arbitrary function and it will be eliminated out of the problem in due course. Substituting this into the momentum equation leads to

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

Berman flow

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

${\displaystyle f^{iv}+\operatorname {Re} (f'^{2}-ff'')'=0}$

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

${\displaystyle f'''+\operatorname {Re} (f'^{2}-ff'')=C}$

with boundary conditions

${\displaystyle f(0)=f''(0)=f(1)-1=f'(1)=0}$

This third order nonlinear ordinary differential equation requires three boundary condition and the fourth boundary condition is to determine the constant ${\displaystyle C}$. and this equation is found to possess multiple solutions.[4][5] The figure shows the numerical solution for low Reynolds number, solving the equation for large Reynolds number is not a trivial computation.

• Taylor–Culick flow