Berman flow

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 $x=0,\; \backslash \; y=0$ 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 becomeDrazin, P. G., & Riley, N. (2006). The Navier-Stokes equations: a classification of flows and exact solutions (No. 334). Cambridge University Press.

:$\backslash begin\{align\}$

\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{align}

with boundary conditions

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

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 $x$.Proudman, I. (1960). An example of steady laminar flow at large Reynolds number. Journal of Fluid Mechanics, 9(4), 593-602. Therefore, Berman introduced the following form,

:$\backslash eta\; =\; \backslash frac\{y\}\{h\},\; \backslash quad\; \backslash psi(x,\backslash eta)\; =\; [h\backslash bar\; u\_o-xV]f(\backslash eta),\; \backslash quad\; u\; =\; \backslash left(\backslash bar\; u\_o\; -\; \backslash frac\{Vx\}\{h\}\backslash right)f\text{'}(\backslash eta),\; \backslash quad\; v=V\; f(\backslash eta)$

where $\backslash bar\; u\_0$ is the average value (averaged cross-sectionally) of $u$ at $x=0$, that is to say

:$\backslash bar\; u\_0\; =\; \backslash frac\{1\}\{2\}\backslash int\_\{-1\}^\{1\}\; u(0,\backslash eta)\; d\backslash eta=\backslash frac\{\backslash bar\; u\_0\}\{2\}\; \backslash int\_\{-1\}^1f\text{'}(\backslash eta)\; d\backslash eta=\backslash frac\{\backslash bar\; u\_0\}\{2\}\; [f(1)-f(-1)]\; =\; \backslash bar\; u\_0$

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

:$\backslash begin\{align\}$

- \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{align}

File:Berman.pdf

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

:$f^\{iv\}\; +\; \backslash operatorname\{Re\}\; (f\text{'}^2\; -\; ff\text{'}\text{'})\text{'}=0$

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

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

with boundary conditions

:$f(0)=f\text{'}\text{'}(0)=\; f(1)-1=f\text{'}(1)=0$

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.Wang, C-A., T-W. Hwang, and Y-Y. Chen. "Existence of solutions for Berman's equation from Laminar flows in a porous channel with suction." Computers & Mathematics with Applications 20.2 (1990): 35–40.Hwang, Tzy-Wei, and Ching-An Wang. "On multiple solutions for Berman's problem." Proceedings of the Royal Society of Edinburgh: Section A Mathematics 121.3-4 (1992): 219–230. The figure shows the numerical solution for low Reynolds number, solving the equation for large Reynolds number is not a trivial computation.

In the limit $Re\backslash rightarrow\; 0$, the solution can be written as

:$f(\backslash eta)\; =\; \backslash frac\{1\}\{2\}(3\backslash eta-\backslash eta^3)\; +\; \backslash frac\{Re\}\{280\}(3\backslash eta^3-2\backslash eta-\backslash eta^7)\; +\; O(Re^2).$

In the limit $Re\backslash rightarrow\; -\backslash infty$, the leading-order solution is given byMorduchow, M. (1957). On laminar flow through a channel or tube with injection: application of method of averages. Quarterly of Applied Mathematics, 14(4), 361-368.

:$f(\backslash eta)\; =\; \backslash sin\; \backslash frac\{\backslash pi\}\{2\}\backslash eta.$

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

The corresponding problem in porous pipe flows was addressed by S. W. Yuan and A. Finkelstein in 1955.Yuan, S. W., Finkelstein, A. (1955). Laminar pipe flow with injection and suction through a porous wall. PRINCETON UNIV NJ JAMES FORRESTAL RESEARCH CENTER.

• Taylor–Culick flow

{{Reflist|30em}}

Category:Flow regimes

Category:Fluid dynamics