Simple examplew of viscous flow
Contents
Simple examplew of viscous flow¶
Channel flow¶
The book goes through many examples: let me discuss a particularly simple one, the steady flow of viscous fluid in two dimensions through a straight channel. Let the walls of the channel be at \(y = 0,L\), extending in the \(x\) direction, while the fluid is flowing entirely along the \(x\) direction, and is \(t\)-independent. Incompressibility sets \(\partial_x u = 0\), so that \(u = u(y)\). We will look for a solution fo which \(p = p(x)\), so that the fluid is being pushed down the channel. Plugging these solutions into he Navier-Stokes equation, we find
Each side depends on an independent variable, so they must be equal to a constant \(\nu c\). We can write \(u(y) = c y^2 + d y + e\). We set \(u = 0\) at \(y = 0\), so that \(e = 0\), and also at \(y = L\), so that \(c L^2 + d L = 0\), or \(d = - c L\), so that \(u(y) = c (y^2 - L y)\). \(c\) is thus set by the pressure gradient. Since \(\partial_x p = \nu \rho c = \mu c\), \(p = \mu c x + p_0\).
Flow down a plane.¶
Anther example, discussed in the book, is one of a layer of water with height \(h\) undergoing steady flow down an incline at angle \(\alpha\) from the ground, with atmospheric pressure \(p_0\) at the top of the fluid. Take the problem to be two-dimensional with \(x,z\) the directions parallel and perpendicular to the incline, and \(z = 0\) the incline and \(z = h\) the upper boundary. Let \(p = p(z)\), and \({\vec u} = u(z) {\hat x}\), where incompressibility kills the \(x\) dependence.
The Navier-Stokes equations in components become:
NOte that \({\vec u} \cdot {\vec\nabla} {\vec u} = 0\) automatically. At \(z = 0\) the boundary conditions are \(u(0) = 0\), At \(z = h\), the no-slip condition is clearly false. At such a fluid interface, we instead impose the “no stress” condition on the tangential stress The tangential stress is taken to be proportional to \(\partial_z v\), so this must vanish. The pressure, as stted above, is \(p(h) = p_0\). The resoluting solution is
This is an idealized situation with a flat bottom. In practice, deviations fom flatness will start to lead to nonvanishing intertial terms, and turbulent flow is possible. If we assume that the hill has a rise angle \(\beta\), and a length scale \(L \sim h \beta\), the natural Reynolds number is
For water \(\nu \sim 10^{-4} m^2/s\). If we cnsider a puddle with \(h \sim 1 mm\), and \(\alpha \sim .01\), we find for laminar flow \(v \sim 5cm/sec\). Add an imperfection and then \(Re \sim 50\beta\).
For a large slow river with \(h \sim 10 m\), \(\alpha \sim 10^{-4}\), we would find \(v \sim 100 km/sec\) which is crazy. Here however \(Re \sim 10^{12} \beta\). Thus, the slightest imperfection leads to turbulence, and our solution above is invalid.