Simple Two- and Three-Dimensional Flow Problems of the Navier-Stokes Equations

Abstract

This chapter begins with studying steady state layer flows through cylindrical conduits (ellipse, triangle, rectangle) and use of the Prandtl membrane analogy. This study of the Navier-Stokes fluids is important in geophysical fluid dynamics and is manifest in Ekman’s theory and its extensions, where non-inertial effects chiefly influence the details of the fluid flow, evidenced in the Ekman spiral in atmospheric and oceanic boundary flows and in free geostrophic flows as their outer solutions. Extensions of the behavior exhibited by the assumption of constant (turbulent) viscosity are based on influences of depth dependence of the viscosity which influences the circulation pattern of such steady flows. Unsteady flows are analyzed for viscous flows along an oscillating wall and the growth of a viscous boundary layer as a response of an initial tangential velocity jump with time. The chapter closes with the study of an axial laminar jet and viscous flows in a converging two-dimensional channel.

Appendix 8.A: Construction of the Solution (8.22) to the Boundary Value Problem (8.8)

Here, we wish to demonstrate how the solution (8.22) to the boundary value problem (8.8) is constructed for a rectangle as shown in Fig.  8.28 .

Fig. 8.28

Construction of the laminar velocity profile through a rectangular chute

Valentin Boussinesq [6] chose a trial solution of the form

$$\begin{aligned} u(y, z) = -\frac{1}{2} \frac{p^{\prime }h^{2}}{\eta } \left\{ \frac{y}{h} - \left( \frac{y}{h}\right) ^{2} \right\} + f(y, z). \end{aligned}$$

(8.178)

The first term on the right of this equation satisfies the Poisson equation (8.8) exactly and the boundary conditions at $y=0$ and $y = h$. Thus, the correcting function f satisfies the Laplace equation , zero boundary conditions at $y = 0$ and $y = h$ and $f(y, 0) = f(y, \ell ) = \textstyle {\frac{1}{2} \frac{p^{\prime }h^{2}}{\eta } \left( y/h - (y/h)^{2}\right) }$. So, the boundary value problem for the function f is given by

$$\begin{aligned} \begin{array}{lll} &{}\displaystyle \frac{\partial ^{2}f}{\partial y^{2}} + \frac{\partial ^{2}f}{\partial z^{2}} = 0, &{}\quad \text {in } {\mathcal D}, \\ &{}\displaystyle f = 0, &{}\quad y = 0, \quad y= h, \\ &{} \displaystyle f = \frac{1}{2}\frac{p^{\prime }h^{2}}{\eta }\left( \frac{y}{h} - \left( \frac{y}{h}\right) ^{2}\right) , &{}\quad z = 0, \quad z = \ell . \end{array} \end{aligned}$$

(8.179)

Trying with the product ansatz $f(y, z) = f_{1}(y)\,f_{2}(z)$, the Laplace Eq. (8.179)$_{1}$ yields

$$\begin{aligned} \frac{f_{1}^{\prime \prime }}{f_{1}} = - \frac{f_{2}^{\prime \prime }}{f_{2}} = - \lambda ^{2} \quad \Longrightarrow \quad \left\{ \begin{array}{l} f_{1}^{\prime \prime } + \lambda ^{2} f_{1} = 0, \\ f_{2}^{\prime \prime } - \lambda ^{2}f_{2} = 0. \end{array} \right. \end{aligned}$$

(8.180)

Of the trigonometric functions

$$\begin{aligned} f_{1}^{(m)} = \sin \left( \frac{m\pi }{h}y\right) ,\; \cos \left( \frac{m \pi }{h}y\right) \end{aligned}$$

the cosine function violates the boundary conditions at $y=0$ and $y=h$. So, the correct solution will be a combination of the sine-functions for some $m = 1,2,3,\ldots ,\infty$. It is also obvious that $\lambda = m\pi /h$. With this value then follows that (8.180)$_{2}$ gives rise to the independent solution functions

$$\begin{aligned} \sinh \left( \frac{m\pi }{h}z\right) , \qquad \sinh \left( \frac{m\pi }{h}(\ell - z)\right) . \end{aligned}$$

(8.181)

These are chosen in this form, since the function f(y, z) must primarily be corrected where the trial solution (8.178) does not satisfy the boundary condition.

It follows that the linear combination

$$\begin{aligned} f =\sum _{m=1}^{\infty }\sin \left( \frac{m\pi }{h}y\right) \left\{ A_{m} \sinh \left( \frac{m\pi }{h}z\right) + B_{m}\sinh \left( \frac{m\pi }{h} (\ell -z)\right) \right\} .\qquad \end{aligned}$$

(8.182)

is in principle able to match the boundary conditions (8.179)$_{3}$; explicitly,

$$\begin{aligned} \sum _{m=1}^{\infty }\sin \left( \frac{m \pi }{h}y\right) B_{m} \sinh \left( \frac{m\pi \ell }{h}\right) = \frac{1}{2} \frac{p^{\prime }h^{2}}{\eta } \left( \frac{y}{h} - \left( \frac{y}{h}\right) ^{2}\right) , \end{aligned}$$

(8.183)

plus the same equation, in which $B_{m}$ is replaced by $A_{m}$. Thus, $B_{m} = A_{m}$ and these coefficients can be determined by standard Fourier series inversion. The result is given in (8.25).