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Stagnation Point flow.

For an ideal fluid the flow against an infinite flat plate in the plane y = 0 is given by

u = Ux,

(122)

v = -Uy,

(123)

where U is a constant. When viscosity is included, it still must be true that u is proportional to x, for small x and for all y.Thus, for small x, at least, we may take

u = kxF(y)

(124)

The governing equations for steady flow in terms of dimensional variables are given by equations (2.1) to (2.3). If we substitute equation (7.124) into the continuity equation (2.3), we obtain

$\begin{displaymath}\frac{\partial v}{\partial y} = -kF(y) \end{displaymath}$

(125)

This suggest that we take

u = kxf'(Ay),

(126)

v = -Bf(Ay).

(127)

Now, we substitute equations (7.126) and (7.127) into the governing equations (2.1) to (2.3). We obtain from

$\begin{displaymath}% latex2html id marker 1479 \mbox{(\ref{eq:ch21a})} \rightarr... ...rac{1}{\rho}\frac{\partial p}{\partial x}+\nu kxA^{2}f'''(Ay), \end{displaymath}$

(128)

$\begin{displaymath}% latex2html id marker 1490 \mbox{(\ref{eq:ch21b})} \rightarr... ...\frac{1}{\rho}\frac{\partial p}{\partial y}-\nu Bf''(Ay)A^{2}, \end{displaymath}$

(129)

$\begin{displaymath}% latex2html id marker 1500 \mbox{(\ref{eq:ch21c})} \rightarrow kf'(Ay)-BAf'(Ay) = 0. \end{displaymath}$

(130)

This last equation implies that BA = k, so we can write equations (7.128) and (7.130) in the form

$\begin{displaymath}% latex2html id marker 1506 \mbox{(\ref{eq:ch238g})} \rightar... ...f=-\frac{1}{\rho}\frac{\partial p}{\partial x}+\nu kxA^{2}f''' \end{displaymath}$

(131)

$\begin{displaymath}% latex2html id marker 1518 \mbox{(\ref{eq:ch238h})} \rightarrow Bkff'=-\frac{1}{\rho}\frac{\partial p}{\partial y}-\nu kAf'' \end{displaymath}$

(132)

We can solve equation (7.131) for the pressure derivative with respect to x to obtain

$\begin{displaymath}-\frac{1}{\rho}\frac{\partial p}{\partial x} = k^{2}x[(f')^{2}-ff''-\frac{\nu}{k}A^{2}f'''] \end{displaymath}$

(133)

Since the the term between brackets in the right side of the equation above is not a function of x, we set

$\begin{displaymath}(f')^{2}-ff''-\frac{\nu}{k}A^{2}f''' = 1, \end{displaymath}$

(134)

which implies, according to equation (7.133), that

$\begin{displaymath}-\frac{1}{\rho}\frac{\partial p}{\partial x} = k^{2}x \rightarrow -\frac{1}{\rho}p = \frac{1}{2}k^{2}x^{2}+C(y) \end{displaymath}$

(135)

If we substitute equation (7.135) into equation (7.132), we have the relation

$\begin{displaymath}Bkff' = \frac{dC}{dy}-\nu kAf'', \end{displaymath}$

(136)

and if we integrate this equation, we obtain that

$\begin{displaymath}C(y) = \frac{1}{2}B\frac{k}{A}f^{2}+\nu kf'-C_{0}. \end{displaymath}$

(137)

Next, we substitute the expression for C(y) above into the equation (7.135) for the pressure, which gives

$\begin{displaymath}-\frac{1}{\rho}p = \frac{1}{2}k^{2}x^{2}+\frac{1}{2}B\frac{k}{A}f^{2}+\nu kf'-C_{0}. \end{displaymath}$

(138)

To simplify equation (7.134), we chose the coefficient of f''' equal to one, which implies that

$\begin{displaymath}\nu \frac{A^{2}}{k} = 1 \rightarrow A = \sqrt{\frac{k}{\nu}}, \end{displaymath}$

(139)

and since BA = k, we have that

$\begin{displaymath}B = \sqrt{\nu k}. \end{displaymath}$

(140)

With equations (7.138) and (7.139), we can write the equation (7.138) as follows

$\begin{displaymath}\frac{p_{0}-p}{\rho} = \frac{1}{2}k^{2}x^{2}+\frac{1}{2}\nu kf^{2}+\nu kf' \end{displaymath}$

(141)

with $C_{0} = p_{0}/\rho$ and p₀ is the pressure as $y\rightarrow \infty$ . The function $f(\eta)$ , where $\eta = \sqrt{k/\nu}y$ satisfies the ordinary differential equation

(f')²-ff''-f'''-1 = 0

(142)

with boundary conditions:

No slip condition at the plate surface.

$\begin{displaymath}u = 0 \mbox{\ at\ } y = 0 \rightarrow f'(\eta) = 0 \mbox{\ at\ } \eta = 0, \end{displaymath}$ (143)
No flux across the plate:

$\begin{displaymath}v = 0 \mbox{\ at\ } y = 0 \rightarrow f(\eta) = 0 \mbox{\ at\ } \eta = 0, \end{displaymath}$ (144)
At $y\rightarrow \infty$ , we have u = kx, which implies that

$\begin{displaymath}f'(\eta) = 1 \mbox{\ as\ } \eta \rightarrow \infty \end{displaymath}$ (145)

In summary, the function $f(\eta)$ is the solution of the boundary value problem given by the equations (7.142) to (7.145), which has no closed form solution. The equation ordinary differential equation (7.142) is non-linear and has to be solved numerically together with the boundary conditions (7.143) to (7.145). Figure 2 below illustarte the result of the numerical evaluation of the boundary value problem given by equations (7.142) to (7.145) for $\eta$ in the range $0 < \eta < 10$ .

**Figure:** Functions $f(\eta ), f'(\eta )$ and $f''(\eta )$ . The horizontal axis represents the range of values of $\eta$ considered, and in the vertical axis we have the values of the functions $f(\eta ), f'(\eta )$ and $f''(\eta )$ .
$\includegraphics[width=5.5in,height=5.5in]{ps/func_ff1f2a.eps}$

Once we know the values of $f(\eta ), f'(\eta )$ and $f''(\eta )$ , we can obtain the velocities u and v given, respectivelly, by equations (7.126) and (7.127) with B and A given, respectively, by equations (7.139) and (7.140), the pressure field is given by equation (7.141).

Next: Two-dimensional Laminar Jet. Up: No Title Previous: Approximate Method Based on

Karl P Burr
2003-03-12