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

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:

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 $f''(\eta )$.

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 up previous
Next: Two-dimensional Laminar Jet. Up: No Title Previous: Approximate Method Based on
Karl P Burr