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Two-dimensional Laminar Jet.

We consider a two-dimensional jet as illustrated in the figure below. x is the horizontal coordinate and y is the vertical coordinate. u and v are, respectively, the horizontal and vertical fluid velocities. The jet in the direction of the x axis generates a flow where the fluid velocity along the y axis tends to zero. We assume that the boundary layer approximation is valid and the governing equation for the fluid motion are equations (2.21) to (2.23), but with $\partial u/\partial t$. The pressure does not vary in the y direction according to equation (2.22), so the pressure is constant across the boundary layer and its gradient is given by the pressure gradient outside the boundary layer. For this problem there is no pressure gradient, so the governing equations for the fluid motion are


 \begin{displaymath}u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} = \nu\frac{\partial^{2}u}{\partial y^{2}},
\end{displaymath} (146)


 \begin{displaymath}\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y} = 0
\end{displaymath} (147)

Next, we integrate equation (8.146) with respect to the y variable from $-\infty$ to $+\infty$, which gives


 \begin{displaymath}\int_{-\infty}^{\infty}(u \frac{\partial u}{\partial x}+v\fra...
...u\int_{-\infty}^{\infty}\frac{\partial^{2}u}{\partial y^{2}}dy
\end{displaymath} (148)

We can write


\begin{displaymath}v\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(vu)-u\frac{\partial v}{\partial y}, \notag
\end{displaymath}  

and if we multiply the continuity equation (8.147) by u, we can write the equation above as


 \begin{displaymath}v\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(vu)-u\frac{\partial u}{\partial x},
\end{displaymath} (149)

Next, we substitute equation (8.149) into equation (8.148), we perform the integration and we use the boundary condition


 \begin{displaymath}u \rightarrow 0 \mbox{\ as\ } y \rightarrow \pm \infty.
\end{displaymath} (150)

Equation (8.148) simplifies to


 \begin{displaymath}\frac{\partial}{\partial x}\int_{-\infty}^{+\infty}u^{2}dy = 0
\end{displaymath} (151)

Thus, the momentum flux is constant in x. In other words,


 \begin{displaymath}\rho\int_{-\infty}^{\infty}u^{2}dy = M
\end{displaymath} (152)

We call $\delta(x)$ the order of the magnitude of the boundary layer thickness at position x. Since the orifice is very small we have $\delta(0) \rightarrow 0$. At the orifice the equation (8.152) can be written as


 \begin{displaymath}\rho u(0)^{2}\delta(0) = M,
\end{displaymath} (153)

which implies that


 \begin{displaymath}u(0) = \delta(0)^{-1/2}
\end{displaymath} (154)

The Mass flux at the orifice is


 \begin{displaymath}\rho u(0)\delta(0) \sim \delta(0)^{1/2} \rightarrow 0,
\end{displaymath} (155)

hence unimportant. A jet is the result of a momentum source, not a volume source. Next, we are going to solve the boundary layer equations for the jet. We introduce the stream function $\psi$ related to the velocities u and v according to the equations


 \begin{displaymath}u = \frac{\partial\psi}{\partial y}
\end{displaymath} (156)


 \begin{displaymath}v = - \frac{\partial\psi}{\partial x}
\end{displaymath} (157)

In terms of the stream function, the x momentum equation (8.146) assume the form


 \begin{displaymath}\frac{\partial\psi}{\partial y}\frac{\partial^{2}\psi}{\parti...
...}{\partial y^{2}} = \nu\frac{\partial^{3}\psi}{\partial y^{3}}
\end{displaymath} (158)

with the boundary conditions


 \begin{displaymath}\psi \rightarrow 0 \mbox{\ as\ } y \rightarrow \pm\infty
\end{displaymath} (159)

and


 \begin{displaymath}\rho\int_{-\infty}^{\infty}\left(\frac{\partial\psi}{\partial y}\right)^{2}dy = M
\end{displaymath} (160)

We are going to solve the boundary value problem given by equations (8.158) to (8.160) by looking for a similarity solution. We look for a one-parameter transformation of variables y, x and $\psi$ under which the equations of the boundary value problem mentioned above are invariant. A particular useful transformation is


 \begin{displaymath}x = \lambda^{a}x', y = \lambda^{b}y' \mbox{\ and\ } \psi = \lambda^{c}\psi'
\end{displaymath} (161)

The requirement of invariance of the boundary value problem (8.158) to (8.160) under such transformation implies that


 
c+b-a = 0 (162)


 
c-b = 0 (163)

and no information is gained from equation (8.160). From the equations (8.162) and (8.163) we obtain


 \begin{displaymath}c = a/3 \mbox{\ and\ } b = 2a/3
\end{displaymath} (164)

This suggest that we take


 \begin{displaymath}\frac{\psi}{Bx^{1/3}} = f(\eta) \mbox{\ with\ } \eta = \frac{Cy}{x^{2/3}},
\end{displaymath} (165)

where the coefficients B and C are chosen to simplify the appearance of the final equation. We substitute (8.165) into equation (8.158), which gives the ordinary differential equation


 
3f'''+(f')2+ff'' = 0, (166)

and for $\psi$ and $\eta $ we have


 \begin{displaymath}\psi = \left(\frac{M\nu x}{\rho}\right)^{1/3}f(\eta) \mbox{\ with\ } \eta = \left(\frac{M}{\rho\nu^{2}x^{2}}\right)^{1/3}
\end{displaymath} (167)

the boundary conditions become


 \begin{displaymath}f'(\pm\infty) \rightarrow 0, f(0) = f''(0) = 0 \mbox{(symmetry)}
\end{displaymath} (168)

and


 \begin{displaymath}1 = \int_{-\infty}^{\infty}\vert f'(\eta)\vert^{2}d\eta
\end{displaymath} (169)

Now we integrate equation (8.166) once with respect to $\eta $, which gives


\begin{displaymath}3f''+ff' = \mbox{\ constant\ } = 0, \notag
\end{displaymath}  

and integrating again


 \begin{displaymath}3f'+\frac{1}{2}f^{2} = c^{2}
\end{displaymath} (170)

We now write $f = \sqrt{2}F$ and $\eta = 3\sqrt{2}\zeta$. then equation (8.170) assumes the form


 \begin{displaymath}\frac{dF}{d\zeta}+F^{2}= c^{2} \rightarrow \frac{dF/c}{1-(F/c)^{2}} = cd\zeta
\end{displaymath} (171)

which can be integrated in closed form, so we have


 \begin{displaymath}c\zeta = \tanh^{-1}(F/c)
\end{displaymath} (172)

since F(0) = 0. Thus


 \begin{displaymath}f = \sqrt{2}F = \sqrt{2}c\tanh\left(\frac{c\eta}{3\sqrt{2}}\right).
\end{displaymath} (173)

We substitute the equation above for f in the boundary condition (8.169), which gives


 \begin{displaymath}1 = \frac{c^{3}\sqrt{2}}{3}\int_{-\infty}^{\infty}\mbox{sech}...
...rac{4\sqrt{2}c^{3}}{9} \rightarrow c^{3} = \frac{9}{4\sqrt{2}}
\end{displaymath} (174)

Finally, we write


 \begin{displaymath}\chi = \left(\frac{M}{48\rho\nu^{2}}\right)^{1/3}\frac{y}{x^{2/3}}
\end{displaymath} (175)

The final solution has the form


 \begin{displaymath}\psi = \left(\frac{9M\nu x}{2\rho}\right)^{1/3}\tanh(\chi)
\end{displaymath} (176)

and for the velocities we have


 \begin{displaymath}u = \frac{\partial \psi}{\partial y} = \left(\frac{3M^{2}}{32\rho^{2}\nu x}\right)^{1/3}\mbox{sech}^{2}(\chi)
\end{displaymath} (177)


 \begin{displaymath}v = -\frac{\partial \psi}{\partial x} = \left(\frac{M\nu}{6\rho x^{2}}\right)^{1/3}(2\chi\mbox{sech}^{2}\chi-\tanh\chi)
\end{displaymath} (178)

Next, we discuss the implications of the results obtained above. The jet width can be defined by $\chi = \pm \chi_{0}$ such that $u \rightarrow 0$. Then from equation (8.177), we realize that the jet width is proportional to x2/3. The centerline velocity, let say umax, according to equation (8.177) is proportional to x-1/3. As $\chi \rightarrow \pm\infty$, according to equation (8.178), $v \rightarrow \mp (M\nu/6\rho x^{2})^{1/3}$. This implies a that there is entrainment from the jet eddges as show in the figure 4. If we define the Reynolds number as $R=u_{max}\delta/nu$, equation (8.177) implies that $R \propto x^{1/3}$. To illustrate the streamlines for the flow generated by the jet, we present figure 3. The velocity field is illustrated in figure 3.


  
Figure: Streamlines obtained from equation (8.176) with $M = 1000 kg/sec^{2}, \nu = 0.01 m^{2}/sec$ and $\rho = 1000 kg/m^{3}$.
\includegraphics[width=5.5in,height=5.5in]{ps/psi.eps}


  
Figure: Velocity field obtained from equations (8.177) and (8.178) with $M = 1000 kg/sec^{2}, \nu = 0.01 m^{2}/sec$ and $\rho = 1000 kg/m^{3}$.
\includegraphics[width=5.5in,height=5.5in]{ps/v_field.eps}


next up previous
Next: About this document ... Up: No Title Previous: Stagnation Point flow.
Karl P Burr
2003-03-12