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Next: Incompressible Aerodynamics Up: Incompressible Boundary Layers Previous: Approximate Solutions of Boundary

Exercises

  1. Fluid flows between two non-parallel plane walls, towards the intersection of the planes, in such a manner that if $x$ is measured along a wall from the intersection of the planes then $U(x)=-U_0/x$, where $U_0$ is a positive constant. Verify that a solution of the boundary layer equation (704) can be found such that $\psi $ is a function of $y/x$ only. Demonstrate that this solution yields

    \begin{displaymath}
\frac{u(x,y)}{U(x)} = F\left[\left(\frac{U_0}{\nu}\right)^{1/2}\,\frac{y}{x}\right],
\end{displaymath}

    where $u=\partial\psi/\partial y$, and

    \begin{displaymath}
F'' - F^{\,2} = -1,
\end{displaymath}

    subject to the boundary conditions $F(0)=0$ and $F(\infty)=1$. Verify that

    \begin{displaymath}
F(z) = 3\,\tanh^2\left(\alpha+\frac{z}{\sqrt{2}}\right)-2
\end{displaymath}

    is a suitable solution of the above differential equation, where $\tanh^2\alpha =2/3$.

  2. A jet of water issues from a straight narrow slit in a wall, and mixes with the surrounding water, which is at rest. On the assumption that the motion is non-turbulent and two-dimensional, and that the approximations of boundary layer theory apply, the stream function satisfies the boundary layer equation

    \begin{displaymath}
\nu\,\frac{\partial^3\psi}{\partial y^3}-\frac{\partial\psi}...
...partial y}\,\frac{\partial^2\psi}{\partial x\,\partial y} = 0.
\end{displaymath}

    Here, the symmetry axis of the jet is assumed to run along the $x$-direction, whereas the $y$-direction is perpendicular to this axis. The velocity of the jet parallel to the symmetry axis is

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

    where $u(x,-y)=u(x,y)$, and $u(x,y)\rightarrow 0$ as $y\rightarrow\infty$. We expect the momentum flux of the jet parallel to its symmetry axis,

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

    to be independent of $x$.

    Consider a self-similar stream function of the form

    \begin{displaymath}
\psi(x,y)=\psi_0\,x^p\,F(y/x^q).
\end{displaymath}

    Demonstrate that the boundary layer equation requires that $p+q=1$, and that $M$ is only independent of $x$ when $2\,p-q=0$. Hence, deduce that $p=1/3$ and $q=2/3$.

    Suppose that

    \begin{displaymath}
\psi(x,y)=-6\,\nu\,x^{1/3}\,F(y/x^{2/3}).
\end{displaymath}

    Demonstrate that $F(z)$ satisfies

    \begin{displaymath}
F''' +2\,F\,F'' + 2\,F'^{\,2} = 0,
\end{displaymath}

    subject to the constraints that $F'(-z)=F'(z)$, and $F'(z)\rightarrow 0$ as $z\rightarrow\infty$. Show that

    \begin{displaymath}
F(z)=\alpha\,\tanh(\alpha\,z)
\end{displaymath}

    is a suitable solution, and that

    \begin{displaymath}
M= 48\,\rho\,\nu^2\,\alpha^3.
\end{displaymath}

  3. The growth of a boundary layer can be inhibited by sucking some of the fluid through a porous wall. Consider conventional boundary layer theory. As a consequence of suction, the boundary condition on the normal velocity at the wall is modified to $v(x,0)=-v_s$, where $v_s$ is the (constant) suction velocity. Demonstrate that, in the presence of suction, the von Kármán velocity integral becomes

    \begin{displaymath}
\nu\left.\frac{\partial u}{\partial y}\right\vert _{y=0} = U...
...ta_2}{dx} + U\,\frac{dU}{dx}\,(\delta_1+2\,\delta_2) + U\,v_s.
\end{displaymath}

    Suppose that

    \begin{displaymath}
u(x,y) = U(x)\left\{\begin{array}{ccl}
\sin(\alpha\,y)&\mbox...
...(2\,\alpha)\\ [0.5ex]
1&&y>\pi/(2\,\alpha)
\end{array}\right.,
\end{displaymath}

    where $\alpha=\alpha(x)$. Demonstrate that the displacement and momentum widths of the boundary layer are
    $\displaystyle \delta_1$ $\textstyle =$ $\displaystyle (\pi/2-1)\,\alpha^{-1},$  
    $\displaystyle \delta_2$ $\textstyle =$ $\displaystyle (1-\pi/4)\,\alpha^{-1},$  

    respectively. Hence, deduce that

    \begin{displaymath}
\frac{\nu\,(\pi/2-1)^2}{\delta_1} = U\,(1-\pi/4)\,\frac{d\delta_1}{dx} + \frac{dU}{dx}\,\delta_1+(\pi/2-1)\,v_s.
\end{displaymath}

    Consider a boundary layer on a flat plate, for which $U(x) = U_0$. Show that, in the absence of suction,

    \begin{displaymath}
\delta_1 = (\pi/2-1)\left(\frac{8}{4-\pi}\right)^{1/2}\left(\frac{\nu\,x}{U_0}\right)^{1/2},
\end{displaymath}

    but that in the presence of suction

    \begin{displaymath}
\delta_1 = \frac{(\pi/2-1)\,\nu}{v_s}.
\end{displaymath}

    Hence, deduce that, for a plate of length $L$, suction is capable of significantly reducing the thickness of the boundary layer when

    \begin{displaymath}
\frac{v_s}{U_0}\gg \frac{1}{{\rm Re}^{1/2}},
\end{displaymath}

    where ${\rm Re} = U_0\,L/\nu$.

next up previous
Next: Incompressible Aerodynamics Up: Incompressible Boundary Layers Previous: Approximate Solutions of Boundary
Richard Fitzpatrick 2012-04-27