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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 (8.35) can be found such that $ \psi $ is a function of $ y/x$ only. Demonstrate that this solution yields

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

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

    $\displaystyle F'' - F^{\,2} = -1,
$

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

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

    is a suitable solution of the previous 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

    $\displaystyle \nu\,\frac{\partial^{\,3}\psi}{\partial y^{\,3}}-\frac{\partial\p...
...rtial\psi}{\partial y}\,\frac{\partial^{\,2}\psi}{\partial x\,\partial y} = 0.
$

    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

    $\displaystyle u(x,y)= -\frac{\partial\psi}{\partial y},
$

    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,

    $\displaystyle M = \rho\int_{-\infty}^{\infty}u^{\,2}\,dy,
$

    to be independent of $ x$ .

    Consider a self-similar stream function of the form

    $\displaystyle \psi(x,y)=\psi_0\,x^{\,p}\,F(y/x^{\,q}).
$

    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

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

    Demonstrate that $ F(z)$ satisfies

    $\displaystyle F''' +2\,F\,F'' + 2\,F'^{\,2} = 0,
$

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

    $\displaystyle F(z)=\alpha\,\tanh(\alpha\,z)
$

    is a suitable solution, and that

    $\displaystyle M= 48\,\rho\,\nu^{\,2}\,\alpha^{\,3}.
$

  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

    $\displaystyle \nu\left.\frac{\partial u}{\partial y}\right\vert _{y=0} = U^{\,2}\,\frac{d\delta_2}{dx} + U\,\frac{dU}{dx}\,(\delta_1+2\,\delta_2) + U\,v_s.
$

    Suppose that

    $\displaystyle u(x,y) = U(x)\left\{\begin{array}{lcl}
\sin(\alpha\,y)&\mbox{\hsp...
...0\leq y\leq \pi/(2\,\alpha)\\ [0.5ex]
1&&y>\pi/(2\,\alpha)
\end{array}\right.,
$

    where $ \alpha=\alpha(x)$ . Demonstrate that the displacement and momentum widths of the boundary layer are

    $\displaystyle \delta_1$ $\displaystyle =(\pi/2-1)\,\alpha^{\,-1},$    
    $\displaystyle \delta_2$ $\displaystyle =(1-\pi/4)\,\alpha^{\,-1},$    

    respectively. Hence, deduce that

    $\displaystyle \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.
$

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

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

    but that in the presence of suction

    $\displaystyle \delta_1 = \frac{(\pi/2-1)\,\nu}{v_s}.
$

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

    $\displaystyle \frac{v_s}{U_0}\gg \frac{1}{{\rm Re}^{1/2}},
$

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

next up previous
Next: Incompressible Aerodynamics Up: Incompressible Boundary Layers Previous: Approximate Solutions of Boundary
Richard Fitzpatrick 2016-03-31