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Next: 3.9 Forces on a Up: 3.8 - Method of Previous: Solution: method of image

Solution: method of image for a circle in a free stream near a wall - boundary condition on the circle.

First, we obtain the $x$ and $y$ components of the velocity vector.


\begin{displaymath}V_{y} = \frac{\partial \phi}{\partial y} = Ux\left(-\frac{2a^... ...^2)^2}-\frac{2a^{2}(y+b)}{(x^{2}+(y+b)^{2})^{2}}\right) \notag \end{displaymath}  

and


\begin{displaymath}V_{x} = \frac{\partial\phi}{\partial x} = U\left( {1 + \frac{... ...y-b)^2)^2}-\frac{2a^{2}x}{(x^{2}+(y+b)^{2})^{2}}\right) \notag \end{displaymath}  

Second, we make the change of variables


\begin{align}x = & r\cos\theta, \notag \\ y = & b+r\sin\theta. \notag \end{align}

We substitute the expressions for $x$ and $y$ in terms of $r$ and $\theta$ in the equations for $v_{x}$ and $v_{y}$. Then, we set $r = a$, and we define $p = \cos\theta, q = \sin\theta$ and $t = b/a$. Now we write


\begin{displaymath}\frac{v_{x}}{U} = \left(1+\frac{1}{p^{2}+q^{2}}+\frac{1}{p^{2... ...2}+q^{2})^{2}}-\frac{2p}{(p^{2}+(q+2t)^{2})^{2}}\right) \notag \end{displaymath}  

and


\begin{displaymath}\frac{v_{y}}{U} = p\left(-\frac{2q}{(p^{2}+q^{2})^{2}}-\frac{2(q+2t)}{(p^{2}+(q+2t)^{2})^{2}}\right). \notag \end{displaymath}  

Third, evaluate the radial velocity $V_{r}$ at the circles surface. The radial velocity $V_{r}$ can be written in terms of the velocities $V_{x}$ and $V_{y}$ according to the equation


\begin{align}V_{r} = & V_{x}(r,\theta)\frac{\partial x}{\partial r}+V_{y}(r,\the... ... \\ = & V_{x}(r,\theta)\cos\theta+V_{y}(r,\theta)\sin\theta \notag \end{align}

With this equation and the equations for $V_{x}$ and $V_{y}$ at the surface of the circle, we have


\begin{displaymath}\begin{split} V_{r}\vert _{r=a} & = p\left(1+\frac{1}{p^{2}+q... ...rac{2(q+2t)}{(p^{2}+(q+2t)^{2})^{2}}\right) \end{split} \notag \end{displaymath}  

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Next: 3.9 Forces on a Up: 3.8 - Method of Previous: Solution: method of image