Non Lifting flow over a circular cylinder

When there is a superposition of a uniform flow with a doublet (Which is a source-sink pair of flows), a non-lifting flow over a circular cylinder can be analysed.

Stream function $'\psi '$ for a uniform flow is  $$\psi  = \left( {{V_\infty }r\sin \theta } \right)\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)$$ velocity field is obtained by differentiating above equation $${V_r} = \left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\cos \theta$$ $$\,{\rm{and}}\,{{\rm{V}}_\theta } =  - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\sin \theta$$

Stagnation points can be obtained by equating ${V_r}$ and ${{\rm{V}}_\theta }$ to zero.Considering an incompressible and inviscid flow pressure coefficient over a circular cylinder is  $${C_p} = 1 - 4{\sin ^2}\theta$$ Example: Calculate locations on the surface of a cylinder where surface pressure equals to free-stream pressure considering a non-lifting flow.

Solution: Pressure coefficient is given as  $${C_p} = \frac{{p - {p_\infty }}}{{{q_\infty }}}$$ Since $p = {p_\infty }$ therefore ${C_p} = 0$.Also,  $${C_p} = 1 - 4{\sin ^2}\theta  = 0$$ $$\Rightarrow \sin \theta  =  \pm \frac{1}{2}$$ $$\Rightarrow \theta  = {30^ \circ },{150^ \circ },{210^ \circ },{330^ \circ }$$ are the locations on the surface of cylinder.
Example: Consider a non-lifting flow over a circular cylinder and derive an expression for the pressure coefficient at any arbitrary point $\left( {r,\theta } \right)$ in the flow.On the surface of the cylinder show that it reduces to ${C_p} = 1 - 4{\sin ^2}\theta$.
Solution: The stream function for a non lifting flow over a circular cylinder is given as $$\psi  = \left( {{V_\infty }r\sin \theta } \right)\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)$$   $${V_r} = \frac{1}{r}\frac{{\partial \psi }}{{\partial \theta }} = \left( {{V_\infty }\cos \theta } \right)\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)$$ $${V_\theta } =  - \frac{{\partial \psi }}{{\partial r}} =  - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\sin \theta$$ $${V^2} = V_r^2 + V_\theta ^2 = \left( {V_\infty ^2{{\cos }^2}\theta } \right){\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)^2} + {\left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right)^2}V_\infty ^2{\sin ^2}\theta$$ Coefficient of pressure ${C_p}$ will be  $${C_p} = 1 - \frac{{{V^2}}}{{V_\infty ^2}} = 1 - {\cos ^2}\theta {\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)^2} + {\left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right)^2}{\sin ^2}\theta$$ $${C_p} = 1 - {\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)^2}{\cos ^2}\theta  - {\left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right)^2}{\sin ^2}\theta$$ This represents coefficient of pressure at any arbitrary point in the flow.At, the surface of cylinder r = R, therefore,  $${C_p} = 1 - \frac{{{V^2}}}{{V_\infty ^2}} = 1 - 4{\sin ^2}\theta$$ Example: Consider the non lifting flow over a circular cylinder of a given radius, where free stream velocity is ${V_\infty }$.If ${V_\infty }$ is doubled,explain if there is any change in the shape of the stream lines.
Solution:  $${V_r} = \frac{1}{r}\frac{{\partial \psi }}{{\partial \theta }} = \left( {{V_\infty }\cos \theta } \right)\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)$$ and  $${V_\theta } =  - \frac{{\partial \psi }}{{\partial r}} =  - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\sin \theta$$ therefore,  $$\frac{{{V_r}}}{{{V_\infty }}} = \left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)\cos \theta$$ $$\frac{{{V_\theta }}}{{{V_\infty }}} =  - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right)\sin \theta$$ So, at any given point $\left( {r,\theta } \right)$, ${V_r}$ and ${{V_\theta }}$ are both directly proportional to${{V_\infty }}$.Therefore, the direction of the resultant,$\mathop V\limits^ \to$ is the same, for any value of ${{V_\infty }}$.This infers that the shape of the streamlines remains the same.