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Lifting flow over a cylinder

Lifting flow over a circular cylinder can be synthesised as superposition of a vortex flow of strength \('\tau '\) with non lifting flow over a cylinder.
The stream function for the flow can be given as\[\psi  = \left( {{V_\infty }r\sin \theta } \right)\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right) + \frac{\tau }{{2\pi }}\ln \frac{r}{R}\]and velocity components in the radial and tangential direction can be obtained as \[{V_r} = \left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\cos \theta \] \[{V_\theta } =  - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\sin \theta  - \frac{\tau }{{2\pi r}}\]The velocity on the surface of the cylinder can be given as  \[V =  - 2{V_\infty }\sin \theta  - \frac{\tau }{{2\pi R}}\] and coefficient of lift as \[{c_l} = \frac{\tau }{{R{V_\infty }}}\]Lift per unit span is given as \[{L^\prime} = {\rho _\infty }{V_\infty }\tau \]
Example: Calculate the peak coefficient of pressure for a lifting flow over a circular cylinder for which lifting coefficient  is 70.
Solution: Maximum velocity for lifting flow over a circular cylinder occurs at top surfaces and is equal to summation of non-lifting value of maximum flow velocity,\( - 2{V_\infty }\) and vortex \(\frac{{ - \tau }}{{2\pi R}}\) therefore,\[V =  - 2{V_\infty } - \frac{\tau }{{2\pi R}}\]coefficient of lift is \[{c_l} = \frac{\tau }{{R{V_\infty }}} = 7\]therefore, \(\frac{\tau }{R} = 7{V_\infty }\); On substitution the value of \(\frac{\tau }{R}\) in the velocity equation, we get \[V =  - 2{V_\infty } - \frac{7}{{2\pi }}{V_\infty } =  - 3.11{V_\infty }\]therefore peak coefficient of pressure will be \[{C_p} = 1 - {\left( {\frac{V}{{{V_\infty }}}} \right)^2} = 1 - {\left( { - \frac{{3.11{V_\infty }}}{{{V_\infty }}}} \right)^2}\]\[ = 1 - {\left( { - 3.11} \right)^2} =  - 8.67\]Example: Calculate the lift per unit span for a lifting flow over a circular cylinder with a diameter of 0.8 m, having a free stream velocity of 35 m/s and maximum velocity on the surface of the cylinder being 80 m/s ,considering freestream sea level conditions.
Solution: At sea level, density of air = \(1.225\,kg/{m^3}\).The maximum velocity for a lifting flow over a circular cylinder occurs at top surface of the cylinder, where \(\theta \) equals to \({90^ \circ }\).\[{V_\theta } =  - 2{V_\infty }\sin \theta  - \frac{\tau }{{2\pi R}}\]\[ \Rightarrow \tau  =  - 2\pi R\left( {{V_\theta } + 2{V_\infty }} \right)\]\({V_\theta }\) is negative in clockwise direction and \('\tau '\) is positive, according to sign conventions.Therefore, \[\tau  =  - 2\pi R\left( {{V_\theta } + 2{V_\infty }} \right) =  - 2\pi \left( {0.4} \right)\left[ { - 80 + 2\left( {35} \right)} \right]\]\[\tau  =  - 2\pi \left( {0.4} \right)\left[ { - 10} \right]\]\[\tau  = 25.133\,{m^2}/s\]Therefore, lift per unit span is \[{L^\prime} = {\rho _\infty }{V_\infty }\tau \]\[{L^\prime} = 1.225 \times 35 \times 25.133\]\[ = 1077.577\,N/m\]Example: If the velocity \({V_\infty }\) of a flow is doubled in the case of lifting flow over a circular cylinder of a given radius and with a given circulation, does the shape of the streamline change on keeping the circulation same.
Solution: Since for the case of lifting flow over a circular cylinder \[\left( {\frac{{{V_r}}}{{{V_\infty }}}} \right) = \left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)\cos \theta \]and\[\left( {\frac{{{V_\theta }}}{{{V_\infty }}}} \right) =  - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right)\sin \theta  - \frac{\tau }{{2\pi r{V_\infty }}}\]Since\(\left( {\frac{{{V_\theta }}}{{{V_\infty }}}} \right)\) is a function of \({V_\infty }\), so as \({V_\infty }\) changes, the direction of the resultant velocity at a given point also changes.Therefore as \({V_\infty }\) is changing the shape of streamlines also changes.
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