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.