# 10.3.1.1: Adding Circulation to a Cylinder

- Page ID
- 1292

The cylinder discussed in the previous sections was made from a dipole in a uniform flow field. It was demonstrated that in the potential flow has no resistance, and no lift due to symmetry of the pressure distribution. Thus, it was suggested that by adding an additional component that it would change the symmetry but not change the shape and hence it would provide the representation cylinder with lift. It turned out that this idea yields a better understanding of the one primary reason of lift. This results was verified by the experimental evidence. The linear characteristic (superposition principle) provides by adding the stream function of the free vortex to the previous the stream function for the case. The stream function in this case (see equation (93)) is

\[

\label{if:eq:doubletVortex}

\psi = U_0\, r \,\sin \theta \left( 1 - \left(\dfrac {r}{a}\right)^2 \right)

+ \dfrac{\Gamma }{2\,\pi} \ln \dfrac{a}{r} \tag{116}

\]

It can be noticed that this stream function (116) on the body is equal to \(\psi(r=a)=0\). Hence, the shape of the body remains a circle. The corresponding radial velocity in cylindrical coordinates (unchanged) and is

\[

\label{if:eq:doubletVortex:UrIni}

U_r = \dfrac{1}{r} \dfrac{\partial \psi}{\partial \theta} =

U_0\, \cos \theta \, \left( 1 - \left(\dfrac {a}{r}\right)^2 \right) \tag{117}

\]

The tangential velocity is changed (add velocity at the top and reduce velocity at the bottom or vice versa depending of the sign of the \(\Gamma\)) to be

\[

\label{if:eq:doubletVortex:Ur}

U_{\theta} = - \dfrac{\partial \psi}{\partial r} =

U_0\, \sin \theta \, \left( 1 + \left(\dfrac {a}{r}\right)^2 \right) +

\dfrac{\Gamma }{2\,\pi\,r} \tag{118}

\]

As it was stated before, examination of the stream function \(\psi=0\) is constructed. As it was constructed and discussed earlier it was observed that the location of stagnation stream function is on \(r=a\). On this line, equation (116) can be written as

\[

\label{if:eq:doubletVortex0}

0 = U_0\, r \,\sin \theta \left( 1 - \left(\dfrac {a}{r}\right)^2 \right)

+ \dfrac{\Gamma }{2\,\pi} \ln \dfrac{a}{r} \tag{119}

\]

or

\[

\label{if:eq:doubletVortex0a}

\sin \theta = - \dfrac{\dfrac{\Gamma }{2\,\pi} \ln \dfrac{r}{a} }

{ U_0\, r\, \left( 1 - \left(\dfrac {a}{r}\right)^2 \right) } =

\dfrac{\Gamma }{ 4\,\pi \,U_0\,\dfrac{r}{a}\, a}

\dfrac{2 \ln \dfrac{a}{r} }

{ \left( 1 - \left(\dfrac {a}{r}\right)^2 \right) } = \\

\dfrac{\Gamma }{ 4\,\pi \,U_0\,\dfrac{r}{a}\, a}

\dfrac{ \ln \left( \dfrac{a}{r} \right)^2 }

{ 1 - \left(\dfrac {a}{r}\right)^2 } =

\dfrac{\Gamma }{ 4\,\pi \,U_0\,\overline{r}\, a}

\dfrac{ \ln \left( \dfrac{1}{\overline{r}} \right)^2 }

{ 1 - \left(\dfrac {1}{\overline{r}}\right)^2 } \qquad \tag{120}

\]

At the point \(r=a\) the ratio in the box is approaching \(0/0\) and to examine what happen to it L'Hopital's rule can be applied. The examination can be simplified by denoting \(\xi= (a/r)^2 = \overline{r}\) and noticing that \(\xi=1\) at that point and hence

\[

\label{if:eq:doubletVortex0b}

\lim_{\xi\rightarrow 1} \dfrac{ \ln \, \xi }{ 1 - \xi } =

\lim_{\xi\rightarrow 1} \dfrac{ \dfrac{1}{\xi} }{ - 1} = - 1 \tag{121}

\]

Hence, the relationship expressed in equation (119) as

\[

\label{if:eq:doubletVortex00}

\sin \theta = \dfrac{-\Gamma}{4\,\pi\, U_0\, a} \tag{122}

\]

This condition (122) limits the value of maximum circulation on the body due to the maximum value of sin function. The doublet strength maximum strength can be The condition

\[

\label{if:eq:doubletVortex:limit}

\left| \Gamma \right| \le 4\,\pi\, U_0\, a \tag{123}

\]

The value of doublet strength determines the stagnation points (which were moved by the free vortex so to speak). For example, the stagnation points for the value \(\Gamma= -2\,\sqrt{2- \sqrt{3}} \,\pi\,U_0\,a\) can be evaluated as

\[

\label{if:eq:doubletVortex:exmpleValue15}

\sin \theta = \dfrac{\overbrace{ box{$

{2\,\sqrt{2- \sqrt{3}} \,\pi\,U_0\,a}$ } }^{-\Gamma} }{4\,\pi\, U_0\, a} = \dfrac{\sqrt{2- \sqrt{3}}}{2} \tag{124}

\]

The solution for equation (theta, \(\theta\)) (124) is \(15^{\circ}\) or \(\pi/12\) and \(165^{\circ}\) or \(11\,\pi/12\). For various stagnation points can be found in similar way. The rest of the points of the stagnation stream lines are found from the equation (120). For the previous example with specific value of the ratio, \(\overline{\Gamma}\) as

\[

\label{if:eq:doubletVortex0a0}

\sin\theta =

\dfrac{ \sqrt{2- \sqrt{3}}\, a}{2\,r}\dfrac{ \ln \left( \dfrac{a}{r} \right)^2 }

{ 1 - \left(\dfrac {a}{r}\right)^2 } \tag{125}

\]

There is a special point where the two points are merging \(0\) and \(\pi\). For all other points stream function can be calculated from equation (116) can be written as

\[

\label{if:eq:doubletVortexPsi}

\dfrac{\psi}{U_0\,a} = \dfrac{r}{a} \,\sin \theta \left( 1 - \left(\dfrac {a}{r}\right)^2 \right)

+ \dfrac{\Gamma }{2\,\pi\,U_0\,a} \ln \dfrac{r}{a} \tag{126}

\]

or in a previous dimensionless form plus multiply by \(\overline{r}\) as

\[

\label{if:eq:doubletVortexPsiInterd}

\dfrac{\overline{r}\,\overline{\psi}}{\sin \theta} = \overline{r}^2 \,

\left( 1 - \left(\dfrac {1}{\overline{r}}\right)^2 \right)

+ \,\dfrac{\Gamma \overline{r}}{2\,\pi\,U_0\,a\,\sin \theta}

\ln {\overline{r}} \tag{127}

\]

After some rearrangement of moving the left hand side to right and denoting \(\overline{\Gamma}=\dfrac{\Gamma}{4\,\pi\,U_0\,a}\) along with the previous definition of \(\overline{\psi}=2\,n\) equation (127) becomes

\[

\label{if:eq:doubletVortexPsiGov}

0 = \overline{r}^2 - \dfrac{\overline{r}\,\overline{\psi}}{\sin \theta} - 1

+ \dfrac{2\,\overline{\Gamma}\, \overline{r} \,\ln\overline{r} }{\sin \theta} \tag{128}

\]

Note the sign in front the last term with the \(\Gamma\) is changed because

the ratio in the logarithm is reversed. The stagnation line occur when \(n=0\) hence equation (128) satisfied for all \(\overline{r}=1\) regardless to value of the \(\theta\). However, these are not the only solutions. To obtain the solution equation (stagnation line) (128) is rearranged as

\[

\label{if:eq:doubletVortexPsiGovR}

\theta = \sin^{-1} \left( \dfrac {2\,\overline{\Gamma}\, \overline{r} \,\ln\overline{r} }

{1 - \overline{r}^2} \right) \tag{129}

\]

Equation (128) has three roots (sometime only one) in the most zone and parameters. One roots is in the vicinity of zero. The second roots is around the one (1). The third and the largest root which has the physical meaning is obtained when the dominate term \(\overline{r}^2\) "takes'' control.

*Figure 10.2 Doublet in a uniform flow with Vortex in various conditions. Typical condition for the dimensionless Vortex below on and dimensionless vortex equal to one. The figures were generated by the GLE and the program will be available on the on–line version of the book.*

Example 10.8

Example 10.9

Expand the GLE provided code to cover the case where the dimensionless vortex is over one (1).

### Pressure Distribution Around the solid Body

The interesting part of the above analysis is to find or express the pressure around the body. With this expression the resistance and the lift can be calculated. The body reacts to static pressure, as opposed to dynamic pressure, and hence this part of the pressure needed to be evaluated. For this process the Bernoulli's equation is utilized and can be written as

\[

\label{if:eq:BernoulliEq}

P_{\theta} = P_0 - \dfrac{1}{2} \rho \left( {U_r}^2 + {U_{\theta}}^2 \right) \tag{130}

\]

It can be noticed that the two cylindrical components were accounted for. The radial component is zero (no flow cross the stream line) and hence the total velocity is the tangential velocity (see equation (118) where \(r=a\)) which can be written as

\[

\label{if:eq:tangentialVortex}

U_{\theta} = 2\,U_0\,\sin\theta + \dfrac{\Gamma}{2\,\pi\,a} \tag{131}

\]

Thus, the pressure on the cylinder can be written as

\[

\label{if:eq:vortex:P}

P = P_0 - \dfrac{1}{2} \, \rho \left( 4\,{U_0}^2\, {\sin^2\theta} +

\dfrac{2\,U_0\,\Gamma\,\sin\theta}{\pi\,a} +

\dfrac{\Gamma^2}{4\,\pi^2\,a^2}

\right) \tag{132}

\]

Equation (132) is a parabolic equation with respect to \(\theta\) (\(\sin\theta\)). The symmetry dictates that D'Alembert's paradox is valid i.e that there is no resistance to the flow. However, in this case there is no symmetry around \(x\) coordinate (see Figure 10.16). The distortion of the symmetry around \(x\) coordinate contribute to lift and expected. The lift can be calculated from the integral around the solid body (stream line) and taking only the \(y\) component. The force elements is

\[

\label{if:eq:dF}

dF = -\mathbf{j} \,\mathbf{\cdot}\,P\, \mathbf{n} dA \tag{133}

\]

where in this case \(\mathbf{j}\) is the vertical unit vector in the downward direction, and the infinitesimal area has direction which here is broken into in the value \(dA\) and the standard direction \(\mathbf{n}\). To carry the integration the unit vector \(\mathbf{n}\) is written as

\[

\label{if:eq:unitVector}

\mathbf{n} = \mathbf{i}\,\cos\theta + \mathbf{j}\,\sin\theta \tag{134}

\]

The reason for definition or split (134) to take into account only the the vertical component. Using the above derivation leads to

\[

\label{if:eq:vortex:dotV}

\mathbf{j}\mathbf{\cdot}\mathbf{n} = \sin\theta \tag{135}

\]

The lift per unit length will be

\[

\label{if:eq:vortex:liftIntegralIni}

L = - \int_0^{2\,\pi} \left[P_0 - \dfrac{1}{2} \, \rho \left( 4\,{U_0}^2\, {\sin^2\theta} +

\dfrac{2\,U_0\,\Gamma\,\sin\theta}{\pi\,a} +

\dfrac{\Gamma^2}{4\,\pi^2\,a^2} \right) \right] \overbrace{\sin\theta}^

{\scriptscriptstyle eq. \eqref{if:eq:vortex:dotV}}

\,a\,d\theta \tag{136}

\]

Integration of the \(\sin\theta\) in power of odd number between \(0\) and \(2\,\pi\)is zero. Hence the only term that left from the integration (136) is

\[

\label{if:eq:vortex:integral}

L = - \dfrac{\rho\,U_0\,\Gamma}{\pi\,a}\,\int_0^{2\,\pi} \sin^2\theta d\theta

= U_0\,\rho\,\Gamma \tag{137}

\]

The lift created by the circulating referred as the Magnus effect which name after a Jewish scientist who live in Germany who discover or observed this phenomenon. In fact, physicists and engineers dismiss this phenomenon is ``optical illusion.'' However, the physical explanation is based on the viscosity and the vortex is the mechanism that was found to transfer the viscosity to inviscid flow.

### Contributors

Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.