Next:
IdealFlow3
Previous:
IdealFlow1
Chapter If (continue)
10.3 Potential Flow Functions Inventory
This section describes several simple scenarios of the flow field.
These flow fields will be described and exhibits utilizition of the potential and
stream functions.
These flow fields can be combined by utilizing superimposing principle.
Uniform Flow
The trivial flow is the uniform flow in which the fluid field moves directly and uniformly
from one side to another side.
This flow is further simplified, that is the coordinates system aligned with
to flow so the $x$–coordinate in the direction of the flow.
In this case the velocity is given by
\begin{align}
\label{if:eq:UniformFlowUx}
\begin{array}{rl}
U_x = U_0 \\
U_y = 0
\end{array}
\end{align}
and according to definitions in this chapter
\begin{align}
\label{if:eq:UFdef}
U_x = \dfrac{\partial \phi}{\partial x} =
\dfrac{\partial \psi}{\partial y} = U_0
\end{align}
Hence, it can be obtained that
\begin{align}
\label{if:eq:UFps}
\begin{array}{rl}
\phi = U_0\,x + f_y(y) \
\psi = U_0\,x + f_x(x)
\end{array}
\end{align}
where $f_y(y)$ is arbitrary function of the $y$ and $f_x(x)$ is arbitrary function of $x$.
In the same time these function have to satisfy the condition
\begin{align}
\label{if:eq:UFcondition}
U_y = \dfrac{\partial \phi}{\partial x} \quad \text{and} \quad
\dfrac{\partial \psi}{\partial x} = 0
\end{align}
These conditions dictate that
\begin{align}
\label{if:eq:UFcondiRes}
\begin{array}{rl}
\dfrac{d\,f_y(y)}{dy} = 0 \\
\dfrac{d\,f_x(x)}{dx} = 0
\end{array}
\end{align}
Hence
\begin{align}
\label{if:eq:UFSola}
f_y(y) = constant \Longrightarrow \phi = U_0 \, x + constant
\end{align}
\begin{align}
\label{if:eq:UFSolb}
f_x(x) = constant \Longrightarrow \psi = U_0 \, y + constant
\end{align}
These lines can be exhibits for various constants as shown in Figure below.
Fig. 10.6 Uniform Flow Streamlines and Potential Lines.
Line Source and Sink Flow
Another typical flow is a flow from a point or a line in a two dimensional field.
This flow is only an idealization of the flow into a single point.
Clearly this kind of flow cannot exist because the velocity approaches infinity at the
singular point of the source.
Yet this idea has its usefulness and is commonly used by many engineers.
This idea can be combined with other flow fields and provide a more realistic situation.
Fig. 10.7 Streamlines and Potential lines due to Source or sink.
The volumetric flow rate (two dimensional) $\dot{Q}$ denotes the flow rate out or in to control
volume into the source or sink.
The flow rate is shown in Figure 10.7 is constant for every potential line.
The flow rate can be determined by
\begin{align}
\label{if:eq:sourceFlowRate}
\dot{Q} = 2\,\pi\,r\, U_r
\end{align}
Where $\dot{Q}$ is the volumetric flow rate, $r$ is distance from the origin and $U_r$ is the velocity
pointing out or into the origin depending whether origin has source or sink.
The relationship between the potential function to velocity dictates that
\begin{align}
\label{if:eq:potentialSource1}
\boldsymbol{\nabla} \phi = \mathbf{\pmb{U}} = U\,\mathbf{\widehat{r}} =
\dfrac{\dot{Q}}{ 2\,\pi\,r}\,\mathbf{\widehat{r}}
\end{align}
Explicitly writing the gradient in cylindrical coordinate results as
\begin{align}
\label{if:eq:sourceGradient}
\dfrac{\partial \phi}{\partial r} \mathbf{\widehat{r}} +
\dfrac{1}{ r} \dfrac{\partial \phi}{\partial \theta} \mathbf{\widehat{\boldsymbol{\theta}} } +
\dfrac{\partial \phi}{\partial z} \mathbf{\widehat{z}} =
\dfrac{\dot{Q}}{ 2\,\pi\,r}\,\mathbf{\widehat{r}} + 0 \, \mathbf{\widehat{\boldsymbol{\theta}} }
+ 0 \, \mathbf{\widehat{z}}
\end{align}
Equation \eqref{if:eq:sourceGradient} the gradient components must satisfy the following
\begin{align}
\label{if:eq:sourceComponents}
\begin{array}{rl}
\dfrac{\partial \phi}{\partial r} = \dfrac{\dot{Q}}{ 2\,\pi\,r}\,\mathbf{\widehat{r}} \\
\dfrac{\partial \phi}{\partial z} = \dfrac{\partial \phi}{\partial \theta} = 0
\end{array}
\end{align}
The integration of equation results in
\begin{align}
\label{if:eq:sourcePotential}
\phi  \phi_0 = \dfrac{\dot{Q}}{ 2\,\pi\,r}\, \ln \dfrac{r}{r_0}
\end{align}
where $r_0$ is the radius at a known point and $\phi_0$ is the potential at that point.
The stream function can be obtained by similar equations that were used or Cartesian coordinates.
In the same fashion it can be written that
\begin{align}
\label{if:eq:cylindricalIni}
d\psi = \bbb{U} \bbb{\cdot} \widehat{s} \,d\ell
\end{align}
Where in this case $d\ell = r\,d\theta$ (the shortest distance between two adjoining stream lines is
perpendicular to both lines) and hence equation \eqref{if:eq:cylindricalIni} is
\begin{align}
\label{if:eq:cylindrical}
d\psi = \bbb{U} \bbb{\cdot} \,r\,d\theta \,\bbb{\widehat{r}} =
\dfrac{\dot{Q}}{ 2\,\pi\,r}\, r\,d\theta = \dfrac{\dot{Q}}{ 2\,\pi}\, \,d\theta
\end{align}
Note that the direction of $\bbb{U}$ and $\bbb{\widehat{r}}$ is identical.
The integration of equation \eqref{if:eq:cylindrical} yields
\begin{align}
\label{if:eq:sourceStream}
\psi  \psi_0 = \dfrac{\dot{Q}}{ 2\,\pi\,r}\, \left( \theta \theta_0\right)
\end{align}
It traditionally chosen that the stream function $\psi_0$ is zero at $\theta=0$.
This operation is possible because the integration constant and the arbitrary reference.
In the case of the sink rather than the source, the velocity is in the opposite direction.
Hence the flow rate is negative and the same equations obtained.
\begin{align}
\label{if:eq:sinkePotential}
\phi  \phi_0 = \dfrac{\dot{Q}}{ 2\,\pi\,r}\, \ln \dfrac{r}{r_0}
\end{align}
\begin{align}
\label{if:eq:sinkStream}
\psi  \psi_0 = \dfrac{\dot{Q}}{ 2\,\pi\,r}\, \left( \theta \theta_0\right)
\end{align}
Free Vortex Flow
Fig. 10.8 Two dimensional Vortex free flow.
In the diagram exhibits part the circle to explain the stream lines and potential lines.
As opposed to the radial flow direction (which was discussed under the source and sink) the flow in the tangential
direction is referred to as the free vortex flow.
Another typical name for this kind of flow is the potential vortex flow.
The flow is circulating the origin or another point.
The velocity is only a function of the distance from the radius as
\begin{align}
\label{if:eq:Uvortisity}
U_{\theta} = f(r)
\end{align}
And in vector notation the flow is
\begin{align}
\label{if:eq:UvecVortisity}
\bbb{U} = \widehat{\boldsymbol{\theta}}\, f(r)
\end{align}
The fundamental aspect of the potential flow is that this flow must be irrotational flow.
The gradient of the potential in cylindrical coordinates is
\begin{align}
\label{if:eq:gradientCyl}
\bbb{U} = \boldsymbol{\nabla}\phi = \dfrac{\partial \phi}{\partial r}\, \widehat{\bbb{r}} +
\dfrac{1}{r}\,\dfrac{\partial \phi}{\partial \theta} \,\widehat{\bbb{\theta}}
\end{align}
Hence, equation \eqref{if:eq:gradientCyl} dictates that
\begin{align}
\label{if:eq:vortisityODE}
\begin{array}{rl}
\dfrac{1}{r}\,\dfrac{\partial \phi}{\partial \theta} = f(r) \\
\dfrac{\partial \phi}{\partial r} = 0
\end{array}
\end{align}
From these equations it can be seem that
\begin{align}
\label{if:eq:vertisityODEsol1}
\phi = \phi (\theta)
\end{align}
and
\begin{align}
\label{if:eq:vertisityODE2}
\dfrac{\partial \phi}{\partial \theta} = r \,f(r)
\end{align}
Equation \eqref{if:eq:vertisityODE2} states that the potential function depends on the angle, $\theta$
while it also a function of the radius.
The only what the above requirement is obtained when the derivative of $\phi$and
the equation are equal to a constant.
Thus,
\begin{align}
\label{if:eq:vertisityODE2sol}
\begin{array}{rl}
r \,f(r) = c \Longrightarrow f(r) = \dfrac{c}{r} \\
\dfrac{\partial \phi}{\partial \theta} = c \Longrightarrow \phi\phi_0 = c_1\, ( 0  \theta_0)
\end{array}
\end{align}
It can be observed from equation \eqref{if:eq:vertisityODE2} that the velocity varies inversely with
the radius.
This variation is referred in the literature as the natural vortex as oppose to forced vortex where
the velocity varies in any different functionality.
It has to be noted that forced vortex flow is not potential flow.
The stream function can be found in the ``standard'' way as
\begin{align}
\nonumber
d\psi = \bbb{U} \,\bbb{\cdot} \, \widehat{s} \,dr
\end{align}
It can observed, in this case, from Figure 10.8 that $\widehat{s}= \widehat{\bbb{\theta}}$
hence
\begin{align}
\label{if:eq:vortexPotentialD}
d\psi = \widehat{\bbb{\theta}} \, \dfrac{c_1}{r} \,\bbb{\cdot} \,(\widehat{\bbb{\theta}}) \,dr
= c_1 \dfrac{dr}{r}
\end{align}
Thus,
\begin{align}
\label{if:eq:vortexPotential}
\psi  \psi_0 = c_1\, \ln \left( \dfrac{r}{r_0} \right)
\end{align}
The source point or the origin of the source is a singular point of the stream function
and there it cannot be properly defined.
Equation \eqref{if:eq:vertisityODE2sol} dictates that velocity at the origin is infinity.
This similar to natural situation such as tornadoes, hurricanes, and whirlpools where
the velocity approaches a very large value near the core.
In these situation the pressure became very low as the velocity increase.
Since the pressure cannot attain negative value or even approach zero value, the physical situation
changes.
At the core of these phenomenon a relative zone calm zone is obtained.
The Circulation Concept
In the construction of the potential flow or the inviscid flow researchers discover
important concept of circulation.
This term mathematically defined as a close path integral around area (in two dimensional flow)
of the velocity along the path.
The circulation is denoted as $\Gamma$ and defined as
\begin{align}
\label{if:eq:circulation}
\Gamma = \oint \bbb{U}_s\, ds
\end{align}
Where the velocity $\bbb{U}_s$ represents the velocity component in the direction of the path.
The symbol $\oint$ indicating that the integral in over a close path.
Fig. 10.9 Circulation path to illustrate varies calculations.
Mathematically to obtain the integral the velocity component in the direction of the path
has to be chosen and it can be defined as
\begin{align}
\label{if:eq:Circulation}
\Gamma = \oint_C \bbb{U} \, \bbb{\cdot}\, \widehat{\bbb{ds}}
\end{align}
Substituting the definition potential function into equation \eqref{if:eq:Circulation} provides
\begin{align}
\label{if:eq:CirculationPotential}
\Gamma = \oint_C \bbb{\nabla}\phi \, \bbb{\cdot}\, \overbrace{\widehat{\bbb{s}} \,ds}^{\widehat{\bbb{ds}}}
\end{align}
And using some mathematical manipulations yields
\begin{align}
\label{if:eq:cpManipulation}
\Gamma = \oint_C \overbrace{\dfrac{d\phi}{ds} }^{{\nabla}\phi \, \bbb{\cdot}\, \widehat{\bbb{s}}} \,ds
= \oint_C d\phi
\end{align}
The integration of equation \eqref{if:eq:cpManipulation} results in
\begin{align}
\label{if:eq:cpManipulationR1}
\Gamma = \oint_C d\phi = \phi_2 (\text{starting point})  \phi_1 (\text{starting point})
\end{align}
Unless the potential function is dual or multi value, the difference between the two
points is zero.
In fact this what is expected from the close path integral.
However, in a free vortex situation the situation is different.
The integral in that case is the integral around a circular path which is
\begin{align}
\label{if:eq:circulationFreeVortex}
\Gamma = \oint \bbb{U} \,\bbb{\cdot}\, \bbb{i} \, \overbrace{r \,d\theta}{ds} =
\oint \dfrac{c_1}{r} \,r\,d\theta
= c_1\,2\,\pi
\end{align}
In this case the circulation, $\Gamma$ is not vanishing.
In this example, the potential function $\phi$ is a multiple value as potential
function the potential function with a single value.
Calculate the circulation of the source on the path of the circle around the origin with
radius $a$ for a source of a given strength.
Solution
The circulation can be carried by the integration
\begin{align}
\label{circulationSource:fromDef}
\Gamma = \oint \overbrace{\bbb{U} \,\bbb{\cdot}\, \bbb{i}}^{=0} \, r \,d\theta\,{ds} = 0
\end{align}
Since the velocity is perpendicular to the path at every point on the path, the integral identically is zero.
Thus, there are two kinds of potential functions one where there are single value and those with
multi value.
The free vortex is the cases where the circulation add the value of the potential function
every rotation.
Hence, it can be concluded that the potential function of vortex is multi value
which increases by the same amount every time, $c_1\,2\,\pi$.
In this case value at $\theta=0$ is different because the potential function did not
circulate or encompass a singular point.
In the other cases, every additional enclosing adds to the value of potential function a value.
It was found that the circulation, $\Gamma$ is zero when
there is no singular point within the region inside the path.}}}}

For the free vortex the integration constant can be found if the circulation is known as
\begin{align}
\label{if:eq:ConstatntVortext}
c_1 = \dfrac{\Gamma}{2\,\pi}
\end{align}
In the literature, the term $\Gamma$ is, some times, referred to as the ``strength'' of the vortex.
The common form of the stream function and potential function is in the form of
\begin{align}
\label{if:eq:typicalPhiVortex}
\phi = \dfrac{\Gamma}{2\,\pi} \left(\theta \theta_0\right) + \phi_0
\end{align}
\begin{align}
\label{if:eq:typicalPsiVortex}
\psi = \dfrac{\Gamma}{2\,\pi} \ln\left( \dfrac{r}{r_0}\right) + \psi_0
\end{align}
Superposition of Flows
For incompressible flow and two dimensional the continuity equation reads
\begin{align}
\label{if:eq:Laplaces}
\bbb{\nabla} \bbb{\cdot} \bbb{U} =
\bbb{\nabla} \bbb{\cdot} \bbb{\nabla} \phi =
\bbb{\nabla}^2 \phi =
\dfrac{\partial^2\phi}{\partial x^2} + \dfrac{\partial^2\phi}{\partial y^2} = 0
\end{align}
The potential function must satisfy the Laplace's equation which is a linear partial differential equation.
The velocity perpendicular to a solid boundary must be zero (boundary must be solid) and hence
it dictates the boundary conditions on the potential equation.
From mathematical point of view this boundary condition as
\begin{align}
\label{if:eq:bcSolid}
\bbb{U}_n = \dfrac{d\phi}{dn} = \bbb{\nabla}{\phi}\bbb{\cdot} \widehat{\bbb{n}} = 0
\end{align}
In this case, $\widehat{\bbb{n}}$ represents the unit vector normal to the surface.
A solution to certain boundary condition with certain configuration geometry and shape is a velocity
flow field which can be described by the potential function, $\phi$.
If such function exist it can be denoted as $\phi_1$.
If another velocity flow field exists which describes, or is, the solutions to a different boundary
condition(s) it is denoted as $\phi_2$.
The Laplacian of first potential is zero, $\bbb{\nabla}^2\phi_1=0$ and the same is true for the second one
$\bbb{\nabla}^2\phi_2=0$.
Hence, it can be written that
\begin{align}
\label{if:eq:laplacianSuperposition}
\overbrace{\bbb{\nabla}^2\phi_1}^{=0} + \overbrace{\bbb{\nabla}^2\phi_2}^{=0} = 0
\end{align}
Since the Laplace mathematical operator is linear the two potential can be combined as
\begin{align}
\label{if:eq:laplacianSuperpositionCombined}
\bbb{\nabla}^2\left(\phi_1 + \phi_2\right) = 0
\end{align}
The boundary conditions can be also treated in the same fashion.
On a solid boundary condition for both functions is zero hence
\begin{align}
\label{if:eq:laplacianSuperpositionBC1}
\dfrac{d\phi_1}{dn} = \dfrac{d\phi_2}{dn} = 0
\end{align}
and the normal derivative is linear operator and thus
\begin{align}
\label{if:eq:laplacianSuperpositionBC2}
\dfrac{d\left(\phi_1+ \phi_2\right)}{dn} = 0
\end{align}
It can be observed that the combined new potential function create a new velocity field.
In fact it can be written that
\begin{align}
\label{if:eq:combinedU}
\bbb{U} = \bbb{\nabla} (\phi_1+\phi_2) = \bbb{\nabla} \phi_1+ \bbb{\nabla} \phi_2 =
\bbb{U}_1 + \bbb{U}_2
\end{align}
The velocities $\bbb{U}_1$ and $\bbb{U}_2$ are obtained from $\phi_1$ and $\phi_2$ respectively.
Hence, the superposition of the solutions is the characteristic of the potential flow.
Source and Sink Flow or Doublet Flow
In the potential flow, there is a special case where the source and sink are combined since it represents
a special and useful shape.
A source is located at point B which is $r_0$ from the origin on the positive $x$ coordinate.
The flow rate from the source is $Q_0$ and the potential function is
\begin{align}
\label{if:eq:doublet:sourcePF}
Q_1 = \dfrac{Q_0}{2\,\pi} \ln \left( \dfrac {r_B}{r_0}\right)
\end{align}
Fig. 10.10 Combination of the Source and Sink located at a distance $r_0$ from the origin
on the $x$ coordinate. The source is on the right.
The sink is at the same distance but at the negative side of the $x$ coordinate and hence it can be represented
by the potential function
\begin{align}
\label{if:eq:doublet:sinkPF}
Q_1 = \dfrac{Q_0}{2\,\pi} \ln \left( \dfrac {r_A}{r_0}\right)
\end{align}
The description is depicted on Figure 10.10.
The distances, $r_A$ and $r_B$ are defined from the points $A$ and $B$ respectively.
The potential of the source and the sink is
\begin{align}
\label{if:eq:doubletPotential}
\phi = \dfrac{Q_0}{2\,\pi} \, \left( \ln r_A  \ln r_B\right)
\end{align}
In this case, it is more convenient to represent the situation utilizing the cylindrical coordinates.
The Law of Cosines for the right triangle ($\overline{OBR}$) this cases reads
\begin{align}
\label{if:eq:cosinesLawR}
{r_B}^2 = r^2 + {r_0}^2  2\,r\,r_0\,cos\theta
\end{align}
In the same manner it applied to the left triangle as
\begin{align}
\label{if:eq:cosinesLawL}
{r_A}^2 = r^2 + {r_0}^2 + 2\,r\,r_0\,cos\theta
\end{align}
Therefore, equation \eqref{if:eq:doubletPotential} can be written as
\begin{align}
\label{if:eq:cosinesLawCombine}
\phi =  \dfrac{Q_0}{2\,\pi} \, \dfrac{1}{2} \, \ln \left(
\dfrac{\dfrac{r^2+{r_0}^2}{2\,r\,r_0\, \cos \theta} + 1}
{\dfrac{r^2+{r_0}^2}{2\,r\,r_0\, \cos \theta}  1}\right)
\end{align}
It can be shown that
the following the identity exist
Caution: mathematical details which can be skipped
\begin{align}
\label{if:eq:mathIdenty}
\coth^{1} (\xi) = \dfrac{1}{2} \ln \left( \dfrac{\xi + 1 }{ \xi  1} \right)
\end{align}
where $\xi$ is a dummy variable.
Hence, substituting into equation \eqref{if:eq:cosinesLawCombine} the identity of
equation \eqref{if:eq:mathIdenty} results in
\begin{align}
\label{if:eq:doubletPotentialFc}
\phi =  \dfrac{Q_0}{2\,\pi} \, \coth^{1} \left( \dfrac{r^2+{r_0}^2}{2\,r\,r_0\, \cos \theta}\right)
\end{align}
The several following stages are more of a mathematical nature which provide minimal
contribution to physical understanding but are provide to interested reader.
The manipulations are easier with an implicit solution and thus
\begin{align}
\label{if:eq:doubletImpliciteIni}
\coth \left(  \dfrac{2\,\pi\,\phi}{Q}\right) = \dfrac{r^2+{r_0}^2}{2\,r\,r_0\, \cos \theta}
\end{align}
Equation \eqref{if:eq:doubletImpliciteIni}, when noticing that
the $\,\cos\theta\, \coth(x) =  \coth(x)$, can be written as
\begin{align}
\label{if:eq:doubletImpliciteInif}
2\,r_0\,r\,\cos\theta\coth \left( \dfrac{2\,\pi\,\phi}{Q}\right)
= {r^2+{r_0}^2}
\end{align}
In Cartesian coordinates equation \eqref{if:eq:doubletImpliciteInif} can be written as
\begin{align}
\label{if:eq:doubletImpliciteInic}
2\,r_0\,\overbrace{x}^{r\,\cos\theta}\,
\coth \left(  \dfrac{2\,\pi\,\phi}{Q}\right) = {x^2+y^2 +{r_0}^2}
\end{align}
Equation \eqref{if:eq:doubletImpliciteInic} can be rearranged by the left hand side to right as and moving ${r_0}^2$ to left side
result in
\begin{align}
\label{if:eq:doubletImpliciteInic1}
 {r_0}^2 = 2\,r_0\,\overbrace{x}^{r\,\cos\theta}\,
\coth \left( \dfrac{2\,\pi\,\phi}{Q}\right) + x^2+y^2
\end{align}
Add to both sides ${r_0}^2\,\coth^2\dfrac{2\,\pi\,\phi}{Q_0}$ transfers equation
qref{if:eq:doubletImpliciteInic1}
\begin{align}
\label{if:eq:doubletImpliciteInic2a}
{r_0}^2\,\coth^2\dfrac{2\,\pi\,\phi}{Q_0}  {r_0}^2 =
{r_0}^2\,\coth^2\dfrac{2\,\pi\,\phi}{Q_0} + 2\,r_0\,\overbrace{x}^{r\,\cos\theta}\,
\coth \left( \dfrac{2\,\pi\,\phi}{Q}\right) + x^2+y^2
\end{align}
The hyperbolic identity
can be written as
\begin{align}
\label{if:eq:doubletImpliciteInic2}
{r_0}^2\,\text{csch}^2\dfrac{2\,\pi\,\phi}{Q_0} =
{r_0}^2\,\coth^2\dfrac{2\,\pi\,\phi}{Q_0} + 2\,r_0\,\overbrace{x}^{r\,\cos\theta}\,
\coth \left( \dfrac{2\,\pi\,\phi}{Q}\right) + x^2+y^2
\end{align}
End Caution: mathematical details
It can be noticed that first three term on the right hand side are actually quadratic
and can be written as
\begin{align}
\label{if:eq:doubletImpliciteInic3}
{r_0}^2\,\text{csch}^2\dfrac{2\,\pi\,\phi}{Q_0} =
\left( {r_0}\,\coth\dfrac{2\,\pi\,\phi}{Q_0} + x \right)^2 + y^2
\end{align}
equation \eqref{if:eq:doubletImpliciteInic3} represents a circle with a radius
${r_0}\,\text{csch}\dfrac{2\,\pi\,\phi}{Q_0}$ and a center at
$\pm r_0\coth \left( \dfrac{2\,\pi \, \phi}{Q_0} \right)$.
The potential lines depicted on Figure 10.11.
For the drawing purposes equation \eqref{if:eq:doubletImpliciteInic3} is transformed
into a dimensionless form as
\begin{align}
\label{if:eq:doubletImpliciteInicDlss}
\left( \coth\dfrac{2\,\pi\,\phi}{Q_0} + \dfrac{x}{r_0} \right)^2 + \left(\dfrac{y}{r_0}\right)^2
= \text{csch}^2\dfrac{2\,\pi\,\phi}{Q_0}
\end{align}
Notice that the stream function has the same dimensions as the source/sink flow rate.
Fig. 10.11 Stream and Potential line for a source and sink.
It can be noticed that stream line (in blue to green) and the potential line are in orange to crimson. This figure is relative distances of $x/r_0$ and $y/r_0$. The parameter that change is $2\,\pi\,\phi/Q_0$ and $2\,\pi\,\psi/Q_0$.
Notice that for give larger of $\phi$ the circles are smaller.
The stream lines can be obtained by utilizing similar procedure.
The double stream function is made from the combination of the source and sink
because stream functions can be added up.
Hence,
\begin{align}
\label{if:eq:doubleStream}
\psi = \psi_1 + \psi_2 = \dfrac{Q_0}{2\,\pi} \left( \theta_1  \theta_2\right)
\end{align}
The angle $\theta_1$ and $\theta_2$ shown in Figure 10.11 related
other geometrical parameters as
\begin{align}
\label{if:eq:doubletAngle1}
\theta_1 = \tan^{1} \dfrac{y}{xr_0}
\end{align}
and
\begin{align}
\label{if:eq:doubletAngle2}
\theta_2 = \tan^{1} \dfrac{y}{x+r_0}
\end{align}
The stream function becomes
\begin{align}
\label{if:eq:doubletStreamIni}
\psi = \dfrac{Q_0}{2\,\pi} \left( \tan^{1} \dfrac{y}{xr_0}  \tan^{1} \dfrac{y}{x+r_0} \right)
\end{align}
Caution: mathematical details which can be skipped
Rearranging equation \eqref{if:eq:doubletStreamIni} yields
\begin{align}
\label{if:eq:doubletStreamIntermid}
\dfrac{2\,\pi\,\psi}{Q_0} =
\tan^{1} \dfrac{y}{xr_0}  \tan^{1} \dfrac{y}{x+r_0}
\end{align}
Utilizing
the identity $\tan^{1} u + \tan^{1} v = \tan^{1} \left( \dfrac{u+v}{1uv} \right)$
Equation \eqref{if:eq:doubletStreamIntermid} transfers to
\begin{align}
\label{if:eq:doubletStreamIntermidT}
\tan \dfrac{2\,\pi\,\psi}{Q_0} =
\dfrac{\dfrac{y}{xr_0}  \dfrac{y}{x+r_0}}
{1 + \dfrac{y^2}{x^2{r_0}^2}}
\end{align}
As in the potential function cases,
Several manipulations to convert the equation \eqref{if:eq:doubletStreamIntermidT}
form so it can be represented in a ``standard'' geometrical shapes are done before to potential function.
Reversing and finding the common denominator provide
\begin{align}
\label{if:eq:doubletStreamIntermidRev}
\cot \dfrac{2\,\pi\,\psi}{Q_0} =
\dfrac { \dfrac{ x^2{r_0}^2 + y^2 }{x^2{r_0}^2} } {\dfrac{y\, (x+r_0)  y\,(xr_0) }{x^2  {r_0}^2} }
= \dfrac{ x^2{r_0}^2 + y^2 }{\underbrace{y\, (x+r_0) + y\,(xr_0) }_{2\,y\,r_0 } }
\end{align}
or
\begin{align}
\label{if:eq:doubletStreamIntermidRev1}
x^2+ y^2{r_0}^2 = 2\, r_0 \, y \cot \dfrac{2\,\pi\,\psi}{Q_0}
\end{align}
End Caution: mathematical details
Equation \eqref{if:eq:doubletStreamIntermidRev1} can be rearranged, into a typical circular representation
\begin{align}
\label{if:eq:doubleStreamF1}
x^2 +\left( y  r_0\cot\dfrac{2\,\pi\,\psi}{Q_0}\right)^2 =
\left( r_0\csc\dfrac{2\,\pi\,\psi}{Q_0} \right)^2
\end{align}
Equation \eqref{if:eq:doubleStreamF1} describes circles with center on the $y$ coordinates at $y=r_0\cot\dfrac{2\,\pi\,\psi}{Q_0}$.
It can be noticed that these circles are orthogonal to the the circle that represents the
the potential lines.
For the drawing it is convenient to write equation \eqref{if:eq:doubleStreamF1} in dimensionless
form as
\begin{align}
\label{if:eq:doubleStreamF2}
\left( \dfrac{x}{r_0}\right)^2 +\left( \dfrac{y} {r_0}  \cot\dfrac{2\,\pi\,\psi}{Q_0}\right)^2 =
\left( \csc\dfrac{2\,\pi\,\psi}{Q_0} \right)^2
\end{align}
Dipole Flow
It was found that when the distance between the sink and source shrinks to zero
a new possibility is created which provides benefits to new understanding.
The new combination is referred to as the dipole.
Even though, the construction of source/sink to a single location (as the radius is reduced
to zero) the new ``creature'' has direction as opposed to the scalar characteristics of source
and sink.
First the potential function and stream function will be presented.
The potential function is
\begin{align}
\label{if:eq:doubleImitIni}
\lim_{r_0\rightarrow 0}\phi =  \dfrac{Q_0}{2\,\pi} \, \dfrac{1}{2}\,\ln \left(
\dfrac{r^2+{r_0}^2  2\,r\,r_0\, \cos \theta}
{r^2+{r_0}^2 + 2\,r\,r_0\, \cos \theta }\right)
\end{align}
To determine the value of the quantity in equation \eqref{if:eq:doubleImitIni} the L'HÃ´pital's rule
will be used.
First the appropriate form will be derived so the technique can be used.
Caution: mathematical details which can be skipped
Multiplying and dividing equation \eqref{if:eq:doubleImitIni} by $2\,r_0$ yields
\begin{align}
\label{if:eq:doubleImit1}
\lim_{r_0\rightarrow 0}\phi = \overbrace{\dfrac{Q_0\,2\,r_0}{2\,\pi}}
^{{ 1^{st}\, part} }
\, \overbrace{\dfrac{1}{\underbrace{2\,\,2}_{4}\,r_0}
\,\ln \left(
\dfrac{r^2+{r_0}^2  2\,r\,r_0\, \cos \theta}
{r^2+{r_0}^2 + 2\,r\,r_0\, \cos \theta }\right) }^{\mathrm 2^{nd} \, part}
\end{align}
Equation \eqref{if:eq:doubleImit1} has two parts.
The first part, $(Q_0\,2\,r_0)/2\,\pi$, which is a function of $Q_0$ and $r_0$ and the second part
which is a function of $r_0$.
While reducing $r_0$ to zero, the flow increases in such way that the combination of $Q_0\,r_0$
is constant.
Hence, the second part has to be examined and arranged for this purpose.
\begin{align}
\label{if:eq:doubleImit2}
\lim_{r_0\rightarrow 0} \dfrac{
\ln \left(
\dfrac{r^2+{r_0}^2  2\,r\,r_0\, \cos \theta}
{r^2+{r_0}^2 + 2\,r\,r_0\, \cos \theta }\right)
}{4\,r_0}
\end{align}
It can be noticed that the ratio in the natural logarithm approach one $r_0\rightarrow 0$.
The L'Hopital's rule can be applied because the situation of nature of $0/0$.
The numerator can be found using a short cut
End Caution: mathematical details
at
\begin{align}
\label{if:eq:doubleImit2Lhopital}
\lim_{r_0\rightarrow 0} \dfrac{ \dfrac{\cancelto{0}{2\,r_0}  2\,r\,\cos\theta}
{r^2+\cancelto{0}{{r_0}^2}  \cancelto{0}{2\,r\,r_0\, \cos \theta} } 
\dfrac{\cancelto{0}{2\,r_0} + 2\,r\,\cos\theta}
{r^2+\cancelto{0}{{r_0}^2} + \cancelto{0}{2\,r\,r_0\, \cos \theta}} }
{4} =  \dfrac{\cos\theta}{r}
\end{align}
Combining the first and part with the second part results in
\begin{align}
\label{if:eq:doubleImitF1}
\phi =  \dfrac{Q_0\,r_0}{\pi} \dfrac{\cos\theta}{r}
\end{align}
After the potential function was established the attention can be turned into the stream function.
To establish the stream function, the continuity equation in cylindrical is used which is
\begin{align}
\nonumber
\bbb{\nabla} \bbb{\cdot} \bbb{U} = \dfrac{1}{r}
\left( \dfrac{\partial \,r\,U_r}{\partial r} + \dfrac{\partial \,U_{\theta}}{\partial \theta} \right)
\end{align}
The transformation of equations \eqref{if:eq:xConclusion} and \eqref{if:eq:yConclusion} to cylindrical
coordinates results in
\begin{align}
\label{if:eq:streamFunCylindricalR}
U_r = \dfrac{1}{r} \dfrac{\partial \psi}{\partial \theta}
\end{align}
\begin{align}
\label{if:eq:streamFunCylindricalTheta}
U_{\theta} =  \dfrac{\partial \psi}{\partial r}
\end{align}
The relationship for the potential function of the cylindrical coordinates was determined before an
appear the relationship \eqref{if:eq:UxpotentianlSteam} and \eqref{if:eq:UypotentianlSteam}
\begin{align}
\label{if:eq:potentionalUCylindricalIni}
U_r = \dfrac{\partial \phi}{\partial r} \quad \text{and} \quad
\end{align}
\begin{align}
\label{if:eq:potentionalUCylindrical}
U_{\theta} = \dfrac{1}{r} \dfrac{\partial \phi}{\partial \theta}
\end{align}
Thus the relationships that were obtained before for Cartesian coordinates
is written in cylindrical coordinates as
\begin{align}
\label{if:eq:psUrCylindrical}
\dfrac{\partial \phi}{\partial r} = \dfrac{1}{r} \dfrac{\partial \psi}{\partial \theta}
\end{align}
\begin{align}
\label{if:eq:psUthetaCylindrical}
\dfrac{1}{r} \dfrac{\partial \phi}{\partial \theta} =  \dfrac{\partial \psi}{\partial r}
\end{align}
In the case of the dipole, the knowledge of the potential function is used to
obtain the stream function.
The derivative of the potential function as respect to the radius is
\begin{align}
\label{if:eq:doublePdr}
\dfrac{\partial \phi}{\partial r} = \dfrac{Q_0}{2\,\pi} \,\dfrac{\cos\theta}{r^2}
\dfrac{1}{r} \dfrac{\partial \psi}{\partial \theta}
\end{align}
And
\begin{align}
\label{if:eq:doubleSdr}
\dfrac{1}{r} \dfrac{\partial \phi}{\partial \theta} = \dfrac{Q_0}{2\,\pi} \,\dfrac{\sin\theta}{r^2}
 \dfrac{\partial \psi}{\partial r}
\end{align}
From equation \eqref{if:eq:doublePdr} after integration with respect to $\theta$ one can obtain
\begin{align}
\label{if:eq:doublePdrSol}
\psi = \dfrac{Q_0}{2\,\pi\,r} \, \sin \theta + f(r)
\end{align}
and from equation \eqref{if:eq:doubleSdr} one can obtain that
\begin{align}
\label{if:eq:doubleDdrSol}
 \dfrac{\partial \psi}{\partial r} = \dfrac{Q_0}{2\,\pi\,r^2} \sin\theta + f'(r)
\end{align}
The only way that these conditions co–exist is $f(r)$ to be constant and thus $f'(r)$ is zero.
The general solution of the stream function is then
\begin{align}
\label{if:eq:doubleStreamF}
\psi = \dfrac{Q_0\,\sin\theta}{2\,\pi\,r}
\end{align}
Caution: mathematical details which can be skipped
The potential function and stream function describe the circles as following:
In equation \eqref{if:eq:doubleStreamF} it can be recognized that $r= \sqrt{x^2+y^2}$
Thus, multiply equation \eqref{if:eq:doubleStreamF} by $r$ and some rearrangement yield
\begin{align}
\label{if:eq:psiCircule1}
\dfrac{2\,\pi\,\psi}{Q_0}\,\left(\overbrace{x^2+y^2}^{r^2}\right) = \overbrace{y}^{r\,sin\theta}
\end{align}
Further rearranging equation \eqref{if:eq:psiCircule1} provides
\begin{align}
\label{if:eq:psiCircule2}
\left(\overbrace{x^2+y^2}^{r^2}\right) = \dfrac{Q_0}{2\,\pi\,\psi}\, \overbrace{y}^{r\,\sin\theta}
 \overbrace{
\left( \dfrac{Q_0}{2\,\pi\,\psi} \right)^2 + \left( \dfrac{Q_0}{2\,\pi\,\psi} \right)^2 } ^{=0}
\end{align}
and converting to the standard equation of circles as
\begin{align}
\label{if:eq:psiCircule3}
\overbrace{
y^2  \dfrac{Q_0}{2\,\pi\,\psi}\,y + \left( \dfrac{Q_0}{2\,\pi\,\psi} \right)^2}^{y
\dfrac{Q_0}{2\,\pi\,\psi}} + \,x^2 = \left(\dfrac{Q_0}{2\,\pi\,\psi} \right)^2
\end{align}
End Caution: mathematical details
The equation \eqref{if:eq:doubleStreamF} (or \eqref{if:eq:psiCircule3}) represents a circle with
a radius of $\dfrac{Q_0}{2\,\pi\,\psi}$ with location at $x=0$ and $y=\pm\dfrac{Q_0}{2\,\pi\,\psi}$.
The identical derivations can be done for the potential function.
It can be noticed that the difference between the functions results from difference
of $r\sin\theta$ the instead of the term is $r\cos\theta$.
Thus, the potential functions are made from circles that the centers are at same distance as their
radius from origin on the $x$ coordinate.
It can be noticed that the stream function and the potential function can have positive and negative values
and hence there are family on both sides of coordinates.
Figure 10.12 displays the stream functions (cyan to green color) and potential functions
(gold to crimson color).
Notice the larger the value of the stream function the smaller the circle and the same for the potential
functions.
Fig. 10.12 Stream lines and Potential lines for Doublet.
The potential lines are in gold color to crimson while the stream lines are cyan to green color. Notice the smaller value of the stream function translates the smaller circle. The drawing were made for the constant to be one (1) and direct value can be obtained
by simply multiplying.
It must be noted that in the derivations above it was assumed that the sink is on the left and source is on the right.
Clear similar results will obtained if the sink and source were oriented differently.
Hence the dipole (even though) potential and stream functions are scalar functions have a direction.
In this stage this topic will not be treated but must be kept in question form.
This academic example is provided mostly for practice of the mathematics.
Built the stream function of dipole with angle.
Start with a source and a sink distance $r$ from origin on the line with a angle $\beta$ from
$x$ coordinates.
Let the distance shrink to zero.
Write the stream function.
10.3.1 Flow Around a Circular Cylinder
After several elements of the potential flow were built earlier, the first use of these elements
can be demonstrated.
Perhaps the most celebrated and useful example is the flow past a cylinder which this section will be dealing
with.
The stream function made by superimposing a uniform flow and a doublet is
\begin{align}
\label{if:eq:streamU+doublet1}
\psi = U_0 \, y + \dfrac {Q_0}{2\,\pi} \dfrac{\sin\theta}{r} =
U_0 \, r\, \sin\theta + \dfrac {Q_0}{2\,\pi} \dfrac{r\,\sin\theta}{r^2}
\end{align}
Or after some arrangement equation \eqref{if:eq:streamU+doublet1} becomes
\begin{align}
\label{if:eq:streamU+doublet2}
\psi = U_0 \, r\, \sin\theta \left(1 + \dfrac {Q_0}{2\,U_0\,\pi\,r^2} \right)
\end{align}
Denoting $\dfrac {Q_0}{2\,U_0\,\pi}$ as $a^2$ transforms equation \eqref{if:eq:streamU+doublet2}
to
\begin{align}
\label{if:eq:streamU+doublet3}
\psi = U_0 \, r\, \sin\theta \left(1  \dfrac {a^2}{r^2} \right)
\end{align}
The stream function for $\psi=0$ is
\begin{align}
\label{if:eq:stream0U+doublet}
0 = U_0 \, r\, \sin\theta \left(1  \dfrac {a^2}{r^2} \right)
\end{align}
This value is obtained when $\theta=0$ or $\theta = \pi$ and/or $r=a$.
The stream line that is defined by radius $r=a$ describes a circle with a radius $a$ with a center
in the origin.
The other two lines are the horizontal coordinates.
The flow does not cross any stream line, hence the stream line represented by $r=a$ can
represent a cylindrical solid body.
For the case where $\psi
e 0$ the stream function can be any value.
Multiplying equation \eqref{if:eq:streamU+doublet3} by $r$ and dividing by $U_0\,a^2$ and some rearranging yields
\begin{align}
\label{if:eq:streamU+doublet3intermidiatIni}
\dfrac{r}{a}\,\dfrac{\psi}{a\,U_0} = \left(\dfrac{r}{a}\right)^2\, \sin\theta  \sin\theta
\end{align}
It is convenient, to go through the regular dimensionalzing process as
\begin{align}
\label{if:eq:streamU+doublet3intermidiat}
\overline{r}\,\overline{\psi} = (\overline{r})^2\, \sin\theta  \sin\theta \quad
\text{or}\quad \overline{r}^2  \dfrac{\overline{\psi}}{\sin\theta} \, \overline{r}1 = 0
\end{align}
The radius for other streamlines can found or calculated for a given angle and given value
of the stream function.
The radius is given by
\begin{align}
\label{if:stream0U+doubletR}
\overline{r} = \dfrac{\dfrac{\overline{\psi}}{\sin\theta}\pm
\sqrt{ \left( \dfrac{\overline{\psi}}{\sin\theta} \right)^2+4} }
{2}
\end{align}
It can be observed that the plus sign must be used for radius with positive values
(there are no physical radii which negative absolute value).
The various value of the stream function can be chosen and drawn.
For example, choosing the value of the stream function as multiply of
$\overline{\psi} = 2\,n$ (where $n$ can be any real number) results in
\begin{align}
\label{if:eq:streamaU0U+doubletRIni}
\overline{r} = \dfrac{\dfrac{2\,n}{\sin\theta}\pm
\sqrt{ \left( \dfrac{2\,n}{\sin\theta} \right)^2+4} }
{2} =
n\, \text{csc}(\theta) + \sqrt{n^2\,\text{csc}^2(\theta) + 1}
\end{align}
The various values for of the stream function are represented by the ratios $n$.
For example for $n=1$ the (actual) radius as a function the angle can be written as
\begin{align}
\label{if:eq:streamaU0U+doubletR}
r = a\,\left( \text{csc}(\theta) + \sqrt{\text{csc}^2(\theta) + 1} \right)
\end{align}
The value csc$(\theta)$ for $\theta= 0 \,\, \text{and} \,\, \theta = \pi$ is equal to infinity
($\infty$) and for values of csc$(\theta=\pi/2) = 1$.
Similar every line can be evaluated.
The lines are drawn in Figure 10.13.
Fig. 10.13 Stream function of uniform flow plus doublet results in solid
body with flow around it. Stream function ($n$ and not $\psi$) starts from 2.0 (green line) to 3 the (purple line). The negative streamlines lines are inside the solid body. The arrows are calculated by trapping the $y$ for given $\psi$ around the end points. Hence, the slight difference between the arrow and the line. The more negative the stream function the smaller the counter. The larger positive stream function the further away the line form the $x$ coordinate. It can be noticed closer the ``solid body'' the lines are more curved. The GLE code is attached in the source code to this book.
The value of $n$ is the bubbles.
The velocity of this flow field can be found by using the equations that were developed so far.
The radial velocity is
\begin{align}
\label{if:eq:unifomDoubletUr}
U_r = \dfrac{1}{r} \dfrac{\partial \psi}{\partial \theta} =
U_0 \,\cos \theta \, \left( 1  \dfrac{a^2}{r^2}\right)
\end{align}
The tangential velocity is
\begin{align}
\label{if:eq:unifomDoubletUtheta}
U_r =  \dfrac{\partial \psi}{\partial r} =
U_0 \,\sin \theta \, \left( 1 + \dfrac{a^2}{r^2}\right)
\end{align}
A sink is placed in a uniform flow field from the left to right.
Describe flow field by the stream lines.
Find the shape of the solid body described by this flow.
Solution
The stream function for uniform flow is given by equation \eqref{if:eq:UFSolb} and the stream
by equation \eqref{if:eq:sinkStream} (with positive sign because it is source).
Hence the stream function is
\begin{align}
\label{uniformSource:streamF}
\psi = U_0\,r\,\sin\theta + \dfrac{\dot{Q}} {2\,\pi}\,\theta
\end{align}
For $\psi=0$ equation \eqref{uniformSource:streamF} becomes
\begin{align}
\label{uniformSource:psi0}
r =  \dfrac{\dot{Q}\,\theta} {2\,\pi\,U_0\,\sin\theta}
\end{align}
or in for any value of stream function, $\psi$ as
\begin{align}
\label{uniformSource:psi}
r = \dfrac{\psi}{U_0\,\sin\theta}  \dfrac{\dot{Q}\,\theta} {2\,\pi\,U_0\,\sin\theta}
\end{align}
The long cigar shape resulted from the combination of the uniform flow with the source is
presented in Figure 10.14.
The black line represents the solid body that created and show two different kind of flows.
The exterior and the interior flow represent the external flow outside and the inside
the black line represents the flow on the enclosed body.
The black line divides the streamline, which separates the fluid coming from the uniform
source the flow due to the inside source.
Thus, these flows represent a flow around semi–infinite solid body and flow from
a source in enclosed body.
Fig. 10.14 Source in the Uniform Flow.
The width of the body at infinity for incompressible flow can be determined by
the condition that the flow rate must be the same.
The velocity can be obtained from the stream function.
Substituting into \eqref{uniformSource:psi0} as
\begin{align}
\label{uniformSource:psi0y}
\overbrace{y}^{r\,\sin\theta} =  \dfrac{\dot{Q}\,\theta} {2\,\pi\,U_0}
\end{align}
An noticing that at $\theta=\pi$ is on the right hand side (opposite to your the intuition)
of the solid body (or infinity).
Hence equation \eqref{uniformSource:psi0y} can be written as
\begin{align}
\label{uniformSource:yatInifinity}
y = \dfrac{\dot{Q}\,\cancel{\pi}} {2\,\cancel{\pi}\,U_0}
= \dfrac{\dot{Q}} {2\,U_0}
\end{align}
It can be noticed that sign in front of $y$ is accounted for and thus removed from the equation.
To check if this analysis is consistent with the continuity equation, the velocity
at infinity must be $U=U_0$ because the velocity due to the source is reduced as $\sim 1/r$.
Hence, the source flow rate must must be balanced (see for the integral mass conservation) flow
rate at infinity hence
\begin{align}
\label{uniformSource:eq:infinityFlowRate}
Q = U_0 \,2\,y = U_0 \,2\, \dfrac{Q_0} {2\,U_0} = Q_0
\end{align}
The stagnation point can be seen from Figure 10.14 by ascertaining the location
where the velocity is zero.
Due to the symmetry the location is on ``solid'' body on the $x$–coordinate at some distance from the origin.
This distance can be found by looking the combined velocities as
\begin{align}
\label{uniformFlowSource:stagnationPoint}
U_0 = \dfrac{Q_0}{2\,\pi\,r} \Longrightarrow r = \dfrac{Q_0}{2\,\pi\,U_0}
\end{align}
Pressure Distribution
One advantage of the inviscid flow approach is the ability to have good estimates of the pressure
and velocity distribution.
These two (pressure and velocity distribution) are related via the Bernoulli's equation.
The explanation and use is based on a specific example and for a specific information.
To illustrate this point the velocity distribution consider a doublet in uniform flow
which was examined earlier.
The velocity field is a function of $x,\,y$ and hence to answer questions
such as the location where the highest velocity or the highest velocity itself is required
to find the maximum point.
This operation is a standard operation in mathematics.
However, in this case the observation of Figure 10.13 suggests
that the height velocity is at the the line of the $y$–coordinate.
The fundamental reason for the above conclusion is that the area symmetry around $y$ coordinate
and the fact that cross area shrink.
Fig. 10.15 Velocity field around a doublet in uniform velocity.
The radial velocity is zero on the $y$–coordinate (due the symmetry and similar arguments)
is zero.
The tangential velocity on the ``solid'' body is
\begin{align}
\label{if:eq:circleUtheta}
U_{\theta} =  2\,U_0 \, \sin\theta
\end{align}
The maximum velocity occurs at
\begin{align}
\label{if:eq:circleUthetaD}
\dfrac{d U_{\theta}}{d\theta} =  2\,U_0 \, \cos\theta = 0
\end{align}
The angle $\pi/2$ and $3\,\pi/2$ are satisfying equation \eqref{if:eq:circleUthetaD}.
The velocity as function of the radius is
\begin{align}
\label{if:eq:circleUthetaRad}
U_{\theta} = \pm \,U_0 \left( 1 + \dfrac{a^2}{r^2\dfrac{}{}} \right)
\end{align}
Where the negative sign is for $\theta = \pi/2$ and the positive sign for $\theta = 3\,\pi/2$.
That is the velocity on surface of the ``solid body'' is the highest.
The velocity profile at specific angles is presented in Figure \eqref{if:fig:doubletU}.
Beside the velocity field, the pressure distribution is a common knowledge needed for
many engineering tasks.
The Euler number is a dimensionless number representing the pressure and is defined as
\begin{align}
\label{if:eq:eulerDef}
Eu = \dfrac{P_0P_{\infty}}{\dfrac{1}{2}\,\rho\,{U_0}^2}
\end{align}
In inviscid flow (Euler's equations) as a sub set of Naiver–Stokes equations
the energy conserved hence (see for discussion on Bernoulli equation),
\begin{align}
\label{if:eq:energyEqIni}
P_0 = P + \dfrac{1}{2}\, \rho \, U^2 \quad \text{or} \quad
P_0  P = \dfrac{1}{2}\, \rho \, U^2
\end{align}
Dividing equation \eqref{if:eq:energyEqIni} by ${U_0}^{2}$ yields
\begin{align}
\label{if:eq:energyEqIni1a}
\dfrac{P_0  P} {{U_0}^2} = \dfrac{1}{2}\, \rho \, \dfrac{U^2}{{U_0}^2} \Longrightarrow
\dfrac{P_0  P} {\dfrac{1}{2}\, \rho \, {U_0}^2} = \dfrac{U^2}{{U_0}^2}
\end{align}
The velocity on the surface of the ``solid'' body is given by equation \eqref{if:eq:circleUtheta}
Hence,
\begin{align}
\label{if:eq:energyEqIni1}
\dfrac{P_0  P} {\dfrac{1}{2}\, \rho \, {U_0}^2} = 4\, \sin^2\theta
\end{align}
It is interesting to point that integration of the pressure results in no lift and no
This ``surprising'' conclusion can by provided by carrying the integration of around
the ``solid'' body and taking the $x$ or $y$ component depending if lift or drag is calculated.
Additionally, it can noticed that symmetry play major role which one side cancel the other side.
10.3.1.1 Adding Circulation to a Cylinder
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 \eqref{if:eq:streamU+doublet3}) is
\begin{align}
\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}
\end{align}
It can be noticed that this stream function \eqref{if:eq:doubletVortex}
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
\begin{align}
\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)
\end{align}
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
\begin{align}
\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}
\end{align}
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 \eqref{if:eq:doubletVortex} can be written as
\begin{align}
\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}
\end{align}
or
\begin{multline}
\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
\end{multline}
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
\begin{align}
\label{if:eq:doubletVortex0b}
\lim_{\xi\rightarrow 1} \dfrac{ \ln \, \xi }{ 1  \xi } =
\lim_{\xi\rightarrow 1} \dfrac{ \dfrac{1}{\xi} }{  1} =  1
\end{align}
Hence, the relationship expressed in equation \eqref{if:eq:doubletVortex0} as
\begin{align}
\label{if:eq:doubletVortex00}
\sin \theta = \dfrac{\Gamma}{4\,\pi\, U_0\, a}
\end{align}
This condition \eqref{if:eq:doubletVortex00} 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
\begin{align}
\label{if:eq:doubletVortex:limit}
\left \Gamma \right \le 4\,\pi\, U_0\, a
\end{align}
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
\begin{align}
\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}
\end{align}
The solution for equation (theta, $\theta$) \eqref{if:eq:doubletVortex:exmpleValue15} 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 \eqref{if:eq:doubletVortex0a}.
For the previous example with specific value of the ratio, $\overline{\Gamma}$ as
\begin{align}
\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 }
\end{align}
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 \eqref{if:eq:doubletVortex}
can be written as
\begin{align}
\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}
\end{align}
or in a previous dimensionless form plus multiply by $\overline{r}$ as
\begin{align}
\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}}
\end{align}
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 \eqref{if:eq:doubletVortexPsiInterd} becomes
\begin{align}
\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}
\end{align}
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 \eqref{if:eq:doubletVortexPsiGov} 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) \eqref{if:eq:doubletVortexPsiGov} is rearranged as
\begin{align}
\label{if:eq:doubletVortexPsiGovR}
\theta = \sin^{1} \left( \dfrac {2\,\overline{\Gamma}\, \overline{r} \,\ln\overline{r} }
{1  \overline{r}^2} \right)
\end{align}
Equation \eqref{if:eq:doubletVortexPsiGov} 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.
The results are shown in Figure 10.16.
Figure \ref{if:fig:uniVortexG02} depicts the stream lines when the dimensionless vortex is below one.
Figure 10.16(b) depicts the limiting case where the dimensionless vortex is exactly one.
Once the dimensionless vortex exceeds one, the stagnation points do touch the solid body.
This question is more as a project for students of Fluid Mechanics or Aerodynamics.
The stream lines can be calculated in two ways.
The first way is for the given $n$, the radius can be calculated from equation \eqref{if:eq:doubletVortexPsiGov}.
The second is by calculating the angle for given $r$ from equation \eqref{if:eq:doubletVortexPsiGovR}.
Examine the code (attached with the source code) that was used in generating Figures 10.16 and
describe or write the algorithm what was used.
What is the ``dead'' radius zones?
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
\begin{align}
\label{if:eq:BernoulliEq}
P_{\theta} = P_0  \dfrac{1}{2} \rho \left( {U_r}^2 + {U_{\theta}}^2 \right)
\end{align}
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 \eqref{if:eq:doubletVortex:Ur} where $r=a$) which
can be written as
\begin{align}
\label{if:eq:tangentialVortex}
U_{\theta} = 2\,U_0\,\sin\theta + \dfrac{\Gamma}{2\,\pi\,a}
\end{align}
Thus, the pressure on the cylinder can be written as
\begin{align}
\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)
\end{align}
Equation \eqref{if:eq:vortex:P} 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
\begin{align}
\label{if:eq:dF}
dF = \bbb{j} \,\bbb{\cdot}\,P\, \bbb{n} dA
\end{align}
where in this case $\bbb{j}$ 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 $\bbb{n}$.
To carry the integration the unit vector $\bbb{n}$ is written as
\begin{align}
\label{if:eq:unitVector}
\bbb{n} = \bbb{i}\,\cos\theta + \bbb{j}\,\sin\theta
\end{align}
The reason for definition or split \eqref{if:eq:unitVector} to take into account only the
the vertical component.
Using the above derivation leads to
\begin{align}
\label{if:eq:vortex:dotV}
\bbb{j}\bbb{\cdot}\bbb{n} = \sin\theta
\end{align}
The lift per unit length will be
\begin{align}
\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
\end{align}
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 \eqref{if:eq:vortex:liftIntegralIni} is
\begin{align}
\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
\end{align}
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.
Fig. 10.17 Schematic to explain Magnus's effect.
In certain ranges the simultaneously translate and rotation movement causes the lift of the
moving object.
This can be observed in a thrown ball with spin over 1000 rpm and speed in over 5 m/sec.
In these parameters, the ball is moving in curved line to the target.
To understand the reason for this curving, the schematic if the ball is drawn (Figure 10.17).
The ball is moving to the right and rotating counter clockwise.
The velocity at the top of the ball is reduced due to the rotation
while the velocity at the bottom of the ball is increased.
According to Bernoulli's equation, reduction or increase of the velocity changes the static pressure.
Hence, the static pressure is not symmetrical and it causes a force perpendicular to the ball movement.
It can be noticed the direction of the rotation changes the direction of the forces.
In addition to the change of the pressure, the resistance changes because it is a function of the velocity.
In many ranges the increase of the velocity increase the resistance.
Hence, there are two different velocities at the top and bottom.
The resistance, as a function of the velocity, is different on the bottom as compared to
the top.
These two different mechanisms cause the ball to move in perpendicular direction to the
flow direction.
The circulation mimics the Magnus's effect and hence it is used in representative flow.
In the above discussion it was used for body of perfect circular shape.
However, it was observed that bodies with a very complicated shape such as airplane wing, the lift can
be represented by of vortex.
from the numerical method of Runge–Kutta and by the Russian
Zhukovsky suggest that the dimensionless nature of vortex is controlling the any shape.
The extension can be done by defining the circulation as
\begin{align}
\label{if:eq:generalGamma}
\Gamma= \oint_{C} \bbb{U} \boldsymbol{\cdot} \bbb{ds}=\oint_{C} U\,\cos\theta\, ds
\end{align}
Fig. 10.18 Wing in a typical uniform flow.
\begin{align}
\label{if:eq:kuttaJoukowski}
L= \rho_\infty\, U_\infty\,\Gamma,
\end{align}
The circulation of a ball or cylinder is easy to imagine.
Yet a typical air plane do not rotate.
Perhaps, the representation of inviscid flow of with vortex can represent the viscous flow.
For example flow airplane wing will have typical stream line such as shown in Figure
10.18.
However, the viscous flow does not behaves in this fashion especially at the trailing part of the
wing.
The flow around the wing sheds vortexes because the sharp turn of the flow.
The sheds vortexes existence is like the free vortexes since integral including these vortexes
can be included in the calculations of the circulation (see equation 10.197).
Next:
IdealFlow3
Previous:
IdealFlow1
