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Mathematics of fluid dynamics University of Bath 25-26, MA32051/MA52129

 

Section 6.3 Axisymmetric flow

In this section, we only covered Definition 6.3.1 and Definition 6.3.3. We did not do examples. For the exam, you therefore need to be aware of the concept of axisymmetric flow and the Stokes streamfunction, but you wouldn’t be asked to do a calculation.
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In the special case in which the flow possesses axisymmetry, you might not be surprised to learn that the equations simplify considerably. In fact, axisymmetry can be viewed as a reduction of the dimension of the problem and axisymmetric flows share some features in common with two-dimensional flows.
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Definition 6.3.1. Axisymmetric flow.

Given an axis of symmetry, an axisymmetric flow must possess:
  • rotational symmetry about the axis, meaning the flow is unaffected by a rotation through any angle about the axis, and
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  • reflectional symmetry in any plane containing the axis.
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Remark 6.3.2.

Note that rotationally symmetric flows are not in general axisymmetric. Working in cylindrical polar coordinates \((r,\theta.z)\) centred on the axis of symmetry, any flow that is independent of \(\theta\) is rotationally symmetric. In order to have the stronger property of being axisymmetric, the flow must also satisfy \(u_{\theta}=0\text{.}\)
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A flow \(\bu=f(r,z,t)\be_{\theta}\) is rotationally symmetric but not axisymmetric, for example a line vortex Example 4.2.10.
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Subsection 6.3.1 The Stokes streamfunction for axisymmetric flow

Definition 6.3.3. The Stokes streamfunction.

In cylindrical polar coordinates, \((r,\theta.z)\text{,}\) we define the Stokes streamfunction \(\Psi(r,z,t)\) by
\begin{equation} \bu = \nabla \times \left(\frac{\Psi}{r}\be_{\theta}\right),\tag{6.3.1} \end{equation}
and in spherical polar coordinates, \((r,\theta.\phi)\text{,}\) we define \(\Psi(r,\theta,t)\) by
\begin{equation} \bu = \nabla \times \left(\frac{\Psi}{r\sin\theta}\be_{\phi}\right).\tag{6.3.2} \end{equation}
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Remark 6.3.4.

Note that you would be given the formula for gradient, divergence and curl in cylindrical or spherical polar coordinates.
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Remark 6.3.5.

The definitions (6.3.1) in cylindrical polar coordinates and (6.3.1) in spherical polar coordinates are equivalent since \(\be_{\theta}/r\) in cylindrical polar coordinates and \(\be_{\phi}/(r\sin\theta)\) in spherical polar coordinates are equal to each other.
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Remark 6.3.6.

Recall the streamfunction for two-dimensional flow that was introduced in Section 4.2:
\begin{equation*} \bu = \nabla \times (\psi \bk). \end{equation*}
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Remark 6.3.7.

In cylindrical polar coordinates, if we had written \(\bu=\nabla\times\psi'\be_{\theta}\) instead of (6.3.1), we would find that \(\psi'\) is inversely proportional to \(r\) on streamlines (instead of being constant). This is why it is defined with the factor \(1/r\) before taking the curl. Similar considerations apply in spherical polar coordinates.
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Remark 6.3.8.

The velocity components corresponding to the definitions Definition 6.3.3 are in cylindrical polar coordinates
\begin{equation} u_r=-\frac1r\pd{\Psi}{z},\quad u_z=\frac1r\pd{\Psi}{r},\tag{6.3.3} \end{equation}
and in cylindrical polar coordinates
\begin{equation} u_r=\frac1{r^2\sin\theta}\pd{\Psi}{\theta},\quad u_{\theta}=-\frac1{r\sin\theta}\pd{\Psi}{r}.\tag{6.3.4} \end{equation}
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Proof.

Since the definitions in cylindrical and spherical coordinates are equivalent, it suffices to prove the theorem in spherical coordinates. In this case,
\begin{equation*} \bu\cdot\nabla\Psi =u_r\pd{\Psi}{r}+\frac{u_{\theta}}r\pd{\Psi}{\theta} \end{equation*}
and, substituting the velocity components (6.3.4), this equals zero. Hence \(\Psi\) is constant along streamlines.
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Subsection 6.3.2 Relationship to vorticity

In cylindrical polar coordinates, the vorticity is given by
\begin{equation} \bomega=\omega\be_{\theta}\tag{6.3.5} \end{equation}
where
\begin{equation} \omega=-\frac1r\left(r\pd{}{r}\left(\frac1r\pd{\Psi}{r}\right)+\pd{^2\Psi}{z^2}\right).\tag{6.3.6} \end{equation}
In spherical polar coordinates, the vorticity is
\begin{equation} \bomega=\omega\be_{\phi}\tag{6.3.7} \end{equation}
where
\begin{equation} \omega=-\frac1{r\sin\theta}\left(\pd{^2\Psi}{r^2}+\frac{\sin\theta}{r^2}\pd{}{\theta}\left(\frac1{\sin\theta}\pd{\Psi}{\theta}\right)\right).\tag{6.3.8} \end{equation}
Note that the two expressions for \(\omega\) are equivalent, since \(\be_{\theta}\) in cylindricals equals \(\be_{\theta}\) in spherical polar coordinates.
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In cylindrical polar coordinates, the vorticity equation for axisymmetric flow is
\begin{equation} \DD{}{t}\left(\frac{\omega}{r}\right)=0,\tag{6.3.9} \end{equation}
and in sphericals it is
\begin{equation} \DD{}{t}\left(\frac{\omega}{r\sin\theta}\right)=0.\tag{6.3.10} \end{equation}
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Example 6.3.10. Smoke rings.

In Exercise 6.5.4 you are asked to prove a formulation of the vorticity equation (3.5.8) in the special case of axisymmetric flow. Note that, in this case, the vorticity points purely in the azimuthal direction.
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A smoke ring is an example of an axisymmetric flow with vorticity, see 6:44-7:53 of Vorticity Part 2. We may think of the flow as being everywhere irrotational, except in the smoke ring, with the smoke ring itself representing a vortex tube.
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As a ring gets wider, conservation of mass implies the vortex tube gets thinner inversely proportional to the distance from the axis, and the second Helmholtz vortex theorem implies the vorticity must increase in proportion to distance from the axis so that the circulation remains constant. In Exercise 6.5.4 you will show that, in cylindrical polar coordinates,
\begin{equation*} \DD{}{t}\left(\frac{\omega}{r}\right)=0, \end{equation*}
which expresses the same result: the vorticity \(\omega\) increases in proportion to distance from axis \(r\text{.}\)
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Note also that, as commented in the movie, the smoke ring self-propates. This is another application of the Helmholtz principle, as follows. If we consider a part of the ring, it is subject to the velocity field arising from all the other parts of the ring. All these parts contribute to a forward motion, and this is true for all parts of the ring.
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Subsection 6.3.3 Irrotational flow past a sphere

We consider a sphere of radius \(a\) travelling at a constant speed \(U\) through a large reservoir of fluid and find the velocity field. We assume the sphere has been travelling for long enough that the flow in a frame of reference moving with the sphere is steady.
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Proof that the flow is irrotational: We work in spherical polar coordinates in the frame of reference moving with the sphere and solve (6.3.10) to find the Stokes streamfunction. Steady flow implies the simplification
\begin{equation*} \bu\cdot\nabla\left(\frac{\omega}{r\sin\theta}\right)=0, \end{equation*}
which means that \(\omega/(r\sin\theta)\) is constant along streamlines. As long as there are no closed streamlines, meaning that all streamlines originate at infinity where \(\omega\) is zero, then \(\omega\) must be zero everywhere and the flow is irrotational.
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Setting the vorticity to zero and using the expression (6.3.8), we have to solve
\begin{equation} \pd{^2\Psi}{r^2} +\frac{\sin\theta}{r^2}\pd{}{\theta}\left(\frac1{\sin\theta}\pd{\Psi}{\theta}\right)=0\tag{6.3.11} \end{equation}
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As \(r\rightarrow\infty\text{,}\) we require
\begin{equation*} u_r\sim-U\cos\theta,\quad u_{\theta}\sim U\sin\theta. \end{equation*}
Using the expressions given in (6.3.4), these conditions are equivalent to
\begin{equation} \Psi\sim-\frac12Ur^2\sin^2\theta\tag{6.3.12} \end{equation}
as \(r\rightarrow\infty\text{.}\)
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This suggests trying
\begin{equation*} \Psi=f(r)\sin^2\theta. \end{equation*}
We substitute this into (6.3.11), and find that it is a solution provided that
\begin{equation*} f''-\frac{2f}{r^2}=0, \end{equation*}
which means that
\begin{equation*} f=Ar^2+\frac{B}{r}, \end{equation*}
where \(A\) and \(B\) are constants of integration.
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We need to satisfy the condition at infinity (6.3.12) which gives \(f\sim Ur^2/2\) as \(r\rightarrow\infty\text{,}\) and no penetration at the surface of the sphere, which, using (6.3.4), is equivalent to \(f=0\) at \(r=a\text{.}\)
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These boundary conditions can be satified by the choice \(A=U/2\text{,}\) \(B=-Ua^2/2\text{,}\) and we have
\begin{equation*} \Psi=-\frac12U\left(r^2-\frac{a^3}{r}\right)\sin^2\theta. \end{equation*}
This can be used to sketch the streamlines.
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If required, we can use (6.3.4) to find the velocity components
\begin{equation*} u_r=-U\left(1-\frac{a^3}{r^3}\right)\cos\theta,\quad u_{\theta}=U\left(1+\frac{a^3}{2r^3}\right)\sin\theta. \end{equation*}
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Remark 6.3.11. Alternative method to find the flow using the velocity potential.

Note that we could also use the velocity potential Theorem 3.5.8 to solve this problem. In this case we assume the flow is irrotational to start with.
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Again we work in spherical polar coordinates moving at speed \(U\) with the sphere with the axis pointing in the direction of travel and assume the flow is axisymmetric. Writing \(/bu=\nabla\phi\text{,}\) where \(\phi\) is the potential, the flow is governed by the Laplace equation:
\begin{equation*} \frac1{r^2}\pd{}{r}\left(r^2\pd{\phi}{r}\right) +\frac1{r^2\sin\theta}\pd{}{\theta}\left(\sin\theta\pd{\phi}{\theta}\right)=0. \end{equation*}
We solve this equation subject to boundary conditions of no penetration at \(r=a\) and assume that, far from the sphere, the flow is \(-U\) along the axis:
  • No penetration at the sphere’s surface implies that \({\bf u}\cdot{\bf n}=0\) there. So \(u_r=\pd{\phi}{r}=0\text{.}\)
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  • That the flow tends to \(-U\) implies that \(u_r\rightarrow-U\cos\theta\) and \(u_\theta\rightarrow U\sin\theta\) as \(r\rightarrow\infty\text{.}\) That is, \(\pd{\phi}{r}\rightarrow-U\cos\theta\) and \(\pd{\phi}{\theta}\rightarrow rU\sin\theta\text{.}\)
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The boundary conditions as \(r\rightarrow\infty\) suggest trying \(\phi=f(r)\cos\theta\text{,}\) and, with this assumption, Equation~\eqref{eq:laplacephisimp} becomes
\begin{equation*} \pd{^2f}{r^2}+\frac2r\pd{f}{r}-\frac{2f}{r^2}=0. \end{equation*}
Trying \(f=r^\alpha\) yields \(\alpha=1\) or~\(-2\text{.}\) Thus
\begin{equation*} \phi=\left(Ar+\frac{B}{r^2}\right)\cos\theta, \end{equation*}
and the boundary conditions at \(r=a\) imply \(A-2B/a^3=0\text{,}\) and those as \(r\rightarrow\infty\) imply \(A=-U\text{.}\) Thus \(A=-U\) and \(B=-Ua^3/2\text{,}\)
\begin{equation*} \phi=-Ur\left(1+\frac{a^3}{2r^3}\right)\cos\theta, \end{equation*}
and
\begin{equation*} u_r=\pd{\phi}{r}=-U\left(1-\frac{a^3}{r^3}\right)\cos\theta,\quad u_\theta=\frac1r\pd{\phi}{\theta}=U\left(1+\frac{a^3}{2r^3}\right)\sin\theta. \end{equation*}
This solution has the correct form as \(r\rightarrow\infty\text{,}\) and at the surface of the sphere. It also agrees with the velocity field calculated using the streamfunction method.
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Remark 6.3.12.

At the spherical surface \(r=a\text{,}\) we have the velocity \(u_\theta=(3U/2)\sin\theta\text{.}\) In practical terms, the large pressure gradients created by this flow would lead to boundary layer separation, and a large wake behind the sphere.
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Remark 6.3.13.

In Example 7.8.15 we will calculate the flow past a sphere in a low-Reynolds-number flow.
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