# Elementary flow

Elementary flow is a collection of basic flows from which is possible to construct more complex flows by superposition. Some of the flows reflect specific cases and constraints such as incompressible, irrotational or both as in the case of Potential flow.[1]

## Two Dimensional uniform flow

Given a uniform velocity of a fluid at any position in space:

${\displaystyle \mathbf {V_{0}} =v_{0}\cos(\theta _{0})\mathbf {e} _{x}+v_{0}\sin(\theta _{0})\mathbf {e} _{y}}$

This flow is incompressible because the velocity is constant, the first derivatives of the velocity components are zero, and the total divergence is zero: ${\displaystyle \nabla \cdot \mathbf {v} =0}$

Given the circulation is always zero the flow is also irrotational, we can derive this from the Kelvin's circulation theorem and from the explicit computation of the vorticity:

${\displaystyle \omega _{z}={\frac {\partial v_{x}}{\partial y}}-{\frac {\partial v_{y}}{\partial x}}=0}$

Being incompressible and two dimensional this flow is constructed from a stream function

${\displaystyle v_{x}=-{\frac {\partial \psi }{\partial y}}}$
${\displaystyle v_{y}={\frac {\partial \psi }{\partial x}}}$

from which

${\displaystyle \psi =-v_{0}\sin(\theta _{0})x+v_{0}\cos(\theta _{0})y}$

And in cylindrical coordinates:

${\displaystyle v_{r}=-{\frac {1}{r}}{\frac {\partial \psi }{\partial \theta }}}$
${\displaystyle v_{\theta }={\frac {\partial \psi }{\partial r}}}$

from which

${\displaystyle \psi =-v_{0}r\sin(\theta -\theta _{0})}$

As usual the stream function is defined up to a constant value which here we take as zero. We can also confirm that the flow is irrotational from

${\displaystyle \nabla ^{2}\psi =0}$

Being irrotational, The potential function is instead:

${\displaystyle v_{x}=-{\frac {\partial \phi }{\partial x}}}$
${\displaystyle v_{y}=-{\frac {\partial \phi }{\partial y}}}$

and therefore

${\displaystyle \phi =-v_{0}\cos(\theta _{0})x-v_{0}\sin(\theta _{0})y}$

And in cylindrical coordinates

${\displaystyle v_{r}=-{\frac {1}{r}}{\frac {\partial \phi }{\partial r}}}$
${\displaystyle v_{\theta }=-{\frac {\partial \phi }{\partial \theta }}}$
${\displaystyle \phi =-v_{0}r\cos(\theta -\theta _{0})}$

## Two Dimensional line source

Dashed Potential flow streamlines for an ideal vertical line source

The case of a vertical line emitting at a fixed rate a constant quantity of fluid Q per unit length is a line source. The problem has a cylindrical symmetry and can be treated in two dimension on the orthogonal plane.

Line sources and line sinks (below) are important elementary flows because they play the role of monopole(s) for incompressible fluids (which can also be considered examples of solenoidal fields i.e. divergence free fields). Generic flow patterns can be also de-composed in terms of multipole expansions, in the same manner as for electric and magnetic fields where the monopole is essentially the first non trivial (e.g. constant) term of the expansion.

This flow pattern is also both irrotational and incompressible flow.

This is characterized by a cylindrical symmetry:

${\displaystyle \mathbf {v} =v_{r}(r)\mathbf {e} _{r}}$

Where the total outgoing flux is constant

${\displaystyle \int _{S}\mathbf {v} \cdot d\mathbf {S} =\int _{0}^{2\pi }(v_{r}(r)\,\mathbf {e} _{r})\cdot (\mathbf {e} _{r}\,r\,d\theta )=\!2\pi \,r\,v_{r}(r)=Q}$

Therefore,

${\displaystyle v_{r}={\frac {Q}{2\pi r}}}$

This is derived from a stream function

${\displaystyle \psi (r,\theta )=-{\frac {Q}{2\pi }}\theta }$

or from a potential function

${\displaystyle \phi (r,\theta )=-{\frac {Q}{2\pi }}\ln r}$

## Two Dimensional line sink

The case of a vertical line absorbing at a fixed rate a constant quantity of fluid Q per unit length is a line sink. Everything is the same as the case of a line source a part from the negative sign.

${\displaystyle v_{r}=-{\frac {Q}{2\pi r}}}$

This is derived from a stream function

${\displaystyle \psi (r,\theta )={\frac {Q}{2\pi }}\theta }$

or from a potential function

${\displaystyle \phi (r,\theta )={\frac {Q}{2\pi }}\ln r}$

Given that the two results are the same a part from a minus sign we can treat transparently both line sources and line sinks with the same stream and potential functions permitting Q to assume both positive and negative values and absorbing the minus sign into the definition of Q.

## Two Dimensional doublet or dipole line source

If we consider a line source and a line sink at a distance d we can reuse the results above and the stream function will be

${\displaystyle \psi (\mathbf {r} )=\psi _{Q}(\mathbf {r} -\mathbf {d} /2)-\psi _{Q}(\mathbf {r} +\mathbf {d} /2)\ \simeq \mathbf {d} \cdot \nabla \psi _{Q}(\mathbf {r} )}$

The last approximation is to the first order in d.

Given

${\displaystyle \mathbf {d} =d[\cos(\theta _{0})\mathbf {e} _{x}+\sin(\theta _{0})\mathbf {e} _{y}]=d[\cos(\theta -\theta _{0})\mathbf {e} _{r}+\sin(\theta -\theta _{0})\mathbf {e} _{\theta }]}$

It remains

${\displaystyle \psi (r,\theta )=-{\frac {Qd}{2\pi }}{\frac {\sin(\theta -\theta _{0})}{r}}}$

The velocity is then

${\displaystyle v_{r}(r,\theta )={\frac {Qd}{2\pi }}{\frac {\cos(\theta -\theta _{0})}{r^{2}}}}$
${\displaystyle v_{\theta }(r,\theta )={\frac {Qd}{2\pi }}{\frac {\sin(\theta -\theta _{0})}{r^{2}}}}$

And the potential instead

${\displaystyle \phi (r,\theta )={\frac {Qd}{2\pi }}{\frac {\cos(\theta -\theta _{0})}{r}}}$

## Two Dimensional vortex line

Dashed Potential flow streamlines for an ideal vertical vortex line

This is the case of a vortex filament rotating at constant speed, there is a cylindrical symmetry and the problem can be solved in the orthogonal plane.

Dual to the case above of line sources, vortex lines play the role of monopoles for irrotational flows.

Also in this case the flow is also both irrotational and incompressible and therefore a case of Potential flow.

This is characterized by a cylindrical symmetry:

${\displaystyle \mathbf {v} =v_{\theta }(r)\,\mathbf {e} _{\theta }}$

Where the total circulation is constant for every closed line around the central vortex

${\displaystyle \oint \mathbf {v} \cdot d\mathbf {s} =\int _{0}^{2\pi }(v_{\theta }(r)\,\mathbf {e} _{\theta })\cdot (\mathbf {e} _{\theta }\,r\,d\theta )=\!2\pi \,r\,v_{\theta }(r)=\Gamma }$

and is zero for any line not including the vortex.

Therefore,

${\displaystyle v_{\theta }={\frac {\Gamma }{2\pi r}}}$

This is derived from a stream function

${\displaystyle \psi (r,\theta )={\frac {\Gamma }{2\pi }}\ln r}$

or from a potential function

${\displaystyle \phi (r,\theta )=-{\frac {\Gamma }{2\pi }}\theta }$

Which is dual to the previous case of a line source

## Generic Two Dimensional potential flow

Given an incompressible two dimensional flow which is also irrotational we have:

${\displaystyle \nabla ^{2}\psi =0}$

Which is in cylindrical coordinates [2]

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial \psi }{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\psi }{\partial \theta ^{2}}}=0}$

We look for a solution with separated variables:

${\displaystyle \psi (r,\theta )=R(r)\Theta (\theta )}$

which gives

${\displaystyle {\frac {r}{R(r)}}{\frac {d}{dr}}\left(r{\frac {dR(r)}{dr}}\right)=-{\frac {1}{\Theta (\theta )}}{\frac {d^{2}\Theta (\theta )}{d\theta ^{2}}}}$

Given the left part depends only on r and the right parts depends only on ${\displaystyle \theta }$, the two parts must be equal to a constant independent from r and ${\displaystyle \theta }$. The constant shall be positive[clarification needed]. Therefore,

${\displaystyle r{\frac {d}{dr}}\left(r{\frac {d}{dr}}R(r)\right)=m^{2}R(r)}$
${\displaystyle {\frac {d^{2}\Theta (\theta )}{d\theta ^{2}}}=-m^{2}\Theta (\theta )}$

The solution to the second equation is a linear combination of ${\displaystyle e^{im\theta }}$ and ${\displaystyle e^{-im\theta }}$ In order to have a single valued velocity (and also a single valued stream function) m shall be a positive integer.

therefore the most generic solution is given by

${\displaystyle \psi =\alpha _{0}+\beta _{0}\ln r+\sum _{m>0}{\left(\alpha _{m}r^{m}+\beta _{m}r^{-m}\right)\sin {[m(\theta -\theta _{m})]}}}$

The potential is instead given by

${\displaystyle \phi =\alpha _{0}-\beta _{0}\theta +\sum _{m\mathop {>} 0}{(\alpha _{m}r^{m}-\beta _{m}r^{-m})\cos {[m(\theta -\theta _{m})]}}}$

## References

• Fitzpatrick, Richard (2017), Theoretical fluid dynamics, IOP science, ISBN 978-0-7503-1554-8 CS1 maint: discouraged parameter (link)
• Faber, T.E. (1995), Fluid Dynamics for Physicists, Cambridge university press, ISBN 9780511806735 CS1 maint: discouraged parameter (link)
Specific
1. ^ Oliver, David (2013-03-14). The Shaggy Steed of Physics: Mathematical Beauty in the Physical World. Springer Science & Business Media. ISBN 978-1-4757-4347-0.
2. ^ Laplace operator