# 0.10 Numerical simulation  (Page 8/8)

 Page 8 / 8

This is the equation when the coordinates have the same reference length. If the coordinates are normalized with respect to different lengths, then the equation is as follows.

$\frac{{\partial }^{2}P}{\partial {x}^{2}}+{\alpha }^{2}\frac{{\partial }^{2}P}{\partial {y}^{2}}=2{\alpha }^{2}\left[\frac{{\partial }^{2}\psi }{\partial {x}^{2}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\frac{{\partial }^{2}\psi }{\partial {y}^{2}}-{\left(\frac{{\partial }^{2}\psi }{\partial x\partial y}\right)}^{2}\right]$

The Poisson equation for pressure needs to have boundary conditions for solution. The equations of motion give an expression for the pressure gradient. The normal derivative of pressure at the boundary can be determined for the Neumann-type boundary condition. The equations of motion usually simplify at boundaries if the boundary has no slip BC or if it has parallel flow in the direction either parallel or normal to the boundary. For example, for no-slip

$\begin{array}{c}\frac{\partial P}{\partial x}=\frac{{\alpha }^{2}}{Re}\phantom{\rule{0.166667em}{0ex}}\frac{{\partial }^{2}u}{\partial {y}^{2}},\phantom{\rule{1.em}{0ex}}\mathrm{at}\phantom{\rule{0.277778em}{0ex}}y=c,\phantom{\rule{0.277778em}{0ex}}u=v=0\\ \alpha \frac{\partial P}{\partial y}=\frac{1}{Re}\phantom{\rule{0.166667em}{0ex}}\frac{{\partial }^{2}v}{\partial {x}^{2}},\phantom{\rule{1.em}{0ex}}\mathrm{at}\phantom{\rule{0.277778em}{0ex}}x=c,\phantom{\rule{0.277778em}{0ex}}u=v=0\end{array}$

If the flow is parallel and in the direction normal to the boundary

$\begin{array}{c}\frac{\partial P}{\partial x}=\frac{{\alpha }^{2}}{Re}\phantom{\rule{0.166667em}{0ex}}\frac{{\partial }^{2}u}{\partial {y}^{2}},\phantom{\rule{1.em}{0ex}}y=c,\phantom{\rule{0.277778em}{0ex}}v=0,\phantom{\rule{0.277778em}{0ex}}u=u\left(y\right)\\ \alpha \frac{\partial P}{\partial y}=\frac{1}{Re}\phantom{\rule{0.166667em}{0ex}}\frac{{\partial }^{2}v}{\partial {x}^{2}},\phantom{\rule{1.em}{0ex}}x=c,\phantom{\rule{0.277778em}{0ex}}u=0,\phantom{\rule{0.277778em}{0ex}}v=v\left(x\right)\end{array}$

## Cylindrical-polar coordinates

The code for the numerical solution of the Navier-Stokes equations in Cartesian coordinates can be easily modified to cylindrical-polar coordinates. The coordinate transformation is first illustrated for the Laplacian operator.

${\nabla }^{2}\psi =\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \psi }{\partial r}\right)+\frac{1}{{r}^{2}}\frac{{\partial }^{2}\psi }{\partial {\theta }^{2}},\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}{r}_{1}\le r\le {r}_{2},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{1.em}{0ex}}0\le \theta \le {\theta }_{o}$

The independent variables are made dimensionless with respect to the boundary parameters.

${r}^{*}=\frac{r}{{r}_{1}},\phantom{\rule{1.em}{0ex}}{\theta }^{*}=\frac{\theta }{{\theta }_{o}},\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}1\le {r}^{*}\le \frac{{r}_{2}}{{r}_{1}}=\beta ,\phantom{\rule{1.em}{0ex}}0\le {\theta }^{*}\le 1$

The Lapacian operator with the dimensionless coordinates after dropping the * is now,

${\nabla }^{2}\psi =\frac{1}{{r}_{1}^{2}}\left[\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \psi }{\partial r}\right)+\frac{1}{{\theta }_{o}^{2}}\frac{1}{{r}^{2}}\frac{{\partial }^{2}\psi }{\partial {\theta }^{2}}\right],\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}1\le r\le \beta ,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{1.em}{0ex}}0\le \theta \le 1$

$\begin{array}{c}z=\frac{lnr}{ln\beta }=\gamma lnr,\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}0\le z\le 1,\phantom{\rule{1.em}{0ex}}\gamma =\frac{1}{ln\left({r}_{2}/{r}_{1}\right)}\hfill \\ r=exp\left(z/\gamma \right)\hfill \\ \frac{\partial }{\partial r}=\frac{dz}{dr}\frac{\partial }{\partial z}=\frac{\gamma }{r}\frac{\partial }{\partial z}\hfill \\ \frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \psi }{\partial r}\right)=\frac{{\gamma }^{2}}{{r}^{2}}\frac{{\partial }^{2}\psi }{\partial {z}^{2}}\hfill \end{array}$

The Laplacian operator is now as follows,

${\nabla }^{2}\psi =\frac{1}{{r}_{1}^{2}}\left[\frac{{\gamma }^{2}}{{r}^{2}}\frac{{\partial }^{2}\psi }{\partial {z}^{2}}+\frac{{\alpha }^{2}}{{r}^{2}}\frac{{\partial }^{2}\psi }{\partial {\theta }^{2}}\right],\phantom{\rule{1.em}{0ex}}0\le z\le 1,\phantom{\rule{1.em}{0ex}}0\le \theta \le 1,\phantom{\rule{1.em}{0ex}}\alpha =\frac{1}{{\theta }_{o}}$

The finite difference expression for the Laplacian will be same as that for Cartesian coordinates with $\left(z,\theta \right)$ substituted for $\left(x,y\right)$ except for ${\gamma }^{2}$ and ${r}^{2}$ factors.

$\begin{array}{c}0\le {z}_{i}\le 1,\phantom{\rule{1.em}{0ex}}0\le {\theta }_{j}\le 1,\phantom{\rule{1.em}{0ex}}i,j=1,2,...,JMAX\hfill \\ \delta =\frac{1}{JMAX-1}\hfill \\ {z}_{i}=\delta \left(i-1\right),\phantom{\rule{1.em}{0ex}}{\theta }_{j}=\delta \left(j-1\right)\hfill \end{array}$

The curl operator is modified from that in Cartesian coordinates.

$\begin{array}{ccc}\hfill {v}_{r}& =& \frac{\alpha }{r}\frac{\partial \psi }{\partial \theta }\hfill \\ \hfill {v}_{\theta }& =& -\frac{\partial \psi }{\partial r}=-\frac{\gamma }{r}\frac{\partial \psi }{\partial z}\hfill \\ \hfill w& =& \frac{1}{r}\frac{\partial \left(r\phantom{\rule{0.277778em}{0ex}}{v}_{\theta }\right)}{\partial r}-\frac{\alpha }{r}\frac{\partial {v}_{r}}{\partial \theta }\hfill \\ & =& \frac{\gamma }{{r}^{2}}\frac{\partial \left(r\phantom{\rule{0.277778em}{0ex}}{v}_{\theta }\right)}{\partial z}-\frac{\alpha }{r}\frac{\partial {v}_{r}}{\partial \theta }\hfill \\ & =& -\frac{{\gamma }^{2}}{{r}^{2}}\frac{{\partial }^{2}\psi }{\partial {z}^{2}}-\frac{{\alpha }^{2}}{{r}^{2}}\frac{{\partial }^{2}\psi }{\partial {\theta }^{2}}\hfill \end{array}$

The vorticity boundary conditions with the transformed coordinates is for the $z$ boundary,

${w}_{1}^{BC}=-\frac{{\gamma }^{2}}{{r}^{2}}\frac{2}{{\delta }^{2}}\left({\psi }_{2}-{\psi }_{1}^{BC}\right)\mp \frac{\gamma }{r}\frac{2}{\delta }{v}_{\theta }^{BC}-\frac{\alpha }{r}\frac{\partial {v}_{r}^{BC}}{\partial \theta }$

and for the $\theta$ boundary,

${w}_{1}^{BC}=-\frac{{\alpha }^{2}}{{r}^{2}}\frac{2}{{\delta }^{2}}\left({\psi }_{2}-{\psi }_{1}^{BC}\right)±\frac{\alpha }{r}\frac{2}{\delta }{v}_{r}^{BC}+\frac{\gamma }{{r}^{2}}\frac{\partial \left(r{v}_{\theta }^{BC}\right)}{\partial z}$

The stream function at the boundaries are expressed different from that in Cartesian coordinates.

$\begin{array}{c}d\psi =\left(r\phantom{\rule{0.277778em}{0ex}}{v}_{r}/\alpha \right)\phantom{\rule{0.166667em}{0ex}}d\theta ,\phantom{\rule{1.em}{0ex}}\mathrm{at}\phantom{\rule{0.277778em}{0ex}}z\phantom{\rule{0.277778em}{0ex}}\mathrm{boundary}\hfill \\ d\psi =-\left(r\phantom{\rule{0.277778em}{0ex}}{v}_{\theta }/\gamma \right)\phantom{\rule{0.166667em}{0ex}}dz,\phantom{\rule{1.em}{0ex}}\mathrm{at}\phantom{\rule{0.277778em}{0ex}}\theta \phantom{\rule{0.277778em}{0ex}}\mathrm{boundary}\hfill \end{array}$

The convective terms are expressed different from that in Cartesian coordinates.

$\begin{array}{ccc}\hfill \frac{\partial \left(u\phantom{\rule{0.166667em}{0ex}}w\right)}{\partial x}& ⇒& \frac{1}{r}\frac{\partial \left(r\phantom{\rule{0.166667em}{0ex}}u\phantom{\rule{0.166667em}{0ex}}w\right)}{\partial r}\hfill \\ & =& \frac{\gamma }{{r}^{2}}\frac{\partial \left(r\phantom{\rule{0.166667em}{0ex}}u\phantom{\rule{0.166667em}{0ex}}w\right)}{\partial z}\hfill \\ & \approx & \frac{\gamma }{{r}^{2}\phantom{\rule{0.166667em}{0ex}}\delta }\left[{\left(r\phantom{\rule{0.166667em}{0ex}}u\phantom{\rule{0.166667em}{0ex}}w\right)}_{i+1/2}-{\left(r\phantom{\rule{0.166667em}{0ex}}u\phantom{\rule{0.166667em}{0ex}}w\right)}_{i-1/2}\right]\hfill \\ \hfill \alpha \frac{\partial \left(v\phantom{\rule{0.166667em}{0ex}}w\right)}{\partial y}& ⇒& \frac{\alpha }{r}\frac{\partial \left(v\phantom{\rule{0.166667em}{0ex}}w\right)}{\partial \theta }\hfill \\ & \approx & \frac{\alpha }{r\phantom{\rule{0.166667em}{0ex}}\delta }\left[{\left(v\phantom{\rule{0.166667em}{0ex}}w\right)}_{j+1/2}-{\left(v\phantom{\rule{0.166667em}{0ex}}w\right)}_{j-1/2}\right]\hfill \end{array}$

The code should be written so one can have either Cartesian coordinates or transformed cylindrical-polar coordinates. A parameter will be needed to identify the choice of coordinates, e.g. $icase=1$ for Cartesian coordinates and $icase=2$ for transformed cylindrical-polar coordinates. Also, another parameter should be specified to identify the choice of boundary conditions, e.g., $ibc=1$ for radial flow, $ibc=2$ for Couette flow, and $ibc=3$ for flow around a cylinder. Test cases with known solutions should be used to verify the code. The first case is radial, potential flow from a line source and the second is Couette flow in the annular region between two cylindrical surfaces.

$\begin{array}{c}\left(\begin{array}{c}{v}_{r}=1/r\\ {v}_{\theta }=0\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\end{array}}\phantom{\rule{1.em}{0ex}}\mathrm{radial}\phantom{\rule{0.277778em}{0ex}}\mathrm{flow}\hfill \\ \left(\begin{array}{c}{v}_{r}=0\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\\ {v}_{\theta }=\left(r-1/r\right)/\left(\beta -1/\beta \right)\end{array}}\phantom{\rule{1.em}{0ex}}\mathrm{Couette}\phantom{\rule{0.277778em}{0ex}}\mathrm{flow}\hfill \end{array}$

Flow around a cylinder needs a boundary condition for the flow far away from the cylinder. The flow very far away may be uniform translation. However, this condition may be so far away that it may result in loss of resolution near the cylinder. Another boundary condition that may be specified beyond the region of influence of the boundary layer is to use the potential flow past a cylinder. This boundary condition will not be correct in the wake of cylinder where the flow is disturbed by the convected boundary layer buts its influence may be minimized if the outer boundary is far enough.

$\left(\begin{array}{c}{v}_{r}=\left(1-1/{r}^{2}\right)\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.277778em}{0ex}}cos\theta \\ {v}_{\theta }=-\left(1+1/{r}^{2}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}sin\theta \phantom{\rule{0.277778em}{0ex}}\end{array}}\phantom{\rule{1.em}{0ex}}\mathrm{Potential}\phantom{\rule{0.277778em}{0ex}}\mathrm{flow}\phantom{\rule{0.277778em}{0ex}}\mathrm{past}\phantom{\rule{0.277778em}{0ex}}\mathrm{cylinder}$

Potential flow will not be a valid approximation along the $\theta$ boundaries close to the cylinder. Here, one may assume a surface of symmetry, at least for the upstream side.

anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Got questions? Join the online conversation and get instant answers!