0.10 Numerical simulation  (Page 5/8)

 Page 5 / 8

If you are not familiar with FORTRAN, you should get a paperback book on FORTRAN-77 or FORTRAN-90. An example is D. M. Etter, Structured FORTRAN 77 for Engineers and Scientists . The following is a tutorial from the CAAM 211 webpage. (External Link)

Computer facilities that have compilers usually have an interactive debugger. However, I have not found how to access the debugger on owlnet . The debugger with the Microsoft FORTRAN Powerstation (now HP FORTRAN) is very easy to use.

Transients and finite reynolds number

The development of numerical simulators should start with a problem for which exact solutions exist so that the numerical algorithm can be verified. Steady, creeping flow has many exact solutions and thus it was used as the starting point for developing a numerical simulator for the Navier-Stokes equations. The student should verify that their code is able to duplicate exact solutions to any desired degree of accuracy.

The next stage in the development of the code is to add transients and finite Reynolds numbers. Finite Reynolds number results in nonlinear terms in the Navier-Stokes equation. Algorithms for solving linear systems of equations such as the conjugate gradient methods solves linear systems in one call to the routine. Nonlinear terms need to be iterated or lagged during the transient solution. Thus nonlinear steady state flow may be solved as the evolution of a transient solution from initial conditions to steady state.

The transient and nonlinear terms now need to be included when making the vorticity equation dimensionless.

$\begin{array}{c}\frac{Dw}{Dt}=\frac{\partial w}{\partial t}+{v}_{x}\frac{\partial w}{\partial x}+{v}_{y}\frac{\partial w}{\partial y}=\frac{\mu }{\rho }\left[\frac{{\partial }^{2}w}{\partial {x}^{2}}+\frac{{\partial }^{2}w}{\partial {y}^{2}}\right]\hfill \\ t*=\frac{t}{{t}_{o}}\hfill \\ \frac{\partial w*}{\partial t*}+\left[\frac{U\phantom{\rule{0.166667em}{0ex}}{t}_{o}}{{L}_{x}}\right]\phantom{\rule{0.166667em}{0ex}}{v}_{x}^{*}\frac{\partial w*}{\partial x*}+\alpha \left[\frac{U\phantom{\rule{0.166667em}{0ex}}{t}_{o}}{{L}_{x}}\right]\phantom{\rule{0.166667em}{0ex}}{v}_{y}^{*}\frac{\partial w*}{\partial y*}=\left[\frac{\mu \phantom{\rule{0.166667em}{0ex}}{t}_{o}}{\rho \phantom{\rule{0.166667em}{0ex}}{L}_{x}^{2}}\right]\phantom{\rule{0.166667em}{0ex}}\left[\frac{{\partial }^{2}w*}{\partial x{*}^{2}}+{\alpha }^{2}\frac{{\partial }^{2}w*}{\partial y{*}^{2}}\right]\hfill \end{array}$

Two of the dimensionless groups can be set to unity by expressing the reference time in terms of the ratio of the characteristic velocity and characteristic length. The remaining dimensionless group is the Reynolds number.

$\begin{array}{c}\left[\frac{U\phantom{\rule{0.166667em}{0ex}}{t}_{o}}{{L}_{x}}\right]=1\phantom{\rule{1.em}{0ex}}⇒\phantom{\rule{0.277778em}{0ex}}{t}_{o}=\frac{{L}_{x}}{U}\hfill \\ \left[\frac{\mu \phantom{\rule{0.166667em}{0ex}}{t}_{o}}{\rho \phantom{\rule{0.166667em}{0ex}}{L}_{x}^{2}}\right]=\left[\frac{\mu \phantom{\rule{0.166667em}{0ex}}}{\rho \phantom{\rule{0.166667em}{0ex}}U\phantom{\rule{0.166667em}{0ex}}{L}_{x}^{}}\right]=\frac{1}{\mathrm{Re}}\hfill \end{array}$

The dimensionless equation for vorticity with the * dropped is

$Re\phantom{\rule{0.166667em}{0ex}}\left[\frac{\partial w}{\partial t}+{v}_{x}\frac{\partial w}{\partial x}+\alpha \phantom{\rule{0.166667em}{0ex}}{v}_{y}\frac{\partial w}{\partial y}\right]=\frac{{\partial }^{2}w}{\partial {x}^{2}}+{\alpha }^{2}\frac{{\partial }^{2}w}{\partial {y}^{2}}$

The convective derivative can be written in conservative form by use of the continuity equation.

$\begin{array}{c}{\left({v}_{i}\phantom{\rule{0.166667em}{0ex}}{w}_{j}\right)}_{,i}={v}_{i,i}\phantom{\rule{0.166667em}{0ex}}{w}_{j}+{v}_{i}\phantom{\rule{0.166667em}{0ex}}{w}_{j,i}\hfill \\ \mathrm{but}\hfill \\ {v}_{i,i}=0,\phantom{\rule{1.em}{0ex}}\mathrm{for}\phantom{\rule{0.277778em}{0ex}}\mathrm{incompressible}\phantom{\rule{0.277778em}{0ex}}\mathrm{flow}\hfill \\ \mathrm{thus}\hfill \\ {v}_{i}\phantom{\rule{0.166667em}{0ex}}{w}_{j,i}={\left({v}_{i}\phantom{\rule{0.166667em}{0ex}}{w}_{j}\right)}_{,i}\hfill \\ \mathrm{or}\hfill \\ \mathbf{v}\phantom{\rule{0.166667em}{0ex}}\nabla \mathbf{w}=\nabla •\mathbf{v}\phantom{\rule{0.166667em}{0ex}}\mathbf{w}\hfill \\ Re\phantom{\rule{0.166667em}{0ex}}\left[\frac{\partial w}{\partial t}+\frac{\partial {v}_{x}w}{\partial x}+\alpha \phantom{\rule{0.166667em}{0ex}}\frac{\partial {v}_{y}w}{\partial y}\right]=\frac{{\partial }^{2}w}{\partial {x}^{2}}+{\alpha }^{2}\frac{{\partial }^{2}w}{\partial {y}^{2}}\hfill \end{array}$

We will first illustrate how to compute the transient, linear problem before tackling the nonlinear terms. Stability of the transient finite difference equations is greatly improved if the spatial differences are evaluated at the new, $n+1$ , time level while the time derivative is evaluated with a backward-difference method.

$\begin{array}{c}\frac{\partial w}{\partial t}\approx \frac{{w}_{i,j}^{n+1}-{w}_{i,j}^{n}}{\Delta t}+O\left(\Delta t\right)=\frac{\Delta {w}_{i,j}}{\Delta t}+O\left(\Delta t\right)\hfill \\ \Delta {w}_{i,j}={w}_{i,j}^{n+1}-{w}_{i,j}^{n}\hfill \end{array}$

Recall that the finite difference form of the steady state equations were expressed as a system of equations with coefficients, $a$ , $b$ , , ect. The vorticity equation for the $i$ , $j$ grid point will be rewritten but now with the transient term included. It is convenient to solve for $\Delta w$ rather than ${w}^{n+1}$ at each time step.

$\begin{array}{c}a\phantom{\rule{0.166667em}{0ex}}\Delta {w}_{i+1,j}^{n+1}+b\Delta \phantom{\rule{0.166667em}{0ex}}{w}_{i-1,j}^{n+1}+c\phantom{\rule{0.166667em}{0ex}}\Delta \phantom{\rule{0.166667em}{0ex}}{w}_{i,j+1}^{n+1}+d\phantom{\rule{0.166667em}{0ex}}\Delta {w}_{i,j-1}^{n+1}+e\phantom{\rule{0.166667em}{0ex}}\Delta {w}_{i,j}^{n+1}-\frac{{\delta }^{2}\mathrm{Re}}{\Delta t}\Delta {w}_{i,j}^{n+1}\hfill \\ =f-a\phantom{\rule{0.166667em}{0ex}}{w}_{i+1,j}^{n}-b\phantom{\rule{0.166667em}{0ex}}{w}_{i-1,j}^{n}-c\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{w}_{i,j+1}^{n}-d\phantom{\rule{0.166667em}{0ex}}{w}_{i,j-1}^{n}-e\phantom{\rule{0.166667em}{0ex}}{w}_{i,j}^{n}\hfill \end{array}$

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!