# 0.10 Numerical simulation  (Page 3/8)

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$\begin{array}{c}i=2\hfill \\ f\left(i,j,1\right)=f\left(i,j,1\right)-b\left(i,j,1,1\right)*{\psi }_{1,j}^{BC}\hfill \\ f\left(i,j,2\right)=f\left(i,j,2\right)-b\left(i,j,2,2\right)*\left(2*{\psi }_{1,j}^{BC}/{\delta }^{2}-2*{v}_{v}^{BC}\left(j\right)/\delta -\alpha \frac{\partial {v}_{x}^{BC}}{\partial y}\right)\hfill \\ e\left(i,j,2,1\right)=e\left(i,j,2,1\right)-b\left(i,j,2,2\right)*2/{\delta }^{2}\hfill \end{array}$

If the boundary condition is that of zero vorticity as on a surface of symmetry, it is necessary to only set to zero the coefficient that multiplies the boundary value of vorticity, i.e.,

$\begin{array}{c}i=2\hfill \\ b\left(i,j,2,2\right)=0,\phantom{\rule{1.em}{0ex}}\mathrm{when}\phantom{\rule{0.277778em}{0ex}}w\left(1,j\right)=0\hfill \end{array}$

The linear system of equations and unknowns can be written in matrix form by ordering the equations with the i index changing the fastest, i.e., $i=2,3, ,JMAX-1$ . The linear system of equations can now be expressed in matrix notation as follows.

$\begin{array}{c}\mathbf{A}•\mathbf{x}=\mathbf{F}\hfill \\ \mathrm{where}\hfill \\ \mathbf{x}=\left[\begin{array}{c}{\mathbf{u}}_{2,2}\\ {\mathbf{u}}_{3,2}\\ ⋮\\ {\mathbf{u}}_{JM1,JM1}\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}{\psi }_{2,2}\\ {w}_{2,2}\end{array}\\ \begin{array}{c}{\psi }_{3,2}\\ {w}_{3,2}\end{array}\\ ⋮\\ \begin{array}{c}{\psi }_{JM1,JM1}\\ {w}_{JM1,JM1}\end{array}\end{array}\right]\hfill \end{array}$

The non-zero coefficients of coefficient matrix A for $JMAX=5$ is illustrated below. Since the first and last grid points of each row and column of grid points are boundary conditions, the number of grid with pairs of unknowns is only 3x3. Only one grid point is isolated from boundaries. The $i$ , $j$ location of the coefficients become clearer if a box is drawn to enclose each $2x2$ coefficient.

Most of the elements of the coefficient matrix are zero. In fact, the coefficient matrix is a block pentadiagonal sparse matrix and only the nonzero coefficients need to be stored and processed for solving the linear system of equations. The non-zero coefficients of the coefficient matrix are stored with the row-indexed sparse storage mode and the linear system of equations is solved by preconditioned biconjugate gradient method (Numerical Recipes, 1992).

The row-indexed sparse matrix mode requires storing the nonzero coefficients in the array, $sa\left(k\right)$ , and the column number of the coefficient in the coefficient matrix in the array, $ija\left(k\right)$ , where $k=1,2,$ . The indices that are needed to identify the coefficients are the $i,j$ grid location $\left(i,j=2,3, ,JMAX-1\right)$ , the $m$ equation and dependent variable identifier ( $m=1$ for stream function; $=2$ for vorticity), and the $IA$ row index of the coefficient matrix $\left(IA=1,2,3,4, ,2\left(JMAX-2\right)2\right)$ . The coefficient matrix has two equations for each grid point, the first for the stream function and the second for vorticity. The total number of equations is $NN=2\left(JMAX-2\right)2$ .

The diagonal elements of the coefficient array are first stored in the $sa\left(k\right)$ array. The first element of $ija\left(k\right)$ , $ija\left(1\right)=NN+2$ and can be used to determine the size of the coefficient matrix. The algorithm then cycles over the pairs of rows in the coefficient matrix while keeping track of the $i$ , $j$ locations of the grid point and cycling over equation $m=1\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}2$ . The off-diagonal coefficients are stored in $sa\left(k\right)$ and the column number in the coefficient matrix stored in $ija\left(k\right)$ . As each row is completed, the $k$ index for the next row is stored in the first $NN$ elements of $ija$ .

A test problem for this code is a square box that has one side sliding as to impart a unit tangential component of velocity along this side. All other walls are stationary. The stream function, vorticity, and velocity contours for this problem with a 20x20 grid is illustrated below.

#### Questions & Answers

can someone help me with some logarithmic and exponential equations.
sure. what is your question?
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
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|
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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
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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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
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Yasmin
what is the function of carbon nanotubes?
Cesar
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
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
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