# 0.10 Numerical simulation  (Page 7/8)

 Page 7 / 8
    ui-1/2 = 0.5*(u(i,j)+u(i-1,j)) if ui-1/2 >0 then (uw)i-1/2 = ui-1/2 wi-1 ; b'=b + $\delta$ Re ui-1/2else (uw)i-1/2 = ui-1/2 wi ; e'=e + $\delta$Re ui-1/2endif  ui+1/2 = 0.5*(u(i,j)+u(i+1,j)) if uI+1/2 >0 then (uw)i+1/2 = ui+1/2 wi ; e'=e - $\delta$  Re ui+1/2else (uw)i+1/2 = ui+1/2 wi+1 ; a'=a - $\delta$  Re ui+1/2endif

The y-direction will be similar except the aspect ratio, , must be included. These modifications to the coefficients must be made before the coefficients are updated for the boundary conditions.

## Calculation of pressure

The vorticity-stream function method does not require calculation of pressure to determine the flow field. However, if the force or drag on a body or a conduit is of interest, the pressure must be computed to determine the stress field. Here we will derive the Poisson equation for pressure and determine the boundary conditions using the equations of motion. If the pressure is desired only at the boundary, it may be possible to integrate the pressure gradient determined from the equations of motion. The dimensionless equations of motion for incompressible flow of a Newtonian fluid are as follows.

$\begin{array}{ccc}\frac{d\mathbf{v}*}{dt*}\hfill & =& \frac{\partial \mathbf{v}*}{\partial t*}+\mathbf{v}*•\nabla *\mathbf{v}*=\frac{\partial \mathbf{v}*}{\partial t*}+\nabla *•\left(\mathbf{v}*\mathbf{v}*\right)=\hfill \\ & & -\nabla *P*+\frac{1}{Re}\nabla {*}^{2}\mathbf{v}*\hfill \\ & & \mathrm{where}\hfill \\ \hfill \mathbf{v}*& =& \frac{\mathbf{v}}{U},\phantom{\rule{1.em}{0ex}}\mathbf{x}*=\frac{\mathbf{x}}{L},\phantom{\rule{1.em}{0ex}}t*=\frac{U}{L}t,\phantom{\rule{1.em}{0ex}}P*=\frac{P}{\rho \phantom{\rule{0.166667em}{0ex}}{U}^{2}},\phantom{\rule{1.em}{0ex}}\nabla *=L\phantom{\rule{0.166667em}{0ex}}\nabla \hfill \\ \hfill P& =& p-\rho \phantom{\rule{0.166667em}{0ex}}g\phantom{\rule{0.166667em}{0ex}}z,\phantom{\rule{1.em}{0ex}}Re=\frac{\rho \phantom{\rule{0.166667em}{0ex}}U\phantom{\rule{0.166667em}{0ex}}L}{\mu }\hfill \end{array}$

Here all coordinates were scaled with respect to the same length scale. The aspect ratio must be included in the final equation if the coordinates are scaled with respect to different length scales. We will drop the * and use $x$ and $y$ for the coordinates and $u$ and $v$ as the components of velocity.

The following derivation follows that of Hoffmann and Chiang (1993). The equations of motion in 2-D in conservative form is as follows.

$\begin{array}{c}\frac{\partial u}{\partial t}+\frac{\partial \left({u}^{2}\right)}{\partial x}+\frac{\partial \left(u\phantom{\rule{0.166667em}{0ex}}v\right)}{\partial y}=-\frac{\partial P}{\partial x}+\frac{1}{Re}\left(\frac{{\partial }^{2}u}{\partial {x}^{2}}+\frac{{\partial }^{2}u}{\partial {y}^{2}}\right)\hfill \\ \frac{\partial v}{\partial t}+\frac{\partial \left(u\phantom{\rule{0.166667em}{0ex}}v\right)}{\partial x}+\frac{\partial \left({v}^{2}\right)}{\partial y}=-\frac{\partial P}{\partial y}+\frac{1}{Re}\left(\frac{{\partial }^{2}v}{\partial {x}^{2}}+\frac{{\partial }^{2}v}{\partial {y}^{2}}\right)\hfill \end{array}$

The Laplacian of pressure is determined by taking the divergence of the equations of motion. We will carry out the derivation step wise by first taking the x-derivative of the x-component of the equations of motion.

$\frac{\partial }{\partial t}\left(\frac{\partial u}{\partial x}\right)+2\frac{\partial u}{\partial x}\frac{\partial u}{\partial x}+2u\frac{{\partial }^{2}u}{\partial {x}^{2}}+\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}+u\frac{{\partial }^{2}v}{\partial x\partial y}+\frac{\partial v}{\partial x}\frac{\partial u}{\partial y}+v\frac{{\partial }^{2}u}{\partial x\partial y}=-\frac{{\partial }^{2}P}{\partial {x}^{2}}+\frac{1}{Re}\frac{\partial }{\partial x}\left({\nabla }^{2}u\right)$

Two pair of terms cancels because of the equation of continuity for incompressible flow.

$\begin{array}{c}\frac{\partial u}{\partial x}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0\hfill \\ u\frac{{\partial }^{2}v}{\partial x\partial y}+u\frac{{\partial }^{2}u}{\partial {x}^{2}}=u\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0\hfill \\ \mathrm{Thus}\hfill \\ \frac{\partial }{\partial t}\left(\frac{\partial u}{\partial x}\right)+{\left(\frac{\partial u}{\partial x}\right)}^{2}+u\frac{{\partial }^{2}u}{\partial {x}^{2}}+\frac{\partial v}{\partial x}\frac{\partial u}{\partial y}+v\frac{{\partial }^{2}u}{\partial x\partial y}=-\frac{{\partial }^{2}P}{\partial {x}^{2}}+\frac{1}{\mathrm{Re}}\frac{\partial }{\partial x}\left({\nabla }^{2}u\right)\hfill \end{array}$

Similarly, the y-component of the equations of motion become

$\frac{\partial }{\partial t}\left(\frac{\partial v}{\partial y}\right)+{\left(\frac{\partial v}{\partial y}\right)}^{2}+v\frac{{\partial }^{2}v}{\partial {y}^{2}}+\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}+u\frac{{\partial }^{2}v}{\partial x\partial y}=-\frac{{\partial }^{2}P}{\partial {y}^{2}}+\frac{1}{Re}\frac{\partial }{\partial y}\left({\nabla }^{2}v\right)$

The x and y-components of the above equations are now added together and several pairs of terms cancels pair-wise.

$\begin{array}{c}\frac{\partial }{\partial t}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0\hfill \\ \frac{{\partial }^{2}u}{\partial {x}^{2}}+\frac{{\partial }^{2}v}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0\hfill \\ \frac{{\partial }^{2}u}{\partial x\partial y}+\frac{{\partial }^{2}v}{\partial {y}^{2}}=\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0\hfill \\ \frac{\partial }{\partial x}\left(\frac{{\partial }^{2}u}{\partial {x}^{2}}+\frac{{\partial }^{2}u}{\partial {y}^{2}}\right)+\frac{\partial }{\partial y}\left(\frac{{\partial }^{2}v}{\partial {x}^{2}}+\frac{{\partial }^{2}v}{\partial {y}^{2}}\right)=\frac{{\partial }^{2}}{\partial {x}^{2}}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+\frac{{\partial }^{2}}{\partial {y}^{2}}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0\hfill \end{array}$

Therefore the equations reduce to

${\left(\frac{\partial u}{\partial x}\right)}^{2}+{\left(\frac{\partial v}{\partial y}\right)}^{2}+2\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}=-\left(\frac{{\partial }^{2}P}{\partial {x}^{2}}+\frac{{\partial }^{2}P}{\partial {y}^{2}}\right)$

The left-hand side can be further reduced by consideration of the continuity equation as follows.

$\begin{array}{c}{\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)}^{2}={\left(\frac{\partial u}{\partial x}\right)}^{2}+{\left(\frac{\partial v}{\partial y}\right)}^{2}+2\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}=0\hfill \\ \mathrm{from}\phantom{\rule{0.277778em}{0ex}}\mathrm{which}\hfill \\ {\left(\frac{\partial u}{\partial x}\right)}^{2}+{\left(\frac{\partial v}{\partial y}\right)}^{2}=-2\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}\hfill \end{array}$

Upon substitution, we have

$-\left(\frac{{\partial }^{2}P}{\partial {x}^{2}}+\frac{{\partial }^{2}P}{\partial {y}^{2}}\right)=2\left(\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}-\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}\right)$

This equation can be written in terms of the stream function.

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

#### 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|
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
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)=
. 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
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
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!