<< Chapter < Page Chapter >> Page >
Finite difference grid for discretizing PDE (grid point formulation)

The unit square is now discretized into JMAX by JMAX grid points where the boundary conditions and dependent variables will be evaluated. The grid spacing is δ = 1 / ( J M A X - 1 ) . The first and last row and column are boundary values.

The partial derivatives will be approximated by finite differences. For example, the second derivative of vorticity is discretized by a Taylor's series.

w i + 1 = w i + δ w x i + δ 2 2 2 w x 2 i + δ 3 6 3 w x 3 i + δ 4 24 4 w x 4 x ¯ w i - 1 = w i - δ w x i + δ 2 2 2 w x 2 i - δ 3 6 3 w x 3 i + δ 4 24 4 w x 4 ¯ ¯ x 2 w x 2 i = w i - 1 - 2 w i + w i + 1 δ 2 - δ 2 24 4 w x 4 x ¯ + 4 w x 4 ¯ ¯ x = w i - 1 - 2 w i + w i + 1 δ 2 + O δ 2

The finite difference approximation to the PDE at the interior points results in the following set of equations.

ψ i + 1 , j + ψ i - 1 , j + α 2 ψ i , j + 1 + α 2 ψ i , j - 1 - 2 ( 1 + α 2 ) ψ i , j + δ 2 w i , j = 0 w i + 1 , j + w i - 1 , j + α 2 w i , j + 1 + α 2 w i , j - 1 - 2 ( 1 + α 2 ) w i , j = 0 i , j = 2 , 3 , J M A X - 1

The vorticity at the boundary is discretized and expressed in terms of the components of velocity at the boundary, the stream function values on the boundary and a stream function value in the interior grid. (A greater accuracy is possible by using two interior points.) The stream function at the first interior point ( i = 2 ) from the x boundary is written with a Taylor's series as follows.

ψ 2 = ψ 1 ± δ ψ x 1 + δ 2 2 2 ψ x 2 1 + O ( δ 3 ) 2 ψ x 2 1 = 2 ( ψ 2 - ψ 1 ) δ 2 2 δ ψ x 1 + O δ = 2 ( ψ 2 - ψ 1 ) δ 2 ± 2 δ v y B C + O δ w 1 B C v y x - α v x y B C = - 2 ψ x 2 1 - α v x B C y = - 2 ( ψ 2 - ψ 1 B C ) δ 2 2 δ v y B C - α v x B C y + O δ

The choice of sign depends on whether x is increasing or decreasing at the boundary. Similarly, on a y boundary,

w 1 B C = - 2 α 2 ( ψ 2 - ψ 1 B C ) δ 2 ± 2 α v x B C δ + v y B C x + O δ

The boundary condition on the stream function is specified by the normal component of velocity at the boundaries. Since we have assumed zero normal component of velocity, the stream function is a constant on the boundary, which we specify to be zero.

The stream function at the boundary is calculated from the normal component of velocity by numerical integration using the trapezodial rule, e.g.,

ψ j = ψ j - 1 + δ 2 v x j - 1 + v x j / α

Solution of linear equations

The finite difference equations for the PDE and the boundary conditions are a linear system of equations with two dependent variables. The dependent variables at a x i , y j grid point will be represented as a two component vector of dependent variables,

u i , j = ψ i , j w i , j

The pair of equations for each grid point can be represented in the following form

a i , j u i + 1 , j + b i , j u i - 1 , j + c i , j u i , j + 1 + d i , j u i , j - 1 + e i , j u i , j = f i , j i , j = 2 , 3 , J M A X - 1

Each coefficient is a 2x2 matrix. For example,

e i , j u i , j = e i , j , 1 , 1 ψ i , j + e i , j , 1 , 2 w i , j e i , j , 2 , 1 ψ i , j + e i , j , 2 , 2 w i , j

The components of the 2x2 coefficient matrix are the coefficients from the difference equations. The first row is coefficients for the stream function equation and the second row is coefficients for the vorticity equation. The first column is coefficients for the stream function variable and the second column is the coefficients for the vorticity variable. For example, at interior points not affected by the boundary conditions,

a i , j = 1 0 0 1 b i , j = 1 0 0 1 c i , j = α 2 0 0 α 2 d i , j = α 2 0 0 α 2 e i , j = - 2 ( 1 + α 2 ) δ 2 0 - 2 ( 1 + α 2 ) f i , j = 0 0

The coefficients for the interior grid points adjacent to a boundary are modified as a result of substitution the boundary value of stream function or the linear equation for the boundary vorticity into the difference equations. The stream function equation is coupled to the vorticity with the e i j coefficient and the vorticity equation is coupled to the stream function through the boundary conditions. For example, at a x = 0 boundary, the coefficients will be modified as follows.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
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
I'm not sure why it wrote it the other way
I got X =-6
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
Idrissa Reply
im all ears I need to learn
right! what he said ⤴⤴⤴
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
I'm not good at math so would you help me
what is the problem that i will help you to self with?
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
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
anybody can imagine what will be happen after 100 years from now in nano tech world
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
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
can nanotechnology change the direction of the face of the world
Prasenjit Reply
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.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Transport phenomena. OpenStax CNX. May 24, 2010 Download for free at http://cnx.org/content/col11205/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Transport phenomena' conversation and receive update notifications?