problem with a flexpde code Log Out | Topics | Search
Moderators | Register | Edit Profile

FlexPDE User's Forum » User Postings » problem with a flexpde code « Previous Next »

Author Message
Top of pagePrevious messageNext messageBottom of page Link to this message

Andrea Cova (andrea25)
Member
Username: andrea25

Post Number: 5
Registered: 05-2006
Posted on Friday, June 09, 2006 - 04:19 am:   

Hello,I have realised the following code to resolve a problem of transport-diffusion. I am working on a Student Flexpde5 and,when I try to run the script,it says "Only specially coded sample problems can be run without license".How can I resolve this fact?Could you give me some suggestions?
-------------------------------------------------
title
'Modellizzazione del disastro ecologico di Bhopal'

coordinates
cartesian3

variables
c(threshold=0.1)

definitions
u=4 { velocità di trasporto del vento pari a 4m s^-1}
Ky=[0.0121*X^2*(1+0.0004*X)^(-1)*u]/(2*X) { coefficiente di diffusione lungo l'asse y}
Kz=[0.0064*X^2*(1+0.00015*X)^(-1)*u]/(2*X) /* coefficiente di diffusione lungo l'asse z */
a=90 /* durata dell'emissione pari a 90 min*/
Q0=40 /* intensità della sorgente emissiva pari a 40 tons in 90 min */
hmin=39.6 /* effettiva altezza del punto di emissione delle sostanze inquinanti pari a 39,6 m*/
Hmax=200 /* altezza massima raggiunta dagli agenti inquinanti pari a 200 m*/
ampx=100 /* NON RISPONDENTEampiezza del dominio spaziale lungo l'asse x pari ad 1 Km*/
semiampy=100 /* semiampiezza del dominio spaziale lungo l'asse y pari a 100 m*/

initial values
c = 0

equations
Ky*DYY(c)+Kz*DZZ(c)-u*DX(c)-Dt(c)=0

extrusion
surface 'livello terrestre' Z=0
layer 'spessore del rilievo orografico'
surface 'livello superiore del rilievo' Z=20
layer 'spessore dell"aria sovrastante'
surface 'estremo superiore del dominio in esame' Z=Hmax

boundaries
region 1 /*regione che identifica l'intero dominio spaziale considerato */
start(0,-semiampy)
value(c)=0 line to (ampx,-semiampy)
natural(c)=0 line to (ampx,semiampy)
value(c)=0 line to (0,semiampy)
natural(c)=0 line to close
limited region 2 /* regione occupata dal rilievo orografico */
start(20,-10) line to (40,-10) to (40,10) to (20,10) to close

time 0 to a

plots
for t = endtime
contour(c) on z=0 as "XY concentration" range=(0,a)
contour(c) on x=0 as "YZ concentration" range=(0,a)
contour(c) on y=0 as "XZ concentration" range=(0,a)

end

Add Your Message Here
Post:
Username: Posting Information:
This is a private posting area. Only registered users and moderators may post messages here.
Password:
Options: Enable HTML code in message
Automatically activate URLs in message
Action:

Topics | Last Day | Last Week | Tree View | Search | Help/Instructions | Program Credits Administration