Author |
Message |
ali tinni (alitinni)
New member Username: alitinni
Post Number: 1 Registered: 11-2009
| Posted on Thursday, November 19, 2009 - 01:02 am: | |
Hi I am trying to solve the following system c: (1-phi-phif)/(1+(b*c)^2)))*dt(c)= phi*ff*(cf-c) cf:dt(cf)=-dxx(vf*cf)+(phi*ff(c-cf)). To do so I used the following code VARIABLES cf { concentration in fracture } global variables c { concentration in the matrice } DEFINITIONS phi = 0.01 phif=0.25 b=0.1 vf=8E-12 ff=2.4 INITIAL VALUES cf= 2 c=2 EQUATIONS c: (1+((1-phi-phif)/(1+(b*c)^2)))*dt(c)= phi*ff*(cf-c) cf: dt(cf)=-dxx(vf*cf)+(phi*ff(c-cf)) But I am stocked on the boundaries part, I cannot figure out how to put the following values at x=0, dx(cf)=4 and at x= 50, dx(cf)= 10 i dont think i will need boundary values for c because the 2 equation are highly coupled Thank you |
Robert G. Nelson (rgnelson)
Moderator Username: rgnelson
Post Number: 1305 Registered: 06-2003
| Posted on Thursday, November 19, 2009 - 03:37 pm: | |
Your Cf equation contains a term dxx(vf*cf) which should probably be dx(vf*dx(cf))]. Integrating this term by parts creates a boundary integrand of normal(vf*dx(cf)). This kind of boundary integral is defined by the Natural boundary condition. At the left end, the normal is negative, so Natural(cf)=-4/vf. At the right end the normal is positive, so Natural(cf)=10/vf. See "Natural Boundary Conditions" in the Help Index. Your C equation does not need boundary conditions, not because it is highly coupled, but because you have defined it as a global and there are no spatial derivatives. Spatial boundary conditions are necessary only to define the integration constants generated by integrating spatial derivatives. |
ali tinni (alitinni)
New member Username: alitinni
Post Number: 2 Registered: 11-2009
| Posted on Monday, November 23, 2009 - 02:07 am: | |
Thank you very much but I am trying to plot my solution but a msgbox is displaying "No plot time specifed", and i dont what to do in this case. About the boundary region, does somebody know how to define a boundary region the plot of the system above will show the variation of cf in the space then in function of time and space |
Marek Nelson (mgnelson)
Moderator Username: mgnelson
Post Number: 160 Registered: 07-2007
| Posted on Monday, November 23, 2009 - 01:20 pm: | |
In time dependent problems the display specifications must be preceded by a display-time declaration statement. See "Plot Time Selection" in the Help index. A CONTOUR plot will show the spatial distribution. You cannot plot as a function of time and space, but you can plot a time HISTORY at a given point. See "HISTORIES" in the Help index.
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