| Author | Message | |||
     C. Reginald Jegathese (regi) Member Username: regi Post Number: 5 Registered: 11-2003 |
For the example problem Samples\steady_state\groundwater { POROUS.PDE } I need some help to get velocity distribution? I need some help to get streamline distribution? Thanks./ | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 93 Registered: 06-2003 |
The velocity should be directly derivable from the flux -k*grad(phi). Consult a texbook on porous flow. Once you have the velocity distribution, you can construct streamlines by observing that the definition of the stream function psi in terms of the velocity components U and V is u = dy(psi) v = -dx(psi) Solve the equation div(grad(psi)) = -dx(u)+dy(v) {= -vorticity} with boundary conditions natural(psi)= -tangential(u,v). Then plot the contours of psi. You can do this simultaneously with the solution for u and v, but it makes the computation more expensive. A better way is to TRANSFER the u,v data to a second run that solves for the trajectories. You can do this in a 3D cut plane, too, if you do a contour plot of each velocity and EXPORT each one, then read them in as a TABLE. Cylindrical coordinates require a slight modification. dzz(psi)/r + drr(psi)/r - dr(psi)/r^2 = -vorticity | |||
| Nestor Rueda Vallejo (ruedanestor) New member Username: ruedanestor Post Number: 1 Registered: 01-2004 |
hello, where can i find more information about pressure correction like in the viscous sample? thanks | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 104 Registered: 06-2003 |
See my long comment in the message "Pressure Formulation in Incompressible Fluid", posted in this bulletin board on 1/20/04 (five topics above this one). | |||
| Nestor Rueda Vallejo (ruedanestor) New member Username: ruedanestor Post Number: 2 Registered: 01-2004 |
hello robert thanks a lot, really is a very good explanation. i try to solve a heat exchanger simulation reported at: [1]. X. Chen and P. Han, "A note on the solution of conjugate heat transfer problems using SIMPLE-like algorithms", International Journal of Heat and Fluid Flow, 21, 2000, pp463-467. this is the script: TITLE ' comparacion con ALGOR' VARIABLES u(0.1,0.2) {vector de velocidad x (m/s)} P(0,1) {campo de presion} T(300,800) {campo de temperatura} DEFINITIONS lx = 1 {longitud del tubo (m)} ly = 0.1 {radio del tubo (m)} ro = 1000 {densidad del fluido (Kg/m^3)} visc=staged(10,1,0.15) {viscosidad absoluta} fp=200000 {factor de correccion de presion} Kcp=0.4 { thermal difusivity difusividad termica K/Cp } pe=0 {presion de entrada} ps=10 {presion de salida} EQUATIONS visc*div(grad(u)) - ro*(u*dx(u)) =(dx(p)) {ecuacion de navier-stokes para U} div(grad(p))=fp*dx(u) { pressure correction correccion de presion} ro*(U*dx(T)) = div(Kcp*grad(T)) { energy equation conservacion de la energia} BOUNDARIES REGION 1 START(0,0) natural(p)=0 line to (lx,0) value(p)=pe natural(u)=0 line to (lx,ly) natural(p)=0 value(u) =0 natural(T)=0 line to (0,Ly) value(p)=ps natural(u) =0.2 value(T)=800 line to finish REGION 2 START(0,-ly) natural(p)=0 line to(lx,-ly) value(p)=ps natural(u) =0.1 value(T)=300 line to(lx,-(ly+ly)) natural(p)=0 natural(T)=0 value(u) =0 line to(0,-(ly+ly)) value(p)=pe natural(u)=0 line to finish REGION 3 Kcp=.1 value(U)=0 START(0,-ly) line to(lx,-ly) natural(T)=0 line to(lx,0) line to(0,0) natural(T)=0 line to finish PLOTS grid(x,y) vector(u) contour( u) painted surface( u) elevation(u) from (0,-(ly+ly)) to (0,ly) elevation(t) from (0,-(ly+ly)) to (0,ly) elevation(t) from (0.5*lx,-(ly+ly)) to (0.5*lx,ly) elevation(p) from (0,0) to (lx,0) contour( u) contour( T) painted surface (T) elevation(T) from (0,0) to (lx,0) HISTORIES history(u) at (0,0) history(T) at (0,0) (0,ly) (lx,ly) END could you help me to decide if fp=20000 is a good election. i have the full problem statement and a comparisson of the results in ALGOR and the original study, if you want i can send you this graphics. thanks | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 105 Registered: 06-2003 |
As I pointed out in my message "Pressure Formulation in Incompressible Fluid", one rationalization of the penalty pressure equation is to posit an "almost incompressible" fluid. This relies on defining a hypothetical equation of state P(rho)=P0+L*(rho-rho0). Then the compressible equation becomes dt(P)=-L*rho*div(V). More arm-waving converts dt(P)=D*C*del2(P), with D a representative distance, and C a sound speed. The final result is del2(P)=(L*rho/(D*C))*div(V), where I have erased a troublesome minus sign. Presumably, one could construct an estimate of the penalty parameter (L*rho)/(D*C) from characteristics of the real material. Not wanting to do this in most cases, a convenient rule of thumb is Penalty = 1e4*visc/D^2. As far as I know, there is no physics in this, it's just dimensionally correct. So the 1e4 is a guess, which can be modified for specific cases. One way to do this is to run a staged problem, varying the 1e4 weight, and observing the smallest value for which the solution does not deviate meaningfully from the next larger value. There may be better ways to do this, but I don't know them. | |||
| C. Reginald Jegathese (regi) Member Username: regi Post Number: 6 Registered: 11-2003 |
I am getting Floating-Point Divide by zero. It is a porous annular with three inlet(s) fluid pressure; three outlet(s) fluid pressure and pressure on inner circle (structure). Thanks.
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| Nestor Rueda Vallejo (ruedanestor) Junior Member Username: ruedanestor Post Number: 3 Registered: 01-2004 |
thanks robert, is very clear your answer. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 107 Registered: 06-2003 |
Jegathese: You have started out with wildly inconsistent initial conditions, and FlexPDE is searching about trying to make some sense of the problem. Your initial pressure distribution implies flow through the boundaries, you have discontinuous initial conditions on saturation, and you set your inlet boundary conditions to K*M*grad(P)=Pinlet. Try starting out with a self-consistent initial condition and turn on the driving terms over some reasonable time interval. | |||
| C. Reginald Jegathese (regi) Member Username: regi Post Number: 7 Registered: 11-2003 |
Thanks for the useful comments. I think I should start like steady state and increment over time the pressure b.c/load. | |||
| Nestor Rueda Vallejo (ruedanestor) Member Username: ruedanestor Post Number: 4 Registered: 01-2004 |
hello robert, i have another cuestion: as you can see in my problem the region 3 is a wall but the program don´t understand it. how can i do to help the program to understand that there are no velocities and pressures thanks.
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 109 Registered: 06-2003 |
Vallejo: Your problem document shows the middle layer as metal, but in your problem descriptor, you have given it the same density and viscosity as the fluid. Try setting the density to the document value and a high viscosity (>1000), and the metal should stop moving. | |||
| angela (angela) New member Username: angela Post Number: 1 Registered: 09-2005 |
Hi! I’m trying to model the flow through a porous medium taking into account the interaction between the solid and gaseous phase by heat and mass transfer caused by phenomena that take place on the solid particles. The gas flow in the void space between the particles of the bed is treated as a flow through porous media according to Darcy’s law: the governing equations include a set of coupled non linear differential equations for conservation of mass, species and energy together with the equation of state. The solid phase is treated as a continuum then the equations of particle movement in the packed bed take similar forms as those for the fluid. The problem is unsteady and one-dimensional. Can flexpde solve this type of PDE system? Thank you in advance for your answer. angela |
Porous annular problem
problem description