The problem is interesting because it is non-trivial. I guess that the problem is symmetric and when considering half of the porous pipe (open at both sides) one can replace the problem by that of a porous pipe segment with injection of fluid at a velocity Vo on one side (at x=0) and open at the other side (x=L). If we neglect viscosity in the pipe and assume a steady incompressible flow (water) we can find an analytical model based on the equation of Bernoulli applied in the pipe. A constant resistance R for flow through the wall is assumed (volume flow per surface area entering the pipe given by R(p_infty -p), where p is the pressure in the pipe at location x and p_infty is the surrounding pressure. At high Reynolds numbers an outflow from the open-pipe outlet at x=L implies p(L)=p_infty. If p(0)>p_infty in the pipe there will be outflow and the local velocity will decrease V(x)

There are many factors here that affect the pressure profile, including the porosity, the intrinsic permeability, fluid viscosity and compressibility.

For instance, i would expect a pressure jump at the interface for very low permeability and high viscosity of the fluid.. Of course, not to forget, numeric-related challenges of the coupled problem..

I recommend you to have a look at many literature, with keyword "Coupling free flow and flow in porous media"

Thank you for your recommendation, I'll read in this part and try again.

The problem is interesting because it is non-trivial. I guess that the problem is symmetric and when considering half of the porous pipe (open at both sides) one can replace the problem by that of a porous pipe segment with injection of fluid at a velocity Vo on one side (at x=0) and open at the other side (x=L). If we neglect viscosity in the pipe and assume a steady incompressible flow (water) we can find an analytical model based on the equation of Bernoulli applied in the pipe. A constant resistance R for flow through the wall is assumed (volume flow per surface area entering the pipe given by R(p_infty -p), where p is the pressure in the pipe at location x and p_infty is the surrounding pressure. At high Reynolds numbers an outflow from the open-pipe outlet at x=L implies p(L)=p_infty. If p(0)>p_infty in the pipe there will be outflow and the local velocity will decrease V(x)

Sorry. There is a typo in my answer above. If R is the resistance of the porous wall, the volume flux per unit surface is (p_infty-p)/R. Furthermore the fluid can be air and behave as an incompressible fluid as long as the pressure differences in the flow are small compared to the atmospheric pressure (in other words the square of the Mach number M=U/c is small compared to unity (1/2 rho U^2

Thank you prof. Avraham Hirschberg,

truly appreciate your time and effort, I'll update my case based on current information and back again with the feedback :).

Dear Ahmed,

Soory, I did make a stupid mistake in the added theory.

The differential equation that should be integrated is:

dV/dx=(p(L)-p)/(aR)=rho/(2aR)(V^2-V(L)^2)

or

int [dV/(V^2-V(L)^2)]=rho (x-L)/(2 aR)

or

ln[(V-V(L)/(V+V(L))]=rho (x-L)/(aR)

Hence for x=0:

ln[(Vo-V(L))/(Vo+V(L))]=-rho L/(aR)

is an equation to determine V(L).

I guess you have to check and re-check this before you use it.

Regards,

Mico