There is the explanation on the Output Boundary condition in the titled "Computational Fluid Dynamics" written by HK VERSTEEG, During the iteration cycles of the SIMPLE algorithm there is no guarantee that these velocities will conserve mass over the computational domain a whole. To make the mass flux out equal to the mass flux Min coming into the domain all the outlet velocity components Uni,j are multiplied by the ratio Min/Mout.

Uui,j=Uni-1,j*(Min/Mout)

Please give me the answer why we need this treatment. If we make this treatment in the model where the solid object is located in the flow field, the eddies in the backside of the solid is vanished.

Please give me the solution on this problem.

Regards,

This has been explained many times. A tentative velocity Uu_{i,j}= Un_{i-1,j} means extrapolcation of velocity at outlet. But outflow mass flux from Uu{i,j}, "Mout", not necessary = Min, a correction Un_{i,j}= Uu_{i,j}*(Min/Mout), make the resulting new "Mout" = \int U_n dy = Min.

Please correct me if i'm wrong at this. The mass flow across the domain boundaries MUST balance for a convergent solution of the pressure correction equation. This the reason why we adjust the outlet velocity to ensure global mass conservation. This is also the reason why we require a mass fraction if more than one outlet boundary are present.

The pressure correction equation is a singular equation which cannot be solved with a direct solver if the pressure correction is not fixed at one of the cells in the mesh. A iterative solver however will solve this singular matrix, but only if there is global mass conservation??????

Dear versi,

Thank you for your answer on my question.

Yes,It is true to use this correction , Un_{i,j}= U_{i-1,j}*(Min/Mout), If we confine the area in the Uni,j.

I believe this Min is the total mass flux in the inlet.

I want to show the part of my source code.

-----------------------------------------------

FLOWIN = 0.0

DO 170 J=2,NJM1

ARDEN = DENSIT*SNS(J)

FLOWIN = FLOWIN+ARDEN*U(2,J)

170 CONTINUE

C-----OUTLET

ARDENT = 0.0

FLOW = 0.0

DO 250 J=2,NJM1

ARDEN = DENSIT*SNS(J)

ARDENT = ARDENT+ARDEN

FLOW = FLOW+ARDEN*U(NIM1,J) 250 CONTINUE

UINC = (FLOWIN-FLOW)/ARDENT

DO 260 J=2,NJM1 260 U(NI,J) = U(NIM1,J)

C 260 U(NI,J) = U(NIM1,J)+UINC

-------------------------------------------------

At present, I put the comment at the line of 260 U(NI,J) = U(NIM1,J)+UINC.

I set X-long rectangle area as the flow field.

I set the inlet velocity 1.0 at West and the Outlet is at East in the rectangle flow field.

As the result from the run,with the deleting UINC,

The velocity is decreasing toward the Outlet. I got almost 0.56 at the outlet. If I back the sentense to add UINC as

260 U(NI,J) = U(NIM1,J)+UINC

I can get almost 1.0 at the outlet, It is the same as the inlet velocity.

I can not understand the reason why we need this correction. And also if we do not this correction, the velocity at outlet is deceasing.

DO 260 J=2,NJM1 260 U(NI,J) = U(NI-1,J) means extraploation of velocity,together with fixed outlet pressure, is a very good, well posed BC. Check if outlet velocity still decrease as mesh is refined doubly. If vecolcity at outlet do not conserve mass well as the mesh number is refined, there is something wrong with the main code.

Thank you ,Versi

Do yo mean that we need not to use the line,

260 U(NI,J) = U(NIM1,J)+UINC

if we make the mesh in the model more fine in addition to the line,

DO 260 J=2,NJM1 260 U(NI,J) = U(NIM1,J) ?

Is it enough to use the line only,

DO 260 J=2,NJM1 260 U(NI,J) = U(NIM1,J)

to maintain the mass conservation ?