open TELEMAC-MASCARET The mathematically superior suite of solvers

I'm modifying the conlit.f file to set the bedload solid discharge at the boundary condition of 6 size-classes of bed material.

In conlit.f I have the following routine:

DO K=1,NPTFR

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LIEBOR%I(K)=KSORT
LIQBOR%I(K)=KENT
if (K .ge. 5510 .and. K .le. 5556) then !5510 and 5556 are the limits node of the inlet boundary condition.
QBOR%ADR(1)%P%R(K)=0.168D0
QBOR%ADR(2)%P%R(K)=0.168D0
QBOR%ADR(3)%P%R(K)=0.168D0
QBOR%ADR(4)%P%R(K)=0.168D0
QBOR%ADR(5)%P%R(K)=0.084D0
QBOR%ADR(6)%P%R(K)=0.084D0

else

QBOR%ADR(1)%P%R(K)=0D0
QBOR%ADR(2)%P%R(K)=0D0
QBOR%ADR(3)%P%R(K)=0D0
QBOR%ADR(4)%P%R(K)=0D0
QBOR%ADR(5)%P%R(k)=0D0
QBOR%ADR(6)%P%R(K)=0D0
end if

But, QBOR%ADR(1)%P%R(K) still being 0. Y include this line: WRITE(LU,*) K , so I can see in the temrinal which values takes K, and the maximum was 2655 when I have 11304 boundary nodes so it will never set the solid discharge rate for each sediment class.

How I can fix this problem.
I attached my *. cli file and the conlit.f file.

Best regards,

Gonzalo

I'm running in parallel.
Could be that a problem?

Hello,

Yes this is the problem. The hardcoded values of boundary nodes are no longer valid in parallel, but you can nearly keep the same algorithm, replacing:

if (K .ge. 5510 .and. K .le. 5556)

by:

if (BOUNDARY_COLOUR%I(K) .ge. 5510 .and. BOUNDARY_COLOUR%I(K) .le. 5556)


You will probably have to add :

USE DECLARATIONS_SISYPHE, ONLY : BOUNDARY_COLOUR

at the beginning of the subroutine.

With best regards,

Jean-Michel Hervouet

Thank you very much,

It works pretty good.

I'm trying to have in the inlet boundary condition, a non erodible bed. I impose LIEBOR=KENT

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LIEBOR%I(K)=KENT
LIQBOR%I(K)=KENT

if (BOUNDARY_COLOUR%I(K) .ge. 5510 .and. BOUNDARY_COLOUR%I(K) .le. 5556) then
QBOR%ADR(1)%P%R(K)=0.013D0
QBOR%ADR(2)%P%R(K)=0.013D0
QBOR%ADR(3)%P%R(K)=0.013D0
QBOR%ADR(4)%P%R(K)=0.013D0
QBOR%ADR(5)%P%R(K)=0.0065106D0
QBOR%ADR(6)%P%R(K)=0.0065106D0
EBOR%ADR(1)%P%R(K)=0D0
EBOR%ADR(2)%P%R(K)=0D0
EBOR%ADR(3)%P%R(K)=0D0
EBOR%ADR(4)%P%R(K)=0D0
EBOR%ADR(5)%P%R(K)=0D0
EBOR%ADR(6)%P%R(K)=0D0

else

QBOR%ADR(1)%P%R(K)=0D0
QBOR%ADR(2)%P%R(K)=0D0
QBOR%ADR(3)%P%R(K)=0D0
QBOR%ADR(4)%P%R(K)=0D0
QBOR%ADR(5)%P%R(k)=0D0
QBOR%ADR(6)%P%R(K)=0D0
end if

But the evolution of the bottom is still being not zero in the boundary condition.
How can I fix that?

Best regards,

Gonzalo