Hi,
I tried to make a modified conlit.f for imposing prescribed time dependent suspended sediment load at the model inlet. I have put comments to show what I understood by the lines I added and where I am unsure. The relevant portion of the associated Sisyhe boundary condition file is also given below.
I am also not sure whether I am supposed to leave the upper portion of the original conlit.f unchanged!
conlit.f subroutine ! LICBOR : BOUNDARY CONDITION FOR SEDIMENT CONCENTRATION
! !
USE DECLARATIONS_SISYPHE, ONLY : BOUNDARY_COLOUR ! I added this
IF(SUSP) THEN
!
DO K=1,NPTFR
!
! SO FAR LICBOR=LIEBOR (WITH KADH CHANGED INTO KLOG, SEE ABOVE,
! BUT CAN BE CHANGED)
!
LICBOR%I(K) = LIEBOR%I(K)
!
LICBOR%I(K) = KSORT
! ENTRANCE : IMPOSED CONCENTRATION
! !
! NOTE JMH: KSORT MUST BE TREATED ALSO BECAUSE SUBROUTINE DIFFIN
! MAY CHANGE A KSORT INTO KENT, DEPENDING OF FLOW
!
! IFRLIQ=NUMLIQ%I(K)
! IF(LIEBOR%I(K).EQ.KENT.OR.LIEBOR%I(K).EQ.KSORT) THEN
! DO I=1,NSICLA
! IRANK=I+(IFRLIQ-1)*NSICLA
! CBOR%ADR(I)%P%R(K) = CBOR_CLASSE(IRANK)
! ENDDO
! ENDIF
! My addition for 2 time steps and for 2 sediment classes
IF(BOUNDARY_COLOUR%I(K).GE.4868.AND.BOUNDARY_COLOUR%I(K).LE.4872) THEN
LICBOR%I(K)=KENT ! NOT SURE IF THIS IS REQUIRED
C
IF(AT.LE.1.DO) ! 1= First time step
CBOR%ADR(1)%P%R%(K)= a1 DO ! a1=volume of sediment fraction 1 in m3 at time step 1
CBOR%ADR(2)%P%R%(K)= b1 DO ! b1=volume of sediment fraction 2 in m3 at time step 1
ELSEIF (AT.GT.1.DO.AND.AT.LE.10.DO) THEN ! 10= time step 10
CBOR%ADR(1)%P%R%(K)= a2 DO ! a2=volume of sediment fraction 1 in m3 at time step 10
CBOR%ADR(2)%P%R%(K)= b2 DO ! b2=volume of sediment fraction 2 in m3 at time step 10
ELSEIF ! Carry on like this for other time steps
ENDIF ! not sure if this the right place for ENDIF
ENDIF
!
! READING BOUNDARY CONDITION FILE
!
IF(LICBOR%I(K).EQ.KENT.AND.
& SIS_FILES(SISLIQ)%NAME(1:1).NE.'MYBCFILE') THEN ! 'MYBCFILE' = My boundary condition file
IF(IFRLIQ.GT.0) THEN
DO I=1,NSICLA
CBOR%ADR(I)%P%R(K)=CGL(IFRLIQ,AT)/XMVS
ENDDO
ENDIF
ENDIF
!
ENDDO
!
ENDIF
!
! !
RETURN
END
Sisyphe boundary condition file
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 9458 4862 #
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 9077 4863 #
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 8689 4864 #
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 8305 4865 #
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 7945 4866 #
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 7587 4867 #
4 5 5 0.000 0.000 0.000 0.000 5 0.000 0.000 0.000 7215 4868 # RiverInlet (7215 - 5837)
4 5 5 0.000 0.000 0.000 0.000 5 0.000 0.000 0.000 6863 4869 # RiverInlet (7215 - 5837)
4 5 5 0.000 0.000 0.000 0.000 5 0.000 0.000 0.000 6515 4870 # RiverInlet (7215 - 5837)
4 5 5 0.000 0.000 0.000 0.000 5 0.000 0.000 0.000 6160 4871 # RiverInlet (7215 - 5837)
4 5 5 0.000 0.000 0.000 0.000 5 0.000 0.000 0.000 5837 4872 # RiverInlet (7215 - 5837)
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 5495 4873 #
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 5156 4874 #
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 4844 4875 #
2 2 2 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 4511 4876 #
Any comment will be appreciated
Cheers,
Joy
Community 
