open TELEMAC-MASCARET The mathematically superior suite of solvers

We wanted to specify the rigid boundary (non erodible) at upstream to be the same as bottom elevation and the following sub routine (noerod.f)

! RIGID BEDS POSITION
!

---

!
! DEFAULT VALUE: ZR=ZF-100.D0
!
CALL OV('X=Y+C ',ZR,ZF,ZF,-100.D0,NPOIN)

DO I=1,NPOIN
IF(BOUNDARY_COLOUR%I(I).GE.77 .AND.
& BOUNDARY_COLOUR%I(I).LE.97) THEN
ZR(I)=ZF(I)
ENDIF
ENDDO
!

We still have erosion at upstream boundry. Can somebody assist me here!

Hi,

Use the different type of boundary for sediment as described in the SISYPHE manual and it can solve this problem:

The following thirteen variables for each edge point are first read out of the boundary conditions file:
X1, X2, X3, X4, X5, X6, X7, LIEBOR, EBOR, X10, X11, N, K.

LIEBOR is an integer providing the type of boundary conditions along the edges :
= KENT (numerical value 5): bottom evolution imposed to EBOR value at a liquid boundary,
= KSORT (numerical value 4): free evolution at a liquid boundary,
= KLOG or KADH (numerical value 2 or 0) : solid boundary.
EBOR provides the values of sedimentary evolutions when they are imposed.

A fixed bottom depth is usually imposed at the boundaries with an inflow (EBOR = 0; LIEBOR = KENT),
whereas a free evolution is imposed at the boundaries with an outflow (LIEBOR = KSORT).

Kind regards,
Qilong

Thanks so much Qilong