My last topic: #27852
Hello,
I have already changed my Steering file and Fortran based on your recommendations.
The Silulation runs, but only until 3 H 24 MIN then I get this error in Listing.File:
==========================================================
ITERATION 2450000 TIME: 3 H 24 MIN 10.0000 S ( 12250.0000 S)
BEDLOAD EQUATION SOLVED IN 10 ITERATIONS
BEDLOAD EQUATION SOLVED IN 10 ITERATIONS
BEDLOAD EQUATION SOLVED IN 10 ITERATIONS
MASS-BALANCE (IN VOLUME, INCLUDING VOID):
SUM OF THE EVOLUTIONS : -0.9605709E-09 M3
BOUNDARY 1 BEDLOAD FLUX = -0.2339509E-06 ( M3/S >0 = ENTERING )
BOUNDARY 2 BEDLOAD FLUX = 0.4183671E-07 ( M3/S >0 = ENTERING )
TOTAL BEDLOAD FLUX = -0.1921142E-06 ( M3/S >0 = ENTERING )
LOST VOLUME : 0.3219596E-17 M3 ( IF <0 EXIT )
MASS BALANCE FOR SEDIMENT CLASS : 1
TOTAL VOLUME: 20.5920777953661
SUM OF THE EVOLUTIONS FOR THIS CLASS: 0.1191736E-09
VOLUME THAT ENTERED THE DOMAIN FOR THIS CLASS: 0.1191736E-09 M3
LOST VOLUME : 0.1474879E-17 M3 ( IF <0 EXIT )
MASS BALANCE FOR SEDIMENT CLASS : 2
TOTAL VOLUME: 11.3350313112397
SUM OF THE EVOLUTIONS FOR THIS CLASS: 0.2364818E-11
VOLUME THAT ENTERED THE DOMAIN FOR THIS CLASS: 0.2364816E-11 M3
LOST VOLUME : 0.2221718E-17 M3 ( IF <0 EXIT )
MASS BALANCE FOR SEDIMENT CLASS : 3
TOTAL VOLUME: 10.0845380256850
SUM OF THE EVOLUTIONS FOR THIS CLASS: -0.1082109E-08
VOLUME THAT ENTERED THE DOMAIN FOR THIS CLASS: -0.1082109E-08 M3
LOST VOLUME : -0.4769548E-18 M3 ( IF <0 EXIT )
MAXIMAL EVOLUTION : 0.1550761E-03 NODE : 85827
MINIMAL EVOLUTION : -0.1604934E-03 NODE : 85952
TOTAL MAXIMAL EVOLUTION : 0.3357706 NODE : 85662
TOTAL MINIMAL EVOLUTION : -0.9478816 NODE : 86018
STOP AFTER AN ERROR IN LAYER
IN 1 PROCESSOR(S)
PLANTE: PROGRAM STOPPED AFTER AN ERROR
RETURNING EXIT CODE: 2
_____________
runcode::main:
:
|runCode: Fail to run
***************************
SUBROUTINE NOEROD
***************************
!
! RIGID BEDS POSITION
!
!
! DEFAULT VALUE: ZR=ZF-100.D0
!
! CALL OV('X=Y+C ',ZR,ZF,ZF,-100.D0,NPOIN)
! > .GT. < .LT. >= .GE. <= .LE.
DO J=1,NPOIN
ZR(J) = ZF(J)
IF( Y(J) .GT. 0.3D0
& .AND. Y(J) .LT. 1.7D0 )THEN
ZR(J) =ZF(J) - 1.0D0
ENDIF
ENDDO
In Sis_Steering_File:(the sohl material is defined as multigrain)
/
/ MULTI GRAIN
/
NUMBER OF SIZE-CLASSES OF BED MATERIAL = 3
SEDIMENT DIAMETERS = 0.01264D0; 0.0036D0; 0.00112D0
INITIAL FRACTION FOR PARTICULAR SIZE CLASS = 0.49D0; 0.27D0; 0.24D0
I need your support please!
Rima