open TELEMAC-MASCARET The mathematically superior suite of solvers

Hello,

i indicated a tracer into a corner of a river by giving the location in my condin-subroutine file:

IF(NTRAC.GT.0) THEN
DO ITRAC=1,NTRAC
CALL OS( 'X=C ' , X=T%ADR(ITRAC)%P , C=TRAC0(ITRAC) )
DO I=1,NPOIN
IF(X(I).GE.560157.D0) THEN
T%ADR(1)%P%R(I)=10.D0
ELSE
T%ADR(1)%P%R(I)=0.D0
ENDIF
ENDDO
ENDDO
ENDIF

This 'bay-area' where the tracer is indicated is surrounded by closed boundaries. The initial quantity of the tracer seems correct but after the first iteration there is an expression which i dont understand:

'SOLUTION X=0 BECAUSE L2-NORM OF B VERY SMALL: 0.000000'



================================================================================
ITERATION 0 TIME: 0.0000 S
PRERES: MAXIMUM COURANT NUMBER: 0.000000

BALANCE OF Concentration (UNIT: [-] * M3)
INITIAL QUANTITY OF TRACER : 0.1018133E+10
THE LIQUID BOUNDARIES FILE CONTAINS
335 LINES WITH:
SL(1)

================================================================================
ITERATION 86400 TIME: 30 D 0 H 0 MN 0.0000 S ( 2592000.0000 S)

---

DIFFUSION-PROPAGATION STEP
CVTRVF_POS_2 (SCHEME 13 OR 14): 5 ITERATIONS
EQUNOR (BIEF) : 19 ITERATIONS, RELATIVE PRECISION: 0.9292851E-04
POSITIVE DEPTHS OBTAINED IN 7 ITERATIONS

---

DIFFUSION OF TRACERS STEP
CVTRVF_POS (SCHEME 13 OR 14): 5 ITERATIONS
GRACJG (BIEF) : SOLUTION X=0 BECAUSE L2-NORM OF B VERY SMALL: 0.000000

BALANCE OF Concentration (UNIT: [-] * M3)

INITIAL QUANTITY : 0.000000
FINAL QUANTITY : 0.000000
BOUNDARY 1 FLUX: -0.000000 ( >0 : ENTERING <0 : EXITING )
BOUNDARY 2 FLUX: -0.000000 ( >0 : ENTERING <0 : EXITING )
TOTAL QUANTITY LOST : 0.000000

---

Thank you for your help !!!

i reduced the listening period 1 minute:

The tracer quantity is set to 'NaN' in the time from 6h 39 to 6h 40...
Can anyone explain this to me? Thank you!!!






================================================================================
ITERATION 798 TIME: 6 H 39 MIN 0.0000 S ( 23940.0000 S)

---

DIFFUSION-PROPAGATION STEP
CVTRVF_POS_2 (SCHEME 13 OR 14): 6 ITERATIONS
EQUNOR (BIEF) : 18 ITERATIONS, RELATIVE PRECISION: 0.8192523E-04
POSITIVE DEPTHS OBTAINED IN 6 ITERATIONS

---

DIFFUSION OF TRACERS STEP
CVTRVF_POS (SCHEME 13 OR 14): 6 ITERATIONS
GRACJG (BIEF) : 0 ITERATIONS, RELATIVE PRECISION: 0.1240303E-07

BALANCE OF Concentration (UNIT: [-] * M3)

INITIAL QUANTITY : 0.1018132E+10
FINAL QUANTITY : 0.1018132E+10
BOUNDARY 1 FLUX: -0.7286381E-46 ( >0 : ENTERING <0 : EXITING )
BOUNDARY 2 FLUX: -0.000000 ( >0 : ENTERING <0 : EXITING )
TOTAL QUANTITY LOST : 2.856215
RELATIVE ERROR : 0.2805349E-08

---

BALANCE OF WATER VOLUME
VOLUME IN THE DOMAIN : 0.1115166E+10 M3
FLUX BOUNDARY 1: -2504.907 M3/S ( >0 : ENTERING <0 : EXITING )
FLUX BOUNDARY 2: 2060.002 M3/S ( >0 : ENTERING <0 : EXITING )
ADDITIONAL VOLUME DUE TO SOURCE TERMS: -215.0294 M3
RELATIVE ERROR IN VOLUME AT T = 0.2394E+05 S : -0.5289945E-16
================================================================================
ITERATION 800 TIME: 6 H 40 MIN 0.0000 S ( 24000.0000 S)

---

DIFFUSION-PROPAGATION STEP
CVTRVF_POS_2 (SCHEME 13 OR 14): 6 ITERATIONS
EQUNOR (BIEF) : 18 ITERATIONS, RELATIVE PRECISION: 0.8634848E-04
POSITIVE DEPTHS OBTAINED IN 6 ITERATIONS

---

DIFFUSION OF TRACERS STEP
CVTRVF_POS (SCHEME 13 OR 14): 10 ITERATIONS
GRACJG (BIEF) : SOLUTION X=0 BECAUSE L2-NORM OF B VERY SMALL: NaN

BALANCE OF Concentration (UNIT: [-] * M3)

INITIAL QUANTITY : 0.1018132E+10
FINAL QUANTITY : 0.000000
BOUNDARY 1 FLUX: -0.8511100E-46 ( >0 : ENTERING <0 : EXITING )
BOUNDARY 2 FLUX: -0.000000 ( >0 : ENTERING <0 : EXITING )
TOTAL QUANTITY LOST : 0.1018132E+10
RELATIVE ERROR : 1.000000

Hello,

The first message can be normal when there is a diffusion coefficient equal to 0 or an explicit diffusion (this could be your case). The NaN is not normal. I assume that you have dry zones, could you check that you have TREATMENT OF NEGATIVE DEPTHS : 2, or send your steering file.

With best regards,

Jean-Michel Hervouet

thank you very much for your quick answer !


my steering file:


/

---

/
STEERING FILE :cas_Ti_evap
BOUNDARY CONDITIONS FILE :conlim.cli
LIQUID BOUNDARIES FILE :liquid_boundary.txt
FORTRAN FILE :condin_e.f
GEOMETRY FILE :geo.sel
RESULTS FILE :res_Ti_8h_evap_test_02
/COMPUTATION CONTINUED :YES
/PREVIOUS COMPUTATION FILE :res1
PARALLEL PROCESSORS :8
INITIAL TIME SET TO ZERO :YES
/
/

---

/ GENERAL INFORMATIONS - OUTPUTS
/

---

/
TITLE = 'Tracer as initial condition'
VARIABLES FOR GRAPHIC PRINTOUTS = 'U,V,S,B,H,Q,M,F,L,T1'
GRAPHIC PRINTOUT PERIOD = 10 /each 5 min
LISTING PRINTOUT PERIOD = 2
TIME STEP = 30
DURATION : 25200 / 7 hours
MASS-BALANCE = YES
INFORMATION ABOUT SOLVER = YES
/
/

---

/ INITIAL AND BOUNDARY CONDITIONS
/

---

/
INITIAL CONDITIONS : 'CONSTANT ELEVATION'
INITIAL ELEVATION : 301.76
/OUTPUT OF INITIAL CONDITIONS = YES
/PRESCRIBED ELEVATIONS : 304.;0.
PRESCRIBED FLOWRATES : 0.;2060.0
/
/

---

/ PHYSICAL PARAMETERS
/

---

/
TURBULENCE MODEL = 1
VELOCITY DIFFUSIVITY = 1.E-4
LAW OF BOTTOM FRICTION : 3
FRICTION COEFFICIENT : 30.
WIND = YES
WIND VELOCITY ALONG X = -4.2
WIND VELOCITY ALONG Y = 3.5
COEFFICIENT OF WIND INFLUENCE = 0.000000126
RAIN OR EVAPORATION = YES
RAIN OR EVAPORATION IN MM PER DAY = -5.479
/
/

---

/ NUMERICAL PARAMETERS
/

---

/
EQUATIONS = 'SAINT-VENANT EF'
NUMBER OF PRIVATE ARRAYS = 0
/TIDAL FLATS = YES
/OPTION FOR THE TREATMENT OF TIDAL FLATS : 1 / if =2, you need to write also SUPG OPTION = 2;0
TREATMENT OF NEGATIVE DEPTHS : 2
TYPE OF ADVECTION = 14;5;14
SUPG OPTION = 0;0
DISCRETIZATIONS IN SPACE = 11;11
SOLVER ACCURACY = 1.E-4 / better between 1.E-6 and 1.E-9
SOLVER = 3 / recommended and default
PRECONDITIONING = 2
IMPLICITATION FOR DEPTH = 0.55 / default
IMPLICITATION FOR VELOCITY = 0.55 / default
/INITIAL GUESS FOR H = 1
TREATMENT OF THE LINEAR SYSTEM : 1
MASS-LUMPING ON H : 1
CONTINUITY CORRECTION : YES
FREE SURFACE GRADIENT COMPATIBILITY : 0.9
/
/

---

/ TRACER
/

---

/
NUMBER OF TRACERS : 1
NAMES OF TRACERS = 'Concentration [-] '
COEFFICIENT FOR DIFFUSION OF TRACERS : 1.E-4
/
/
&FIN

Hello,

You are working with the primitive equations option, which costs longer execution times and is less stable. Try instead:

TREATMENT OF THE LINEAR SYSTEM : 1
SOLVER : 1

with your initial time step. You can also have a more stable model with:

IMPLICITATION FOR DEPTH = 1.
IMPLICITATION FOR VELOCITY = 1.

Knowing whether you have dry zones or not is important, especially with evaporation, since a special care is necessary near dry zones with evaporation, to avoid negative depths.

With best regards,

Jean-Michel Hervouet

Hello,

thank you for your efforts!
Please excuse me i if understand you wrong:

treatment of negative depths is/was set to '2'.
Your suggestions for a more stable model didnt show any changes regarding the 'error' NaN.

As a test i left out the evaporation: The 'error' disappeared...

Is there anything i can change in the parameters, to make it work correctly with evaporation?

Hello,

The best would be that you send me your case so that I can see where the error occurs with a compiler that tells me at what line we have a NaN. I suspect that evaporation on dry land is not well tested somewhere.

With best regards,

Jean-Michel Hervouet

Hello,

i set Treatment of tidal flats = 2 and it worked !!

Thank you very much for yor help