Problem Set 2 is basedon a problem in the MT3Dmanual; also discussed inZ&B, p. 228-231.
2D steady state flow in a confined aquifer
We want to predict thebreakthrough curve at thepumping well. The transportproblem is transient.
Units in MT3D – see p. 6-8 in the manual
Recommended: use ppm= mg/l gm/m3
That is, use meters; mass is reported in grams.
Mass = c Q t
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0.00 0.20 0.40 0.60 0.80 1.00 1.20
Time (years)
Co
nce
ntr
atio
n
TVD
HMOC
Upstream weighting
Central FD
Upstream FD
Central FD
TVD
NOTE. These results were produced using an old version of MT3DMS.Please run again with the latest version of the code.
Central Difference Solution
Time step multiplier = 141 time steps
Time step multiplier = 1.213 time steps
See information on solution methodologies underthe MT3DMS tab on the course homepage for moreabout these parameters.
Courant number
Need to designatethese boundarycells as inactiveconcentration cells.Use zone 10 inthe diffusionsproperties menu ofGroundwater Vistas.
Mass Balance Considerations in MT3DMS
Sources of mass balance information:*.out file*.mas filemass balance summary in GW Vistas
See supplemental information for PS#2 postedon the course homepage for more information onmass balance options.
Mass Balance states that:Mass IN = Mass OUTwhere changes in mass storageare considered either as contributionsto mass IN or to mass OUT.
Water Flow:IN= through upper boundary; injection wellOUT= pumping well; lower boundary
Mass Flux:IN= through injection well; changes in storageOUT= pumping well; lower boundary; changes in storage
wells
IN - OUT = S where S = 0 at steady state conditions
Mass Storage: Water
Consider a cell in the model
IN - OUT = S where change in storage isS = S(t2) – S(t1)
If IN > OUT, the water level rises andthere is an increase in mass of water in the cell.IN = OUT + S, where S is positive.Note that S is on the OUT side of the equation.
If OUT > ININ – S = OUT, where S is negativeS is on the IN side of the equation.
Mass Storage: Solute
IN - OUT = S where change in storage isS = S(t2) – S(t1)
If IN > OUT, concentration in cell increases andthere is an increase in solute mass in the cell.IN = OUT + S, where S is positive.Note that S is on the OUT side of the equation.There is an apparent “sink” inside the cell.
If OUT > IN, the concentration in cell decreases andthere is a decrease in solute mass in the cell.IN – S = OUT, where S is negative and S is on the IN side of the equation. There is anapparent “source” inside the cell.
From the *.out file (TVD solution)
S
IN – OUT = 0(INsource+SIN) - (OUTsource + SOUT)= 0
SIN - SOUT = Storage
From the *.out file (TVD solution)
S
IN – OUT = 0(INsource+SIN) - (OUTsource + SOUT)= 0
SIN - SOUT = Storage