Transcript Document

Problem Set 2 is based
on a problem in the MT3D
manual; also discussed in
Z&B, p. 228-231.
2D steady state flow
in a confined aquifer
We want to predict the
breakthrough curve at the
pumping well. The transport
problem is transient.
Zone of low
hydraulic conductivity
Peclet numbers = 5 and 25
GWV screen
Note: the heterogeneity
is not present in the first
column
Units in MT3D
– see p. 6-8 in the manual
Concentration units do not have to be consistent with the units
used for other parameters. It is permissible, for example,
to use “ft” for the system parameters and mg/l for concentration.
However, in that case the units calculated in the mass balance
will have inconsistent units and the mass balance numbers will
need to be manually corrected.
Recommended: use ppm= mg/l  gm/m3
That is, use meters; mass is reported in grams.
Mass = c Q t
Cs = 57.87 ppm
Cs = 0
1.20
Central FD
TVD
1.00
HMOC
TVD
Upstream weighting
Central FD
Concentration
0.80
Upstream FD
0.60
0.40
0.20
0.00
0.00
0.20
0.40
0.60
0.80
1.00
1.20
Time (years)
NOTE. These results were produced using an old version of MT3DMS.
Please run again with the latest version of the code.
MT3DMS Solution Options
PS#2
1
3
4
2
Central Difference Solution
Time step multiplier = 1
41 time steps
Time step multiplier = 1.2
13 time steps
Courant number
See information on solution methodologies under
the MT3DMS tab on the course homepage for more
about these parameters.
Boundary Conditions
---for flow problem
---for transport problem
Head solution-Flow problem is
steady state
Transport Problem
Need to designate
these boundary
cells as inactive
concentration cells.
Use zone 10 in
the diffusions
properties menu of
Groundwater Vistas.
Cells in first row are in zone 10
in the diffusion properties menu
This is necessary to prevent loss of
mass through the boundary by diffusion.
Solution at t=1 year
Mass Balance Considerations in MT3DMS
Sources of mass balance information:
*.out file
*.mas file
mass balance summary in GW Vistas
See supplemental information for PS#2 posted
on the course homepage for more information on
mass balance options.
wells
IN - OUT = S
Water Flow:
IN= through upper boundary;
injection well
OUT= pumping well;
lower boundary
where S = 0 at steady state conditions
Mass Balance states that:
Mass IN = Mass OUT
where changes in mass storage
are considered either as contributions
to mass IN or to mass OUT.
Mass Flux:
IN= through injection well;
changes in storage
OUT= pumping well;
lower boundary;
changes in storage
From the *.out file
(TVD solution)
Mass Storage: Water
Consider a cell in the model
IN - OUT = S
where change in storage is
S = S(t2) – S(t1)
If IN > OUT, the water level rises and
there 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 > IN
IN – S = OUT, where S is negative
S is on the IN side of the equation.
From the *.out file
(TVD solution)
S
S =  c (x y z )
Mass Storage: Solute
IN - OUT = S
where change in storage is
S = S(t2) – S(t1)
If IN > OUT, concentration in cell increases and
there 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 and
there 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 an
apparent “source” inside the cell.
From the *.out file
(TVD solution)
S
IN – OUT = 0
(INsource+SIN) - (OUTsource + SOUT)= 0
SIN - SOUT =  Storage
HMOC *.mas file
q’s =
General form of the ADE:
Expands to 9 terms
Expands to 3 terms
(See eqn. 3.48 in Z&B)
Where does the extra term
come from?
q’
s=
Assume local chemical equilibrium (LEA):
Isotherms
HMOC *.mas file
Mass Balance error for FD solutions should be less
than 1%.
MOC methods typically report high mass balance errors,
especially at early times.
TVD Solution
From the *.out file
(TVD solution)
S
IN – OUT = 0
(INsource+SIN) - (OUTsource + SOUT)= 0
SIN - SOUT =  Storage
From the *.out file
(TVD solution)
Last t = 0.0089422 yr
Mass Flux = (mass at t2 - mass at t1) / t