1. 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.

2. Peclet numbers = 5 and 25 Zone of low hydraulic conductivity

3. Units in MT3D – see p. 6-8 in the manual Recommended: use ppm= mg/l  gm/m 3 That is, use meters; mass is reported in grams. Mass = c Q  t

4. Cs = 57.87 ppm Cs = 0

5. NOTE. These results were produced using an old version of MT3DMS. Please run again with the latest version of the code.

6. MT3DMS Solution Options 1 2 3 4 PS#2

7. Central Difference Solution Time step multiplier = 1 41 time steps Time step multiplier = 1.2 13 time steps

8. See information on solution methodologies under the MT3DMS tab on the course homepage for more about these parameters. Courant number

9. Boundary Conditions

11. Need to designate these boundary cells as inactive concentration cells. Use zone 10 in the diffusions properties menu of Groundwater Vistas.

12. Solution at t=1 year

13. 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.

14. 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. Water Flow: IN= through upper boundary; injection well OUT= pumping well; lower boundary Mass Flux: IN= through injection well; changes in storage OUT= pumping well; lower boundary; changes in storage wells IN - OUT =  S where  S = 0 at steady state conditions

15. From the *.out file (TVD solution)

17. 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.

18. From the *.out file (TVD solution) SS  S =   c (  x  y  z  )

19. 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.

20. From the *.out file (TVD solution) SS IN – OUT = 0 (INsource+  S IN ) - (OUTsource +  S OUT )= 0  S IN -  S OUT =  Storage

21. HMOC *.mas file q’ s =

22. MOC methods typically report high mass balance errors, especially at early times.

23. TVD Solution

24. From the *.out file (TVD solution) SS IN – OUT = 0 (INsource+  S IN ) - (OUTsource +  S OUT )= 0  S IN -  S OUT =  Storage

25. From the *.out file (TVD solution)

26. Last  t = 0.0089422 yr Mass Flux = (mass at t2 - mass at t1) /  t