MOPR(1) = 4 (constant velocity)

Hi, I know most of simulation uses the first option of MOPR(1) as

0: reactive transport

or

2: only flow with no chemical reaction

but when do we use

4: reactive transport using  constant velocity filed? I think we can also generate constant velocity field just using fixed pressures.

By the way, I compared this option 0 and 4 for diffusive transport case in 1-D column and found out the results are quite different.

does this option ignores the (hydrostatic) pressure gradient during the simulation? please give me some idea on when do we use this option.

Thank you : )

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  • Hi Jaeshung,

    MOPR(1)=2 turns of all transport and reaction -- it runs just flow with the TOUGH2 core. MOPR(1)=4 calculates the velocity field using the TOUGH2 core at the first time step, and then keeps the velocities constant (does not solve flow again), but solves full reactive transport.

    So the options are completely different, and don't have anything to do specifically with any pressure conditions or diffusion. One just solves flow and the other reactive transport (including diffusion if turned on), with a constant velocity field.

    regards,

    Eric

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  • Thank you. Dr. Sonnenthal,

    Actually I made a typo (now fixed). I was comparing 0 vs. 4, not 2 vs. 4.

    But I think I understand your answer somehow. 

    So, for the 100% diffusive transport in 1-D vertical column without advection, is it better to use 4 rather than 0? (for example, well known problem about the diffusive metal transport in lake Coeur d’Alen sediments case in TOUGHREACT sample problem)

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