How to apply this unique Mix of Neuman + Dirichlet Boundary conditions, for Temperature & Pressure, 1-d Problem

Hello everyone

I am trying to setup a 1-D validation problem from a research paper for validating experimental results.

In short, I need to apply Neuman & Dirichlet Boundary conditions, for Pressure & Temperature

For Neuman: I need to set the block as Inactive, its done Neuman!!! (P,T both )

For Dirichlet i need to set INCON

But the problem is At one boundary: I need to set Temp as Neuman & Pressure as Dirichlet, How to do this???

Kindly see the pic...attached

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  • Well, the left side is easy.  You just don't add any more elements beyond that boundary, and the no-flux condition gets set by default.  The second is a little trickier.  Give me a little bit to consider it...

  • I think the trick on the right side is to connect that boundary to an element with VERY large volume (~1.0E50).  This will hold both the pressure and temperature constant at that boundary.  Let me ask you this before I fully commit to this response... is this an isothermal simulation?  I see there is no heat flux entering or exiting the system through the boundaries.  Does the temperature within the elements also remain fixed over time?  If so, there's a much easier (and more efficient) way of solving your problem.

    • Mikey Hannon I have tried so many combinations that i am confused now..literally.

      Here's the paper(attached). in this paper, validation is done against the experiments around 4-5 details are also giiven

      THE SETUP: Its actually a horizontal vessel. one circular end is sealed by closed metal face. other end is subjected to depressurization. so in my opinion, 

      the closed end is mimmicked by neumann,

      the depressurization end has pressure which is not going to change(Hence dirichlet), but the temperature is free to change.

      in TOUGH+HYDRATE there are following conditions for INCON block....i guess, i cannot use anyone with fixed temperature, so that leaves me with one choice only.(highlighted one)

    • No, its not an isothermal simulation, i think as the gas is removed as hydrate dissociates, heat is released, so temperature will rise...

  • Stefan Finsterle I am sorry for referring to you for this. but can you please provide your valuable comments and help?

  • So what you're aiming to get is an isobaric, adiabatic boundary on the other side.  This trick can be done.  This trick can be done by assigning the large volume element imposing Dirichlet conditions to a material type with a large heat capacity as well.  You'll still get a heat flux through into that element, but it will be absorbed with negligible temperature changes by the large heat capacity.  This setting can be changed in the SPHT variable at the end of the first line of the ROCKS block listing.   Make sure this element has a very large volume (1.0E50) in the ELEME block and very small (1.0E-10) distance to the interface on the CONNE block.  I cannot tell you what to do about anything specific to TOUGH+Hydrate, but I believe this should address the original issue you brought up here.

    • Mikey Hannon if i am understanding correctly, you mean that i should create

      1. two layers (upper layer with large volume, small interface distance, manually edit, right?

      2. give the upper layer very large volume and high SPCHT say 10^4

  • Yes, but choose 10^5 or higher on the SPHT.  That’ll flag to the program not to include that element in global calculations of fluid mass, etc.

    • Mikey Hannon 

      The Neuman BC on the left side is OK, but I'm not sure the Dirichlet condition for P and Neuman condition for T can be both obtained by adding a large (inactive) element with conditions equal to the initial ones.

      This is OK for the P BC, which will be held at the initial constant value,  P=Po.

      But this choice will enforce the same boundary condition (Dirichlet) for the T too. To suppress the heat conduction the thermal conductivity of the rock domain to which the inactive boundary element belongs should be set to zero. Depending on the interpolation method used by the code for the thermal conductivity along a connection, it could be necessary to set to zero also the thermal conductivity of the last element on the right boundary of the 1D linear grid.

      If the T changes within the 1D grid, the T derivative dT/dx along the last right connection will not be necessarily zero, but there will be no heat conduction because of the heat conductivity assignment. 

    • Alfredo Battistelli , I've also been thinking this over some, too.  I did think of setting the thermal conductivity to zero to suppress heat conduction, but convection would still go on unabated, correct? 

    • Mikey Hannon Sure, flow would not be inhibited through the right boundary. If fluid flux is going outside of the system with no conduction,  the BC for T should be close to Neuman.

      I have some doubt if fluid flux is entering the system from the right boundary.

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    • Alfredo Battistelli I agree, but you would still have fluid exiting the system, so it's not perfect. telegramboy I would suggest following Alfredo's suggestion and assign a zero thermal conductivity to the large element boundary.  If you keep the node distance small, it should make the thermal conductivity of the connection (boundary) zero as well.  That'll get you closer to the boundary conditions you're looking for.

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    • Mikey Hannon 

       a. Fluid exiting the system cannot be avoided, as it is a consequence of the constant P boundary condition.

      b. Using a short node distance from the interface for the inactive element is necessary to have the const P boundary at the right place, but it should  not affect the interpolation of the thermal conductivity. I do not know  what is done in the specific code used, but in TOUGH2 the thermal conductivity along the connection  is weighted considering the fractional lenght of each nodal distance. So, it should be necessary to assign zero thermal conductivity on both the last element on the right boundary and in the inactive element. A small active element on the right boundary could be added and used for this purpose.

    • Alfredo Battistelli Oh.....i am sorry to enter this discussion a bit late. I see that you have added a suggestion which is acknowledged by  Mikey Hannon .

      Mikey Hannon  I will try to adapt my case file to these conditions and let you guys know about the results.

      i will keep this thread updated

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