5. C modeling#

5.1. Characteristics of C modeling#

Plane modeling: D_ PLAN_HHM

_images/Object_163.svg

2 meshes DPTR6 of the D_ PLAN_HHM modeling: HHM_ DPTR6

5.2. Result of the C modeling#

Discretization in time: Several time steps (16) to study the evolution of pressure during the transition phase until it stabilizes. The time pattern is implicit \((\theta \mathrm{=}1)\).

List of calculation times in seconds:

\(\mathrm{1,}\mathrm{5,}\mathrm{10,}\mathrm{50,}\mathrm{100,}\mathrm{500,}{10}^{3},5.\times {10}^{3},{10}^{4},5.\times {10}^{4},{10}^{5},5.\times {10}^{5},{10}^{6},5.\times {10}^{6},{10}^{7},{10}^{10}\mathrm{.}\)

Nodal unknowns: fluid pressures evaluated in Code_Aster are variations from initial pressures, which is why this table shows pressure variations in our comparison between the*Code_Aster* calculation and the reference solution. In addition, the pressure variables used in*Code_Aster* to evaluate the laws of behavior are total gas pressure and capillary pressure.

Node/point

Order number/instant \((s)\)

Value

Press \((\mathrm{Pa})\)

Tolerance

1.2/\(A\) and \(B\)

\(1(t=1s)\)

\(\mathrm{PRE1}\)

-8,565.10-3

\({10}^{-4}\)

\(2(t=5s)\)

\(\mathrm{PRE1}\)

-4,282.10-2

\({10}^{-4}\)

\(3(t=10s)\)

\(\mathrm{PRE1}\)

-8,565.10-2

\({10}^{-4}\)

\(4(t=50s)\)

\(\mathrm{PRE1}\)

-4,282.10-1

\(1\text{\%}\)

\(8(t={5.10}^{3}s)\)

\(\mathrm{PRE1}\)

-4,26.10+1

\(1\text{\%}\)

\(16(t={10}^{10}s)\)

\(\mathrm{PRE1}\)

-4,996.10+3

\(1\text{\%}\)

\(1(t=1s)\)

\(\mathrm{PRE2}\)

6,796.10-6

\({10}^{-4}\)

\(2(t=5s)\)

\(\mathrm{PRE2}\)

3,398.10-5

\({10}^{-4}\)

\(3(t=10s)\)

\(\mathrm{PRE2}\)

6,796.10-5

\({10}^{-4}\)

\(4(t=50s)\)

\(\mathrm{PRE2}\)

3,398.10-4

\({10}^{-4}\)

\(8(t={5.10}^{3}s)\)

\(\mathrm{PRE2}\)

3,384.10-2

\({10}^{-4}\)

\(16(t={10}^{10}s)\)

\(\mathrm{PRE2}\)

3,964

\({10}^{-3}\)

3,4/\(C\) and \(D\)

\(1(t=1s)\)

\(\mathrm{PRE1}\)

8,565.10-3

\({10}^{-4}\)

\(2(t=5s)\)

\(\mathrm{PRE1}\)

4,282.10-2

\({10}^{-4}\)

\(3(t=10s)\)

\(\mathrm{PRE1}\)

8,565.10-2

\({10}^{-4}\)

\(4(t=50s)\)

\(\mathrm{PRE1}\)

4,282.10-1

\(1\text{\%}\)

\(8(t={5.10}^{3}s)\)

\(\mathrm{PRE1}\)

4,26.10+1

\(1\text{\%}\)

\(16(t={10}^{10}s)\)

\(\mathrm{PRE1}\)

4,996.10+3

\(1\text{\%}\)

\(1(t=1s)\)

\(\mathrm{PRE2}\)

-6,796.10-6

\({10}^{-4}\)

\(2(t=5s)\)

\(\mathrm{PRE2}\)

-3,398.10-5

\({10}^{-4}\)

\(3(t=10s)\)

\(\mathrm{PRE2}\)

-6,796.10-5

\({10}^{-4}\)

\(4(t=50s)\)

\(\mathrm{PRE2}\)

-3,398.10-4

\({10}^{-4}\)

\(8(t={5.10}^{3}s)\)

\(\mathrm{PRE2}\)

-3,384.10-2

\({10}^{-4}\)

\(16(t={10}^{10}s)\)

\(\mathrm{PRE2}\)

-3,964

\({10}^{-3}\)