问答中心分类: 数值计算如何解决岩心尺度组份模拟CO2驱出现的Repeat time step: unusual pressure in block: 1 ( 1 )问题
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Debin Kong asked 3年 ago

— Repeat time step: unusual pressure in block: 1 ( 1 )
— Repeat time step: unusual pressure in block: 1 ( 1 )
1
G E M T I M E S T E P S U M M A R Y
====================================================================================================================
Time Step Time Maximum Changes Mat Cum Solver
—————– —————– ——————————————————— Bal time ———–
Size Newton Pressure Saturation Composition Err step Iter/ Cum
No. days cycls Days Date Block kpa Block Fraction Block Fraction % cuts Cycl Fail
—- —— —- —— ———- ——— —— ——— ——— ——— ———- —– —- —– —–
1 2.5e-3 22 2.5e-3 1980.01.01 1 3089 1 .1688 (g) 1 .6317 ( 1) 74e-5 2 1.0 0
2 1.7e-3 3 4.2e-3 1980.01.01 7 1729 1 .1145 (g) 3 .2188 ( 1) 42e-5 2 1.0 0
3 2.9e-3 4 7.2e-3 1980.01.01 6 1382 6 .3108 (g) 6 .4692 ( 1) 26e-5 2 1.0 0
— Repeat time step: unusual pressure in block: 1 ( 1 )
4 1.4e-3 12 8.7e-3 1980.01.01 9 383.5 7 .4121 (g) 7 .4236 ( 1) 33e-5 3 1.0 0
5 1.7e-3 3 1.0e-2 1980.01.01 10 252.4 8 .6059 (g) 8 .4997 ( 1) 14e-4 3 1.0 0
6 1.7e-3 7 1.2e-2 1980.01.01 1 -168 7 .6388 (g) 9 .4536 ( 1) 13e-4 3 1.0 0
7 1.2e-3 5 1.3e-2 1980.01.01 12 176.6 10 .3835 (g) 10 .4174 ( 1) 13e-4 3 1.0 0
8 1.2e-3 3 1.4e-2 1980.01.01 1 -105 11 .4041 (g) 11 .4040 ( 1) 14e-4 3 1.0 0
9 1.5e-3 3 1.6e-2 1980.01.01 13 104.8 12 .5653 (g) 12 .4498 ( 1) 17e-4 3 1.0 0
10 1.7e-3 4 1.8e-2 1980.01.01 12 -148 13 .6641 (g) 13 .5916 ( 1) 17e-4 3 1.0 0
11 1.4e-3 6 1.9e-2 1980.01.01 1 -160 14 .5416 (g) 14 .5619 ( 1) 18e-4 3 1.0 0
12 1.2e-3 3 2.0e-2 1980.01.01 1 -156 15 .4325 (g) 15 .5209 ( 1) 18e-4 3 1.0 0
13 1.1e-3 3 2.1e-2 1980.01.01 1 -150 16 .4154 (g) 16 .5137 ( 1) 18e-4 3 1.0 0
14 1.1e-3 3 2.2e-2 1980.01.01 1 -145 17 .4112 (g) 17 .5080 ( 1) 18e-4 3 1.0 0
15 1.1e-3 3 2.4e-2 1980.01.01 1 -141 18 .4178 (g) 18 .5048 ( 1) 18e-4 3 1.0 0
16 1.0e-3 3 2.5e-2 1980.01.01 1 -138 19 .4321 (g) 19 .5044 ( 1) 18e-4 3 1.0 0
17 1.0e-3 3 2.6e-2 1980.01.01 1 -137 20 .4506 (g) 20 .5060 ( 1) 18e-4 3 1.0 0
18 1.0e-3 3 2.7e-2 1980.01.01 1 -136 21 .4699 (g) 21 .5087 ( 1) 46e-5 3 1.0 0
19 1.0e-3 3 2.8e-2 1980.01.01 22 149.2 22 .4859 (g) 22 .5098 ( 1) 54e-5 3 1.0 0
20 1.0e-3 3 2.9e-2 1980.01.01 23 161.2 23 .5036 (g) 23 .5112 ( 1) 53e-5 3 1.0 0
1
G E M T I M E S T E P S U M M A R Y
====================================================================================================================
Time Step Time Maximum Changes Mat Cum Solver
—————– —————– ——————————————————— Bal time ———–
Size Newton Pressure Saturation Composition Err step Iter/ Cum
No. days cycls Days Date Block kpa Block Fraction Block Fraction % cuts Cycl Fail
—- —— —- —— ———- ——— —— ——— ——— ——— ———- —– —- —– —–
21 1.0e-3 3 3.0e-2 1980.01.01 24 175.0 24 .5223 (g) 24 .5162 ( 1) 52e-5 3 1.0 0
22 1.0e-3 3 3.1e-2 1980.01.01 25 191.0 25 .5555 (g) 25 .5322 ( 1) 51e-5 3 1.0 0

Jacobian-based Allocations: 939 (kB)

Current Total Memory Allocation: 2019 (kB)
2 (MB)

Total number of storage reallocation: 10
Total number of time steps: 22
Total number of Newton cycles: 105
Total number of solver iterations: 105
Total number of time step cuts: 3
Total number of solver failures: 0
Material Balance Error; Weighted by OrigMatInPlace+Inj: 5.1596E-04 %
Solver/CPUs/threads/nlv1ca/cdir/dplanes/host: AIMSOL/ 8/ 1/ 0/ -/ -1/DK
Implicitness, average/peak: 33.0576 %/ 68.75 %
Memory usage, average/peak/vm size: 44 MB/ 44 MB/ 39 MB
Date and time at end of run: 2018-Sep-26 12:44:12
CPU second(s) used: 0.484375
Elapsed second(s): 0.554

1 Answers
0
strength windy 管理员 answered 3年 ago

建议在specify property 中设置 implicit flag 对应的“whole grid”上设置为3.然后在 numerical中 将Adaptive implicit method(AIM)选择‘off’