1. Simulation Study of Miscible Water-Alternating-Gas Injection in a Stratified Reservoir Model
Input parameters based on literature data
Jan Inge Nygård (author)
Pål Østebø Andersen (supervisor)
Kenny Walrond (co-supervisor)

2. 1 WAG cycle
WAG ratio
Half-cycle of gas
Injection rate
(rm3/d)
Duration
(days)
Volume of gas
injected
Half-cycle of water
Volume of water
Gas
Water

3. 𝑊𝐴𝐺 𝑐𝑦𝑐𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ= 𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑤𝑎𝑡𝑒𝑟 + 𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑔𝑎𝑠
1 WAG cycle
WAG ratio
Half-cycle of gas
Injection rate
(rm3/d)
Duration
(days)
Volume of gas
injected
Half-cycle of water
Volume of water
For all scenarios studied in this work;
WAG cycle length is 90 days
Total volume injected per WAG cycle is also constant
Injection occurs at miscible conditions
What is the effect of changing injection rates and duration
of half-cycles, while WAG ratio is kept constant?
Is this effect present in different types of reservoirs?
𝑊𝐴𝐺 𝑟𝑎𝑡𝑖𝑜= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑔𝑎𝑠 = (𝐼𝑛𝑗.𝑟𝑎𝑡𝑒∗𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛) 𝑤𝑎𝑡𝑒𝑟 (𝐼𝑛𝑗.𝑟𝑎𝑡𝑒∗𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛) 𝑔𝑎𝑠 = 𝑅 𝑤 ∗ 𝑇 𝑤 𝑅 𝑔 ∗ 𝑇 𝑔

4. Visual representation of WAG ratios
We study WAG ratios 1:1, 1:2 and 2:1
Water
Gas
WAG ratio 1:2
Water
Gas
WAG ratio 1:1
Water
Gas
WAG ratio 2:1
Water
Gas half-cycle
Water half-cycle
Water half-cycle
Gas half-cycle
One WAG cycle consists of one water half-cycle and one gas half-cycle

5. Waterflood vs Miscible WAG
Sorw = 0.31
Residual oil saturation (Sor)
after waterflood
Soi = 0.81
Initial Oil In Place
(IOIP)
Sorg = 0.00
Residual oil saturation (Sor)
after miscible gasflood
Soi = 0.81
Initial Oil In Place
(IOIP)
Distribution of oil saturation at point of gas breakthrough
Homogeneous model
Case 82

6. Gasflood vs WAG Sorg = 0.00 Soi = 0.81
Residual oil saturation (Sor)
after miscible gasflood
Soi = 0.81
Initial Oil In Place
(IOIP)
Distribution of oil saturation at point of gas breakthrough
Homogeneous model
Case 82

7. Oil Recovery Efficiency Factor
Gas- and waterflood vs WAG
Gasflood
WAG
Waterflood
Increased recovery
Oil Recovery Efficiency Factor
Time

8. Ranking of WAG cases
A sample of the dataset
Case
Figure name
WAG ratio
Total simulation time (days)
Oil Recovery Efficiency (FOE)
FOE per year (%)
Gas produced, cum (10^6 Sm3)
Water produced, cum (10^6 Sm3)
Gas-Oil-Ratio (GOR)
PVs injected
Performance
Number of WAG cycles
102
w _g _h
2:1
5478
0,65755
4,38
1873,70
1,058
232
0,800
25,58 60
100
w _g _h
5844
0,65226
4,07
1919,92
1,004
240
0,810
23,47 64
101
w _g60-750_h
7121
0,64724
3,32
2167,25
0,875
273
0,860
17,98 79 82
w _g _h
1:1
6574
0,66027
3,67
2994,93
1,090
370
0,989
17,93 73 84
w _g _h
6847
0,65846
3,51
2795,54
1,088
346
1,028
16,85 76 83
w _g _h
7213
0,65323
3,31
2980,59
1,045
372
1,040
15,48 80 92
w _g _h
1:2
9496
0,68073
2,62
5451,92
1,095
653
1,435
8,16
105 93
w60-750_g _h
9586
0,66279
2,52
4951,60
1,092
609
1,446
7,96
106 91
w _g _h
9677
0,65923
2,49
4917,40
1,093
608
1,460
7,75
107
𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝐹𝑂𝐸+𝐹𝑂𝐸∗ 𝑃𝑉 𝑖𝑛𝑗𝑒𝑐𝑡𝑒𝑑 𝑑𝑎𝑦𝑠+𝑑𝑎𝑦𝑠∗( 𝑃𝑉 𝑖𝑛𝑗𝑒𝑐𝑡𝑒𝑑 −1) ∗ 10 5 − 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑔𝑎𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝐹𝑂𝐸∗2000
The performance factor values oil recovery differently depending on
Pore-volume (PV) size injected
Number of days of simulation to reach given result
Gas production
Example: Same oil recovery achieved with a lower PV size is ranked higher

9. Homogeneous model
Increasing up model
Permeability ranges from 10 to 500 milli Darcy
Kv/Kh = 0.5, ie. vertical permeability 50 % to that of horizontal permeability
30 % porosity
Dimensions: 1050m*1050m*60m
Variable model
Increasing down model

10. Gas saturation distribution at gas breakthrough
Homogeneous model
Increasing up model
Permeability ranges from 10 to 500 milli Darcy
Here showcasing four different models at WAG ratio 1:1
500 mD
10 mD
250 mD
Variable model
Increasing down model
10 mD
500 mD
500 mD
100 mD
20 mD

11. Comparing all four models with WAG ratio 1:1
Cumulative gas production
Oil Recovery Efficiency Factor
Increasing down
Models:
Homogeneous
Variable
Increasing up
Time
Time

12. Oil Recovery Efficiency Factor
Comparing performance impact of different WAG ratios
Oil Recovery Efficiency Factor
Increasing up model
2:1
1:1
1:2
Time

13. Oil Recovery Efficiency Factor
Gas- and waterflooding compared to different WAG ratios
Oil Recovery Efficiency Factor
Increasing up model
2:1
1:1
1:2
Waterflood
Gasflood
Time

14. The effect of changing gas injection rate, for a WAG ratio of 1:1
Increasing up model
The effect of changing gas injection rate, for a WAG ratio of 1:1
Water
Gas
WAG ratio 1:1
Cumulative gas production
Oil Recovery Efficiency Factor
Lower gas production
Higher oil recovery
Low gas rate
High gas rate
Low gas rate
High gas rate
Time
Time

15. The effect of changing gas injection rate, for a WAG ratio of 1:2
Increasing up model
The effect of changing gas injection rate, for a WAG ratio of 1:2
Cumulative gas production
Oil Recovery Efficiency Factor
Water
Gas
WAG ratio 1:2
Lower gas production
Unchanged oil recovery
Low gas rate
High gas rate
Low gas rate
High gas rate
Time
Time

16. The effect of changing gas injection rate, for a WAG ratio of 2:1
Increasing up model
The effect of changing gas injection rate, for a WAG ratio of 2:1
Water
Gas
WAG ratio 2:1
Lower gas production
Higher oil recovery
Faster production
Cumulative gas production
Oil Recovery Efficiency Factor
Increasing gas rate
Decreasing water rate
Time
Time

17. The effect of changing gas injection rate, for a WAG ratio of 1:1
Increasing down model
The effect of changing gas injection rate, for a WAG ratio of 1:1
Water
Gas
WAG ratio 1:1
Cumulative gas production
Oil Recovery Efficiency Factor
Higher gas rate;
Lower gas production
Faster production
Equal gas and water rate;
Higher gas production
Higher oil recovery
Low gas rate
Equal rates
High gas rate
Low gas rate
Equal rates
High gas rate
Time
Time

18. The effect of changing gas injection rate, for a WAG ratio of 1:2
Increasing down model
The effect of changing gas injection rate, for a WAG ratio of 1:2
WAG ratio 1:2
Cumulative gas production
Oil Recovery Efficiency Factor
Water
Gas
Higher gas rate;
Unchanged gas production
Higher oil recovery
Equal gas and water rate;
Higher gas production
Highest oil recovery
Equal rates
Low gas rate
High gas rate
Equal rates
Low gas rate
High gas rate
Time
Time

19. The effect of changing gas injection rate, for a WAG ratio of 2:1
Increasing down model
The effect of changing gas injection rate, for a WAG ratio of 2:1
Cumulative gas production
Oil Recovery Efficiency Factor
Water
Gas
WAG ratio 2:1
Lower gas production
Higher oil recovery
Faster production
Increasing gas rate
Decreasing water rate
Time
Time

20. Conclusion part 1 of 2 For a fixed WAG ratio Injection scheme
We have seen how changing injection rate and duration of half-cycles can alter production performance
This means that it is important to consider how the WAG ratio is used
This effect is strongest for WAG ratio 2:1
This effect is observed throughout all four reservoir models
Injection scheme
Optimal WAG ratio was not 1:1
Optimal WAG ratio was 2:1 and 1:2, when equal injection rates were used
2:1 Highest performance factor rating (ie. faster production over 5000 days)
1:2 Highest oil recovery efficiency factor (ie. slower production over 9000 days)
Changing injection rate and duration of injection, while keeping the same WAG ratio, does affect the performance of Water-Alternating-Gas technique

21. Conclusion part 2 of 2 Performance factor
Was developed by me to rank over 100 different injection schemes
These are successfully ranked, considering:
Oil recovery efficiency factor
Pore-volume (PV) size injected
Number of days of simulation
Gas production
Other factors may also be implemented in order to alter the rating behaviour
Changing injection rate and duration of injection, while keeping the same WAG ratio, does affect the performance of Water-Alternating-Gas technique

22. References Eclipse Technical Description, version 2014.2
Fifth Comparative Solution Project (SPE 5), for PVT properties
Data from pore-network simulation data of Skauge (2011), for relative permeability and capillary pressure curves

23. 𝑊𝐴𝐺 𝑐𝑦𝑐𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ= 𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑤𝑎𝑡𝑒𝑟 + 𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑔𝑎𝑠
Backup
𝑊𝐴𝐺 𝑐𝑦𝑐𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ= 𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑤𝑎𝑡𝑒𝑟 + 𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑔𝑎𝑠
1 WAG cycle
WAG ratio
Half-cycle of gas
Injection rate
(rm3/d)
Duration
(days)
Volume of gas
injected
Half-cycle of water
Volume of water
𝐼𝑛𝑗.𝑟𝑎𝑡𝑒 𝑔𝑎𝑠 =1500 𝑟𝑚 3 /𝑑
𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑔𝑎𝑠 =45 𝑑𝑎𝑦𝑠 (𝑑)
𝐼𝑛𝑗.𝑟𝑎𝑡𝑒 𝑤𝑎𝑡𝑒𝑟 =1500 𝑟𝑚 3 /𝑑
𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑤𝑎𝑡𝑒𝑟 =45 𝑑𝑎𝑦𝑠 (𝑑)
Example 1
𝑊𝐴𝐺 𝑟𝑎𝑡𝑖𝑜= 1500∗ ∗45 = 𝑟𝑚 𝑟𝑚 3 => 1:1 ratio
𝑊𝐴𝐺 𝑐𝑦𝑐𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ=45𝑑+45𝑑=90𝑑
𝐼𝑛𝑗.𝑟𝑎𝑡𝑒 𝑔𝑎𝑠 =1125 𝑟𝑚 3 /𝑑
𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑔𝑎𝑠 =60 𝑑𝑎𝑦𝑠 (𝑑)
𝐼𝑛𝑗.𝑟𝑎𝑡𝑒 𝑤𝑎𝑡𝑒𝑟 =2250 𝑟𝑚 3 /𝑑
𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑤𝑎𝑡𝑒𝑟 =30 𝑑𝑎𝑦𝑠 (𝑑)
Example 2
𝑊𝐴𝐺 𝑟𝑎𝑡𝑖𝑜= 2250∗ ∗60 = 𝑟𝑚 𝑟𝑚 3 => 1:1 ratio
𝑊𝐴𝐺 𝑐𝑦𝑐𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ=30𝑑+60𝑑=90𝑑
𝑊𝐴𝐺 𝑟𝑎𝑡𝑖𝑜= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑔𝑎𝑠 = (𝐼𝑛𝑗.𝑟𝑎𝑡𝑒∗𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛) 𝑤𝑎𝑡𝑒𝑟 (𝐼𝑛𝑗.𝑟𝑎𝑡𝑒∗𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛) 𝑔𝑎𝑠 = 𝑅 𝑤 ∗ 𝑇 𝑤 𝑅 𝑔 ∗ 𝑇 𝑔