Imbibition curves

[vinomarky]

I would guess that imbibition capillary pressure effect really only would start to be relevant when (a) flow velocities are very low (viscous forces don't dominate) and (b) when imbibition curves are lazy, and not almost vertical as in many cases with higher perm, where practically all the capillary pressure potential is released with very little change in saturation

The problem is that very low perm (low velocity) reservoirs are usually not great candidates for water injection (with gas or not)

While no doubt an interesting academic exercise, I'd be surprised if the use of imbibition curves would yield significant additional insights into commercial WAG fields, and when you take into account the difficulty in getting the data as well as the additional degree of freedom you are introducing into the non-unique history match process, I'm not surprised that most of what I read focuses instead on rel perm hysterisis

Happy to be proven wrong! :-)

[Shakespear]

You are correct, rates will effect whether cap. pressure, gravity or viscous forces will have significant effect on flow of individual phase. Plenty of papers that discuss this issue as regards core and reservoir scale flow.

However this paper does show you that you would need to think about the process you are modeling as the real rock does not behave like the simple differential equations we use in our simulators. In their minds they were interested in sequestering CO2. That implies putting it in and make sure it does not move out of there somewhere else. Hence they were interested in knowing how much CO2 would be trapped once the CO2 injection process was stopped. This turns out to be a bit more complicated.

CO2 being a gas has the following options:
1) Go into solution with the brine
2) Due to buoyancy drift upwards
3) Get trapped in the pore throats due to snap off

Now they looked at 2) and 3) and saw that without hysteresis you will not be modeling the real world. The hysteresis is known to occur in lab hence you have the option to do the same in ECLIPSE. They show the difference in Fig. 5 and discuss it in section 4.1. There is are significant differences, however the time scale is HUGE !!!! They are injecting for 10 yrs. and observing the migration of the CO2 for 490 yrs. !!!!

So we can see that time scale of the mechanism is an important consideration, if it take to see a noticeable difference after 20 yrs. I can ignore the effect if my field will deplete in 10 yrs.

So if I shut-in 6 wells for 6 months, 1 yr. ,will my model account for the REAL processes going on in their region? It is a question that is relevant and sort of speaks to the paper,because I bet it will not be like your simulator shows you. You will have a saturation reversal. Depending again on fluids, rates, rock types, structure relief, wettability, capillary forces etc. it may be necessary to model this if you want to get closer to REALITY. But ....

Same goes for reservoir scale, if our grid blocks are 100*100*3 meters, then pore scale effects will not be modeled by such a model. Notes that capillary pressure is a pore scale effect and has a dispersive effect on the moving fluid front (vertical saturation variation). IT MAY BE important to do it but you will need to do some additional work to take this into account. Plenty of papers of how people try to do and it is not simple.

In the mean time, time is running and the company wants an estimate of well production rates for future economics. So again, scale of the company, scale of the project, scale of $$$, scale of time available involved will decide HOW MUCH you can realistically do.

For Ivan I would say, follow their approach of running a case without and with hysteresis, to show what difference you see if you turn this option ON.

Regardless the paper is a good one.

Here is one example of how "effects" are considered or ignored, BY checking if a difference is noticed when the option is used.

Tbe use of a five-point operator for the grid in Problem 2A (see
Appendix A) results in severe grid-orientation effects. The model
unrealistically predicts steam breakthrough at the far well (one eighth
producer) first. A small capillary Pressure was included to
introduce the minimum amount of dispersion that would damp out
incipient numerical fingers. Straight-line capilltwy pressure curves
were used, and tbe endpoint (i.e., maximum) values were 7 psi
[48 !&a] for tie liquid/gin system and 5 psi W kPa] for the oil/water
system. These values are well within acceptable limits. Bwaus.e
the gridblock are so large (about 30 II [9 m]), this amount of capillary
pressure has little effect on front thickness. Rerunning Problem
1A with the capillary pressure present produced results little
different from the mn with no capillary pressure.

[link Point to another website Only the registered members can access]

[Shakespear]

WAG process research page for our loved Heriot-Watt Univ.

[link Point to another website Only the registered members can access]

[reservoir_engineer]

interesting attaches Shakespear,
thanks

[paolomaldini]

thanks

[paolomaldini]

FOR ALL JUNIORS RES ENG THAT HAVE PROBLEMS WITH PC:

PLEASE I HAVE THIS DATA of lab:
DEPTH: 932.3 m

I need convert to Height vs Sw for work with the petrel function xy aplication.
please can you explain me one workflow to solve my problem.

thanks in advance, I hope your help pmaldini85@hotmail.com
DATA:

DRAINAGE IMBIBITION depth: 932.2 m
PC SHG D HG
KG/CM2 % %
0.03 100
0.09 100
0.18 99.43
0.38 85.18
0.63 63.89
0.88 54.38
1.03 51.43 25.66
1.30 47.46 25.34
1.5 45.12 25.07
2 40.70 24.08
3 35.84 23.25
4 32.86 22.58
5 30.88 22.12
7.5 28.13 21.51
10 26.58 21.10
15 24.83 20.61
25 23.20 20.17
40 21.99 19.89
60 21.15 19.8
80 20.55 19.75
100 20.12 19.73
120 19.72 19.72

i hope your help and if is possible share a spreadsheet for the calcs.
Thanks

[vinomarky]

1. Convert lab pressure to reservoir conditions capillary pressure - see

[link Point to another website Only the registered members can access]
2. Height above free water level = Reservoir conditions pressure / (Density difference x water gradient)
Example - you have Pc = 10psi, oil water system with oil SG = 0.81 and water SG = 1.01, the Height = 10 / ((1.01 - 0.81)*0.443 psi/ft) = 112 ft

[paolomaldini]

Thanks a lot.