TRATAMIENTO PERCUTANEO DEL INFARTO AGUDO Y PERFUSION

It is seen that capillary pressure depends both on wetting phase saturation and the direction of its variation. A typical curve of the capillary pressure in case of two-phase flow, is shown in the Fig. 6.19.

P_cb P(S)cw

S_nc S_w S_wc

1.0 0.0

1

2 3

Figure 6.19: Typical type of capillary pressure curve for a two-phase flow problem: 1 drainage, 2 imbibition and 3 secondary drainage.

Two ways in which one phase can be substituted by the other in a porous medium are usu-ally considered. The first is the process of displacement where the wetting phase is displaced by the non-wetting one, and the second is the process of imbibition, where the non-wetting phase is displaced by the wetting one. The value Pcbis defined as threshold capillary pressure which should be exceeded to provide displacement. If displacement is preceded by imbibition the capillary pressure curve is as the curve 3 in the Fig. 6.19, which is different from the curve 1. The presence of two different curves of imbibition and displacement is called capillary hys-teresis. The presence of the negative capillary pressure near the saturation pointS_w= S_nc was first discovered by Welge (1949). The hysteresis effect i demonstrated by the two curves 2 and 3 in Fig. 6.19.

6.11 Exercises

1. Calculate the energy needed to transform 1 cm³ of water into droplets with an average radius of 1µm. In analogy with displacement processes in porous media, assume an interfacial tensionσo,w= 0.025 N/m.

2. Show that the general expression for capillary pressure

Pc=σ 1 R1

+ 1 R2

 , could be written

P_c=2σcosθ

r ,

for a cylindrical tube. Define the parameters; r, R1,2,θandσ.

3. A capillary glass-cylinder is positioned vertically in a cup of water. Calculate the height of water inside the cylinder when the inner diameter is 0.1 cm. The surface tension between water and air is 72 dyn/cm, and the water is assumed to wet the glass 100 %.

4. In order to displace water by air form a porous plate, a pressure of 25 psig is needed.

Find the diameter, given inµm, of the largest pore channel disconnecting the porous plate, when the surface tensionσair,wis 72 dyn/cm.

5. A horizontal cylinder, filled with oil, is 0.1 m long and has a inner diameter of 0.01 mm.

The oil has a viscosity similar to water, 1 m Pa · s. What is the pressure drop along the cylinder, when the average flow velocity is found to be 0.01 mm/s ?

An equal amount of water and oil is pumped through the cylinder and the water and oil is assumed to move through the tube as droplets, with an average length of 0.03mm pr.

droplet. The advancing contact angle is 40^oand the receding angle is 20^o. Calculate the pressure drop through the tube, assuming the same flow velocity as above. The interfacial tension between water and oil is 25 mN/m.

6. A core sample is placed in a core holder in a centrifuge. The radial distance to the core sample is given by the two position vectors r₁ and r₂. The length of the sample is therefore; r2− r1. At a rotation frequencyω, air will displace some of the water in the sample. The radial distance rd, corresponds to the threshold pressure Pdat that particular rotation frequency. See the figure below.

a) Show that the pressure difference for one phase is given by:

P₂− P₁=1

2ρω²(r₂²− r₁²), when P₂= P(r₂), P₁= P(r₁) b) Show that

Pc(r) =1

2∆ρω²(r₂²− r²)

ω

r Sw

r₁ r2

rd

1.0

0.0

when its known that

P_c1= P_c(r₁) =1

2∆ρω²(r₂²− r²₁) and when

Pd= 1

2∆ρω²(r²₂− r²_d) It is assumed that Pc2= Pc(r2) = 0.

c) The water saturation can be written,

Sw1= Sw(r1) =d( ¯SwPc1) dPc1

when r1∼ r2, assuming the length of core sample to be short compared to the radius of rotation. (This is an approximation, only partly true.)

Use the equations above to derive the a formula giving the capillary pressure as function of the water saturation (P_c–curve), when the capillary pressure is given in kPa.

The following data is given for a core sample, saturated with sea-water and rotated in air.

r1= 4.46 cm r₂= 9.38 cm

∆ρ = 1.09 g/cm³ Vp= 8.23 cm³

RPM^∗ 415 765 850 915 1005 1110 1305

∆V [cm³] 0.00 0.00 0.10 0.15 0.30 0.50 1.10

RPM 1550 1835 2200 2655 3135 3920 4850

∆V [cm³] 2.20 2.90 3.61 4.21 4.72 5.24 5.75 RPM^∗: Rotation pr. Minute and∆V : produced volume.

7. In the laboratory, a capillary pressure difference of 5 psi has been measured between water and air in a core sample. Calculate the corresponding height above the OWC in the reservoir from where the core originates, when the following information is given (assume capillary pressure at the OWC to be zero).

Laboratory Reservoir

σ= 75 dyn/cm σ= 25 dyn/cm

∆ρw/air= 1.0 g/cm³ ∆ρo/w= 0.2 g/cm³

8. In a laboratory experiment, capillary data from two water saturated core samples was obtained by using air as the displacing fluid.

1000 mD core sample 200 mD core sample Pc[psi] Sw Pc[psi] Sw

1.0 1.00 3.0 1.00

1.5 0.80 3.6 0.90

1.8 0.40 4.0 0.60

2.2 0.20 4.5 0.30

3.0 0.13 5.5 0.20

4.0 0.12 7.0 0.18

5.0 0.12 10.0 0.18

Calculate the distribution of vertical water saturation in the stratified reservoir given by the figure below, i.e. determine Swas function hight in the reservoir.

12 ft 4 ft 6 ft3 ft 4 ft FWL5 ft

Additional data:

Laboratory: σw/air= 50 dyn/cm Reservoir: σo/w= 23 dyn/cm

ρo= 0.81 g/cm³ ρw= 1.01 g/cm³

9. Use the air - water capillary pressure curve for laboratory conditions, below, to calculate the saturations; S_o, S_g and S_w at the reservoir level (hight) 120 f t above the oil-water contact (assume P_c = 0 at this level). The distance between the contacts (OWC and GOC) is 70 f t.

Additional data:

Laboratory: σw/air= 72 dyn/cm Reservoir: σo/g=50 dyn/cm

σw/o= 25 dyn/cm ρo= 53 lb/ f t³ ρw= 68 lb/ f t³ ρg= 7 lb/ f t³

20 30 40 50 60 70 80 90 100

Water saturation [%]

0 10 20 30 40 50 60 70 80 90

Capillary pressure [psi]

10. An oil water capillary pressure experiment on a core sample gives the following results:

P_c,o/w [psi] 0 4.4 5.3 5.6 10.5 15.7 35.0

S_w[%] 100 100 90.1 82.4 43.7 32.2 29.8

Given that the sample was taken from a point 100 ft above the oil-water contact, what is the expected water saturation at that elevation? If the hydrocarbon bearing thickness from the crest (top) of the structure to the oil-water contact in 175 ft, what is the average water saturation over this interval? (ρw= 64 lbs/ f t³andρo= 45 lbs/ f t³)

11. If we assume an interfacial tension;σcosθ= 25 dyn/cm and a permeability and porosity respectively 100 mD and 18 %, in the exercise above, we may construct the capillary curve for a laboratory experiment using mercury as non-wetting phase. In the laboratory experiments one assume the lithology to be unchanged, but the permeability and porosity to be respectively 25 mD and 13 %. Find laboratory capillary curve when the interfacial tension to mercury is 370 dyne/cm.

Answers to questions:

1. ∆E = 0.075J, 3. h = 3 cm, 4. d = 0.5µm, 5. ∆p = 3.2 m bar,∆p = 29 bar, 7. h = 5.8 m, 9.

S_o= 0.2, S_g= 0.62 , S_w= 0.36, 11. S_w= 0.41.

Relative Permeability