Dake L.P. Fundamentals of reservoir engineering

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IMMISCIBLE DISPLACEMENT 361

Case No.

(exercise 10.1)

1 100 .28 .006 .520 1.40 37.50

2 10 .45 .051 .220 .91 3.75

3 .4 .80 .300 0 .15 0.15

TABLE 10.3(a)

Values of the shock front and end point relative permeabilities calculated using the data

of exercise 10.1

the results of exercise 10.1 can be analysed as follows:

a) Case 1 - this displacement is unstable due to the very high value of the oil/water

viscosity ratio. This results in the by-passing of oil and consequently the

premature breakthrough of water. The oil recovery at breakthrough is very small

and a great many pore volumes of water will have to be injected to recover all the

movable oil. Under these circumstances oil recovery by water injection is hardly

feasible and consideration should be given to the application of thermal recovery

methods with the aim of reducing the viscosity ratio.

b) Case 2 - the oil/water viscosity ratio is an order of magnitude lower than in case 1

which leads to a stable and much more favourable type of displacement (M

< 1).

This case will be analysed in greater detail in exercise 10.2, in which the oil

recovery after breakthrough is determined as a function of the cumulative water

injected and time.

c) Case 3 - for the displacement of this very low viscosity oil (

= .4 cp) both the

end point and shock front mobility ratios are less than unity and piston-like

displacement occurs. The tangent to the fractional flow curve, from

= S

, f

= 0, meets the curve at the point

bt bt

worw

S1S,f1=− = and therefore

wor

SS1S==−. The total oil recovery at breakthrough is

wc or wc

SS1SS==−−, which is the total movable oil volume.

EXERCISE 10.2 OIL RECOVERY PREDICTION FOR A WATERFLOOD

Water is being injected at a constant rate of 1000 b/d/well in a direct line drive in a

reservoir which has the following rock and fluid properties.

=0.18

=0.20

=5 cp

= 0.5 cp

IMMISCIBLE DISPLACEMENT 362

The relative permeabilities for oil and water are presented in table 10.1 and the flood

pattern geometry is as follows

Dip angle = 0°

Reservoir thickness = 40 ft

Distance between injection wells = 625 ft

Distance between injectors and producers = 2000 ft

Assuming that diffuse flow conditions prevail and that the injection project starts

simultaneously with oil production from the reservoir

1) determine the time when breakthrough occurs

2) determine the cumulative oil production as a function of both the cumulative

water injected and the time.

EXERCISE 10.2 SOLUTION

The relative permeabilities and viscosities of the oil and water are identical with those

of "Case 2" in exercise 10.1. Therefore, the fractional flow curve is the same as drawn

in fig. 10.16, for, which the breakthrough occurs when

bt bt

id pd

S0.45

f0.70

and W N 0.35

1) Calculation of the breakthrough time

For a constant rate of water injection the time is related to the dimensionless water

influx by the general expression

W (one pore volume)

(cu.ft)

q 5.615 365 (cu.ft / year)

××

W 625 40 2000 .18

t(years)

1000 5.615 365

××× ×

××

t4.39W(years)= (10.35)

Therefore breakthrough will occur after a time

= 4.39 × 0.35 = 1.54 years

2) Cumulative oil recovery

The oil recovery after breakthrough, expressed in pore volumes, can be calculated

using the equation

= (S

– S

) + (1 – f

) W

(10.32)

IMMISCIBLE DISPLACEMENT 363

where

= (10.27)

Allowing S

, the water saturation at the producing end of the block, to rise in

increments of 5% (for S

≥

S ) the corresponding values of W

are calculated in

table 10.4 using the data listed in table 10.2 for Case 2.

∆S

∆f

/∆S

∗

.45 (bt) .699

.05 .122 2.440 .475 .410

.50 .821

.05 .072 1.440 .525 .694

.55 .893

.05 .049 .980 .575 1.020

.60 .942

.05 .029 .580 .625 1.724

.65 .971

.05 .016 .320 .675 3.125

.70 .987

.05 .009 .180 .725 5.556

.75 .996

.05 .004 .080 .775 12.500

.80 1.000

TABLE 10.4

In this table, values of ∆f

/∆S

have been calculated rather than determined

graphically as suggested in the text. The values of

∗

in column 6 are the mid points

of each saturation increment, at which discrete values of W

have been calculated

using equ. (10.27). The oil recovery as a function of both W

and time can now be

determined using equ. (10.32), as listed in table 10.5.

∗

we wc

∗

−

∗

−

()

WPV

()

NPV

time (yrs)

equ (10.35)

.475 .275 .765 .235 .410 .371 1.80

.525 .325 .870 .130 .694 .415 3.05

.575 .375 .925 .075 1.020 .452 4.48

.625 .425 .962 .038 1.724 .491 7.57

.675 .475 .982 .018 3.125 .531 13.72

.725 .525 .993 .007 5.556 .564 24.39

TABLE 10.5

IMMISCIBLE DISPLACEMENT 364

The values of

∗

in column 3 of table 10.5 have been obtained from fig. 10.16

(Case 2), for the corresponding values of

∗

. The oil recovery, in reservoir pore

volumes, is plotted as a function of W

and time in fig. 10.17. The maximum possible

recovery is one movable oil volume, i.e. (1 − S

− S

) = 0.6 PV.

In the general case in which the displacement takes place at a fixed pressure, which is

above the bubble point pressure, then

W (PV)

time (yrs)

q = 1000 rb/d

N (PV)

Fig. 10.17 Dimensionless oil recovery (PV) as a function of dimensionless water

injected (PV), and time (exercise 10.2)

pd wc

oil production (rb)

N(1S)

one pore volume (rb) N B

==−

and the conventional expression for the oil recovery is

ppd

owc

(stb.oil)

NB(1S)STOIIP(stb)

−

In exercise 10.2, B

= B

, since displacement occurs at the initial reservoir pressure,

/ N = N

/ (1−S

)

10.6 DISPLACEMENT UNDER SEGREGATED FLOW CONDITIONS

In the previous two sections a one dimensional displacement theory was presented

which relied on the assumption of diffuse flow. Now, precisely the opposite will be

assumed, namely, that displacement occurs under the segregated flow condition

shown in fig.10.18.

In the flooded part of the reservoir water alone is flowing, in the presence of residual

oil, with effective permeability k

′

, where

′

is the end point relative permeability

IMMISCIBLE DISPLACEMENT 365

to water. Similarly, in the unflooded zone oil is flowing in the presence of connate water

with effective permeability k

′

, where

′

is the end point relative permeability to

oil. Furthermore, at any point on the interface between the fluids the pressures in the

oil and water are assumed to be equal. This means that there is a distinct interface with

no capillary transition zone. Segregated flow also assumes that the displacement is

governed by vertical equilibrium, as discussed in

WATER

S = 1 - S

S= S

wor

o or

OIL

S = S

S= 1 - S

wwc

o wc

Fig. 10.18 Displacement of oil by water under segregated flow conditions

sec. 10.2. In this case, since there is no capillary transition zone, gravity forces alone

are responsible for the instantaneous distribution of the fluids in the dip-normal

direction

Dietz has investigated this type of displacement

and, in particular, the conditions

under which it can be regarded as stable; the difference between stable and unstable

displacement in a dipping reservoir being illustrated in fig.10.19.

The condition for stable displacement is that the angle between the fluids' interface and

the direction of flow should remain constant throughout the displacement, fig.10.19(a),

(b), such that

tan

=− =constant

This is only satisfied at relatively low injection rates when the gravity force, arising from

difference in density between the fluids, will act to try and maintain the interface

horizontal and, in the limiting case in which the rate is reduced to zero, a horizontal

interface would result. At high injection rates the viscous forces, driving the fluids

through the reservoir, will prevail over the component of the gravity force, acting in the

downdip direction, resulting in the unstable displacement shown in fig. 10.19(c). Due to

the density difference the water will underrun the oil in the form of a water tongue,

leading to premature water breakthrough. Unstable displacement will occur for the

limiting condition that

tan 0

=− =

IMMISCIBLE DISPLACEMENT 366

WATER

OIL

(a)

(b)

(c)

Fig. 10.19 Illustrating the difference between stable and unstable displacement, under

segregated flow conditions, in a dipping reservoir; (a) stable: G > M

−

−−

−1; M > 1;

ββ

θθ

. (b) stable: G > M−

−−

−1; M < 1;

ββ

θθ

. (c) unstable: G < M−

−−

−1.

If the incompressible displacement is stable then, at all points on the interface, the oil

and water must have the same velocity and, applying Darcy's law at any point on the

interface for displacement in the x direction

ro o o

kk p g sin

1.0133 10

′

∂

æö

==− +

ç÷

∂

èø

and

rw w w

kk p g sin

1.0133 10

′

∂

æö

==− +

ç÷

∂

èø

where u

, u

and u

are the oil, water and total flow velocities, respectively. These

equations can be combined to give

tow

ro rw

gsin

u(pp)

kk kk x 1.0133 10

µµ

æö

∂∆

−=−−+

ç÷

′′

∂×

èø

(10.36)

IMMISCIBLE DISPLACEMENT 367

where ∆

−

. Also, applying the capillary pressure equation, (10.5),

cow

gcos

dP d(p p ) dy

1.0133 10

ρθ

∆

=−=

and for stable displacement (dy/dx—negative)

g cos dy

xdx1.0133 10

ρθ

∂

∆

=−

∂×

which, when substituted in equ. (10.36) gives

ro rw

gdy

ucossin

kk kk dx1.0133 10

µµ

θθ

æö

∆

æö

−= +

ç÷

′′

èø

This equation can be alternatively expressed in terms of the total flow rate, q

, as

rorw rw

wo wt

kkkkAgsin

dy 1

dx tan1.0133 10 q

ρθ

µµ θµ

′æö

′′

∆

æö

−= +

ç÷

èø

dy 1

M1 G 1

dx tan

æö

−= +

ç÷

èø

(10.37)

where M is the end point mobility ratio and G the dimensionless gravity number, which

in Darcy units is

kk A g sin

1.0133 10 q

′

∆

(10.38)

and in field units can be deduced from equ. (10.10) to be

kk A sin

G4.910

−

′

∆

=×

(10.39)

Equation (10.37) can be solved to give the slope of the interface for stable flow as

dy M 1 G

tan tan

dx G

−−

æö

=− =

ç÷

èø

(10.40)

In this equation M is a constant and, when displacing oil by water at a fixed rate in the

updip direction, G is a positive constant. Therefore, the inclination of the interface dy/dx

assumes a fixed value. For stable displacement, as already mentioned, dy/dx must be

a negative constant and this imposes the condition for stability that

G > M−1

The limiting case is when dy/dx = 0 then, as shown in fig. 10.19(c), the water will

underrun the oil in the form of a water tongue. This will occur when

IMMISCIBLE DISPLACEMENT 368

G = M−1

which, using equ. (10.38), can be solved to determine the so-called critical rate for by-

passing as

crit

kk A g sin

qr.cc/sec

1.0133 10 (M 1)

ρθ

′

∆

×−

(10.41)

or in field units

crit

4.9 10 kk A sin

qrb/d

(M 1)

γθ

−

′

×∆

−

(10.42)

Provided the injection rate is maintained below q

crit

gravity forces will stabilize the

displacement.

The magnitude of the mobility ratio also influences the displacement. This can be

appreciated from equ. (10.40), as detailed below.

M > 1

This is the most common physical condition. The displacement is stable if G > M−1, in

which case

, fig. 10.19(a), and unstable if G < M−1.

M = 1

This is a very favourable mobility ratio for which there is no tendency for the water to

by-pass the oil (refer Chapter 4, sec. 9). For M = 1 the displacement is unconditionally

stable. Furthermore,

, and the interface rises horizontally in the reservoir.

M < 1

This mobility ratio also leads to unconditionally stable displacement but in this case,

, fig. 10.19(b).

If the displacement is stable the oil recovery, as a function of cumulative water injected

and time, can be calculated from simple geometrical considerations, as will be

demonstrated in exercise 10.3. An alternative approach is to attempt to reduce the

description of segregated displacement to one dimension and then calculate the oil

recovery using the Buckley-Leverett displacement theory. It is worthwhile pursuing this

idea because it is quite general and can be applied whether the displacement is stable

or not. Consider then the general segregated displacement in a linear, homogeneous

reservoir as shown in fig.10.20.

IMMISCIBLE DISPLACEMENT 369

WATER

S = 1 - S

S= S

wor

o or

OIL

S = S

S= 1 - S

wwc

o wc

1 - b

(x, y)

Fig. 10.20 Segregated displacement of oil by water

Unlike the oil displacement under the assumed diffuse flow condition, described in the

previous section, which could be accounted for using one dimensional mathematics;

the segregated flow depicted in fig. 10.20 is very definitely a two dimensional problem.

In attempting to reduce the mathematical description to one dimension it is necessary

to average the saturations and saturation dependent relative permeabilities over the

reservoir thickness. Flow can then be described as occurring along the centre line of

the reservoir.

At any point x in the linear displacement path, let b be the fractional thickness of the

water, fig. 10.20, thus b = y/h. The thickness averaged water saturation at x is then

S

= b(1−S

) + (1−b)S

which can be solved for b to give

or wc

1S S

−

−−

(10.43)

and, since S

and S

are constants, equ. (10.43) indicates that b is directly

proportional to the average saturation. The thickness averaged relative permeability to

water can be similarly derived as

k

()

S = b k

= 1 − S

) + (1 − b) k

= S

)

and since k

= S

) is zero and k

= 1 − S

) = k´

this can be reduced to

k

()

S = b

′

where

′

is the end point relative permeability to water. For the oil, the same argument

gives the thickness averaged relative permeability as

k

()

S = b k

= 1 − S

) + (1 − b) k

= S

)

IMMISCIBLE DISPLACEMENT 370

k

()

S = b

′

where

′

is the end point relative permeability to oil.

Substituting for b in these expressions, using equ. (10.43), gives

or wc

k(S) k

1S S

æö

−

′

ç÷

−−

èø

(10.44)

and

or wc

1S S

k(S ) k

1S S

æö

−−

′

ç÷

−−

èø

(10.45)

These equations indicate that the thickness averaged relative permeabilities, for

segregated flow, are simply linear functions of the thickness averaged water saturation,

as shown by the solid lines in fig. 10.21.

ro w

()

equ. (10.45)

rw w

()

equ. (10.44)

1 - S

Fig. 10.21 Linear, averaged relative permeability functions for describing segregated

flow in a homogeneous reservoir

Also shown in fig. 10.21, as dashed lines, are the rock relative permeability curves

obtained by measuring relative permeabilities in the laboratory using thin core plugs.

They are measured under conditions which correspond to diffuse flow and represent

point relative permeabilities in the reservoir. As mentioned already, they can only be

used directly in displacement calculations if the water saturation is the same at all

points throughout the thickness. In this unique case point relative permeabilities are

equal to thickness averaged relative permeabilities. In contrast, the linear functions

shown in fig. 10.21, result from the thickness averaging process required to facilitate