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Chapter III Incompressible Two-Phase Reservoirs.
89
CHAPTER
INCOMPRESSIBLE
I -
111
TWO-PHASE
RESERVOIRS
INTRODUCTION
We shall consider in this chapter the flow of two incompressible immiscible fluids through a porous medium. Though this problem is of great practical importance, as it corresponds to the simplest case of secondary oil recovery technique, where the resident oil is displaced by injected water,
it
has received only recently attention from the mathematical
community, despite the huge amount of oil engineering technical literature on the subject. The reason for this may be that even this "simple" (for oil engineers) model has such a complicated and non-standard structure that the usual mathematical tools cannot be applied in an evident manner. The key to the reduction of these two-phase equations to the more familiar system elliptic
of one parabolic saturation equation coupled with an
pressure
equation
is
a
mathematical
transformation of
the
equations, which replaces the two pressure unknowns (one per phase) by only one
pressure
unknown,
called
the
global, or
the
reduced,
or
the
intermediate pressure. This transformation was discovered independently by CHAVENT [ l ] , authors
in 1975 (q9globalpressure") and by two several some Russian
("reduced
ANTONCEV-MONAHOV).
pressure"
see
the
references
1
though
3
of
A detailed explanation is given in section I1 below,
together with a careful discussion of the choice of boundary conditions. The resulting system of equations is summarized in section 111.
As we already mentioned it, rather few mathematical results have been available for this problem. In order to organize logically these results,
we
emphasize
first
the
importance of
the
notion of water
breakthrough time, which corresponds to the time at which the injected water first starts being produced with the oil at a given production well.
Ch.III: Incompressible Two-Phase Reservoirs
90
This time is economically important, as the water oil ratio (WOR) increases very quickly after the water breakthrough time, so that the production well has to be turned off. Depending on the boundary conditions which are used for the model, the breakthrough phenomenon may or may not be well represented. One other mathematical
difficulty
associated
with
this
problem
is
that, under
standard conditions, the parabolic saturation equation is degenerate, (and practically
very
close
to
a
first order hyperbolic equation).
This
degeneracy may or may not be taken into account. Among the papers which do not take properly into account the breakthrough phenomenon, we find CHAVENT [l],
1975 (existence theorem for a
degenerate parabolic equation with simple Dirichlet and Neumann conditions, coupled with a family of elliptic pressure equations), KRUZKOV-SUKORJANSKI,
1977 (existence of classical solutions for the non degenerate problem with Dirichlet and Neumann boundary conditions), ANTONCEV-MONAHOV, 1978 (existence of weak solutions for the degenerate problem with Dirichlet and Neumann boundary conditions, plus some regularity and stability results for simplified problems), ALT-DIBENEDETTO, 1983 (existence of a weak solution for
the
degenerate
problem
with
two
unilateral
overflow
boundary
conditions, these conditions do not allow simultaneous production of oil and water) KROENER-LUCKHAUS, 1984 (existence of solutions for the partially degenerate problem with Dirichlet and Neumann boundary conditions, this author works with the original set of equations, not the transformed one). Concerning the models that take properly into account the water breakthrough phenomenon, an adequate unilateral boundary condition was formulated in CHAVENT [lbis], without an existence theorem (the main part of the paper was devoted to mathematical problems related to the estimation of the non-linearities appearing in the saturation equation). Existence theorems for the resulting degenerate variational inequality were given in CHAVENT [ Z ] for one-dimensional problems (where the pressure and saturation equations decouple), and will be given in section V of this chapter for the general multidimensional case. Regularity results, a description of the asymptotical behaviour, a precise definition and some properties of the water breakthrough time can be found in the work of GAGNEUX [l]-[4]
in the
case where the saturation and pressure equations decouple; part of these results are recalled in this chapter (still in section V).
So we b e l i e v e t h a t t h e material i n s e c t i o n V below is t o d a y ' s most
comprehensive
mathematical
treatment
of
the
two-phase
equations,
a s it
takes i n t o a c c o u n t t h e l a r g e s t number o f r e l e v a n t p h y s i c a l p r o p e r t i e s .
The reader may h a v e n o t i c e d t h a t none of t h e p a p e r s was c i t e d f o r a u n i q u e n e s s theorem. T h i s is b e c a u s e o f t h e c o u p l i n g between t h e p r e s s u r e and
saturation
r e g u l a r i t y of still have
equations,
which
the solution
t o get
such
a
theorem,
it
difficult
t h e uniqueness.
to
obtain
enough
T h i s problem is open
case, t h o u g h KRUZKOV-SUKORJANSKI claim t h e y
t h e non-degenerate
for
makes
because
they
suppose
i n the h y p o t h e s i s
that
the
s o l u t i o n is i n d e e d r e g u l a r .
Tn o r d e r t o be d e f i n i t e , we make p r e c i s e now t h e way i n which t h e reservoir described
i n c h a p t e r I w i l l b e p r o d u c s d t h r o u g h o u t Lhis c h a p t e r
(cf. Figure 1 ) : ( 1 .l )
r II
The l a t e r a l boundary
is s u p p o s e d c l o s e d ;
-
water is i n j e c t e d t h r o u g h t h e wells 1 , 2 , . . . k ,
-
(1.2)
and we d e n o t e by
k
:
=
U k= 1
rk
t h e i n j e c t i o f ? boundary.
...K, and
o i l is prodilned til-ouyh t h e r e m a i n i n g wells k+l , K
(1.3)
we d e n o t e by
rs
=
U
-
T k : t h e p r o d u c t i o n boundary.
k= k+ 1
Figure 1 : Secondary recovery of an o i l reservoir
92
Ch. III: Incompressible nYo-Phase Reservoirs
For the sake of simplicity, we will consider in the sequel the case of one
injection well and one production well
11-
CONSTRUCTION
We
first
give
OF
the
(z=l,K = Z ) .
THE
STATE
characteristics
EQUATIONS
of
the
two
fluids
(see
paragraphs 111.3 and IV.3 of chapter I) They (2.1 1
are
both
supposed
11,
chapter
where
incompressible (in opposition to
compressibility
phenomenon). Hence B = B =1 1 2-
was
.
the
driving
They are immiscible, so two distinct phases are present in the pores of the porous medium. Hence we have pressures
(2.2)
P1
and
P2
corresponding
two
respectively
distinct to
the
pressure in the wetting phase (water) and tp the pressure in the non-wetting phase (oil).
pj
> 0 is the mass of a unit volume of the
jth fluid,
j=l ,2,
(2.3)
>
!+J
0
is the viscosity of the jth fluid, j=1,2.
Let us define at each point x
6 Ci
:
@ . = flow vector of the jth fluid, j=1,2
(cf. ( 3 . 1 2 ) in chapter I for a precise definition).
s(x)
=
saturation in fluid 1 at x
=
vol of fluid 1 around x, vol of fluid 1 + 2
(2.5) 1 - s(x)
=
saturation in fluid 2 at x.
So we have now three dependant variables
:
93
I1 Construction of the State Equations
(2.6)
11.1
-
pressure of the wetting phase (water),
P1
=
P2
=
pressure of the non wetting phase (oil),
?.
=
wetting phase (water) saturation.
THE EQUATIONS INSIDE
Inside 52,
Sl :
THE NOTION OF GLOBAL PRESSURE
the Darcy law applies separately to each of the two
fluids (cf. Chapter I 3111.3.1) with a reduction
k . rJ
where the relative permeabilities
k , of permeability : rJ
depend on
S as shown in figure 7
of chapter I, and also generally on x .
-
A s can be seen in figure 7 of chapter I, the saturation S
remains always,
as long as only displacement phenomena are considered, which is the case in this chapter, between
5,
and
:
is the water residual saturation : for S 6 Sm' there is m' so little water that it is "trapped" by the capillary forces in the pores
.
of the porous medias and can no longer be displaced.
-
similarly to
1-s,
oil
is the
residual saturation and
is
interpreted
'ma
A s the rock and the fluids are supposed incompressible, the only
accumulation term is due to a change in saturation so the conservation laws
for each of the two fluids are (compare with (1.46) of chapter I)
(2.9)
O(X)
@(x)
(2.10)
a ( x ) $(x)
+
xa
i,
div = 0, V xcn, (water conservation law)
-
-t
div $2 = 0, fF x (oil conservation law)
(13)
+
:
V t c 10,TC,
c 52,
V
t E l0,TC.
Ch.III: Incompressible Two-PhaseReservoirs
94
At this point we have, in n, only two equations (2.9), the three unknown
P 1 P2
s.
and
(2.10) for
The missing equation is given by the
capillary pressure law (see (3.15) of chapter I) (2.11 1 Pl - P2 = PC(S,X), depends on
where P
-
as shown i n figure 8 of chapter I.
S
Two things are to be noted concerning these capillary pressure curves :
- They always have a positive derivative
:
(2.12)
-
S [,
- The capillary pressure vanishes for only one value
-
interval :
S ],
- P 1 - P = P =o<=> s = s . 2 c Usually Sc = SM (resp. Sc = Sm) when
(2.13) (resp.
-
Sc of the
-
non-wetting
< 5e < m
S
phase)
saturation.
However
-
is the wetting phase
S
it
can
happen
that
the reduced
water
zM.so we shall distinguish in the sequel between -Sc and -sM.
In order to simplify notation, we shall now use saturation S instead of the actual saturation S :
so that equations (2.7), (2.9),
as at
div
i,
...,(2.11)
(2.15)
@(x)
(2.16)
Q(x) ~ ( 1 - S +) div $2
(2.17)
P1
-
P*
(2.18)
$.J
=
-@(x) k.(S,x) grad [ P . - p .
+
a
=
:
0, +
=
become
=
0,
PC(S,X),
J
J
J
g 21,
j
=
1,2,
I1 Construction of the State Equations
95
where we have defined (caution than in chapter 11)
:
Q and J, represent different quantities
:
kr.(Sm+S(zM-zm),x) (2.21)
k.(S,x) J
=
=
u.
mobility of the
jthfluidj=1,2,
J
P (s,x)
(2.22)
=
P~(S,+S(Z~-S~), x)
=
capillary pressure.
In all the sequel we shall make the following assumptions on the spatial dependance of
k. and
P
J
:
The mobilities, as functions of the reduced saturation, are independant of x
(2.23)
and
k.(S,x)
=
:
k.(S),
+XER,
J
:
the
(2.24)
capillary
P (S,X) where (2.25)
(2.26)
pressure,
as
saturation, is independent of
I
I
P
=
a x
fuhction up
of
the
reduced
to a scaling factor
:
PCM(X) PC(S),
(x) 2 0 is the maximum of the absolute value of the
capillary pressure at the point x, CM pc(S) is a dimensionless function such that -1 2 pc(S) 51 and pc(Sc) = 0. With the choice (2.25) we see in figure 8 of is always an increasing function of S . chapter I that p
The hypotheses
(2.23),
(2.24)
are
very
usual, as
they
are
sufficient to provide a good model of the important phenomena f o r the oil engineers
(including
heterogeneous media).
for
example
counter-flows
by
imbibition
in
Ch. III: Incompressible nYo-Phase Reservoirs
96
However, equations (2.151, (2.16) are not suited for a proper mathematical study
:
for example
disappears as
k (1)
2
where S EO.
in
the zones of
where S = 1 ,
equation (2.16)
0; similarly equation (2.15) disappears in the zones
=
So we are now going to transforme equations (2.15) through (2.18), under hypotheses (2.23), (2.241, into a more tractable form. We shall carry out this transformation in 3 steps
Step 1 (2.27)
:
Sum equation (2.15), (2.16) to get
:
:
+ + div ($1+$2) = 0.
Use then the algebraic identity
and (2.15) to get
(2.29)
Q
:
as
- div { $ +
div {
k t k2 grad kl+k2
-
+
1: (P - P gz 1- ( P 2 - ~ 2 g z) 1 )
4
+
Using (2.17) in (2.29) and adding (2.28) to (2.31) gives
:
(2.30)
In order to simplify notation we define
:
S
(2.32)
a(S)
(2.33)
bO(S)
=
\
b
a(s) ds kl (S)
= kl
(S)+k2(S)
increasing (viscosity-’) “fractional flow1’,increasing (dimensionless)
I1 Construction of the State Equations
97
(2.34)
(2.35)
Pc (x)
(2.36)
r
-p
m g Z(x)
=
gravity potential (same dimension as a pressure).
We rewrite now equations (2.301,
(2.37)
I
=
(2.38)
@
as +
at
[PCM(x) grad a(S)+bl(S) grad PCM(X)
b2 ( S ) grad PG(x)I]
+
div ($1
[$(XI
div
+
+
G2)
=
+
+
+
div {bO(S)(01+02)l
We see that in equations ( 2 . 3 7 ) , S
= 0,
(2.38)
the only dependant + + $1+$2,
and the water + oil flow Vector
which itself depends on S, P1 and
P2
through the definition ( 2 . 1 8 ) .
The notion of global pressure (CHAVENT
:
+
0.
variables are the saturation
Step 2
-
(2.27) with the above notation
[11).
We show now that it is possible to introduce a new unknown called the global pressure, which is a point function of
so that the water+oil flow
+
+
$1+$2
S,
P1
P
and P 2 ,
can be expressed in terms of S, P and grad
-P only. Thus the number of unknowns reduces to two. We define first some notation
-
corresponding to Sc (2.391
is the reduced saturation
defined in ( 2 . 1 3 1 , so pc(Sc) S
Y(S) =
(Sc
J
1 dPc (bo(S)- 7 ) d~ ( S ) d s
=
0) :
(dimensionless),
sC
(dimensionless). One has obviously (2.41)
:
‘f(S)+Y1(S) = (bo(S)
- 71 ) pc(S).
Ch.III: Incompressible nYo-Phase Reservoirs
98
A o(=/oSa (T) dT
1 Figure 2
Figure 3
:
:
S b
: 0
1
The functions a(S) and a(S)
The shapes of the functions ,,b
b,
and
b2
s
99
I1 Constructionof the State Equations
pressure P by : P(x,t)
(2.42)
=
7 1 CP,(x,t)
+
P2(x,t)l
One checks then easily, as
pc(Sc)
+
=
Y(S(X,t))
0 and
PCM(X).
O
so
(2.44)
Let us now calculate the gradient of (2.45)
gradP
=
1
P :
grad(P1+P2) + ( b o ( S ) 2
dpc gradS + Y ( S ) grad PCM. 71 ) PCM dS
But, (2.46)
PCM
dPC dS
grad S
=
PCM grad pc(S)
=
grad [PcM P~(S)I-P, grad PCM.
Using (2.46), (2.45) and using (2.24), (2.27), (2.41) we get 1
:
1
grad P = - grad (P +P ) + ( b ( S ) - -1 grad (P1-P2)-Y,(S)grad PCM. 2 1 2 0 2 which using the definition (2.33) of (2.47)
( k l + k 2 ) grad P
Multiplying (2.47) by (2.48)
Go
=
=
bo
reduces to :
k l grad P 1 + k2 grad P2 - ( k l + k 2 ) Y1
(s) grad
PG.
I$, we obtain
il+i2-I$(x) =
d(S) [grad P+Y1(S) grad PCM+Y2(S)grad Pc}
where we have introduced the following functions
:
100
Ch. III: Incompressible nYo4hase Reservoirs
(2.59)
I”=
k,Pltk2 P2
.-1 Pm
k1+k2
I
I
I I
I
I
Figure 4 Remark 1 :
We
obtain
global
:
from
-2
equation in
c a p i l d r y pressure (2.48) redlJCRS t o
the
b
1
The d and Y
gresnure,
S
I
0 functions
(2.48)
the
special
i n t e r p r e t a t i o n of
case
where
PCM(x) is c o n s t a n t o v e r
the
the
maximum
0 ; i n t h a t case
:
(2.51 1 which,
compared
to
the
one-phase
Darcy
Law
(3.3)
of
c h a p t e ? T, shows t h a t t h e g l o b a l p r e s s u r e f i e l d , a p p l i e d t o a f i c t i t i o u s f l u i d of v i s c o s i t y !J s u c h t h a t krl kr 2 k l P l + k 2 P 2 , yields - = d = -+ -and of d e n s i t y p = k l + k2 !J 9 !J2 e q u a l t o t h e sun of t h e o i l a n d water flows. flow
Step 3 : Let u s d e f i n e t h e f o l l o w i n g vect7r’ f i e l d s (2.52) (2.53)
G1
=
-+(x) g r a d PCM
=
- + ( X I g r a d PG
:
(flow vector
x
m o b i l i t y -1 1,
(flow vector
x mobility-’
),
a
101
II. Construction of the State Equations
which are known (as
@,
PCM
and PG are given).
Then equations (2.37).
as at
(2.54)
Q
(2.55)
r' =
(2.56)
div qo
(2.57)
Go
+ div {r
+
+
2
1
j=o
=
=
(2.48) can be r e w r i t t e L
GJ. }
b.(S) J
-$(x) PCM(x) grad +
=
0,
a(S),
0,
-$(x) d(S) gradP
+
d(S)
2
1
Gj.
Yj(S)
j=l
which are the sought equations in The
(2.38),
saturation
0
equation
for S and P. (2.541,
(2.55)
is
a
non-linear
diffusion-convection evolution equation. The diffusion term (second term of (2.54) is degenerate (as a' = a vanishes for S=O and S=l as one can see in figure 2) or even missing when the capillary pressure is neglected. The transport term (last term of (2.54) is non linear and not necessarily monotone (the fractional flow bo is monotone increasing, but bl
and
b2
are not as shown in figure 3). The pressure equation (2.56),
(2.57) is a family (one for every
t c lO,T[) of elliptic equations.
The
saturation
equation
(2.54)
and
the
+
pressure
equation
(2.56),(2.57) are coupled through the presence of qo in the 1 in the right hand side of (2.54), and the presence of S in the coefficients of (2.57). Remark 2:
+
One tempting thing to do is to replace in (2.54) qo by its expression (2.57) and to rearrange the terms. This manipulation has to be avoided, in view of the forthcoming numerical resolution of (2.54) through (2.57)
:
in order to
be able to solve properly the pressure equation (2.56) we shall use a mixed finite element method (see chapter V), + which computes an approximation of the global flow qo. Oh
6
102
Ch. In: Incompressible nvo-t'hase Reservoirs
This
approximation
+
qOh
will
be
simply
plugged
in
the
saturation equation (2.54). Hence we shall keep the above form of the equations. Moreover, notice that in the 1-D case,
6o
is
constant
in space
(cf.
(2.56)) and the proposed
manipulation has obviously no interest in that case. 0
+
+
The water and oil flow vectors are then related to S, r, q j , j
=
0,1,2 by +
(2.58)
0,
+
+
+
02 = q0'
(2.59)
and the water and oil pressures (2.62)
P1
=
P - [YG) -
(2.63)
P2
=
P
11.2
-
-
[Y(S)
+
P,
and
P2
are then given by
:
1
7 Pc(S)1 PCM, 1
7 PC(S)I
PCM.
THE PRESSURE BOUNDARY CONDITIONS
III (see figure 1 ) we
As we indicated at the beginning of chapter
have drawn in figure 5 the partition of the boundary into three parts re,
r
of the porous medium
rll and r corresponding respectively to injection well
boundary, lateral boundary and production well boundary.
II. Consmction of the State Equations
103
injection we1I
Figure 5 : The pressure boundary conditions.
As
the global
i~r*~?,si~r’eP
i r governed by a family of elliptic
equations, and is related to the global (water+oil) flow, the corresponding boundary
conditions are
monophasic case
similar to thora [used in chapter I1 for the
:
- Bzundary conditions on the lateral boundary The field being supposed closed on vanishes on
:
the (oilcwater) flow
i.e. using (2.58)
+ + q o - v = O
(2.64)
Remark 3
rR
r,,
TL:
on
rR
(Neumann Condition for ( 2 . 5 6 ) , ( 2 . 5 7 ) ) .
Water drive with given pressure or flow conditions can be taken into account by straightforward modification of
the
boundary conditions.
- Boundary boundaries As
a
conditions
on _the injection and
production well
r e -and rs: for the monophasic case, we shall not distinguish, for the
pressure boundary conditions, bet,.ieerl i rijection and production wells
:
we
Ch. III: Incompressible no-Phase Reservoirs
104
rw
s h a l l d e n o t e by
rw and
re
either
rs,
or
by
rw
s E
t h e c u r r e n t p o i n t on
Iw = rwx]O,T[.
oil
The
injection
and
field
is
production
coupled, wells,
through t h e
for
which
we
boundaries, shall
t o the
suppose
a
that
mathematical model is a v a i l a b l e . Such a well model e n a b l e s u s t o c a l c u l a t e , a t every t i m e
t,
the trace
up t o a c o n s t a n t ,
on
Pe
Tw
of t h e p r e s s u r e i n s i d e t h e well,
knowing o n l y t h e f l o w - r a t e h i s t o r y of t h e two f l u i d s
t h r o u g h every p o i n t of t h e well boundary
rw
:
unknown c o n s t a n t , V ( s , t )
6
Z
W.
We g i v e f i r s t some s i m p l e examples o f such models, which u s e t h e h y d r o s t a t i c p r e s s u r e law i n s i d e t h e well b o r e h o l e
:
- water injection w e l l : t h e d e n s i t y of t h e f l u i d i n t h e well is t h a t of w a t e r ; hence t h e h y d r o s t a t i c p r e s s u r e law y i e l d s P ( s ) = p 1 g Z(x)
(2.66)
-
:
unknown c o n s t a n t .
+
production w e l l : h e r e t h e d e n s i t y o f t h e f l u i d i n s i d e t h e well
depends on t h e l o c a l p r o p o r t i o n of the two f l u i d s . We t a k e f o r d e n s i t y i n t h e well b o r e h o l e t h e water
and o i l d e n s i t i e s weighted by t h e i r
local
p r o d u c t i o n r a t e t h r o u g h t h e boundary : +
+
+
P1S1 ' k + P 2 O 2 ' V (2.67)
P(X,t)
@,
=
+
+
+
.V+O2.V
and
r',
Usually
+ qo
+
+ + + + Pl$l ' v + P 2 $ 2 " J
+
-
+ + q0-v +
is very l a r g e n e a r t h e well i n comparison t o ql,
s o t h a t (2.67)
is g e n e r a l l y approximated by ( c f .
(2.601,
+ q2
(2.61)
and (2.33)) : (2.68)
p(s,t)
k l ( S (3, =
t ) )p l + k 2 ( S ( s , t )
p2
k l ( S ( s , t ) ) + k 2 ( S ( 3 , t) )
Note t h a t t h i s d e n s i t y is t h e d e n s i t y , on t h e well boundary f i c t i t i o u s e q u i v a l e n t f l u i d d e f i n e d i n remark 1 .
rw,
Of
the
II. Constructionof the State Equations
L e t , so e Tw
105
be any r e f e r e n c e p o i n t on t h e well boundary ( a t t h e bottom
f o r example f o r
t h e v e r t i c a l models).
One can t a k e as p r o d u c t i o n well
model :
-
( s is t h e n t h e c u r v i l i n e a r a b s c i s s a on
f o r 2D models S
(2.69)
Pe(s,t)
-
1
=
Tw)
p ( s , t ) g d Z ( s ) + unknown c o n s t a n t ,
f o r 3D models
(2.69bis)
Z(S)
Pe(s,t)
1
=
p ( Z , t ) g dZ
+
unknown c o n s t a n t ,
Z(So)
where
t
(2.70)
p(Z,t)
rz
=
=
1 measr
rz
p ( s , t ) ds,
{ s ~ r ~ I z (= sz)] .
I n summary, we see
t h a t , once t h e c o n s t a n t is f i x e d i n ( 2 . 6 6 ) ,
(2.69) o r ( 2 . 6 9 b i s ) , t h e well model e n a b l e s u s t o c a l c u l a t e t h e p r e s s u r e Pe everywhere on rw assuming t h a t t h e d e n s i t y p i n t h e w e l l is known. T h i s is a c t u a l l y t h e case
for
i n j e c t i o n wells,
where p
is t h e i n j e c t e d f l u i d
d e n s i t y , and f o r 2D-horizontal models, where Z ( s ) is c o n s t a n t and hence Pe t o o , and t h i s is a l s o a p p r o x i m a t e l y t h e case f o r p r o d u c t i o n wells, where p is commonly e v a l u a t e d u s i n g (2.68) w i t h t h e s a t u r a t i o n S e v a l u a t e d a t t h e p r e v i o u s time s t e p . Hence we w i l l assume from now on t h a t t h e p r e s s u r e P up t o a c o n s t a n t , on a w e l l boundary
is known,
r
.
We g i v e now t h e boundary c o n d i t i o n s which c a n be used f o r
Ew
e-
Tw:
i f we know t h e bottom p r e s s u r e
i n t h e w e l l , we know
u s i n g t h e well model
The c o n t i n u i t y of t h e p r e s s u r e i n one
(2.65).
P
everywhere on
i n j e c t e d o r produced f l u i d (water f o r example) a c r o s s t h e well boundary shows t h a t most
the
P,
Irw
=
capillary
Pe.
But t h e g l o b a l p r e s s u r e P d i f f e r s from
pressure
(cf.
(2.44)),
which
is u s u a l l y
P,
by a t
s m a l l when
compared w i t h t h e p r e s s u r e drop a c r o s s t h e f i e l d . Hence we s h a l l n e g l e c t t h i s d i f f e r e n c e and t a k e as boundary c o n d i t i o n
:
106
Ch.III: Incompressible TwoPhase Reservoirs
P Remark 4 :
=
on
P
Tw
, a.e.
on lO,T[
(Dirichlet).
If one does not want to make this approximation, then the above condition has to be replaced, to ensure continuity of the water pressure, by (cf. (2.62)) : p
=
1
pe
+
[Y(S) - 7
PC(S)I PCM. 0
-
if we know at every
(s,t)
E
Zw the (water+oil)
flow rate density
Q(s,t) then we take as boundary condition : + + qo-v
=
, a.e
Q on rw
on
]O,T[
(Neumann).
(This condition can be used practically only for 1-D models, or 2-D models with an hypothesis on the repartition of the well production rate on - if we know only the overall injection or production rate
well, the
boundary condition is + + qo-v
QT(t)
=
'
QT(t) of the
:
a.e. on
lO,T[,
rW
P
rw).
(well-type condition) =
Pe
+
unknown constant,
rw
is given by the well model (2.65).
where P
Finally we have only three types of global pressure boundary conditions boundary rD,
rN
(2.71)
r
of Q (as in the monophasic case) into three corresponding parts
and
i
Dirichlet, Neumann and well-type ; so we shall partition the
:
rw
:
rD
=
the part of
r
rN
=
the part of
r where Neumann conditions hold for P,
rw
=
the part of
r
where Dirichlet conditions hold for P,
where well-type conditions hold for P.
107
I1 Construction of the State Equations The global pressure boundary conditions are then :
(2.72)
where
P
is the given exterior pressure
Q(s,t) (2.73)
=
:
TN
lO,TC,
Z"
overall (water+oil) production rate through at time t
Remark 5
r D X
(water+oil) production rate density at the point
=
(s,t) Q (t) T
on
E
rW
l0,TC.
includes generally the lateral boundary, where Q=O; Q is
positive when something is produced through the boundary, and negative when something is injected through the boundam. Remark 6 :
As
the global pressure equation is stationnary, the global
pressure is well defined only if (2.74)
the r-measure of
rD
is non zero,
which we shall generally assume in the following. If (2.74) is not satisfied, then one has to assume that Q(s) d s
(2.75)
+
QT
=
0
rN
in order to be compatible with the incompressibility of the two fluids; then the pressure
P
is only defined up to a
constant. 0
Ch. III: Incompressible nvuo-phaseReservoirs
108
11.3 - THE SATURATION BOUNDARY CONDITIONS The p o s s i b l e c h o i c e s f o r t h e s a t u r a t i o n boundary c o n d i t i o n a t a point s of
r
are d i f f e r e n t d e p e n d i n g o n w h e t h e r t h e g l o b a l ( o i l c w a t e r ) f l o w
is d i r e c t e d i n w a r d or o u t w a r d i.e.
+ + whether q O * J is n e g a t i v e
u s i n g (3.58),
or p o s i t i v e . So we i n t r , , i I i - ~ e r i p s t ( c f . f i g . 7 ) :
(2.76)
r-
=
{s e
r
I
r+
=
{s
r
I
E
+
+
+
+
qo-'j 2 O ] = i n j e c t i o n boundary,
qo-v
>
01 = p r o d u c t i o n bolundary.
V
R
F i g u r e 6 : The p a r t i t i o n of Remark 7 :
r
into
r-
For a f i e l d w i t h c l o s e d o u t e r b o u n d a r y , made up of t h e o u t e r boundary boundary well
re
and
boundary
imbibiLion
or
r+
rs.
r
r+
and
r-
is g e n e r a l l y
and t h e i n j e c t i o n w e l l
.?
is g e n e r a l l y made up of t h e p r o d u c t i o n Xowever,
laboratory
in
some e x p e r i e n c e
displacements,
the
such
notion
as of
i n j e c t i o n and p r o d u c t i o n boundary may n o t be c l e a r l y d e f i n e d ,
so we u s e ( 2 . 7 6 ) One
key
for,
i,+le
dilicil
choice
:?.?not, g i v e r i s e t o a n y a m b i g u i t y .
of
physically
admissible
saturation
boundary c o n d i t i o n s is t o remember t h a t t h e y have t o s a t i s f y t h e r u l e of p r e s s u r e c o n t i n u i t y f o r e a c h o f t h e f ' l ? l i d ? ?'Loding d w o s s t h e boundary
:
II. Construction of the State Equations
109
- if only one fluid is flowing across the boundary, its continuity is ensured by the pressure boundary condition, and the boundary saturation is unimportant. - if two fluids are simultaneously flowing across the boundary, their two pressures unique pressure
P1
Pe
and
P2
r
on
have to be both equal to the
so that using the property (2.13) of the
existing outside of the porous medium Sl,
on I", i.e. P -P = O 1 2 capillary pressure curve, this implies
necessarily
s=SC-
(usually S = 1
if
is the
S
wetting fluid saturation). So a necessary condition for our saturation boundary conditions to be physically admissible is that they satisfy
11.3.1
-
Saturation boundary conditions on the injection boundary
One can use for
-
11.3.1.1
where
:
r- two types of boundary conditions
r-
:
Dirichlet condition :
Se is a given boundary saturation.
Remark 8
The boundary condition S=S
C'
which satisfies (2.77), can be
used, in the usual case where
S
=1,
for the modelling of
water injection. However, when the capillary diffusion term div
f
in
(2.54)
is
non-zero,
this
boundary
condition
generally leads to a production of oil through the water injection
boundary;
though
this
oil
production
can
be
observed under certain experimental circumstances, it is not present under the usual field conditions. So condition (2.78) has to be used for the modelling of water injection at high injection rates only, for which the parasitic oil production occurs only during the very short period when the porous
Ch.HI: Incompressible f i d h a s e Reservoirs
110
medium is not yet saturaLed with water in t h e vinicity of the injection boundary I#--( cf. fig.")
- oil
0
-
4-'p2
9'
+
+ oil +water
'0
st
no oil flow
- -water -
,water
'1
,oil+water
'0
oil +water
!
1
0
r-
spacell
spacer,
AT INITIAL
DURING
SHORT PERIOD
TIME Figure 7
A
:
space
LATER
Approximate modelling of water injection by a S=1
Dirichlet boundary condition in case of high injection rates Remark 9 :
Conditions ( 2 . 7 8 ) with
Se
does not necessarily satisfy (2.77). However the condition S=O
can be used for,the modelling of oil injection high
injection rates, as the p a r a s i t i c water production (during when (2.77) is violated) occurs only during the short period where the porous medium is not yet saturated in oil in the vicinity of 11.3.1.2
-
r-.
Given water injection rate :
0
111
II. Construction of the State Equations where
:
Q.(s,t)
jth fluid production rate density at point
=
(2.80)
Remark 10
s and
is negative in case of injection).
:
On the part of
fluid), -
+ + qo-v
T- where
< 0 (global injection of
one can model injection of
pure water or Oil by
taking the following values for Q , in (2.79)
+
(2.81)
Q1
=
(2.82)
Q
=
1
+
qo-w < 0 0
<=>
<=>
Q2
0
=
:
water injection,
oil injection.
A water injection rate
Q1
different from (2.81) or (2.82)
would generally lead to simultaneous flow of water and oil
Remark 1 1
:
and hence violate (2.77) as we could not ensure that
S
would remain equal to 1.
0
On the part of
r-
where
+ + qo*v
=
0
(globally closed
boundary) one can model closure conditions by taking Q 1=0, which automatically ensures Q2=0. Remark 12
:
In the case where the overall water injection rate Ql,(t) known through only a part
rw
of
is
r-, (2.79) can be replaced
by
(2.83) IS
=
unknown constant on
rW.
The second condition of (2.83) is mathematically and practically convenient, but has, to my knowledge, no physical justification.
0
112
Ch. III: Incompressible llvo-f'hase Reservoirs
-
11.3.2.
One c a n u s e f o r
-
11.3.2.1
where
r+
Saturation boundary conditions on the production boundary
r+
three t y p e s of boundary c o n d i t i o n s :
Dirichlet conditions :
Se
is a g i v e n boundary s a t u r a t i o n .
Remark 13 :
For
a
displacement
s a t u r a t i o n is S
=
can
0
without
of
used
be
oil
over
0
violating
by
water,
the f i e l d ,
where
condition
the
initial
(2.84)
with
b e f o r e t h e w a t e r b r e a k t h r o u g h time,
(2.77).
However
condition
this
is
not
a d m i s s i b l e a f t e r t h e b r e a k t h r o u g h time, when o i l and w a t e r flow simultaneously. 0
Remark 14 :
The D i r i c h l e t c o n d i t i o n ( 2 . 8 4 ) w i t h :
always s a t i s f i e s (2.77) However,
generally leads, term,
to
and t h u s is p h y s i c a l l y a d m i s s i b l e .
b e f o r e t h e water b r e a k t h r o u g h time t h i s c o n d i t i o n i n t h e p r e s e n c e of
intrusion
p r o d u c t i o n boundary experimentally
of
water
r+ ( c f . f i g .
observed.
in
a capillary diffusion field
the
through
the
8 ) , which is g e n e r a l l y n o t
However,
this
s i m p l e and l e a d s , t o g e t h e r w i t h S=S
on
condition
r-
is
very
( c f . Remark 8 ) t o
s i m p l e homogeneous D i r i c h l e t boundary c o n d i t i o n s f o r t h e o i l saturation
Remark 15 :
1-S
i n t h e u s u a l c a s e where Sc
We have s e e n t h a t t h e water
S=O
=
1.
is a good c o n d i t i o n on
b r e a k t h r o u g h time (remark 1 3 ) and t h a t
0
r+
before S=S
is
s a t i s f a c t o r y a f t e r t h e b r e a k t h r o u g h time (remark 1 4 ) . Thus a more r e a l i s t i c c o n d i t i o n would be t o t a k e t o 1 a t t h e b r e a k t h r o u g h time. Dirichlet
But
this
Se jumping from 0
is n o t a s i m p l e
c o n d i t i o n , as t h e b r e a k t h r o u g h time is n o t known
II. Construction of the State Equations
113
-
oil
oil%
2 ‘
4
water c1‘-
no water flow
oi i+water -3L
-
&water
n
1
D
Oh
initial time Figure 8
:
r+r-
space
stationary curve
space
transitory period
The effect of an
production boundary
S-Sc
T+
b e f o r e h a i d . 31,
L 4
oil +water
r+
later
Dirichlet boundary condition on the
f o r a sufficciently long 1-D field.
W H Iirtvc? t n
u s e t h e u n i l a t e r a l c o n d i t i o n which
we describe now.
0
- Unilateral boundary condition
11.3.2.2
On t h e p r o d u c t i o n boundsry w e t t i n g phase
S
Y+, a s long a s t h e s a t u r a t i o n of t h e is s t r i c t l y less t h a n Sc, we see from (2.77) t h a t
e i t h e r water o r o i l , but n o t b o t h , can flow o u t of forces,
the w e t t i n g phase
(irater,
i n c a s e of
i n s i d e t h e porous media as l o n g a s t h e s a t u r a t i o n less t h a n
(2.85)
Sc,
s o t h a t we g e t
s < sc
+
+
$, .LI
>
=
:
0
on
R.
Due t o c a p i l l a r y
w a t e r + o i l flow) is r e t a i n e d
r+.
S
remains s t r i c t l y
114
Ch. III: Incompressible W o P h a s e Reservoirs
S becomes equal to Sc, then water is allowed to get out of
When
the field : this is the breakthrough time. We shall suppose that no water
r+,
is available outside the porous medium on only be directed outward from +
(2.86)
$l'v
+
so that the water flow can
:
r+.
on
2 0
Thus conditions (2.85) (2.86) give the unilateral production boundary condition :
(2.87)
on (Sc-S)
t (s) BT
(2.88)
+
$,*V
=
+
=
r+,
f
t
6
10,T[,
0,
first time when S becomes equal to S at s c
r+.
J
We have drawn for the case of boundary condition (2.871, typical variations of S with time at s
E
r+
(figure 9) and with space for given
times t (figure 10). Remark 16 :
In the usual case where
S
=1, C-
one has simultaneously on
r+
after the breakthrough time : (2.89) which implies, as condition boundary
a(l)=O,
introduces a
r+
that
boundary
as au
=
~
3
.
So the unilateral
layer on
the production
after the breakthrough time (cf. figure 1 0 ) . 0
I1 Constructionof the State Equations
115
s4
Figure 9 : Evolution of the water saturation on the production boundary
r+
with the unilateral boundary condition (2.87).
+-
P1.v >o
4 -
P1.v = 0
space phase @ of figure 9
r+
space
phase @ of figure 9
r+
space
phase @ of figure 9
Figure 10 : Profiles of the water saturation inside the porous medium D near the production boundary
r+
at different times, when
the unilateral boundary condition (2.87) is used.
r+
116
11-3.2.3
Ch. III: Incompressible n v d h a s e Reservoirs
- Given water/oil production ratio (MR). A
r+
widely used condition for
is obtained by neglecting the
boundary layer introduced by the unilateral condition, and by requiring
r+ be proportional to their respective
that the water and oil flow through mobilities on
l‘+
:
+ - +
$l.v - = + + @2*v
(2.90)
kl(S)
a~ t
E
lO,TC,
k2(S)
or equivalently : + +
r-v +
(2.91)
1
+
+
b.(S) qj.v J
j = l ,2
=
0 on
r+
fF t c 10,TC.
This boundary condition does not satisfy ( 2 . 7 7 ) . +
Since the water+oil flow field
qo
becomes very large near the
well as the diameter of this latter is very small, one checks easily from (2.60),
(2.61)
that near the well (but outside of the boundary layer) one
always has, whatever the saturation boundary condition is, that + kl( S )
$1
(2.92)
i,
z
:
k2(S)
which shows that condition ( 2 . 9 0 ) or ( 2 . 9 1 )
and the unilateral condition
become equivalent for small well diameters. Remark 17
:
It can be a l s o convenient to use a slight variant of ( 2 . 9 0 ) or ( 2 . 9 1 ) . Let
(2.93)
can take
:
1 (:-;
+
1
‘w
j=1,2
S
=
rw
b.(S) 3
c
r+
be one production well boundary. One
+ - +
qjw)
unknown constant on
=
0
‘W
t t c l0,TC V t
E
10,TC.
We shall use this condition to find the equivalent point source model for the saturation equation.
0
III. &Summary of Equations
111-
117
OF
SUMMARY
EQUATIONS
INCOMPRESSIBLE
111.1
-
=
rock porosity,
(3.2)
K(x)
=
rock permeability,
(3.3)
o(x)
=
section of field,
(3.4)
Z(x)
=
depth.
P
(3.6)
P
111.3
1 2
S
-
FLOWS
ROCK
Q
.. .at point x
Q
PHYSICAL U"0WWS
(3.5)
(3.7)
AND
CHARACTERISTICS DEPENDING ONLY ON THE RESERVOIR
@(x)
-
THO-PHASE
FLUIDS
(3.1)
111.2
OF
=
pressure inside the wetting phase,
=
pressure inside the non-wetting phase,
=
wetting-phase saturation.
CHARACTERISTICS DEPENDING ONLY ON THE FLUIDS
of fluid j , (3.9)
(3.11)
p. =
j = l,2
viscosity
The two fluids are incompressible, i.e.
p1 =
cste,
p2 =
CSte
FOR
118
Ch. III: Incompressible nvliuoPhase Reserwoirs
111.4 CHARACTERISTICS DEPENDING BOTH ON FLUIDS AND ROCK
(3.12)
(3.13)
-
Sm(x)
-
SM(x)
=
wetting phase residual saturation at point x
=
1-non-wetting phase residual saturation at point x E R ,
ZM(x)]
k .(s,x) : [sm(x),
x
R
[O,l]
-f
E
R,
relative permeability
=
(3.14) of fluid j,
(3.15)
P1
-
P
2
=
j=1,2,
P (2.x) : [sm(x), sM(x)] c
x
R
+
R
=
capillary pressure.
Hypothesis : If one introduces : S(x,t) - s,(x) (3.16) S(x,t) = then we suppose that k.(S)
(3.17)
J
=
reduced saturation of the
- Sm(X)
SM(X)
wetting phase,
:
-
kr.(Sm(x) =
+
S(FM(x) - sm(x)), x) =
u J.
mobility
is a function of the reduced saturation S only and that there exist functions PCM(x) and pc(S) such that :
where : PCM(x) 2 0
(3.19)
=
maximum of the absolute value of the capillary pressure at
(3.20)
I
x
E
R,
reduced capillary function (dimensionless, increasing), with -1
S pc(S) 5 1,
pc(Sc)
=
0 (where usually
Sc=l).
III. Summary ofEquations
(3.21)
119
PCM (x)
maximum capillary pressure at x c i2 (defined in
=
(3.19)),
(3.22)
PG (x)
(3.23)
Q(X)
=
(ZM(XI
(3.24)
$(x)
=
o(x) K(x),
(3.25)
;,(x)
=
-$(x) grad PCM(x) (governs the effects of capillary
(3.26)
G2(x)
=
-$(x) grad PG (x) (governs the effects of gravity).
-p
=
m @(x)
=
gravity potential at x
- Sm(X)l
6 0,
o(x) $(X),
pressure heterogeneity),
111.5
-
AUXILIARY DEPENDANT VARIABLES +
(3.27)
$
1
=
flow vector of the wetting phase,
=
flow vector of the non-wetting phase,
(3.28)
+ $2
(3.29)
r
=
part of
+ qo
=
+ + $ l + O2
+
(3.30) 111.6
-
+
-t
$
=
1
and of -$2 due to capillary diffusion,
global flow vector.
TRACES ON r=an OF THE DEPENDANT VARIABLES They may be known o r unknown depending on the type of the boundary
condition used.
(3.31)
Pe
=
trace of pressure, (index e stands for "exterior")
(3.32)
Se
(3.33)
Q1
(3.34)
Q2
=
trace of " * v
(3.35)
Q
=
trace of qo * =~ global production rate density
=
trace of saturation, trace of
=
+
$1
+
+
-
=~
=
+
wetting-phase production rate density through r , non-wetting phase production rate density through r , through
Caution
'
Q1
9
Q2
r.
and Q are negative in case of injection into R .
Ch. III: Incompressible ThwPhase Reservoirs
120
r
111.7-PARTITIONS OF THE BOUNDARY
(3.36)
r
=
(3.37)
rD
=
OF THE POROUS MEDIUM Q
r D u rN u r W
where (cf. fig. 1 1 ) :
r where the pressure is specified (D stands
part of
f o r Dirichlet),
rN = part of r where the global flow is specified (N
(3.38)
stands for Neumann),
rw = part of r through which the overall global flow is
(3.39)
specified (W stands for well),
r- u r+ with r-
(3.40)
r
(3.41)
r-
=
[
s
6
rl
Q
=
+ + q o - L 5 0 }=global injection boundary,
(3.42)
r+
=
{
s
6
rl
Q
=
+ + qo-v
111.8
=
n
>
r+
0 where (cf. Fig. 1 2 )
0 }
=
:
global production boundary.
- FUNCTIONS AND COEFFICIENTS DEPENDING ON -
REDUCED SATURATION S
ONLY
(3.43) (3.44)
k.(S)
p (S)
mobility of jth fluid (defined in (3.17)
=
J
reduced capillary function defined in ( 3 . 1 8 ) , ( 3 . 2 0 ) ,
=
k l k2 dPc kl k2 dS ’
(3.45)
a(S)
=
(3.46)
a(S)
=
(3.47)
bo(S)
=
kl ,
(3.48)
bl(S)
=
-PCW
positive ,
+
S
a(t) dt,
increasing,
0
increasing (fractional flow),
k t + k2 k l k2
I
k\+k2
(3.49)
j=1,2,
bZ(S)
=
k1
k2
k1 + k2
(P,
- P2) Pm
’
121
III. Summary of Equations
reduced saturation f o r which pc(Sc)
(3.50)
S
(3.53)
Y2(S)
(3.54)
d(S)
=
=
kl p1
+ k 2 p2
(3.56)
-
k l + k2
1 Pm
(cf. 3.20)),
0
’
(kl + k2).
=
111.9 - MAIN DEPENDANT VARIABLES
(3.55)
x
=
:
S
=
reduced saturation (defined in (3.16)),
P
=
-21
(P +P 1
2
) +
PCM Y(S)
=
global pressure.
111.10 - EQUATIONS FOR PRESSURE, SATURATION AND PLOW VECTORS
Equations governing the global pressure P f o r every t
- Inside
C0,Tl
k l :
+
(3.57)
div qo
(3.58)
Go
-
E
0,
=
2 =
-$d(S) gradP
On the boundary
r
P on r
P
(3.60)
qo*v = Q on
+
e
1
j=l
Yj(S)
rN
=
P
+
=
TD u TN u
rw
:
(Neumann),
r’ P
r
Gj.
(Dirichlet),
D
+
d(S)
(we use the partition
(3.59)
=
+
unknown constant on
(well-type)
rw.
:
Ch. III: Incompressible nvvo4hase Reservoirs
122
Equations governing the saturation S inside Q
(3.63)
-
-
=
OxlO.T]
r’ =
:
-$PcM grad a ( S ) .
On the boundary
r
(3.65)
-
S
=
=
(3.66)
S
=
S
+
:
(Neumann);
Q1
on the global production boundary
or
r,) :
(Dirichlet),
S
+ + @,-v
r = rr- one can take
(we use the partition
on the global injection boundary
(3.64) or
:
r+ one can take
:
(Dirichlet), +
+
+
(unilateral), (3.67) S 5 Sc, @1. V , ( S c - S ) q , * L = 0 or + + + + (3.68) r-v + 1 b. qv: = 0 (WOR equal to mobility ratio). j=1,2
- At the initial time t=O (3.69)
S
=
So(x)
:
on a.
III. Summary of Equations
123
Figure 12 : Example of saturation boundary conditions compatible with the pressure conditions of fig. 1 1 when Pe is constant over
rD-
Ch. III: Incompressible nyO-Phose Reservoirs
124
Equations for separate phase pressures and flows (3.70) (3.71) (3.72) (3.73)
P1
=
P
-
[Y(S)
-
1
PC(S) 1 PCM’
:
IV. An Alternative Model for Diphasic Wells
IV - A N
125
ALTERNATIVE MODEL F O R D I P H A S I C YELLS
As we have done i n c h a p t e r I1 f o r t h e case of monophasic wells, we
s h a l l t r y here t o r e p l a c e t h e boundary c o n d i t i o n s used i n s e c t i o n s I1 and
I11 f o r t h e model f o r d i p h a s i c wells by p o i n t s o u r c e s a p p e a r i n g i n t h e r i g h t hand s i d e of t h e e q u a t i o n s ( d i s t r i b u t e d s o u r c e s can t h e n be o b t a i n e d by approximating t h e D i r a c f u n c t i o n s ) . The d i p h a s i c e q u a t i o n s being much more i n t r i c a t e t h a n t h e monophasic o n e , we s h a l l proceed f o r m a l l y o n l y .
We c o n s i d e r t h e t y p i c a l s i t u a t i o n of a c l o s e d f i e l d i n j e c t i o n well DE- and one p r o d u c t i o n well D
E+
that :
QE
w i t h one
( c f . f i g u r e 1 3 ) , and suppose
where a , b are t h e ' l c e n t e r s l l of t h e i n j e c t i o n and p r o d u c t i o n wells.
We production
take well
as p r e s s u r e the
boundary c o n d i t i o n s a t t h e
well-type
condition
(3.61),
where
( w a t e r + o i l ) i n j e c t i o n and p r o d u c t i o n r a t e , and where on
r
EL
of a given regular function.
boundary
e
Q T ( t )is t h e e q IS t h e t r a c e
We t a k e t h e n as s a t u r a t i o n boundary c o n d i t i o n on t h e p r o d u c t i o n TE+
aDE+
=
the
r a t i o " c o n d i t i o n (2.90).
variant
(2.93)
of t h e l'WOR e q u a l t o m o b i l i t y
which becomes e q u i v a l e n t ( a t l e a s t f o r m a l l y ) t o
t h e u n i l a t e r a l c o n d i t i o n when
E+O.
On t h e i n j e c t i o n boundary
The e q u a t i o n s g o v e r n i n g t h e p r e s s u r e the saturation SE
-
( d e f i n e d up t o a c o n s t a n t ) and
are t h e n : 0 with
=
qoE
PE
rE-=aDE- we
QIT (condition 2.83)).
t a k e as g i v e n t h e o v e r a l l water i n j e c t i o n r a t e
(4.2)
P =P
i n j e c t i o n and
=
0
;
OE
on
r.
=
- @ d E gradPt
+t
6
]O,T[,
+
dE
/ J=1
+ YjEqjE,
i n RE,
126
Ch. 111:Incompressible n v d h a s e Reservoirs
i
J
(4.3)
sly
-Q ,T ,
=
*
$,€
I
+.+ (r.v
+
E+
s
Let
so
=
ii
=
c o n s t a n t on
zE-,
E-
I. E-
=
2 + + 1 b. 9 . j=1 J C J E
on R
SI
=
c o n s t a n t on L
E+
E+
be t h e r e s e r v o i r ; we u s e t h e same kind of
DE+
t e c h n i q u e a s i n c h a p t e r 11, s e c t i o n 111, but f o r m a l l y o n l y .
We go f i r s t t o t h e l i m i t i n t h e p r e s s u r e e q u a t i o n ( 4 . 2 ) : The p r e s s u r e P
where
v
satisfies, f o r every
depends on
w
r E J.
$dE
gradvE) av
av
>=
8VE *dE
a\,
t
E
]O,T[
:
by :
- d i v ($d
=
’
at t=O.
-
D
R~
o ,
=
=
o ,
w,
vElr
,
=
constant,
EJ
0 , on
r.
One can go t o t h e l i m i t f o r m a l l y i n (4.7) i f
j =
+,-,
IV. An Alternotive Model for Diphasic Wells
Figure 13
:
127
The field used to determine equivalent point sources
which is expected if
v.+
and the functions whose
Pet+
are the traces are
regular enough. We get then
(4.10)
-1 Pw
=
Q,(t)
v(a
E
LZ(Sz)
s.t
n
Y
w
-R
w
=
0.where v depends on w by
:
Equations (4.10) and (4.11) inem that P is the (ultra weak) Solution
Of :
128
Ch. III: Incompressible nYoPhase Reservoirs
-
+ divqO = Q , ( t ) (4.12)
Go
=
+
+
6(x-a) - Q,(t)
-+d gradP
+
d
2
1
j=l
6(x-b)
in
i,
on
r,
+ Yj qj,
(lo'" = 0
which is very similar t o t h e monophasic r e s u l t .
We go now t o t h e l i m i t saturation
where v
in
s a t u r a t i o n equation (4.3).
the
satisfies :
SE
depends on w by :
av -a 2 at
-
'CM
d i v ( $ PCM a E gradv avE
= W
=
in
QEp
on 1,
(4.14)
\
vE(T)
=
0.
Going f o r m a l l y t o t h e l i m i t i n ( 4 . 1 3 ) , ( 4 . 1 4 )
where
v
depends on -0
(4.16)
J,
w
2at
by : d i v ( + PCM a g r a d v )
P c M a av -=O
v(T)
av
=
0.
yields :
=
w
-
i n Q, on X,
The
129
N.An Alternative Model for Diphasic Wells This means t h a t
S
is f o r m a l l y t h e
( u l t r a - w e a k ) s o l u t i o n of
:
(4.17)
s
=
so
on
ii
We check now t h a t ,
at
t=O.
d e s p i t e the d e l t a - f u n c t i o n s appearing i n the
r i g h t hand s i d e of t h e s a t u r a t i o n e q u a t i o n (4.17).
its s o l u t i o n
S
always
satisfies 0 I S(x,t) I 1
(4.18) a s soon a s (4.19)
I
0
s SO(X) I 1
0 I Q I T ( tS) Q T ( t )
The f u n c t i o n s a and b .
for
S E [0,1],
.I'
a . e . on
G,
a.e. on
ii
aF
t
E
10,TC.
j=O,l,2, which a r e p h y s i c a l l y d e f i n e d only
have t o be c o n t i n u e d o u t s i d e of t h e i n t e r v a l [0,11, as we
d o n ' t know a p r i o r i t h a t ( 4 . 1 8 ) h o l d s . So we choose
:
f o r r, C [0,13 f o r r, M u l t i p l y i n g t h e f i r s t e q u a t i o n of ( 4 . 1 7 ) by Green f o r m u l a , we g e t
x=(S-l )
+
e
[0,11.
and u s i n g a
Ch.III: Inrompressible Tiuo-Phase Reservoirs
130
where B
:
lR
+
R is defined as one primitive
of
bo
and using the pressure equation (4.12), we obtain
or
(4.24)
which yields
x
0 i.e.
S 5
1
using the hypothesis (4.19) and (4.20).
x
=
-(S
Similarly, multiplying by
yields
:
which yields S 20. Remark 18 :
0
The formulation (4.12), DOUGLAS-EWING-WHEELER
(4.17) is used by some authors, cf.
[l],
EWING-WHEELER for approximation
studies. We shall prefer in the following the boundary-source formulation of sections I1 and I11 which is more general and well suited to our mathematical and numerical tools. 0
V
-
MATHEMATICAL
STUDY OF THE INCOMPRESSIBLE
TWO-PHASE FLOW PROBLEMS The aim of this section is to obtain some rigorous mathematical results on the existence of solutions to the two-phase incompressible flow model summarized i n section 111.
131
V. Mathematical Study V.l
-
SETTING OF THE PROBLEM
Let,
Rn be a bounded, convex domain, with regular boundary r re u rll u rs (referred to respectively as
R E
partitioned into (5.1)
entry, lateral and output boundaries), and with normal unit v
pointing outwards from Q,
T >O
]O,T[, Q=nx]O,T[
(5.2)
be
the given
time
interval of interest,
be the space time domain, and
Z = r ~ l o , T [ : (resp. Ze,
I t * Zs).
We
consider
equations (5.3)
in
Qx]O,T[
the
following system of partial differential
:
div
*)
in Q,
0
q =
(5.4) (5.5) (5.6) (5.7)
+ +
q.v
=
0
on XL,
q.:
=
A(P-Pe)
on zs,
(5.8)
(5.9) (5.10)
on 1 e'
(5.11
on
(5.12)
on zs,
(5.13)
where (5.14)
U(X,O)
=
uo(x)
equations"
zL,
on n at t=O,
Ch. III: Incompressible n v d h a s e Reservoirs
132
and q,b stand for
outside the summation sign Z.
qo, bo
We make the following assumptions on the coefficients in (5.3) through (5.14) (5.15)
(5.16) (5.17)
Q
6
L2(Ze)
Q 2 0 a.e. on Ze,
(5.18)
h
E
L"(ZS)
h 2 m2
d,u,a,b,b.,Y.,
IR (5.19)
R
+
J
on 1S'
are continuous bounded functions of
such that
d(c) t m,
a(c)
bo(c)
=
b.(c)
= 0 , .tF
J
j=1,2
J
a.e.
b(c)
E
a'(c)
=
t 0,
Y 5
f
R, u ( 0 )
=
0;
Y 5 E R; b(<)Zl, Y ,C 2 1 ; ] - ~ , O l u [l,+w[, fF j=l,2;
[O,l], E
a.e. on R.
(5.20)
0 5 uo(x) 5 1
Remark 19 :
If one thinks of u as being the non-wetting phase saturation and of P as being the global pressure, and if one affects the subsript 1
to the non-wetting phase and 2 to the wetting
phase (with the corresponding changes in the definition of the non-linearities), then equations (5.3) through (5.14) correspond, for large 1, to the two-phase flow equations (3.57)
through
(3.69)
of
section I11 with the boundary
conditions indicated in figure 11, 12. One can however notice that the unilateral condition (5.12) is now taken on C
9'
whereas it was taken on Z+ in (3.67) in
section 111. As we noticed in Remark 7, the two boundaries and
r+
r
may differ.
So we take a closer l o o k at the meaning of the unilateral
condition (5.12) on Ts.
133
V. Mathematical Study
s
A t point
*
+ +
Ts where q - v L O(i.e. s
E
model
t h e n we have
t h a t t h e u n i l a t e r a l c o n d i t i o n (5.12) is an
s e e n i n 311.3.2.2 accurate
r,)
E
which
takes
into
account
the
capillary
boundary l a y e r on t h e p r o d u c t i o n boundary.
.At
point s
rs
E
+ +
q-v
where
<
0 (i.e. s
E
r-)
we check now
t h a t , under t h e h y p o t h e s i s (5.21)
<
b(0)
(which
1
satisfied
is
in
practice
where
b(0)
=
O),
the
c o n d i t i o n (5.12) becomes (5.22)
>
u
+
and
0
+
lp2-.o
=
0.
Suppose f o r a w h i l e t h a t +
c o n d i t i o n $I 2'v
+
L 0
and t h e f a c t t h a t b l ( 0 ) +
Q,-V
+
=
u(s)
=
0
f o r such an
s. Then t h e
g i v e s , u s i n g t h e d e f i n i t i o n (5.14) of =
b2(0)
=
0 :
+
$2
+ + + + ( l - b ( ~ ) )q - v - r - b L 0.
Hence from ( 5 . 9 ) and (5.21) we g e t
-
--
i.e.,
6
+ +
( 1 - b ( 0 ) ) q.d
as J, L m > 0 ,
which is a c o n t r a d i c t i o n t o t h e h y p o t h e s i s
u(s)
=
0 because
t h e s o l u t i o n u o f ( 5 . 3 ) t h r o u g h (5.14) s a t i s f i e s , as we s h a l l
see, 0 6 u ( x , t ) 51.
In c o n c l u s i o n t o t h i s remark, we see t h a t t h e e q u a t i o n s ( 5 . 3 ) t o (5.14)
c a n model a n experiment of d i s p l a c e m e n t of a non
w e t t i n g phase by a w e t t i n g phase i n j e c t e d t h r o u g h c l o s e d boundary
r,,
and a n o u t p u t boundary
surrounded by non-wetting f l u i d
:
rs
re,
with a
which is
Ch.III: Incompressible lbo-Phase Reservoirs
134
- on the (global) production part
r+
of
rs
one observes
first production of non-wetting phase only, followed by production of both phases,
r- of rs, the surrounding
- on the (global) injection part
non-wetting f l u i d enters the porous medium, and no wetting phase escapes.
Remark 20 :
0
It is worth noting that equations (5.3) through (5.14) cover the problem of the water dam : the wetting phase
exactly")
(index 2) is the water, the non-wetting phase (index 1 ) is the air, and the boundary conditions are indicated i n figure 14. This problem has been studied by BAIOCCHI in the case where one supposes that the air pressure is constant (i.e. the mobility of air is infinite) and that the air saturation takes only
the values 0 and
1
(which implies that the
capillary pressure is neglected).
Remark 21
:
0
It would be very interesting i f we could know a priori from only the pressure equations (5.3) through (5.7)
rs
of +
q2
=
0
rS
c
-t
and
principle,
r+,
boundary.
the partition
r- and r + . This is for example the case if q1
into
Pe
=
P C P i.e.
all
=
constant, which gives, using a maximum + + on rs and hence q-v 2 0 on r : here
a.e.
the
output
boundary
is
a
production 0
(1) up to the incompressibility hypothesis, uhich can be released - cf. section I of chapter IV.
V.Mathematical Study
135
V.2 - VARIATIONAL FORMULATIONS
Let
(5.23)
re
and
V
{ v
=
II VII
=
[v]
=
rS
be of non-zero measure,
H ' ( Q ) lvlr
=
e
} ,
0
H=L2(a),
Ia lgradv12 ( I Igradvl' I v2 a (
J
1/ 2
which is a norm on V, which is a norm on H ' ( Q ) ,
+
rS
equivalent norms on H ,
K
(5.24)
=
{ v
E
lvlr L 0 ]
VI
be a closed convex set of V.
S
We identify H
with
H'
using the scalar product
I I$
associated with the norm
on H ,
so
:
that we have the following
embeddings :
(5.26)
I
VcHcV' with dense inclusions and continuous injections.
We denote by transported from the
11 ll* and norm 11 11
canonical isomorphism from
((
,
))*
the norm and scalar product on V'
and scalar product ( (
V onto V'.
,
))
on V by the
and we remark that, due to the
chosen method for the identification of H with H ' , we have (5.27)
50
4 v
we shall use (
E
V,
4 f
E
H c V'
VIV= (f,V)$
to denote the duality between V
and V'
.
136
Ch.III: Incompressible Tivo-PhaseReservoirs
.
The pressure boundary conditions
...
. . , . . . . . . . . . . . . . .. . .. . . . Q ..' * . .. .:.. . .. .r\ .* .. .., ... . .... . . . . .. . . . . . . . ... ... .. ... .. ... ... .. .. .. .. . .. .. .. . .. . . . . .. .. .' .. . . '.. * .,
- - - / .
1
i.
.
,
A
.
The (air) saturation boundary conditions (the shorn partition of
Figure 14 :
r
into
r-
and
r+
is completely hypothetical)
Boundary conditions for the water dam model
137
V. Mathematical Study
Let u s d e f i n e :
(5.
The s t r o n g v a r i a t i o n a l f o r m u l a t i o n of e q u a t i o n s ( 5 . 3 ) through (5.14) is then : Problem
(
g ): f i n d
(5.29)
P
(5.30)
-(Ggradw+h
(5.31)
4
=
(5.32)
u
E
(5.33)
(*( t ) , v - u ( t ) ) ,
(5.34)
u(0)
f
L‘(0.T;
such t h a t
H’(SZ)),
a
d(u)
(u,P)
I (P-Pe)w=I r
2
1-1) gradP
+
a . e on l O , T C ,
Qw aF w f H ’ ( Q ) ,
re
1
Y . ( u ) Gjl,
j=1
J
and
where (5.35)
w, +
dt
=
u
Gj-r‘=G
Indeed, t h e equations (5.30), t o get
.grad ( v - u ( t ) ) t 0
f f
au - d i v oat E
v E K t lO,TC,
2
-1
j=l
bj(u)Gj-
( 5 . 3 1 ) a r e o b t a i n e d from ( 5 . 3 ) through ( 5 . 5 ) f i r s t rewrite ( 5 . 8 ) u s i n g
:
i2
=
K,
gradcc(u)+ ( l - b ) ( u ) ) G
inequality (5.33),
(5.14) and ( 5 . 3 ) which g i v e s
M u l t i p l y i n g t h e n by v
2
0’
2 + + @ * = q- 1 bj(u) j=O
i n a s t a n d a r d way;
(
a
0.
i n t e g r a t i n g over SZ,
and u s i n g a Green’s formula
and t h e boundary c o n d i t i o n s ( 5 . 1 2 ) , (5.13) we o b t a i n
:
Ch. III: Incompressible Two-Phase Reservoirs
138
d (u,v), +
(5.36)
J
i2.grad v
=
n
J
+2 - + 4 v v L
o
rS
V V E K aF t E lO,T[.
Replacing v by u(t) in (5.36) and substracting it from (5.36) yields the sought inequality (5.33). 0 The resolution of problem ( p
$ )
with the hypotheses (5.15) through
(5.20) is not possible, because the lack of V-ellipticity of the diffusion term in (5.33) (remember that (5.19) says only that c t ' ( c ) the diffusion term u
E
=
a(r,) 2 0, i.e.
is degenerate) is contradictory to the fact that
L'(0,T;V). So we want to weaken the formulation of our problem so that it can
have a solution even in the degenerate case. We remark for that purpose gradu in (5.33) is V-elliptic for the that the diffusion term ,' Ggradu(u)
a
function B(u) where 8 is defined by
:
We define the weak variational formulation of equation (5.3) through (5.14) by : -
Y. Mathematical Study
In
problem
139
the
( 3 )i n
next
paragraphs,
we
t h e non-degenerate
s h a l l g i v e e x i s t e n c e theorems f o r c a s e and f o r probleio
( 2 ' )i n
the
d e g e n e r a t e c a s e and t h e n make a d e t a i l e d s t u d y , due t o GAGNEUX, of t h e case where t h e p r e s s u r e and s a t u r a i t i o n e q u a t i o n s can be s o l v e d s e p a r a t e l y . We begin w i t h some p r e l i m i n a r y lemmas.
V.3
- SOHE PRELIMINARY L W S
Lemma 1 : by
Under h y p o t h e s i s (5.19)
on b , t h e f u n c t i o n B d e f i n e d
:
(5.41)
B
6
v(lR)
s.t. B ' ( C )
h a s t h e f o l l o w i n g graph :
=
1 -b(c),
B(0)
=
0
Ch. III: Incompressible Tho-Phase Reservoirs
140
-Lemma 2
:
(compactness lemma for the non-degenerate case)
Suppose that a satisfies the hypothesis (5.19) and
3
(5.42)
>
q
Then for every M>O (5.43)
S,
such that
0
11 B(V)llqy
is relatively compact in L'(Q), L 2 ( Z ) (Y denotes the "trace on
Proof : -
This
r7
>
:
+ F € E .
0
the set
{v e L2(Q)I
=
a(5) 2
:
lemma
11 gll.1;.
M,
2
and Y(S,)
}
2 M
is relatively compact in
r'' operator).
follows simply
from the
compactness of
injection of W defined in (5.28) into L2(0,T;H'-E(i2)) f o r any
I
:
(compactness lemma for the degenerate case)
-
Suppose now that a satisfies hypothesis (5.26) and
3e
(5.44)
l0,ll
E
s.t.
c(a)
=
Sup
5
OS5<521
sM
(5.45)
{ v
=
6
L ~ ( Q ) ~ o2 v(x,t) S 1 6 M
The relative compactness of
SH
2
5- 5
<
+-.
5
llglb
11 B(V)IIQ
:
[I /a(?) d-rl'
Then for every M>O the set
M,
a.e. on Q,
1
and 'Y(S M ) is relatively compact in (Y still denotes the "trace on rrr operator).
is relatively compact in LZ(Q),
Proof
the
> 0. 0
Lemma 3
I
E
:
in CHAVENT [l].
in L'fQ)
was already proven
We give here the proof for the sake of completeness. Define
(cf. 5.37)) -1
F = B
.
V. Mathematical Study (5.44)
Then [O,
B(1)I
141
just means that F
is an HBlder function of order
v
IF(tl)-F(t2)1
0
on
:
t,, t2
E
CO,B(l)I
e
5 c(a) Itl-t21
.
We divide the proof into three steps : Step 1 : Properties of F ( Y ) for
< s <
~ S ~ P ( Q )o,
Y
1 ,I
< p <
+-
.
The following characterization of WS” holds (cf. LIONS-MAGENES, tome 1 p.59)
and the quantity
is a norm equivalent to the usual one on WS”(n). One checks then easily that, for any function satisfying 0 5 w(x) 2 1 a.e. on 9, F(w) satisfies : F(w)
€
Ws‘ ”‘(n)
with
(5.46)
II F(w) IIwsl*PI(n) Step 2 : Compactness of
Let v v(t)
=
F(w(t))
S,
2 c(a)
s’
es
=
, p’
P e
E
WS”(5l)
’-
e
II WII ws,p(Q)’
in a subspace of L’(Q).
SM, hence w=B(v)
E
9 and 0 5
a.e. on Q. Hence
w 5 B(1)
with w(t) c V c H’(Q) f o r almost every t
As
=
w
H’(R) c Ws’2(Q) f o r any s
E
]O,l[
f
lO,T[.
we get from (5.46)
:
142
Ch. HI: Incompressible nYo.Phase Reservoirs
From (5.47) one gets easily, using the continuity of the injection of H ~ ( R ) into wS'P(n)
which proves that
:
As 0 is bounded and regular, we have, for
ws'
(5.50)
0
<
s"
<
s'
:
( a ) c ws"*p' ( a ) c LP' ( a ) c LZ(Q) c V'
with compact injection from Ws' " ' (a)
into Ws"*p' ( a ) ,
so that, from a standard compactness argument (cf. LIONS [21 p.58) we get the relative compactness of WM in Lp' (O,T;WS""' Thus we get (5.51
(0))
for 0 <
s"
<
s'
.
:
SM is relatively compact in L ~ / ~ ( o , T ; w ~ ' " ( a~ )/)~for O
Step 3 : Final compactness results.
First, as
218
>
0 and as Ws "'2'e(a)
c L2(R) with continuous
injection, we get the relative compactness of SM in L2(Q). Secondly, the s"-a/z,2/e ( r ) (cf. wsl* ,2/e ( a ) into W trace operator 'i maps continuously LIONS [l] sff
> 2'
p.87), So
we get from (5.51) used with 0 < 2e < srr<
compactness of Remark 22 :
which itself injects continuously into L2(r) as soon as
'i(S )
M
in L2(z).
8
the relative 0
It is possible to give simple sufficient conditions on the function a so that the condition (5.44) holds. From the HBlder inequality 5 5 e/2]i/edT \ e 5-5 = 1 x dT 2 { [a(?) 5
5 5
[a(T)
5 -812 111-8
1
d.r ]l-e
143
V. Mathematical Study
and the definition (5.44) of c(a) we get :
Hence a first sufficient condition on a for (5.44) to hold is that there exists on c0.11; then e
p>O
such that l/a’ is Lebesgue integrable
]0,1] of (5.44) is given by
e
=
2V 2’+1.
From this we deduce a practical sufficient condition
:
(5.52)
which 0
implies
< e < Min { -2,
that
2+P0
(5.44)
2 -1. 2+P1
holds
for
any
e
satisfying
This contains the physical case of two-phase flow in porous media where a(<) is given by (2.31) and has the aspect shown in figure 2 .
0
+
+
We study now in Lemma 4 the continuity of P. q, b 2
to u on SM.
with respect
144
'
Ch. III: Incompressible T w d h a s e Reservoirs
Lemma 4 : Suppose now that the hypotheses of either lemma 2 (i.e. (5.42) and (5.43)) or lemma 3 (i.e. (5.44) and (5.45)), with the corresponding notation, are satisfied, that (5.12) through (5.17), (5.23) through (5.28) and (5.37) hold, and that :
For a given M > O , (5.54
uk
let uk, u
->
+ +
Gk~$2k
Uk +
(5.56)
'kl'
(5.57)
B(Uk)
(5.58)
duk dt ->
(5.59)
Pk
(5.61) (5.62)
qk
+ '2k
-+
strongly in L2(Q), strongly in L 2 ( 2 ) ,
'12
->
weakly in
B(u)
du dt
weakly in
+
->
->
, ,
weakly in L2(0,T; H'(a)),
->P
lim inf k+a
:
be the corresponding unique solutions of
u
(5.55)
(5.60)
be given such that
u weakly in L2(Q).
Let pkt (resP. p,q,@,) (5.55) through (5.57). Then we have :
-+
S,
-
q
weakly in [L2(Q)In,
+
weakly in [L'(Q)]",
2 '
I ;2k-graduk 1 ;2-grad u.
Q
2
Q
Proof : (5.55) and (5.56) result directly from the relative compactness of SM and Y(SM) shown in lemma 2 or 3, and (5.58) follows simply from (5.54) (5.59) and the definition of SM. We turn now to the proof of (5.57) through (5.61 ).
145
V. Mathematical Study
- A priori estimates :
From the definition of S,,, we get (5.63)
wk
=
B(uk)
is bounded in (%
.
From the elliptic equations (5.30) and (5.31) and hypotheses (5.12) through (5.16) we get Pk
is bounded in L2(0,T;H1(Q)).
qk
is bounded in [L2(Q)ln.
(5.64)
Rewriting (5.35) as (5.65) and using (5.63) and (5.64) and the hypotheses (5.12) through (5.161, we obtain
i,,is bounded
(5.66)
-
in [L2(Q)ln.
Extraction of subsequence :
From (5.55), (5.56), (5.63), (5.64), (5.66) we get the existence of a subsequence u u
P
such that
:
a.e. on Q,
'U
a.e. on 1, iii)
w
=
B(u
->
B
weakly in (%
,
weakly in L2(0,T; H'(R)), weakly in [L2(Q)ln, weakly in [L2(Q)ln. Using (5.67) i) and ii) and the Lebesgue convergence theorem we obtain that
:
146
Ch.III: Incompressible Two-Phase Reservoirs
- Passing t o the limit
:
Using the weak convergences of (5.67) and the strong convergences of (5.68). we can pass to the limit in (5.30) through (5.31) and (5.65)
,,,
+
+
+
+
( a l l written with u P qp, $211), which shows that = P, = q and + + + p' $ 2 = 4'. As P , q, $ 2 are uniquely defined by (5.301, (5.31) and (5.65), +
w
and as we have seen in (5.68) i) that = B(u) which is also uniquely + + defined, we get the convergence of the whole sequences B(uk ) , Pk, qk, $2k in (5.67) iii) and iv), which proves (5.571, (5.59) through (5.61). We prove now (5.62). From (5.65) and (5.30) with w=B(uk) we get :
GZk
(5.70)
grad u
Q
k
1I O
:
lim inf k+m
I Glgrad
Q
From (5.59) we see that with (5.69) gives :
2
Q
From (5.57) we get
J, lgrad B(uk)Iz
=
PkIZ
QB(uk) -
re
B(uk)Iz 2
>-
PI z
The last term of (5.70) can be rewritten
where B . ( c ) J
b.(c) =
l;z
a
(c)
Q
1
T
1 h(Pk-Pe)
O
rs
+
1 bj(uk)
j=l Q
B(uk)
Gj.grad uk.
]grad B(u)Iz. weakly in L2(Z), which together
:
is continuous and bounded (cf. (5.53)).
V. Mathematical Study
147
Similarly as for
B.(uk)
(5.68) one can prove that
J
strongly in L2(Q), which together with (5.57) shows that : 2
1
lim
Q
k + m j=1
B.(uk) :.-grad B(uk) . I
J
I Bj(u)
2 =
1
Q
j-1
;:grad J
+
B.(u) J
B(u)
which ends the proof of (5.62) and of lemma 4. Remark
23
:
0
on the functions bj, j=1,2 is not
The hypothesis (5.53)
constraining from a practical point of view
:
- for a non-degenerate problem it is always satisfied
for a degenerate problem coming from two-phase flow, we get from the definitions of a, b, and b2 in terms of the *
mobilities k. and the capillary pressure J
p,
. *
(5.72)
As p,
is usually a bounded function with positive, bounded
below derivative (cf. figure 8 of Chapter I), the hypothesis (5.53) is practically always satisfied. 0
V.4
-
RESOLUTION IN THE NON DEGENERATE CASE
We
suppose throughout this
paragraph
that hypothesis
(5.42)
holds :
3 q>O
(5.42)
s.t. a(5) 2
q
>
0 a.e. on
R
and we want to show the existence of a solution of problem general hypotheses (5.28).
We
use
essentially that problem. Let
(5.79)
(5.1)
(5.2),
(5.6) through (5.18),
( g under ) (5.23)
the
through
for this a penalization technique. The proof follows given
in CHAVENT
[2]
for
a simpler one-dimensional
Ch. III: Incompressible nve-Phase Reservoirs
148
be g i v e n , and d e f i n e t h e p e n a l i z e d problem
( g c :)
(5.85 (5.86
I t s s o l u t i o n is g i v e n by t h e f o l l o w i n g theorem. .THEOREM 1 : Suppose t h a t
t h r o u g h (5.28)
(gE) admits
(5.87)
and ( 5 . 3 7 ) ,
(5.1), (5.42)
a t least a s o l u t i o n
(5.2),
(5.6)
and (5.79)
through (5.11),
(5.23).
h o l d . Then t h e problem
u,P s a t i s f y i n g
:
V. Mathematical Study
Proof : -
149
- Existence : let u
W be given, and define P,
6
through (5.31) unchanged, and u by :
4
by (5.29)
(5.88)
The family of elliptic equations (5.29) through (5.31) admits (for +
a given u ) a unique solution P , q. Then the non linear parabolic equation (5.88) admits ( u and theorem 1.2 p. 1 6 2 1 ,
being now given) a unique solution u (cf. LIONS [ 2 ] , as it is driven by an operator which is the sum of a
linear elliptic operator and of a penalization operator, this latter being monotone, bounded, and semi-continuous. Using standard bounds, we get
:
So we *ave defined a mapping u+u
from S,
into itself. As SM is
convex and weakly compact in W, we get from the Kakutani theorem the existence of a fixed point of this mapping (i.e. the existence of u ) as
soon as the mapping
u+p
is continuous on
SM
for the weak topology of
W, which can be proved with the same techniques as in lemma 4. - Majorations on
:
(5.87) i) is obtained in a standard way by
i n (5.81) and using the general hypotheses (5.16) through
taking
w=u
(5.20).
Taking then
E-
u
v=-u. in (5.84), using (5.81) through (5.82) with
w=B(v) and integrating between 0 and T we obtain :
which gives, using (5.87) i) and lemma 1
:
150
Ch. III: Incompressible lk-Phase Reservoirs
and (5.87)
ii) follows then immediately.
Taking then v=u in (5.84) and still using (5.81) through (5.82) (with W=B(UE)) we get, with the notation (5.53) :
(5.89)
B ( u + ) 5 1 and IB(-u;)I
Using lemma 1 we note that integrating from 0 to t we obtain
(5.90)
i
2
u-
and
:
2
+
+
1 2m j=1 1 IIgjII: II GjII;L2(Q),n
II Allcn
which yields (5.87)
I(pc-Pe)+lL*(ls) IUJL iii) and
2
(
xs)
iv) using (5.87) i)
One gets then from (5.84)
and
ii).
:
which, with (5.87) i) and ii) and the general hypotheses (5.18) through (5.20), yields the sought result. This ends the proof of Theorem 1. 0 We come back now to problem
(@)
and give
V. Mathematical Study 2
-THEOREM
through
:
151
Suppose t h a t ( 5 . 1 ) ,
(5.28).
and
(5.42)
(5.2),
hold.
( 5 . 6 ) t h r o u g h (5.111,
Then problem ( @ ) (5.29)
(5.35), admits a t l e a s t a s o l u t i o n (u,P) s a t i s f y i n g (5.92)
a.e.
0 5 u(x,t) 2 1
on
We s h a l l o b t a i n
:
solutions
(YE,
of
PE)
q
such t h a t
:
as t h e l i m i t of a subsequence of t h e
(u,P) the
through
:
Q,
and t h e r e e x i s t v a r i o u s c o n s t a n t s c independant of
Proof
(5.23+
penalized
problem
using
(PE),
standard
t e c h n i q u e s . We g i v e t h e proof f o r t h e s a k e of completeness. From (5.42) and ( 5 . 8 7 ) i v ) such t h a t
uE e SM
existence of through
a subsequence
(5.62),
uk
uk
->
u
E
v) we s e e t h a t t h e r e e x i s t s M > O From lemma 2 and 4 , we g e t t h e
and of @2 k ( r e s p . P ,
S
5k,
[where P k ,
( r e s p . u ) ] , and s a t i s f y i n g a l s o (5.94)
and
defined i n (5.43).
+M
% q, 9
+€
2
s a t i s f y i n g (5.54)
1 c o r r e s p o n d t o uk
:
weakly i n W.
We check f i r s t t h a t
(u,P)
is a s o l u t i o n of
(8 ) :
- ( 5 . 2 9 ) t h r o u g h (5.31) and (5.35) are s a t i s f i e d by (5.59) through (5.61 1.
uk
+
u
-
- From ( 5 . 5 6 ) and t h e Lebesgue theorem, we g e t t h e convergence of i n L 2 ( Z ) ( f o r a subsequence a t l e a s t ) , which t o g e t h e r w i t h (5.87)
ii) gives
-
u =O
on
Ts, s o t h a t (5.32) is s a t i s f i e d .
-Replacing v by v-uk, v e K
i n ( 5 . 8 4 ) (hence v-11 =o) y i e l d s
(5.95)
IY v
e K,
a . e . on 1 0 , ~ ~ .
:
Ch.III: Incompressible Tbc-Phase Reservoirs
152
Take v=v(t), v c
a
in (5.95)
and integrate from 0 to T :
(5.96)
Using the weak lower semi-continuity of v
-f
-21
Iv(T)Ii on W we see that:
du
(5.97)
k+-
which together with (5.58), limit in (5.96).
Thus du u-v$@
(z,
(5.61).
(5.62)
makes it possible to pass to the
6 i2
grad (u-v) I 0,
VVE&
I
which is (5.33). - The initial condition (5.3'1) follows from (5.851, and from the continuity of the linear mapping u+u(O) from W into H. So (u,P) is a solution to problem
obviously from (5.87). principle
- Taking
(5.92)
(5.93)
result
using a maximum
:
get, as for (5.89)
v = u ~ v with V = -u- in (5.33)
(v
E
K as v(
-
v
=
(u-l)+ in (5.33),
which is permissi 1
: =
+ -
(ulr - I ) + e
=
o
=o) we
'curs
:
Then v=O, and u>O a.e. on Q. Taking v = u ~ P with as
(8). The bounds
There remains to prove
(5.9'1)
as u
Ire
=
o
153
V. Mathematical Study
we g e t
:
f I bj
j = l 51
(P+1)
GJ.
g r a d P S 0.
The l a s t term v a n i s h e s , u s i n g ( 5 . 1 9 ) and t h e f a c t t h a t P+1 2 1 . Using t h e n
(5.30) w i t h w
=
B ( P + l ) we o b t a i n :
4 $ I'+'(t)Ii +
m
11 B(9(t))l12 I +
X(P-Pe)
B(9+1)
rS
-
f,
Q B(9+1)
0.
=
'e
Noting t h a t B is monotone i n c r e a s i n g and ( t a k e wEl i n ( 5 . 3 0 ) ) t h a t :
we s e e t h a t (5.100)
:
+ 2 I@(t)Ii 11 +
m
B(9(t))l12s
J
x(P-P,)-
[B(9+1)-B(1)1
TS
where t h e r i g h t hand s i d e term v a n i s h e s a s B ( 5 ) is c o n s t a n t f o r 5>1 ( c f . lemma 1 ) . Hence P 3 , i . e .
us1
a.e.
on Q. T h i s completes t h e proof of
theorem 2 . Remark 24
0
:
Suppose t h a t we r e l a x t h e h y p o t h e s e s
i n ( 5 . 1 9 ) . Then we have t o make t h e f o l l o w i n g changes i n t h e p r o o f s of theorems 2 and 3 :
-
To o b t a i n t h e energy bounds on u - i n ( 5 . 8 9 ) through (5.901,
we have t o suppose t h a t :
-either B ( c ) is bounded f o r < > O , which is a c h i e v e d i f (5.102)
0 S 1 - b(<) S
4 5
with 6
>
1
Clt III: Incompressible TwoPhase Reservoirs
154
(and then
B(5) 5 B ( 1 )
-) 1
+
5-1
-or (5.1 0 3 )
one knows a p r i o r i t h a t P 2 P
-
rs.
on
To o b t a i n t h a t us1 by t h e maximum p r i n c i p l e i n (5.100) we
have t o suppose t h a t ( 5 . 1 0 3 ) h o l d s , so t h a t t h e l a s t term can be dropped o u t of ( 5 . 1 0 0 ) .
In c o n c l u s i o n , theorems 2 and 3 h o l d -with (5.101) replaced by (5.102)
:
[but then t h e conclusion
u61 of theorem 3 must be r e p l a c e d by t h e weaker one
where
c(q)
-without
->
+
.
(5.101)
when
11
* 0.
1
if one knows that
P
2 P
.
i n s t a n c e b y t h e maximum p r i n c i p l e of remark 21 Remark 25 :
One s e e s
in
(5.90)
s o u r c e term f o r exactly,
as
and
:
(5.100)
that
(P-Pe)-l
on
TS
Is, f o r 0
acts as a
t h e s a t u r a t i o n e q u a t i o n , which c o r r e s p o n d s
formally
h(P-Pe)
=
+ + q*,
on
rs,
to
the
i n t e r p r e t a t i o n of t h e u n i l a t e r a l boundary c o n d i t i o n g i v e n i n remark 19. Remark 26 :
0
The uniqueness of t h e s o l u t i o n of t h e non-degenerate problem h a s n o t y e t been proved. The d i f f i c u l t y f o r t h i s a r i s e s from
the
coupling
between
the
pressure
and
saturation
e q u a t i o n s . As soon as t h e c o u p l i n g f a i l s , t h e n u n i q u e n e s s can be o b t a i n e d by s t a n d a r d methods ( c f . SV.6 below). 0
K Mathematical Study Remark 27 :
155
Asymptotical behaviour of u(t) in the special case where
G2
=
(no gravity and
0
no
G1
=
capillary heterogeneity effects)
As we already noticed in and where Pe = constant on rs. remark 21, one gets in this case using a maximum principle P 2 P
on
rS, so
that the corresponding source term (cf.
remark 25) vanishes. Passing to the limit in (5.891, we obtain
:
which using the Poinearre inequality gives
Hence
:
which shows that
u(t)+O
in L2(Q), i.e.
one can recover all
the mobile o i l of the field by injecting water for a long enough time. This property may fail as soon as gravity or capillary heterogeneity effects are present. V.5
-
RESOLUTION I N THE DEGENERATE CASE We suppose throughout this paragraph that the following hypotheses
hold
:
(5.44)
(5.53)
Ch. III: Incompressible nYo-Phase Reservoirs
156
,THEOREM
3
Suppose t h a t ( 5 . 1 ) t h r o u g h ( 5 . 2 1 ,
:
(5.23)
( 5 . 2 8 ) , ( 5 . 3 7 ) , (5.44) and ( 5 . 5 3 ) h o l d , and t h a t moreover
u0
(5.105)
K.
E
Then t h e problem (5.40),
through
:
(5.35),
( P I
admits a t
),
(5.29) t h r o u g h ( 5 . 3 1 ) , ( 5 . 3 8 ) through
least
a solution
(u,P),
which moreover
satisfies : (5.106)
u
/
E
L"(0,T;H).
( 5 . 2 9 ) t h r o u g h (5.31) unchanged,
i) ii)
~ ( u c ~ ~) n % ,
iii)
[s,
(5.1 07 ) +
du
v-u ( t ) ) , +
I
u (0)
v)
-f 02ri
ri
=
q
I 5 grad u
n
g r a d (v-u ( t ) )2 0
R ;2n
iv)
17
17
=
17
V v
uo,
grad u ( u ) rl
+
B' ( u q )
-
grad(v-u ) ( t ) ) 17
6
2
1
j=l
K , a.e.
on 1 0 , T [ ,
bj(uq) ;j*
w i t h t h e bounds ( j u s t t r a n s l a t e d F r o m ( 5 . 9 2 ) t h r o u g h ( 5 . 9 3 ) ) :
-
157
V. Mathematical Study
where the c are constants independant of n. From (5.108) i), iv), v) we see that there exists an M>O such that u € SM defined in (5.45). From lemma 3 and 4 we get the existence of a rl subsequence uk satisfying (5.54) through (5.62). We check now that the couple (u,P) is a solution of @I ) :
v
E
Wn
(5.29) through (5.31) result from (5.59) through (5.60), (5.107) ii), (5.56) and (5.57) -> (5.58) --> zduf q l ' , (5.108) i) and (5.55) -> (5.40), (5.107) v) and (5.61) ->(5.35).
B(u)
,
There remains to prove (5.39) : take in (5.107) iii) v=v(t) with and v(0) = u and integrate from 0 to T :
&
0
2
which gives
nk
t
1 J, grad uk grad v
0
:
2 E L2(Q), one can pass to the limit i n (5.109) by using (5.58), dt (5.61) and (5.108) iv), which yields (5.39). This completes the Proof of
If
theorem 4 .
V.6
-
0
THE CASE OF DECOUPLED PRESSURE AND SATURATION EQUATIONS
The coupling between the pressure and saturation equation makes the mathematical study of the whole system very difficult. For example, it has not yet been possible to prove the uniqueness of the solution of the coupled system of equations, even i n the case of a non degenerate saturation equation with simpler boundary conditions. Similarly, the
Ch.III: Incompressible nYo-Phase Reservoirs
158
demonstration of the existence of a strong solution of the coupled system relieves on the L"-regularity of the transport field not achieved when
+
Go,
which is usually
qo is given as the solution of the elliptic pressure
equation with coefficient d(u). In order to give further results in these two directions, we suppose throughout this paragraph that :
i
the
transport
(5.291,
field
...,(5.31)
is
+ qo
given by
the
pressure
equation
independant of u and of time t and
satisfies
(5.111)
which is the most restrictive assumption, and that
:
(5.112)
Gj
(5.113)
b.
Remark 28 :
Hypothesis (5.111) is satisfied in two cases
J
f
f
CL"(n)ln
j=1,2
W''"(IR)
j=O,l , 2 .
- one-dimensional problems
(q
0
:
is then constant in space)
with constant given global injection rate Q(t)
5
Q on
r
(but
one dimensional problems with Dirichlet pressure conditions on I' and I' do not satisfy (5.111)).
- multidimensional
problems
with
neither +
+
gravity
nor
capillary pressure heterogeneity (hence q1=q2=O) and such that d(u)
1 (cross-mobility curves).
The following results are due to G. GAGNEUX [ll, [2].
0
We shall
follow his proof and hence give only the main steps of the demonstrations.
159
V. Mathematical Study
V.6.1
-
Regularity and asymptotic behaviour for the non degenerate case
4
-THEOREM
:
(Regularity of the non degenerate case with compatible
-
initial data). We assume the hypotheses of theorem 2, (5.111) through (5.113) and that
:
(5.114)
uo
K.
6
Then the solution u
of (5.32) through (5.35) given by theorem
2 is unique and satisfies moreover
u
(5.115)
t
L"(0,T; V),
du
(5.116) (5.117)
div
L2(Q),
r' =
-div ({grad
and for almost every t, t-.i
xd
(5.118)
:
a
t
cx(u))
€
C0,Tl and
L2(Q),
uo,
Ci0 E
K
:
lu(x,t) - G(x, t--r)(+ I 0
where u (resp. il) is the solution corresponding to uo (resp. f i g ) ; this implies (5.118bis) Proof
d
lu(t) - Ci(t-T)IL,(a) 2 0.
:
(5.119)
du (E(t), v-a(u(t)), Let then s c ,
c
>
+
f a
i2 grad
(v-a(u(t))
0 be the following approximation of the (sign)+
function for 5 I 0 s
p
=
v E K, a.e. on l0,TC.
2 0 f
for 0 2 5 I for
I 5
E
Ch.III: Incompressible no-Phase Reservoirs
160
and l e t u and G be two s o l u t i o n s of (5.119) c o r r e s p o n d i n g t o t h e i n i t i a l c o n d i t i o n s uo and Go. For
(TI
<
T l e t w be d e f i n e d as :
One checks t h e n t h a t : (5.121)
v
=
cr(u(t))
(5.122)
v
=
s.(w(t))
E
a(a(t))
Using (5.121)
+
E
K f o r a.e.
t s.t.
t , t - T
E
CO,T[.
s ( w ( t ) ) c K.
i n ( 5 , 1 1 9 ) , and (5.122) i n (5.119) w i t h
t-T
and 0
i n s t e a d of t and u, and t a k i n g t h e d i f f e r e n c e y i e l d s :
which t o g e t h e r (5.123) y i e l d s
The
Lebesgue
w i t h t h e Cauchy-Schwartz
inequality i n the
l a s t term of
:
convergence
theorem
shows
that
sc(w)
+
sgn(w)
-
s g n ( u ( t ) - i l ( t + r ) ) i n L 2 , which t o g e t h e r w i t h (5.18) y i e l d s t h e sought r e s u l t (5.18).
V. Mathematical Study
161
. Regularity of u
Let N be a positive integer, t
:
following time discretization of (5.33) 1 n n-1 , - ( u -u
V - U ~ )+ ~
(5.125)
I
n
=
- ,
and consider the
:
$; &ad(v-un)
2 0, aF
v K, n=l ,2,...,N ,
u = u
0'
For uniquely un
E
given in K, the first equation of (5.125) defines
un-'
K (the existence can be proved using a fixed point theorem as
in Theorem 1, and the uniqueness is obtained as above). Taking v=O in (5.125) and using the same majorations as for (5.89), (5.90), we have lun1; 2
N
1
lun - u
n=1
(5.1 26)
c
n=l,2,...,N
c
=
~u
n-1 4T
+
12
0 0
T
As
IIBjll:
J=1
5 c
11 ~ 1 1 ,
2
-m .1
+
1;
N
,
,
T
1 11 BCU")~~' 2
n= 1
c/m
where
(Meas
rs) 1 /2 I ( P - P ~ ) - ~ ~ ~ ( ~ ~ )
+
lq;2(R)-
was done in (5.119) for the continuous case, (5.125) is equivalent to 1 n n-1 , v-a(un))O + j grad (v-a(un)) 2 o - ( u -u
n
aF
Taking v
=
v
c
K
,
n=l ,2,...,N.
a(u"-l) we get
But the b.'s are Lipschitzian and the J
&'
;.Is
J
belong to L"'(Q)
SO
that
:
Ch. III: Incompressible Tbo-Phase Reservoirs
162
Summing then (5.127) (with (5.128)) from n=l to M 5 N we obtain
:
From the bounds (5.126) and (5.129) one proves easily, using the same techniques as in the proof of theorem 2, that Uh
->
weakly in L 5 ( 0 , T ; V) and weakly* in L*(OT; V),
u
weakly in LZ(Q), where u is the solution of (5.32) through (5.35) and where uh (resp. a h is the piecewise constant (resp. continuous piecewise linear) function 0 1 N "interpolating" the (u , u ,..., u ) sequence defined by (5.125). This ends the proof of theorem 4.
-
THEOREM 5 :
0
(regularity
for
the
non
degenerate
case
with
non
compatible initial data). We assume the hypotheses of theorem 2, plus (5.111) through (5.113) and that 0
(5.20)
I u (x) 4 1. 0
Then the solution u of (5.32) through (5.35) given by theorem 2
is unique and satisfies u
(5.136)
:
e
L"(0,T; V),
6
LZ(Q),
$
(5.137)
fi
(5.138)
f i div
=
-div
[igrad
a ( u ) ] e L 2 ( 0 , T ; H),
and satisfies also (5.118) and (5.118bis) for almost every t,-r f ]OT[ and uo,
0 such that (5.20) holds.
-
163
V. lclathematieal Study
Proof :
and let
Let
K, k=l,2,... be chosen such that
uOk E
be the corresponding solution of (5.32) through (5.35) given
U,
by theorem 4. Using the same techniques as in the proof of theorem 2, one can prove that uk
:
u
+
in L " ( 0 , T ;
H) weak* and in L 2 ( Q ) strong,
a(uk) [ resp 8 ( u k ) ]
+
a(u)
[resp. B(u)]
weakly in L 2 ( 0 , T ; V ) ,
We have now to obtain additional estimations on uk in order to prove (5.136) and (5.137). also
(ak =
We first remark that
etc...)
a(u,)
du (-,
(5.140)
k
V-CL,)~
dt
U,
satisfies (5.119) and hence
:
+
I a
-+ (J
grad v-irad (v-a,)
aF
v
E
+
K, a.e. on l O , T [ .
For any positive integer p, we define then solution of
:
(t)
+
(5.141 1 Up(0) which, as
dak E dt
u (5.142)
=
a (0) k
=
u,(t),
n ( u o k ) < K,
L Z ( O , T ; H), has the following properties
'a
~k
u;.
uP (t) =
+
dak dt
strongly in L 2 ( 0 , T ; V ) , strongly in L Z ( O , T ; HI.
U (t) to be the P
164
Ch.III: Incompressible hv-Phase Reservoirs
U (t) E K f o r e v e r y t , one can t a k e v=U ( t ) i n ( 5 . 1 4 0 ) . P P M u l t i p l y i n g t h e n by s and i n t e g r a t i n g between s=O and s = t y i e l d s : As moreover
But,
where
Using
then
(5.145)
in
the
As we know from (5.142)
that
i n t e g r a t i n g by p a r t s y i e l d s
e x i s t s a subsequence, s t i l l denoted by
]O,T[.
Since
right
hand
side
dt
+
(5.143)
U +a s t r o n g l y i n L 2 ( 0 , T ; P k
such t h a t
U
P' One can t h e n p a s s t o t h e l i m i t i n (5.147) when
duk -(s)
of
and
:
+
H k - g r a d CY
k
=
dgk
-(s) dt
+
Hk(s)*&ad
V), t h e r e
U ( t ) + a , ( t ) a.e.
P
on
p-'", which y i e l d s :
gk(S),
we o b t a i n
K Mathematical Study
165
which, u s i n g (5.1391, (5.1441, (5.146) shows t h a t (5.1 49 )
1
f i g B(u) fi
E
:
L'(Q),
a(u) e L"(0,T; V),
E1=a1'2, a'
which p r o v e s ( 5 . 1 3 6 ) , (5.137) as Taking i n t e g r a t i n g over
then
in
C0,TI
yields
(5.33)
v=u(t)
=
a 2
n >
0.
with
B(t)
8
e
.@ ( Q )
and
:
au _ at
(5.150)
in which t o g e t h e r w i t h
is possible i n
a'
(5.137) p r o v e s
@I
(Q),
(5.138) as t h e m u l t i p l i c a t i o n by
fi
(Q).
F i n a l l y , (5.118) and ( 5 . 1 1 8 b i s )
(and hence t h e uniqueness of u )
a r e proved by t h e same t e c h n i q u e s as i n theorem 4 .
-
0
THEOREM 6 : (Asymptotic behaviour of u in t h e non-degenerate case)Let t h e h y p o t h e s i s of theorem 5 h o l d ; i f t h e i n i t i a l data u 0
s a t i s f i e s moreover u
0
E
:
H'(Q) [ r e s p . uo e K 1 ,
G20-&ad v t 0 [ r e s p . 201
s.t. v t o , a.e. on Q,
V V E V
Q where
- + i2, J, g r a d =
a(uo)
+
C1-b(uo)l
2
io - 1
j=l
bj(uo)
ij,
aul
+ = - 2 0 [ r e s p . 201 on ?S and (which f o r m a l l y means t h a t d i v $ 20 at t-o + + t h a t $ 2 . . , l t = 0 t 0 [ r e s p . 2 01 on r k u r s ) , t h e n one h a s :
(5.152)
au
(x,t) 2
o
[ r e s p . 2 01
a.e. on nxCO,+nC
I
Ch. MI: Incompressible nvo4hase Reservoirs
166
J u(x,t)dx in the field at time R t is a decreasing [resp. increasing] function of time), (which implies that the amount of oil
(which implies that the amount of oil produced per unit time is a decreasing [resp. increasing] function of time) and (5.154) u(t) + u _ strongly in LP(Q) for every p L 1 and weakly in V where um is among the solutions of
:
the only one which satisfies : (5.156)
ucd=
u', [resp. ub,=
Sup Ul,'Uo
Proof
:
Inf u:,]. u' >u 0
cn
The proof of (5.118) in theorems 4 and 5 requires only that
where
s (w)
s,(u(u(t))-a(ii(t))
=
E
V
is positive a.e. on n.
Under hypothesis (5.151) these inequalities are satisfied for the following two choices of u and & : i)
u(t)
=
solution of (5.32),
ii) u(t) E u fF t, 0
Choosing i ) [resp. ii)] (5.157)
..., (5.35),
&(t)
G(t) solution of (5.32),
uo fF t ,
...,(5.35).
we shall show that
u(x,t) 2 u (x) [resp. u(x,t) 2 u (x)l 0
0
a.e. on ~ x l 0 , ~ ~ C .
V. Mathematical Study
167
From now on we c o n s i d e r o n l y t h e f i r s t case i n ( 5 . 1 5 7 ) ( t h e second b e i n g t r e a t e d i n t h e same way). Taking t h e n i n (5.118) G=u w i t h O < T < t < T y i e l d s
which, as
U(T)
(5.1 58 ) which
5 u
0
proves
a.e. on R u s i n g ( 5 . 1 5 7 ) , shows t h a t : 2
U(X,t)
a . e . on R , f o r 0 2
U(X,t-T)
(5.152),
decreasing function Hence
:
and
shows
t--u(x,t)
that,
a.e.
for
has a l i m i t ,
x
E
2 t, R,
the
positive
which we d e n o t e by
uw(x).
:
0
(5.159 )
s
u_(x) i 1
u(t)
+
Let then f (5.160)
f(t)
i n L P ( Q ) f o r every p > t .
U<,
: [O,
=
a . e . on R ,
-.[
7
R be d e f i n e d by
I u ( x , t ) dx
f t t O .
a
The f u n c t i o n f is c o n t i n u o u s and r e p r e s e n t s t h e amount of o i l i n t h e f i e l d a t time t. From ( 5 . 1 5 8 ) ,
(5.159) and (5.118) w i t h 9
f ( t ) h f(t+T)
aF
f(t)
when t
1 u _ ( x ) dx f(t)-f(t+T) f f(t')-f(t'+I)
(5.161)
+
f_
=
(5.162)
, I
, f'_(t)
+
OSt6t'
exist f t
>
0,
a . e . on lo,+-.[
f' ( t ) 2 f'
f o r a.e. t , r h 0
Sup t>O
Ess f'(t)
Hence, n o t i n g t h a t f ' ( t ) = -
2 0 =
0.
I- ddut ( t ) IL ' ( 0 )
, one g e t s
-,
and
f' ( t ) = f ' _ ( t ) = f ' t ) ( t + L )
u we g e t
2 0,
t , T
which p r o v e s (5.153) and shows t h a t fl(t)
=
:
f T 2 0,
Ch. III: Incompressible 7boPhase Reservoirs
168
Taking then v
=
0 in (5.119) yields :
Hence :
2m . J=1
which, using (5.163) shows that
stays in a bounded set of V when
a(u(t))
t-,. Hence there exists a subsequence a(u(t'))
Using
then
(5.159),
(5.163),
such that
(5.165)
and
the
weak
lower
semicontinuity of the norm, one can pass to the limit in (5.119) (for v given) when uo
t-,
which shows that
E
K
u~,,necessarily satisfies (5.155).
In order to prove (5.156) one just remarks that (5.118) holds for uc, solution of (5.155), with act) E urn
satisfying (5.20) and for any
+ t, which
shows that
io u
(XI
t u_(x)
a.e. on
(5.166)
:
Cl
which ends the proof of theorem 6. Remark 29
:
u(x,t) 2 U_(X) a.e. on a
x
lo,.,[
0
In the special case where q1 = q, = 0 ( n o gravitational or capillary heterogeneity effects) and where the given exterior
169
V. Morhemoticol Study
pressure P that PLP m
2
rs,
is constant on
we have seen (cf. remark 21 )
on Ts; we get then from (5.164)
11 a(u(t))l12 s
E(t)
+
:
o
which proves that necessarily ucd= 0. Remark 30 :
The asymptotical behaviour of
0
u(t)
in the case where it
does not evolve monotonically (i.e. when the initial data does not
satisfy
however,
noticing
(5.151)) that
u0 open problem. One has
is an (5.166) holds
hypothesis that 0 5 u (x) 2 1 0
under
the
sole
:
(5.1 67 )
Remark 31
An
example of multiple steady-state solutions u-. Consider a
vertical
1-D porous slab
Q
= ]O,II[,
with insulated lateral
boundary, and with 7Tinjection"boundary and "production" boundary Ts
=
fe
that
:
q1
0 (no capillary heterogeneity effects)
=
{ O } at the top
=
1 1 ) at the bottom. We suppose
(5.168)
Then
u
represents the oil
saturation in one imbibition
experiment ( q 3 ) in a vertical sample maintained in contact with water at the top ( u ( 0 )
=
0), with insulated bottom end
(as the unilateral condition resumes to
+
+
$2'1)
=
0 when u>O),
Ch. III: Incompressible f i o q h a s e Reservoirs
170
and with oil and water mobilities kl and k2 such that
51
J u(1-u) and a capillary pressure curve kl+k2 p (u) = arccos (1-2u).
-=
The steady-state equation (5.155) becomes now : $2
=
u(0)
(5.169 )
constant on lO,n[ =
0
,
u(n) 2 0
$,(n)
,
0
2
- The class of initial data u0 ax]O,-[
u(n)$2(II) E
K
=
0.
such that
au
2 0
a.e. on
contains only the stationary states : uo = ,:u 0 2 a a . urn is defined in figure 15 (if (5.151) holds in
2 II, where
the bracket case, then necessarily necessarily
@20
(il)
$20(II)
2
0 ; hence
0 and uo is one of the stationary
=
solutions). One checks in this example that the equilibrium profile
0
um
is exactly that of the capillary pressure. This property, which is always true, under condition (5.168), in 1-D samples with both ends insulated, is used as a physical definition of the capillary pressure law. Remark 32
:
0
If we replace, in the last remark, the function b2(U) by
e),
(which corresponds to the (unbounded) capillary pressure law p,(u) = Log then the steady-state equation (5.155) has o n l y one solution urn E 0. This comes from the fact the "equilibrium profile''
x= Log
boundary condition u(0)
=
0.
1 -u
does not satisfy the 0
V. Mathematical Study
V.6.2
-
171
Regularity and asymptotic behaviour for the degenerate case We turn now to the degenerate case: in order to handle this case,
we had replaced in the case of the coupled system of equations, the
8)by the weaker (8 1, which required that
saturation equations (5.32) through (5.35) of problem ( formulation (5.38)
u,,
E
through (5.40)
of problem
K in order to get a solution (theorem 3 ) . We shall now treat this case in another way, and, still following
CACNEUX [ l ] , [2], we introduce the following variational formulation :
Problem (@)
6%
,
(5.174)
B(u)E
(5.175)
($ (t), v-a(u(t))),
(5.176)
~ ( 0 =) u 0 ,
(5.177)
0 2 u(x,t) 2 1
:
find u such that
$Eq'p +
(v-a(u(t.1,) L 0
;,(t).&ad
a
Y. v
E
K
a.e. on lO,TC,
a.e. on Q.
we snail be able to show the existence of a solution without the -compatibility condition from [O,T]
u
0
E
K. Since (5.174) implies that u is continuous
into H equipped with the weak topology, the equations (5.175),
(5.176) make sense.
One a'
=
checks
a L 6 > 0, the
easily problems
that,
in
the
non
degenerate case where
( 3 )and (2") are
equivalent. In the
degenerate case, the inequality (5.175) can be formally shown to satisfy the saturation equations ( 5 . 8 ) through (5.13) with the boundary conditions on has a trace on El.
Ee
and
E
a(u)
instead of u in
(which is satisfying because a ( u )
172
Ch. III: Incompressible nuo-Phase Reservoirs
1
(
The corresponding \ solution converges \
a
b
a
Asymptotical behaviour?
n X F i g u r e 15 : Exaaples of initial data
yielding monotonic and uo non-monotonic evolution of the saturation profile.
U
V. Mathematical Study -THEOREM
1 73
7 : ( r e g u l a r i t y f o r t h e d e g e n e r a t e c a s e w i t h non compatible
i n i t i a l data).
-
We make t h e h y p o t h e s e s of theorem 3 b u t w i t h (5.105) r e p l a c e d by (5.20)
Then t h e problem one
"entropy"
(PI)(5.174)
solution
u
v i s c o s i t y ) s a t i s f y i n g moreover
$
(5.178)
fi
(5.179)
Proof
:
a.e. i n R.
o < u 0( x ) s l
Let
B(u)
E
t h r o u g h (5.177) a d m i t s a t l e a s t
(defined
by
addition
of
a
vanishing
:
L2(Q),
a(u) e L w ( O , T ; V).
u
n
be
the
s o l u t i o n of
theorem 5
a ( s ) = a ( C ) + q . We know from theorems 2 and 5 t h a t u
( c f . ( 5 . 1 0 8 ) , and (5.148) ( 5 . 1 4 9 ) )
:
n
corresponding t o
is bounded as f o l l o w s
where t h e c o n s t a n t s C are independant o f n. Using t h e same compactness argument a s i n t h e proof of theorem 3 , one c a n p a s s t o t h e l i m i t when u
n
n+O,
and hence show t h a t a subsequence of
converges toward one s o l u t i o n of problem
(5.179 )
.
which s a t i s f i e s ( 5 . 1 7 8 ) , 0
Ch. III: Incompressible TboPhase Reservoirs
174
Remark 33 :
The c h a r a c t e r i z a t i o n of t h e e n t r o p y s o l u t i o n s of t h e proof
results
in
, and
@It)
of t h e i r u n i q u e n e s s , have n o t y e t been done. For that
direction
diffusion-convection
in
the
case of
the
degenerate
e q u a t i o n s ( i n s t e a d o f i n e q u a l i t i e s ) one
can see VOLPERT-HUDJAEV and B R E N I E R . 0
-
THEOREM 8 :
(Asymptotic behaviour i n t h e d e g e n e r a t e case)-
Let t h e h y p o t h e s e s of theorem 7 h o l d , and suppose t h a t
s a t i s f i e s moreover (5.151) and (5.181)
n
j, g r a d uo g r a d v t
0
aF
vfV
(which f o r m a l l y means t h a t d i v ( $ g r a d u,)
on
rk
and
rs).
u(t)
-t
S 0
on R and t h a t
Then t h e "entropy" s o l u t i o n u of (
urn
0
s . t . v t 0 a.e. on $2
strong
j
where uw i s , among t h e s o l u t i o n s of :
t h e o n l y one which s a t i s f i e s (5.156).
i n L'(Q)
au
5 2 0
gTt) defined
theorem 7 ( b y a d d i t i o n of a v a n i s h i n g v i s c o s i t y ) s a t i s f i e s
(5.184)
u
:
:
f o r every, p L 1,
in
175
V. Mathematical Study
Proof : -
u
Let
be t h e s o l u t i o n of t h e non d e g e n e r a t e problem i n t r o d u c e d i n
n
t h e proof o f theorem 7 . One knows t h e n t h a t a ( u ) is bounded i n L " ( 1 6 , T C ; V ) n d a ( u ) is bounded i n L 2 ( ] 6 , T , [ ; H ) when 6+0. Then a ( u ) and a ( u )
and t h a t
n
n
are c o n t i n u o u s from
i n t o V (equipped w i t h t h e weak t o p o l o g y ) , and
[O,m]
and u are c o n t i n u o u s from ] O , f > [
We check f i r s t t h a t , f o r a subsequence, s t i l l denoted by u f o l l o w i n g convergence p r o p e r t i e s
u
u
n
i n t o H (equipped w i t h t h e weak t o p o l o g y ) .
n'
one h a s t h e
:
t>O
a ( u (t))
+
a ( u ( t ) ) weakly i n V , s t r o n g l y i n H and a.e. on Q,
un(x,t)
+
u(x,t)
n
U
p
-f
a.e.
u(t)
in
on
Q , and hence
LP(Q)
...when
r~
Y -f
p 2 1
0.
I n o r d e r t o prove t h i s , we i n t r o d u c e t h e c a n o n i c a l isomorphism A from V o n t o V' D(A)
=
{ v
E
,
associated with t h e s c a l a r product ( (
VIA v
[U
E
6>0, V
)),
and i t s domain
}. Then we g e t from ( 5 . 2 7 ) and (5.178) t h a t
H c V
t , t Ob 6 ,
U v
E
:
D(A) C V ,
and t h e same p r o p e r t y h o l d s f o r a(u ) .
n
Let then { a E the Dirac function
E
6 ( t ) 2 0,
Then Hence, as
L'(IR),
bE(t-tO)
6E(t)
Av
n
L
>
]
be an a p p r o x i m a t i o n sequence of
=
i f It1 2
0
V')
L,
1
W
for 0
6 (t)dt = 1. E
<
6 S t0-E
and v
E
0 given) :
to+(1 s E ( t - t 0 ) ( ( a ( u ~ ( t ) ) , v )-)
0
0
L'(6,T;
E
to+E -E
>
V.
u ( u ) i s , f o r a subsequence, converging toward u(u) i n L m ( 1 6 , T [ ;
V ) weak s t a r , we g e t ( f o r
t
t
:
t
6 ~ ( t - t o ) ( ( a ( u ( t ) ) , v ) ) d=t & f E ( n ) 0
-E
+
n -to
0.
Ch. III: Incompressible %@Phase Reservoirs
116
E>O be g i v e n ; l e t t i n g
Let
which is t r u e f o r any
q-0 we o b t a i n :
L 0 ; hence we have proven t h a t
I
The f i r s t a s s e r t i o n o f (5.187) f o l l o w s t h e n from ( 5 . 1 8 9 ) , from t h e
f a c t t h a t a ( u ) is bounded i n L"'([G,T];
V ) and from t h e f a c t t h a t
n
D(A) is
dense i n V . The remainder of (5.187) f o l l o w s t h e n immediately. From (5.151) and (5.181) w e see t h a t uo s a t i s f i e s , f o r a l l (5.151) w i t h a
6 applies t o u
n
17
i n s t e a d of a (where a ( c )
=
Hence we have
:
n
for all
+t,T
(5.1 90 1
>
q>O.
5
0
[ n + a ( r ) ] d r ) , s o t h a t theorem
u ( x , t ) 2 u (x,t+T)
0,
n
n
for a.e
x
E
a.
Using (5.187) one can p a s s t o t h e l i m i t i n (5.190) when
+ t,
(5.191)
so that,
for
u(x,t) 2 u ( x , ~ + T )
T>O
almost
every
x
f
R,
u(x,t)
->
U_(X)
f o r a.e. x
E
q+O:
a.
t + u ( x , t ) is a p o s i t i v e d e c r e a s i n g
f u n c t i o n : l e t u s d e n o t e by uc,(x) its l i m i t (5.192)
n>O,
:
a.e. on a.
t 4 m
T h e f u n c t i o n f d e f i n e d by (5.160) is c o n t i n u o u s on 1 0 ,
c o n t i n u o u s from [O,-[ [O,@[
+
w [ ,
a s u is
i n t o H equipped w i t h t h e weak topology. Let t h e n f :
IR be d e f i n e d as
:
n
1I7
VI. The Case of Fields with Different Rock Types
Yt50. Then we get from (5.187)
:
Y t>O
(5.194)
fn(t)
We see from (5.192)
+
when n+O.
f(t)
and (5.194)
that f satisfies (5.161),
and the
rest of the proof is the same as in theorem 6, of course with (5.155) replaced by (5.186). Remark 34 :
0
Theorem 8 covers the practical case of a field initially saturated with oil, when u (x) E 1 on il
:
0
in this case, the
saturation tends to a stationary profile
uo,
which is
identically zero in the special case where no gravity or capillary heterogeneity effects are present and where the imposed exterior pressure is constant on
r
.
0
V I
-
THE
CASE
OF
F I E L D S WITH
DIFFERENT
ROCK
TYPES
Up to now, we have always supposed that the shapes of the non linearities a, b.
J'
d,
the hypotheses (3.17 )
'I.
J
were the same all over il. This was the result of
through (3.20 )
relative permeabilities and
capillary
concerning the dependance of the pressure
laws
upon the spatial
variable x. In petroleum engineering terms, this would be rephrased by saying that we have considered a field containing a single rock type. This notion of V o c k
type?' thus appears as an hypothesis simplifying the spatial
dependance of the relative permeabilities and capillary pressure curves
:
inside a given rock type, (argiles, or sandstone or...), the porosity $, the permeability K and the maximum capillary pressure PCM may vary from one place to the other, but, at a given point x and for a given (actual) saturation
5, the
relative permeabilities and the capillary pressure are
perfectly determined once one has been given the residual saturations
-
S (x) and m
l-sM(x) at that point.
Ch. III: Incompressible Tivo-Phase Reservoirs
178
VI.1
-
THE DIFFERENT ROCK MODELS
Define : d. J
1 V.
=
j=l,2 =
krl kr2 + d=-
(6.1)
krl dl
=
=
krl dl
p2
p1
v
mobility of jth fluid,
3
=
+
kr d 2 2
global mobility,
=
fractional flow.
Then choosing values for k r l , kr2 and p choosing values for d,w and p,. terms of d, v and p
.
clearly amounts to
Hence the rock-models will be specified in
In all rock models for incompressible fluids, we take as given the following functions of the reduced saturation S (6.2)
(6.3)
S
->
S
->
:
pc(S) satisfying ( 3 . 2 0 1 , w(S) satisfying
w(0) = 0 ,
v(l)
=
1,
L
an increasing function of S.
As for the choice of the global mobility function d , we shall distinguish two cases Case 1 :
:
Rock model of t h e f i r s t kind : we take as given the following
function of the reduced saturation S
S (6.4)
->
d(S) satisfying
d(0) 2 d2,
d(1) 2 d l , d(S) > 0 .
The relative permeabilities generated by this model are (6.5) These
relative
permeabilities
depend
only
on
the
reduced
saturation S , which is the assumption made in sections I1 and 111. All
equations in these paragraphs have been established using a rock model
of the first kind.
VI. The Case of Fields with Different Rock Types
179
This rock model can be used for two phase flows without exchanges between phases, where
-
sM remain away from 0
Sm and
and 1.
However, when exchanges between phases take place (see chapter IV, 8111 and IV). The actual saturation
s
may take values outside of the
and sm(x) may approach to 0 and sM(x) may approach
interval [zm(x,) sM(x)],
1 as one tends toward the critical point. So one will need to calculate the
relative permeabilities (but not the capillary pressure) for values
s of
the
But,
actual saturation lying outside the interval [sm(x), ZM(x)].
continuing the relative permeabilities given by (6.5) outside the interval (x), S (x)] would lead to discontinuous relative permeabilities, as kr 1 m(x,SM ) = - I 1, whereas the physics indicates that krl(x,l) = 1 (cf. So rock models of the first kind are not valid in situations figure::61 [S
El(1)
where the residual saturations ?m(x)sl-sM(x)
approach zero.
Rock model of the second kind : we take as given the following function of the actual saturation :
Case 2 :
l
(6.6)
-
S -> d(0)
d(S) =
d2,
satisfying d(1)
=
d l , d(S)
>
0
and we suppose that
(6.7)
the function
1 1
given in (6.3) is continued by 0 for S 2 0 and
by 1 for S t 1. Now we can generate relative permeabilities over the entire range of actual saturations
by setting
(6.8) where S is the reduced saturation corresponding to
5
at the point x (given
by (6.16). The formula (6.8) yields continuous relative permeabilities when
-
Sm
0 and
-
Sm + 1 , as shown in figure 17, and hence this rock model has to be chosen when exchanges between phases take place. +
Of course, the shapes of the relative permeability curves, when expressed as functions of the reduced saturation S, will slightly change from one point x to the other.
Ch.III: Incompressible Tbo-Phase Reservoirs
180
I
Figure 16 : The discontinuous limiting relative permeabilities (dashed lines) obtained with a rock model of the first kind when the residual saturations tend towards zero.
0 Figure 17
The continuous limiting relative permeabilities (dashed
line) obtained with a rock model of the second kind when the residual saturations tend towards zero.
181
VI. The Case of Fields with Different Rock Q p e s
Remark 35 :
For the practical determination of functions, it is enough to know
the d(5)
and v ( S )
:
for one point xo, i.e. for one rock sample,
-
one set of relative permeability curves z+kr.( x o , S ) , j = l,2
(6.9)
J
over the whole interval of (non reduced) saturation and the residual saturations sm(xo) and l-zM(xo),
(6.10)
the viscosities 11, Then
d(2)
and
and
\,(S)
u2 of the two fluids. are
determined by
(6.1) without
ambiguity.
0
The two-phase equations developped in sections I1 and I11 for the (implicit) case of a rock model of the first kind, remain valid for a rock model of the second kind, with the following modifications d(S) has to be replaced by d(x,S)
=
:
d(S),
(with an evident abuse of notation), (6.11)
b ( S ) is equal to > ( S ) , 0 b ( S ) , b (S) become bl(x,S), b 2 ( X , S ) . 1
2
and (6.12)
has to be replaced by
The theoretical results developped in section V remain valid with a rock model of the second kind, as the supplementary dependance of d, b l , b2 on x does not change the proofs.
Ch.III: Incompressible no-Phase Reservoirs
182
VI.2
-
THE CASE OF A FIELD WITH M DIFFERENT ROCK TYPES
Let us now consider the case of a field R , which contains M
.
different rock types. Let Om, m=l ..M be the spatial domain occupied by convention that
rmk=
rmll be
m
the boundary between R and R', with the Q 0 if Rm and R do not meet, or meet only on a line
each type of rock, let
or a point for n=3 or meet only on a point for n=2, and attach a superscript m to each quantity related to Rm. According to the notion of rock-type, not only the shape of relative permeabilities and of capillary pressure curves, but also the maximum capillary pressure PCM may differ in each Rm.
It is hence
necessary to allow for discontinuities of the maximum capillary pressure P (x) across the boundaries rmQ; so PCM (x) will consist of M regular CM m functions PCM(x), defined over Q m' and which do not necessarily meet continuously at the internal boundaries TmQ. We look now for the equation in R , appropriate when for each rock-type, a rock-model of the first kind is used a)
Inside each of the Q",
:
we can proceed as we have done in
m and Pm the saturation and global
sections I1 and 111. We denote by S m pressure in R , which satisfy :
The
equations +m
capillary flow r
,
governing,
in
the global pressure
Qm
the
saturation Sm,
+m
Pm and the (water+oil) flow q
are (cf. (3.571, (3.58), (3.62), (3.63) and (3.72)) : +m Y x e om, (6.15) div qo = 0
and,
(6.18)
the
v
x e Rm'
183
VI. The Gme of Fields with Different Rock Types
Figure 18 : An example of a field with four different types of rock
R 2 and we have
(here il
+m r
(6.19)
We
13)
boundaries
=
m
r 13
0 and
=
m
-$PcM
r2,,
=
0 with our convention).
m
t x
grad a ( S )
€
am.
have then to choose continuity conditions at the
rmll between
different types of rock.
For convenience, we shall denote the jump of any quantity 0 across
by
mR
[ e
1;
em - e p.
=
On any T m R , Il=1 conditions
...M ,
. m=l.
..L, we have to satisfy the following
:
Continuity of pressure : each of the pressure to be continuous (6.20)
[
Pj
1;
=
P
1
and P 2 has
:
0.
- Conservation of masses has to be continuous
:
:
the flux of each of the two fluids
Ch. III: Incompressible Tluo-PhaseReservoirs
184 +
+ m
(6.21)
t q o * v l I 1 = 0,
(6.22)
[
where
;,.;I;
=
0,
is any normal t o TmL. 'I)
Now we want to obtain from (6.15) to (6.22) equations
valid over all P. We f i r s t i n t r o d u c e f o r t h a t purpose f u n c t i o n s d e f i n e d
........as
(6.24)
+ d i v qo
(6.25)
0
(6.26) !
as at
$,(x,t)
=
0
d i v $1 =
x
E
a, t t >
t x
€
a, t t > 0
f
+
+
=
r'(x,t)
0 2 +
1
j=o
soon as x
b.(x,S(x,t)) G.(x,t) J J
f X E
0,
a, t > O .
E
nm
VI. The Case of Fields with Different Rock Q p e s
185
(6.27) (6.28)
So we first get rid of those gradients by using a variationnal
formulation of (6.16), (6.19). Let the test function
?
be any (regular)
mapping from Om into 1". Multiplying (6.16) and (6.19) scalarly by integrating over Om and using a Green's formula we obtain :
?,
(6.30)
+
i m is the exterior normal to the boundary m of Qm. We combine now (6.291, (6.30) in the following ways : +m to * For every regular z:Q -+ R n , we set s = restriction of
where
m'
9
and sum up equations (6.29) and (6.30) for all m , which yields
:
But the above equations alone are not equivalent to (6.291,
(6.30). We have to use also other combinations.
Ch. III: Incompressible nYo Phase Reservoirs
186
For every m , II
= 1,2.
zm
..M, m < I I , for every regular
g
: Q+lRy
with
support in the interior of u 5, , set gm = the restriction of s' to Qm, = the restriction of s' to QII, and take the difference between ( 6 . 2 9 )
;'
(resp. ( 6 . 3 0 ) ) on ilm and on O R . This yields
1
s'.:
(pm+pR)
Pm div
=
rmL
s'
:
P R div . +s
-
%
nm
(6.33)
Y m,L=1 aF
1
s'
: Q
...M, m < R , +
lRn, supp
2.;
(am(Sm)+aL(SR))
:is in the interior of -Qm crm(Sm)div
=
m'
rml
-
4
r - s -j -
(6.34)
Qm "CM
aF
m,R=l...M,
aF ;:Q+IRn,
s'
+I
-
+
1
%+
s'
I I R
(S )
-
QII,
div
5
r - s -
Q R "CM
m
supp
CI
U
is in the interior of
-
'ilm
u QRR'
Equation ( 6 . 2 4 ) through ( 6 . 2 8 ) and ( 6 . 3 1 ) through ( 6 . 3 4 ) are the sought equations governing the evolution of S and P over a field containing different types of rock. We
see
that
taking
into
account more
than
one rock
type
introduces, besides the forseeable dependance on x of all non linearities, integrals over the boundaries
rmII separating different rock types in the
right-hand side of the variational equations (6.31) and (6.32) determining
Go
and
r'
(caution : the flux of
r'
across the boundaries
contiquous ! ) , plus additional equations ( 6 . 2 7 1 , associated with the boundaries TmR
.
rmII is not
( 6 . 2 8 ) and (6.331, (6.34)
VI. The Case of Fields with Different Rock Types
187
We shall see in chapter V, for the case of a single rock type (i.e.
when the last term in (6.31) and (6.32)
disappears), that equations
(6.24) through (6.26) and (6.31), (6.32) are the basis for the numerical calculation of P and S by mixed and discontinuous finite elements. The numerical techniques described in chapter V can be generalized'to the case of multiple rock-types, provided that the finite element mesh is chosen in such a way that the TmR boundaries consist of edges of elements of the mesh,
CHAPTER
INCOMPRESSIBLE
I -
111
TWO-PHASE
RESERVOIRS
INTRODUCTION
We shall consider in this chapter the flow of two incompressible immiscible fluids through a porous medium. Though this problem is of great practical importance, as it corresponds to the simplest case of secondary oil recovery technique, where the resident oil is displaced by injected water,
it
has received only recently attention from the mathematical
community, despite the huge amount of oil engineering technical literature on the subject. The reason for this may be that even this "simple" (for oil engineers) model has such a complicated and non-standard structure that the usual mathematical tools cannot be applied in an evident manner. The key to the reduction of these two-phase equations to the more familiar system elliptic
of one parabolic saturation equation coupled with an
pressure
equation
is
a
mathematical
transformation of
the
equations, which replaces the two pressure unknowns (one per phase) by only one
pressure
unknown,
called
the
global, or
the
reduced,
or
the
intermediate pressure. This transformation was discovered independently by CHAVENT [ l ] , authors
in 1975 (q9globalpressure") and by two several some Russian
("reduced
ANTONCEV-MONAHOV).
pressure"
see
the
references
1
though
3
of
A detailed explanation is given in section I1 below,
together with a careful discussion of the choice of boundary conditions. The resulting system of equations is summarized in section 111.
As we already mentioned it, rather few mathematical results have been available for this problem. In order to organize logically these results,
we
emphasize
first
the
importance of
the
notion of water
breakthrough time, which corresponds to the time at which the injected water first starts being produced with the oil at a given production well.
Ch.III: Incompressible Two-Phase Reservoirs
90
This time is economically important, as the water oil ratio (WOR) increases very quickly after the water breakthrough time, so that the production well has to be turned off. Depending on the boundary conditions which are used for the model, the breakthrough phenomenon may or may not be well represented. One other mathematical
difficulty
associated
with
this
problem
is
that, under
standard conditions, the parabolic saturation equation is degenerate, (and practically
very
close
to
a
first order hyperbolic equation).
This
degeneracy may or may not be taken into account. Among the papers which do not take properly into account the breakthrough phenomenon, we find CHAVENT [l],
1975 (existence theorem for a
degenerate parabolic equation with simple Dirichlet and Neumann conditions, coupled with a family of elliptic pressure equations), KRUZKOV-SUKORJANSKI,
1977 (existence of classical solutions for the non degenerate problem with Dirichlet and Neumann boundary conditions), ANTONCEV-MONAHOV, 1978 (existence of weak solutions for the degenerate problem with Dirichlet and Neumann boundary conditions, plus some regularity and stability results for simplified problems), ALT-DIBENEDETTO, 1983 (existence of a weak solution for
the
degenerate
problem
with
two
unilateral
overflow
boundary
conditions, these conditions do not allow simultaneous production of oil and water) KROENER-LUCKHAUS, 1984 (existence of solutions for the partially degenerate problem with Dirichlet and Neumann boundary conditions, this author works with the original set of equations, not the transformed one). Concerning the models that take properly into account the water breakthrough phenomenon, an adequate unilateral boundary condition was formulated in CHAVENT [lbis], without an existence theorem (the main part of the paper was devoted to mathematical problems related to the estimation of the non-linearities appearing in the saturation equation). Existence theorems for the resulting degenerate variational inequality were given in CHAVENT [ Z ] for one-dimensional problems (where the pressure and saturation equations decouple), and will be given in section V of this chapter for the general multidimensional case. Regularity results, a description of the asymptotical behaviour, a precise definition and some properties of the water breakthrough time can be found in the work of GAGNEUX [l]-[4]
in the
case where the saturation and pressure equations decouple; part of these results are recalled in this chapter (still in section V).
So we b e l i e v e t h a t t h e material i n s e c t i o n V below is t o d a y ' s most
comprehensive
mathematical
treatment
of
the
two-phase
equations,
a s it
takes i n t o a c c o u n t t h e l a r g e s t number o f r e l e v a n t p h y s i c a l p r o p e r t i e s .
The reader may h a v e n o t i c e d t h a t none of t h e p a p e r s was c i t e d f o r a u n i q u e n e s s theorem. T h i s is b e c a u s e o f t h e c o u p l i n g between t h e p r e s s u r e and
saturation
r e g u l a r i t y of still have
equations,
which
the solution
t o get
such
a
theorem,
it
difficult
t h e uniqueness.
to
obtain
enough
T h i s problem is open
case, t h o u g h KRUZKOV-SUKORJANSKI claim t h e y
t h e non-degenerate
for
makes
because
they
suppose
i n the h y p o t h e s i s
that
the
s o l u t i o n is i n d e e d r e g u l a r .
Tn o r d e r t o be d e f i n i t e , we make p r e c i s e now t h e way i n which t h e reservoir described
i n c h a p t e r I w i l l b e p r o d u c s d t h r o u g h o u t Lhis c h a p t e r
(cf. Figure 1 ) : ( 1 .l )
r II
The l a t e r a l boundary
is s u p p o s e d c l o s e d ;
-
water is i n j e c t e d t h r o u g h t h e wells 1 , 2 , . . . k ,
-
(1.2)
and we d e n o t e by
k
:
=
U k= 1
rk
t h e i n j e c t i o f ? boundary.
...K, and
o i l is prodilned til-ouyh t h e r e m a i n i n g wells k+l , K
(1.3)
we d e n o t e by
rs
=
U
-
T k : t h e p r o d u c t i o n boundary.
k= k+ 1
Figure 1 : Secondary recovery of an o i l reservoir
92
Ch. III: Incompressible nYo-Phase Reservoirs
For the sake of simplicity, we will consider in the sequel the case of one
injection well and one production well
11-
CONSTRUCTION
We
first
give
OF
the
(z=l,K = Z ) .
THE
STATE
characteristics
EQUATIONS
of
the
two
fluids
(see
paragraphs 111.3 and IV.3 of chapter I) They (2.1 1
are
both
supposed
11,
chapter
where
incompressible (in opposition to
compressibility
phenomenon). Hence B = B =1 1 2-
was
.
the
driving
They are immiscible, so two distinct phases are present in the pores of the porous medium. Hence we have pressures
(2.2)
P1
and
P2
corresponding
two
respectively
distinct to
the
pressure in the wetting phase (water) and tp the pressure in the non-wetting phase (oil).
pj
> 0 is the mass of a unit volume of the
jth fluid,
j=l ,2,
(2.3)
>
!+J
0
is the viscosity of the jth fluid, j=1,2.
Let us define at each point x
6 Ci
:
@ . = flow vector of the jth fluid, j=1,2
(cf. ( 3 . 1 2 ) in chapter I for a precise definition).
s(x)
=
saturation in fluid 1 at x
=
vol of fluid 1 around x, vol of fluid 1 + 2
(2.5) 1 - s(x)
=
saturation in fluid 2 at x.
So we have now three dependant variables
:
93
I1 Construction of the State Equations
(2.6)
11.1
-
pressure of the wetting phase (water),
P1
=
P2
=
pressure of the non wetting phase (oil),
?.
=
wetting phase (water) saturation.
THE EQUATIONS INSIDE
Inside 52,
Sl :
THE NOTION OF GLOBAL PRESSURE
the Darcy law applies separately to each of the two
fluids (cf. Chapter I 3111.3.1) with a reduction
k . rJ
where the relative permeabilities
k , of permeability : rJ
depend on
S as shown in figure 7
of chapter I, and also generally on x .
-
A s can be seen in figure 7 of chapter I, the saturation S
remains always,
as long as only displacement phenomena are considered, which is the case in this chapter, between
5,
and
:
is the water residual saturation : for S 6 Sm' there is m' so little water that it is "trapped" by the capillary forces in the pores
.
of the porous medias and can no longer be displaced.
-
similarly to
1-s,
oil
is the
residual saturation and
is
interpreted
'ma
A s the rock and the fluids are supposed incompressible, the only
accumulation term is due to a change in saturation so the conservation laws
for each of the two fluids are (compare with (1.46) of chapter I)
(2.9)
O(X)
@(x)
(2.10)
a ( x ) $(x)
+
xa
i,
div = 0, V xcn, (water conservation law)
-
-t
div $2 = 0, fF x (oil conservation law)
(13)
+
:
V t c 10,TC,
c 52,
V
t E l0,TC.
Ch.III: Incompressible Two-PhaseReservoirs
94
At this point we have, in n, only two equations (2.9), the three unknown
P 1 P2
s.
and
(2.10) for
The missing equation is given by the
capillary pressure law (see (3.15) of chapter I) (2.11 1 Pl - P2 = PC(S,X), depends on
where P
-
as shown i n figure 8 of chapter I.
S
Two things are to be noted concerning these capillary pressure curves :
- They always have a positive derivative
:
(2.12)
-
S [,
- The capillary pressure vanishes for only one value
-
interval :
S ],
- P 1 - P = P =o<=> s = s . 2 c Usually Sc = SM (resp. Sc = Sm) when
(2.13) (resp.
-
Sc of the
-
non-wetting
< 5e < m
S
phase)
saturation.
However
-
is the wetting phase
S
it
can
happen
that
the reduced
water
zM.so we shall distinguish in the sequel between -Sc and -sM.
In order to simplify notation, we shall now use saturation S instead of the actual saturation S :
so that equations (2.7), (2.9),
as at
div
i,
...,(2.11)
(2.15)
@(x)
(2.16)
Q(x) ~ ( 1 - S +) div $2
(2.17)
P1
-
P*
(2.18)
$.J
=
-@(x) k.(S,x) grad [ P . - p .
+
a
=
:
0, +
=
become
=
0,
PC(S,X),
J
J
J
g 21,
j
=
1,2,
I1 Construction of the State Equations
95
where we have defined (caution than in chapter 11)
:
Q and J, represent different quantities
:
kr.(Sm+S(zM-zm),x) (2.21)
k.(S,x) J
=
=
u.
mobility of the
jthfluidj=1,2,
J
P (s,x)
(2.22)
=
P~(S,+S(Z~-S~), x)
=
capillary pressure.
In all the sequel we shall make the following assumptions on the spatial dependance of
k. and
P
J
:
The mobilities, as functions of the reduced saturation, are independant of x
(2.23)
and
k.(S,x)
=
:
k.(S),
+XER,
J
:
the
(2.24)
capillary
P (S,X) where (2.25)
(2.26)
pressure,
as
saturation, is independent of
I
I
P
=
a x
fuhction up
of
the
reduced
to a scaling factor
:
PCM(X) PC(S),
(x) 2 0 is the maximum of the absolute value of the
capillary pressure at the point x, CM pc(S) is a dimensionless function such that -1 2 pc(S) 51 and pc(Sc) = 0. With the choice (2.25) we see in figure 8 of is always an increasing function of S . chapter I that p
The hypotheses
(2.23),
(2.24)
are
very
usual, as
they
are
sufficient to provide a good model of the important phenomena f o r the oil engineers
(including
heterogeneous media).
for
example
counter-flows
by
imbibition
in
Ch. III: Incompressible nYo-Phase Reservoirs
96
However, equations (2.151, (2.16) are not suited for a proper mathematical study
:
for example
disappears as
k (1)
2
where S EO.
in
the zones of
where S = 1 ,
equation (2.16)
0; similarly equation (2.15) disappears in the zones
=
So we are now going to transforme equations (2.15) through (2.18), under hypotheses (2.23), (2.241, into a more tractable form. We shall carry out this transformation in 3 steps
Step 1 (2.27)
:
Sum equation (2.15), (2.16) to get
:
:
+ + div ($1+$2) = 0.
Use then the algebraic identity
and (2.15) to get
(2.29)
Q
:
as
- div { $ +
div {
k t k2 grad kl+k2
-
+
1: (P - P gz 1- ( P 2 - ~ 2 g z) 1 )
4
+
Using (2.17) in (2.29) and adding (2.28) to (2.31) gives
:
(2.30)
In order to simplify notation we define
:
S
(2.32)
a(S)
(2.33)
bO(S)
=
\
b
a(s) ds kl (S)
= kl
(S)+k2(S)
increasing (viscosity-’) “fractional flow1’,increasing (dimensionless)
I1 Construction of the State Equations
97
(2.34)
(2.35)
Pc (x)
(2.36)
r
-p
m g Z(x)
=
gravity potential (same dimension as a pressure).
We rewrite now equations (2.301,
(2.37)
I
=
(2.38)
@
as +
at
[PCM(x) grad a(S)+bl(S) grad PCM(X)
b2 ( S ) grad PG(x)I]
+
div ($1
[$(XI
div
+
+
G2)
=
+
+
+
div {bO(S)(01+02)l
We see that in equations ( 2 . 3 7 ) , S
= 0,
(2.38)
the only dependant + + $1+$2,
and the water + oil flow Vector
which itself depends on S, P1 and
P2
through the definition ( 2 . 1 8 ) .
The notion of global pressure (CHAVENT
:
+
0.
variables are the saturation
Step 2
-
(2.27) with the above notation
[11).
We show now that it is possible to introduce a new unknown called the global pressure, which is a point function of
so that the water+oil flow
+
+
$1+$2
S,
P1
P
and P 2 ,
can be expressed in terms of S, P and grad
-P only. Thus the number of unknowns reduces to two. We define first some notation
-
corresponding to Sc (2.391
is the reduced saturation
defined in ( 2 . 1 3 1 , so pc(Sc) S
Y(S) =
(Sc
J
1 dPc (bo(S)- 7 ) d~ ( S ) d s
=
0) :
(dimensionless),
sC
(dimensionless). One has obviously (2.41)
:
‘f(S)+Y1(S) = (bo(S)
- 71 ) pc(S).
Ch.III: Incompressible nYo-Phase Reservoirs
98
A o(=/oSa (T) dT
1 Figure 2
Figure 3
:
:
S b
: 0
1
The functions a(S) and a(S)
The shapes of the functions ,,b
b,
and
b2
s
99
I1 Constructionof the State Equations
pressure P by : P(x,t)
(2.42)
=
7 1 CP,(x,t)
+
P2(x,t)l
One checks then easily, as
pc(Sc)
+
=
Y(S(X,t))
0 and
PCM(X).
O
so
(2.44)
Let us now calculate the gradient of (2.45)
gradP
=
1
P :
grad(P1+P2) + ( b o ( S ) 2
dpc gradS + Y ( S ) grad PCM. 71 ) PCM dS
But, (2.46)
PCM
dPC dS
grad S
=
PCM grad pc(S)
=
grad [PcM P~(S)I-P, grad PCM.
Using (2.46), (2.45) and using (2.24), (2.27), (2.41) we get 1
:
1
grad P = - grad (P +P ) + ( b ( S ) - -1 grad (P1-P2)-Y,(S)grad PCM. 2 1 2 0 2 which using the definition (2.33) of (2.47)
( k l + k 2 ) grad P
Multiplying (2.47) by (2.48)
Go
=
=
bo
reduces to :
k l grad P 1 + k2 grad P2 - ( k l + k 2 ) Y1
(s) grad
PG.
I$, we obtain
il+i2-I$(x) =
d(S) [grad P+Y1(S) grad PCM+Y2(S)grad Pc}
where we have introduced the following functions
:
100
Ch. III: Incompressible nYo4hase Reservoirs
(2.59)
I”=
k,Pltk2 P2
.-1 Pm
k1+k2
I
I
I I
I
I
Figure 4 Remark 1 :
We
obtain
global
:
from
-2
equation in
c a p i l d r y pressure (2.48) redlJCRS t o
the
b
1
The d and Y
gresnure,
S
I
0 functions
(2.48)
the
special
i n t e r p r e t a t i o n of
case
where
PCM(x) is c o n s t a n t o v e r
the
the
maximum
0 ; i n t h a t case
:
(2.51 1 which,
compared
to
the
one-phase
Darcy
Law
(3.3)
of
c h a p t e ? T, shows t h a t t h e g l o b a l p r e s s u r e f i e l d , a p p l i e d t o a f i c t i t i o u s f l u i d of v i s c o s i t y !J s u c h t h a t krl kr 2 k l P l + k 2 P 2 , yields - = d = -+ -and of d e n s i t y p = k l + k2 !J 9 !J2 e q u a l t o t h e sun of t h e o i l a n d water flows. flow
Step 3 : Let u s d e f i n e t h e f o l l o w i n g vect7r’ f i e l d s (2.52) (2.53)
G1
=
-+(x) g r a d PCM
=
- + ( X I g r a d PG
:
(flow vector
x
m o b i l i t y -1 1,
(flow vector
x mobility-’
),
a
101
II. Construction of the State Equations
which are known (as
@,
PCM
and PG are given).
Then equations (2.37).
as at
(2.54)
Q
(2.55)
r' =
(2.56)
div qo
(2.57)
Go
+ div {r
+
+
2
1
j=o
=
=
(2.48) can be r e w r i t t e L
GJ. }
b.(S) J
-$(x) PCM(x) grad +
=
0,
a(S),
0,
-$(x) d(S) gradP
+
d(S)
2
1
Gj.
Yj(S)
j=l
which are the sought equations in The
(2.38),
saturation
0
equation
for S and P. (2.541,
(2.55)
is
a
non-linear
diffusion-convection evolution equation. The diffusion term (second term of (2.54) is degenerate (as a' = a vanishes for S=O and S=l as one can see in figure 2) or even missing when the capillary pressure is neglected. The transport term (last term of (2.54) is non linear and not necessarily monotone (the fractional flow bo is monotone increasing, but bl
and
b2
are not as shown in figure 3). The pressure equation (2.56),
(2.57) is a family (one for every
t c lO,T[) of elliptic equations.
The
saturation
equation
(2.54)
and
the
+
pressure
equation
(2.56),(2.57) are coupled through the presence of qo in the 1 in the right hand side of (2.54), and the presence of S in the coefficients of (2.57). Remark 2:
+
One tempting thing to do is to replace in (2.54) qo by its expression (2.57) and to rearrange the terms. This manipulation has to be avoided, in view of the forthcoming numerical resolution of (2.54) through (2.57)
:
in order to
be able to solve properly the pressure equation (2.56) we shall use a mixed finite element method (see chapter V), + which computes an approximation of the global flow qo. Oh
6
102
Ch. In: Incompressible nvo-t'hase Reservoirs
This
approximation
+
qOh
will
be
simply
plugged
in
the
saturation equation (2.54). Hence we shall keep the above form of the equations. Moreover, notice that in the 1-D case,
6o
is
constant
in space
(cf.
(2.56)) and the proposed
manipulation has obviously no interest in that case. 0
+
+
The water and oil flow vectors are then related to S, r, q j , j
=
0,1,2 by +
(2.58)
0,
+
+
+
02 = q0'
(2.59)
and the water and oil pressures (2.62)
P1
=
P - [YG) -
(2.63)
P2
=
P
11.2
-
-
[Y(S)
+
P,
and
P2
are then given by
:
1
7 Pc(S)1 PCM, 1
7 PC(S)I
PCM.
THE PRESSURE BOUNDARY CONDITIONS
III (see figure 1 ) we
As we indicated at the beginning of chapter
have drawn in figure 5 the partition of the boundary into three parts re,
r
of the porous medium
rll and r corresponding respectively to injection well
boundary, lateral boundary and production well boundary.
II. Consmction of the State Equations
103
injection we1I
Figure 5 : The pressure boundary conditions.
As
the global
i~r*~?,si~r’eP
i r governed by a family of elliptic
equations, and is related to the global (water+oil) flow, the corresponding boundary
conditions are
monophasic case
similar to thora [used in chapter I1 for the
:
- Bzundary conditions on the lateral boundary The field being supposed closed on vanishes on
:
the (oilcwater) flow
i.e. using (2.58)
+ + q o - v = O
(2.64)
Remark 3
rR
r,,
TL:
on
rR
(Neumann Condition for ( 2 . 5 6 ) , ( 2 . 5 7 ) ) .
Water drive with given pressure or flow conditions can be taken into account by straightforward modification of
the
boundary conditions.
- Boundary boundaries As
a
conditions
on _the injection and
production well
r e -and rs: for the monophasic case, we shall not distinguish, for the
pressure boundary conditions, bet,.ieerl i rijection and production wells
:
we
Ch. III: Incompressible no-Phase Reservoirs
104
rw
s h a l l d e n o t e by
rw and
re
either
rs,
or
by
rw
s E
t h e c u r r e n t p o i n t on
Iw = rwx]O,T[.
oil
The
injection
and
field
is
production
coupled, wells,
through t h e
for
which
we
boundaries, shall
t o the
suppose
a
that
mathematical model is a v a i l a b l e . Such a well model e n a b l e s u s t o c a l c u l a t e , a t every t i m e
t,
the trace
up t o a c o n s t a n t ,
on
Pe
Tw
of t h e p r e s s u r e i n s i d e t h e well,
knowing o n l y t h e f l o w - r a t e h i s t o r y of t h e two f l u i d s
t h r o u g h every p o i n t of t h e well boundary
rw
:
unknown c o n s t a n t , V ( s , t )
6
Z
W.
We g i v e f i r s t some s i m p l e examples o f such models, which u s e t h e h y d r o s t a t i c p r e s s u r e law i n s i d e t h e well b o r e h o l e
:
- water injection w e l l : t h e d e n s i t y of t h e f l u i d i n t h e well is t h a t of w a t e r ; hence t h e h y d r o s t a t i c p r e s s u r e law y i e l d s P ( s ) = p 1 g Z(x)
(2.66)
-
:
unknown c o n s t a n t .
+
production w e l l : h e r e t h e d e n s i t y o f t h e f l u i d i n s i d e t h e well
depends on t h e l o c a l p r o p o r t i o n of the two f l u i d s . We t a k e f o r d e n s i t y i n t h e well b o r e h o l e t h e water
and o i l d e n s i t i e s weighted by t h e i r
local
p r o d u c t i o n r a t e t h r o u g h t h e boundary : +
+
+
P1S1 ' k + P 2 O 2 ' V (2.67)
P(X,t)
@,
=
+
+
+
.V+O2.V
and
r',
Usually
+ qo
+
+ + + + Pl$l ' v + P 2 $ 2 " J
+
-
+ + q0-v +
is very l a r g e n e a r t h e well i n comparison t o ql,
s o t h a t (2.67)
is g e n e r a l l y approximated by ( c f .
(2.601,
+ q2
(2.61)
and (2.33)) : (2.68)
p(s,t)
k l ( S (3, =
t ) )p l + k 2 ( S ( s , t )
p2
k l ( S ( s , t ) ) + k 2 ( S ( 3 , t) )
Note t h a t t h i s d e n s i t y is t h e d e n s i t y , on t h e well boundary f i c t i t i o u s e q u i v a l e n t f l u i d d e f i n e d i n remark 1 .
rw,
Of
the
II. Constructionof the State Equations
L e t , so e Tw
105
be any r e f e r e n c e p o i n t on t h e well boundary ( a t t h e bottom
f o r example f o r
t h e v e r t i c a l models).
One can t a k e as p r o d u c t i o n well
model :
-
( s is t h e n t h e c u r v i l i n e a r a b s c i s s a on
f o r 2D models S
(2.69)
Pe(s,t)
-
1
=
Tw)
p ( s , t ) g d Z ( s ) + unknown c o n s t a n t ,
f o r 3D models
(2.69bis)
Z(S)
Pe(s,t)
1
=
p ( Z , t ) g dZ
+
unknown c o n s t a n t ,
Z(So)
where
t
(2.70)
p(Z,t)
rz
=
=
1 measr
rz
p ( s , t ) ds,
{ s ~ r ~ I z (= sz)] .
I n summary, we see
t h a t , once t h e c o n s t a n t is f i x e d i n ( 2 . 6 6 ) ,
(2.69) o r ( 2 . 6 9 b i s ) , t h e well model e n a b l e s u s t o c a l c u l a t e t h e p r e s s u r e Pe everywhere on rw assuming t h a t t h e d e n s i t y p i n t h e w e l l is known. T h i s is a c t u a l l y t h e case
for
i n j e c t i o n wells,
where p
is t h e i n j e c t e d f l u i d
d e n s i t y , and f o r 2D-horizontal models, where Z ( s ) is c o n s t a n t and hence Pe t o o , and t h i s is a l s o a p p r o x i m a t e l y t h e case f o r p r o d u c t i o n wells, where p is commonly e v a l u a t e d u s i n g (2.68) w i t h t h e s a t u r a t i o n S e v a l u a t e d a t t h e p r e v i o u s time s t e p . Hence we w i l l assume from now on t h a t t h e p r e s s u r e P up t o a c o n s t a n t , on a w e l l boundary
is known,
r
.
We g i v e now t h e boundary c o n d i t i o n s which c a n be used f o r
Ew
e-
Tw:
i f we know t h e bottom p r e s s u r e
i n t h e w e l l , we know
u s i n g t h e well model
The c o n t i n u i t y of t h e p r e s s u r e i n one
(2.65).
P
everywhere on
i n j e c t e d o r produced f l u i d (water f o r example) a c r o s s t h e well boundary shows t h a t most
the
P,
Irw
=
capillary
Pe.
But t h e g l o b a l p r e s s u r e P d i f f e r s from
pressure
(cf.
(2.44)),
which
is u s u a l l y
P,
by a t
s m a l l when
compared w i t h t h e p r e s s u r e drop a c r o s s t h e f i e l d . Hence we s h a l l n e g l e c t t h i s d i f f e r e n c e and t a k e as boundary c o n d i t i o n
:
106
Ch.III: Incompressible TwoPhase Reservoirs
P Remark 4 :
=
on
P
Tw
, a.e.
on lO,T[
(Dirichlet).
If one does not want to make this approximation, then the above condition has to be replaced, to ensure continuity of the water pressure, by (cf. (2.62)) : p
=
1
pe
+
[Y(S) - 7
PC(S)I PCM. 0
-
if we know at every
(s,t)
E
Zw the (water+oil)
flow rate density
Q(s,t) then we take as boundary condition : + + qo-v
=
, a.e
Q on rw
on
]O,T[
(Neumann).
(This condition can be used practically only for 1-D models, or 2-D models with an hypothesis on the repartition of the well production rate on - if we know only the overall injection or production rate
well, the
boundary condition is + + qo-v
QT(t)
=
'
QT(t) of the
:
a.e. on
lO,T[,
rW
P
rw).
(well-type condition) =
Pe
+
unknown constant,
rw
is given by the well model (2.65).
where P
Finally we have only three types of global pressure boundary conditions boundary rD,
rN
(2.71)
r
of Q (as in the monophasic case) into three corresponding parts
and
i
Dirichlet, Neumann and well-type ; so we shall partition the
:
rw
:
rD
=
the part of
r
rN
=
the part of
r where Neumann conditions hold for P,
rw
=
the part of
r
where Dirichlet conditions hold for P,
where well-type conditions hold for P.
107
I1 Construction of the State Equations The global pressure boundary conditions are then :
(2.72)
where
P
is the given exterior pressure
Q(s,t) (2.73)
=
:
TN
lO,TC,
Z"
overall (water+oil) production rate through at time t
Remark 5
r D X
(water+oil) production rate density at the point
=
(s,t) Q (t) T
on
E
rW
l0,TC.
includes generally the lateral boundary, where Q=O; Q is
positive when something is produced through the boundary, and negative when something is injected through the boundam. Remark 6 :
As
the global pressure equation is stationnary, the global
pressure is well defined only if (2.74)
the r-measure of
rD
is non zero,
which we shall generally assume in the following. If (2.74) is not satisfied, then one has to assume that Q(s) d s
(2.75)
+
QT
=
0
rN
in order to be compatible with the incompressibility of the two fluids; then the pressure
P
is only defined up to a
constant. 0
Ch. III: Incompressible nvuo-phaseReservoirs
108
11.3 - THE SATURATION BOUNDARY CONDITIONS The p o s s i b l e c h o i c e s f o r t h e s a t u r a t i o n boundary c o n d i t i o n a t a point s of
r
are d i f f e r e n t d e p e n d i n g o n w h e t h e r t h e g l o b a l ( o i l c w a t e r ) f l o w
is d i r e c t e d i n w a r d or o u t w a r d i.e.
+ + whether q O * J is n e g a t i v e
u s i n g (3.58),
or p o s i t i v e . So we i n t r , , i I i - ~ e r i p s t ( c f . f i g . 7 ) :
(2.76)
r-
=
{s e
r
I
r+
=
{s
r
I
E
+
+
+
+
qo-'j 2 O ] = i n j e c t i o n boundary,
qo-v
>
01 = p r o d u c t i o n bolundary.
V
R
F i g u r e 6 : The p a r t i t i o n of Remark 7 :
r
into
r-
For a f i e l d w i t h c l o s e d o u t e r b o u n d a r y , made up of t h e o u t e r boundary boundary well
re
and
boundary
imbibiLion
or
r+
rs.
r
r+
and
r-
is g e n e r a l l y
and t h e i n j e c t i o n w e l l
.?
is g e n e r a l l y made up of t h e p r o d u c t i o n Xowever,
laboratory
in
some e x p e r i e n c e
displacements,
the
such
notion
as of
i n j e c t i o n and p r o d u c t i o n boundary may n o t be c l e a r l y d e f i n e d ,
so we u s e ( 2 . 7 6 ) One
key
for,
i,+le
dilicil
choice
:?.?not, g i v e r i s e t o a n y a m b i g u i t y .
of
physically
admissible
saturation
boundary c o n d i t i o n s is t o remember t h a t t h e y have t o s a t i s f y t h e r u l e of p r e s s u r e c o n t i n u i t y f o r e a c h o f t h e f ' l ? l i d ? ?'Loding d w o s s t h e boundary
:
II. Construction of the State Equations
109
- if only one fluid is flowing across the boundary, its continuity is ensured by the pressure boundary condition, and the boundary saturation is unimportant. - if two fluids are simultaneously flowing across the boundary, their two pressures unique pressure
P1
Pe
and
P2
r
on
have to be both equal to the
so that using the property (2.13) of the
existing outside of the porous medium Sl,
on I", i.e. P -P = O 1 2 capillary pressure curve, this implies
necessarily
s=SC-
(usually S = 1
if
is the
S
wetting fluid saturation). So a necessary condition for our saturation boundary conditions to be physically admissible is that they satisfy
11.3.1
-
Saturation boundary conditions on the injection boundary
One can use for
-
11.3.1.1
where
:
r- two types of boundary conditions
r-
:
Dirichlet condition :
Se is a given boundary saturation.
Remark 8
The boundary condition S=S
C'
which satisfies (2.77), can be
used, in the usual case where
S
=1,
for the modelling of
water injection. However, when the capillary diffusion term div
f
in
(2.54)
is
non-zero,
this
boundary
condition
generally leads to a production of oil through the water injection
boundary;
though
this
oil
production
can
be
observed under certain experimental circumstances, it is not present under the usual field conditions. So condition (2.78) has to be used for the modelling of water injection at high injection rates only, for which the parasitic oil production occurs only during the very short period when the porous
Ch.HI: Incompressible f i d h a s e Reservoirs
110
medium is not yet saturaLed with water in t h e vinicity of the injection boundary I#--( cf. fig.")
- oil
0
-
4-'p2
9'
+
+ oil +water
'0
st
no oil flow
- -water -
,water
'1
,oil+water
'0
oil +water
!
1
0
r-
spacell
spacer,
AT INITIAL
DURING
SHORT PERIOD
TIME Figure 7
A
:
space
LATER
Approximate modelling of water injection by a S=1
Dirichlet boundary condition in case of high injection rates Remark 9 :
Conditions ( 2 . 7 8 ) with
Se
does not necessarily satisfy (2.77). However the condition S=O
can be used for,the modelling of oil injection high
injection rates, as the p a r a s i t i c water production (during when (2.77) is violated) occurs only during the short period where the porous medium is not yet saturated in oil in the vicinity of 11.3.1.2
-
r-.
Given water injection rate :
0
111
II. Construction of the State Equations where
:
Q.(s,t)
jth fluid production rate density at point
=
(2.80)
Remark 10
s and
is negative in case of injection).
:
On the part of
fluid), -
+ + qo-v
T- where
< 0 (global injection of
one can model injection of
pure water or Oil by
taking the following values for Q , in (2.79)
+
(2.81)
Q1
=
(2.82)
Q
=
1
+
qo-w < 0 0
<=>
<=>
Q2
0
=
:
water injection,
oil injection.
A water injection rate
Q1
different from (2.81) or (2.82)
would generally lead to simultaneous flow of water and oil
Remark 1 1
:
and hence violate (2.77) as we could not ensure that
S
would remain equal to 1.
0
On the part of
r-
where
+ + qo*v
=
0
(globally closed
boundary) one can model closure conditions by taking Q 1=0, which automatically ensures Q2=0. Remark 12
:
In the case where the overall water injection rate Ql,(t) known through only a part
rw
of
is
r-, (2.79) can be replaced
by
(2.83) IS
=
unknown constant on
rW.
The second condition of (2.83) is mathematically and practically convenient, but has, to my knowledge, no physical justification.
0
112
Ch. III: Incompressible llvo-f'hase Reservoirs
-
11.3.2.
One c a n u s e f o r
-
11.3.2.1
where
r+
Saturation boundary conditions on the production boundary
r+
three t y p e s of boundary c o n d i t i o n s :
Dirichlet conditions :
Se
is a g i v e n boundary s a t u r a t i o n .
Remark 13 :
For
a
displacement
s a t u r a t i o n is S
=
can
0
without
of
used
be
oil
over
0
violating
by
water,
the f i e l d ,
where
condition
the
initial
(2.84)
with
b e f o r e t h e w a t e r b r e a k t h r o u g h time,
(2.77).
However
condition
this
is
not
a d m i s s i b l e a f t e r t h e b r e a k t h r o u g h time, when o i l and w a t e r flow simultaneously. 0
Remark 14 :
The D i r i c h l e t c o n d i t i o n ( 2 . 8 4 ) w i t h :
always s a t i s f i e s (2.77) However,
generally leads, term,
to
and t h u s is p h y s i c a l l y a d m i s s i b l e .
b e f o r e t h e water b r e a k t h r o u g h time t h i s c o n d i t i o n i n t h e p r e s e n c e of
intrusion
p r o d u c t i o n boundary experimentally
of
water
r+ ( c f . f i g .
observed.
in
a capillary diffusion field
the
through
the
8 ) , which is g e n e r a l l y n o t
However,
this
s i m p l e and l e a d s , t o g e t h e r w i t h S=S
on
condition
r-
is
very
( c f . Remark 8 ) t o
s i m p l e homogeneous D i r i c h l e t boundary c o n d i t i o n s f o r t h e o i l saturation
Remark 15 :
1-S
i n t h e u s u a l c a s e where Sc
We have s e e n t h a t t h e water
S=O
=
1.
is a good c o n d i t i o n on
b r e a k t h r o u g h time (remark 1 3 ) and t h a t
0
r+
before S=S
is
s a t i s f a c t o r y a f t e r t h e b r e a k t h r o u g h time (remark 1 4 ) . Thus a more r e a l i s t i c c o n d i t i o n would be t o t a k e t o 1 a t t h e b r e a k t h r o u g h time. Dirichlet
But
this
Se jumping from 0
is n o t a s i m p l e
c o n d i t i o n , as t h e b r e a k t h r o u g h time is n o t known
II. Construction of the State Equations
113
-
oil
oil%
2 ‘
4
water c1‘-
no water flow
oi i+water -3L
-
&water
n
1
D
Oh
initial time Figure 8
:
r+r-
space
stationary curve
space
transitory period
The effect of an
production boundary
S-Sc
T+
b e f o r e h a i d . 31,
L 4
oil +water
r+
later
Dirichlet boundary condition on the
f o r a sufficciently long 1-D field.
W H Iirtvc? t n
u s e t h e u n i l a t e r a l c o n d i t i o n which
we describe now.
0
- Unilateral boundary condition
11.3.2.2
On t h e p r o d u c t i o n boundsry w e t t i n g phase
S
Y+, a s long a s t h e s a t u r a t i o n of t h e is s t r i c t l y less t h a n Sc, we see from (2.77) t h a t
e i t h e r water o r o i l , but n o t b o t h , can flow o u t of forces,
the w e t t i n g phase
(irater,
i n c a s e of
i n s i d e t h e porous media as l o n g a s t h e s a t u r a t i o n less t h a n
(2.85)
Sc,
s o t h a t we g e t
s < sc
+
+
$, .LI
>
=
:
0
on
R.
Due t o c a p i l l a r y
w a t e r + o i l flow) is r e t a i n e d
r+.
S
remains s t r i c t l y
114
Ch. III: Incompressible W o P h a s e Reservoirs
S becomes equal to Sc, then water is allowed to get out of
When
the field : this is the breakthrough time. We shall suppose that no water
r+,
is available outside the porous medium on only be directed outward from +
(2.86)
$l'v
+
so that the water flow can
:
r+.
on
2 0
Thus conditions (2.85) (2.86) give the unilateral production boundary condition :
(2.87)
on (Sc-S)
t (s) BT
(2.88)
+
$,*V
=
+
=
r+,
f
t
6
10,T[,
0,
first time when S becomes equal to S at s c
r+.
J
We have drawn for the case of boundary condition (2.871, typical variations of S with time at s
E
r+
(figure 9) and with space for given
times t (figure 10). Remark 16 :
In the usual case where
S
=1, C-
one has simultaneously on
r+
after the breakthrough time : (2.89) which implies, as condition boundary
a(l)=O,
introduces a
r+
that
boundary
as au
=
~
3
.
So the unilateral
layer on
the production
after the breakthrough time (cf. figure 1 0 ) . 0
I1 Constructionof the State Equations
115
s4
Figure 9 : Evolution of the water saturation on the production boundary
r+
with the unilateral boundary condition (2.87).
+-
P1.v >o
4 -
P1.v = 0
space phase @ of figure 9
r+
space
phase @ of figure 9
r+
space
phase @ of figure 9
Figure 10 : Profiles of the water saturation inside the porous medium D near the production boundary
r+
at different times, when
the unilateral boundary condition (2.87) is used.
r+
116
11-3.2.3
Ch. III: Incompressible n v d h a s e Reservoirs
- Given water/oil production ratio (MR). A
r+
widely used condition for
is obtained by neglecting the
boundary layer introduced by the unilateral condition, and by requiring
r+ be proportional to their respective
that the water and oil flow through mobilities on
l‘+
:
+ - +
$l.v - = + + @2*v
(2.90)
kl(S)
a~ t
E
lO,TC,
k2(S)
or equivalently : + +
r-v +
(2.91)
1
+
+
b.(S) qj.v J
j = l ,2
=
0 on
r+
fF t c 10,TC.
This boundary condition does not satisfy ( 2 . 7 7 ) . +
Since the water+oil flow field
qo
becomes very large near the
well as the diameter of this latter is very small, one checks easily from (2.60),
(2.61)
that near the well (but outside of the boundary layer) one
always has, whatever the saturation boundary condition is, that + kl( S )
$1
(2.92)
i,
z
:
k2(S)
which shows that condition ( 2 . 9 0 ) or ( 2 . 9 1 )
and the unilateral condition
become equivalent for small well diameters. Remark 17
:
It can be a l s o convenient to use a slight variant of ( 2 . 9 0 ) or ( 2 . 9 1 ) . Let
(2.93)
can take
:
1 (:-;
+
1
‘w
j=1,2
S
=
rw
b.(S) 3
c
r+
be one production well boundary. One
+ - +
qjw)
unknown constant on
=
0
‘W
t t c l0,TC V t
E
10,TC.
We shall use this condition to find the equivalent point source model for the saturation equation.
0
III. &Summary of Equations
111-
117
OF
SUMMARY
EQUATIONS
INCOMPRESSIBLE
111.1
-
=
rock porosity,
(3.2)
K(x)
=
rock permeability,
(3.3)
o(x)
=
section of field,
(3.4)
Z(x)
=
depth.
P
(3.6)
P
111.3
1 2
S
-
FLOWS
ROCK
Q
.. .at point x
Q
PHYSICAL U"0WWS
(3.5)
(3.7)
AND
CHARACTERISTICS DEPENDING ONLY ON THE RESERVOIR
@(x)
-
THO-PHASE
FLUIDS
(3.1)
111.2
OF
=
pressure inside the wetting phase,
=
pressure inside the non-wetting phase,
=
wetting-phase saturation.
CHARACTERISTICS DEPENDING ONLY ON THE FLUIDS
of fluid j , (3.9)
(3.11)
p. =
j = l,2
viscosity
The two fluids are incompressible, i.e.
p1 =
cste,
p2 =
CSte
FOR
118
Ch. III: Incompressible nvliuoPhase Reserwoirs
111.4 CHARACTERISTICS DEPENDING BOTH ON FLUIDS AND ROCK
(3.12)
(3.13)
-
Sm(x)
-
SM(x)
=
wetting phase residual saturation at point x
=
1-non-wetting phase residual saturation at point x E R ,
ZM(x)]
k .(s,x) : [sm(x),
x
R
[O,l]
-f
E
R,
relative permeability
=
(3.14) of fluid j,
(3.15)
P1
-
P
2
=
j=1,2,
P (2.x) : [sm(x), sM(x)] c
x
R
+
R
=
capillary pressure.
Hypothesis : If one introduces : S(x,t) - s,(x) (3.16) S(x,t) = then we suppose that k.(S)
(3.17)
J
=
reduced saturation of the
- Sm(X)
SM(X)
wetting phase,
:
-
kr.(Sm(x) =
+
S(FM(x) - sm(x)), x) =
u J.
mobility
is a function of the reduced saturation S only and that there exist functions PCM(x) and pc(S) such that :
where : PCM(x) 2 0
(3.19)
=
maximum of the absolute value of the capillary pressure at
(3.20)
I
x
E
R,
reduced capillary function (dimensionless, increasing), with -1
S pc(S) 5 1,
pc(Sc)
=
0 (where usually
Sc=l).
III. Summary ofEquations
(3.21)
119
PCM (x)
maximum capillary pressure at x c i2 (defined in
=
(3.19)),
(3.22)
PG (x)
(3.23)
Q(X)
=
(ZM(XI
(3.24)
$(x)
=
o(x) K(x),
(3.25)
;,(x)
=
-$(x) grad PCM(x) (governs the effects of capillary
(3.26)
G2(x)
=
-$(x) grad PG (x) (governs the effects of gravity).
-p
=
m @(x)
=
gravity potential at x
- Sm(X)l
6 0,
o(x) $(X),
pressure heterogeneity),
111.5
-
AUXILIARY DEPENDANT VARIABLES +
(3.27)
$
1
=
flow vector of the wetting phase,
=
flow vector of the non-wetting phase,
(3.28)
+ $2
(3.29)
r
=
part of
+ qo
=
+ + $ l + O2
+
(3.30) 111.6
-
+
-t
$
=
1
and of -$2 due to capillary diffusion,
global flow vector.
TRACES ON r=an OF THE DEPENDANT VARIABLES They may be known o r unknown depending on the type of the boundary
condition used.
(3.31)
Pe
=
trace of pressure, (index e stands for "exterior")
(3.32)
Se
(3.33)
Q1
(3.34)
Q2
=
trace of " * v
(3.35)
Q
=
trace of qo * =~ global production rate density
=
trace of saturation, trace of
=
+
$1
+
+
-
=~
=
+
wetting-phase production rate density through r , non-wetting phase production rate density through r , through
Caution
'
Q1
9
Q2
r.
and Q are negative in case of injection into R .
Ch. III: Incompressible ThwPhase Reservoirs
120
r
111.7-PARTITIONS OF THE BOUNDARY
(3.36)
r
=
(3.37)
rD
=
OF THE POROUS MEDIUM Q
r D u rN u r W
where (cf. fig. 1 1 ) :
r where the pressure is specified (D stands
part of
f o r Dirichlet),
rN = part of r where the global flow is specified (N
(3.38)
stands for Neumann),
rw = part of r through which the overall global flow is
(3.39)
specified (W stands for well),
r- u r+ with r-
(3.40)
r
(3.41)
r-
=
[
s
6
rl
Q
=
+ + q o - L 5 0 }=global injection boundary,
(3.42)
r+
=
{
s
6
rl
Q
=
+ + qo-v
111.8
=
n
>
r+
0 where (cf. Fig. 1 2 )
0 }
=
:
global production boundary.
- FUNCTIONS AND COEFFICIENTS DEPENDING ON -
REDUCED SATURATION S
ONLY
(3.43) (3.44)
k.(S)
p (S)
mobility of jth fluid (defined in (3.17)
=
J
reduced capillary function defined in ( 3 . 1 8 ) , ( 3 . 2 0 ) ,
=
k l k2 dPc kl k2 dS ’
(3.45)
a(S)
=
(3.46)
a(S)
=
(3.47)
bo(S)
=
kl ,
(3.48)
bl(S)
=
-PCW
positive ,
+
S
a(t) dt,
increasing,
0
increasing (fractional flow),
k t + k2 k l k2
I
k\+k2
(3.49)
j=1,2,
bZ(S)
=
k1
k2
k1 + k2
(P,
- P2) Pm
’
121
III. Summary of Equations
reduced saturation f o r which pc(Sc)
(3.50)
S
(3.53)
Y2(S)
(3.54)
d(S)
=
=
kl p1
+ k 2 p2
(3.56)
-
k l + k2
1 Pm
(cf. 3.20)),
0
’
(kl + k2).
=
111.9 - MAIN DEPENDANT VARIABLES
(3.55)
x
=
:
S
=
reduced saturation (defined in (3.16)),
P
=
-21
(P +P 1
2
) +
PCM Y(S)
=
global pressure.
111.10 - EQUATIONS FOR PRESSURE, SATURATION AND PLOW VECTORS
Equations governing the global pressure P f o r every t
- Inside
C0,Tl
k l :
+
(3.57)
div qo
(3.58)
Go
-
E
0,
=
2 =
-$d(S) gradP
On the boundary
r
P on r
P
(3.60)
qo*v = Q on
+
e
1
j=l
Yj(S)
rN
=
P
+
=
TD u TN u
rw
:
(Neumann),
r’ P
r
Gj.
(Dirichlet),
D
+
d(S)
(we use the partition
(3.59)
=
+
unknown constant on
(well-type)
rw.
:
Ch. III: Incompressible nvvo4hase Reservoirs
122
Equations governing the saturation S inside Q
(3.63)
-
-
=
OxlO.T]
r’ =
:
-$PcM grad a ( S ) .
On the boundary
r
(3.65)
-
S
=
=
(3.66)
S
=
S
+
:
(Neumann);
Q1
on the global production boundary
or
r,) :
(Dirichlet),
S
+ + @,-v
r = rr- one can take
(we use the partition
on the global injection boundary
(3.64) or
:
r+ one can take
:
(Dirichlet), +
+
+
(unilateral), (3.67) S 5 Sc, @1. V , ( S c - S ) q , * L = 0 or + + + + (3.68) r-v + 1 b. qv: = 0 (WOR equal to mobility ratio). j=1,2
- At the initial time t=O (3.69)
S
=
So(x)
:
on a.
III. Summary of Equations
123
Figure 12 : Example of saturation boundary conditions compatible with the pressure conditions of fig. 1 1 when Pe is constant over
rD-
Ch. III: Incompressible nyO-Phose Reservoirs
124
Equations for separate phase pressures and flows (3.70) (3.71) (3.72) (3.73)
P1
=
P
-
[Y(S)
-
1
PC(S) 1 PCM’
:
IV. An Alternative Model for Diphasic Wells
IV - A N
125
ALTERNATIVE MODEL F O R D I P H A S I C YELLS
As we have done i n c h a p t e r I1 f o r t h e case of monophasic wells, we
s h a l l t r y here t o r e p l a c e t h e boundary c o n d i t i o n s used i n s e c t i o n s I1 and
I11 f o r t h e model f o r d i p h a s i c wells by p o i n t s o u r c e s a p p e a r i n g i n t h e r i g h t hand s i d e of t h e e q u a t i o n s ( d i s t r i b u t e d s o u r c e s can t h e n be o b t a i n e d by approximating t h e D i r a c f u n c t i o n s ) . The d i p h a s i c e q u a t i o n s being much more i n t r i c a t e t h a n t h e monophasic o n e , we s h a l l proceed f o r m a l l y o n l y .
We c o n s i d e r t h e t y p i c a l s i t u a t i o n of a c l o s e d f i e l d i n j e c t i o n well DE- and one p r o d u c t i o n well D
E+
that :
QE
w i t h one
( c f . f i g u r e 1 3 ) , and suppose
where a , b are t h e ' l c e n t e r s l l of t h e i n j e c t i o n and p r o d u c t i o n wells.
We production
take well
as p r e s s u r e the
boundary c o n d i t i o n s a t t h e
well-type
condition
(3.61),
where
( w a t e r + o i l ) i n j e c t i o n and p r o d u c t i o n r a t e , and where on
r
EL
of a given regular function.
boundary
e
Q T ( t )is t h e e q IS t h e t r a c e
We t a k e t h e n as s a t u r a t i o n boundary c o n d i t i o n on t h e p r o d u c t i o n TE+
aDE+
=
the
r a t i o " c o n d i t i o n (2.90).
variant
(2.93)
of t h e l'WOR e q u a l t o m o b i l i t y
which becomes e q u i v a l e n t ( a t l e a s t f o r m a l l y ) t o
t h e u n i l a t e r a l c o n d i t i o n when
E+O.
On t h e i n j e c t i o n boundary
The e q u a t i o n s g o v e r n i n g t h e p r e s s u r e the saturation SE
-
( d e f i n e d up t o a c o n s t a n t ) and
are t h e n : 0 with
=
qoE
PE
rE-=aDE- we
QIT (condition 2.83)).
t a k e as g i v e n t h e o v e r a l l water i n j e c t i o n r a t e
(4.2)
P =P
i n j e c t i o n and
=
0
;
OE
on
r.
=
- @ d E gradPt
+t
6
]O,T[,
+
dE
/ J=1
+ YjEqjE,
i n RE,
126
Ch. 111:Incompressible n v d h a s e Reservoirs
i
J
(4.3)
sly
-Q ,T ,
=
*
$,€
I
+.+ (r.v
+
E+
s
Let
so
=
ii
=
c o n s t a n t on
zE-,
E-
I. E-
=
2 + + 1 b. 9 . j=1 J C J E
on R
SI
=
c o n s t a n t on L
E+
E+
be t h e r e s e r v o i r ; we u s e t h e same kind of
DE+
t e c h n i q u e a s i n c h a p t e r 11, s e c t i o n 111, but f o r m a l l y o n l y .
We go f i r s t t o t h e l i m i t i n t h e p r e s s u r e e q u a t i o n ( 4 . 2 ) : The p r e s s u r e P
where
v
satisfies, f o r every
depends on
w
r E J.
$dE
gradvE) av
av
>=
8VE *dE
a\,
t
E
]O,T[
:
by :
- d i v ($d
=
’
at t=O.
-
D
R~
o ,
=
=
o ,
w,
vElr
,
=
constant,
EJ
0 , on
r.
One can go t o t h e l i m i t f o r m a l l y i n (4.7) i f
j =
+,-,
IV. An Alternotive Model for Diphasic Wells
Figure 13
:
127
The field used to determine equivalent point sources
which is expected if
v.+
and the functions whose
Pet+
are the traces are
regular enough. We get then
(4.10)
-1 Pw
=
Q,(t)
v(a
E
LZ(Sz)
s.t
n
Y
w
-R
w
=
0.where v depends on w by
:
Equations (4.10) and (4.11) inem that P is the (ultra weak) Solution
Of :
128
Ch. III: Incompressible nYoPhase Reservoirs
-
+ divqO = Q , ( t ) (4.12)
Go
=
+
+
6(x-a) - Q,(t)
-+d gradP
+
d
2
1
j=l
6(x-b)
in
i,
on
r,
+ Yj qj,
(lo'" = 0
which is very similar t o t h e monophasic r e s u l t .
We go now t o t h e l i m i t saturation
where v
in
s a t u r a t i o n equation (4.3).
the
satisfies :
SE
depends on w by :
av -a 2 at
-
'CM
d i v ( $ PCM a E gradv avE
= W
=
in
QEp
on 1,
(4.14)
\
vE(T)
=
0.
Going f o r m a l l y t o t h e l i m i t i n ( 4 . 1 3 ) , ( 4 . 1 4 )
where
v
depends on -0
(4.16)
J,
w
2at
by : d i v ( + PCM a g r a d v )
P c M a av -=O
v(T)
av
=
0.
yields :
=
w
-
i n Q, on X,
The
129
N.An Alternative Model for Diphasic Wells This means t h a t
S
is f o r m a l l y t h e
( u l t r a - w e a k ) s o l u t i o n of
:
(4.17)
s
=
so
on
ii
We check now t h a t ,
at
t=O.
d e s p i t e the d e l t a - f u n c t i o n s appearing i n the
r i g h t hand s i d e of t h e s a t u r a t i o n e q u a t i o n (4.17).
its s o l u t i o n
S
always
satisfies 0 I S(x,t) I 1
(4.18) a s soon a s (4.19)
I
0
s SO(X) I 1
0 I Q I T ( tS) Q T ( t )
The f u n c t i o n s a and b .
for
S E [0,1],
.I'
a . e . on
G,
a.e. on
ii
aF
t
E
10,TC.
j=O,l,2, which a r e p h y s i c a l l y d e f i n e d only
have t o be c o n t i n u e d o u t s i d e of t h e i n t e r v a l [0,11, as we
d o n ' t know a p r i o r i t h a t ( 4 . 1 8 ) h o l d s . So we choose
:
f o r r, C [0,13 f o r r, M u l t i p l y i n g t h e f i r s t e q u a t i o n of ( 4 . 1 7 ) by Green f o r m u l a , we g e t
x=(S-l )
+
e
[0,11.
and u s i n g a
Ch.III: Inrompressible Tiuo-Phase Reservoirs
130
where B
:
lR
+
R is defined as one primitive
of
bo
and using the pressure equation (4.12), we obtain
or
(4.24)
which yields
x
0 i.e.
S 5
1
using the hypothesis (4.19) and (4.20).
x
=
-(S
Similarly, multiplying by
yields
:
which yields S 20. Remark 18 :
0
The formulation (4.12), DOUGLAS-EWING-WHEELER
(4.17) is used by some authors, cf.
[l],
EWING-WHEELER for approximation
studies. We shall prefer in the following the boundary-source formulation of sections I1 and I11 which is more general and well suited to our mathematical and numerical tools. 0
V
-
MATHEMATICAL
STUDY OF THE INCOMPRESSIBLE
TWO-PHASE FLOW PROBLEMS The aim of this section is to obtain some rigorous mathematical results on the existence of solutions to the two-phase incompressible flow model summarized i n section 111.
131
V. Mathematical Study V.l
-
SETTING OF THE PROBLEM
Let,
Rn be a bounded, convex domain, with regular boundary r re u rll u rs (referred to respectively as
R E
partitioned into (5.1)
entry, lateral and output boundaries), and with normal unit v
pointing outwards from Q,
T >O
]O,T[, Q=nx]O,T[
(5.2)
be
the given
time
interval of interest,
be the space time domain, and
Z = r ~ l o , T [ : (resp. Ze,
I t * Zs).
We
consider
equations (5.3)
in
Qx]O,T[
the
following system of partial differential
:
div
*)
in Q,
0
q =
(5.4) (5.5) (5.6) (5.7)
+ +
q.v
=
0
on XL,
q.:
=
A(P-Pe)
on zs,
(5.8)
(5.9) (5.10)
on 1 e'
(5.11
on
(5.12)
on zs,
(5.13)
where (5.14)
U(X,O)
=
uo(x)
equations"
zL,
on n at t=O,
Ch. III: Incompressible n v d h a s e Reservoirs
132
and q,b stand for
outside the summation sign Z.
qo, bo
We make the following assumptions on the coefficients in (5.3) through (5.14) (5.15)
(5.16) (5.17)
Q
6
L2(Ze)
Q 2 0 a.e. on Ze,
(5.18)
h
E
L"(ZS)
h 2 m2
d,u,a,b,b.,Y.,
IR (5.19)
R
+
J
on 1S'
are continuous bounded functions of
such that
d(c) t m,
a(c)
bo(c)
=
b.(c)
= 0 , .tF
J
j=1,2
J
a.e.
b(c)
E
a'(c)
=
t 0,
Y 5
f
R, u ( 0 )
=
0;
Y 5 E R; b(<)Zl, Y ,C 2 1 ; ] - ~ , O l u [l,+w[, fF j=l,2;
[O,l], E
a.e. on R.
(5.20)
0 5 uo(x) 5 1
Remark 19 :
If one thinks of u as being the non-wetting phase saturation and of P as being the global pressure, and if one affects the subsript 1
to the non-wetting phase and 2 to the wetting
phase (with the corresponding changes in the definition of the non-linearities), then equations (5.3) through (5.14) correspond, for large 1, to the two-phase flow equations (3.57)
through
(3.69)
of
section I11 with the boundary
conditions indicated in figure 11, 12. One can however notice that the unilateral condition (5.12) is now taken on C
9'
whereas it was taken on Z+ in (3.67) in
section 111. As we noticed in Remark 7, the two boundaries and
r+
r
may differ.
So we take a closer l o o k at the meaning of the unilateral
condition (5.12) on Ts.
133
V. Mathematical Study
s
A t point
*
+ +
Ts where q - v L O(i.e. s
E
model
t h e n we have
t h a t t h e u n i l a t e r a l c o n d i t i o n (5.12) is an
s e e n i n 311.3.2.2 accurate
r,)
E
which
takes
into
account
the
capillary
boundary l a y e r on t h e p r o d u c t i o n boundary.
.At
point s
rs
E
+ +
q-v
where
<
0 (i.e. s
E
r-)
we check now
t h a t , under t h e h y p o t h e s i s (5.21)
<
b(0)
(which
1
satisfied
is
in
practice
where
b(0)
=
O),
the
c o n d i t i o n (5.12) becomes (5.22)
>
u
+
and
0
+
lp2-.o
=
0.
Suppose f o r a w h i l e t h a t +
c o n d i t i o n $I 2'v
+
L 0
and t h e f a c t t h a t b l ( 0 ) +
Q,-V
+
=
u(s)
=
0
f o r such an
s. Then t h e
g i v e s , u s i n g t h e d e f i n i t i o n (5.14) of =
b2(0)
=
0 :
+
$2
+ + + + ( l - b ( ~ ) )q - v - r - b L 0.
Hence from ( 5 . 9 ) and (5.21) we g e t
-
--
i.e.,
6
+ +
( 1 - b ( 0 ) ) q.d
as J, L m > 0 ,
which is a c o n t r a d i c t i o n t o t h e h y p o t h e s i s
u(s)
=
0 because
t h e s o l u t i o n u o f ( 5 . 3 ) t h r o u g h (5.14) s a t i s f i e s , as we s h a l l
see, 0 6 u ( x , t ) 51.
In c o n c l u s i o n t o t h i s remark, we see t h a t t h e e q u a t i o n s ( 5 . 3 ) t o (5.14)
c a n model a n experiment of d i s p l a c e m e n t of a non
w e t t i n g phase by a w e t t i n g phase i n j e c t e d t h r o u g h c l o s e d boundary
r,,
and a n o u t p u t boundary
surrounded by non-wetting f l u i d
:
rs
re,
with a
which is
Ch.III: Incompressible lbo-Phase Reservoirs
134
- on the (global) production part
r+
of
rs
one observes
first production of non-wetting phase only, followed by production of both phases,
r- of rs, the surrounding
- on the (global) injection part
non-wetting f l u i d enters the porous medium, and no wetting phase escapes.
Remark 20 :
0
It is worth noting that equations (5.3) through (5.14) cover the problem of the water dam : the wetting phase
exactly")
(index 2) is the water, the non-wetting phase (index 1 ) is the air, and the boundary conditions are indicated i n figure 14. This problem has been studied by BAIOCCHI in the case where one supposes that the air pressure is constant (i.e. the mobility of air is infinite) and that the air saturation takes only
the values 0 and
1
(which implies that the
capillary pressure is neglected).
Remark 21
:
0
It would be very interesting i f we could know a priori from only the pressure equations (5.3) through (5.7)
rs
of +
q2
=
0
rS
c
-t
and
principle,
r+,
boundary.
the partition
r- and r + . This is for example the case if q1
into
Pe
=
P C P i.e.
all
=
constant, which gives, using a maximum + + on rs and hence q-v 2 0 on r : here
a.e.
the
output
boundary
is
a
production 0
(1) up to the incompressibility hypothesis, uhich can be released - cf. section I of chapter IV.
V.Mathematical Study
135
V.2 - VARIATIONAL FORMULATIONS
Let
(5.23)
re
and
V
{ v
=
II VII
=
[v]
=
rS
be of non-zero measure,
H ' ( Q ) lvlr
=
e
} ,
0
H=L2(a),
Ia lgradv12 ( I Igradvl' I v2 a (
J
1/ 2
which is a norm on V, which is a norm on H ' ( Q ) ,
+
rS
equivalent norms on H ,
K
(5.24)
=
{ v
E
lvlr L 0 ]
VI
be a closed convex set of V.
S
We identify H
with
H'
using the scalar product
I I$
associated with the norm
on H ,
so
:
that we have the following
embeddings :
(5.26)
I
VcHcV' with dense inclusions and continuous injections.
We denote by transported from the
11 ll* and norm 11 11
canonical isomorphism from
((
,
))*
the norm and scalar product on V'
and scalar product ( (
V onto V'.
,
))
on V by the
and we remark that, due to the
chosen method for the identification of H with H ' , we have (5.27)
50
4 v
we shall use (
E
V,
4 f
E
H c V'
to denote the duality between V
and V'
.
136
Ch.III: Incompressible Tivo-PhaseReservoirs
.
The pressure boundary conditions
...
. . , . . . . . . . . . . . . . .. . .. . . . Q ..' * . .. .:.. . .. .r\ .* .. .., ... . .... . . . . .. . . . . . . . ... ... .. ... .. ... ... .. .. .. .. . .. .. .. . .. . . . . .. .. .' .. . . '.. * .,
- - - / .
1
i.
.
,
A
.
The (air) saturation boundary conditions (the shorn partition of
Figure 14 :
r
into
r-
and
r+
is completely hypothetical)
Boundary conditions for the water dam model
137
V. Mathematical Study
Let u s d e f i n e :
(5.
The s t r o n g v a r i a t i o n a l f o r m u l a t i o n of e q u a t i o n s ( 5 . 3 ) through (5.14) is then : Problem
(
g ): f i n d
(5.29)
P
(5.30)
-(Ggradw+h
(5.31)
4
=
(5.32)
u
E
(5.33)
(*( t ) , v - u ( t ) ) ,
(5.34)
u(0)
f
L‘(0.T;
such t h a t
H’(SZ)),
a
d(u)
(u,P)
I (P-Pe)w=I r
2
1-1) gradP
+
a . e on l O , T C ,
Qw aF w f H ’ ( Q ) ,
re
1
Y . ( u ) Gjl,
j=1
J
and
where (5.35)
w, +
dt
=
u
Gj-r‘=G
Indeed, t h e equations (5.30), t o get
.grad ( v - u ( t ) ) t 0
f f
au - d i v oat E
v E K t lO,TC,
2
-1
j=l
bj(u)Gj-
( 5 . 3 1 ) a r e o b t a i n e d from ( 5 . 3 ) through ( 5 . 5 ) f i r s t rewrite ( 5 . 8 ) u s i n g
:
i2
=
K,
gradcc(u)+ ( l - b ) ( u ) ) G
inequality (5.33),
(5.14) and ( 5 . 3 ) which g i v e s
M u l t i p l y i n g t h e n by v
2
0’
2 + + @ * = q- 1 bj(u) j=O
i n a s t a n d a r d way;
(
a
0.
i n t e g r a t i n g over SZ,
and u s i n g a Green’s formula
and t h e boundary c o n d i t i o n s ( 5 . 1 2 ) , (5.13) we o b t a i n
:
Ch. III: Incompressible Two-Phase Reservoirs
138
d (u,v), +
(5.36)
J
i2.grad v
=
n
J
+2 - + 4 v v L
o
rS
V V E K aF t E lO,T[.
Replacing v by u(t) in (5.36) and substracting it from (5.36) yields the sought inequality (5.33). 0 The resolution of problem ( p
$ )
with the hypotheses (5.15) through
(5.20) is not possible, because the lack of V-ellipticity of the diffusion term in (5.33) (remember that (5.19) says only that c t ' ( c ) the diffusion term u
E
=
a(r,) 2 0, i.e.
is degenerate) is contradictory to the fact that
L'(0,T;V). So we want to weaken the formulation of our problem so that it can
have a solution even in the degenerate case. We remark for that purpose gradu in (5.33) is V-elliptic for the that the diffusion term ,' Ggradu(u)
a
function B(u) where 8 is defined by
:
We define the weak variational formulation of equation (5.3) through (5.14) by : -
Y. Mathematical Study
In
problem
139
the
( 3 )i n
next
paragraphs,
we
t h e non-degenerate
s h a l l g i v e e x i s t e n c e theorems f o r c a s e and f o r probleio
( 2 ' )i n
the
d e g e n e r a t e c a s e and t h e n make a d e t a i l e d s t u d y , due t o GAGNEUX, of t h e case where t h e p r e s s u r e and s a t u r a i t i o n e q u a t i o n s can be s o l v e d s e p a r a t e l y . We begin w i t h some p r e l i m i n a r y lemmas.
V.3
- SOHE PRELIMINARY L W S
Lemma 1 : by
Under h y p o t h e s i s (5.19)
on b , t h e f u n c t i o n B d e f i n e d
:
(5.41)
B
6
v(lR)
s.t. B ' ( C )
h a s t h e f o l l o w i n g graph :
=
1 -b(c),
B(0)
=
0
Ch. III: Incompressible Tho-Phase Reservoirs
140
-Lemma 2
:
(compactness lemma for the non-degenerate case)
Suppose that a satisfies the hypothesis (5.19) and
3
(5.42)
>
q
Then for every M>O (5.43)
S,
such that
0
11 B(V)llqy
is relatively compact in L'(Q), L 2 ( Z ) (Y denotes the "trace on
Proof : -
This
r7
>
:
+ F € E .
0
the set
{v e L2(Q)I
=
a(5) 2
:
lemma
11 gll.1;.
M,
2
and Y(S,)
}
2 M
is relatively compact in
r'' operator).
follows simply
from the
compactness of
injection of W defined in (5.28) into L2(0,T;H'-E(i2)) f o r any
I
:
(compactness lemma for the degenerate case)
-
Suppose now that a satisfies hypothesis (5.26) and
3e
(5.44)
l0,ll
E
s.t.
c(a)
=
Sup
5
OS5<521
sM
(5.45)
{ v
=
6
L ~ ( Q ) ~ o2 v(x,t) S 1 6 M
The relative compactness of
SH
2
5- 5
<
+-.
5
llglb
11 B(V)IIQ
:
[I /a(?) d-rl'
Then for every M>O the set
M,
a.e. on Q,
1
and 'Y(S M ) is relatively compact in (Y still denotes the "trace on rrr operator).
is relatively compact in LZ(Q),
Proof
the
> 0. 0
Lemma 3
I
E
:
in CHAVENT [l].
in L'fQ)
was already proven
We give here the proof for the sake of completeness. Define
(cf. 5.37)) -1
F = B
.
V. Mathematical Study (5.44)
Then [O,
B(1)I
141
just means that F
is an HBlder function of order
v
IF(tl)-F(t2)1
0
on
:
t,, t2
E
CO,B(l)I
e
5 c(a) Itl-t21
.
We divide the proof into three steps : Step 1 : Properties of F ( Y ) for
< s <
~ S ~ P ( Q )o,
Y
1 ,I
< p <
+-
.
The following characterization of WS” holds (cf. LIONS-MAGENES, tome 1 p.59)
and the quantity
is a norm equivalent to the usual one on WS”(n). One checks then easily that, for any function satisfying 0 5 w(x) 2 1 a.e. on 9, F(w) satisfies : F(w)
€
Ws‘ ”‘(n)
with
(5.46)
II F(w) IIwsl*PI(n) Step 2 : Compactness of
Let v v(t)
=
F(w(t))
S,
2 c(a)
s’
es
=
, p’
P e
E
WS”(5l)
’-
e
II WII ws,p(Q)’
in a subspace of L’(Q).
SM, hence w=B(v)
E
9 and 0 5
a.e. on Q. Hence
w 5 B(1)
with w(t) c V c H’(Q) f o r almost every t
As
=
w
H’(R) c Ws’2(Q) f o r any s
E
]O,l[
f
lO,T[.
we get from (5.46)
:
142
Ch. HI: Incompressible nYo.Phase Reservoirs
From (5.47) one gets easily, using the continuity of the injection of H ~ ( R ) into wS'P(n)
which proves that
:
As 0 is bounded and regular, we have, for
ws'
(5.50)
0
<
s"
<
s'
:
( a ) c ws"*p' ( a ) c LP' ( a ) c LZ(Q) c V'
with compact injection from Ws' " ' (a)
into Ws"*p' ( a ) ,
so that, from a standard compactness argument (cf. LIONS [21 p.58) we get the relative compactness of WM in Lp' (O,T;WS""' Thus we get (5.51
(0))
for 0 <
s"
<
s'
.
:
SM is relatively compact in L ~ / ~ ( o , T ; w ~ ' " ( a~ )/)~for O
Step 3 : Final compactness results.
First, as
218
>
0 and as Ws "'2'e(a)
c L2(R) with continuous
injection, we get the relative compactness of SM in L2(Q). Secondly, the s"-a/z,2/e ( r ) (cf. wsl* ,2/e ( a ) into W trace operator 'i maps continuously LIONS [l] sff
> 2'
p.87), So
we get from (5.51) used with 0 < 2e < srr<
compactness of Remark 22 :
which itself injects continuously into L2(r) as soon as
'i(S )
M
in L2(z).
8
the relative 0
It is possible to give simple sufficient conditions on the function a so that the condition (5.44) holds. From the HBlder inequality 5 5 e/2]i/edT \ e 5-5 = 1 x dT 2 { [a(?) 5
5 5
[a(T)
5 -812 111-8
1
d.r ]l-e
143
V. Mathematical Study
and the definition (5.44) of c(a) we get :
Hence a first sufficient condition on a for (5.44) to hold is that there exists on c0.11; then e
p>O
such that l/a’ is Lebesgue integrable
]0,1] of (5.44) is given by
e
=
2V 2’+1.
From this we deduce a practical sufficient condition
:
(5.52)
which 0
implies
< e < Min { -2,
that
2+P0
(5.44)
2 -1. 2+P1
holds
for
any
e
satisfying
This contains the physical case of two-phase flow in porous media where a(<) is given by (2.31) and has the aspect shown in figure 2 .
0
+
+
We study now in Lemma 4 the continuity of P. q, b 2
to u on SM.
with respect
144
'
Ch. III: Incompressible T w d h a s e Reservoirs
Lemma 4 : Suppose now that the hypotheses of either lemma 2 (i.e. (5.42) and (5.43)) or lemma 3 (i.e. (5.44) and (5.45)), with the corresponding notation, are satisfied, that (5.12) through (5.17), (5.23) through (5.28) and (5.37) hold, and that :
For a given M > O , (5.54
uk
let uk, u
->
+ +
Gk~$2k
Uk +
(5.56)
'kl'
(5.57)
B(Uk)
(5.58)
duk dt ->
(5.59)
Pk
(5.61) (5.62)
qk
+ '2k
-+
strongly in L2(Q), strongly in L 2 ( 2 ) ,
'12
->
weakly in
B(u)
du dt
weakly in
+
->
->
, ,
weakly in L2(0,T; H'(a)),
->P
lim inf k+a
:
be the corresponding unique solutions of
u
(5.55)
(5.60)
be given such that
u weakly in L2(Q).
Let pkt (resP. p,q,@,) (5.55) through (5.57). Then we have :
-+
S,
-
q
weakly in [L2(Q)In,
+
weakly in [L'(Q)]",
2 '
I ;2k-graduk 1 ;2-grad u.
Q
2
Q
Proof : (5.55) and (5.56) result directly from the relative compactness of SM and Y(SM) shown in lemma 2 or 3, and (5.58) follows simply from (5.54) (5.59) and the definition of SM. We turn now to the proof of (5.57) through (5.61 ).
145
V. Mathematical Study
- A priori estimates :
From the definition of S,,, we get (5.63)
wk
=
B(uk)
is bounded in (%
.
From the elliptic equations (5.30) and (5.31) and hypotheses (5.12) through (5.16) we get Pk
is bounded in L2(0,T;H1(Q)).
qk
is bounded in [L2(Q)ln.
(5.64)
Rewriting (5.35) as (5.65) and using (5.63) and (5.64) and the hypotheses (5.12) through (5.161, we obtain
i,,is bounded
(5.66)
-
in [L2(Q)ln.
Extraction of subsequence :
From (5.55), (5.56), (5.63), (5.64), (5.66) we get the existence of a subsequence u u
P
such that
:
a.e. on Q,
'U
a.e. on 1, iii)
w
=
B(u
->
B
weakly in (%
,
weakly in L2(0,T; H'(R)), weakly in [L2(Q)ln, weakly in [L2(Q)ln. Using (5.67) i) and ii) and the Lebesgue convergence theorem we obtain that
:
146
Ch.III: Incompressible Two-Phase Reservoirs
- Passing t o the limit
:
Using the weak convergences of (5.67) and the strong convergences of (5.68). we can pass to the limit in (5.30) through (5.31) and (5.65)
,,,
+
+
+
+
( a l l written with u P qp, $211), which shows that = P, = q and + + + p' $ 2 = 4'. As P , q, $ 2 are uniquely defined by (5.301, (5.31) and (5.65), +
w
and as we have seen in (5.68) i) that = B(u) which is also uniquely + + defined, we get the convergence of the whole sequences B(uk ) , Pk, qk, $2k in (5.67) iii) and iv), which proves (5.571, (5.59) through (5.61). We prove now (5.62). From (5.65) and (5.30) with w=B(uk) we get :
GZk
(5.70)
grad u
Q
k
1I O
:
lim inf k+m
I Glgrad
Q
From (5.59) we see that with (5.69) gives :
2
Q
From (5.57) we get
J, lgrad B(uk)Iz
=
PkIZ
QB(uk) -
re
B(uk)Iz 2
>-
PI z
The last term of (5.70) can be rewritten
where B . ( c ) J
b.(c) =
l;z
a
(c)
Q
1
T
1 h(Pk-Pe)
O
rs
+
1 bj(uk)
j=l Q
B(uk)
Gj.grad uk.
]grad B(u)Iz. weakly in L2(Z), which together
:
is continuous and bounded (cf. (5.53)).
V. Mathematical Study
147
Similarly as for
B.(uk)
(5.68) one can prove that
J
strongly in L2(Q), which together with (5.57) shows that : 2
1
lim
Q
k + m j=1
B.(uk) :.-grad B(uk) . I
J
I Bj(u)
2 =
1
Q
j-1
;:grad J
+
B.(u) J
B(u)
which ends the proof of (5.62) and of lemma 4. Remark
23
:
0
on the functions bj, j=1,2 is not
The hypothesis (5.53)
constraining from a practical point of view
:
- for a non-degenerate problem it is always satisfied
for a degenerate problem coming from two-phase flow, we get from the definitions of a, b, and b2 in terms of the *
mobilities k. and the capillary pressure J
p,
. *
(5.72)
As p,
is usually a bounded function with positive, bounded
below derivative (cf. figure 8 of Chapter I), the hypothesis (5.53) is practically always satisfied. 0
V.4
-
RESOLUTION IN THE NON DEGENERATE CASE
We
suppose throughout this
paragraph
that hypothesis
(5.42)
holds :
3 q>O
(5.42)
s.t. a(5) 2
q
>
0 a.e. on
R
and we want to show the existence of a solution of problem general hypotheses (5.28).
We
use
essentially that problem. Let
(5.79)
(5.1)
(5.2),
(5.6) through (5.18),
( g under ) (5.23)
the
through
for this a penalization technique. The proof follows given
in CHAVENT
[2]
for
a simpler one-dimensional
Ch. III: Incompressible nve-Phase Reservoirs
148
be g i v e n , and d e f i n e t h e p e n a l i z e d problem
( g c :)
(5.85 (5.86
I t s s o l u t i o n is g i v e n by t h e f o l l o w i n g theorem. .THEOREM 1 : Suppose t h a t
t h r o u g h (5.28)
(gE) admits
(5.87)
and ( 5 . 3 7 ) ,
(5.1), (5.42)
a t least a s o l u t i o n
(5.2),
(5.6)
and (5.79)
through (5.11),
(5.23).
h o l d . Then t h e problem
u,P s a t i s f y i n g
:
V. Mathematical Study
Proof : -
149
- Existence : let u
W be given, and define P,
6
through (5.31) unchanged, and u by :
4
by (5.29)
(5.88)
The family of elliptic equations (5.29) through (5.31) admits (for +
a given u ) a unique solution P , q. Then the non linear parabolic equation (5.88) admits ( u and theorem 1.2 p. 1 6 2 1 ,
being now given) a unique solution u (cf. LIONS [ 2 ] , as it is driven by an operator which is the sum of a
linear elliptic operator and of a penalization operator, this latter being monotone, bounded, and semi-continuous. Using standard bounds, we get
:
So we *ave defined a mapping u+u
from S,
into itself. As SM is
convex and weakly compact in W, we get from the Kakutani theorem the existence of a fixed point of this mapping (i.e. the existence of u ) as
soon as the mapping
u+p
is continuous on
SM
for the weak topology of
W, which can be proved with the same techniques as in lemma 4. - Majorations on
:
(5.87) i) is obtained in a standard way by
i n (5.81) and using the general hypotheses (5.16) through
taking
w=u
(5.20).
Taking then
E-
u
v=-u. in (5.84), using (5.81) through (5.82) with
w=B(v) and integrating between 0 and T we obtain :
which gives, using (5.87) i) and lemma 1
:
150
Ch. III: Incompressible lk-Phase Reservoirs
and (5.87)
ii) follows then immediately.
Taking then v=u in (5.84) and still using (5.81) through (5.82) (with W=B(UE)) we get, with the notation (5.53) :
(5.89)
B ( u + ) 5 1 and IB(-u;)I
Using lemma 1 we note that integrating from 0 to t we obtain
(5.90)
i
2
u-
and
:
2
+
+
1 2m j=1 1 IIgjII: II GjII;L2(Q),n
II Allcn
which yields (5.87)
I(pc-Pe)+lL*(ls) IUJL iii) and
2
(
xs)
iv) using (5.87) i)
One gets then from (5.84)
and
ii).
:
which, with (5.87) i) and ii) and the general hypotheses (5.18) through (5.20), yields the sought result. This ends the proof of Theorem 1. 0 We come back now to problem
(@)
and give
V. Mathematical Study 2
-THEOREM
through
:
151
Suppose t h a t ( 5 . 1 ) ,
(5.28).
and
(5.42)
(5.2),
hold.
( 5 . 6 ) t h r o u g h (5.111,
Then problem ( @ ) (5.29)
(5.35), admits a t l e a s t a s o l u t i o n (u,P) s a t i s f y i n g (5.92)
a.e.
0 5 u(x,t) 2 1
on
We s h a l l o b t a i n
:
solutions
(YE,
of
PE)
q
such t h a t
:
as t h e l i m i t of a subsequence of t h e
(u,P) the
through
:
Q,
and t h e r e e x i s t v a r i o u s c o n s t a n t s c independant of
Proof
(5.23+
penalized
problem
using
(PE),
standard
t e c h n i q u e s . We g i v e t h e proof f o r t h e s a k e of completeness. From (5.42) and ( 5 . 8 7 ) i v ) such t h a t
uE e SM
existence of through
a subsequence
(5.62),
uk
uk
->
u
E
v) we s e e t h a t t h e r e e x i s t s M > O From lemma 2 and 4 , we g e t t h e
and of @2 k ( r e s p . P ,
S
5k,
[where P k ,
( r e s p . u ) ] , and s a t i s f y i n g a l s o (5.94)
and
defined i n (5.43).
+M
% q, 9
+€
2
s a t i s f y i n g (5.54)
1 c o r r e s p o n d t o uk
:
weakly i n W.
We check f i r s t t h a t
(u,P)
is a s o l u t i o n of
(8 ) :
- ( 5 . 2 9 ) t h r o u g h (5.31) and (5.35) are s a t i s f i e d by (5.59) through (5.61 1.
uk
+
u
-
- From ( 5 . 5 6 ) and t h e Lebesgue theorem, we g e t t h e convergence of i n L 2 ( Z ) ( f o r a subsequence a t l e a s t ) , which t o g e t h e r w i t h (5.87)
ii) gives
-
u =O
on
Ts, s o t h a t (5.32) is s a t i s f i e d .
-Replacing v by v-uk, v e K
i n ( 5 . 8 4 ) (hence v-11 =o) y i e l d s
(5.95)
IY v
e K,
a . e . on 1 0 , ~ ~ .
:
Ch.III: Incompressible Tbc-Phase Reservoirs
152
Take v=v(t), v c
a
in (5.95)
and integrate from 0 to T :
(5.96)
Using the weak lower semi-continuity of v
-f
-21
Iv(T)Ii on W we see that:
du
(5.97)
k+-
which together with (5.58), limit in (5.96).
Thus du u-v$@
(z,
(5.61).
(5.62)
makes it possible to pass to the
6 i2
grad (u-v) I 0,
VVE&
I
which is (5.33). - The initial condition (5.3'1) follows from (5.851, and from the continuity of the linear mapping u+u(O) from W into H. So (u,P) is a solution to problem
obviously from (5.87). principle
- Taking
(5.92)
(5.93)
result
using a maximum
:
get, as for (5.89)
v = u ~ v with V = -u- in (5.33)
(v
E
K as v(
-
v
=
(u-l)+ in (5.33),
which is permissi 1
: =
+ -
(ulr - I ) + e
=
o
=o) we
'curs
:
Then v=O, and u>O a.e. on Q. Taking v = u ~ P with as
(8). The bounds
There remains to prove
(5.9'1)
as u
Ire
=
o
153
V. Mathematical Study
we g e t
:
f I bj
j = l 51
(P+1)
GJ.
g r a d P S 0.
The l a s t term v a n i s h e s , u s i n g ( 5 . 1 9 ) and t h e f a c t t h a t P+1 2 1 . Using t h e n
(5.30) w i t h w
=
B ( P + l ) we o b t a i n :
4 $ I'+'(t)Ii +
m
11 B(9(t))l12 I +
X(P-Pe)
B(9+1)
rS
-
f,
Q B(9+1)
0.
=
'e
Noting t h a t B is monotone i n c r e a s i n g and ( t a k e wEl i n ( 5 . 3 0 ) ) t h a t :
we s e e t h a t (5.100)
:
+ 2 I@(t)Ii 11 +
m
B(9(t))l12s
J
x(P-P,)-
[B(9+1)-B(1)1
TS
where t h e r i g h t hand s i d e term v a n i s h e s a s B ( 5 ) is c o n s t a n t f o r 5>1 ( c f . lemma 1 ) . Hence P 3 , i . e .
us1
a.e.
on Q. T h i s completes t h e proof of
theorem 2 . Remark 24
0
:
Suppose t h a t we r e l a x t h e h y p o t h e s e s
i n ( 5 . 1 9 ) . Then we have t o make t h e f o l l o w i n g changes i n t h e p r o o f s of theorems 2 and 3 :
-
To o b t a i n t h e energy bounds on u - i n ( 5 . 8 9 ) through (5.901,
we have t o suppose t h a t :
-either B ( c ) is bounded f o r < > O , which is a c h i e v e d i f (5.102)
0 S 1 - b(<) S
4 5
with 6
>
1
Clt III: Incompressible TwoPhase Reservoirs
154
(and then
B(5) 5 B ( 1 )
-) 1
+
5-1
-or (5.1 0 3 )
one knows a p r i o r i t h a t P 2 P
-
rs.
on
To o b t a i n t h a t us1 by t h e maximum p r i n c i p l e i n (5.100) we
have t o suppose t h a t ( 5 . 1 0 3 ) h o l d s , so t h a t t h e l a s t term can be dropped o u t of ( 5 . 1 0 0 ) .
In c o n c l u s i o n , theorems 2 and 3 h o l d -with (5.101) replaced by (5.102)
:
[but then t h e conclusion
u61 of theorem 3 must be r e p l a c e d by t h e weaker one
where
c(q)
-without
->
+
.
(5.101)
when
11
* 0.
1
if one knows that
P
2 P
.
i n s t a n c e b y t h e maximum p r i n c i p l e of remark 21 Remark 25 :
One s e e s
in
(5.90)
s o u r c e term f o r exactly,
as
and
:
(5.100)
that
(P-Pe)-l
on
TS
Is, f o r 0
acts as a
t h e s a t u r a t i o n e q u a t i o n , which c o r r e s p o n d s
formally
h(P-Pe)
=
+ + q*,
on
rs,
to
the
i n t e r p r e t a t i o n of t h e u n i l a t e r a l boundary c o n d i t i o n g i v e n i n remark 19. Remark 26 :
0
The uniqueness of t h e s o l u t i o n of t h e non-degenerate problem h a s n o t y e t been proved. The d i f f i c u l t y f o r t h i s a r i s e s from
the
coupling
between
the
pressure
and
saturation
e q u a t i o n s . As soon as t h e c o u p l i n g f a i l s , t h e n u n i q u e n e s s can be o b t a i n e d by s t a n d a r d methods ( c f . SV.6 below). 0
K Mathematical Study Remark 27 :
155
Asymptotical behaviour of u(t) in the special case where
G2
=
(no gravity and
0
no
G1
=
capillary heterogeneity effects)
As we already noticed in and where Pe = constant on rs. remark 21, one gets in this case using a maximum principle P 2 P
on
rS, so
that the corresponding source term (cf.
remark 25) vanishes. Passing to the limit in (5.891, we obtain
:
which using the Poinearre inequality gives
Hence
:
which shows that
u(t)+O
in L2(Q), i.e.
one can recover all
the mobile o i l of the field by injecting water for a long enough time. This property may fail as soon as gravity or capillary heterogeneity effects are present. V.5
-
RESOLUTION I N THE DEGENERATE CASE We suppose throughout this paragraph that the following hypotheses
hold
:
(5.44)
(5.53)
Ch. III: Incompressible nYo-Phase Reservoirs
156
,THEOREM
3
Suppose t h a t ( 5 . 1 ) t h r o u g h ( 5 . 2 1 ,
:
(5.23)
( 5 . 2 8 ) , ( 5 . 3 7 ) , (5.44) and ( 5 . 5 3 ) h o l d , and t h a t moreover
u0
(5.105)
K.
E
Then t h e problem (5.40),
through
:
(5.35),
( P I
admits a t
),
(5.29) t h r o u g h ( 5 . 3 1 ) , ( 5 . 3 8 ) through
least
a solution
(u,P),
which moreover
satisfies : (5.106)
u
/
E
L"(0,T;H).
( 5 . 2 9 ) t h r o u g h (5.31) unchanged,
i) ii)
~ ( u c ~ ~) n % ,
iii)
[s,
(5.1 07 ) +
du
v-u ( t ) ) , +
I
u (0)
v)
-f 02ri
ri
=
q
I 5 grad u
n
g r a d (v-u ( t ) )2 0
R ;2n
iv)
17
17
=
17
V v
uo,
grad u ( u ) rl
+
B' ( u q )
-
grad(v-u ) ( t ) ) 17
6
2
1
j=l
K , a.e.
on 1 0 , T [ ,
bj(uq) ;j*
w i t h t h e bounds ( j u s t t r a n s l a t e d F r o m ( 5 . 9 2 ) t h r o u g h ( 5 . 9 3 ) ) :
-
157
V. Mathematical Study
where the c are constants independant of n. From (5.108) i), iv), v) we see that there exists an M>O such that u € SM defined in (5.45). From lemma 3 and 4 we get the existence of a rl subsequence uk satisfying (5.54) through (5.62). We check now that the couple (u,P) is a solution of @I ) :
v
E
Wn
(5.29) through (5.31) result from (5.59) through (5.60), (5.107) ii), (5.56) and (5.57) -> (5.58) --> zduf q l ' , (5.108) i) and (5.55) -> (5.40), (5.107) v) and (5.61) ->(5.35).
B(u)
,
There remains to prove (5.39) : take in (5.107) iii) v=v(t) with and v(0) = u and integrate from 0 to T :
&
0
2
which gives
nk
t
1 J, grad uk grad v
0
:
2 E L2(Q), one can pass to the limit i n (5.109) by using (5.58), dt (5.61) and (5.108) iv), which yields (5.39). This completes the Proof of
If
theorem 4 .
V.6
-
0
THE CASE OF DECOUPLED PRESSURE AND SATURATION EQUATIONS
The coupling between the pressure and saturation equation makes the mathematical study of the whole system very difficult. For example, it has not yet been possible to prove the uniqueness of the solution of the coupled system of equations, even i n the case of a non degenerate saturation equation with simpler boundary conditions. Similarly, the
Ch.III: Incompressible nYo-Phase Reservoirs
158
demonstration of the existence of a strong solution of the coupled system relieves on the L"-regularity of the transport field not achieved when
+
Go,
which is usually
qo is given as the solution of the elliptic pressure
equation with coefficient d(u). In order to give further results in these two directions, we suppose throughout this paragraph that :
i
the
transport
(5.291,
field
...,(5.31)
is
+ qo
given by
the
pressure
equation
independant of u and of time t and
satisfies
(5.111)
which is the most restrictive assumption, and that
:
(5.112)
Gj
(5.113)
b.
Remark 28 :
Hypothesis (5.111) is satisfied in two cases
J
f
f
CL"(n)ln
j=1,2
W''"(IR)
j=O,l , 2 .
- one-dimensional problems
(q
0
:
is then constant in space)
with constant given global injection rate Q(t)
5
Q on
r
(but
one dimensional problems with Dirichlet pressure conditions on I' and I' do not satisfy (5.111)).
- multidimensional
problems
with
neither +
+
gravity
nor
capillary pressure heterogeneity (hence q1=q2=O) and such that d(u)
1 (cross-mobility curves).
The following results are due to G. GAGNEUX [ll, [2].
0
We shall
follow his proof and hence give only the main steps of the demonstrations.
159
V. Mathematical Study
V.6.1
-
Regularity and asymptotic behaviour for the non degenerate case
4
-THEOREM
:
(Regularity of the non degenerate case with compatible
-
initial data). We assume the hypotheses of theorem 2, (5.111) through (5.113) and that
:
(5.114)
uo
K.
6
Then the solution u
of (5.32) through (5.35) given by theorem
2 is unique and satisfies moreover
u
(5.115)
t
L"(0,T; V),
du
(5.116) (5.117)
div
L2(Q),
r' =
-div ({grad
and for almost every t, t-.i
xd
(5.118)
:
a
t
cx(u))
€
C0,Tl and
L2(Q),
uo,
Ci0 E
K
:
lu(x,t) - G(x, t--r)(+ I 0
where u (resp. il) is the solution corresponding to uo (resp. f i g ) ; this implies (5.118bis) Proof
d
lu(t) - Ci(t-T)IL,(a) 2 0.
:
(5.119)
du (E(t), v-a(u(t)), Let then s c ,
c
>
+
f a
i2 grad
(v-a(u(t))
0 be the following approximation of the (sign)+
function for 5 I 0 s
p
=
v E K, a.e. on l0,TC.
2 0 f
for 0 2 5 I for
I 5
E
Ch.III: Incompressible no-Phase Reservoirs
160
and l e t u and G be two s o l u t i o n s of (5.119) c o r r e s p o n d i n g t o t h e i n i t i a l c o n d i t i o n s uo and Go. For
(TI
<
T l e t w be d e f i n e d as :
One checks t h e n t h a t : (5.121)
v
=
cr(u(t))
(5.122)
v
=
s.(w(t))
E
a(a(t))
Using (5.121)
+
E
K f o r a.e.
t s.t.
t , t - T
E
CO,T[.
s ( w ( t ) ) c K.
i n ( 5 , 1 1 9 ) , and (5.122) i n (5.119) w i t h
t-T
and 0
i n s t e a d of t and u, and t a k i n g t h e d i f f e r e n c e y i e l d s :
which t o g e t h e r (5.123) y i e l d s
The
Lebesgue
w i t h t h e Cauchy-Schwartz
inequality i n the
l a s t term of
:
convergence
theorem
shows
that
sc(w)
+
sgn(w)
-
s g n ( u ( t ) - i l ( t + r ) ) i n L 2 , which t o g e t h e r w i t h (5.18) y i e l d s t h e sought r e s u l t (5.18).
V. Mathematical Study
161
. Regularity of u
Let N be a positive integer, t
:
following time discretization of (5.33) 1 n n-1 , - ( u -u
V - U ~ )+ ~
(5.125)
I
n
=
- ,
and consider the
:
$; &ad(v-un)
2 0, aF
v K, n=l ,2,...,N ,
u = u
0'
For uniquely un
E
given in K, the first equation of (5.125) defines
un-'
K (the existence can be proved using a fixed point theorem as
in Theorem 1, and the uniqueness is obtained as above). Taking v=O in (5.125) and using the same majorations as for (5.89), (5.90), we have lun1; 2
N
1
lun - u
n=1
(5.1 26)
c
n=l,2,...,N
c
=
~u
n-1 4T
+
12
0 0
T
As
IIBjll:
J=1
5 c
11 ~ 1 1 ,
2
-m .1
+
1;
N
,
,
T
1 11 BCU")~~' 2
n= 1
c/m
where
(Meas
rs) 1 /2 I ( P - P ~ ) - ~ ~ ~ ( ~ ~ )
+
lq;2(R)-
was done in (5.119) for the continuous case, (5.125) is equivalent to 1 n n-1 , v-a(un))O + j grad (v-a(un)) 2 o - ( u -u
n
aF
Taking v
=
v
c
K
,
n=l ,2,...,N.
a(u"-l) we get
But the b.'s are Lipschitzian and the J
&'
;.Is
J
belong to L"'(Q)
SO
that
:
Ch. III: Incompressible Tbo-Phase Reservoirs
162
Summing then (5.127) (with (5.128)) from n=l to M 5 N we obtain
:
From the bounds (5.126) and (5.129) one proves easily, using the same techniques as in the proof of theorem 2, that Uh
->
weakly in L 5 ( 0 , T ; V) and weakly* in L*(OT; V),
u
weakly in LZ(Q), where u is the solution of (5.32) through (5.35) and where uh (resp. a h is the piecewise constant (resp. continuous piecewise linear) function 0 1 N "interpolating" the (u , u ,..., u ) sequence defined by (5.125). This ends the proof of theorem 4.
-
THEOREM 5 :
0
(regularity
for
the
non
degenerate
case
with
non
compatible initial data). We assume the hypotheses of theorem 2, plus (5.111) through (5.113) and that 0
(5.20)
I u (x) 4 1. 0
Then the solution u of (5.32) through (5.35) given by theorem 2
is unique and satisfies u
(5.136)
:
e
L"(0,T; V),
6
LZ(Q),
$
(5.137)
fi
(5.138)
f i div
=
-div
[igrad
a ( u ) ] e L 2 ( 0 , T ; H),
and satisfies also (5.118) and (5.118bis) for almost every t,-r f ]OT[ and uo,
0 such that (5.20) holds.
-
163
V. lclathematieal Study
Proof :
and let
Let
K, k=l,2,... be chosen such that
uOk E
be the corresponding solution of (5.32) through (5.35) given
U,
by theorem 4. Using the same techniques as in the proof of theorem 2, one can prove that uk
:
u
+
in L " ( 0 , T ;
H) weak* and in L 2 ( Q ) strong,
a(uk) [ resp 8 ( u k ) ]
+
a(u)
[resp. B(u)]
weakly in L 2 ( 0 , T ; V ) ,
We have now to obtain additional estimations on uk in order to prove (5.136) and (5.137). also
(ak =
We first remark that
etc...)
a(u,)
du (-,
(5.140)
k
V-CL,)~
dt
U,
satisfies (5.119) and hence
:
+
I a
-+ (J
grad v-irad (v-a,)
aF
v
E
+
K, a.e. on l O , T [ .
For any positive integer p, we define then solution of
:
(t)
+
(5.141 1 Up(0) which, as
dak E dt
u (5.142)
=
a (0) k
=
u,(t),
n ( u o k ) < K,
L Z ( O , T ; H), has the following properties
'a
~k
u;.
uP (t) =
+
dak dt
strongly in L 2 ( 0 , T ; V ) , strongly in L Z ( O , T ; HI.
U (t) to be the P
164
Ch.III: Incompressible hv-Phase Reservoirs
U (t) E K f o r e v e r y t , one can t a k e v=U ( t ) i n ( 5 . 1 4 0 ) . P P M u l t i p l y i n g t h e n by s and i n t e g r a t i n g between s=O and s = t y i e l d s : As moreover
But,
where
Using
then
(5.145)
in
the
As we know from (5.142)
that
i n t e g r a t i n g by p a r t s y i e l d s
e x i s t s a subsequence, s t i l l denoted by
]O,T[.
Since
right
hand
side
dt
+
(5.143)
U +a s t r o n g l y i n L 2 ( 0 , T ; P k
such t h a t
U
P' One can t h e n p a s s t o t h e l i m i t i n (5.147) when
duk -(s)
of
and
:
+
H k - g r a d CY
k
=
dgk
-(s) dt
+
Hk(s)*&ad
V), t h e r e
U ( t ) + a , ( t ) a.e.
P
on
p-'", which y i e l d s :
gk(S),
we o b t a i n
K Mathematical Study
165
which, u s i n g (5.1391, (5.1441, (5.146) shows t h a t (5.1 49 )
1
f i g B(u) fi
E
:
L'(Q),
a(u) e L"(0,T; V),
E1=a1'2, a'
which p r o v e s ( 5 . 1 3 6 ) , (5.137) as Taking i n t e g r a t i n g over
then
in
C0,TI
yields
(5.33)
v=u(t)
=
a 2
n >
0.
with
B(t)
8
e
.@ ( Q )
and
:
au _ at
(5.150)
in which t o g e t h e r w i t h
is possible i n
a'
(5.137) p r o v e s
@I
(Q),
(5.138) as t h e m u l t i p l i c a t i o n by
fi
(Q).
F i n a l l y , (5.118) and ( 5 . 1 1 8 b i s )
(and hence t h e uniqueness of u )
a r e proved by t h e same t e c h n i q u e s as i n theorem 4 .
-
0
THEOREM 6 : (Asymptotic behaviour of u in t h e non-degenerate case)Let t h e h y p o t h e s i s of theorem 5 h o l d ; i f t h e i n i t i a l data u 0
s a t i s f i e s moreover u
0
E
:
H'(Q) [ r e s p . uo e K 1 ,
G20-&ad v t 0 [ r e s p . 201
s.t. v t o , a.e. on Q,
V V E V
Q where
- + i2, J, g r a d =
a(uo)
+
C1-b(uo)l
2
io - 1
j=l
bj(uo)
ij,
aul
+ = - 2 0 [ r e s p . 201 on ?S and (which f o r m a l l y means t h a t d i v $ 20 at t-o + + t h a t $ 2 . . , l t = 0 t 0 [ r e s p . 2 01 on r k u r s ) , t h e n one h a s :
(5.152)
au
(x,t) 2
o
[ r e s p . 2 01
a.e. on nxCO,+nC
I
Ch. MI: Incompressible nvo4hase Reservoirs
166
J u(x,t)dx in the field at time R t is a decreasing [resp. increasing] function of time), (which implies that the amount of oil
(which implies that the amount of oil produced per unit time is a decreasing [resp. increasing] function of time) and (5.154) u(t) + u _ strongly in LP(Q) for every p L 1 and weakly in V where um is among the solutions of
:
the only one which satisfies : (5.156)
ucd=
u', [resp. ub,=
Sup Ul,'Uo
Proof
:
Inf u:,]. u' >u 0
cn
The proof of (5.118) in theorems 4 and 5 requires only that
where
s (w)
s,(u(u(t))-a(ii(t))
=
E
V
is positive a.e. on n.
Under hypothesis (5.151) these inequalities are satisfied for the following two choices of u and & : i)
u(t)
=
solution of (5.32),
ii) u(t) E u fF t, 0
Choosing i ) [resp. ii)] (5.157)
..., (5.35),
&(t)
G(t) solution of (5.32),
uo fF t ,
...,(5.35).
we shall show that
u(x,t) 2 u (x) [resp. u(x,t) 2 u (x)l 0
0
a.e. on ~ x l 0 , ~ ~ C .
V. Mathematical Study
167
From now on we c o n s i d e r o n l y t h e f i r s t case i n ( 5 . 1 5 7 ) ( t h e second b e i n g t r e a t e d i n t h e same way). Taking t h e n i n (5.118) G=u w i t h O < T < t < T y i e l d s
which, as
U(T)
(5.1 58 ) which
5 u
0
proves
a.e. on R u s i n g ( 5 . 1 5 7 ) , shows t h a t : 2
U(X,t)
a . e . on R , f o r 0 2
U(X,t-T)
(5.152),
decreasing function Hence
:
and
shows
t--u(x,t)
that,
a.e.
for
has a l i m i t ,
x
E
2 t, R,
the
positive
which we d e n o t e by
uw(x).
:
0
(5.159 )
s
u_(x) i 1
u(t)
+
Let then f (5.160)
f(t)
i n L P ( Q ) f o r every p > t .
U<,
: [O,
=
a . e . on R ,
-.[
7
R be d e f i n e d by
I u ( x , t ) dx
f t t O .
a
The f u n c t i o n f is c o n t i n u o u s and r e p r e s e n t s t h e amount of o i l i n t h e f i e l d a t time t. From ( 5 . 1 5 8 ) ,
(5.159) and (5.118) w i t h 9
f ( t ) h f(t+T)
aF
f(t)
when t
1 u _ ( x ) dx f(t)-f(t+T) f f(t')-f(t'+I)
(5.161)
+
f_
=
(5.162)
, I
, f'_(t)
+
OSt6t'
exist f t
>
0,
a . e . on lo,+-.[
f' ( t ) 2 f'
f o r a.e. t , r h 0
Sup t>O
Ess f'(t)
Hence, n o t i n g t h a t f ' ( t ) = -
2 0 =
0.
I- ddut ( t ) IL ' ( 0 )
, one g e t s
-,
and
f' ( t ) = f ' _ ( t ) = f ' t ) ( t + L )
u we g e t
2 0,
t , T
which p r o v e s (5.153) and shows t h a t fl(t)
=
:
f T 2 0,
Ch. III: Incompressible 7boPhase Reservoirs
168
Taking then v
=
0 in (5.119) yields :
Hence :
2m . J=1
which, using (5.163) shows that
stays in a bounded set of V when
a(u(t))
t-,. Hence there exists a subsequence a(u(t'))
Using
then
(5.159),
(5.163),
such that
(5.165)
and
the
weak
lower
semicontinuity of the norm, one can pass to the limit in (5.119) (for v given) when uo
t-,
which shows that
E
K
u~,,necessarily satisfies (5.155).
In order to prove (5.156) one just remarks that (5.118) holds for uc, solution of (5.155), with act) E urn
satisfying (5.20) and for any
+ t, which
shows that
io u
(XI
t u_(x)
a.e. on
(5.166)
:
Cl
which ends the proof of theorem 6. Remark 29
:
u(x,t) 2 U_(X) a.e. on a
x
lo,.,[
0
In the special case where q1 = q, = 0 ( n o gravitational or capillary heterogeneity effects) and where the given exterior
169
V. Morhemoticol Study
pressure P that PLP m
2
rs,
is constant on
we have seen (cf. remark 21 )
on Ts; we get then from (5.164)
11 a(u(t))l12 s
E(t)
+
:
o
which proves that necessarily ucd= 0. Remark 30 :
The asymptotical behaviour of
0
u(t)
in the case where it
does not evolve monotonically (i.e. when the initial data does not
satisfy
however,
noticing
(5.151)) that
u0 open problem. One has
is an (5.166) holds
hypothesis that 0 5 u (x) 2 1 0
under
the
sole
:
(5.1 67 )
Remark 31
An
example of multiple steady-state solutions u-. Consider a
vertical
1-D porous slab
Q
= ]O,II[,
with insulated lateral
boundary, and with 7Tinjection"boundary and "production" boundary Ts
=
fe
that
:
q1
0 (no capillary heterogeneity effects)
=
{ O } at the top
=
1 1 ) at the bottom. We suppose
(5.168)
Then
u
represents the oil
saturation in one imbibition
experiment ( q 3 ) in a vertical sample maintained in contact with water at the top ( u ( 0 )
=
0), with insulated bottom end
(as the unilateral condition resumes to
+
+
$2'1)
=
0 when u>O),
Ch. III: Incompressible f i o q h a s e Reservoirs
170
and with oil and water mobilities kl and k2 such that
51
J u(1-u) and a capillary pressure curve kl+k2 p (u) = arccos (1-2u).
-=
The steady-state equation (5.155) becomes now : $2
=
u(0)
(5.169 )
constant on lO,n[ =
0
,
u(n) 2 0
$,(n)
,
0
2
- The class of initial data u0 ax]O,-[
u(n)$2(II) E
K
=
0.
such that
au
2 0
a.e. on
contains only the stationary states : uo = ,:u 0 2 a a . urn is defined in figure 15 (if (5.151) holds in
2 II, where
the bracket case, then necessarily necessarily
@20
(il)
$20(II)
2
0 ; hence
0 and uo is one of the stationary
=
solutions). One checks in this example that the equilibrium profile
0
um
is exactly that of the capillary pressure. This property, which is always true, under condition (5.168), in 1-D samples with both ends insulated, is used as a physical definition of the capillary pressure law. Remark 32
:
0
If we replace, in the last remark, the function b2(U) by
e),
(which corresponds to the (unbounded) capillary pressure law p,(u) = Log then the steady-state equation (5.155) has o n l y one solution urn E 0. This comes from the fact the "equilibrium profile''
x= Log
boundary condition u(0)
=
0.
1 -u
does not satisfy the 0
V. Mathematical Study
V.6.2
-
171
Regularity and asymptotic behaviour for the degenerate case We turn now to the degenerate case: in order to handle this case,
we had replaced in the case of the coupled system of equations, the
8)by the weaker (8 1, which required that
saturation equations (5.32) through (5.35) of problem ( formulation (5.38)
u,,
E
through (5.40)
of problem
K in order to get a solution (theorem 3 ) . We shall now treat this case in another way, and, still following
CACNEUX [ l ] , [2], we introduce the following variational formulation :
Problem (@)
6%
,
(5.174)
B(u)E
(5.175)
($ (t), v-a(u(t))),
(5.176)
~ ( 0 =) u 0 ,
(5.177)
0 2 u(x,t) 2 1
:
find u such that
$Eq'p +
(v-a(u(t.1,) L 0
;,(t).&ad
a
Y. v
E
K
a.e. on lO,TC,
a.e. on Q.
we snail be able to show the existence of a solution without the -compatibility condition from [O,T]
u
0
E
K. Since (5.174) implies that u is continuous
into H equipped with the weak topology, the equations (5.175),
(5.176) make sense.
One a'
=
checks
a L 6 > 0, the
easily problems
that,
in
the
non
degenerate case where
( 3 )and (2") are
equivalent. In the
degenerate case, the inequality (5.175) can be formally shown to satisfy the saturation equations ( 5 . 8 ) through (5.13) with the boundary conditions on has a trace on El.
Ee
and
E
a(u)
instead of u in
(which is satisfying because a ( u )
172
Ch. III: Incompressible nuo-Phase Reservoirs
1
(
The corresponding \ solution converges \
a
b
a
Asymptotical behaviour?
n X F i g u r e 15 : Exaaples of initial data
yielding monotonic and uo non-monotonic evolution of the saturation profile.
U
V. Mathematical Study -THEOREM
1 73
7 : ( r e g u l a r i t y f o r t h e d e g e n e r a t e c a s e w i t h non compatible
i n i t i a l data).
-
We make t h e h y p o t h e s e s of theorem 3 b u t w i t h (5.105) r e p l a c e d by (5.20)
Then t h e problem one
"entropy"
(PI)(5.174)
solution
u
v i s c o s i t y ) s a t i s f y i n g moreover
$
(5.178)
fi
(5.179)
Proof
:
a.e. i n R.
o < u 0( x ) s l
Let
B(u)
E
t h r o u g h (5.177) a d m i t s a t l e a s t
(defined
by
addition
of
a
vanishing
:
L2(Q),
a(u) e L w ( O , T ; V).
u
n
be
the
s o l u t i o n of
theorem 5
a ( s ) = a ( C ) + q . We know from theorems 2 and 5 t h a t u
( c f . ( 5 . 1 0 8 ) , and (5.148) ( 5 . 1 4 9 ) )
:
n
corresponding t o
is bounded as f o l l o w s
where t h e c o n s t a n t s C are independant o f n. Using t h e same compactness argument a s i n t h e proof of theorem 3 , one c a n p a s s t o t h e l i m i t when u
n
n+O,
and hence show t h a t a subsequence of
converges toward one s o l u t i o n of problem
(5.179 )
.
which s a t i s f i e s ( 5 . 1 7 8 ) , 0
Ch. III: Incompressible TboPhase Reservoirs
174
Remark 33 :
The c h a r a c t e r i z a t i o n of t h e e n t r o p y s o l u t i o n s of t h e proof
results
in
, and
@It)
of t h e i r u n i q u e n e s s , have n o t y e t been done. For that
direction
diffusion-convection
in
the
case of
the
degenerate
e q u a t i o n s ( i n s t e a d o f i n e q u a l i t i e s ) one
can see VOLPERT-HUDJAEV and B R E N I E R . 0
-
THEOREM 8 :
(Asymptotic behaviour i n t h e d e g e n e r a t e case)-
Let t h e h y p o t h e s e s of theorem 7 h o l d , and suppose t h a t
s a t i s f i e s moreover (5.151) and (5.181)
n
j, g r a d uo g r a d v t
0
aF
vfV
(which f o r m a l l y means t h a t d i v ( $ g r a d u,)
on
rk
and
rs).
u(t)
-t
S 0
on R and t h a t
Then t h e "entropy" s o l u t i o n u of (
urn
0
s . t . v t 0 a.e. on $2
strong
j
where uw i s , among t h e s o l u t i o n s of :
t h e o n l y one which s a t i s f i e s (5.156).
i n L'(Q)
au
5 2 0
gTt) defined
theorem 7 ( b y a d d i t i o n of a v a n i s h i n g v i s c o s i t y ) s a t i s f i e s
(5.184)
u
:
:
f o r every, p L 1,
in
175
V. Mathematical Study
Proof : -
u
Let
be t h e s o l u t i o n of t h e non d e g e n e r a t e problem i n t r o d u c e d i n
n
t h e proof o f theorem 7 . One knows t h e n t h a t a ( u ) is bounded i n L " ( 1 6 , T C ; V ) n d a ( u ) is bounded i n L 2 ( ] 6 , T , [ ; H ) when 6+0. Then a ( u ) and a ( u )
and t h a t
n
n
are c o n t i n u o u s from
i n t o V (equipped w i t h t h e weak t o p o l o g y ) , and
[O,m]
and u are c o n t i n u o u s from ] O , f > [
We check f i r s t t h a t , f o r a subsequence, s t i l l denoted by u f o l l o w i n g convergence p r o p e r t i e s
u
u
n
i n t o H (equipped w i t h t h e weak t o p o l o g y ) .
n'
one h a s t h e
:
t>O
a ( u (t))
+
a ( u ( t ) ) weakly i n V , s t r o n g l y i n H and a.e. on Q,
un(x,t)
+
u(x,t)
n
U
p
-f
a.e.
u(t)
in
on
Q , and hence
LP(Q)
...when
r~
Y -f
p 2 1
0.
I n o r d e r t o prove t h i s , we i n t r o d u c e t h e c a n o n i c a l isomorphism A from V o n t o V' D(A)
=
{ v
E
,
associated with t h e s c a l a r product ( (
VIA v
[U
E
6>0, V
)),
and i t s domain
}. Then we g e t from ( 5 . 2 7 ) and (5.178) t h a t
H c V
t , t Ob 6 ,
U v
E
:
D(A) C V ,
and t h e same p r o p e r t y h o l d s f o r a(u ) .
n
Let then { a E the Dirac function
E
6 ( t ) 2 0,
Then Hence, as
L'(IR),
bE(t-tO)
6E(t)
Av
n
L
>
]
be an a p p r o x i m a t i o n sequence of
=
i f It1 2
0
V')
L,
1
W
for 0
6 (t)dt = 1. E
<
6 S t0-E
and v
E
0 given) :
to+(1 s E ( t - t 0 ) ( ( a ( u ~ ( t ) ) , v )-)
0
0
L'(6,T;
E
to+E -E
>
V.
u ( u ) i s , f o r a subsequence, converging toward u(u) i n L m ( 1 6 , T [ ;
V ) weak s t a r , we g e t ( f o r
t
t
:
t
6 ~ ( t - t o ) ( ( a ( u ( t ) ) , v ) ) d=t & f E ( n ) 0
-E
+
n -to
0.
Ch. III: Incompressible %@Phase Reservoirs
116
E>O be g i v e n ; l e t t i n g
Let
which is t r u e f o r any
q-0 we o b t a i n :
L 0 ; hence we have proven t h a t
I
The f i r s t a s s e r t i o n o f (5.187) f o l l o w s t h e n from ( 5 . 1 8 9 ) , from t h e
f a c t t h a t a ( u ) is bounded i n L"'([G,T];
V ) and from t h e f a c t t h a t
n
D(A) is
dense i n V . The remainder of (5.187) f o l l o w s t h e n immediately. From (5.151) and (5.181) w e see t h a t uo s a t i s f i e s , f o r a l l (5.151) w i t h a
6 applies t o u
n
17
i n s t e a d of a (where a ( c )
=
Hence we have
:
n
for all
+t,T
(5.1 90 1
>
q>O.
5
0
[ n + a ( r ) ] d r ) , s o t h a t theorem
u ( x , t ) 2 u (x,t+T)
0,
n
n
for a.e
x
E
a.
Using (5.187) one can p a s s t o t h e l i m i t i n (5.190) when
+ t,
(5.191)
so that,
for
u(x,t) 2 u ( x , ~ + T )
T>O
almost
every
x
f
R,
u(x,t)
->
U_(X)
f o r a.e. x
E
q+O:
a.
t + u ( x , t ) is a p o s i t i v e d e c r e a s i n g
f u n c t i o n : l e t u s d e n o t e by uc,(x) its l i m i t (5.192)
n>O,
:
a.e. on a.
t 4 m
T h e f u n c t i o n f d e f i n e d by (5.160) is c o n t i n u o u s on 1 0 ,
c o n t i n u o u s from [O,-[ [O,@[
+
w [ ,
a s u is
i n t o H equipped w i t h t h e weak topology. Let t h e n f :
IR be d e f i n e d as
:
n
1I7
VI. The Case of Fields with Different Rock Types
Yt50. Then we get from (5.187)
:
Y t>O
(5.194)
fn(t)
We see from (5.192)
+
when n+O.
f(t)
and (5.194)
that f satisfies (5.161),
and the
rest of the proof is the same as in theorem 6, of course with (5.155) replaced by (5.186). Remark 34 :
0
Theorem 8 covers the practical case of a field initially saturated with oil, when u (x) E 1 on il
:
0
in this case, the
saturation tends to a stationary profile
uo,
which is
identically zero in the special case where no gravity or capillary heterogeneity effects are present and where the imposed exterior pressure is constant on
r
.
0
V I
-
THE
CASE
OF
F I E L D S WITH
DIFFERENT
ROCK
TYPES
Up to now, we have always supposed that the shapes of the non linearities a, b.
J'
d,
the hypotheses (3.17 )
'I.
J
were the same all over il. This was the result of
through (3.20 )
relative permeabilities and
capillary
concerning the dependance of the pressure
laws
upon the spatial
variable x. In petroleum engineering terms, this would be rephrased by saying that we have considered a field containing a single rock type. This notion of V o c k
type?' thus appears as an hypothesis simplifying the spatial
dependance of the relative permeabilities and capillary pressure curves
:
inside a given rock type, (argiles, or sandstone or...), the porosity $, the permeability K and the maximum capillary pressure PCM may vary from one place to the other, but, at a given point x and for a given (actual) saturation
5, the
relative permeabilities and the capillary pressure are
perfectly determined once one has been given the residual saturations
-
S (x) and m
l-sM(x) at that point.
Ch. III: Incompressible Tivo-Phase Reservoirs
178
VI.1
-
THE DIFFERENT ROCK MODELS
Define : d. J
1 V.
=
j=l,2 =
krl kr2 + d=-
(6.1)
krl dl
=
=
krl dl
p2
p1
v
mobility of jth fluid,
3
=
+
kr d 2 2
global mobility,
=
fractional flow.
Then choosing values for k r l , kr2 and p choosing values for d,w and p,. terms of d, v and p
.
clearly amounts to
Hence the rock-models will be specified in
In all rock models for incompressible fluids, we take as given the following functions of the reduced saturation S (6.2)
(6.3)
S
->
S
->
:
pc(S) satisfying ( 3 . 2 0 1 , w(S) satisfying
w(0) = 0 ,
v(l)
=
1,
L
an increasing function of S.
As for the choice of the global mobility function d , we shall distinguish two cases Case 1 :
:
Rock model of t h e f i r s t kind : we take as given the following
function of the reduced saturation S
S (6.4)
->
d(S) satisfying
d(0) 2 d2,
d(1) 2 d l , d(S) > 0 .
The relative permeabilities generated by this model are (6.5) These
relative
permeabilities
depend
only
on
the
reduced
saturation S , which is the assumption made in sections I1 and 111. All
equations in these paragraphs have been established using a rock model
of the first kind.
VI. The Case of Fields with Different Rock Types
179
This rock model can be used for two phase flows without exchanges between phases, where
-
sM remain away from 0
Sm and
and 1.
However, when exchanges between phases take place (see chapter IV, 8111 and IV). The actual saturation
s
may take values outside of the
and sm(x) may approach to 0 and sM(x) may approach
interval [zm(x,) sM(x)],
1 as one tends toward the critical point. So one will need to calculate the
relative permeabilities (but not the capillary pressure) for values
s of
the
But,
actual saturation lying outside the interval [sm(x), ZM(x)].
continuing the relative permeabilities given by (6.5) outside the interval (x), S (x)] would lead to discontinuous relative permeabilities, as kr 1 m(x,SM ) = - I 1, whereas the physics indicates that krl(x,l) = 1 (cf. So rock models of the first kind are not valid in situations figure::61 [S
El(1)
where the residual saturations ?m(x)sl-sM(x)
approach zero.
Rock model of the second kind : we take as given the following function of the actual saturation :
Case 2 :
l
(6.6)
-
S -> d(0)
d(S) =
d2,
satisfying d(1)
=
d l , d(S)
>
0
and we suppose that
(6.7)
the function
1 1
given in (6.3) is continued by 0 for S 2 0 and
by 1 for S t 1. Now we can generate relative permeabilities over the entire range of actual saturations
by setting
(6.8) where S is the reduced saturation corresponding to
5
at the point x (given
by (6.16). The formula (6.8) yields continuous relative permeabilities when
-
Sm
0 and
-
Sm + 1 , as shown in figure 17, and hence this rock model has to be chosen when exchanges between phases take place. +
Of course, the shapes of the relative permeability curves, when expressed as functions of the reduced saturation S, will slightly change from one point x to the other.
Ch.III: Incompressible Tbo-Phase Reservoirs
180
I
Figure 16 : The discontinuous limiting relative permeabilities (dashed lines) obtained with a rock model of the first kind when the residual saturations tend towards zero.
0 Figure 17
The continuous limiting relative permeabilities (dashed
line) obtained with a rock model of the second kind when the residual saturations tend towards zero.
181
VI. The Case of Fields with Different Rock Q p e s
Remark 35 :
For the practical determination of functions, it is enough to know
the d(5)
and v ( S )
:
for one point xo, i.e. for one rock sample,
-
one set of relative permeability curves z+kr.( x o , S ) , j = l,2
(6.9)
J
over the whole interval of (non reduced) saturation and the residual saturations sm(xo) and l-zM(xo),
(6.10)
the viscosities 11, Then
d(2)
and
and
\,(S)
u2 of the two fluids. are
determined by
(6.1) without
ambiguity.
0
The two-phase equations developped in sections I1 and I11 for the (implicit) case of a rock model of the first kind, remain valid for a rock model of the second kind, with the following modifications d(S) has to be replaced by d(x,S)
=
:
d(S),
(with an evident abuse of notation), (6.11)
b ( S ) is equal to > ( S ) , 0 b ( S ) , b (S) become bl(x,S), b 2 ( X , S ) . 1
2
and (6.12)
has to be replaced by
The theoretical results developped in section V remain valid with a rock model of the second kind, as the supplementary dependance of d, b l , b2 on x does not change the proofs.
Ch.III: Incompressible no-Phase Reservoirs
182
VI.2
-
THE CASE OF A FIELD WITH M DIFFERENT ROCK TYPES
Let us now consider the case of a field R , which contains M
.
different rock types. Let Om, m=l ..M be the spatial domain occupied by convention that
rmk=
rmll be
m
the boundary between R and R', with the Q 0 if Rm and R do not meet, or meet only on a line
each type of rock, let
or a point for n=3 or meet only on a point for n=2, and attach a superscript m to each quantity related to Rm. According to the notion of rock-type, not only the shape of relative permeabilities and of capillary pressure curves, but also the maximum capillary pressure PCM may differ in each Rm.
It is hence
necessary to allow for discontinuities of the maximum capillary pressure P (x) across the boundaries rmQ; so PCM (x) will consist of M regular CM m functions PCM(x), defined over Q m' and which do not necessarily meet continuously at the internal boundaries TmQ. We look now for the equation in R , appropriate when for each rock-type, a rock-model of the first kind is used a)
Inside each of the Q",
:
we can proceed as we have done in
m and Pm the saturation and global
sections I1 and 111. We denote by S m pressure in R , which satisfy :
The
equations +m
capillary flow r
,
governing,
in
the global pressure
Qm
the
saturation Sm,
+m
Pm and the (water+oil) flow q
are (cf. (3.571, (3.58), (3.62), (3.63) and (3.72)) : +m Y x e om, (6.15) div qo = 0
and,
(6.18)
the
v
x e Rm'
183
VI. The Gme of Fields with Different Rock Types
Figure 18 : An example of a field with four different types of rock
R 2 and we have
(here il
+m r
(6.19)
We
13)
boundaries
=
m
r 13
0 and
=
m
-$PcM
r2,,
=
0 with our convention).
m
t x
grad a ( S )
€
am.
have then to choose continuity conditions at the
rmll between
different types of rock.
For convenience, we shall denote the jump of any quantity 0 across
by
mR
[ e
1;
em - e p.
=
On any T m R , Il=1 conditions
...M ,
. m=l.
..L, we have to satisfy the following
:
Continuity of pressure : each of the pressure to be continuous (6.20)
[
Pj
1;
=
P
1
and P 2 has
:
0.
- Conservation of masses has to be continuous
:
:
the flux of each of the two fluids
Ch. III: Incompressible Tluo-PhaseReservoirs
184 +
+ m
(6.21)
t q o * v l I 1 = 0,
(6.22)
[
where
;,.;I;
=
0,
is any normal t o TmL. 'I)
Now we want to obtain from (6.15) to (6.22) equations
valid over all P. We f i r s t i n t r o d u c e f o r t h a t purpose f u n c t i o n s d e f i n e d
........as
(6.24)
+ d i v qo
(6.25)
0
(6.26) !
as at
$,(x,t)
=
0
d i v $1 =
x
E
a, t t >
t x
€
a, t t > 0
f
+
+
=
r'(x,t)
0 2 +
1
j=o
soon as x
b.(x,S(x,t)) G.(x,t) J J
f X E
0,
a, t > O .
E
nm
VI. The Case of Fields with Different Rock Q p e s
185
(6.27) (6.28)
So we first get rid of those gradients by using a variationnal
formulation of (6.16), (6.19). Let the test function
?
be any (regular)
mapping from Om into 1". Multiplying (6.16) and (6.19) scalarly by integrating over Om and using a Green's formula we obtain :
?,
(6.30)
+
i m is the exterior normal to the boundary m of Qm. We combine now (6.291, (6.30) in the following ways : +m to * For every regular z:Q -+ R n , we set s = restriction of
where
m'
9
and sum up equations (6.29) and (6.30) for all m , which yields
:
But the above equations alone are not equivalent to (6.291,
(6.30). We have to use also other combinations.
Ch. III: Incompressible nYo Phase Reservoirs
186
For every m , II
= 1,2.
zm
..M, m < I I , for every regular
g
: Q+lRy
with
support in the interior of u 5, , set gm = the restriction of s' to Qm, = the restriction of s' to QII, and take the difference between ( 6 . 2 9 )
;'
(resp. ( 6 . 3 0 ) ) on ilm and on O R . This yields
1
s'.:
(pm+pR)
Pm div
=
rmL
s'
:
P R div . +s
-
%
nm
(6.33)
Y m,L=1 aF
1
s'
: Q
...M, m < R , +
lRn, supp
2.;
(am(Sm)+aL(SR))
:is in the interior of -Qm crm(Sm)div
=
m'
rml
-
4
r - s -j -
(6.34)
Qm "CM
aF
m,R=l...M,
aF ;:Q+IRn,
s'
+I
-
+
1
%+
s'
I I R
(S )
-
QII,
div
5
r - s -
Q R "CM
m
supp
CI
U
is in the interior of
-
'ilm
u QRR'
Equation ( 6 . 2 4 ) through ( 6 . 2 8 ) and ( 6 . 3 1 ) through ( 6 . 3 4 ) are the sought equations governing the evolution of S and P over a field containing different types of rock. We
see
that
taking
into
account more
than
one rock
type
introduces, besides the forseeable dependance on x of all non linearities, integrals over the boundaries
rmII separating different rock types in the
right-hand side of the variational equations (6.31) and (6.32) determining
Go
and
r'
(caution : the flux of
r'
across the boundaries
contiquous ! ) , plus additional equations ( 6 . 2 7 1 , associated with the boundaries TmR
.
rmII is not
( 6 . 2 8 ) and (6.331, (6.34)
VI. The Case of Fields with Different Rock Types
187
We shall see in chapter V, for the case of a single rock type (i.e.
when the last term in (6.31) and (6.32)
disappears), that equations
(6.24) through (6.26) and (6.31), (6.32) are the basis for the numerical calculation of P and S by mixed and discontinuous finite elements. The numerical techniques described in chapter V can be generalized'to the case of multiple rock-types, provided that the finite element mesh is chosen in such a way that the TmR boundaries consist of edges of elements of the mesh,