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8/12/2019 Cours BO 2014.Pdf1
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Simulation of oil recovery with miscible gas injection
using a black oil model
R. Masson
04/06/2014
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1 Black oil model
2 Discretization of Black Oil models
3 Solution of the discrete system of equations
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Motivations of compositional two phase oil gas models
Saturated or under saturated oil reservoirs: volatile HydrocarbonComponents (HC) evaporate with pressure drop
Oil recovery by miscible gas injection
Gas condensate reservoirs: liquid phase appears with pressure drop
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Black Oil Model: definition
Two phases: oil and gasTwo components: heavy HC (h), and light HC: (l)
The light component can dissolve into the oil phase
The heavy component cannot evaporate into the gas phase
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Black Oil Model: conservation equations
t
o (1 c) So
+div
(1 c) o
kr,o(So)
o Vo
= 0,
to c So +g Sg+ divc o kr,o(So)
o Vo
+ div
g kr,g(Sg)
g Vg
= 0,
Vo = KP og
, Vg = K
P gg
,
So + Sg = 1.
Oil phase properties: o(P, c), o(P, c)
Gas phase properties: g(P), g(P)
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Black Oil Model: thermodynamical equilibrium
If the oil and gas phases are both present, the thermodynamical equilibriumimposes that c= c(P).
Sg (c(P) c) = 0,Sg 0,
(c(P) c) 0.
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Black Oil Model
t
o(P, c) (1 c) So
+div
(1 c) o(P, c)
kr,o(So)
o(P,c)Vo
= 0,
t
o(P, c) c So +g(P) Sg
+div
c o(P, c)
kr,o(So)
o(P,c) Vo
+ div
g(P)
kr,g(Sg)
g(P) Vg
= 0,
Vo = KP o(P, c)g
, Vg = K
P g(P)g
,
So + Sg = 1,Sg (c(P) c) = 0,Sg 0,(c(P) c) 0.
Remarks:
Only one set of unknowns P, So, Sg, cand equations because the oil phaseis always present.
No flash needed.
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Black Oil Model: Bo, Bg and Rs
Bo
= Vo
Vo,h,s=
h,s
o(P, c) (1 c), B
g=
Vg
Vg,l,s=
l,s
g(P)
Rs= Vo,l,s
Vo,h,s=
1 c
c
l,s
h,s
h,s and l,sare the oil and gas densities at surface conditions. T is fixed.
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Black Oil Model: phase properties in terms ofBo, Bg andRs
Bo(P), Rs(P), o
(P) are given at equilibrium (saturated oil).Bg(P) and
g(P) are given.
For saturated oil (c= c(P) orP=Pb(c))and undersaturated oil (c< c(P) orP>Pb(c))
Bo(P, Pb) = Bo(Pb) + cBo(P Pb), o(P, Pb) = o(Pb) + co(P Pb)
o(P, c) = h,s
(1 c) Bo(Pb(c)) + cBo(P Pb(c)),
o
(P, c) = o
(Pb(c)) + co(P Pb(c)),c(P) =
l,s
h,s Rs(P) +l,s,
Pb(c) = c1(c),
g(P) = l,s
Bg(P).
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Discretization of Black Oil model: 2D fivespots test case
2D horizontal reservoir with 4 gas injectors at the corners and 1 producerat the center.Prescribed pressures Pinj for the injection wells and Pprod for theproduction wellInitial state: under saturated oil at Pprod
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Discretization of Black Oil model: diffusion fluxes andupwinding (without gravity)
Diffusion fluxes (without gravity for simplicity)
Fint1(), = F
int2(),
=Tint p1() p2(), int,Fprodw = WIprodw
pprod(w) pprodw
, w prod,
Finjw = WIinjw
pinj(w) p
injw
, w
inj,
Upwinding (without gravity): for all int
up() =
1() = si Fint1(), >0,
2() = si Fint1(),
0.
Five Spots example: prod = one producer, inj = the four injectors.
Di i i f Bl k Oil M d l di
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Discretization of Black Oil Model: discrete system(without gravity)
Heavy HC residual
Rh,
= o(P
, c
) So
(1 c
) mn1h, ||
t
+
int
(1 cup()) o(Pup(), cup())
kro(Soup()
)
o(Pup(), cup()) Fint,
+
wprod|prod(w)=
o(P, c) (1 c) kro(So)
o(P, c)
Fprodw
+ = 0,
Di i i f Bl k Oil M d l di
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Discretization of Black Oil Model: discrete system(without gravity)
Light HC residual
Rl, =
o(P, c) So c+
g(P) Sg m
n1l,
||t
+ int
cup() o(Pup(), cup())
kro(Soup()
)
o
(Pup(), cup())
Fint,
+
int
g(Pup())krg(S
g
up())
g(Pup()) Fint,
+
wprod|prod(w)=
c
o(P, c) kro(So)
o(P, c)+g(P)
krg(Sg )
g(P)
Fprodw
+= 0,
+ winj|inj(w)=
g(Pinjw
) krg(1)
g(Pinj
w)Finjw
= 0,
Di i i f Bl k Oil M d l di
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Discretization of Black Oil Model: discrete system(without gravity)
Local closure laws (volume conservation an thermodynamical equilibrium)
So
+ Sg
= 1,
Sg (c(P) c) = 0,
Sg 0,
(c(P) c) 0.
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Discretization of Black Oil Model: Newton type algorithm
The oil phase is always present in all cells (Sor >0)
Gas phase indicator for all cell K
I = 1 if gas phase is present0 if gas phase is absent
Given I for all cells, the closure laws are eliminated at the non linear level: ifI= 1 : S
o = 1 S
g , c = c(P),
ifI= 0 : Sg = 0, S
o = 1,
the unknowns X for the Newton linearization depend on I for all cells:
ifI= 1 : X =
P, S
g
,
ifI= 0 : X =
P, c
.
S l ti f th di t t f ti N t t
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Solution of the discrete system of equations: Newton typealgorithm
The conservation equations are linearized with respect to the set of
unknownsX =
X
K
taking into account the previous elimination of
the closure laws for a given I:
R(X) =
Rh,(X)
Rl,(X)
R= RK, dX = RX
1
R
On each cell update of the unknowns P, S, c and of the Gas phaseindicatorI to satisfy the inequality constraints S
g 0 and c c(P),
K.
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Solution of the discrete system of equations: algorithm
P= (P)K
, I= (I)K
, S= (S =Sg
)K
,c= (c)K
.
R=
Rh,, Rl,
K
.
Initialization ofP=P0,I= I0, S=S0, c=c0 and t at t= 0.
Time loop: while t then Restart the time step: t= tt,
t= t/2, P= P0,I=I0,S= S0, c= c0
Else new time step: update t
and setP
0
=P
,I
0
=I,S
0
=S
,c
0
=c
Solution of the discrete system of equations: update of
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Solution of the discrete system of equations: update ofP, S, c andI
dX = J1 R
Newton update: for = 1, , N:
P P+ dX(21),ifI = 1 then S S+ dX(2), c = c(P)else c c+ dX(2), S = 0.
Gas phase indicator update: for = 1, , N:IfI = 1 then
ifS < 0 thenI 0, S 0, c c(P)
else ifS > 1Sor thenS 1 Sor,
Else ifI = 0 then
ifc > c(P) then
I 1, S 0, c c(P)
Solution of the discrete system of equations: computation
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Solution of the discrete system of equations: computationof the residual R(P, S, c,P0, S0, c0, t)
numbering of the cells: = 1, , N
numbering of the equations:
Rh R(2k 1), = 1, , N
Rl
R(2
k)
, = 1
,
, N
Loop on cells = 1, , N
o = o(P, c),
g = g(P),
R(2 1) R(2 1) +
o (1S) (1c)o(P0,c0) (1S0) (1c0)
||
t
R(2) R(2) +
o
(1S) c+g S
o(P0,c0
) (1S0
) c0g(P0
) S0
||
t
Solution of the discrete system of equations: computation
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Solution of the discrete system of equations: computationof the residual R(P, S, c,P0, S0, c0, t)
Loop on interior faces:
Fh = (1 cup()) oup()
kro(1Sup())
o Fint1(),
Fl =
cup() oup()
kro(1Sup())
o +g
up()
krg(Sup())
g
Fint1(),
R(21() 1) R(21() 1) + Fh
R(22() 1) R(22() 1) Fh
R(21()) R(21()) + FlR(22()) R(22()) F
l
Solution of the discrete system of equations: computation
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Solution of the discrete system of equations: computationof the residual R(P, S, c,P0, S0, c0, t)
Loop on injector wells w:
R(2inj(w)) R(2inj(w)) +g(Pinj
w)
krg(1)
gFinj
w
Producer well w
R(2prod(w) 1) R(2prod(w) 1) +
o (1 c)
kro(1S)o F
prodw
+
R(2prod(w)) R(2prod(w)) +
c o kro(1S)
o +g krg(S)g
Fprodw
+
Solution of the discrete system of equations: Newton
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Solution of the discrete system of equations: Newton
linearization R(X) +
RX
dX = 0
Unknowns numbering for the Newton linearization: = 1, , N
dX(2 1) = dP,
dX(2) =
dS if I = 1,
dc if I = 0.
Notation X1, X2
=
P, S
if I= 1 c= c(P),
P, c
if I= 0 S= 0.
Derivatives: example ofo
(P, c):
o
X1 =o
P(P, c) +
o
c (P, c)
c
P(P)I +
oP
(P, c)
(1 I)
o
X2 = o
c (P, c)(1 I)
Solution of the discrete system of equations: Newton
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Solution of the discrete system of equations: Newton
linearization R(X) +
RX
dX = 0
Jacobian: example of the accumulation term for Rh:o (1 S) (1 c)
o(P0, c0) (1 S
0) (1 c
0)||
t
Loop on cells
J(2 1, 2 1) J(2 1, 2 1) +I
o (1 S)
cP
(P)
+(1 S) (1 c)
o
P(P, c) +
o
c(P, c)
cP
(P)
||t
+(1 I)oP(P, c) (1 S) (1 c)
||t
J(2 1, 2) J(2 1, 2) +I
o (1 c)
||t
+(1 I)
o
c(P, c) (1 S) (1 c)
o (1 S)
||t
Solution of the discrete system of equations: Newton
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Solution of the discrete system of equations: Newton
linearization R(X) +
RX
dX = 0
Jacobian: example of the interior face flux term:
Fh = (1 cup()) oup()
kro(1Sup())
o Fint1(),
Loop on interior faces : 1 =1(), 2 =2(), up=up()
Gh = (1 cup) oup
kro(1Sup )
o
J(21 1, 21 1) J(21 1, 21 1) + Gh T
int
J(22 1, 22 1) J(22 1, 22 1) + Gh T
int
J(21 1, 22 1) J(21 1, 22 1) Gh T
int
J(22 1, 21 1) J(22 1, 21 1) Gh Tint
to be continued ...
Solution of the discrete system of equations: Newton
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y q
linearization R(X) +
RX
dX = 0
suite of the interior flux term Fh
=I
(1 cup)
o
P(Pup , cup) + (1 cup)
o
c(Pup , cup)
cP
(Pup)
cP
(Pup)oup
+ (1 I) (1 cup)
o
P(Pup , cup)
kro(1Sup )
o Fint1,
= I
(1cup )oup
okroS
((1 Sup)
Fint1,
+(1 I)oup + (1 cup)
o
c(Pup , cup)
kro(1Sup )
o Fint1,
J(21 1, 2up 1) J(21 1, 2up 1) +J(22 1, 2up 1) J(22 1, 2up 1) J(21 1, 2up) J(21 1, 2up) +J(22 1, 2up) J(22 1, 2up)
to be continued for all remaining terms of the Jacobian ...
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