UNIVERSITI TEKNOLOGI PETRONAS

PAB3053
RESERVOIR MODELING AND SIMULATION
MAY 2017

Dr. Mohammed Abdalla Ayoub

 Outlines
Introduction

Governing Equations for black oil (Beta) models

Discretization formula and solution Difference

Oil-water Formulation

Concluding remarks

2
 Main Components of A Simulator
Geological model

Reservoir model

Fluid model

Petropysical model

Mathematical model.

3
 Rock
1
Hydrocarbon

water

Rock matrix
1

Mass of phase m into CV over t Mass of phase out of CV over t ----(1)

= Mass of Phase m accumulated in CV over t

This is a general continuity equation 4

 Difference in systems
Components: Gas, oil, and water (usually received at the
surface)
Phases: Liquid, vapor, aqua (those formed at the reservoir
conditions)
Oil Volume + Gas Volume + Water Volume = Pore Volume

Components thought to correspond to chemical molecules

Phases are homogeneous mixtures of these molecules
separated from other phases by a fluid interface.

5
 Continuity equation in field units
Let vm,x be the velocity of phase
m (m=o, w), in the x-direction in
bbl/(ft2-D).
To convert vm,x to ft/D we
multiply it by 5.615 ft3/bbl
Let m be the density of phase
m in lbm/ft3.
Let Sm be the saturation of
Mass of phase m entering the
phase m
CV through the surface at
x-x/2 over t=

5.615 ( m vm , x )x x / 2 Ax x / 2 t (lbm) ----(2)

6
 Continuity equation in field units

We may have a source/sink term where the phase m may be

added to or removed from the system.
Let qm,sc be volumetric flow rate of phase m at surface
conditions in STB/D, then mass entering into CV by a source is
then:
Mass of phase m entering into the CV by a source over t =

5.615 qm , sc m , sc t (lbm) ----(3)

I define qm,sc > 0 for production and qm,sc < 0 for

7
injection.
 Continuity equation in field units

Mass of phase m leaving the CV through the surface at x+x/2

over t =Mass of phase m entering into the CV by a source over
t = 5.615 ( m v m , x )x + x / 2 Ax + x / 2 t (lbm) ----(4)

In Eqs. 2 and 3, Ax + x / 2 represents the cross-sectional area

perpendicular to flow at x x 2 For the CV considered, the
cross-sectional area perpendicular to flow in the x-direction
does not change with x and is equal to A x x = zy
2

8
 Continuity equation in field units

Mass of accumulated in CV over t = Mass of phase m present

in CV at t+t - Mass of phase m present in CV at t ----(5)

where
Mass of phase m present in CV at t+ t = + (lbm )
Mass of phase m present in CV at t = (lbm )
Here Vb is the bulk volume of CV=x.y.z

9
 Continuity equation in field units

We assume that bulk volume Vb does not change with time so

using the equations in the preceding slide, we have
Mass of accumulated in CV over t
----(6)

Using Eqs. 2, 3, 4, and 5 in Eq. 1 gives

5.615 ( m vm , x )x x / 2 yzt 5.615 ( m vm , x )x + x / 2 yzt

----(7)
5.615 qm , sc m , sc t = [( m S m )t + t ( m S m )t ]xyz
10
 Continuity equation in field units
Dividing both sides of Eq. 6 by x y z t and rearranging the
resulting equation gives

( m vm , x )x + x / 2 ( m vm , x )x x / 2 5.615qm , sc m , sc
5.615
x Vb
----(8)
( m S m )t + t ( m S m )
=
t t

11
 Continuity equation in field units
Now, taking limits in Eq. 7 as x and t approaching zero, and
then using the limit equation below:
f
= lim x 0
( 2
) (
f x + x , t f x x , t
2
)
x x ----(9)

f f (x, t + t ) f (x, t )
= lim x 0
x t
gives
( m vm , x )x + x / 2 ( m vm , x )x x / 2 5.615qm , sc m , sc
5.615
x Vb
( m S m )t + t ( m S m )t
----(10a)
=
t
12
 Continuity equation in field units
Or

( m vm , x ) qm , sc m , sc ( m S m )
5.615 5.615 = ----(10b)
x Vb t

for m=o, w. Eq. 10b is called the continuity

equation for 1-D linear flow.

13
 Continuity equation in field units
If we consider flow in all three directions (in x-,y-, and z-
directions), we can derive the 3-D continuity equation, similarly
to 1-D linear case. The 3-D continuity equation is given by

( m vm , x ) ( m vm , y ) ( m vm , z )
5.615 + +
x y z
qm , sc m , sc ( m S m ) ----(11)
5.615 =
Vb t
for m = o, w.
14
 Continuity equation in vector notion

In x-y-z coordinate system, we vm , x

define velocity vector for a vm = vm , y ----(12)

phase m as vm , z

f
x
From calculus, we define the gradient of a f
scalar function f(x,y,z) in x-y-z coordinate f = ----(13)
y
system as f

z 15
 Continuity equation in vector notion

If is a vector function, which changes with x,y,z, then its

gradient of (also called the divergence of ) is a scalar and is given
by

x
v x
v x v y v z
.v = v y = + + ----(15)
y x y z
v z

z
16
 Continuity equation in vector notion

Mass flux of phase m is given by m vm (which has units of
Mass/(Area-Time)). Note that mass flux is a vector. Then its
divergence in x-y-z coordinate system is given by

( m vm , x ) ( m vm , y ) ( m vm , z )


.( m vm ) =
----(14)
+ +
x y z

17
 Hints
The mass flow rate per unit area in a given
direction for any fluid is the product of the
density of the fluid and its velocity in that
direction.

Cp is the concentration of any phase (oil, water,

and gas) in the block defined in mass per unit
volume
18
 Continuity equation in vector notion

Comparing the left-hand side of Eq. 11 with Eq.14, then using

Eq. 14 in Eq. 11, we can write Eq. 11 in terms of divergence of
mass flux as

qm , sc m , sc ( m S m )
5.615.( m vm ) 5.615 = ----(14)
Vb t

19
 Continuity equation in vector notion

Eq. 15 is the continuity equation that applies for any coordinate

system. So, it is general. By using coordinate transformation, we
in a given orthogonal
can find the expression for .( m vm )
coordinate system and use it in Eq. 15 to obtain the continuity
equation in that coordinate system. For example, in cylindrical
coordinate system (r--z), is given as

1 1

.( m vm ) = (r mvm,r ) + ( mvm, ) + ( mvm, z ) ----(16)
r r r z
20
 Continuity equation in vector notion
Using Eq. 16 in Eq. 15 gives the continuity equation in cylindrical
coordinates:
1 1
5.615 (r mvm,r ) + ( mvm, ) + ( mvm, z )
r r r z
----(17)
5.615qm , sc m , sc
= ( m S m )
Vb t

If we assume that flow occurs only in radial (r-) direction, i.e.,

, = , = 0 in Eq.17, then Eq. 17 reduces to

21
 Continuity equation in vector notion

1 5.615qm , sc m , sc
5.615 (r mvm,r ) = ( m S m ) ----(18)
r r Vb t

which is the continuity equation for radial flow.

22
 Darcys Equation in Differential Form

In field units, velocity of phase m in x, y or z direction is given

by Darcys equation
ku k rm pm D
vm ,u = 1.127 10 3
m
m u u ----(19)

for u = x, y, or z , and m = o, w

where ku is the absolute permeability in the u direction,

u = x,y, or z.
23
Combining Darcys Law with Material
Balance
Continuity equation with accumulation terms
in 3D :
(v x ) ( ) ( )
= +Q .( v ) = +Q
x t t

in 3D with gravity :
k ( )
. (P D ) = +Q
t
where
= g
24
 Black Oil Model

=

Surface Volume Reservoir Volume

1 B
1 1

S

u

25
 Beta Model for Multiphase Flow

The beta model can consider flow up to three

phases (liquid hydrocarbon or oleic phase,
hydrocarbon vapor or gaseous phase, and water or
aqueous phase).

It assumes that each hydrocarbon phase (oleic or

gaseous phase) consists of only two components -
the oil and gas components.
26
 Beta Model for Multiphase Flow

In beta model:

Oil phase: Liquid hydrocarbon at any pressure and

temperature.

Oil component: Liquid hydrocarbon at stock tank

(standard) conditions.

27
Beta Model for Multiphase Flow in a
Reservoir
In beta model:

Gas Phase: Hydrocarbon vapor at any

pressure and temperature.

Gas Component: Hydrocarbon vapor at

standard conditions.

28
Beta Model for Multiphase Flow in a
Reservoir
In beta model:

Water phase: Aqueous water at any pressure and

temperature.

Water Component: Liquid H2O at standard conditions.

Because water component cannot exist in oil and gas

phases, water phase = water component in beta model.
29
Beta Model for Multiphase Flow in a
Reservoir
Note that in fact,
Oil component can exist in either the oil or gas phase at reservoir
conditions.
Gas Component can exist in either the oil or gas phase at
reservoir conditions.

Note: if we wish, we can allow hydrocarbon gas dissolved in the

water phase, in this case gas component can also exist in water
phase at reservoir conditions, but traditional beta model assumes no
gas component exists in water and assumes that oil component
exists only in the oil phase.
30
 Simplified Beta Model

In a simplified (traditional) beta model assumes that

Each hydrocarbon phase consists of only two components.

The gas component may exist in either the oil or gas phase.

The oil component exists only in the oil phase; i.e., oil cannot be
vaporized into gas phase.

Water component does not exist in hydrocarbon phases.

Dispersion of components are negligible.

Instantaneous thermodynamic equilibrium thru reservoir.

31
 Applicability of the Simplified Beta
Model
This model is appropriate for low-volatility oil systems, consisting
of mainly of methane and heavy components.

Note that oil component exists only in the oil phase, i.e., no volatility
of oil.

In this simplified two-component hydrocarbon model, it is assumed

that no mass transfer occurs between the water and the other two
phases. In other words, this means that water component cannot
exist in oil or gas phase.
32
 Model Assumptions, cont.
Beta models assume that PVT properties are only
function of pressure, while compositional models
assume that PVT properties are function of both
pressure and composition, in the case of isothermal flow
(i.e., flow at constant temperature throughout the
reservoir).

33
 Model Assumptions
The basic black oil model assumes multi-phase,
isothermal flow of three phases; 2 hydrocarbon phases
(oil and gas) and water. The hydrocarbon system is
approximated by 2 components:
(1) a non-volatile (black) oil, and

(2) a volatile gas which is soluble in the oil phase.

There is also a water component.

34
Basic Black Oil Model Assumptions (-Model)

Component Phase
Oil liquid
Gas vapor
Water aqua

Water and oil are immiscible, they do not exchange mass or

change phase. The gas component is soluble in the oil phase but
not the water phase. Water is usually assumed to be the wetting
phase, with oil having intermediate wettability and gas being
non-wetting.
35
P-T Diagram

36
 PVT Data, Bo

Gas out of
Solution
Surface
(Ps,Ts)

Pb

Bob
Oil
Reservoi
r (P,T)

In field units, Bo is typically expressed in Hg

Rbbl/STB, Bg is Rbbl/Scf or cuft/Scf. 37
 Beta Model Data, Rs

Dissolved Gas Oil Ratio:

In field units, Rs has a unit of Scf/STB

38
 Beta Model Data, densities
o

w
o =
1
Bo
[
o , sc + Rso g , sc ]
Pb P

w =
1
[ w,sc ] This assumes no oil component
exists in the gas phase
Bw

P
g

g =
1
Bg
[
g , sc ] This assumes no gas component
is dissolved in the water phase

P 39
 PVT Data, Oil density (o)
is almost always reported in terms of a stock tank gravity (which is a dead
oil); most simulators adjust this value to reservoir conditions using the
following relationship below the bubble point
o , Sc + 13.56 g Rso
o =
Bo
where
o = oil density, lb/ft 3
oS = stock tank oil density, lb/ft 3
g = gas gravity
Rso = dissolved gas, MCF/STB
Bo = oil formation volume factor, RVB/STB.

Above the bubble point, Rsobp is used in place of Rso. A normal range of API
gravities is from 45 to 10 corresponding to densities of 50.0 and 62.4 in lb/ft3
respectively. 40
 PVT Data, Gas density (g)
is usually input as a gas gravity (g) or in units of lb/MCF. The relationship
between these two quantities at standard conditions is:
(28.9)(14.7 )(1000) g Rso
g , Sc =
(10.73)(460 + 60)

g , Sc = gas density, lb/MCF

g = gas gravity
1000 g , Sc
and gas densities are generated from g = ,
5.615 Bg
where
Bg = gas formation volume factor, RVB/MCF
and density gradients are calculated with g / 144000 in psi/ft .
A normal range for gas gravities is from 0.6 to 1.2 which corresponds
to values of 45.7 to 91.4 in lb/MCF.
41
 PVT Data, Water density (w)

is required as either a density in lb/ft3 or as a specific gravity (w). The

relationship between the two is:
w = 62.4 w
and due to salinity the water density at standard conditions may be
estimated from:
w,Sc = 62.4 + 0.465xS,
where
S = salinity,%.
Finally, the standard density may be corrected to reservoir conditions using

wS
w =
Bw
and the gradient calculated from w/144. Most oilfield waters have densities
slightly greater than 62.4 lb/ft3.
42
 PVT Data, Oil compressibility (co)
Can be or bubble point; however, the only value(s) required in simulators are for
undersaturated conditions where the compressibility is used to adjust the oil
formation volume factor from bubble point conditions, using either:
co (P Pbp )
Bo = Bobp e
or [ ]
Bo = Bobp 1 co (P Pbp )

Co may be measured in the lab or obtained from correlations. Standard units

for oil compressibility are /psi which yields oil compressibility values ranging
from 6 106 to 20 106;
a more recent unit is the microsip which is 106 times greater.

43
Beta Model-Continuity Equation (details)
Volumes of oil, water and gas components in the CV at the
standard or stock tank conditions at a given time t:
xyzSo
Volume of Oil (STB) = Oil component only in Oil
5.615Bo phase
xyzSw
Volume of Water(STB) = Water component only in water
5.615B w phase
xyz Sg So Gas component exists in
Volume of Gas (Scf ) = + Rs both oil and gas phases
5.615 Bg Bo

44
Free gas Dissolved gas
Beta Model-Continuity Equation (details)

Let vo,x, vw,x and vg,x be the velocities of oil, water, and gas phases in

the x-direction:

Bo is in Rbbl/STB, Bg is in Rbbl/Scf, Bw is in Rbbl/STB.

Velocities (vo,x, vw,x and vg,x) are in Rbbl/(ft2-D).

Then, the volumes of oil, water and gas components entering

and leaving the element in the x-direction.

45
Beta Model-Continuity Equation (details)
v
Volume of Oil entering the CV = yzt o,x
Bo x
v
Volume of Water entering the CV = yzt w,x
Bw x
yzt v g,x v o,x
Volume of Gas entering the CV = +
5.615 Bg Bo
x

v
Volume of Oil leaving the CV = yzt o,x
Bo x + x
v
Volume of Water leaving the CV = yzt w,x
B w x + x
yzt v g,x v o,x
Volume of Gas leaving the CV = +
5.615 Bg Bo
x + x 46
Beta Model-Continuity Equation (Oil)
Accumulation of oil component in CV =
Oil component entering into CV - Oil component leaving the CV
xyz S o xyz S o v v
= yzt o,x yzt o,x
5.615 Bo t + t 5.615 Bo t Bo x Bo x + x

Dividing both sides by x y z t and then taking limits as x 0 and

t 0

1 So v o,x q o,sc
=
5.615 t Bo x Bo Vb
qO,sc > 0 for prod., and qO,sc < 0 for injection,
qO,sc = 0 for shut-in
47
Beta Model-Continuity Equation (Water)
Accumulation of water component in CV =
water component entering into CV - water component leaving the CV
xyz S w xyz S w v v
= yzt w,x yzt w,x
5.615 Bw t + t 5.615 Bw t Bw x Bw x + x

Dividing both sides by x y z t and then taking limits as x 0 and

t 0

1 Sw v w,x q w,sc
=
5.615 t B w x B w Vb
qw,sc > 0 for prod., and qw,sc < 0 for injection,
qw,sc = 0 for shut-in
48
 Beta Model-Continuity Equation (Gas)
Accumulation of Gas component in CV =
Gas component entering into CV Gas component leaving the CV

xyz S g S o xyz S g S o
+ Rs + Rs

=

5.615 B g Bo
5.615 B g Bo
t + t t

v g,x vo, x v g,x vo, x

yzt + Rs + Rs
B
B g Bo x g Bo x + x

Dividing both sides by x y z t and then taking limits as x 0 and

t 0
1 Sg So v g,x v o , x q g,sc

+ Rs = + Rs
5.615 t Bg Bo x Bg Bo Vb

qg,sc > 0 for prod., and qg,sc < 0 for injection,

qg,sc = 0 for shut-in 49
 Darcys Equation (Oil, Water, and Gas)
The superficial velocity of each phase in the previous equations is
described by Darcys equation:

k x k ro po D
vo , x = 1.127 10 3
o
o x x
k k p g
3 x rg D
v g , x = 1.127 10 g
g x x
3 k x k rw p w D
vw, x = 1.127 10 w
w x x
50
Diffusivity Equation (Oil, Water, and Gas)
Combining the continuity equation for each component by the Darcys velocity
for each phase, then we obtain the diffusivity Equations for each component:
1 So 3 k x k ro p o D q o,sc
Oil : = 1.127 10 o
5.615 t Bo x o x x Vb

1 Sg So
Gas : + Rs
=
5.615 t Bg Bo

3
k x k rg p g D q g,sc
1.127 10 g
x g x x Vb
1 Sw 3 k x k rw p w D q w,sc
Water : = 1.127 10 w
5.615 t B w x w x x Vb
51
 Number of Variables in Beta Model
Suppose we have three-phase flow (e.g., o, w, and g).

Then, the number of variables are only (6) six po, pw, and pg, So,
Sw, and Sg

Then, the number of equations should be six:

3 (three) mass balance equations for each comp.

1 saturation constraint So+Sw+ Sg = 1 equation

2 capillary pressure equations

pcow=po-pw and pcgo=pg-po

52
 Beta Model (Recap Equations)

vw q w,sc 1 Sw
Water Component : =
Bw Vb 5.615 t B w
v o q o,sc 1 So
Oil Component := =
Bo Vb 5.615 t Bo
Gas Component :
Rs 1 q g,sc 1 Sg S
vo + vg = + Rs o
B B V 5 . 615 t Bg B
o g b o

53
 Beta Model data Entry
The input needed by a black-oil simulator is a table of
physical properties of Bo, Bw, Bg, Rs, viscosities as a
function pressure.

Relative permeability data and also the capillary

pressure between the phases as a function of saturation
are also input.

54
 Flow Equations
If we consider Cartesian coordinate system (x-y-z) and use Darcys
equation in the continuity equation, we obtain the flow equation
for phase m as (in field units)
k x k rm pm D
m m
x m x x
5.615q
k k p D ( m S m m )
c1 m x rm


m
m

m , sc m , sc
=
y m y y Vb t
---(20)

k k p D
m x rm m m
z m z z

where c1= 6.33x10-3

55
 Simulation Equations Black Oil
We have 3 component balances and six dependent variables which
are the unknowns:
Po , So , Pw , Sw, Pg , Sg

Needs 3 additional relationships:

So + Sw + Sg = 1
Pcow = Po -Pw = f(Sw)
Pcog = Pg -Po = f(Sg)

56
 Flow Equations
For oil/water system, we have two flow equations one for oil
(Eq. 20 with m=o) and the other for water (Eq. 20 with m=w).
To solve these two equations for oil and water phase pressure
and saturations, we need two more equations.

These are (assume water-wet):

Pcow = Po -Pw = f(Sw) ---(21)

and

So + Sw = 1 ---(22)
57
 Flow Equations
Eq. 20 applies only for immiscible fluids. There is no mass transfer
considered between oil and water phases. So, masses of oil and
water are conserved.
Of course, due to compressibility of oil and water, their volumes do
change, but their masses do not change.
We define formation volume factors to relate reservoir volumes to
standard (or stock tank) conditions:

(assumes no mass transfer )

Vm ,res m ,res
Bm = =
Vm , sc m , sc ---(23)

58
 Flow Equations
If we divide both sides of Eq. 20 by m,sc (note that m,sc is
constant) and use the definition of formation volume factor for
phase m (Eq. 23) in the resulting equation, we obtain:

k x k rm pm D
m
x m x x

k x k rm pm D 5.615q m , sc S m
c1 m = ---(24)
y m y y Vb t Bm

k k p D

x rm
m
m
z m z z

59
 1-D Linear Flow Equations
If we assume that flow of phase m occurs only in the x-direction,
i.e., vm,y = vm,y= 0, and take kx = k in Eq.24, then Eq. 24 reduces to:

kk rm pm z 5.615q m , sc S m
c1 m = ---(25)
x m Bm x x Vb t Bm

where c1= 6.33x10-3

60
 1-D Linear Flow Equations
If we assume that we have a horizontal reservoir (or core)
where the flow occurs, then Eq. 25 reduces to

kk rm pm 5.615q m , sc S m
c1 =
x Bm m x Vb t Bm ---(26)
for m = o, w

where c1= 6.33x10-3

61
 1-D Linear Flow Equations
For oil (Eq. 26 with m=o):

kk ro po 5.615q o , sc S o
c1 =
x Bo o x Vb t Bo ---(27)

For water (Eq. 26 with m=w), we have

kk rw pw 5.615q w, sc S w
c1 = ---(28)
x Bw w x Vb t Bw

where c1= 6.33x10-3

62
63
 Finite Difference Solutions for
Oil/Water Systems
Fully Implicit Method
Computationally more rigorous and stable

Iterative Newton-Raphson Method is required

Requires large computational times

IMPES (Implicit Pressure and Explicit Saturation)

Can suffer from stabilities

Computationally simple to solve

64
 Formulations-General

IMPES Type
Solve pressure and saturation separately
Adaptive implicit
In each time step solve few grid-blocks implicitly
and the rest explicitly
Fully implicit
Solve for pressure, saturation, and composition
simultaneously

65
 Differences between solutions
If single phase; assume pressure gradients in space do not change much
between timesteps. Solution at time n+1 can be estimated from existing
values of pressure (explicit solution).
Implicit solution:
1. Guess Pressures & Saturations.
2. Calculate PVT properties, PC, and krel (for Flow Coefficients).
3. Calculate New Pressures & Saturations.
4. Do New Pressures = Guessed Pressures
and New Saturations = Guessed Saturations'?
No Go back to Step 1 but SKIP Step 2 (Inner Iteration).
No Go back to Step 1 (Outer Iteration).
Yes - Go on.
5. Go to the next timestep and start Step 1 again.
66
 IMPLLICIT Solution Algorithm
3 Single Phase Equations
- Oil- Gas- Water

Simultaneous Solution
Implicit Pressure-Implicit Solution

3 Partial Differential
Equations

Express Solution Terms

as Capillary Pressure

Solve For Pressures to Oil,

Gas, Water Individually

New Capillary Pressures

Compute Solutions

Stop
67
 Differences between solutions

The difference between these two types of calculations can be

summarized as follows.

Explicit solutions are based on the solution at the last timestep, and
only one leap forward in time is made. Implicit solutions are based

on both the previous and current solution and involve multiple

estimates of the solution at time n + I.

68
 Finite Difference Solutions for
Oil/Water Systems

We are interested in the solution of this problem

due to water flooding applications.

Performance prediction for water flooding (breakthrough

time, effect of reservoir heterogeneity, etc.)

69
Water flooding

70
 Simulation of 1-D Linear Water/Oil
Displacement
For oil (Eq. 26 with m=o):

kk ro po 5.615q o , sc S o
c1 =
x Bo o x Vb t Bo ---(29)

For water (Eq. 26 with m=w), we have

kk rw pw 5.615q w, sc S w
c1 = ---(30)
x Bw w x Vb t Bw

where c1= 6.33x10-3

71
 Simulation of 1-D Linear Water/Oil
Displacement
We wish to solve Eqs. 29 and 30 for phase
pressures (po and pw) as well as phase saturations
(So and Sw).
Assuming water wet rock, the oil and water phase
pressures are related to each other through the
capillary pressure, i.e.,
pc (S ) = po pw ---(31)

72
 Simulation of 1-D Linear Water/Oil
Displacement
The oil and water phase saturations are related to each other
by the material equation:
So + Sw = 1 ---(32)

Eqs. 31 and 32 indicate that if we solve water (or oil) phase

pressure and water (or oil) phase saturations, we can then
determine oil (or water) phase pressures and saturations using
Eqs. 31 and 32.

73
1-D Linear Water/Oil Displacement

The common approach is to solve for the oil phase pressures,

and the water saturation. We will use a block-centred grid
system as shown

Producer Injector No-flow

No-flow
Boundary
Boundary

x =x1/2= 0 x = xNx+1/2=Lx 74
1-D Linear Water/Oil Displacement

We can use a uniform grid or non-uniform grid in the x-direction.

For generality, lets use non-uniform block-centred grid.

Gridblock i

xi-1 xi xi+1
xi-1/2 xi+1/2
75
 Simulation of 1-D Linear Water/Oil
Displacement
The finite difference equation for the oil equation (Eq. 29) is
given by

c1 kk ro po ,i +1 po ,i kk ro po ,i po ,i 1
n +1 n +1 n +1 n +1 n +1 n +1


xi Bo o 1+1 2 xi +1 xi Bo o 11 2 xi xi 1
---(33)
5.615qon,+i ,1sc 1 S o ,i i S on,iin
n +1 n +1

= n +1 n +1
n
xi wh t Bo ,i Bo ,i

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Simulation of 1-D Linear Water/Oil
Displacement
Similarly, the finite difference equation for the water equation
(Eq. 30) is given by

c1 kk rw pw,i +1 pw,i kk rw pw,i pw,i 1

n +1 n +1 n +1 n +1 n +1 n +1


xi Bw w 1+1 2 xi +1 xi Bw w 11 2 xi xi 1

---(34)
5.615qwn +,i1,sc 1 S w,i i S wn ,iin
n +1 n +1

= n +1 n +1
n
xi wh t Bw,i Bw,i

77
 Simulation of 1-D Linear Water/Oil
Displacement
Multiply both sides of Eqs. 33 and 34 by the bulk volume and introduce
transmissibility and accumulation terms, then we have the following
difference forms for the oil and water phases:

~ n +1 S m ,i i S mn ,iin
n +1 n +1

( ) ( )
Tmn,+i +11 2 pmn +,i1+1 pmn +,i1 Tmn,+i 11 2 pwn +,i1 pwn +,i11 qmn +,i1,sc = Vi n +1
n ---(35)
Bm ,i Bm ,i
for m = o, w
k i +1 2 k nrm+1,i +1 2 wh
T n +1
= 2 1.127 10 3
m ,i +1 2 Bn +1 x + x
m m av. i i +1

k i 1 2 k nrm+1,i 1 2 wh
T n +1
= 2 1.127 10 3
m ,i 1 2 Bn +1 x + x
m m av. i i 1

~ x i wh
VIn +1 =
5.615t n +1
78
Simulation of 1-D Linear Water/Oil
Displacement
As oil and water phases are related through the oil/water
capillary pressure,
pcow = po pw,
so,
- pw = po-pcow
Also, oil and water saturations are related to each other by
So + Sw = 1,
- So = 1 - Sw
79
Simulation of 1-D Linear Water/Oil
Displacement
So, we can solve Eq. 35 with m = o and m = w for oil phase

pressures po,i, for i = 1, 2,, Nx,

and water saturations Sw,i, for i = 1, 2,, Nx, at time tn+1.

We will have 2Nx equations and 2Nx unknowns.

80
Simulation of 1-D Linear Water/Oil
Displacement
The system of equations described by Eq. 35 are nonlinear.
It can be solved by the use of Newton-Raphson method (discussed
earlier) to solve these nonlinear system .
An easy approach is to use a method called IMPES (Implicit
Pressure Explicit Saturation) to solve this system.
In IMPES method, we linearize Eq. 35 by backdating the saturation
dependent relative permeability terms; i.e., evaluate these terms at
old time level n.

81
 Concluding Remarks
Why we derived the continuity equation and the flow equations
for a black oil simulator???
Answer: they were presented to emphasize the point that the
development of reservoir simulators are based on the
fundamental principles of mass balances and the Darcy flow
equation, which are the bedrock of most calculations in
petroleum reservoir engineering.

In other types of simulators, mass balances are applied on the

components within the phases, and other fundamental flow
equations may be included in the derivations of the relevant
equations.
82
 Concluding Remarks
In a compositional simulator, mass balances are applied to the defined
components of the hydrocarbon phases, and the equilibrium between
components in the vapor and liquid phases are determined with an equation of
state.

For thermal simulators, an energy balance on the system in addition to mass

balances on water, steam, and oil are included in the relevant equations.

The point here is to reassure potential users of reservoir simulation that

reservoir simulators are based on fundamental principles of mass balance
energy balance, Darcy flow equations, etc. that form the basis of most
analytical reservoir engineering calculations.
83
Questions?

84