Simulation Examples (2015)

1

RESERVOIR SIMULATION EXAMPLE AND HOMEWORK

PROBLEMS

Matthew T. Balhoff

Petroleum and Geosystems Engineering

University of Texas at Austin

August 2014

CONFIDENTIAL, DO NOT DISTRIBUTE

2

Table of Contents

1) Explicit Solution to 1D Flow ...................................................................................................3

2) Implicit Solution to 1D Flow ...................................................................................................7

3) Explicit/Implicit Solution to 1D Flow in terms of transmissibility ...................................10

4) Crank-Nicholson Solution to 1D flow in terms of transmissibility ...................................13

5) Implicit Solution to 1D flow with constant rate wells ........................................................16

6) Implicit solution to heterogeneous reservoir .......................................................................19

7) Implicit solution to heterogeneous reservoir with variable grid sizes ..............................22

8) Implicit Solution to 1D flow with a constant BHP well ......................................................24

9) Implicit Solution to 1D flow with gravity.............................................................................29

10) Implicit Solution to flow in 2D ...........................................................................................31

11) Implicit Solution to flow in 2D with horizontal wells .......................................................35

12) Implicit Solution to flow in 3D ...........................................................................................39

13) Implicit Solution to flow in 3D with wells .........................................................................42

14) IMPES solution to multiphase flow in 1D without capillary pressure

(Bulkley-Leverett) Aziz and Settari example ............................................................45

15) SS solution to multiphase flow in 1D without capillary pressure

(Bulkley-Leverett) Aziz and Settari example ............................................................49

16) 1D Dual porosity/permeability model of naturally fractured reservoir...52

17) IMPEC Solution to compositional flow (a tracer) in 1D55

3

Example 1. Explicit Solution to 1D flow

Consider a 1D reservoir with the following reservoir and fluid properties:

= 0.2 k = 50 mD

=1 cp Bw = 1 rb/stb (assume constant)

ct = 10-6 psi-1.

L= 10,000 ft (Reservoir Length)

A = 200,000 ft2 (Cross Sectional Area)

The initial condition is P = 1000 psi. The boundary conditions are P = 2000 psi at x = 0 and no

flow (q = 0) at x = L. Determine the pressure field in the reservoir using 4 uniform blocks. Use a

time step of t = 1.0 days.

Solution:

The pressure is governed by the 1D diffusivity equation with the following boundary conditions:

2

2

1;

: p( ,0) 1000

1: p(0, ) 2000

2 : ( , ) 0

t

p p k

t x c

IC x psi

BC t psi

pBC L t

x

The explicit finite difference solution is given by the formula:

1

1 122n n n n ni i i i i

tP P P P P

x

Where the dimensionless diffusivity is written as:

2 2 6 1 2 2

50 mD 1.0 days mD-days-psi40 0.2532

( ) ( ) (1 cp)(0.2)(1.0 10 psi ) (2500 ft) cp-ftt

t k t

x c x

Note that we are guaranteed stability with the explicit method since

4

At t = 1.0 days (n = 1), we solve explicitly for the pressures

1 0 0 0 0 0

1 1 0 1 2 0

1 0 0 0 0

2 2 1 2 3

1 0 0 0 0

3 3 2 3 4

1 0 0 0 0

4 4 3 4 5

2 1000 0.2532 2 1000 1000

2 1000 0.2532 1000 2 1000 1000

2 1000 0.2532 1000 2 1000 1000

2 1000

P P P P P P

P P P P P

P P P P P

P P P P P

0

50.2532 1000 2 1000 P

There is no block #0 or block #5. We must use our boundary conditions:

1) At x = 0, we have a constant pressure boundary P=Pin=2000 psi

0 10 12

2in in

P PP P P P

2) At x = L, we have a no-flow condition

4 55 40

P PP P

x

Substitute these values into the pressure equations:

1 0 0 0

1 1 1 2

1 0 0 0 0

2 2 1 2 3

1 0 0 0

0

1

0

3 3 2 3 4

2 1000 0.2532 2 1000 1000

2 1000 0.2532 1000 2 1000 1000 1000 psi

2 1000 0.2532 1000 2 1000 1

2 2 200

00

0 1000 1506.4 i

0

psinP P P P

P P P P

P P

P

P P P P P

1 0 0 040

4 43 4

1000 psi

2 1000 0.2532 1000 2 1000 1000 psi1000P P P P P

At t = 2.0 days (n = 2)

2 1 1

1 1 2

2 1 1 1

2 2 2 3

2 1

1 1

1

1 1 1

3 3 2 3 4

1

1

1

2 0.2532 2 2000 1000

2 1000 0.2532 2 1000 1000

2 1000 0.2532 1000 2 10

2 1506.4 1506.4 2 1506.4 1628.1 psi

1506.4 1128.2 psi

0

in P P

P

P P P P

P P P P

P P P P P

2 1 1 1 14 4 3 4 4

0 1000 1000 psi

2 1000 0.2532 1000 2 1000 1000 1000 psiP P P P P

5

At t = 3.0 days (n = 3)

23 2

1 1

3 2 2

2 2

1 1 2

2 2

2 2 3

3 2 2 2

3 3 3 4

1 2

2

2

2 1628.1 1628.1 2 1628.1 1128.2 1689.9 psi

2 1128.2 1628.1 2

2 0.2532 2 2000

0.2532 1000

2 1000 0

1128.2 1223.3 psi

112.2532

inP P P

P P P

P P P

P P

PP P P P

3 2 2 2 24 4 3 4 4

2 1000 1000

2 1000 0.2532 1000 2 1000 100

8.2 1032.5

0 1000 p

si

s

p

iP P P P P

At t=, the reservoir comes to equilibrium at P = 2000 psi in all blocks

Table 1.1 Summary of block pressures using the explicit method for example 1.

Block #1 (psi)

1250 ft

Block #2 (psi)

3750 ft

Block #3 (psi)

6250 ft

Block #4 (psi)

8750 ft

0 days 1000 1000 1000 1000

1 days 1506.4 1000 1000 1000

2 days 1628.1 1128.2 1000 1000

3 days 1689.9 1222.3 1032.5 1000

2000 2000 2000 2000

The solution makes physical sense. The pressure in the reservoir increases over time as a result

of the dirichlet (constant pressure) boundary condition of P = 2000 psi, until the reservoir comes

to steady state at 2000 psi. Pressure increases fastest in block #1 because it is near the boundary

and slowest in block #4 because it is far from the boundary.

A plot of the numerical (explicit) solution to the problem is shown in figure 1.1a, 1.1b, and 1.1c

using 4, 8, and 12 blocks, respectively. A plot of the actual (analytical) solution is given in

figure 1.1d. Recall the analytical solution for this problem is given by:

2

2

(2 1)

41

0

4 ( 1) (2 1), cos

2 1 2

nn tn

init LB

n

p n xp x t p e

n L

The numerical solution demonstrates the same trends as the analytical solution, but is not as

smooth because only a finite # of grid blocks are used. The numerical solution is particularly

poor for only 4 grid blocks, but we could have increased accuracy by using more grids and

smaller time steps as long as the stability criterion (

6

Figure 1.1. Solutions to pressure vs. reservoir distance at various times. (a) 4 grids (x = 2500 feet), t = 1 day; (b)

8 grids (x = 1250 feet), t = 0.1 day; (c) 12 grids (x = 833 feet), t = 0.01 day; (d) analytical solution. For the numerical solutions, the points are the grid pressures and the continuous lines are linearly interpolated values.

7

Example 2. Implicit Solution to 1D flow


= 0.2 k = 50 mD


ct = 10-6 psi-1.

L= 10,000 ft (reservoir length)




time step of t = 1.0 days.

Solution:


2

2

1;

: p( ,0) 1000

1: p(0, ) 2000

2 : ( , ) 0

t

p p k

t x c

IC x psi

BC t psi

pBC L t

x

The implicit finite difference solution is given by the formula:

1 1 1

1 1(1 2 )n n n n

i i i iP P P P

Where the dimensionless diffusivity is written as:

2 2 6 1 2 2

50 mD 1.0 days mD-days-psi40 0.2532

( ) ( ) (1 cp)(0.2)(1.0 10 psi ) (2500 ft) cp-ftt

t k t

x c x

L=10000 ft x = 2500 ft

2000

psi

No

flow Pinitial = 1000 psi

8

The equations for each block can now be written as a system of linear equations

1 1 1 0

0 1 2 1

1 1 1 0

1 2 3 2

1 1 1 0

2 3 4 3

1 1 1 0

3 4 5 4

(1 2 )

(1 2 )

(1 2 )

(1 2 )

P P P P

P P P P

P P P P

P P P P

Once again we have to use boundary conditions to eliminate the blocks outside of the domain:

1) At x = 0, we have a constant pressure boundary P=Pin=2000 psi

0 10 1

22i nn i

P PP P PP

2) At x = L, we have a no-flow condition

44 5

50P P

Px

P

Substituting the boundary conditions into the equations and writing them in matrix form:

1 0

1 1

1 0

2 2

1 0

3 3

1 0

4 4

1 0 0

1 2 0 0

0 1 2 0

0 0 1 0

3 2 inP P

P

P P

P

P

P P

At t = 1 days, the solution can be found by solving the system of equations: 1

1

1

2

1

3

1

4

1 3(0.2532) 0.2532 0 0 1000 2 0.2532 2000

0.2532 1 2(0.2532) (0.2532) 0 1000 0

0 0.2532 1 2(0.2532) 0.2532 1000 0

0 0 0.2532 1 0.2532 1000 0

P

P

P

P

Which has the solution P1 = [1295.1; 1051.1; 1009.9; 1001.8] psi

At t = 2 days, use the solution at the previous time step. Note that the matrix remains the same,

but the right hand side (RHS) vector changes from the previous timestep

2

1

2

2

2

3

2

4

1 3(0.2532) 0.2532 0 0 1295.1 2 0.2532 2000

0.2532 1 2(0.2532) (0.2532) 0 1051.1 0

0 0.2532 1 2(0.2532) 0.2532 1009.9 0

0 0 0.2532 1 0.2532 1001.8 0

P

P

P

P

9


At t = 0.03 days, use the solution at the previous time step. Note that the matrix remains the same.

3

1

3

2

3

3

3

4

1 3(0.2532) 0.2532 0 0 1472.5 2 0.2532 2000

0.2532 1 2(0.2532) (0.2532) 0 1117.9 0

0 0.2532 1 2(0.2532) 0.2532 1026.9 0

0 0 0.2532 1 0.2532 1006.9 0

P

P

P

P


Eventually, the solution reaches steady state and Pss = [2000; 2000; 2000; 2000] psi. Table 2.1

summarizes the results obtained using the implicit method as shown above and compares them to

the results for the same problem. As expected, the results are nearly identical. The very slight

differences are caused by the fact that CMG does not allow for dirichlet boundary conditions, so

one was fabricated by placing a constant bottomhole pressure well near the edge and introducing

a slight amount of additional error. In a sense, the hand solution is slightly more accurate than CMG, but both have noticeable discrepancies compared to the true (analytical solution).

Table 2.1 Comparison of numerical solution by hand (example 2) to the same problem in CMG

Method Block #1 (psi)

1250 ft

Block #2 (psi)

3750 ft

Block #3 (psi)

6250 ft

Block #4 (psi)

8750 ft

Initial 1000 1000 1000 1000

1 day Implicit 1295.1 1051.1 1008.9 1001.8

CMG 1294.9 1051.1 1008.9 1001.8

2 days

Implicit 1472.5 1117.9 1026.9 1006.9

CMG 1472.2 1117.8 1026.8 1006.9

3 days

Implicit 1582.9 1184.9 1051.6 1015.9

CMG 1582.6 1184.8 1051.5 1015.9

2000 2000 2000 2000

10

Example 3.Explicit/Implicit Solution to 1D flow in terms of transmissibility (flow units)


= 0.2 k = 50 mD


ct = 10-6 psi-1.

L= 10000 ft (reservoir length)




time step of t = 1 days.

Solution:


2

2

1;

: ( ,0) 1000

1: (0, ) 2000

2 : ( , ) 0

t

P P k

t x c

IC P x psi

BC P t psi

PBC L t

x

The explicit and implicit finite difference solution in flow rate units is given by the matrix

equations

1 1

1

n n n

n n

t

t t

P P B TP Q

B BT P P Q

Or, (the implicit equations)

1

11 1

122 2

133 3

144 4

3 0 0 0 0 0 0 0 0

2 0 0 0 0 0 0 01 1

0 2 0 0 0 0 0 0

0 0 0 0 0 0 0 0

n n

n n

n n

n n

QT T B BP P

QT T T B BP P

t t QT T T B BP P

QT T B BP P

L=10000 ft x = 2500 ft

2000

psi

No


11

Where:

3

2 1

1

50 2000004000

1 1 2500

200000 2500 0.2 1 6 100.0

2 2 4000 2000 1.6 7

i i t

in

kA mD ftT

x cp

ftB V c ft ft E psi

psi

mD ft psiQ TP E

cp

Therefore the matrices and vectors can be written as:

3

3

30

12000 4000 0 0

4000 8000 4000 06.33 3

0 4000 8000 4000

0 0 4000 4000

100

100

100

100

1000 2 4000 2000

1000 0; 6.33 03

1000 0

1000 0

ftE

day psi

ft

psi

ftpsi E

day

T

B

P Q

Note that T, B, and Q never change with time. Only the pressure from the previous time step

changes. The solution for block pressures (explicitly or implicitly) is summarized in Table 3.1

Table 3.1 Summary of block pressures for various solution techniques


1250 ft

Block #2 (psi)

3750 ft

Block #3 (psi)

6250 ft

Block #4 (psi)

8750 ft

Initial 1000 1000 1000 1000

1 days

Explicit 1506.4 1000 1000 1000

Implicit 1295.1 1051.1 1008.9 1001.8

12

CMG 1294.9 1051.1 1008.9 1001.8

Analytical 1482.3 1035.0 1000.4 1000.0

2 days

Explicit 1628.1 1128.2 1000 1000

Implicit 1472.5 1117.9 1026.9 1006.9

CMG 1472.2 1117.8 1026.8 1006.9

Analytical 1619.3 1136.1 1013.0 1000.5

3 days

Explicit 1689.9 1222.3 1032.5 1000

Implicit 1582.9 1184.9 1051.6 1015.9

CMG 1582.6 1184.8 1051.5 1015.9

Analytical 1685 1223.6 1042.5 1004.8

2000 2000 2000 2000

A few important notes about table 3.1

Solutions for explicit and implicit method are exactly the same as example 1 and 2, respectively. Example 3 just worked in different units (flow rate instead of pressure)

We get different answers for the explicit and implicit solution, but neither is more accurate than the other.

CMG is nearly identical to the solution obtained using the implicit method by hand. This is because CMG uses the implicit method and same equations. The very slight differences

are a result of the fact that CMG doesnt allow for true constant pressure boundary conditions

We could improve the accuracy of both the explicit and implicit method by using smaller time steps and smaller grids and both would converge to the analytical solution

However, the explicit method would become unstable if the grid size were reduced without reducing the time steps in an appropriate fashion.

13

Example 4. Crank-Nicholson Solution to 1D flow in terms of transmissibility (flow rate

units)


= 0.2 k = 50 mD

=1 cp Bw=1 rb/stb (assume constant)

ct = 10-6 psi-1.





time step of t = 1 days.

Solution:


2

2

1;

: p( ,0) 1000

1: p(0, ) 2000

2 : ( , ) 0

t

p p k

t x c

IC x psi

BC t psi

pBC L t

x

The Crank-Nicholson solution is a hybrid of the explicit and implicit methods with = .

11 n nt t

B BT P T P Q

Where:

3

2 1

1

50 2000004000

1 1 2500

200000 2500 0.2 1 6 100

2 2 4000 2000 1.6 7

w

i i t

in

kA mD ftT

B x cp

ftB V c ft ft E psi

psi

mD ft psiQ TP E

cp

Plugging in the numbers we get:

L=10000 ft x = 2500 ft

2000

psi

No


14

1

1 1

1

2 2

1

3 3

1

4 4

138 12.7 0 0 62 12.7 0 0 1.6 7

12.7 125.3 12.7 0 12.7 74.7 12.7 0 0

0 12.7 125.3 12.7 0 12.7 74.7 12.7 0

0 0 12.7 112.7 0 0 12.7 87.3 0

n n

n n

n n

n n

EP P

P P

P P

P P

6.33 03E

Table 4.1 summarizes the solution for the C-N method at various times and compares it to the

explicit, implicit, CMG, and analytical solutions.

Table 4.1 Summary of block pressures for example 4 using various solution techniques


1250 ft

Block #2 (psi)

3750 ft

Block #3 (psi)

6250 ft

Block #4 (psi)

8750 ft

Initial 1000 1000 1000 1000

1 days

Explicit 1506.4 1000 1000 1000

Implicit 1295.1 1051.1 1008.9 1001.8

CMG 1294.9 1051.1 1008.9 1001.8

C-N 1370.5 1037.8 1003.9 1000.4

Analytical 1482.3 1035.0 1000.4 1000.0

2 days

Explicit 1628.1 1128.2 1000 1000

Implicit 1472.5 1117.9 1026.9 1006.9

CMG 1472.2 1117.8 1026.8 1006.9

C-N 1547.8 1117.5 1018.3 1002.8

Analytical 1619.3 1136.1 1013.0 1000.5

3 days

Explicit 1689.9 1222.3 1032.5 1000

Implicit 1582.9 1184.9 1051.6 1015.9

CMG 1582.6 1184.8 1051.5 1015.9

C-N 1642 1196.5 1043.9 1009.2

Analytical 1685 1223.6 1042.5 1004.8

2000 2000 2000 2000

The results are not surprising; the mixed (Crank-Nicholson) method gives an answer in between the explicit and implicit methods. Moreover, we expect the answer to be more accurate

because it is an O(t2) method. However, it does require more computational effort per time step than either explicit or implicit method. Alternatively, we could have used smaller time steps in

the explicit/implicit methods instead to get more accuracy.

15

Figure 4.1 Comparison of the explicit, implicit, and C-N methods (4 grid blocks, t= 1 day) to the analytical solution after 3 days. As expected, none of the three numerical approaches

perfectly match the analytical solution, but all would converge to it if more (an infinite number)

grids were employed.

16

Example 5. Implicit Solution to 1D flow with constant rate wells.


= 0.2 k = 50 mD


ct = 10-6 psi-1.



The initial condition is P = 1000 psi. The boundary conditions are no flow (q = 0) at x = 0 and P

= 2000 psi at x = L (note: these boundary conditions are different than in the previous examples).

There is an injection well of 1000 ft3/day at x = 0 and a producer of 1000 ft3/day at x = L.

Determine the pressure field in the reservoir using 4 uniform blocks. Use a time step of t = 1.0 days.

Solution:

The pressure is governed by the 1D diffusivity equation with sources and sinks and the following

boundary conditions:

2

2

1;

: p( ,0) 1000

1: (0, ) 0

2 : p( , ) 2000

t

p p kq

t x c

IC x psi

pBC t

x

BC L t psi

There are 2 key differences between this problem and the previous examples (1-4).

(1) The boundary conditions are flipped; no flux at x = 0 and constant pressure at x = L. This will affect both the T matrix and Q vector

(2) There are constant rate wells (sources and sinks) in the reservoir. These are treated by including in the source (Q) vector

Recall the Control Volume approach shows that a mass balance on each block is given by:

Writing the equation implicitly for each block we get a system of equations:

L=10000 ft x = 2500 ft

2000

psi No

flow

Pinitial 1000 psi

q=1000 ft3/d q=-1000 ft3/d

11 1 n ni i i i i i iB

T P P T P P P P qt

17

1 1 1 1 1

0 1 2 1 1 1 1

1 1 1 1 1

1 2 3 2 2 2 2

1 1 1 1 1

2 3 4 3 3 3 3

1 1 1 1 1

3 4 5 4 4 4 4

n n n n n n

n n n n n n

n n n n n n

n n n n n n

BT P P T P P P P q

t

BT P P T P P P P q

t

BT P P T P P P P q

t

BT P P T P P P P q

t

Sources and sinks

The position of the wells is used to determine which blocks they reside in. Here, there is an

injector in block #1 and a producer in block #4. Therefore q1=-q4 = 1000 ft3/day (Recall our

convention that an injector is positive). There are no wells in block #2 or #3 (q2=q3=0).

Boundary conditions

In block #1 we have a no-flow boundary condition at x = 0. That means that the first term of

equation # vanishes, i.e.

1 10 1 0n nT P P In block #4 we have a constant pressure at the boundary. We dont know P5 (block #5 doesnt exist), but we know P = Pout at the boundary. The boundary is half-way in between blocks 4 and 5 and that would mean the effective transmissibility is twice as large. Therefore we can say:

1 1 14 5 42 0n n n outT P P T P P In matrix form we get the usual:

1n n

t t

B BT P P Q

Where

3

30

4000 4000 0 0

4000 8000 4000 06.33 3

0 4000 8000 4000

0 0 4000 12000

100 1000 1000

100 1000 0; ;

100 1000 0

100 1000 6.33E-03 2 2000 4000 1000

ftE

day psi

ft ftpsi

psi

T

B P Q3

day

18

Table 5.1 Summary of block pressures for example 5


1250 ft

Block #2 (psi)

3750 ft

Block #3 (psi)

6250 ft

Block #4 (psi)

8750 ft

Initial 1000 1000 1000 1000

1 day

Implicit 1010 1010.1 1050.3 1289.4

CMG 1010 1010.1 1050.3 1289.2

2 days

Implicit 1022 1030 1116.3 1463.3

CMG 1022 1030 1116.3 1463

3 days

Implicit 1037.1 1056.9 1182.9 1571.7

CMG 1037 1056.8 1182.8 1571.4

The pressure in block #1 begins to rise immediately above the initial pressure (1000 psi) as a

result of the injector well at x = 0. Eventually it also begins to feel the effects of the constant

pressure boundary condition at x = L. At x = L, the pressure rises quickly because if the P = 2000

psi boundary condition, but the increase is somewhat mitigated by the producer well also at

x = L.

The implicit solution obtained by hand is again almost identical to CMG, with very small

differences near the dirichlet boundary condition.

19

Example 6. Implicit solution to heterogeneous reservoir

Consider a 1D reservoir with the following reservoir properties:

0.2 L = 10,000 ft

A = 200,000 ft2.

= 1 cp, Bw = 1 rb/stb (assume constant)

ct = 10-6 psi-1.

Heterogeneous permeability (k1 = 10 mD; k2 = 100 mD; k3 = 50 mD; k4 = 20 mD)


flow (q = 0) x = L. Determine the pressure field in the reservoir using 4 uniform-sized blocks.

Use a time step of t = 1.0 days.

Note: the permeability is a function of position, k(x), so the PDE must keep the permeability

inside the spatial derivative.

: p( ,0) 1000

1: p(0, ) 2000

2 : ( , ) 0

t

p k pc

t x x

IC x psi

BC t psi

pBC L t

x

Recall that a mass balance on each block can be written as:

1 1 1 1 11 1 1 12 2

n n n n n nii i i i i ii i

BT P P T P P P P

t

Writing the equation for all blocks leads to the familiar matrix equations for implicit solution to

pressure.

1n n

t t

B BT P P Q

Transmissibilities must come from a Harmonic mean of permeability

L=10000 ft x = 2500 ft

2000

psi

No

flow k2 = 100 mD

k3 = 50 mD

k4 = 20 mD

k1 = 10 mD

Pinitial = 1000 psi

20

1 3 32 2 2

3 3 5 52 2 2 2

5 5 7 72 2 2 2

7 7 92 2 2

0 0

0

0

0 0

T T T

T T T T

T T T T

T T T

T

First off, calculate the inter-block transmissibility using the harmonic mean

1 1

32

1 2

1 1

52

2 3

1 1

72

3 4

1 1 1 12 2 18.2

10 100

1 1 1 12 2 66.7

100 50

1 1 1 12 2 28.6

50 20

k mDk k

k mDk k

k mDk k

The half transmissibilities can now be calculated:

32

32

52

52

72

72

18.2 2000001454

1 1 2500

66.7 2000005333

1 1 2500

28.6 2000002286

1 1 2500

w

w

w

k A mD ftT

B x cp

k A mD ftT

B x cp

k A mD ftT

B x cp

The boundary transmissibilities can be calculated using the boundary conditions

(1) At x = 0, the pressure is constant. T1/2=2T1 = 1600 mD-ft/cp (2) At x = L, there is no flow (no transmissibility). Therefore T9/2 = 0

Therefore we get:

3

30

2 800 1454 1454 0 0

1454 1454 5333 5333 06.33 3

0 5333 5333 2286 2286

0 0 2286 2286

100 1000 2 2000 800 6.33 03

100 1000 0; ;

100 1000 0

100 1000 0

ftE

day psi

E

ftpsi

psi

T

B P Q3ft

day

21

Table 6.1 Summary of block pressures for example 6 using implicit method


1250 ft

Block #2 (psi)

3750 ft

Block #3 (psi)

6250 ft

Block #4 (psi)

8750 ft

Initial 1000 1000 1000 1000

1 day

Implicit 1085.3 1005.8 1001.3 1000.2

CMG 1085.3 1005.8 1001.3 1000.2

2 days

Implicit 1157.5 1015.3 1004.3 1000.7

CMG 1157.5 1015.3 1004.4 1000.7

3 days

Implicit 1219 1027 1009.3 1001.8

CMG 1218.9 1027 1009.3 1001.8

2000 2000 2000 2000

22

Example 7. Implicit solution to heterogeneous reservoir with variable grid sizes

0.2 L = 10000 ft

A = 200,000 ft2.

= 1 cp, Bw = 1 rb/stb (assume constant)

ct = 10-6 psi-1.

Heterogeneous permeability (k1 = 10 mD; k2 = 100 mD; k3 = 50 mD; k4 = 20 mD)

Non-uniform uniform-sized blocks (x1 = 2000 ft; x2= 3000 ft; x3= 1500 ft; x4= 3500 ft).


flow (q = 0) x = L. Determine the pressure field in the reservoir. Use a time step of t = 1.0 days.

Note: the permeability is a function of position, k(x), so the PDE must keep the permeability

inside the derivative.

: p( ,0) 1000

1: p(0, ) 2000

2 : ( , ) 0

t

p k pc

t x x

IC x psi

BC t psi

pBC L t

x

The harmonic mean of permeabilities now involves weighting of grid block sizes

1 23

2 1 2

1 2

2 35

2 32

2 3

3 47

2 3 4

3 4

2000 300021.74 mD

2000 3000

10 100

3000 150075 mD

3000 1500

100 50

150 350

1500

50

x xk

x x

k k

x xk

xx

k k

x xk

x x

k k

24.39 mD3500

20

L=10000

ft

x4= 3500 ft

2000

psi

No

flow k2 = 100 mD

k3 = 50 mD

k4 = 20 mD

k1 = 10 mD

Pinitial = 1000 psi

x3 = 1500 ft

x2 = 3000 ft

x1 = 2000 ft

23

Half transmissibilities can now be computed:

32

32

1 2

52

52

2 3

72

72

3 4

21.7 200000 mD-ft1739

2 1 1 2500 cp

75 200000 mD-ft6666.7

2 1 1 2250 cp

24.4 200000 mD-ft1951

2 1 1 2500 cp

w

w

w

k AT

B x x

k AT

B x x

k AT

B x x

The boundary transmissibilities can be calculated using the boundary conditions:

(1)At x = 0, the pressure is constant. T1/2=2T1 = 2000 mD-ft/cp

(2)At x = L, there is no flow (no transmissibility). Therefore T9/2 = 0

So the matrices and vectors can now be computed. Note that the B matrix has also changed

because x (and therefore block volume) varies.

3

30

2 1000 1739 1739 0 0

1739 1739 6667 6667 06.33 3

0 6667 6667 1951 1951

0 0 1951 1951

80 1000 2 1000 2000 6.33 03

120 1000 0; ;

60 1000 0

140 1000 0

ftE

day psi

E

ftpsi

psi

T

B P Q3ft

day

Table 7.1 Summary of block pressures for example 7 using implicit method


1000 ft

Block #2 (psi)

3500 ft

Block #3 (psi)

5750 ft

Block #4 (psi)

8250 ft

Initial 1000 1000 1000 1000

1 day

Implicit 1123.0 1008.6 1003.2 1000.3

CMG 1122.9 1008.6 1003.2 1000.3

2 days

Implicit 1219.4 1022.3 1010 1001.1

CMG 1219.3 1022.3 1010 1001

3 days

Implicit 1295.6 1039.1 1019.9 1002.6

CMG 1295.4 1039.1 1019.9 1002.6

2000 2000 2000 2000

24

Example 8. Implicit Solution to 1D flow with a constant BHP well

Consider a 1D reservoir with the following properties:

= 0.2 k = 50 mD


ct = 10-6 psi-1.

L= 10000 ft (length)


The initial condition is P = 1000 psi. The boundary conditions are no flow (q = 0) at x = 0 and

constant pressure, P = 2000 psi, at x = L. There is a constant-rate injection well of 1000 ft3/day at

x = 0. There is a constant BHP producer with P = 800 psi at x = 6250 feet.

Determine the pressure field in the reservoir using 4 uniform blocks. Use a time step of t = 1 days. Assume the radius of both wells is 0.25 feet, that skin factor is negligible, and that the

reservoir thickness is 250 feet.

Solution:



2

2

1;

: ( ,0) 1000

1: (0, ) 0

2 : ( , ) 2000

t

p p kq

t x c

IC p x psi

pBC t

x

BC p L t psi

Note that this problem is similar to example #5 in terms of boundary conditions (no flow at x = 0

and constant P at x = L). Also like example #5 we have 2 wells in the problem, but now the

producer is a constant BHP well of 800 psi at x= 6250 feet. For the 4-block system here, the

constant BHP well is in block #3.

We need to calculate the productivity index (J) of the BHP well in block #3, where the height

can be found from the area and thickness and r0 is found from the Peaceman correction

20000080

2500

0.2 0.2 2500 500eq

Ah ft

W

r x ft ft

L=10000 ft x = 2500 ft

2000

psi No

flow Pinitial 1000 psi

q=1000 ft3/d Pbh= 800 psi

W =2500 ft

h

25

2 50 8023306.5

5001 cp 1 ln 0ln

0.25

w

i

eq

w

w

mD ftkh mD ftJ

r cpB s

r

Recall the Control Volume approach shows that a mass balance on each block is given by:

11 1 n ni i i i i i iB

T P P T P P P P qt

And for a constant BHP well we have:

wi i i wq J P P Writing for each block and noting we have wells in blocks 1 and 3

1 1 1 1 1

0 1 2 1 1 1 1

1 1 1 1 1

1 2 3 2 2 2

1 1 1 1 1 1

2 3 4 3 3 3 3 3

1 1 1 1 1

3 4 5 4 4 4

0

0

n n n n n n

n n n n n n

n n n n n n w n

w

n n n n n n

BT P P T P P P P q

t

BT P P T P P P P

t

BT P P T P P P P J P P

t

BT P P T P P P P

t

The boundary conditions are treated as usual for constant pressure and no flux. In matrix form

we get the usual matrix equations:

1n n

t t

B BT J P P Q

Where

30

0 0 4000 4000 0 0

2 0 4000 8000 4000 06.33 3

0 2 0 4000 8000 4000

0 0 3 0 0 4000 12000

0 100 1000

0 100 10006.33 3; ;

3306.5 100 1000

0 100 1000

T T

T T TE

T T T

T T

ftE P

psi

T

J B

1

32

3 3

4

;

1000

0

3306.5 800 6.33 03

2 2 4000 2000 6.33 03

w

w

out

psi

q

q ft

q J P E day

q TP E

Q

26



1250 ft

Block #2 (psi)

3750 ft

Block #3 (psi)

6250 ft

Block #4 (psi)

8750 ft

Initial 1000 1000 1000 1000

1 day

Implicit 1008.9 1004.7 1019.2 1290.6

CMG 1008.9 1004.7 1019.1 1290.3

2 days

Implicit 1018.3 1015.8 1057.2 1461.1

CMG 1018.3 1015.8 1057.1 1460.8

3 days

Implicit 1029 1031.6 1096.8 1563.8

CMG 1029 1031.6 1096.6 1563.4

1441.2 1401.7 1362.2 1787.4

The pressure in the reservoir increases with time as a result of the injector well in block #1 and

the constant pressure boundary condition (2000 psi) at x = L. However, this pressure increase is

somewhat offset by the low-pressure bottomhole pressure well (800 psi) that acts as a producer.

The wells prevent the steady solution from equilibrating at 2000 psi. Notice that at steady state,

the pressure is highest in block #4 which is adjacent to the 2000 psi boundary and the second

highest pressure is in block #1 which contains an injector well. The pressure is lowest in block

#3 which contains the 800 psi BHP well.

Well Rates and Pressures

We can also calculate the production rate in the constant BHP well and the bottom hole pressure

in the constant rate wells using the productivity index.

i i BHP iQ J P P

Block #3 constant BHP well (calculate rates)

3

3

mD-ft ftt = 0: 3306.5 6.33 03 800 psi 1000 psi 4186.03

cp day

mD-ft ftt = 1: 3306.5 6.33 03 800 psi 1019.2 psi 4587.89

cp day

mD-ftt = 2: 3306.5 6.33 03 8

cp

i i BHP i

i i BHP i

i i BHP i

Q J P P E

Q J P P E

Q J P P E

3

3

3

ft00 psi 1057.2 psi 5383.23

day

mD-ft ftt = 3: 3306.5 6.33 03 800 psi 1096.8 psi 6212.07

cp day

mD-ft ftt = : 3306.5 6.33 03 800 psi 1362.2 psi 11766.90

cp day

i i BHP i

i i BHP i

Q J P P E

Q J P P E

27

Block #1 constant rate well (calculate BHP) 3

3

3

ft1000day

t = 0: 1000 psi 1047.78 psimD-ft3306.5 6.33 03

cp

ft1000day

t = 1: 1008.9 psi 1056.68 psimD-ft3306.5 6.33 03

cp

ft1000day

t = 2: mD-ft3306.5 6.

cp

iBHP i

i

iBHP i

i

iBHP i

i

QP P

J E

QP P

J E

QP P

J

3

3

1018.3 psi 1066.08 psi33 03

ft1000day

t = 3: 1029.0 psi 1076.78 psimD-ft3306.5 6.33 03

cp

ft1000day

t = : 1441.2 psi 1488.98 psimD-ft3306.5 6.33 03

cp

iBHP i

i

iBHP i

i

E

QP P

J E

QP P

J E

Figure 8.1. Production rate of the constant BHP well placed at 6250 feet (block #3). The rate of the well increases

with time because the reservoir pressure increases with time. Fluids are being injected both through the injector well

(x = 0 feet) and from the boundary condition (x = L). Eventually, the reservoir reaches steady state; fluid produced

equals that injected and no change/increase in rate are observed with time

28

Figure 8.2. Well pressure of the constant rate well placed at 0 feet (block #1). Since the reservoir pressure (and block

#1 pressure) increase with time, the well BHP must increase with time to maintain the constant rate of 1000 ft3/day.

The bottomhole pressure reaches steady state at late times.

29

Example 9: Implicit Solution to 1D flow with gravity

= 0.2 k = 50 mD

=1 cp ct = 10-6 psi-1.

A = 200,000 ft2 (cross sectional area)

Bw = 1 rb/stb (assume constant)

= 60o (angle between the reservoir and the ground) L= 10000 ft (reservoir length) Area)

The initial condition is P = 1000 psi. The boundary conditions are P = 1800 psi at x = 0 and no flow

(q = 0) at x = L. Determine the pressure field in the reservoir using 4 uniform blocks. Use a time step of t = 1 day.

Solution:


: ( ,0) 1000

1: (0, ) 1800

2 : ( , ) 0

p p zg

t x x x

IC p x psi

BC p t psi

pBC L t

x

The implicit finite difference solution in flow rate units is given by the matrix equations

1n n

t t

B BT P P Q G

Where:

z4 = sin(60)* x

z3 = 2*sin(60)* x

z2 = 3*sin(60)* x

z1 = 4*sin(60)* x

30

3

32 1

1

3 2

m

3 2

50 20000 ft4000 6.33 3

1 1 2500 day

ft200000 ft 2500 ft 0.2 1 6 psi 100.0

psi

2 dirichlet boundary condition

lbft ft62.4 sin 60

day-psi ft 144 in

w

i i t

B B

o

kAT E

B x

B V c E

Q T P gz

G T gz T x

3ft ft

dayi

Note that 62.4/144 is 0.433 psi/ft which is the hydrostatic head gradient for a vertical column of water.

Here the reservoir is at an angle , so we multiply by sin (60) which is a hydrostatic head of 0.375 psi/ft. So at equilibrium (steady state) we expect that the pressure to be higher at lower depths, in fact an extra

0.375 psi for every foot of the reservoir. Note that zB = 0 in this problem


3

0

4000 4000 0 0

4000 8000 4000 06.33 3

0 4000 8000 4000

0 0 4000 12000

100

100;

100

100

1000 1250 2 4000 (1800 0.4

1000 3750; ;

1000 6250

1000 8750

E

ft

psi

psi psi

T

B

P z Q3

33 0)

0

0

0

4000 4000 0 0 1082.5 0

4000 8000 4000 0 3247.6 04.33 6.33 3

0 4000 8000 4000 5412.7 0

0 0 4000 12000 7577.7 27409

ft

day

E

G

The solution to this problem is summarized as follows:

1 2 3 3

1294.3 1469.3 1575.5 1531.4

1045.3 1098.8 1145.8 594.3psi; psi; psi; psi;

974.9 940.0 901.7 342.9

805.5 643.3 506.1 1280.2

P P P P

32

Example 10. Implicit Solution to flow in 2D


= 0.2 k = 50 mD

= 1 cp Bw = 1 rb/stb (assume constant)

ct = 110-6 psi-1.

L= 10,000 ft (horizontal length, x)

W=10,000 ft (horizontal width, y)

h = 20 ft (vertical thickness, z)

The initial condition is P = 1000 psi and the boundary conditions are no flow (q = 0) at x = 0,

y = 0, and y = H. There is a constant pressure boundary of P = 2000 psi, at x = L.

There is a constant-rate injector of 1000 ft3/day at x = 5000 ft, y = 5000 ft and a constant BHP

well (producer) with P = 800 psi at x = 9000 ft, y = 9000 ft. Both wells have a radius of 0.25 ft

and no skin factor.

Determine the pressure field in the reservoir using 9 uniform blocks (3 in each direction). Use a

time step of t = 1 day.

Solution:



2 2

2 2;

: p( , ,0) 1000

1: (0, , ) 0; 2 : p( , , ) 2000

3: ( ,0, ) 0; 4 : ( , , ) 0

t

w w

c p k p p q

B t B x y V

IC x y psi

pBC y t BC L y t psi

x

p pBC x t BC x H t

y y

The numerical solution is the same as always in matrix form:

No

flow

No flow

No flow

2000

psi

h=20 ft

L=10,000 ft

W= 10,000 ft

x=y 3,333 ft

q=1000 ft3/d

Pbh=800 psi

33

1n n

t t

B BT J P P Q

The mass balance for a general block i,j is given

1 1 1 1 1 1 1 1 11 1, , 1 1, , 1 , 1 , 1 , 1 , , , ,, , , ,2 2 2 2

n n n n n n n n n nii j i j i j i j i j i j i j i j i j i j i ji j i j i j i j

BT P P T P P T P P T P P P P q

t

The Transmissibility matrix is now pentadiagonal. Note that the transmissibity is uniform

because permeability is constant throughout the reservoir and x = y

1 1, ,

2 2

1000i j i j

w

kA mD ftT T T

B x cp

Note the boundary conditions. On the 3 no-flow boundaries, T = 0. On the constant pressure boundary, we add 2T. Therefore we get:

2 0 0 0 0 0 0

3 0 0 0 0 0

0 4 0 0 0 0 0

0 0 3 0 0 0

0 0 4 0 0

0 0 0 5 0 0

0 0 0 0 0 2 0

0 0 0 0 0 3

0 0 0 0 0 0 4

T T T

T T T T

T T T

T T T T

T T T T T

T T T T

T T T

T T T T

T T T

T

Computing the productivity (note that x = y) in this problem.

33,3

0.2 666.7

2 50 mD 20ft2 mD-ft ft796.5 5.042

666.7 cp psi-day1 cp lnln

0.25

eq

w

eq

w

w

r x ft

khJ

rB s

r

Plugging into the transmissibility matrix we get:

34

2000 1000 0 1000 0 0 0 0 0

1000 3000 1000 0 1000 0 0 0 0

0 1000 4000 0 0 1000 0 0 0

1000 0 0 3000 1000 0 1000 0 0

0 1000 0 1000 4000 1000 0 1000 0

0 0 1000 0 1000 5000 0 0 1000

0 0 0 1000 0 0 2000 1000 0

0 0 0 0 1000 0 1000 3000 1000

0 0 0 0 0 1000 0 10

TmD-ft

cp

00 4000

0

0

0

0mD-ft

0cp

0

0

0

796.5

J

The other matrices and vectors of interest are:

3

-13333.3 ft 3333.3 ft 20 ft 0.2 1.0 6 psi 44.4i i i tft

B V c Epsi

3

0

44.4 0 0 0 0 0 0 0 0

0 44.4 0 0 0 0 0 0 0

0 0 44.4 0 0 0 0 0 0

0 0 0 44.4 0 0 0 0 0ft

0 0 0 0 44.4 0 0 0 0psi

0 0 0 0 0 44.4 0 0 0

0 0 0 0 0 0 44.4 0 0

0 0 0 0 0 0 0 44.4 0

0 0 0 0 0 0 0 0 44.4

1000

1000

1000

1000

1000

1000

1000

1000

1000

B

P

3

3,3

3,3

0 0

0 0

2 25320

0 0ft

1000psi;psi-day

2 25320

0 0

0 0

2 5.042 800 25320

e

e

wf e

TP

Q

TP

J P TP

Q

35

Solving the system of equations gives the solution at the first time step and the steady-state value

P1 =

1003.2

1024.2

1202.0

1004.3

1037.1

1200.9

1002.9

1021.4

1174.7

psi; P2 =

1010.6

1060.7

1347.0

1013.1

1080.7

1343.8

1009.6

1053.3

1293.2

psi; P3 =

1022.6

1102.5

1452.4

1026.0

1126.1

1446.6

1020.3

1089.4

1375.4

psi; Pss =

1968.0

1977.3

1980.9

195

1957.3

1942.0

1939.7

1887.4

1765.2

psi


Method Block #1

(psi)

Block #4

(psi)

Block #5

(psi)

Block #6

(psi)

Block #9

(psi)

Initial 1000 1000 1000 1000 1000

1 day

Implicit 1003.2 1004.3 1037.1 1200.9 1174.7

CMG 1003.2 1004.3 1037.0 1200.6 1174.4

2 days

Implicit 1010.6 1013.1 1080.7 1343.8 1293.2

CMG 1010.6 1013.1 1080.6 1343.4 1292.8

3 days

Implicit 1022.6 1026.0 1126.1 1446.6 1375.4

CMG 1022.5 1026.0 1125.9 1446.1 1375.0

1968.0 1958.7 1982.9 1946.1 1766.9

36

Example 11. Implicit Solution to flow in 2D with horizontal wells


= 0.2 kx = ky = 50 mD; kz = 5 mD

= 1 cp Bw = 1 rb/stb (assume constant)

ct = 110-6 psi-1.

L= 10,000 ft (horizontal length, x)

W=10,000 ft (horizontal width, y)

h = 20 ft (vertical thickness, z)

The initial condition is P = 1000 psi. The boundary conditions are no flow (q = 0) at x = 0, y = 0, and

y = W. There is a constant pressure boundary of P = 2000 psi, at x = L. There is a horizontal well at the

center of the reservoir y = W/2 that spans then entire length L of the reservoir. It has a constant bottom

hole pressure of 800 psi. The well has a radius of 0.25 ft and a skin factor of -0.75.

Determine the pressure field in the reservoir using 9 uniform blocks (3 in each direction). Use a time step

of t = 1 day.

Solution:



2 2

2 2;

: p( , ,0) 1000

1: (0, , ) 0; 2 : p( , , ) 2000

3: ( ,0, ) 0; 4 : ( , , ) 0

t

w

p k p p qc

t B x y V

IC x y psi

pBC y t BC L y t psi

x

p pBC x t BC x H t

y y


1n n

t t

B BT J P P Q

7 8 9

4 5 6

1 2 3

No

flow

No flow

No flow

2000

psi

h=20 ft

L=10,000 ft

W= 10,000 ft

x=y 3,333 ft

37

The mass balance for a general block i,j is given

1 1 1 1 1 1 1 1 11 1, , 1 1, , 1 , 1 , 1 , 1 , , , ,, , , ,2 2 2 2

n n n n n n n n n n

i j i j i j i j i j i j i j i j i j i j i ji j i j i j i j

BT P P T P P T P P T P P P P q

t

The Transmissibility matrix is now pentadiagonal. Note that the transmissibity is uniform because

permeability is constant throughout the reservoir and x = y

Ti 1

2, j=T

i, j 12

=T =kA

mBwDx

=103mD- ft

cp

Note the boundary conditions. On the 3 no-flow boundaries, T = 0. On the constant pressure boundary, we add 2T. Therefore we get:

4

5

6

2 0 0 0 0 0 0 0

3 0 0 0 0 0 0

0 4 0 0 0 0 0 0

0 0 3 0 0 0

0 0 4 0 0

0 0 0 5 0 0

0 0 0 0 0 2 0 0

0 0 0 0 0 3 0

0 0 0 0 0 0 4 0

T T T

T T T T

T T T

T T T T J

T T T T T J

T T T T J

T T T

T T T T

T T T

T J

The horizontal well is treated differently from a vertical in 3 ways: (1) It spans multiple, 3, grid

blocks and a productivity index is needed in each block (effectively there is a well in each

block); (2) the well is horizontal so our equation for productivity index is altered for direction;

and (3) permeability is anisotropic, so the equation for the equivalent radius is more complicated.

Computing the productivity (note that x = y) in this problem.

e=

0.28 kzky( )

12 Dy2 + k

ykz( )

12 Dz2

12

kzky( )

14 + k

ykz( )

14

=

0.28 0.1( )1

2 3333.3 ft( )2

+ 10( )1

2 20 ft( )2

12

0.1( )1

4 + 10( )1

4

= 224.28 ft

J =2pDx k

ykz

m lnre

rw

+ s

=2p 3333.3 ft( ) 50 5

1 cp( ) ln224.28

0.25

-0.75

= 54742.81mD-ft

cp= 346.52

ft3

psi-day

Plugging into the transmissibility matrix we get:

38

T =

2000 -1000 0 -1000 0 0 0 0 0

-1000 3000 -1000 0 -1000 0 0 0 0

0 -1000 4000 0 0 -1000 0 0 0

-1000 0 0 3000 -1000 0 -1000 0 0

0 -1000 0 -1000 4000 -1000 0 -1000 0

0 0 -1000 0 -1000 5000 0 0 -1000

0 0 0 -1000 0 0 2000 -1000 0

0 0 0 0 -1000 0 -1000 3000 -1000

0 0 0 0 0 -1000 0 -1000 4000

mD-ft

cp

J =

0

0

0

54742.8

54742.8

54742.8

0

0

0

mD-ft

cp


3

-13333.3 ft 3333.3 ft 20 ft 0.2 1.0 6 psi 44.4i i i tft

B V c Epsi

B =

44.4 0 0 0 0 0 0 0 0

0 44.4 0 0 0 0 0 0 0

0 0 44.4 0 0 0 0 0 0

0 0 0 44.4 0 0 0 0 0

0 0 0 0 44.4 0 0 0 0

0 0 0 0 0 44.4 0 0 0

0 0 0 0 0 0 44.4 0 0

0 0 0 0 0 0 0 44.4 0

0 0 0 0 0 0 0 0 44.4

ft3

psi

P0 =

1000

1000

1000

1000

1000

1000

1000

1000

1000

psi; Q =

0

0

2TPe

J4Pwf

J5Pwf

J6Pwf

+ 2TPe

0

0

2TPe

=

0

0

25320

277218

277218

277218+ 25320

0

0

25320

ft3

psi-day

39

Solving the system of equations gives the solution at the first time step and the steady-state value

P1 =

980.7

997.9

1169.5

827.7

828.8

868.5

980.7

997.9

1169.5

psi;P2 =

964.1

1003.6

1276.9

808.2

810.3

857.6

964.1

1003.6

1276.9

psi; P3 =

952.0

1013.2

1346.3

805.7

808.6

858.5

952.0

1013.2

1346.3

psi P ss =

942.4

1079.6

1485.7

805.1

810.7

863.3

942.4

1079.6

1485.7

psi

The initial pressure of the reservoir was 1000 psi. The horizontal well in blocks 4, 5, and 6 are bringing

the reservoir pressure down (towards 800 psi) and the constant pressure boundary conditions adjacent to

blocks 3, 6, and 9 are raising it towards 2000 psi. Blocks 3 and 9 are the highest pressure (but below 2000

psi) because they have no well but are adjacent to the constant-pressure boundary). Block # 6 has the low

pressure well and is adjacent to the constant pressure boundary. Also note the symmetry, blocks #1, 2,

and 3 are symmetric to 7,8, and 9 and therefore have the same pressures.


Method Block #1

(psi)

Block #4

(psi)

Block #5

(psi)

Block #6

(psi)

Block #9

(psi)

Initial 1000 1000 1000 1000 1000

1 day

Implicit 980.7 827.7 828.8 868.5 1169.5

CMG 980.7 827.7 828.8 868.5 1169.2

2 days

Implicit 964.1 808.2 810.3 857.6 1276.9

CMG 964.1 808.2 810.3 857.6 1276.6

3 days

Implicit 952.0 805.7 808.6 858.5 1346.3

CMG 952.0 805.7 808.6 858.5 1345.9

942.4 805.1 810.7 863.3 1485.7

40

12. Implicit Solution to flow in 3D


= 0.2 k = 50 mD

= 1 cp Bo = 1 rb/stb (assume constant)

ct = 110-6 psi-1

L= 10000 ft (horizontal length, x)

W = 10000 ft (horizontal width, y)

h = 20 ft (vertical height, z)

x = y=5000 ft z = 10 ft

The initial condition is P = 1000 psi. The boundary conditions on all faces are no flow. Determine the

pressure field in the reservoir using 8 uniform blocks (222). Because this problem reaches steady state

quickly, use a time step of t = 0.0001 day.

Solution:



2 2 2

2 2 2

1;

: ( ,0) 1000

: 0 on all

t

p p p p kq

t x y z c

IC p x psi

BCs q

The numerical solution is the same as always in matrix form; there is a gravity term since there is a

vertical component of the reservoir.

1n n

t t

B BT P P Q G

Note that the transmissibity is not uniform because although permeability is constant throughout the

reservoir, x = y is not equal to z. Therefore, we will have different transmissibilities for flow in the horizontal and vertical directions Th and Tv respectively.

Th=kDxDz

mBwDx

= 500mD- ft

cp

Tv=kDxDy

mBwDz

=125*106mD- ft

cp

41

Note the boundary conditions. On the no-flow boundaries, T = 0 (and since this is such a small problem, all eight blocks are on all 3 boundaries!). Since the problem is 3D each block has 6 neighbors

(plus itself) so the transmissibility matrix is heptadiagonal. Therefore we get:

T =

2Th+T

v-T

h-T

h0 -T

v0 0 0

-Th

2Th+T

v0 -T

h0 -T

v0 0

-Th

0 2Th+T

v-T

h0 0 -T

v0

0 -Th

-Th

2Th+T

v0 0 0 -T

v

-Tv

0 0 0 2Th+T

v-T

h-T

h0

0 -Tv

0 0 -Th

2Th+T

v0 -T

h

0 0 -Tv

0 -Th

0 2Th+T

v-T

h

0 0 0 -Tv

0 -Th

-Th

2Th+T

v


Bi=V

ifict= 5000 ft5000 ft10 ft( ) 0.2( ) 1.0E -6 psi-1( ) = 50

ft3

psi

B =

50 0 0 0 0 0 0 0

0 50 0 0 0 0 0 0

0 0 50 0 0 0 0 0

0 0 0 50 0 0 0 0

0 0 0 0 50 0 0 0

0 0 0 0 0 50 0 0

0 0 0 0 0 0 50 0

0 0 0 0 0 0 0 50

ft3

psi

P0 =

1000

1000

1000

1000

1000

1000

1000

1000

psi; Q =

0

0

0

0

0

0

0

0

ft3

day; z =

5

5

5

5

15

15

15

15

ft G = 0.433Tz =

-3426100

-3426100

-3426100

-3426100

3426100

3426100

3426100

3426100

ft3

psi

Solving the system of equations gives the solution at various times:

42

P1 =

998.4

998.4

998.4

998.4

1001.6

1001.6

1001.6

1001.6

psi; P2 =

998.0

998.0

998.0

998.0

1002.0

1002.0

1002.0

1002.0

psi; P3 =

997.9

997.9

997.9

997.9

1002.1

1002.1

1002.1

1002.1

psi; P ss =

997.8

997.8

997.8

997.8

1002.2

1002.2

1002.2

1002.2

psi;

The solution in somewhat boring since the reservoir is homogeneous, has no wells, and all boundaries are sealed (no flow). The only driving force is gravity, so all bottom blocks (5, 6, 7 and 8) increase with

time (above 1000 psi) and top blocks (1, 2, 3, and 4) decrease with time. The steady state solution

CANNOT be found directly from T\G, because the matrix T is singular. Instead it must be found through

time stepping. The steady state solution is quickly reached (~0.0005 days) and the pressure gradient at

equilibrium is 0.433 psi/ft with the centerline (z=10 ft) equal to the initial condition (P=1000 psi)


Method Block #1

(psi)

Block #4

(psi)

Block #5

(psi)

Block #6

(psi)

Block #8

(psi)

Initial 1000 1000 1000 1000 1000

Steady

State

Implicit 997.8 997.8 1002.2 1002.2 1002.2

CMG 997.8 997.8 1002.2 1002.2 1002.2

Note that timesteps 1, 2 and 3 are too small to be represented in CMG.

43

Example 13. Implicit Solution to flow in 3D with wells

Consider a 3D reservoir with 2 wells and the following properties:

= 0.2 For top layer, kx =ky = 50 mD ;

For bottom layer, kx =ky = 40 mD

kz = 5 mD

= 1 cp Bo = 1 rb/stb (assume constant)

ct = 110-6 psi-1

L= 10000 ft (horizontal length x)

W=10000 ft (horizontal width y)

= 20 ft (vertical height, z) x=y= 5000 ft z = 10 ft

The initial condition is P = 1000 psi and the boundary conditions are no flow. There is a constant-rate

injector of 1000 ft3/day at x = 2500 ft, y=2500 ft and a constant BHP well (producer with P = 800 psi at x

= 7500 ft, y = 7500 ft. Both wells have a radius of 0.25 ft and are in the top layer of the reservoir.

Determine the pressure field in the reservoir using 8 uniform blocks (222). Use a time step of t = 1 day.

Solution:



2 2 2

2 2 2

1;

: ( ,0) 1000

: 0 on all

t

p p p p kq

t x y z c

IC p x psi

BCs q


1n n

t t

B BT J P P Q G

The productivity matrix, J, has been decoupled from the T matrix.

Note that the transmissibity is not uniform because permeability is anisotropic and x = y is not equal to z. Therefore, we will have different transmissibilities for flow in the horizontal and vertical directions Th and Tv respectively.

44

Th-top

=kxDxDz

mBwDx

= 500mD- ft

cp

Th-bottom

=kxDxDz

mBwDx

= 400mD- ft

cp

Tv=kzDxDy

mBwDz

=125*105mD- ft

cp

Note the boundary conditions. On the no-flow boundaries, T = 0 (and since this is such a small problem, all eight blocks are on all 3 boundaries!). Since the problem is 3D each block has 6 neighbors

(plus itself) so the transmissibility matrix is heptadiagonal. Therefore we get:

T =

2Th+T

v-T

h-T

h0 -T

v0 0 0

-Th

2Th+T

v0 -T

h0 -T

v0 0

-Th

0 2Th+T

v-T

h0 0 -T

v0

0 -Th

-Th

2Th+T

v0 0 0 -T

v

-Tv

0 0 0 2Th+T

v-T

h-T

h0

0 -Tv

0 0 -Th

2Th+T

v0 -T

h

0 0 -Tv

0 -Th

0 2Th+T

v-T

h

0 0 0 -Tv

0 -Th

-Th

2Th+T

v

There is a BHP well in block #4. Since the well is vertical and horizontal permeability is isotropic,

re= 0.2Dx = 0.2*5000 =1000 ft

J =2pDz k

ykx

m lnre

rw

+ s

=2p 10 ft( ) 50 50

1 cp( ) ln1000

0.25

= 378.78mD-ft

cp= 2.4

ft3

psi-day


Bi=V

ifict= 5000 ft5000 ft10 ft( ) 0.2( ) 1.0E -6 psi-1( ) = 50

ft3

psi

45

B =

50 0 0 0 0 0 0 0

0 50 0 0 0 0 0 0

0 0 50 0 0 0 0 0

0 0 0 50 0 0 0 0

0 0 0 0 50 0 0 0

0 0 0 0 0 50 0 0

0 0 0 0 0 0 50 0

0 0 0 0 0 0 0 50

ft3

psi J =

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 J 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

mD-ft

cp

P0 =

1000

1000

1000

1000

1000

1000

1000

1000

psi; Q =

Q

0

0

J Pwf

0

0

0

0

=

1000

0

0

1918.13

0

0

0

0

ft3

day z =

5

5

5

5

15

15

15

15

ft G = 0.433Tz =

-342610

-342610

-342610

-342610

342610

342610

342610

342610

ft3

psi

Solving the system of equations gives the solution at various times:

P1 =

1006.8

998.1

998.1

993.7

1011.2

1002.4

1002.4

998.0

psi; P2 =

1015.0

998.5

998.5

990.1

1019.3

1002.9

1002.9

994.4

psi; P3 =

1022.3

999.2

999.2

987.0

1026.7

1003.5

1003.5

991.3

psi; P ss =

1392.6

1304.8

1304.8

1217.1

1396.9

1309.2

1309.2

1221.4

psi

A bottom-hole pressure well of 800 psi is in block #4 and is producing fluid. Therefore block #4 is always

the lowest pressure in the top layer of the reservoir, but above 800 psi. The blocks in the lower layer at

equilibrium have higher pressures than blocks in the upper layer because of hydrostatic forces. Block #1

is relatively high because it has an injector well.


Method Block #1

(psi)

Block #2

(psi)

Block #4

(psi)

Block #5

(psi)

Block #8

(psi)

Initial 1000 1000 1000 1000 1000

1 day

Implicit 1006.8 998.1 993.7 1011.2 998.0

CMG 1006.8 998.1 993.7 1011.2 998.0

2 days

Implicit 1015.0 998.5 990.1 1019.3 994.4

CMG 1015.0 998.5 990.1 1019.3 994.4

3 days

Implicit 1022.3 999.2 987.0 1026.7 991.3

CMG 1022.3 999.2 987.0 1026.7 991.3

1392.6 1304.8 1217.1 1396.9 1221.4

46

Example 14.IMPES solution to multiphase flow in 1D without capillary pressure (Bulkley-

Leverett) Aziz and Settari example. Consider a 1D reservoir undergoing a water flood (water displaces oil) with the following

properties:

= 0.2 k = 100 mD w=o = 1 cp, cfw =cfo =110

-5 psi-1

Bo = Bw = 1.

L = 1000 ft (Length)

A = 10000 ft2 (Cross Sectional Area)

The initial condition is P = 1000 psi and Swi = Swr = 0.2.

The boundary conditions are no flow (q = 0) on both ends. Water is injected a constant-rate of

426.5 ft3/day at x = 0 ft. Reservoir fluids are produced at the same rate at x = L. Both wells have

a radius of 0.25 ft and no skin. Determine the pressure and saturation field in the reservoir using

3 uniform blocks and the IMPES method. Use a time step of t = 1 day.

Note that the relative permeability can be described by a Brooks-Corey type relationship

330.2 ; 1

1

rw ro

w wi

wi wr

k S k S

S SS

S S

At each time step we solve the pressure equation (overall mass balance) implicitly

No

flow

L=1000 ft

x = 333 ft

No

flow

Pinitial 1000 psi

q=426.5 ft3/d q=-426.5 ft3/d

47

1 1 1 1 11 1 1 12 2

1

n n n n n ni ti i i i i i ii i

n n

V cT P P T P P P P q

t

or

t t

B BT P P Q

and then the water saturation (water phase balance) explicitly

1

1 1 1 1 1

1 1 1 12 2

nn n w n n w n n www w i i i i ii i

B tS S T P P T P P q

V

Note that unlike in single phase flow, our T matrix changes in every time step because saturation (and therefore relative permeability) change.

time = 1 days

A t = 0 the water saturation is equation the residual saturation:

3

2

0.2 0.0

1 1.01

w wr rw

w wiro

wi

S S k

S Sk

S

Calculating the water and oil transmissibilities (Which are constant throughout the reservoir

because permeability is constant and saturation is uniform at t = 0).

3000 0

3000 3000

w rw rw

w w

o ro ro

o o

kA mD ftT k k

B x cp

kA mD ftT k k

B x cp

In matrix form, the water, oil, and total transmissibilities can be written as:

0 0 0 3000 3000 0

0 0 0 3000 6000 3000

0 0 0 0 3000 3000

3000 3000 0

3000 6000 3000

0 3000 3000

mD ft

cp

w o

w o

T T

T T T

The B, P, and Q arrays are then computed. The total compressibility is a weighted average of the

phase compressibilities:

5 11.0 10t o o w wc S c S c psi

48

,1 3

,2

,3

30

0 0 6.66 0 0

0 0 0 6.66 0

0 0 0 0 6.66

1000 426.5

1000 0

1000 426.5

f

f

f

V cft

V cpsi

V c

ftP psi Q

day

B

Solving for the pressure implicitly we get:

1

1017

P = 1000 psi

983

The saturation can then be found explicitly. Note that water is injected but no water is produced

at t = 0.

1 1

30

0.2 426.5

0.2 ; 0.0

0.2 0.0

n n w n www w

w

w

B t

V

ftS Q

day

1

w

S S T P Q

0.2006

S = 0.2

0.2

time = 2 days (Upwinding is used to calculate the relative permeability)

1. Block 1: Sw1= 0.2006, P1= 1017 . krw,1 = 2.4210-6 and kro,1 = 0.9968 2. Block 2: Sw2= 0.2, P2= 1000 . krw,2 = 0.0 and kro,2 = 1.0 3. Block 3: Sw3= 0.2, P3= 1017 . krw,3 = 0.0 and kro,3 = 1.0

The T matrices can be calculated as:

7

7.2 7.2 0 2990 2990 0

7.2 7.2 0 10 2990 5990 3000

0 0 0 0 3000 3000

mD ft

cp

w oT T

The B and Q arrays remain unchanged in this problem (although they might change in another

problem). Our new solution is:

49

2 2

1021 0.2013

P = 1000 psi; S = 0.20000001

979 0.2

time = 3 days (Upwinding is used to calculate the relative permeability)

The T matrices can be calculated as:

5.62 5 5.2 05 0 2981 2981 0

5.82 05 5.82 05 9.1 16 2981 5981 3000

0 9.1 16 9.1 16 0 3000 3000

E EmD ft

E E Ecp

E E

w oT T

The B and Q arrays remain unchanged in this problem (although they might change in another

problem). Our new solution is:

3 3

1022 0.2019

P = 1000 psi; S = 0.20000003

978 0.2


Method Block #1

167 ft

Block #2

500 ft

Block #3

833 ft

P, psi Sw P, psi Sw P, psi Sw

Initial 1000 0.2 1000 0.2 1000 0.2

0.01 day

IMPES 1017 0.2006 1000 0.2 983.4 0.2

CMG 1017 0.2006 1000 0.2 983.5 0.2

0.02 day

IMPES 1021 0.2013 1000 0.20000001 979 0.2

CMG 1021 0.2012 1000 0.2 979.2 0.2

0.03 day

Implicit 1022 0.2019 1000 0.20000003 978 0.2

CMG 1022 0.2019 1000 0.2 978.1 0.2

1115 0.8 1000 0.8 884 0.8

50

Example 15. SS solution to 1D multiphase flow w/o Pc (Bulkley-Leverett; Aziz and Settari

example).

Consider a 1D reservoir undergoing a water flood (water displaces oil) with the following

properties:

= 0.2 k = 100 mD

w=o = 1 cp cfw =cfw =110

-5 psi-1

Bo = Bw = 1.

L = 1000 ft (Length)

A = 10000 ft2 (Cross Sectional Area)

The initial condition is P = 1000 psi and Swi = Swr = 0.2.

The boundary conditions are no flow (q = 0) on both ends. Water is injected a constant-rate of

426.5 ft3/day at x = 0 ft. Reservoir fluids are produced at the same rate at x = L. Both wells have

a radius of 0.25 ft and no skin. Determine the pressure and saturation field in the reservoir using

3 uniform blocks. Use a time step of t = 1 day. Use the SS method, but handle the nonlinearities by using the solution for P and Sw in the previous time step.

Note that the relative permeability can be described by a Brooks-Corey type relationship

330.2 ; 1

1

rw ro

w wi

wi wr

k S k S

S SS

S S

A t=0 the water saturation is equation the residual saturation

3

2

0.2 0.0

1 1.01

w wr rw

w wiro

wi

S S k

S Sk

S

Calculating the water and oil transmissibilities (Which are constant throughout the reservoir

because permeability is constant and saturation is uniform at t = 0).

3000 0

3000 3000

w rw rw

w

o ro ro

o

kA mD ftT k k

x cp

kA mD ftT k k

x cp

51

t = 1 day

In matrix form the transmissibilities can be written as the following (recall the block-nature in

the SS method):

3000 0 3000 0 0 0

0 0 0 0 0 0

3000 0 6000 0 3000 0

0 0 0 0 0 0

0 0 3000 0 3000 0

0 0 0 0 0 0

T

The saturations did not change in blocks 2 and 3 and those values of D will stay the same.

However, block 1 will change slightly because of the change in saturation

1 3,

11,1

3

12,1 1

,

1

,

21,1

3.33 6 0.2 0.2 1 5 ft1.332

1 day-psi

3.33 6 0.2 ft666000

1 1 day

1 3.33 6 0.2 0.8 1 5 ft5.328

1

i

i

n

i i w i w

i i

l

w i

n

i i w i o

V S c E Ed

t

V Ed

B t

V S c E Ed

t

3

3

22,1 1

,

day-psi

3.33 6 0.2 ft666000

1 1 day

i i

n

o i

V Ed

B t

Which gives a slightly different value of D from the previous timestep

1.332 666000 0 0 0 0

5.328 666000 0 0 0 0

0 0 1.332 666000 0 0

0 0 5.328 666000 0 0

0 0 0 0 1.332 666000

0 0 0 0 5.328 666000

D3

1000 426.5

0.2 0

1000 0;

0.2 0

1000 0

0.2 426.5

ft

day

X Q

Calculating the solution: 1n n T D X DX Q

1017

0.2006

1000X =

0.2

983

0.2

Which is extremely close to the solution obtained from IMPES. Unless we took large toke steps

we anticipate the methods to give nearly identical results.

52

t = 2 days (we need to use upwinding for transmissibilities)

In matrix form the transmissibilities can be written as the following (recall the block-nature in

the SS method):

2990 0 2990 0 0 0

7.2 7 0 7.2 7 0 0 0

2990 0 5990 0 3000 0

7.2 7 0 7.2 7 0 0 0

0 0 3000 0 3000 0

0 0 0 0 0 0

E E

E E

T

The saturations did not change in blocks 2 and 3 and those values of D will stay the same.

However, block 1 will change slightly because of the change in saturation

1 3,

11,1

3

12,1 1

,

1

,

21,1

3.33 6 0.2006 0.2 1 5 ft1.336

1 day-psi

3.33 6 0.2 ft666000

1 1 day

1 3.33 6 0.2006 0.8 1 55

1

i

i

n

i i w i w

i i

n

w i

n

i i w i o

V S c E Ed

t

V Ed

B t

V S c E Ed

t

3

3

22,1 1

,

ft.344

day-psi

3.33 6 0.2 ft666000

1 1 day

i i

n

o i

V Ed

B t

Which gives a slightly different value of D from the previous timestep

1.336 666000 0 0 0 0

5.344 666000 0 0 0 0

0 0 1.332 666000 0 0

0 0 5.328 666000 0 0

0 0 0 0 1.332 666000

0 0 0 0 5.328 666000

D3

1017 426.5

0.2006 0

1000 0;

0.2 0

983 0

0.2 426.5

ft

day

X Q

Calculating the solution: n 1 n T D X DX Q

1021

0.2012

9999.8X =

0.20000004

979

0.20004

The solution is slightly different than that found using IMPES. One can continue future

time steps using the above procedure.

53

Matthew T. Balhoff, University of Texas at Austin (Dept of Petroleum and Geosystems Engineering) Reservoir Simulation Example and Homework Problems (2011), CONFIDENTIAL

Example 16. 1D Dual Porosity/Permeability Model of Naturally-Fractured Reservoir

Consider a 1D fractured reservoir with the following reservoir and fluid properties:

m= 0.2 km = 0.01 mD

f= 0.01 kf = 50 mD


ct = 10-6 psi-1.

L= 10000 ft (reservoir length)


Tmf = 100 mD-ft/cp

A reservoir has a low permeability of 0.01 mD and 20% porosity. It has natural fractures with

effective permeability of 50 mD and 0.01% porosity. The initial condition is P = 1000 psi. The

boundary conditions are P = 500 psi at x = 0 and no flow (q = 0) at x = L. Determine the pressure

field in the reservoir using a dual porosity/permeability model and 4 uniform blocks. Use a time

step of t = 1 days.

Solution:


2

2

2

2

1;

1;

: ( ,0) 1000

1: (0, ) 500

2 : ( , ) 0

m m mm f m

m m t

f f f

f m f

f f t

P P kP P

t x c

P P kP P

t x c

IC P x psi

BC P t psi

PBC L t

x

In the dual porosity/permeability model, we solve the matrix and fracture as separate domains

but flow is coupled trough a transfer function.

, , , ,i mf i mf i m i fQ T P P

Note that in the dual porosity model we assume no flow in the matrix (km=0), but in the dual permeability model km is finite and flow can flow through the matrix, fracture, and interchange.

L=10000 ft x = 2500 ft

500

psi

No


54


The mass balance (implicit finite difference equations) can be written as follows:

,1 1 1 1 1 1 1

1 , 1 , 1 , 1 , , , , ,, ,2 2

,1 1 1 1 1 1 1

1 , 1 , 1 , 1 , , , , ,, ,2 2

m in n n n n n n n

m i m i m i m i m i m i mf m i f im i m i

f in n n n n n n n

f i f i f i f i f i f i mf m i f if i f i

BT P P T P P P P T P P

t

BT P P T P P P P T P P

t

Where:

32 1

,

2

,

0.01 200000 mD-ft0.8 (Dual Permeability; = 0 dual porosity)

1 1 2500 cp

50 200000 mD-ft4000

1 1 2500 cp

ft200000 2500 0.2 1 6 100.0

psi

200000 2500 0.0

mm

f

f

m i i m t

f i i f t

k AT

x

k AT

x

B V c ft ft E psi

B V c ft ft

3

1 ft1 1 6 5.0psi

mD-ft100

cpmf

E psi

T


0

3

2

2

3

2

2

m mf m mf

m m mf m mf

m m mf m mf

m m mf m mf

mf f mf f

mf f f mf f

mf f f mf f

mf f f mf

m

m

m

m

f

f

f

f

T T T T

T T T T T

T T T T T

T T T T T

T T T T

T T T T T

T T T T T

T T T T

B

B

B

B

B

B

B

B

T

B P

1000 2

1000

1000

1000psi;

1000 2

1000

1000

1000

m B

f B

T P

T P

Q

55


Dual Porosity Solution: Km = Tm = 0

1 2 3

1 2

994.3 988.4 982.4 500

995.2 989.5 983.8 500

995.7 990.3 984.7 500

995.9 990.7 985.2 500

546.7 517.9

612.7 546

653.7

673.8

ss

m m m m

f f

P P P P

P P

3

512.8 500

.3 532.3 500

566.0 545.3 500

576.0 551.9 500

ss

f fP P

Dual Porosity/ Permeability Solution: Km = 50 mD

1 2 3

1 2

994.1 988.0 981.9 500

995.2 989.5 983.8 500

995.7 990.3 984.7 500

995.9 990.7 985.2 500

546.7 517.9

612.1 546

653.7

673.8

ss

m m m m

f f

P P P P

P P

3

512.8 500

.3 532.2 500

566.0 545.3 500

576.0 552.0 500

ss

f fP P

56


Example 17: IMPEC Solution to compositional flow (a tracer) in 1D


= 0.2 k = 50 mD


ct = 10-6 psi-1.

L= 10000 ft (length)


D = 100 ft2/day

Fluid flow in a 1D reservoir is driven by 2 dirichlet boundary conditions, PB1== 2000 psi at x=0

and PB2=1000 psi at x = L. The initial condition is P =1000 psi.

At time = 0, a tracer is injected at x=0 with mass concentration, w= 1 lbm/lbsolution. The initial

concentration is 0.0 throughout the reservoir and the boundary condition at x = L is no species

flux, dw/dx=0. The dispersion coefficient, D, is taken to be a constant (although in reality should

be a function of velocity and velocity varies with x before reaching steady state)

Solution:


2

2

3

m

3

m

;

: ( ,0) 0 lb /ft ; 1000 psi

1: (0, ) 1.0 lb /ft ; 2000 psi

2 : ( , ) 0; 1000 psi

p p

t x

c c cD u

t x x x

IC c x p

BC c t p

cBC L t p

x

There two coupled PDEs, one for pressure and one for tracer concentration. We can solve the

coupled PDEs using a fully implicit method, sequential implicit method (implicit pressure,

implicit concentration), or IMPEC approach (implicit pressure, explicit concentration). Here,

well used IMPEC.

57


The implicit pressure equation is given in the usual form:

1n n

t t

B BT P P Q

The explicit equation for concentration is given as:

1 1 1 112n n n n n n nc tdiag c

w w d T P Uw Q w c P P

Where the arrays are given as:

3 3 3 32 2 2 2 3 3

2 2

3 3 3 3 5 5 5 5 3 3 5 52 2 2 2 2 2 2 2 2 2 2 2

5 5 7 72 2 2 25 5 5 5 7 7 7 7

2 2 2 2 2 2 2 2

7 72 27 7 7 7

2 2 2 2

,1

,1

1

12

,1

w

i i

w

i i

w

T w T wU U

T w T w T w T w U U U U

U U U UT w T w T w T w

U UT w T w

B t

V

B t

V

B t

V

cT U

d

,11

,22

,33

,44

,1

1 1

2 2

Q

t

tn

t

t

ti i

w

i i

in in

c

cw

cwdiag

cw

cw

B t

V

T P U w

w c

And the block and interblock constants are determined by:

1 12 2

1 1 1 12,,2 2 2

,

1 1

12

1

if upwinding

if

i ii i

ic i i iw w w i

i i i

ii i i

k A D A VT w U d

B x B x B t

w P Pw

w P P

58


The following flow arrays are constants and do not change with time.

3

3

0

12000 4000 0 0

4000 8000 4000 06.33 3

0 4000 8000 4000

0 0 4000 12000

100

100

100

100

1000 2 4000 2000

1000 0; 6.33 03

1000 0

1000 2 4000 1000

ftE

day psi

ft

psi

psi E

T

B

P Q3ft

day

The following transport arrays are constants and do not change with time.

3

5

5 3

5

5

0

24000 8000 0 0

8000 16000 8000 0 ft6.33 3

0 8000 16000 8000 day-psi

0 0 8000 24000

2 10

2 10 ft

psi2 10

2 10

0 2 4000 2000 2 8000 1

0 0lb;

0 lb_soln 0

0 0

c

E

12

U

d

w Q3ft

6.33 03day

E

The Tc matrix depends on the upwinded block concentration and therefore changes with time.

59


time = 1

Initially, the concentration of tracer is zero throughout the reservoir but is injected at x=0. The

interblock values for concentration are upwinded and (in this problem) flow always goes from

left to right. So only w1/2 is non-zero. The Tc matrix can be written as follows:

3(0)

8000 0 0 0

0 0 0 06.33 3

0 0 0 0

0 0 0 0

c

ftE

day psi

T

The implicit solution for pressure and subsequent explicit solution for concentration gives:

1 1

1827.7 0.0124

1548.5 0;

1312.7 0

1101.6 0

P w

time = 2

Since a small amount of tracer is present in block #1, the upwinded concentration w3/2 is finite.

The new Tc matrix can be written as follows:

3(0)

8049.4 49.4 0 0

49.4 49.4 0 06.33 3

0 0 0 0

0 0 0 0

c

ftE

day psi

T


1 1

1870.9 0.0234

1616.2 8.93 05;

1366.8 0

1121.8 0

EP w

time = 3

Since a small amount of tracer is present in block #2, the upwinded concentration w5/2 is finite.

The new Tc matrix can be written as follows:

60


3(0)

8049.4 49.4 0 0

49.4 49.4 0 06.33 3

0 0 0 0

0 0 0 0

c

ftE

day psi

T


2 2

1874.5 0.0343

1624.0 2.56 04;

1374.0 6.39 07

1124.5 0

EP w

E

Time =

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