exam07

Solution to Final Exam TPG4160 Reservoir Simulation, May 29, 2007

page 2 of 11

Question 1 (11 points) a) Make sketches of Black Oil fluid properties wgosowgo RBBB µµµ ,,,,,, . Label bubble

point pressure, and saturated and undersaturated regions. b) Express reservoir densities for the three fluids in terms of wSgSoSsowgo RBBB !!! ,,,,,, . c) Express the density of the part of the reservoir oil that remains liquid at the surface. d) Express the density of the part of the reservoir oil that becomes gas at the surface. e) Express the gas density using real gas equation. f) Write the definition for fluid compressibility. g) Write an expression for pore compressibility a)

b)

�

!o =!oS + !gsRso

Bo

!g =!gS

Bg

!w =!wS

Bw

c)

�

!oL

=!oS

Bo

d)

�

!oG =!gsRso

Bo

e) PV = nZRT .

�

! "g = "gS

P

Z

ZS

PS

f)

�

c f = !(1

V)("V

"P)T g)

�

cr

= (1

!)("!

"P)T

P P P

P P P

Bw Bg Bo Rso

µw µg µo

P


page 3 of 11

Question 2 (12 points) For a completely water-wet system, make sketches of saturation functions (including labels for important points/areas)

a) Oil-water system: imbibition and drainage

�

krw, k

row, P

cow vs.

�

Sw

b) Oil-gas system: imbibition and drainage

�

krg , krog , Pcog vs.

�

Sg c) Typical contours of three-phase

�

kro

in a ternary (triangular) diagram (axes

�

So, Sw, Sg ) For a mixed-wet oil-water system, sketch

d) imbibition

�

Pcow

vs.

�

Sw. Also label areas of spontaneous and forced imbibition.

a)

K

Sw

Swir1.0

r

Sw

Swir1.0

oil

water

Drainage curves

Pcd

Pcow

Sw

Swir

Kr

Sw

Swir

1-Sor

oil

water

Imbibition curves

1-Sor

Pcow

b)

So1.0

Kr

SoSorg 1-Sgc

gas

oilSo=1

Drainage

process

Sorg

Pdog

Pcog


page 4 of 11

oil

So

Kr

SoSorg 1-Sgro

oil

gasSo=Sor

Imbibition

process

Sorg

Pcog

1-Sgro c)

100% gas

100% oil100% water

minimum liquid

saturation under

gas displacement

minimum oil plus

gas saturation under

water displacement

Swir

Sor

0.2

0.4

0.6

0.8

0.1

Question 3 (12 points) Show all steps in the derivation of the following two partial differential equations for one-phase flow in a one-dimensional, linear, horizontal porous material. Start with the one-dimensional, one-phase, continuity equation and define the models used for fluid and rock behavior, and state all assumptions made.

a)

�

! 2P

!x 2="µc

k

!P

!t

b)

�

!!x

k

µB

!P!x

"

# $

%

& ' = (

cr

B+d(1/B)

dP

"

# $

%

& ' !P!t

For constant cross sectional area, the continuity equation simplifies to:

�

!"

"x#u( ) =

"

"t$#( )

Darcy's equation, which for one dimensional, horizontal flow is:

�

u = !k

µ

"P

"x.

Rock compressibility:

�

cr

= (1

!)("!

"P)T

�

!d"

dP= "c

r


page 5 of 11

a) Substitution:

�

!!x

"k

µ

!P!x

#

$ %

&

' ( =

!!t

)"( )

Fluid compressibility:

�

c f = !(1

V)("V

"P)T = (

1

#)("#

"P)T $ #c f = (

d#

dP)

Right side

�

!

!t"#( ) = #

d"

dP

!P

!t+ "

d#

dP

!P

!t= #"cr + "#c f( )

!P

!t= #"c

!P

!t

Left side: Assume

�

k and µ are constants:

�

!!x

"k

µ

!P!x

#

$ %

&

' ( =

k

µ

!!x

"!P!x

#

$ %

&

' (

Then

�

!!x

"!P!x

#

$ %

&

' ( = "

!!x

!P!x

#

$ %

&

' ( +

!P!x

#

$ %

&

' ( d"dP

!P!x

= "! 2P!x 2

#

$ %

&

' ( +

!P!x

#

$ %

&

' (

2

d"dP

= "! 2P!x 2

#

$ %

&

' ( +

!P!x

#

$ %

&

' (

2

"c f

We make the assumption that

�

! 2P!x 2

"

# $

%

& ' >>

!P!x

"

# $

%

& '

2

c f

Then

�

!!x

"!P!x

#

$ %

&

' ( ) "

! 2P!x 2

#

$ %

&

' (

And the equation becomes:

�

! 2P

!x 2="µc

k

!P

!t

b) Substitution:

�

!!x

"k

µ

!P!x

#

$ %

&

' ( =

!!t

)"( )

Fluid density:

�

!o =!oS + !gSRso

Bo

=constant

Bo

Right side

�

!!t

"B

#

$ %

&

' ( =

1

B

!"!P

+ "!(1/B)!t

=1

B

d"dP

!P!t

+ "d(1/B)

dP

!P!t

=

"cr

B

!P!t

+ "d(1/B)

dP

!P!t

or

�

!!t

"B

#

$ %

&

' ( = constant "

cr

B+d(1/B)

dP

)

* +

,

- .

!P!t

Left side

�

!!x

"k

µ

!P!x

#

$ %

&

' ( =

!!x

constant

B

k

µ

!P!x

#

$ %

&

' ( = constant

!!x

k

Bµ

!P!x

#

$ %

&

' (

Thus, the flow equation becomes:

�

!!x

k

µB

!P!x

"

# $

%

& ' = (

cr

B+d(1/B)

dP

)

* + ,

- . !P!t


page 6 of 11

Question 4 (10 points) Use Taylor series and show all steps in the discretization of the following two equations:

a)

�

! 2P

!x 2="µc

k

!P

!t

b)

�

!!x

k

µB

!P!x

"

# $

%

& ' = (

cr

B+d(1/B)

dP

"

# $

%

& ' !P!t

a) Right side:

�

P(x, t) = P(x, t + !t) +"!t

1!# P (x,t + !t) +

("!t)2

2!# # P (x, t + !t) +

("!t)3

3!# # # P (x,t + !t) + .....

Solving for the time derivative, we get:

�

(!P

!t)i

t+"t=Pi

t+"t# P

i

t

"t+ O("t) .

Left side:

�

P(x + !x,t + !t) = P(x, t) +!x

1!" P (x,t + !t) +

(!x)2

2!" " P (x,t + !t) +

(!x)3

3!" " " P (x,t + !t) + .....

�

P(x !"x,t + "t) = P(x, t) +(!"x)

1!# P (x, t + "t) +

(!"x)2

2!# # P (x,t + "t) +

(!"x)3

3!# # # P (x, t + "t) + .....

By adding these two expressions, and solving for the second derivative, we get the following approximation:

�

(!2P

!x2)i

t+"t=Pi+1

t+"t# 2P

i

t+"t+ P

i#1

t+"t

("x)2

+ O("x2)

Substituting into the equation, we get:

�

Pi+1

t+!t " 2Pi

t+!t+ P

i"1

t+!t

!x 2# (

$µc

k)Pi

t+!t " Pi

t

!t

b) Right side: We use the same approximation for the pressure derivative as in a):

�

!cr

B+d(1/B)

dP

"

# $

%

& ' ((P(t)

)

* +

,

- .

i

t+/t

0 !cr

B+d(1/B)

dP

"

# $

%

& '

)

* +

,

- .

i

t+/tPi

t+/t 1 Pi

t

/t

Left side:

�

(k

µB)!P

!x

"

# $

%

& ' i+1/ 2

= (k

µB)!P

!x

"

# $

%

& ' i

+(x /2

1!

!

!x(k

µB)!P

!x

"

# $

%

& ' i

+((x /2)2

2!

! 2

!x 2(k

µB)!P

!x

"

# $

%

& ' i

+ .....

�

(k

µB)!P

!x

"

# $

%

& ' i(1/ 2

= (k

µB)!P

!x

"

# $

%

& ' i

+()x /2

1!

!

!x(k

µB)!P

!x

"

# $

%

& ' i

+(()x /2)2

2!

! 2

!x 2(k

µB)!P

!x

"

# $

%

& ' i

+ .....

combination yields

�

!

!x(k

µB)!P

!x

"

# $

%

& ' i

=

(k

µB)!P

!x

"

# $

%

& ' i+1/ 2

( (k

µB)!P

!x

"

# $

%

& ' i(1/ 2

)x+ O()x 2) .

Using similar central difference approximations for the two pressure gradients:


page 7 of 11

�

!P!x

"

# $

%

& '

i+1/ 2

=Pi+1

( Pi

)x+ O()x)

and

�

!P!x

"

# $

%

& '

i(1/ 2

=Pi( P

i(1

)x+ O()x) .

the expression becomes:

�

!

!x(k

µB)!P

!x

"

# $

%

& ' i

(

(k

µB)Pi+1 ) Pi*x

"

# $

%

& ' i+1/ 2

) (k

µB)Pi) P

i)1

*x

"

# $

%

& ' i)1/ 2

*x

or

�

!

!x(k

µB)!P

!x

"

# $

%

& ' i

( (k

µB)i+1/ 2

Pi+1 ) Pi*x 2

) (k

µB)i)1/ 2

Pi) P

i)1

*x 2

Thus, the difference equation becomes:

�

(k

µB)i+1/ 2

Pi+1 ! Pi"x 2

! (k

µB)i!1/ 2

Pi! P

i!1

"x 2# $

cr

B+d(1/B)

dP

%

& '

(

) *

+

, -

.

/ 0 i

t+"tPi

t+"t ! Pi

t

"t

Question 5 (12 points) For two-phase flow of oil and water in a horizontal, one-dimensional porous medium, the flow equations can be written (including well terms):

�

!!x

kkro

µoBo

!Po

!x

"

# $

%

& ' ( ) q o =

!!t

*So

Bo

"

# $

%

& '

�

!!x

kkrw

µwBw

!Pw

!x

"

# $

%

& ' ( ) q w =

!!t

*Sw

Bw

"

# $

%

& ' ,

where

�

Pw

= Po! P

cow

�

So

+ Sw

=1 a) Write the two flow equations on discretized forms in terms of transmissibilities, storage

coefficients and pressure and saturation differences (Do not derive). b) List the assumptions for IMPES solution, and outline briefly how we solve for pressures

and saturations c) What are the limitations of the IMPES solution?

a)

�

Txoi+1 2 Poi+1 ! Poi( ) + Txoi!1 2 Poi!1 ! Poi( ) ! " q oi

= Cpooi Poi ! Poi

t( ) + Cswoi Swi ! Swi

t( ), i =1,N

�

Txwi+1 2 Poi+1 ! Poi( ) ! Pcowi+1 ! Pcowi( )[ ] + Txwi!1 2 Poi!1 ! Poi( ) ! Pcowi!1 ! Pcowi( )[ ] ! " q wi

= Cpowi Poi ! Poi

t( ) + Cswwi Swi ! Swi

t( ), i =1,N

b) In the IMPES solution, all coefficients and capillary pressures are evaluated at time=t. The two equations are combined so that the saturation terms are eliminated. The resulting equation is the pressure equation:

�

aiPo

i!1+ b

iPo

i+ c

iPo

i+1= d

i, i =1,N


page 8 of 11

which may be solved for pressures in all grid blocks by Gaussian Elimination Method, or some other method. Then, the saturations may be solved for explicitly by using one of the equations. Using the oil equation, yields:

�

Swi = Swi

t+

1

Cswoi

tTxoi+1/ 2

tPoi+1 ! Poi( ) + Txoi!1/ 2

tPoi!1 ! Poi( ) ! " q oi !Cpooi

tPoi ! Poi

t( )[ ] , i =1,N

c) The approximations made in the IMPES method, namely the evaluation of coefficients at old time level when solving for pressures and saturations at a new time level, puts restrictions on the solution which sometimes may be severe. Obviously, the greatest implications are on the saturation dependent parameters, relative permeability and capillary pressure. These change rapidly with changing saturation, and therefore IMPES may not be well suited for problems where rapid variations take place. IMPES is mainly used for simulation of field scale systems, with relatively large grid blocks and slow rates of change. It is normally not suited for simulation of rapid changes close to wells, such as coning studies, or other systems of rapid changes. However, provided that time steps are kept small, IMPES provides accurate and stable solutions to a long range of reservoir problems. Question 6 (23 points) For a one-dimensional, horizontal, 3-phase oil, water, gas system, the general flow equations are (including well terms):

�

!!x

kkro

µoBo

!Po

!x

"

# $

%

& ' ( ) q o =

!!t

*So

Bo

"

# $

%

& ' ,

�

!!x

kkrg

µgBg

!Pg

!x+ Rso

kkro

µoBo

!Po

!x

"

# $ $

%

& ' ' ( ) q g ( Rso

) q o =!!t

*Sg

Bg

+ Rso

*So

Bo

"

# $ $

%

& ' '

�

!!x

kkrw

µwBw

!Pw

!x

"

# $

%

& ' ( ) q w =

!!t

*Sw

Bw

"

# $

%

& '

a) Explain briefly the physical meaning of each term in all three equations. b) What are the criteria for saturated flow? What are the functional dependencies of

�

Rso

and Bo?

c) What are the primary unknowns when solving the saturated equations? d) What are the criteria for undersaturated flow? What are the functional dependencies of

�

Rso

and Bo?

e) What are the primary unknowns when solving the undersaturated equations? f) Rewrite the equations above for undersaturated flow conditions.

a)

�

!!x

kkro

µoBo

!Po

!x

"

# $

%

& ' ( ) q o =

!!t

*So

Bo

"

# $

%

& '

transport of oil well potential accumulation of oil

�

!!x

kkrg

µgBg

!Pg

!x+ Rso

kkro

µoBo

!Po

!x

"

# $ $

%

& ' ' ( ) q g ( Rso

) q o =!!t

*Sg

Bg

+ Rso

*So

Bo

"

# $ $

%

& ' '

transport of transport of gas well oil well pot. accumulation. accumulation free gas sol. gas potential (solution gas) of free gas of solution gas


page 9 of 11

�

!!x

kkrw

µwBw

!Pw

!x

"

# $

%

& ' ( ) q w =

!!t

*Sw

Bw

"

# $

%

& '

transport of water well potential accumulation of water b) Saturated flow: criteria

�

Po = Pbp and

�

Sg > 0.

dependencies

�

Bo = f Po( ) and

�

Rso = f Po( ) . c)

�

Po and Sg d) Undersaturated flow: criteria

�

Po > Pbp and

�

Sg = 0.

dependencies

�

Bo = f Po,Pbp( ) and

�

Rso = f Pbp( ) .

e)

�

Po and Pbp

f)

�

!!x

kkro

µoBo

!Po

!x

"

# $

%

& ' ( ) q o =

!!t

*So

Bo

"

# $

%

& '

�

!!x

Rso

kkro

µoBo

!Po

!x

"

# $

%

& ' ( ) q g ( Rso

) q o =!!t

Rso

*So

Bo

"

# $

%

& ' ,

�

!!x

kkrw

µwBw

!Pw

!x

"

# $

%

& ' ( ) q w =

!!t

*Sw

Bw

"

# $

%

& '


page 10 of 11

Question 7 (10 points) For a one-dimensional (z), 3-phase oil, water, gas system, outline how initial pressures and saturations may be computed in a simulation model, assuming that equilibrium conditions apply. Sketch the grid system, including gas-oil-contact (GOC) and water-oil-contact (WOC). Also sketch the oil-gas and oil-water capillary pressure curves, and show the steps involved in the calculations of initial pressures, gas and water saturations


page 11 of 11

Question 8 (10 points) Using compositional fluid formulation, the flow equation for component

�

k of a system with

�

Nc components may be written as:

�

!!x

Ckg"g

kkrg

µg

!Pg!x

+ Cko"o

kkro

µo

!Po!x

#

$ % %

&

' ( ( =

!!t

) Ckg"gSg + Cko"oSo( )[ ], k =1,Nc

The Black Oil fluid formulation results in the following flow equations for oil and gas:

!!x

kkrg

Bgµg

!Pg!x

+kkroRso

Boµo

!Po!x

"

#$

%

&' =

!!t

(Sg

Bg+SoRso

Bo

"

#$

%

&'

)

*++

,

-..

�

!!x

kkro

Bo

µo

!Po

!x

"

# $

%

& ' =

!!t

(So

Bo

"

# $

%

& '

If we define the Black Oil model to be a pseudo-compositional model with two components, efine the components and the fractions needed to convert the compositional equations to Black-Oil equations. component 1: oil – k=o

�

!!x

Cog"g

kkrg

µg

!Pg!x

+ Coo"o

kkro

µo

!Po!x

#

$ % %

&

' ( ( =

!!t

) Cog"gSg + Coo"oSo( )[ ]

component 2: gas – k=g

�

!!x

Cgg"g

kkrg

µg

!Pg!x

+ Cgo"o

kkro

µo

!Po!x

#

$ % %

&

' ( ( =

!!t

) Cgg"gSg + Cgo"oSo( )[ ]

Question: what are the fractions needed to get the Black Oil equations:

�

!!x

kkrg

Bgµg

!Pg!x

+kkroRso

Boµo

!Po!x

"

# $ $

%

& ' ' =

!!t

(Sg

Bg

+SoRso

Bg

"

# $ $

%

& ' '

)

*

+ +

,

-

.

.

�

!!x

kkro

Bo

µo

!Po

!x

"

# $

%

& ' =

!!t

(So

Bo

"

# $

%

& '

Answer:

�

fraction of "gas in gas": Cgg = 1

fraction of "oil in gas": Cog = 0

fraction of "gas in oil": Cgo =!gSRso

!oBo

fraction of "oil in oil": Coo =!oS

!oBo

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