Reservoir Engineering I Course Notes - Shariff University.pdf

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Reservoir Engineering I, Mohsen Masihi

Sharif University of Technology

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Reservoir Engineering I 0 ·~

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1 .5 :r

2007

Course Notes No. 26133

Mohsen Masihi

Department of Chemical and Petroleum Engineering Sharif University of Technology, Tehran, Iran

[email protected]

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Table of contents

Course overview ....... ........................... .................................................................................................................. .. .............. 3 )-Introduction ............... ............................................. ............. .................... ............................................................................ 5

1-1 Porous media and its characteristics ............................................................................................................................ 8 1-2 Darcy's law and its range of validity ......................................................................................................................... 17 1-3 Porosity and permeability relationship ...................................................................................................................... 23 1-4 Classification of reservoir flow systems ......................... ........................................................................................... 26 1-5 Reservoir boundary conditions .................................................................................................................................. 33 Units ................................................................................................................................................................................ 37

2-Steady state single phase flow ................................................................................. .... ......................... .......................... .. 38 2-1 Steady state flow in linear reservoir geometry .......................................................................................................... 38 2-2 Steady state flow in redial reservoir geometry .......................................................................................................... 46 2-3 Steady state flow in spherical reservoir geometry ..................................................................................................... 50 2-4 Average permeabi I ity in heterogeneous reservoir ..................... ................................................................................ 50 2-5 Pressure drawdown in a well ......................................................................................... ............................................ 54 2-6 Average reservoir pressure ........................................ ......... ............................................................................. .......... 57 2-7 Altered permeability zone ......................................................................................................................................... 58 2-8 Flow dependent skin .................................................................................................................................................. 63

3-Unsteady state single phase flow ...................................................................................................................................... 66 3-1 Governing flow equations ........................ ................................................................................................................. 66 3-2 Radial oil flow in a fully penetrating vertical well .................................................................................................... 72 3-3 Radial gas flow in a fully penetrating vertical well ..... ............ .................... ....................................... ~ ............... ....... 80

4-New flow solutions using superposition .......................................................... ................................................................ 84 4-1 Principal of superposition ................................................... ................. ... ....................................................... ... ......... 84 4-2 Effect of multiple wells ......................................................................................................................... .................... 85 4-3 Effect ofvariable flow rate .................... .. ........................................................................................................... ~ ...... 87 4-4 Effect of nearby boundaries ...................................................... .. .............................................................................. 95

5-Pseudo steady state single phase flow ............................. ..... ............................................ ................. .... ...... ....... ............ I 05 5-I Slightly compressible fluid and redial geometry ........... ...................... ......... .. ...................................................... .... ! 06 5-2 Effect of non -circular drainage area ....................................................................................................................... 11 0 5-3 Compressible fluid and redial geometry ...................... ..................................... ..... ........... ......... ........................... ... l.l6 5-4 Productivity index and IPR curves .......................................................................................................................... II6

6-Two phase fluid flow in porous media ........................................................................................................................... 124 6-.f Buckley Leverett analysis .......................................... ................................ .............................. ... .................. ... ....... 128 6-2 Welge construction ........... .. .. ..... ............................. ................. .......... ... ...................................... ....................... ...... 138

References and further reading .......................................................................................................................................... l49

•

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Course overview

This course provides an introduction to the basics of reservoir engineering. This is started with an introduction to the different classification of flow regimes in a reservoir. Pressure drop within the reservoir as a result of production for finite and infinite reservoir will be discussed. Many important concepts such as steady state, transient and semi steady state flow, skin effect, Dietz shap factor, productivity index, IPR etc will be introduced. Moreover, some simple models of single phase liquid flow using diffusion equation will be set up and the solution to the model equations will ·be found . Basics ingredients of the diffusion equation such as mobility and storativity will be introduced. We will also introduce other useful technique such as superposition concept to find the solution of complicated well configurations using the previously derived solution of the simple well configurations. These ideas will . also be extended to gas flow in porous media. Finally the analysis of flow behavior of two immiscible fluids in one dimension will be presented.

The style of the course will be a mix of geology, physics and mathematics in order to work out the governing diffusion equations in porous media and to find the solution for them. For this, a good understanding of calculus and differential equations are recommended.

At the end of the course, you will learn how to model different flow systems in porous media. You will also be familiar with many important concepts used in reservoir engineering.

These prepared course notes follow closely the sequence of material that will be presented. However, these are further suggested materials:

1-L. P. Dake, Fundamentals of reservoir engineering (Elsevier, 1998)·

2-B. C. Craft and M. F. Hawkins, Applied Petroleum Engineering (Prentice Hall, 1991)

3-R. E. Collins, Flow of fluids through porous materials (REC Publishers, 1991)

4-C. S. Matthew and D.G. Russell, Pressure build up and flow test in wells (SPE, 1967)

5- G. de Marsily, Quantitative hydrogeology (Academic Press, 1986)

Syllabus

I. Introduction to the flow behaviour in porous media. Basic definitions such as porosity, permeability, definition of Representative Elementary Volume (REV), Darcy's law, conservation of mass. Different flow systems based on geometry, compressibility of fluids, phases and time dependence, different boundary conditions.

2. Steady state single phase flow. Solution to these systems under different fluid compressibility assumptions and various geometries. Pressure drawdown into a well. Definition of skin and its various types.

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3. Unsteady state single phase flow. Deriving the governing flow e·quati.on for slightly compressible and presenting the solution to this known as line source solution. Logarithmic approximation to line source solution. Finding solution to the compressible fluids .

4. New solution using superposition. Definition of superposition principal. Using this to find new solutions in the case of multiple wells and for a well producing with variable flow rates. Also

- this helps to find the effect of the nearby boundaries.

5. Semi-steady state single phase flow. Solution to these systems under different fluid compressibility assumptions in radial geometry. The effect of non-circularity of the drainage area and definition of the Dietz shape factor. Concepts of productivity index (PI) and inflow performance relationship (IPR)

6. Two phase flow. One dimensional diffuse flow conditions. Theory of Buckley Leverett used to analyze this. Breakthrough and recovery calculation

Course structure

The course will consist of approximately 16 sessions (each one 3 hours) of lectures. The course will be assessed by a series of homework as well as a Mid term and a final examinations.

The notes in some places cover more material than can reasonably be covered during the course and in other places have deliberate gaps for more discussions. Please feel free to ask questions during the course.

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1-Introduction Reservoir engineering is at the heart of the other activities (such as geology, geophysics, petrophysics, production process, economics etc) that receive information, process it and then pass it on to others.

In reservoir engineering courses we aim is to: a) understand the nature of rock-fluid interaction. b) investigate the nature of fluid displacement. c) develop a flow model consistent with geology, rock and

fluid property and past performance characteristics. d) predict the future reservoir recovery. e) develop improved recovery methods of fluids.

In "reservoir engineering (II)" course we will use a very sitnplified reservoir model called Material Balance :Equation bfll~I~_ct~fll!hl__!}owjrls.ig~~JJ~servoj_r_j!LQider ~ify ~~s~~--~r:_i_y~_lJleGha_riisms_-alid_ ~ However, we often want to have reservoir behavior due to fluid injection or production from a well as a function of time which requires a time and/or flow rate dependent model of reservoir and this is the things that we will discuss in this course, "reservoir engineering (I)".

The subject of flow in porous media is an important subject in reservoir engineering. Note that this is a very complex phenomenon and cannot easily be compared with

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flow throl!gh pipes or conduits. It is rather easy to measure the length and diameter of a pipe and compute its flow -:-capacity as a function of pressure; however, there are no clear flow paths in porous media to be evaluated.

Understanding the flow characteristic within a reservoir is necessary for: a) determining the productivity of a reservoir. b) deciding the optimum strategy to maximize recovery. c) interpretation of the well test data.

The analysis of flow in porous media has extensively investigated either ~xperimentally (using([ac!£~ or real

. cores) or analytically in various disciplines including ~~--

hydrology and petroleum engineering. The aim was to . formulate laws and correlations that can then be utilized to

make reliable predictions. 'LA- el-. ·l2t

. /' For example, as a result of experimental studies on the flow of water through unconsolidated sand · filter b~_~s _in 1856, ~nry-:_ Darcy found _____ tiiafn --t~,o~ ___ a

homogenous fll!id __ in ~- porous medium is proportional _to the driving_ force -a~d-" InverselY -proportloilaltofluid viscosity (Darcy's law}\ However, this applle~ only--m --------·---------- · -· ··-··-· ···· .. ····· ···. ~ ,:

laminar. flow regime; --In turbulent flow, which occurs at higher velocities, the pressure gradient increase at a__gr~~ter rC!_t~ thal}_g_oe.s_the flq_w. _ _r~t~~ Fortunately, except for large production or injection rates in vicinity of the well bore, Department of Chemical & Pwoleum Engint<ering, Sharif University ofTechnology, Tehran, IRAN 6

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flow in the reservoir and in most laboratory tests is by design, streamlined and Darcy's law is valid.

To investigate the flow behavior in a reservoir we need an appropriate flow model (governing differential equation) along with necessary initial and boundary conditions that can be solved(ip order to Imd the flow_chara.cJ:ecistii) The governing flow equation can be developed by using the conservation of m.ass, transport phenomena (Navier-Stokes eq~atiQ11S or simple Darcy 1aw_)___and tlle_~gy_(ltion __ <?f ~-

-----

The resulting partial differential equation especially in the case of multi phase flow may not be solved analytically, so \V~ _may n~ed _:nl1m~rical methods such as fin!!_e differ~ (this is the subject of "reservoir modeling. and simulation" course). However, il_l case of single phase_flQW __ withsi_mple boundary_ condit~~ns one can _ us~-il~~ly:tiQ.al rn._e_thods in

- ·· ·· ···· ···-- -· -- · I

order to find the solutionl

The resulting mathematical formulas will vary depending . -- :::: c:: . --. - ~ . - ... - S=:::::;::::

upon the characteristics of the reservoir (such as reservoir geoJ11~!xy_, _ f!g~ ____ re._gi~-~~-of fluids in the reservoir). As can be seen later there exists 3 cases for reservoir geometries 3 _ types of fluids and 3 cases for reservoir regimes;' hence there rna exist _27 types _ gf reservoir flow formulations. Before describing various ~---··- · -- · .. -reservoir flow fonnulations let us first review the main characteristics of a porous medium (such as porosity and Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN 7

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. permeability), Darcy's law and its limitations and various flow systems.

1-1 Porous media and its characteristics The reservoir rocks may range from very loose and unconsolidated sand to a very hard and dense sandstone, limestone, or dolomite. The grains may be bonded together with a number of materials (such as silica, calcite, or clay). Knowledge of the physical properties of the rock and the existing interaction between the hydrocarbon system and the formation is critical in understanding the flow behavior in the reservoir. Porous media refers to any geological rock that comprised of grains and void spaces between grains as shown in figure below,

flow

~

Non-interconnected pore space

A ~'Pical1D pore structure where the pore space is black

This exhibits two essential characteristics: a) capacity for storage of fluids: needs void space within

the rock (i.e. porosity). b) transmissvity capability to the fluids: needs continuity

of the void spaces (permeability).


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Porosity The porosity of a rock is a measure of the storage capacity (pore volume) that is capable of holding fluids] It is related to the void spaces that are the inter-granular pore spaces between sedimentary particles at the microscopic scale. Mathematically porosity is defined as,

V V. -V t/J = pore X 1 OO = bulk grain X 1 00 ~ulk ~ulk

The porosity · of a cubical packing (the least compact a!~a~_~ent) of spheiiCal-grau:ls with the_ sa_me , .. $!?;e i_s 47.6% (i.e. the maximum porosity). In reality reseryoir

( ----------------- --- --- --------- .

rocks are not regular. 1 pegree _of irreg!llari~y_ _ _r~-~~ts th~ §~_sediments, I?_h~a!_ __ a.p.d __ ch~mica_l___proc~sses _to \ylilChlliey ·are exposed.\ Hence grains sorting/shape and ---------=-------------------------------- . size distribution, compacting due to overburden and cementation at grain contacts affect the porosity.

As the sediments were deposited and the rocks were being formed, some void ypace§. bec~Jn~-i~9lated from the _ other~ by_~~-Y-~ ~~ri1eiltati~n) Thus, many of the void spaces are interconnected -while some of the pore spaces are completely isolated. This leads to two types of porosity, absolute porosity and effective porosity.

Absolute porosity: it is defined as the ratio of the total ------ . ....,

p~re spst_ce illlhe ro~k to the bulk volume regardless of the pore interconn~ft~<?AlA--~rock type- may have--conside~able

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absolute porosity and yet have no conductivity to fluid for lack of pore.

Absolute porosity . ¢ = ~otal void space

a ~ulk

Effective porosity: it is the ratio of interconnected pore space to the bulk volume,

Effective porosity A. = T';nterconnected void 'f'eff

Vz,ulk

· The ~ffective porosity is the value that is ____ ll~ed in all reservoir engineering calculations because it represents the i~~d ~ore space that con!_~ins _ _tbe __ J~_GQY.era~le ·.

-·hydrocarbon flu1 s. \See the table below for the porosity ranges of various rock types.

35-45% Sandstone more consolidated 20-35%

15-35% 5-20% 10-30% 5-40%

One application of the effective porosity is its use in detennining the original hydrocarbon volume in place. Consider a reservoir with an areal extent of A acres and an

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average thickness of h feet. Bulk volume of the reservoir and the reservoir pore volume can be determined from, Total reservoir bulk volume= 43560 Ah (fe) or 7758 Ah (bbl)

Reservoir pore volume PV = 43560 Ah¢ (fe) or 7758 Ah¢ (bbl)

Example: Calculate the initial oil in place for an oil reservoir with the following available data: A= 640 acres, h = I 0 ft, Swc = 0.25, ¢ = 15%, Boi = 1.306 bbl /STB

Solution: Pore volume= 7758 (640) (10) (0.15) = 7,447,680 bbl Initial oil in place=l2,412,800 (1-0.25)/1.306=4,276,998 STB

Representative Elementary Volume When we talk about the flow in porous media we do not assign the property to a point which may lie in a sand grain or void spaces. Pro erties such as p~y~e'!biJity and pressure are assigne - in averaged form to a certain region of porous media (typlcally--abourro: grains), called representative elementary volume (REV). The REV is tak~ to be large enough in comparison with a pore, -~et

s~all in com_]2_~!ison with regional variations in properties of the medium. - ----- - -- ---- -------------------~---

See figures below which show the effect of volume considered on computed porosity.

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Adopted Pn:.n B<:u 11972)

Permeability .

............................... .................................. ............................. ...... , ............... ~ .... . .................................. ................................ ................................... .............................. .......................... •••••••••••••••••••••••• .......................... ······················~· ............................ •••••••••••••••••••••••• .......................... •••••••••••••••••••••••• ............................. ......................... .......................... •••••••••••••••••••••••• ............................. ......................... ......................... .........................

Pel}11~~b!_lity is a measure of the flow cap(:lg_ity __ Qf a PO!QYS medium. To defiiie--tlll:s-;-coiislder the macroscopic flow in a porous medium. Historically, Henry Darcy, 1856 was the first person who foun~ a_n Ci_mPfrica~ for:. this by _exam~~ing vertical filtration through sand b~ It end up with a linear mathematical equation called Darcy's la\v (analogous to other linear laws such as Ohm's law in electrical conduction, Fourier's law in heat conduction and Pick's law in solute diffusion),

J.vtvt'I'J f (0 rciV

Q l11h Q kM -oc--- :::::> -----A J.1 L A J.1 L

jllliifjl!m ' L :

' ' '

_y_~

water


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Darcy also changed the type of sand packs, which affected the value of the proportionality constant. Later the effect of the fluid viscosity was also included. Permeability is a proportionality constant relating th_~ __ t}g_~_r'!!~- -~H!~Lpr~ss:ure differential across the sample. It is a rock ·property (not depend on the fluid).

Remember that the porosity varies between 0-48% . . -.

However, there is a 2-4 orders of magnitude variation in the permeability (see table below). For example, the value of permeability in sandstones depends on s~~d grain,_si~e, s~e, orientatio_n, arrang~~~Dt-..19-mina.tions/clay contentscementation and water composition.

fractured rock masses 1-10000 non consolidated > 1000 sandstones 100-500 carbonates < 10 shales < 1

Using dimensions of parameters Q (L3/T ), A (L2), L(L),

P(M/T2L) and JL (MILT) into Darcy's equation results ip, M

L3 = kL2 T2L ~ k = Li~ T ~ L ,,~~

He~ce pe~~eability. h_~_tl~imensjQ_11 si!filar to the area ( unlt--Df m_In SI system). Also a Darcy un1t can be used; a

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Reservoir Engineering l, Mohsen Masihi ----·--·- ------- -------- - .----------------------------------- . .

---··· ------ ,

/ ·rock mass having permeability of 1 Darcy under 1 atm/cm l 1 pressure gradient would transmit fluid of 1 cp viscosity

with velocity of 1 crn/s. The interchange of these units can be seen by, - ---

( 3) L 10-6~ (10-3 Pa.s)(10-2m) k (Dare ) = QJ.l = s = 0.987 x 10-12 m 2 = (1~Im) 2 . y !J.PA (101325Pa)(10-4m2) __ ____ ---------- --- --- -- -"

To get insight into these units, let compare Darcy law (u = ~ t.P) with the Poiseulle' s equation (u = R2 ap) of the fluid

Jl L BJI az

flow in a pipe which gives, k = R82

m 2 = 0.1266 X 1012 R2 Darcy . In other words, a pipe with a radius of 1 J.lffi has permeability of 126.6 mD! ~ Note that one Darey-Ts-aliigllpermeability

. - --- ---- ---------~ ---- --·-- -- ·- ··-

value~' In practice, the .-permeability of most rocks much lower than this so the term ·millidarcy, 1 mD is used.

Example: Brine of JL = 1 cp is flowing at a··rate of 0.5 cm3 /sec under a 2 atm pressure differential into a core plug of 4 em long and . 3 cm2 cross section. Calculate the absolute permeability.

Solution: we use Darcy units for the calculation. Q = -k (P2 -PI) "' ----- ~~--- ---- 1'

A j.J L /// // I.

05 k 2 1/ [/. i"- = lx 4 => k = 0.333 Darcy ~. =======c=======:::;!

Now let's find the Darcy equation in field units with Q( ::~), k (mD), A (ft2

), J.l (cp), tJ.P (Psi) and L(ft).

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Q(cm3 1 s) = k(D)A(cm2

) ~(atm) J.L(cp) L(cm)

. k(D)(lOOOmD)A(cm 2)( 1 ft

2

2 J !1p(atm)(14.7 psi)

Q(cm3 1 s)( 1 bbl/day ) = lD 929 em 1 atm ) 1.84 cm

3/s J.L(cp) L(cm)( 1ft

30.48 em

Q = 1.127 X 10-3 JcA f1p J.l L

Example: Brine of J.l = 1 cp is flowing at a rate of 0.3 bbl/day under a 30 psi pressure differential into a core plug of 0.1 ft long and 0.0215 ft2 cross section. Calculate the absolute permeability.

Solution: k= QpL =413 md 0.001127 t¥J.A

~~-~tens1on-of-the-·Darcy's law can be obtained for an inclined reservoir) In this case the ~ffiressure difference (P1 - P2) is notthe only driving force in .£!Jilted res~~YQiL~$. the gra~it~tio!!~t force is -al~g·- ~ntfi_b_ilf~_d-_and.s.Q_tll~xectur

--·· ·- ' ···- .. ___:.--

sum of!~ese two are- necessary. Note that the gravitational forcels always 1rec ve H~ally downward while the force that results from an applied pressure drop may be in any direction. In practice, we introduce a new parameter called fluid potential, <D. The fluid potential at any point in the reservoir is defined as, ([). = p.- (L) LlZ·

L L 144 l


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where ~zi is the vertical distance from a point i in the reservoir to the datum level. In terms of the specific gravity of fluids this is <Pi= Pi- oA33 r Llzi. Now we can write the Darcy' equation in the potential form as,

kA cp - <P = -1.127 X 10-3 2 1

q ~ L

Replacing the potentials at the upstream and downstream flow into the above equation gives the Darcy' law in field units for a tilted reservoir,

U = ~ = 0.001127 ~ [ P,P'"•am -: dowm~am 0.4335ysi.,J

u: apparent velocity (bbVd.ft2)

L : distance along flow path ( ft)

y: fluid specific gravity

L

Lsina

---,.-~~----l- Datumlcrel

a : angle of fluid flow .measured counter-clockwise

Exercise#! A fluid is flowing in the following inclined porous sample ofdiameter 183ft, dip angle 8.43° with the given data ( r = 0.693psil ft,k = I45md,J1 =0.32cp,Swc = 0.25,¢ = 20% ).

Calculate the flow rate, apparent velocity and actual velocity.

F' = 3400J'sia

P = 3380Psio

l(!(i'

i '"~--- II ~0 _ ___.J


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a


1-2 Darcy's law and its range of validity Darcy's law has been established based the following assumptions:

a) steady state flow: i.e. no transient flow as is usually the case in reservoirs (it take months to even years). For lab measurement can be assumed.

b) One p_hase~~lO~ _ ~tur~i~}_l_j.e. one fluid flow!QS.J this can be true in the laboratory test by cleaning cores; however, in the reservoir there will be a certain connate water saturation, or gas, oil and mobile water flowing through the same pore space).

c) No reaction- between the fluid and rock: i.e. inert and no ~ ~-- -- f . - --

change to the p9re structure with timeT note that hydraulic fracturing, reaction of COz_ in nresence of water with --------------- - -~--~----------- -·-----~ c~bonate rockin_C02 flooding .may alter this).

d)~~ i.e. the pore structure and the material properties are similar everywhere and not vary (in reality, the layered nature and large areal extent of a reservoir rock will produce variations in the vertical and horizontal permeability).

e)( Laminar flo~: Darcy's law is valid for laminar flow regime i.e. no turbulent flow. In practice, Reynolds number Re = p.u.d I J1, a dimensionless number which Depanment of Chemical & Petroleum Engineering, Sharif University ofTechnol~gy, Tehran, IRAN 17

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compares the strength of inertial and viscous forces is used to characterize the laminar flow regime, where d i~ the mean pore __ di~~-!~E~- ~~_4,_ ~-1~ __ the me_an yel_~c!_!Y· From ~arii_}~;Kozeny _!~~~~~!!_ dis~ussed _later th~£riterion _for

1 · · b · ~-u-1k < 1 o . \ am1nar regnne can _ e wntten as, \ = · ~ :__/ --~~

f) Q~~~~ndition~ assumed over the period of the reservoir development. -

Later we will discuss some remedies to critical violation from Darcy's law assumption such as the case of turbulent -----==---------·-----flow around a well in the reservoir. --

Differential form of Darcy's law The differential form of Dare 's law for one di ensiQnal, horizontal flow in artesian coordinates in field units is:

---- -~--~~----~-~

U=Q=~0.001127!dP ~ .~ ' ' A _ J1 dx ~foS5 wt .... d ~~

The velocity u that is not the actual velocity of the flowing fluid but the apparent velocity determined by dividing the flow rate b~_c,ross-sectional area acros_s which the fluid

, _ __ ·-------------- --.... -

is flowing. However actual velocity was de~d ___ by dJviding of apllli[ent velocity_Qv~rthe=ro~k- -poro~ity. Note that the negative sign shows that the pressure increases in one direction while the length increases in the opposite direction.

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·In radial coordinates the difference ap should be changed to ax aP and also since the radius is measured positive in the ar

direction opposite to the flow (see below figure) so the negative sign in the Darcy equation disappears.

Cen!e< ~llh<>well

Hence the differential form of the Darcy's law for one radial flow in cylindrical coordinateS/in field units is:

, _______ .... ·----- ----------· --~-------- ~ ·- -~ ·---- ------ ·· .

u = Q = 0.001127~ dP A JL dr

Darcy's law in higher dimensions Darcy's law described in one dimension can be extended to two and three dimensions by using projection of velocity vector in each direction as shown below, · .

ux = - ~: aa~, u = - ky 8P, uz =- kz ( 8P + p gJ r A y JL By f.1 az

z f

X

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Due to directional nature of permeability, the fl~w can be modeled (see figure below) considering a permeability tensor with 9 scalar components

----ax-- --c=J high penneability lamina

- low penneability lamina

In commercial simulators the nondiagonal elements are often ignored (for efficient solution) so its form becomes:

Remember that reservoir is called homogenous if· rock· properties are the same at different points otherwise it is called heterogeneous. Also a reservoir is called isotropic if rock properties in the single point of the reservoir are the same in all direction otherwise it is called anisotropic. In terms of permeability property these can be shown as,

Isotropic reservoir . Anisotropic reservoir Homogeneous reservoir

kxi -=I= kyi * kzi

kxi = kxj' kyi = kyj' kzi = kzj

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Heterogeneous reservoir

Where ~_4_i are two different points_ in th~ __ _reservoir. Notethat the environment of deposition, diagenesis (e.g. solution and dissolution) and fracturing are major factors that control the anisotropy.

Deviation from Darcy's law Darcy's law was established based on some assumptions. Deviation from Darcy's law behavior occurs if those assumptions(violated .

...__,~~

Let's consider the first case; in most areas of the reservoir the laminar flow ·regime, Re < ~d;lio~-to

- _ the well bore when yelocity is high (such as gas production -·or flow through fractures) then there may turbulent flow

occur which also termed non- Darcy flow.

Darcy's law is replaced_ by a non linear Forchheimer equation which is al_so an empirical law,

dP f1 2 --= -u + (J.p.u dx k

where fJ is inertial (non-Darcy) flow term which shows an ~

additiona_L12!~~-~~drop due 1Q-E~n-D~rcy effects. p.u~ __ js related to kinetic energy per unit volurrieCii fluid which is considerable at high flow rates.

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. , , .,

, , , ,

. -., ' . .

,'

,~ ~;'

' . "'

Now consider the second case. Darcy's law considers a continuum type behavior with mean velocity (average over a large number of molecules) in a porous medium.

In normal situation fluid molecules collide with the pore wall a!}d due to friction velocity of the fluid at the pore·

, "' wall is zero. Low pressure ases have longer mean free c,;J ' - ··--

/ (AtP( path i.e~ _ ~he -~-~~ JE_~!~cu~e~ ollig~ \\'_i!_h the pore -~~!-~~- ~i~re ~r//~ frequently tlian ~hde ~Ith) ot er mo~ec~l:s. Thts 1mphes / I ·that the gas flow through the pores as Individual molecules

and not as a fluid continuum. This is called slip flow in which the gas velocity at pore wall is not zero which leads to an additional flux and increase in permeability.

LiQuid Gas

V(wall) =0 Finite Velocity at Wall

Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN · 22

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Klinkenberg ( 1941) showed k measured during gas flow · - through a capillary tube can be corrected by,

kooreotOO = 1( I+!) where b is directly proportional to temperature and inversely proportional to molecular diameter and capillary radius. This effect is noticeable if b/P>O.l (see figures be low). ---------------------------

300 1 I I

250 ~ I

- i ~200 1 .~ i ~150 ' m I CU I

E : - ' ~100 j !

501

. intercept = k,o"

0 200

0

400 600 800 1000

11Pressure(1/atm)

Klinkenberg effect is usually negligible for flow in the reservoir (b in the range 0.2-2 psi) while in laboratory condition as we often use low pressure gas (safer/quicker/ cheaper than reservoir condition) thi~e.

1-3 Porosity and permeability relationship Macroscopic flow in porQ:t~_~-l!l~4i~ can be compared with flow in microscopic ca:-pillary tubes. Assume a bundle of capillary tubes representing a rock.


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Actually, tubes are not ·straight 1.:/&f#Mtewe+t;~U¢1)

Note that surface chemistry such as capillary force, surface tension may be dominant in small tubes especially when there is multi-phase flow. Flow through a tube is governed by Poiseulle's law,

du f------ ---------- --- ---..--f-rt/,\ Fy = J-LA-dr

Laminar flow

\-----------------------------\V

L

F p(2;rrrL)du => u= 2MJ1L(R22_r2J. A D = P.. _ P. _ _ Y _ . . . dr ur 2 1- -

nr2 nr2

· dQ=u.dA=- · · 21rrdr ~ Q=--M (R2 -r

2) 1rr

4M

2pL 2 8J1L

For n capillaries with the mean radius · r and tortuosity coefficient t,

Q = n1rr4

M Poiseulle's law 8J1(tL)

This is comparable with Darcy's law with: -4

k = n!l.r 8At

This comparison helps finding k, ¢ relationship


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From definition of Porosity, V -2L -z

¢ = void = nnr t = nnr t

~ulk A·L A Use specific surface M of the solid material, M =As= n ·(2nr)Lt = 2¢

~ (l-¢)·A·L r.(l-¢)

Combination of this with permeability relation gtves Carmen-Kozeny relationship, ______

. ~·~) . "'~ M2 .(1- ¢)2

where C depends on lith~ing, tortuosity. There may exist, for example, a rock with high porosity of fine due to the good sorting but low permeability as M is high.

Note that surface to volume ratio Mrxl/d and so k -~-!!2

• \ ~· ---- -----~ ---- --·· -- ~----- -- - ~ -- - ----~ --~

See figure below for the relation of permeability and porosity in different rock types .

,

Sue rosie Dolomite

E 100~~--~+H~~~~~~

~

:0 liS Cll

E .... : 1o~-r4-~~~+-~~~_,


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1-4 Classification of reservoir flow systems The mathematical relationships that are used to describe the flow behavior in the reservoir will vary depending upon the characteristics of the reservoir. Various flow systems can be considered based on, types of fluid in the reservoir, flow regimes, the reservoir geometry and the existing phases in the reservoir.

Types of fluids in the reservoir The isgtherm 1 corn ressibility coefficient is con!roJling factor in identifying the type of the reservoir fluid. In general, reservoir fluids are classified into incompressible fluids, slightly compressible fluids and compressible fluids. The isothermal compressibility coefficient C is described mathematically/

In terms of fluid volume1;c=-~(~;1

In terms of fluid density,~~(~; j) Where V and P are the vOlume and density of the fluid, respectively ..

a) Incompressible fluids: An incompressible fluid is defined as the fluid, whose vq_lume (or dens it does not -

~- ----...__ .. ___ ___ dV = dp =O

c~ge ___ ~th pres~~ i.e., dP dP In reality, incompressible fluids do not exist; this behavior, however, may be assum~d in some cases to simplify the derivation


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and the final form of many flow equations. We may assume this for oil-water displacement in reservoirs.

b) Slightly compressible fluids: These exhibit small c~ volume or density, with__9lilln~~~· Most reservoir crude o1ls and water systems are of tli1s type. Knowing the volume Vi of a slightly compressible liquid at a reference (usually initial) pressure pi, the changes in the volumetric behavior of this fluid as a function of pressure P can be mathematically described. Consider reservoir pressure drops from ~ to P ,

P v dV - Jc·dP ~- J-

Pi V V I

~ V = V; · ec·(P;-P) ~ V; ( 1 + c · ( ~ - P)) Vl ~r~: ( Ytfer~V\.c~} · ·

c) Compressible fluids: These ~xperience large changes in ~e _as a function of pressure. fAll gasesare .consig_er~~ con1 ressibJe fluid~~In these cases by using real gas law, v =znRTi f.J, the isothermal con1pressibility of a compressible·

.. . --- · ·----------·· ~ fluid is described by this expression, .

c,=-~(~~l=~->: with a simple cg = 11 p for ideal gas. Typical values for the compressibility of various fluids are:

cw ~ cf ~ 10-6 psia

Co~ lOx 10-6 psia

cg ~ 500xl0-6 psia

Department of Chemical & Petroleum Engineering, SharifUniversity ofTeclmology, Tehran, IRAN 27

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Figures below show schematically volume and ~ensity

changes against pressure for the above types of fluids.

Slightly tumpressiblc

Fluid flow regime

~· t ""

Hilghly Cornpreuible

SGgl•tly compressible

Ltcomprusible

Prcssw·c

There are basically three t~pes of flow regime~ __ !h~~'" .. must . be recognized in order __ !Q __ 9e~~ril?~_!be~_f!.!:lid _fl9w be~-avior ·.· and reservoir p~essu~~_gis~r~ funct!Qn_ of tirpe. These are steady-state flow, Pseudosteady-state flow and unsteady-state flow.

a) Steady state flow: this is the case if the pressure at every location in the reservoir remains constant, i.e., does not change with time. Mathematically, this condition IS

expressed as, dP . -=0 dt

This states that the rate of change of pressure P with respect to time t at any location is zero. In reservoirs, the steady-state flow condition can only occur when the reservoir is completely recharged and supported by a


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strong aquifer or press~t~11f!D.:_g~ qp~r.Cltions with withdrawal rate matches injection rate.

b) Pseudosteady state flow: when the pressure at different locations in the reservoir is declining linearly as a function of time, __ i.e._, at a constant declining rate, the flowing condition is characterized as the pseudosteady-state flow. Mathematically, this states that the rate _ of change of pressure with r~~t to tilne~at~~~iti~_n}~ ~onstant, dP I dt = cte ~- ----

It should be pointed out that the pseudosteady-state t19_w_is con1monly referred to as setni steady-state flow and quasi steady-state flo~. --An -exanipfe -is -a well -pr~duclng at -a

suffi~ient period •of t~me at s_ort~t~trat~e middl~jpf~L~ the hnear reservOir w1th the~~Il~~ryy l>"' I"~+ t

c) Unsteady state flow: this is also called transient flow and is defined as the fluid flowing condition at which the rate of change of pre~sure with respect to time at any position in the reservoir is not zero or constant. This definition suggests that the pressure derivative with respect to ti1ne is essentially a function of both position and time, thus, dP I dt = f(x,t) ..

An exa1nple is a short period after a well producing at fixed rate in the middle0f linearreservoir before the n0=

- -~ --------·--·-·

flow boundarx. is felt- re~ervoir appears infinite in extent. -·- ·-·- - ----- ---·


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Figure below compares. the pressure declines as a function of time of these three flow regitnes.

Time

Reservoir geometry

Steady-State Flow

Sernisteady-State Flow

Unsteady-State Flow

The shape of a reservoir has a significant effect on the flow behavior. Most reservoirs have irregular boundaries and a rigorous tnathematical description of the geometry is often possible only with the use of numerical simulators· . .For many engineering purposes, however, the actual flow geometry may be represented by radial flow, linear and spherical or hemispherical geometries.

a) Linear fluid flow: this occurs when flow paths are parallel and the fluid flows in a single direction. In addition, the cross sectional area to flow must be constant. Figure below shows an idealized linear flow systen1.

~ ~ -------------------~

··-. ·--·-- . --~--.-----:-····· ------·

-- - - -----· .....

·· ·---~ - --·- ....... ············· --· ......... ,. ........


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A co1n1non application of linear flow equations is the fluid flow into a vertical hydraulic fracture as shown in figures below.

Isomeoic View

Han View

Fracture

! i · -----

--'+ i • --~---~- t- j

c::> wellbore

___,. l .-----+I

I -4···----"'-

'-----+ 1

b) Radial fluid flow: in the absence of severe reservoir het~~_Qge~~ities, flow into or away from a wellbore will follow radTaf flow lines from a substantial distance from the well bore. Becaus-e -fluids-move toward the welffrom all directions and coverage at the well bore, the term ·radial flow is given to characterize the flow of fluid into the

· wellbore.

Figures below show idealized flow lines and iso-potential lines for a radial flow system

weBbore ·

-~------------ . . ... . --:_~J ~---. ---. ····-----~~--··-···-Side View • ·"i P w! tc FlowLines

------·- ~>-: !oc----------·-·-·-----10-i ~---------·--·----------.-; 4-----


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c) Spherical and hemispherical flow: depending upon the type of well bore completion or a very thick formation may · be better described by using the spherical geometry; it is possible to have a spherical or hemispherical flow near the · wellbore (see figure below).

""Ubore

Sid.eView

A well that only partially penetrates the pay zone (figure below) could result in hemispherical flow. The condition could arise where coning of bottom water is important.

WeHhore

Side View FlowLines

Existing phases in the reservoir The mathematical formulations of the flow vary in forms and cotnplexity depending upon the number of mobile fluids in the reservoir. There are three cases of flowing systems, single-phase flow (oil, water, or gas), two-phase flow (oil-water, oil-gas, or gas·-water) and three-phase flow (oil, water and gas). The description of the fluid flow and the subsequent analysis of pressure data becomes more difficult as the number of mobile fluids increases.

DepartJ11ent of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN 32

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A relevant parameter related to the problem of multi phase flow is fluid saturation. Saturation is defined as that fraction, or percent, of the pore volume occupied by a particular fluid (e.g. the oil, gas, or water). This property is expressed mathematically by the following relationship,

Total volume of fluid fluid saturation = p l

ore vo ume

Applying above relationship concept of saturation to each reservoir fluid gives, S = Volume of oil S =Volume of gas S = Volume of water

0 Pore volume ' 9 Pore volume ' w Pore volume

~tion the sum of saturation is l.Oth;;ref~ ------ -------- ---- _ 50 + Sw +59 = 1 _

Note that we mostly·us · g e phase in this c9urse and analyzing multi phase flow will be discussed in "numerical simulation" course.

1-5 Reservoir boundary conditions The flow equation to describe the flow behavior in the reservoir, as can be seen later, is a POE of a second order in position but first order in the time therefore it ne~_ds _ _!_~9 6oundary conditions and one initial conditiop_ tQ __ ~omr-J~te the formulation. The two boundary conditions are Inner boundary condition and outer boundary condition.

a) Inner boundary condition: two different physical conditions may exist: i) constant terminal rate (CTR) such as the case of a constant wellbore production,


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kAaPI qS/ = ---

jiB ar T=rw

where, A =2nrh in the case of radial systems and qs1 is the flow rate at the sandface level.

ii) constant terminal pressure (CTP) as applied to the wellbore. P = Pwf atr =rw

where, Pwr is a constant bottom hole pressure or a fixed pressure at initial water oil contact aquifer flow rate varies.

b) Outer boundary condition: the specification of the pressure. behavior at the external boundary depends on the type of reservoir considered. i) Volumetric reservoir: a closed system characterized by no fluid flow across the outer boundary aPI = o.

or r . e

ii) ~eservoir associated wit~_ 9-_g_~~ c~_p_ or_ bq~t.Q!P_ ~§tter is featured by constant pressure at the external boundaries, p =pi at ··-;--···Te. In this ···case .. the ·-pressure distribution-does

not change with time i.e. steady state condition is reached.

iii) Infinitely_J~rge re.s_ervoir: This is a mathematical -----------

concept which is very useful during the initial condition stages of well production when the pressure disturbances do not travel far enough to reach the boundary limr-?oo P = Pi.


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In p~otice, the reservoir boundary may __ he_g~om~t~_ic(;llly vgy complex (see_ figure below). There may exist various faults such as €~ faults.

Complex boundaries (Cross Section)

(Upper and lower boundaries)

~---CONSTANT PRESSURE

MOBILITY CHANGE

(lrom no-flow to const~t pressure)

NO FLOW

However, some simple boundary conditions with which we can find the· analytical solutions can be used in this course (figures below).

A well close to a fault

~" ... ,.,

~ d -~-

• • • "' • ~ -~ - w • ·-""

well ~ ·. " "" :~

I

A well inside a corrzer '\)

'.\'tll ·-;·_---

Department of Chemical & Petrolew1i Engineering, Sharif University ofTeclmology, Tehran, IRAN 35

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A well in a channel

A well in a rectangular reservoir

A well in an open rectangle

I

.a~! I t

d, I

--- --- -. we11 a~:

'

Two wells close to a fault

Note that, in reality, for a more complex configuration we have to use numerical simulation. Generally, actual flow pattern in the reservoir is very complex due to:

a) shape of oil bearing formations and aquifers are quite irregular.

b) heterogeneity ( ¢, k, Swc) in oil-bearing and water- bearing formations.

c) deviated well bore result in an irregular well pattern through the pay zone.


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d) different drainage area due to different production rate in wells.

e) partially penetrated the pay zone due to not fully perforated wells.

Units There are different set of units including SI units and field units. The following conversion factors is used to convert different units.

Parameter SI unit Field unit Dare~ unit Length, L · · m 3.28 ft ?em Area, A mz 1o.16 fe ?cm2

Volume, V m3 6.293 barrel ? cm3

Time, t sec hr ? sec Pressure, p Pa

. 6 1.45x10-4 psi 9.87xl o- atm Flow rate, Q m3/s 7.41x10-5 bbl/day ? cm3/s Mass, m kg 2.2lb ? gr Density, p kg/m3 6.24 x10-2 lbm/ft3 ? gr/cm3

Permeability, k m2 1.01 x10 12 Darcy 1.01 x109 mD Viscosity, 1.1 Pa.s 1000 cp 1000 cp Compressibility, c 1/Pa 6896 1/psi 1 Acres.ft=43560 ft3

1 bbl=5.615 ft3


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2-Steady state single P.hase flow This is the simplest flow regime in which the pressure throughout the reservoir does not change with time. Note that steady state condition holds after a sufficient time to let the external sources come to action. Before that it is either unsteady state or pseudo or semi steady state.

This section uses the steady-state flow to describe the flow behavior of several fluid types in the reservoir. These include steady state flow in linear reservoir geometry (for incompressible, slightly compressible and compressible fluids), in radial reservoir geometry (for incompressible, slightly compressible and compressible fluids) and also in spherical reservoir geometry (for incompressible, slightly compressible and compressible fluids).

2-1 Steady state flow in linear reservoir geometry a) Incompressible fluid: In the linear system, it is assutned the flow occurs through constant cross-sectional area A, where both ends are entirely open to flow. It is also assumed that no flow crosses the sides, top, or bottom as shown in figure below.

~(If: [71 '"! · - ··-- L - - ...


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If an incompressible fluid is flowing across the element dx, then the fluid velocity and the flow rate are constants at all points also viscosity is pressure independent. The flow behavior in this system can be given by Darcy's equation. Then separating the variables and integrating over the length ofthe.linear system gives,

u = (j_ = -k dp => (j_ r dx = -k r dp A J.1 dx A fL Pt

q = kA (p1

- p2

) . or q = 0.001127 kA(p1 - p2 ) (STB/D) pL BpL

A : fe, L : ft, K : md, p : psi, p : cp

The variation of pressure against x is found by, qpB

p =PI- 0.001127kA X

Note that in the case of inclined reservoir the extended ......__ .··-·-~-arcy' s law using_1he fluid potent~aL§_~9-~!4._t?_~-~~~~· )

b) Slightly compressible fluid: . Previously we found that for slightly compressible fluids there is linear dependency of volume (or flow rate) to pressure p. Substituting this equation in Darcy's equation, separating the variables and integrating gives,

{v = v, [ 1 + c(p, - p) J q = q, [1 + c(p, - p) J


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kA ln(l+c(pr-P2)J or qr=0.001127 kA ln(l+c(p,-pz)J . qr =peL l+c(p,- p

1) BpcL l+c(p,- P

1)

~ \ _,_ ' \f'l4C- I) - - p -- \) I ~rs+r·cn..-.- I I - Y'"')-

~electing th~ upstream pressure p1 and q1 as reference pressure and flowrate respectively and substituting gives,

~ q1 = [0.00~~~7kA] !n(l + C(P1 - Pz)]

Then usingJaylor-serieS)approximation, In(l + x) ::: x gives, = [0.001127kA] (P1 - P2 )

q ~B L

This is exactly in the form of the related equation derived for incompressible fluids.

Example: For a linear reservoir, compare the flow rate under steady state regime assuming a) incompressible b) slightly compressible fluid and conclude. A= 45 ft 2

, Bo = 1.127 bbl/STB, L _ 450ft, J-1 = 2.5 cp,

c = 65 x 10-6 psr1, M=IOO psi, k = 250 mD, p1 =ref

Solution: Using available data for incompressible fluid then q = 1.123 bbl/d; · for slightly compressible fluid we get q = 1.131 bbl/d. This shows that the volume is not a strong function of p for slightly compressible fluid.

Department ofC~emical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN 40

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Exercise#2 assume a vertical well .with bottom hole pressure of 27 50 Psia in a horizontal layer under steady state flow regime as shown below. The available data are: r = 0.693 psi I ft' k = 145md, J1 = 0.32cp, swc = 0.25,¢ = 20%, c = 5 X 1 o-4

psC1

' Bo = 1.356

Determine the flowrate and pressure distribution in the reservoir considering the reference pressure as a) external pressure, b) well bore pressure and c) 1000 Psia.

c) Compressible fluid: For the flow of gas in a homogeneous-linear system, the real-gas equation-of-state can be applied . to rela!_~ __ !h~ --- ----- -- -~---------------~----- ·-

pressure, temperature, and volume at__ av_y state to the -------~--~----------- - --- - · ·--·------

CQrresponding value~ at standard conditions (as flow rate has a unit of bbl in field un{i -·system~ .so5 .615 appears in

Day . \,V

the formula), 6 f(jt~ s rn n v t r / d

V -- z p sc sc ' z p sc q sc -"") or q =-----'~-

PT:c P 5.6l5~c

Where q is gas flow rate at pressure ( bbl ), qsc: gas flow Day

rate at standard conditions c~:~), z: gas compressibility

factor, Tsc: standard temperature, R and Psc: standard


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pressure, Psia. Replacing the gas flow rate in the Darcy's law, and separating the variables gives,

q == (Psc) (zT) ( Qsc ) == -0.001127 kA dP Tsc P 5.615 · J1 dx

zT Pscqsc = - 0.001127! dp => qscTPsc f dx =- fP2 _f!_dp p 5.615AT:c J1 dx 0.00633T:cAk P1 ZJl

The gas viscosity and compressibility factor are pressure dependent (as .can be seen in the following figures); hence, appropriate techniques for evaluating the above integral is necessary.

Gas co~npressibility f<lctol v~IS41S pleSStue

1 ~ l l ; l ~ ~ ~ 1.1 .• •. ----··j -- .... ----- i···. ---··-··( .... ----- --~ - - -- - ~ - - - - ·l··-·---·-··i···--; _____ _

1.05 •••..•.... ; •..•. •.• . ..; .••.•. .. : .. ~----- ---- -- ! ........ .) ........... ] ........ . ; : : : - : : : ] 1 -·- ---- ---~-----------L .. _______ _ j __ ·-- - ---- - -~----·--····i---- ---- - - - : __________ _ ~- 1 1· 1 i io-t.m i

1 .95 ....... +······· --·]--.. .... .. t----··--·+·· --· ··-+-- -----+-·-··--·-= : : : : I :

; 0.9 -------- ~ ---- --- --- -1·-----------t'· ---H··--r------- .. l·------- -··t·--·------t! : : : : : :

0.85 · ····---- · :-·· - ·· ····j··--·-----·-i·-----·--·-·: ··---·-- -~-~--· · ·- - ·---:---- · - -----

0.8 ·---·--··+·····-----:--- ·: ··------··+----····-+-·----·-:+---·--·--·

0.015 -····--·-·:--·-·······j···· .. ·····j······--- -- -:··-----·-··:···········j······ ... o.u ·-·--···-·;·-···-- --··j··---·-····t·····-----·t·····-·····j····-· ···j···--····--

~:JJJ.II l ; : ~ ; j

o.oz ··----·-··:-----·--···;-·- ---- -· -:····-··-----r-·---···--·r--·· ·-··--·:··--·--···· O.Oli ·--- · ··--·;- . . -.••. i··-- ... ····t ...... ···-. r.- ... --· ... ~ .......... ·t·· ....... -.

~ ~ ; l - l l : : : : : 0'75o·'------::,ooo::----::-'2ooo::---:-:':-JOOo:--_~-,-:-:L:4ooo,--_ ,--=soo:-:':-o---:-6oo:L-o ---:lrooo

~f~O.

1000 2000 lOOO ~000 5000 6000 7000 Pressure (Psia)

We can use either the averaging technique (p2 method) or real pseudo pressure variable to deal with this.

6'_ tbJ..e.t<do r;{fRjjv.,y(. --\'itrtft\bJ(e / ";f; ./ _:.. . CD r .. ----------- --- -~ Jr J? tfdtJ>if/ ~-J? metho.<J;') 1 ''------.. . . --~ For---a low pressure gas ( < 2000psi) we have ideal gas

behavior, also at high pressure the variation of JlZ is not too much so we expect to see a linear relation as shown in figure below. Department of Chemical & Petroleum Engineering, Sharif University of Technology, Tehran, IRAN 42

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p

Exercise#3 consider typical values for gas molecular weight and temperature then by using appropriate correlations plot gas compressibility, gas viscosity andpz versus pressure.

The above integral may then be approximated by, P2 1 2 2

f_p_dp = _ P2 - P1 z II z II - J p2 + p2 2

PI r r p = 2 I

-----~-2- --------------/ 0.00316T,cAk(p~- p~) q sc = r;::;- -L

Psc 1 z J1 ''----------------

This means that in the case of gas flow there is a similar solution to _that pre_y_~ously derived for liquid cases except_ we use p2 instead _ ~fp.

Example: A natural gas with a specific gravity of 0.72 is flowing in linear porous media. The available data are: T = 140° f,p1 = 2100psi,p2 = 1894.73psi,A = 4200 fe,L = 2500ft,k = 60 mD

Calculate the gas flow rate in scf/day.

Department of Chemical & Petroleum Engmeering, Sharif University of Technology, Tehran, IRAN 43

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Solution:

P== p2 + p2

1 2

2

2100 2 + 1894.732

------ == 2000 Psia 2

Using the following equation, gas pseudo-critical properties are, Tpc . 168 + 325y9 - 12.5yJ == 168 + 325(0.72)- 12.5(0.72)2

== 395.5 °R Ppc == 677 + 15.0y9 - 37.5yJ == 677 + 15(0.72)- 37.5(0.72) 2

== 668.4 Psia The pseudo-reduced pressure and temperature are,

2000 Ppr ==

668.4

= 2. 99

600 Tpr == 395.-5 == 1.52

. Using Standing-Katz chart for gas compressibility factor gives, z=0.78. Then using Lee-Gonzales-Eakin 6orrelation for gas viscosity gives,

J1 = 0.0173 Cp 0.003164(520)(4500)(60)(2100 2

- 1894.73 2)

qsc == (14.7)(600)(2500)(0.78)(0.173) Scf A1A1Scf

= 1224242 -D = 1.22 ay Day

LPseu~~ssure meth~-~~ . /For intermediate pressure the linear assumption used

/

/ previously may not be accurate enough. In this region, we can define pseudo pressure variable as (see figure below),


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p

lP(P) = 2 J !_dP f.lZ

0

Plz

The previous integral equation can be separated in two parts as,

[ qscPscT ] JPz P 1 ( JP

1

P JP2

P ) 0.006328kT5 cA L = - P, ZJl.g dP. · · 2 2

0 ZJl.g dP - 2

0 ZJl.g dP

Inserting the new pseudo pressure variable results in, 0.003164kTscA ( ljJ(P1) - $(Pz))

qsc = PscT L .

This is similar to the solution observed for liquid flow provided except we use pseudo pressure.

Exercise#4 for the previous example, evaluate the gas flow rate using pseudo pressure technique in your calculations and compare the results.

Example: To account for non-laminar flow near the wellbore, use Forchheimer's equation to find 11p by assuming steady state, incompressible, single phase flow into a fully penetrating well. Department of Chemical & Petroleum Engineering, Sharif University of Technology, Tehran, IRAN 45

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Solution:

u = q & __ dP_ = _JL u + f3.p.u2 21lrh dr k

dP _ JL ( q ) f3 ( q )2

_ JLq (_!_) + f3.pq2 (-1 )

- dr - k 21lrh + .p. 21lrh - k21lh r 47r2h2 • r 2

P r ( A B ) JLq ln r fJ.pq2 ( 1 1 J -J dP = J ---;+--;> dr => Pw- p 2:rkh rw + 4:r2h2 rw -;

Pw r,.

2-2 Steady state flow in redial reservoir geometry a) Incompressible fluid: In a radial flow system, all fluids move toward the producing well from all directions so the .. pressure in the formation at the wellbore (known as the bottom-hole flowing pressure BHP., Pwt) must be less than the pressure in the formation at some distance from the well.

The formation is considered to have a uniform thickness and a constant permeability. As fluid is incompressible; the flow rate and the fluid viscosity must be constant at all radii. Due to the steady-state flowing condition, the pressure profile around the wellbore is maintained constant with time. Let Pwr represent the maintained bottom-hole flowing preSSUre at the Wellbore radiUS rw and Pe donate the external pressure at the external or drainage radius. Darcy's equation for the flow rate at any radius is,

u = !L = !5:_ dp where: A = 27rrh A fL dr

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f I I I i ! - t - H t-t f J I v I

Taking integral form both sides simplifying it gives,

of above equations and

re q dr - _-k r dp w 21rrh f1 Pw

27rkh(pe- Pw) q = ---'----

,u In ( re I r w )

or q = 0.00708 kh Pe - Pw (STB/D) B ,U In ( re I rw) ·

The external (drainage) radius re can be determined from the well spacing (A: in acres) as,

Te = J435:0A In ·practice, neither the external radius nor the wellbore radius is generally known with the precision. Fortunately, they enter the equation as a logarithm, so that the error in the equation is not that much.

Finally, the previous equation of flow rate can be arranged to find the pressure distribution,

p - p + [ q B 11 ] In (r I r ) - ·. wf 0.00708kh w


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Example: For an incompressible fluid flowing under steady state conditions and with Pw = 100 atm and Pe = 500 atm, rw = 0.25 ft, re = 1500 ft, k = 70.6 mD,

bbl f1 = lcp, h = 150 ft and Bo=1.306- calculate the flow

STB rate in the reservoir and pressure distribution (P vs. r) and the velocity distribution (V vs. r) in the reservoir.

Solution: _ _

3 (70.6)(150) (500- 100) x 14.7 _ STB

q - 7.08 X 10 06 1500

- 38803.27 -D 1.3 1 ( ) ay

n 0.25 Pressure distribution in the reservoir,

. [(38803.27)(1)(1.306)] ( ( r )) P(Psia) - 1470 + 0.00708(70.6)(150) ln 0.25 -. _

P (atm) = 100 + 45.97ln (-r-) . . 0.25

The f]QW __ r~~e for the jp.comQress~J?le fluid ___ _ is ___ QQnstant through reservoir--so -the velocity is, -----

q q 38803.27 5.615 2.05 X 10-3 ft v--- - x -----

- AB -:-_ 2rrrhB - 2rr(150)(1.306)r 86400 - r s

Pt*"-111& ve1sus I•Hiius iu 101di4l !JeOineuy

m~---:----~~~~ : :

~50 ----------·--·--····- ·-- - ; ------··· ···--·-··-····· ·j······ ·····················

toO ·---· · · ·····-········- -+ -·· ··-·-·-- -············-··1·······-···········-·'· ···· / : : rr: ·_::::::·::··:·::-·:·::.::r::·-_::_:::::::::::::::::::1::::.:::::::::::::.::.:::::

\~---·· ···· -·-- ······· ···· ··-······ · · - -- · -- - ·- ··· ··· ·, ····1· · · ·· ·· ··· ·····'··-· -- ····-200 • - ·-- .•• ·-- .• -- --- .- •• •:. ·i· ... -- . -- .. -... -... ------ -~ -! ~ .- .. --. -.. :. _-. ~ ........ ~ - ~ -: .. ---ISO ·-······ ·- ·· --··- -----···-j-· ·--·-·-·-····-·· ·-:-.. ---·1·--··-···-····-·······-···-

100 0 --'i,.-.uoc;____..o_~w.,;h.;;;~

..-'

1100

X 10-J Veftlclfy \'81SII5 lo'tliUS

g~~~~~~~~~ ! l l : : : : :

8 -----~- --····-i- -- -----:--------i--...... : ..... - ~ -~-- ..... ··--·-----:- ... ···-i· ..... . : : : : : : : : :

1 --- --~- •••• -.- i--------l- .... ---~--·- - ---I----- - - - ~-- ----·-~---·----~ - ·- - ---~--- -- --: : : : : ; : : :

, - -- .. -i---- · ... ; ·····---i--------~------- -I- --- ··· - ~---- - - -- ~ --- -· ·- -1- -----·- i ...... .

a~ I,I.IJIFI z ··---·: .... ...j ...... ), .... ) .... .... ~ -- -··-~j ....... . ; ....... ~- --· --·1·-····· f ••••. J. ..... ! ...... L. .. ..i. ....... l ...... J ..... ) ....... l ...... 1 ...... .

~ l : ; : : : : : 9 10

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This shows that 25.56 percent of pressure drop occurs only in 2.25 ft away from the well center which show that the most of the pressure drog occurs near the _ well bore: Also, the nearer tg __ the wellb9_re the highcr_the velocity.

b) Slightly compressible fluid: The flow equation for radial flow of slightly compressible fluids at steady state conditions using linear pressure dependency of the flow rate can be derived as follows.

c) Compressible fluid: The flow equation for radial flow of highly compressible (gases) fluids at steady state conditions using the p2

formulation can be derived as follows.

f.1 dr

or


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2-3 Steady state flow in spherical reservoir geometry Depending upon the type of wellbore completion configuration, we may have a spherical or hemispherical flow near the wellbore. The method used to develop the flow rate equation is the same as that of the radial flow except for cross-sectional area of the flow which is A= 4;rr

2 •

Exercise#5 derive the equation for steady state flow rate of an incompressible, slightly compressible and compressible fluid in the spherical coordinate system.

2-4 Average permeability in heterogeneous reservoir It is rare to have a homogeneous reservoir in reality. Often, reservoir contains distinct. layers, blocks, or . concentric rings . of varying porosity and permeability so we need appropriate averaging methods. As porosity is scalar so a simple average method e.g arithmetic average can be used.

However, as permeability is a tensor so we need an appropriate average of permeability for the reservoir depending on how the permeability values were distributed as the rock was deposited. This section describes how we can use Darcy's equation in such cases in spite of assumption of homogeneity of rock in Darcy's equation. Consider the following simple cases:

a) ow perpendicu~o/b;ddi-;;~ ·) -- ---····-----... -------------------- . --r


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Suppose there ex-ist tlrree horizontal layers in series In which an incompressible fluid flows through them,

As considered fluid is incompressible, so flowrate which passes through the layers are equal, Q1 = Q2 = Q3 but total pressure drop is f).Pt = flP1 + f).P2 + f).P3 • We can replace the ·three layers with a single layer with the same fluid condition shown above. Writing Darcy's equation for the layers and simplifying gives,

or K =~L-~~~ ave L...J 1 L...J K

I I i

K = L,+L2+L3 ave .. L. L2 L3

-+-+-Kl K2 K3

Hence for constant thickness beds, the harmonic average is obtained.

Example: Determine the average permeability of three beds placed in series with equal width and thickness the following length (permeability) are 6 (1 0), 18 (50), 36 (1000 mD).

Solution: - L1 + L2 + L3 6 + 18 + 36 k- - = 60.24mD - L1 L2 L3 - 6 18 36

k1 + k 2 + k 3 10 +SO+ 1000

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Exercise#6 show that the average permeability for flow through radial beds in series is given by,

Center oflh~Woll

Example: Determine that the average permeability for flow through spherical beds in series.

Solution: Similar approach is used as before.

1 1 - - -k _ rw re

- 1 1

. b J:!low paratkifobe~ · S ifnilar approach can be used here except pressure drops are the same but flow rate are different.

Q~

Department of Chemical & Petroleum Engineering, Sharif University ofTeclmology, Tehran, IRAN 52

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n n

~K;A; ~K;h; Q=IQi => KH= I = n n

i=l IA; I hi

Thickness weighted average of layer permeability g1ves permeability in direction of flow.

Example: Determine that the average permeability for flow through radial beds in parallel.

Solution: ·. Similar approach is used as before . . k = L~1 kihi

2:f=1 hi

. Geometric mean V$ .

Pressur<! solvcr

k 2 , hl k,.h,

k p "~

In this case both geometric average or pressure solver (fine grain) is applicable


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n

I( hi lnki} or lnK = --=1---

Example: A reservoir has five distinct formation segments with the same thickness that are connected in series. The length (permeability) are: 150 (80), 200 (50), 300 (30), 500 (20), 200 ft (1 0 md). Use these to calculate the reservoir average permeability by assuming (i) linear flow and (ii) radial flow system.

Solution: with linear flow Kav==22.18 md with radial flow Kav==52. 72 md

2-5 Pressure draw down hi a well Let's consider the problem of steady state flow into a well producing at fixed rate q. The pressure at outer boundary re is assumed asp Pi· This may be realistic ·where pressure maintenance exists such as water injection/aquifer feed. Again using Darcy's law in appropriate form gives,

Q= 27rrhk dp

f.1 dr =>

Jdr = J 2Jrkh dp r JLQ r,.. Pw

r+dr r I I

Q


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Reservoir Engineering I, Mohsen Masihi ~

P - p = pBQ In (r I r ) w 0.00708kh w

d ~ _) _9 (- j_. Lsr-Dupuit-Thiem equation( 1865)

This shows that most of the pressure drawdown (p-pz) occurs close to the well (logarithmic behavior)

Example: To increase the well productivity, a well is stimulated with the steam injection for some times. This

h . . . .t::: T. T causes t e reservoir temperature to Increase 1rorn ro to s

and oil viscosity to change from f-Loo to fls for rw < r < rs '

Assume steady state radial flow condition holds and the

productivity index to be defined asPI=qi(Pe-Pw) where q

is the well flow rate and Fe and Pw are the pressure at the external boundary and the wellbore respectively.

Plstimulated well . Determine the ratio PI t. 1 t d 11 use the following data uns 1mu a e we

T00 = 110° f, T5 = 500° f, Jloo = 600 cp, Jls = 3 cp, rw = 0.2 ft, r5 = 70 ft, re = 400 ft .


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Solution:

P,- Pw = :;;:h In(r, I rw) for rw < r < r,

p,- p, = ;;:h In(r.f r,) for r, < r < r,

Pe - Pw = Jlooq ( lls In~+ ln re J Stimulated well 2trkh Jloo rw rs

Pe- Pw = qJloo ln(~J Unstimulated well 2trkh ~v

PJStimulated = --=-q- - ltrkh l well Pe- Pw fLoo lls ln rs + ln re

floo rw rs q . 2trkh 1

p J Unstimulated = - ----well Pe- Pw Jloo In re

ln ~ ln 400 P/Stimulatedwell . rw 0 2 4 29 - __ __:..:.__- 3 70 . 400 = .

P/Unstimulatedwell lls ln rs +ln re -ln-+ln- . Jloo rw rs 600 0.2 70

Exercise#7 Consider the following two reservoirs: (left) a single layer of thickness and permeability of 25 ft and 7.5 md and (right) two distinct horizontal layers. Assuming steady state flow regime compares the oil production rate for 2000 psi drawdown pressure in these two reservoirs. Available data are:

¢ = 20%, B0

= 1.2 bbl/stb, fL = 5cp,rw = 10 in,re =660ft

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Reservoir Engineering l, Mohsen Masihi

7.5md

1,\ I

15ft: I I

'!I 7.\

10ft: ~

2-6 Average reservoir pressure

q2

10 md

5 md

Outer boundary pressure is not economic to be measured by drilling another well so average reservoir pressure is often used, which can be estimated from well tes~~ This is ~obtained using Dupuit-Thiem equatiol) (figure below).

h

element or vOlume. dV,. at radius, rand at pressure. P


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Hence the average

Example: A well produces oil at a fixed rate of 13 m3/d.

from a reservoir with P = 13.2 bar at bottorri hole flowing pressure Pw=125 bar. Use these data to calculate the permeability.

. 3 ¢=23%, Bo =1.36, h=23m, p=l4x10- Pa.s, rw =0.15m, re =210m

Solution:

J5- P = JLQ [In~-_!_] w 2Jrkh rw 2

=>

2-7 Altered permeability zone In reality permeability around the wellbore may be higher or lower than the rest of reservoir due to:

a) formation damage during drilling and completion. b) Infiltration of drilling mud into the well bore, change of

wettability, pore blockage. c) . during production (sand or precipitation from the

hydrocarbon fluid or from formation brine) alter wettability or plug pores.

d) well bore intersect fractures enhance permeability. e) hydraulic fracturing/acidizing during \Vork over


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Skin effect is a term that refers to the permeability alteration around the wellbore. This effect is referred to as a wellbore damage and the region of altered permeability is called the skin zone. This zone can extend from a few -----inches to several feet from the w__cljb_ur_e. Thus, the

- ·- --------permeability near the wellbore is always different from the permeability away from the well &!here the formation has not been affected by drilling or stimylation~

Damaged zone

Undamaged Zone

K

Prenwe-Proflle

~fo.v ,_____ ___ 1$.:c~ r.

The total skin is considere~. to be m;de1 ue ~f t~~ . fo)!c;_~ing

components: ~· -....___/1:r;J!t/cfv;J// 1(/"~ -+:.

Jgamage s@ this is caused by mechanical reduction of the formation in the vicinity of a wellbore. The mud filtrate invades the formation adjacent to the wellbore carrying fine materials responsible for plugging the pores. Also, some clays may swell upon contacting the filtrate.

~rtial penetratin~ skin); many oil and gas wells are completedover a fraction of the total productive interval. Perforation over a portion of the producing zone restricts


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the flow entry and causes pressure reduction at the producing well sandface.

Mobilit skin: Lower pressure within the vicinity of an oil well producing from a satura~e servoir cause~on gas libration in that region. -~si()n around ____ the wellbore thus is dependent on distance from the ;ciL

Consider an altered zone of uniform permeability ks with

an outer radius of rs . b.fls is the difference in the bottom hole flowing pressure at the well when skin is present and not present then from Dupuit- Thiem equation (1865):

( ) pq 1n rs p-pw noskin = -. 2trkh rw

P - ( · h k" ) Jlq In rs - Pw wzt s m = -. 2trksh rw

11pskin = Pw( with skin)- Pw(no skin)

11p . = pq In!£- Jlq In!£= flq (!_-lJln!£ skm 2trksh rw 2trkh . rw 21lkh ks rw

S = J-Lq (~-lJln~x 21rkh 21fkh ks rw J-Lq

S = ( kk -IJ In !L in dimensionless form \ s rw


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Note that the positive skin indicates that the permeability around wellbore has been reduced. The skin effect may be interpreted by using a conce ~ective wellbore radius, which is defined r~ = rwex~-:j2. Incorporating this into the flow equation gt s,-----------,

2trkh p- Pw q=

J-1 I rs n- , rw

·Example: Calculate the skin factor due to radial damage for wellbore radius 0.328ft, permeability impairment

k I ks = 5 and amage _penetration . . 5 ft. 0

1<,J

0. . ;:/ , ·~VI 0/

Solution: ~,/~

S=(~-1Jln~=(5-l)ln °·328+0.S =3.7

ks rw 0.328

For this system as the pressure drops are additive then the


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Exercise#8 calculate oil production rate ~n bbl/d from a

well with the following data: re = 500', Pw = 500 psia, k = 250md

Pe =1500psia,J.io =2cp,h=l0',rw =O.S',Bo =1.35bbl/stb The rock is

fractured to improve permeability. The fracturing job resulted in an effective ten fold increase of k around the well out to a radius of 201

• What is the flow rate after fracturing? Calculate the average reservoir permeability.

~uv~ ~Pv 1- c~<=); ~~-~~ f .. ~.. r .\ / · '\'\· {""~·~), .:\~9. I ~ 7 .• - 0 ,~ - <-'- ' I .;)' ' -)<

- - ~.::_!__ s· · i'l)''v ' - · L>jlf/u' - . -} • 0 tf 1</. PI { • -.1 V

Exercise#9 show that the skin factor in spherical flow

. . . . (( k ) ( 1 1 )) regime lS S . ks - 1 rw - rs

Note that once a numerical value of the skin factor is determined, it must be converted into actual pressure loss across the zone of altered permeability as the final impact of skin on the well productivity cannot be judged by the absolute value of the skin factor itself. A· _p_arameter

~ -···-· ~ ..

showin_g_Jhe alteration of well productivity due to well s~tiQ.IL or datnage is __ fto.w efficiency, FE, damage f~!Qr_{I)E)_amLd a m.agexat-ia(DR) __ whi~h_js defined as, FE == Pe-Pwt-llPskin DR == __!__ DF == 1- FE

Pe-Pwt ' FE'

If 11Pskin is positive FE is less than one but in the case of negative value of the pr_essure__drop FE is positive.


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Exercise#lO A well is producing 100 srs oil at a measured . Day

flowing bottom hole pressure of 17 50 Psia. Recent pressure survey and well test analysis show that the external reservoir pressure is 2000 Psia. Logs indicate a net sand thickness of 10ft. the well has a radius of 0.25 ft and produces from 90 acres drainage area. Fluid satnples indicate that oil viscosity is 0.5 cp and FVF is 1.5 bbL and

STB

finally core analysis showed that permeability is ~.Ql?!P· Determine the average permeability of the reservoir, effective wellbore radius, skin factor, damage zone radius, additional pressure drop because of the skin, flow efficiency, damage ratio and damage factor. t ·

.:-.. "t •

2-8 Flow dependent skin As mentioned before, the flow around wellbore may become turbulent results in an increase of the total pre~re drop. In such cases nonlinear Forchheimer's equatioiJ can be used. Assuming· radial steady state, I incompressible, ~le phase flo}V into a f!!llY penetrating well have,

dP ... 11 ( q ) ( q )2 dr == 0.001127k 2rrrh + {Jp 2rrrh

Separating the variable and integrating gives, Pe re

J dP = J fo.oo:127k (2:rh) + flp (2;h) 2

} dr Pw rw

re re

J 11 (q J q 2 == 0.00112 7 k 2rrrh) dr + {J p (2rrrh) dr


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P.-P. = Jlq In(re)+flpqz(~-~) e w 0.00708kh rw 4rr 2 h 2 rw re .

P. -P. = /lQ ln(re)+F 2 e w 0.00708kh rw q

Pe - Pw = 11q8 In (re) + F B2 q2 with flow rate in STB,

0.00708kh rw Day

- JlqB ( (re) FBq) Pe- Pw = 0.0070Bkh In rw + 0.00708kh-;;-

. kh Pe -Pw · q ·= 0.00708 ( (r: ) ·.· .. )

J1B ln _g_ + Dq rw

Where, D is called turbulancy factor. -Incorporating damage skin into above equation results,

. -kh Pe- Pw u ~ q = 0.00708 ( (r. ) )

:P ~ft ~ r-:Jf~o ,'t=tf<4,JlB In r: + S + Dq -:- fb Ycr D jt I? (o .. ~'-t.:~lk)}( ;rP . ) J ,( ~~ !! . ~ ·Yf. . ~71'l0,_/ .::1. ~ 1

· Tl).1nk about the aimension oft e@rbu~ncy factll!! " And the apparent skin factor is defines as, S' = s + Dq.

T~~~_efore g~ne.raLfQflll_ of steady state and incompress_ible for radial geoll;letry .oJflu_Lq··ftow--is;---------- --. ··-.

// kh Pe- Pw · . ...

~ 0.00708 JiB (zn (;e) + S')) ~---- .... ' 11. ,/ ~- ---------- / · ··

jfo>-f-.tW( rl'~ Department of Chemical & Petroleum Engineering, Sha{ifUniversity ofTechnology, Tehran, IRAN 64

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Exercise#ll A new drilled well completed at the depth .of 3000 ft from the surface with a 7.5"casing in a high permeable gas reservoir with gas specific gravity of 0.78 and reservoir temperature of 190°f. Recent pressure survey and well test analysis shows that the external reservoir

· pressure is 5000 Psia. Log analysis indicates a net sand thickness of 200 ft. The well produces from 450 acres drainage area. Surface temperature and pressure are 60 oF

and 14.7 Psia respectively. And primary core analysis showed that gas relative permeability is 200mD. Some cores were taken from drilled well ·by gun shooter and showed near wellbore permeability changes to lOOmD. A simple test was done on the well for evaluation of approximately required acid to remove skin caused by well completion and drilling fluids. The well flowrate versus wellbore pressure for the well after the test is,

Flow rate ( MM~:t) da

Wellbore pressure (Psia)

4.103 2000 3.774 2500 3.384 3000 2.929 3500 2.385 4000 1.68 4500 0.748 4900 0 5000

Evaluate the skin and turbulancy factors, gas flowrate if the desired wellhead pressure is 3600, flow efficiency, damage factor and damage ratio for the well. Finally calculate the value of required acid if acid dissolving power is 0. 082 ~ock vo~ume •

Reqwred aczd volume


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3-Unsteady state single phase flow This regitne, at which the rate of change of pressure with respect to time at any position in the reservoir is a function of time and position, occurs for example for the flow movement in the middle of a reservoir before the no-flow boundary is felt- or when the reservoir appears as infinite in extent. Here we start first with deriving the governing flow equations in Cartesian and cylindrical coordinates. We shall ~~s.ume that the reservoir is homogenous and isotropic, reservoir thickness is constant, flow is horizoJ?.tal and laminar ~d pressure gradient is small (Dar~y~_§__ law applica e . / !:li 1s s 1 tl com ressible but porosity, permeability ~n fluid viscosity are pressure independent.

3-1 Governing flow equation·s Let's start with. deriving the flow· equation in Cartesian coordinates.

Cartesian coordinates: Under the steady-state flowing condition, the same quantity of fluid enters the system as leaves it. In unsteagy -§tat~ regime.,__ the flow ~ate iqtQ _an __ el_ement of volum~ of a porous media may not be the same as the flow rate .out of that element in the Other ~' the fluid content of the ----porous medium changes with time. The mathematical formulation of the transient-flow equation is based on cotnbining the following equations/laws and setting of boundary and initial conditions: Department of Chemical & Petroleum Engineering, Sharif University ofTcchnology, Tehran, IRAN 66

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a) ~()ntinui!_L __ ~-~- The continuity equation is essentially a material balance equation that accounts for every pound mass of fluid produced, injected, or remaining in the reservoir.

b) Transport equation: The continuity equation is combined with the equation for fluid motion (transport equation such as Darcy's equation) to describe the fluid flow rate "in" and "out" of the reservoir.

c) Equation of state: The fluid compressibility equation is used which describes the changes in .the fluid volume as a function of pressure. ·

The~nservation of m~(i.e. continuity equation) is: . . Input- Output+ Gen- Cons=Accumulation

,'-:1 A

I o

' ' 1 I 1 I

x' x'+dx

p.u.Aix dt- p.u.AL+dx dt = p.A.dx.¢1t+dt - p.A.dx.¢11

p.u.AI dx - p.u.AI p.A.dx.¢1 d - p.A.dx.¢1 _ x+ x = 1+ t t

dx dt

Department of Chemical & Petroleum Engineering, Sharif Univers ity ofTechnology, Tehran, IRAN 67

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a(p.u) a(p.¢) for one dimension ax at

In three dimensions,

a(p.u) a(p.u) a(p.u) a(p.¢) ·· . ( ) __ ..;._x + Y + z =- =dzv p.u

ax ay az . at

This is called the continuity equation and it provides the principle of conservation of mass in Cartesian coordinates. The transport equation must be introduced into the continuity equation to relate the fluid velocity to the pressure gradient within the control volume.

We use Darcy's law, k aP

u=---.-.f.l ax

For equation of state, assume constant compressibility (i.e. assuming slightly compressible fluids) .

. 1 av 1 ap c = ---- = --- oil compressibility

0 v ap pap 1 8¢

c = - -- rock compressibility f ¢ ap

Inserting these into the diffusivity equation and expanding g1ves,

Department of Chemical & Petrolewn Engineering, Sharif University ofTeclmology, Tehran, IRAN 68

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a(p.u) _ a(p.¢) ax at

-(p au +u ap) = (p 8¢ +¢ ap) ax ax at at

-(p au+ u ap apJ = (p a¢ ap +¢ op apJ (~)(t~ ax ap ax ap at ap at (~ :/

-(-p!_ a2; _!.p.co(a~x)'_ (p.¢.cf ap +¢.p.co ap) p~ p k)) & &

/ \ / ·\ . \ · ~.l:cn·b~ / ~ '"~"' jc,<;..• \{(IJV' t5 )'"\..}) J

Assume small pressure gr~, small fluid CQ_ll!Qressibilit and constant k, ¢, f.1 . The second term is l).egligible. Hence the nQ!liilearj:erm is negligible in terms ' ___ .. ...---... _____ . --....:..__....

of-total compressibility ct = ( c 1 +co). . . . . . ~- . ·S•OVl

!..._ a2 p _ .~. ( ) ap ~· a

2 P _ JL.¢~Jap 1 i]\~/"" .... \·· P 2 - P·cr· c J +co --r 2 - · \ \ L "~ J_...

J1 ~"\ at x ~/a~_; . ~~

;ll =·. k · 1is hydraulic diffusivity(m 2 Is) /T

\,, f.1.¢.ct ) / ..._~-......__ ___ ~ /

This is called the diffusivity equation. The next step in completing the formulation of the problem is to choose the appropriate initial and boundary conditions.

Radial coordinates: Consider the flow element shown in figure below. The element has a width of dr and is located at a distance of r

from the center of the well. The porous element has a differential volume of dV.


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Center ~r tlte well

h

According to the concept of the material-balance equation: Input- Output+ Gen- Cons Accumulation p.u.A j dt- p.u.Aj d dt == p.A.dr.¢1 d - p.A.dr.¢j

r r+ r t+ t t

A== 2:rrh a(r.p.u) a(p.¢)

- == r ,....--_ -=----=--ar . at

This is called the continuity equation and it provides the principle of conservation of mass · in radial coordinates. The transport equation (Darcy's Law) must be introduced into the continuity equation to relate the fluid velocity to the pressure gradient within the control volume dV.

- B(r.p.u) o(p.¢) - ==r---

or at

-~~(r.p.!!__ BpJ == (p 8¢ Bp +¢ BpJ r or J1 Br ap Bt at

1 o ( · k BpJ ( Bp BpJ - r Br r.p. J1 or == p.¢.c1 at+¢ at

This is the general partial differential equation used to describe the flow of any fluid flowing in a radial direction Department of Chemical & Petroleum Engineering, Sharif University ofTc:chnology, Tehran, IRAN 70

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Reservoir Engineering I, Mohscn Masihi

in porous media. Compressible and slightly compressible fluids, however, must be treated separately in order to develop practical equations that can be used to describe the flow behavior of these two fluids. Let us now continue our derivation for slightly compressible fluids (oils).

( a(r.u) apJ ( · ap apJ - p +r.u.p.co- =r p .¢.c1 -+¢.p.co-

8r ar at at

( ( J ( J2J ka ap k ap ap - --p- r.- --r.p.C

0 - =r.p.¢(c +co)-

J.l or ar J.1 or f at

Assume a small pressure gradient, and a small fluid

compressibility and constant k, ¢, J.l, h the second therm will be negligible and then by using the definition of total .compressibility we· can have.

_!_~(r. ap J = J-1.¢.ct ap r or or k ot

This is called the diffusivity equation. It is one of the most important equations in petroleum engineering. The equation is particularly used in analysis well testing data. The form of this in field units with time t recorded in hours becomes ·· ·· -· · ·· · · · -

// ' ····---.

1/r· a ( ap J J-L.¢ .c

1 ap -, __ ~

lr ar r. ar = 0.000264k_9~ -----·---------- --·

with c:Psr1, k:mD, r: ft, P: psi, t: hr, J-L: cp


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The term 1J == k/ fUf!C is called the diffusivity constant or hydr(!uli_g __ _d_i_ffl!sivity that is the ratio of the mobility to stor~tivity. -------- -~--

Exercise#12 derive the diffusivity equation 1n spherical geometry.

3-2 Radial oil flow in a fully penetrating vertical well Consider a shut-in well that is centered in a homogeneous circular reservoir of radius re with a uniform pressure Pi throughout the reservoir. This initial reservoir condition represents the zero producing time (see figures below).

q=O

A)Shutin

~~~ll__is allowed to _flow_at a constant :Q_9~_ rate ofq, a pressure disturbance will" be created a~ ___ !P.~ -~!i!:J-d face. The pressure at the well bore, P wr, will drop instantaneously as ------- ---- - ····--······---· ·· ·· - ..

the well is opened. The pressu~~ disturbance will move a~ay "froni-the -wellbore at a ~e that--is ___ deieffilmed by the rock and fluid properties. Figure below shows that at time t r: the pressure disturbance has moved a distance r 1 into the reservoir. Notice that the pressure disturbance radius is continuously increasing with time. This radius is called commonly radius of investigation and referred to as rinv· It is also important to point out that as long as the radius of

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investigation has not reached the reservoir boundary, i.e., . re, the reservoir will be acting as if it is infinite in size. During this ti111e we say that the reservoir is infinite acting bec~use the outer drainage radius re can be mathematically infinite.

,, ... ,, P; -~,----::::._~---=::- ~

I ~;L----~-~-~-----

· ~ v--,, \

. ~'

CB)Constant Flow R~ r~

A similar discussion to the above can be used to describe a ~tEll that is producing at a constant bottom-hole flowing :R~ Figure below illustrates the propagation of the radius of investigation with respect to time-.;~ At time -t4, the pressure disturbance reaches the boundary, Le., rinv = re. This causes the pressure behavior to change.

P;

r~ C) Constant Pv.f

For the problem of a single _phase flow In a ful!_y penetrating vertical weq located in an infinite ho-Elogeneous and isotropic reservoir of fixed thickness, the governing flow equation in cylindrical coordinates is,

_!_~(r. apJ = JL.¢.cl ap or _!_~(r. apJ = JL.¢.c{ op r ar ar k at r ar ar 0.000263k at


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To obtain a solution to this it is necessary to specify an initit;tl condition und impooc t"YY·o boundcuy vvuU.itivu~. ·Tlle

initial condition simply states that the reservoir is at a ..._________ ·--···---------····· · · ·-~---··--·--·--·--··--· . . . -

uniform __I?.ressu~~- Pi when productiol!_!?egins. The ·two boundary conditions require t~well is Rroducing at a constant production rate_and that the reservoir behaves as if it were infinite in size, i:e.;-0oo. Mathematically these are: r

I ~_;V p t=O = P; -I.C:

/ fi. -1-re.· /; . . - . B C#l

PL~oo P; · f'i'/;.1" ·.r:!J ll7,,-f._..~~Q/

<. . : / k (2 h) -~ . / I _,-- ~ .

trr p \ , J'~, . ~ l/_~. (fl B C#2. q = - ' / u . ./d {(. .::> . . a I - .

J1 --.Y-cv-/ A simpler version of this problem is .when well bore radius is infinitely €11 rw =saO) In practice this is applicable wh~n radius tion of ressure ulse is reater than

;:-_::;;:;. . ·--·--

w llbore radius. Solution to this in terms of Ei-function is called line source /Kelvin/Theis solution which- is work for infinite reservoirs (Matthews and Russell 1967).

= . - qJ.L Ei(¢J.Lcr2 J ~:--~;~6 qpB E'( ¢J.lcr2 J p p, 47rkh 4kt or ~~·---~----~--.Q.·_~_9~t p: psi,q: stb I d,_fl: cp, B: bbl I stb, k: md, h,r: ft,c: psr1 ,t~_ hr J

--The exponential integral function is defined by,

00

J e-udu Ei(-x) ==- u

X


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And can be approximated as below,

[ x xz x3 ]

Ei( -x) = ln(x) + 0.5772- l! + 2(2!)- 3(3!) +etc

Values of this exponential integral function can be found from a table or a graph. Graph of the Ei-function for 0.01 < x < 10 is shown below (Craft 1991).

10:··-~~ EXPONENTIAl. INT£GRAL VALUES

al -u , , ' ' Ej{-XI = -J e (,1 du

lC. i

43 [~'~,·~·- ~'·i1

~~!!~~~ii~~~·~~~~Fo~r~x~<0~.~0~2~~~~ . ·~· · . Ej(-)(}~lt~.{xl•0.577

-......f I """""_,..,..._

2 I ........ f I l ' i ~ I i ._ f ~ I • I ; I I I _._ I ! ( j I . I I I I j ! I j ,-+-I i

t---1-':''<H-+-+-+- o -.o2 -.04 -.os -.oa -.1o + I.OIII'II'II'i'illj II' I E;t-xl 0.8

0.6

0.4·--~ 0.3

' ' 0.2 1

..... I I ! I I

l ! I I I I

O.lf~~~~~~~~~~~~!l~~~~!i~~~~ .08

~ ~

" ;

.0

411111'111' ~~ 1"1; 11'1 .03

.02 • • ,, \

I i I i \ t i l I I !

-1.5 -2.0 -2.5 -3.0 -3..5 Ejl-xl


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The exponential integral function can be approximated by a logarithmic function as for x<O.Ol

Ei(-x)=r+lnx+ LC-lf xn ( ~~oJ n!n n!n

When the parameter (x) in the Hi-function is less than 0.01, the semi-log radial flow ·approximation can be found,

P = P;- Jlq (1n( kt 2 ]+0.8091] Jit/Jc1r

2

< 0.01 4iCkh . Jl¢c

1r 4kt or:

p(r,t) P;-70

·6qBp(ln( kt

2 J-4.254J for 39.51 ¢Jlc,r

2

<0.01 ~ ~sr kt

j:d( IJ lA.-J) ·-? t ~ J W]

Applying skin damage effect results in,

l:lpto~al= 70.6qBJ1(ln( . kt 2J-4.254+2SJ kh ¢f1c1r

Note that in the well test analysis where units are th~ same as field unit except for the time that is in hours, the above ------- ....

equatio~ ___ Qecome·s, ~-lJ ~tJ oClfrt. ,~r-

P = P;-162

·6qB f1 (log( kt

2 J- 3.23 + 0.87 sJ for ¢f1c1r

2

<0.0 1 kh ¢pc1r 0.001 05kt

Range of application and limitation: 1-_~on cannot repre~PL-the initial flow into a welJpor~ ( represente~ . ~y a .Ii11e) so some !i.l}!e has to be elapsed "for Hie feiat1ve ·- -s1ze- -o{. ihe·--r~voir--to- have a

negligible effect on flow so:

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J.'he reservoir must be infinite acting.1 From real reservoir

~~----~--~~~--~~~~ rmance behavior Ei is va 1d or:

Example: Use semi-log approximation with these data to estimate k and ~c, r w = 4 in, J1 = 0. 3 c p, q = 2 0 0 b b 11 d, h = 15 ft

t( min) 1 5 10 30 60

Pw (psi) 4761 4673 4633 4573 4535

Solution: First we plot the pressure against In (t),

4800 ~

!-Ri __________ ___ ___________ ib"-------- . . . __ .-- ol 4750 1 : -.__ ,/ ----, \

1 j ·. __ __ Y = -55.284Ln(x) \5J---1 4700 ~ i -.

·r· " ., ; ·- i...-\b (/) I - ·- -- - - ;,_·-

I-"-~ - -,e; 4650 ~ 5: !

Q_ ' I

4600 ~ !

c)t~ .. t - \ oj iu/ o{--

y f)

i 4550 ~ r-t 4500 -'---'---'--'-..__._,.._,_.u..____,__..l-L..L.LU...u...___._...L.-L..LJ...J..JLJ...I..-_--'--'---'-...L...Jj...L.LU

0.01 0.1 1

t(rnin)

qJ-L ( 4kt J Pw = Pi - In 2 + 2S 4Jrkh r¢pcrw

~ 10 ,. b•

Department of Chemical & Petroleum Engineering, Sharif University of Technology, Tehran, IRAN

100

77

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1 -jlq

55 3 . 6895 Pa

s ope= =- . pst---4:rrkh psi

-0.3cp 0.001 Pa.s 200 bbl 0.1589m3 day cp d bbl 3600 x 24s

---------~---------------------

4x3.14xkxl5ft 0

·3048

m ft

kh= 0.3x0.001x200x0.1589 m2

4x 3.14x 55.3x 6895x3600x 24x15x 0.3048

-t5 2 · mD kh = 5 x 1 0 m

15 2 = 5.1 mD

0.987xlo- m

At t* we should have Pw =pi

p(r,t) pi 70

'6qBp[ln( kt

2J-4.254J for 39.51 ¢pc1r

2

<0.01 ~ ~sr kt

k t*==e4·254 => t/Jc1 ~12.1x10-4psr1

f1¢c1r},

Example: An acidized well was produced for 6 days at a rate of 400 srB. The following information on well,

Day

formation and fluid properties are available, rw = 0.25 ft, B 0 = 1.12, pi =2800psi, ¢=0.2, k=30mD, f.1 = 0.4 cp, c, = 3 X w-s psi-!' h = 40 ft

Calculate P(rw) and P(r=800 ft)

Solution: (fEst check the validi!v of Ei functTqn?

(30)t

(0.4)(3 X 10-5)(0.20)(0.252) > 3950·69

> 1.97 x lo-s Days

Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN

yields -----7 t

78

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check the validity of logarithmic approximation. (3 0) t . yields

(0.4)(3 X 10-5 )(0.20)(0.25 2 ) > 15802·8 ~

> 7.90 x lo-s Days

t

As t = 6 days so logarithmic approximation can be used. (400)(0.4)(1.12) { ( (30)(6) )

p = 2800- 70.6 (30)( 40) ln (0.4)(3 x 10-5)(0.20)(0.252)

-4.2537 + 2(0:)} [l·L.·;·l ~ --------~ ' ~7 C" vl V ~) l> c"

At a distance of 800 ft the Ei function is valid to so~ _ _thj_s - --··- .---·

problem. ··~ (400)(0.4)(1.12) . ( - (800 2)(0.4)(3 X 10-5 )(0.2))

p- 2800 + 70.6 . (30)(40) . EL 39.51 (30)(6)

. . (400)(0.4)(1.12) P _;_· 28oo + 7o 6 E ·c-o 3371) . (30) ( 40) l .

(400)(0.4)(1.12) . P = 2800- 70.6 (

30) (

40) (0.821) = 2791.34 Psra

Exercise#l3 A well is producing 100 stb/ d oil. The reservoir initial pressure is 2000 psi. Logs indicate a net sand thickness of 50 ft. The well has a radius of 0.25 ft and drains an area of 1000 ft. Other data are: ¢ = 20%, B

0 = 1.5 bbl/ stb, J.l = 0.5 cp, c = 2 x 10-9 Pa-1

, k = 50md.

Calculate the pressure after 0.1 hr, 1 hr and 10 hr at the well bore, the points r= 10 ft, 100 ft. Then plot the pressure profile within the reservoir versus time


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Exercise#14 A well with the following data producing at 266 sTm

3

faces a declin~ well pressure from initial value at Day

2 77 bar to 192 bar after 6 min. calculate the skin factor. rw = 0.15ft, re = 61 OOft, Bo = 1.39, ¢=0.28, k=lOOmD, Jl = 0.8 cp, ct = 2.3 X 10-9 pa"1

' h = 8.5 ft

3-3 Radial gas flow in a fully penetrating vertical well ~s visco~ity al_!__d density vary _§jgnificantly ~ith ~~~s~re and theJ:d'ore the q._~sumptions used in derivation of liquid flow equation are not valid for gas systems. Equation of state (EOS) and gas compressibility for a gas IS,

pM · 1 1 dz zd(p!z) p = zRT and cg = p --; · dp = p dp

c~~bin.in.g_ the above two basic gas eguatj.Ql)~ _~it~ that of general partial differential equation useq __ _!o describe the flow of any fluid flowing in a radial direction gives, ·-~ - - - ·· ···----- ---· -··

_!_~(_E_r opJ = _E_¢c op r or z Jl or zk t at

To resolve the nonlinearity in flow equation, there are two methods:

a) Ci@i:C8aSQSeudo meth~] (Al-Hussainy, Ramey, Crawford 1966) btp-squared me!~9:9XRussell, Goodrich et al) Department of Chemical & Petroleum Engineering, Sharif University ofT~chnology, Tehran, IRAN 80

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Real gas pseudo pressure: Using pseudo-pressure defined as,

p2 d lf(p) = J1dp or __1!_ = _p_ o~

o ZJ.l dp ZJ-1

S_~Q_stituting the PDE with pseudo-pressure gives, 1 a ( a lf) ¢f-1c1 a lf/ 1 a ( a lf/ ¢f-1c1 a lf/ -;or r or = k at or -;or r or = 0.000633k at

This is the governing diffusivity equation for a gas flow in porous media. J

This differential equation relates the real gas pseudo pressure (real gas potential) to the time t and the radius r.

Al~Hussainy, Ramey, and Crawford 1966 ointed g_u~ that in gas well tes_ting analysis; the constant-rate solutio

------------····-·--

has more practical applications than that provided by the constant pressure solution. jThe authors provided with the exact solution to this that is commonly referred to as the solution method. !~posing the constant-rate condition, AlHussainy, et al. (1966) p!Q_Q9sed the following exa<?t s,o.tut!on to the3fiffusivity equatiffi? .

.>jV /lil.J'c./-/ . 0{11

[57895.3Psc] [c(9 T] [- ( kt ) ]

1/J(Pwf) = l/J(Pi)- _ T kh log cpfliCtir~ - 3.23 + 0.875

Pwt< Psia, Pe: Psia\qsc ~t: ·~ k: mD, Psc= Psia, T5J0~ rw: ft,~ ~1, ~ f1i:_9.?_, Cti: Psi::T_ -~:<~~ . .._____:/ . - ~~-X~-.. . "' ~ ~-- · /

/), \1/ \"" 'l ~ /T'


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Exercise#lS A well with a wellbore radius of 0.25 ft is producing at a constant flow rate of 120 MMscr under

Day

transient flow conditions from a gas reservoir with gas specific gravity of 0.78 and skin factor of 2.67. The initial reservoir pressure (shut-in pressure) is 5000 Psia at 240°F. The formation permeability and thickness are 100 mD and 10 ft, respectively. The porosity is recorded as 1 7 .5%. Standard temperature and pressure are 60 oF and 14.7 Psia respectively. Plot sandface bottom hole ~~ssure versus

------·------ ~

time for one day us!_gg exact solution metho__d. Initial rock compressibility and turbulancy factor are Psia-1

.

. . ~Q121,,>_~1~S ·S~,~ T} ···· .5

-~ --,(!!' "-"-/Is· • . .

There ts also other solution that approximate the exact solution (the pressure-squared method)

p-squared method: The first approximation here is to remove the pressuredependent term (11z) outside the integral that defines l/J(Pwr)

and 1/J(PJ in previous problem to give, 2 p

lf/(p) = =-= f pdp ZJl 0

The bars over ,~. and z represent the values of the gas viscosity and compressibility factor as evaluated at the

p?-+p2 average pressure p = t 2 wf

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Combining this with the previous exact solution gives, 2 2 [1637q9 Tjlz] [ ( kt ) ]

Pwf =Pi - kh log q;fliCtir.J - 3.23 + 0.875

The abov~ap~~~!!()n solution forms indi~ate_!h~Ltbe product 11z) is assumed constant at the avera e res sure?. l;his effective y 1m1ts the applicability of the P method to r~§~oir pressures (P < 20QQ, _.discuss~g_be_fo!e ). It should be pointed out that when the P2 method is used to determine p wf an iterative procedure should be done, do you know how?

/ _., \

S~"" ;,.I

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4-New flow solutions using superposition The solutions to the diffusivity equation presented previously appear to be applicable only for describing the pressure distribution in an infinite reservoir that was caused by a constant production from a single wei[ Since --- --·----

~eal reservoir s stems usually have several wells th'!!_are ------······--- ------- -- ·

OQerating at Erying r~tes_,_ a m2Ee generalized approach is needed to study the fluid flow behavior during the unsteady state flow period. The principle ()f -siip-eq)osition -------is a powerful concept that can be applied to remove the restrictions that have been imposed on various forms of solution to the transient flow equation.

4-1 Principal of superposition Mathematically the superposition theorem states that any sum of individual solutions of a second order linear differential equation is also a solution of the equation. This concept can be applied to account for the ef(ect~_ of tpJl!ti£!~_wells, ,ra~e change and nearby boundary.

A differential operator L that operates on function y is linear if it satisfies the following conditions: L(y,,y2 ) = L(y~.) + L(y2 )

L ( cy) = cL ( y) An equation is nonlinear if it contains pressure or its derivative to a power higher than one and/or multiplies by one another.


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SEperposition allows · us to find new solution to the diffusivity equation by adding the solution of two P!~~usly-known solutions provided that the both I.C and B.C are satisfied. Note that if we superpose the pressure then, Pinew = p; + p;' = 2 Pi. However, if we use drawdown then they have the same initial and outer boundary condition.

Hence we superpose (Pi - P) instead of P .

4-2 Effect of multiple wells Frequently, it is desired to account for the effects of more than one well on the pressure at some point in the reservoir. The superposition concept states that the total .rressure drop at any point in the reservoir is the sum of the pressure changes at that point caused by flow in each of --the wells in the reservoir. ---Ap·------------·-·-----

Ll Total at r* .

. ·= b.Pnrop due to well (1) + b.Pnrop due to well (2) + · ·· + b.Pnrop due to well (n)

u~ng either Ei-function,~ · q11B . ( r

2) 11P = P·- P = -70.6-Et --

l kh 41}t

or logarithmic approximation, in the range of their validity, for pressure drop of each well

qJ1B { ( kt \ ~ ,Q !JP == 70.6 kh ln J1cf>Ctr2) -~+ ~


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For better understanding let's consider below field w~th three producing wells and one shut-in well we are going to see what would be the pressure at the shut-in well when well (1), well (2) and well (3) produce for t1 , t 2 and t3

with the flowrate of q1 , q2 and q3 respectively, _. -----------... ...

~ ' " ... " ' , '

/, _..:"<=l!Y~...... ,\ .. ·················· ................................................ ,,,_ I / ' wei~) ''-. I ,/ I / "'-, . 1:'

. I "- t:' \ 1. · ' 3 ····-.....~· . b~n·ed

./\, . .. __ )~1_,{ . •. • ... d, ... Actin

• . .·· · w'RI (1) ' Actin· .•.. .12 " ; · ' . ' ;,.-··/ , / d . '~'ell(-~ ., ;.

"-------"" .1\ l \. \

'\ / ··. ' / '~ ./ ....... . .._._ . .....-· ······- ....................... ·

flPTotal at observed well

== flPvrop due to well (1) + flPDrop due to wel_l (2)

+ 11PDrop due to well (3)

flPTotal at observed well

·~ ' v •

r)' ; ,\ ~ i \

J-- t\ 'J;

B ( d2

) B ( d2

) == 70.6 q1 J1 Ei - 1 + 70.6 qz!l Ei - 2

. kh 47Jtl kh 47Jt2

B ( d2

) + 70.6 q3 J1 Ei - 3

kh 41Jt3

I

Note that when the weU is a producing well; the flow arte 9 appears in abov; form . the ositiv~---.i~~l'l Qtherwi_se 1n the case of injection well; negative sign must be~d. For example assume well (2) was an injection well in the previous figure !1P must be evaluated as below,


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Reservoir Engineering I, Mohsen Masihi /

-·· - ---- ----- ------~- J/

·~~~ ( d2 ) · B ( d 2 ) - = 70.6 q1 /1 Ei - 1 - 70.6 q2 f1 Ei - 2

kh 4ryG kh 4ryG

+ 70.6 q3 f1B Ei (- d~ ) kh 4ryt3

Exercise# 16 Suppose an injector well with flow rate q located between two producers with rates q/n. For wells of the same radii prove that the following expression is true for the relation between q and the injector and producer pressure wells.

kh 0.00708-(P1 -P2)

q - J.l - - ( n+l) l 1 d ~ !1: nJZCw)

~ffect of variable flow rate . li!!_Prncfice-:--n1e wells produce at varying rates --and so 1fls im ortant that we be ab e ·to redict the res sure behavior wh~n rate changes. concept of superposition states "?ve~x_ flow rate change in a well wi rresult in a pressure response which is independent of the ~essure responses caused by other previous rate changes/ Accordingly, the total pressure drop that has occurred at any time is the summation of pressure changes causes separately by each net flow rate change.


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Consider the case of a sh~t-in well, i.e., q . 0, that was then allowed to produce at a series of constant rates for the different time periods shown in the figure below. To ---calculate the total pressure drop at~stance of d~_fr_o_m_the wellbore after t, the composite solution is obtained by adding the individual constant-rate solutions at the specified rate-time sequence, or,

Flow Rate

t1Protal == ~Pql + ll.Pqz + ... + D.Pqn

l

I I

t, t~ Time

To eval~ate 11Pq; first of all assume there exists only flowrate (i.e.,q1) therefore piessure __ drop due to tht~ flow rate is, --As the flow rate changes from q1 to q2 at t 1 , we can assume that the well continues to produce with the previous flowrate but another well with the same characteristics and with the flowrate of q 2 - q1 starts to flow at the same location (see figure below). Department of Chemical & Petroleum Engineering, SharifUniversity ofTechnoiQgy, Tehran, !RAN 88

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,,

Therefo~e 11Pq2 is evaluated as below,

~p == 70.6 (qz - q1)J1B Ei (- dz ) qz kh 41](t- tl)

Using this idea for the other periods gives,

~p == 70.6 (q3 - qz)JlB Ei (- dz ) q3 kh 4r](t - t 2 )

~P. == 70.6 (qn - qn-1)J1B Ei (- d2 ) qn . kh 41](t- tn-1)

Now as the total pressure drop is the sum of these then by substituting we get,

q1 J1B . ( d2

) l:lPTotal == 70.6 kh EL - 47Jt

+ 70.6 (qz- q1)J1B Ei (- dz ) kh 4ry(t- t1)

+ 70.6 (q3- qz)JlB Ei (- dz ) + ... kh 4ry(t - t2 )

Department of Chemical & Petrolewu Engineering, Sharif University ofTeclmology, Tehran, IRAN 89

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This idea can be simply extended to any rate history (as shown below) when we approximate the rate history into

some time intervals ( qt for 0 < t < ti ; q2 for ti < t < t2 ).

time

..

Note that with this the cumulative production must be maintained. In the limit of infinitely small time intervals the· drawdown can be found by using convolution integral (Duhamel's principal).

Exercise#17 A well is opened to flow at 150 srB for 24 Day

hours. The flow rate is then increased to 360 STB and Day

lasted for another 24 hours. The well flow rate is then STB

reduced to 310 - for 16 hours. Calculate the pressure Day

drop in a shut-in well 700ft away from the well given:

Department of. Chemical & Petroleum Engineering, Sharif University of Technology, Tehran, IRAN 90

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~ = 3000 psi,¢= 15%, Bo = 1.2 bbl/ stb, J1 = 2 cp,.

c = 12 X 10-6 Psi"1' k = 1 OOmd, h=20ft, rw = 0.25 ft

Example: A special case of the variable rate is known as buildup test when n = 2, q! = q, qz = 0, b.tl = t (figure below). Find the solution for this case.

Solution:

\ 'v0~ Pw , ,i.~ 'vr

. '<------------:;> : At .. t time

First consider the semi-log approximation for drawdown due to production period from time zero up to time equal to t + ~~, 2

1rkh (pi- p 1) = !_(ln ( k 2

(t + L1t )J + 0.8091] pq 2 p¢ct~v

TQ~~jmagine we start at timet injecting fluid at the rate q. The drawdown due to this has the same form as before ex_~~t that the elapse-timeisTMj and -q must be used for injection.

-2trkh (pi- P2 )=!_(ln( kL1t 2 J + 0.8091] jlq 2 p¢ctr..v


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Now we superimpose the two solutions,

(p;-p)= JLq (ln(k(t+!i:)J+0.8091)~ -JLq (In( k!it 2]+0.8091]

4Jrkh JltPC1rw 4Jrkh JltPC1rw

Simplifying this results in,

Jlq ( (k(t+L1t)J ( kl1t JJ p.- p = In -In ( ' } 41ikh J.l¢c,r~ J.l¢c,r~

= _ JLq (rn t + 11t) p Pi 4Jrkh At

Horner (1951)

Exercise#18 Consider two producing wells which are 50 ft

apart. Well (A)flows at 50~:: for 8 hours, then the rate is

increased to 80 STB. Well (B) produces at the initial rate of Day · ·

·. STB · . STB · 70 - for 7 hours and then the rate IS decreased to 25 -.

Day Day

Estimate the pressure at the well (A) after 9 hours. The available data are: ¢ ~ 30%, Ba = 1.2 bbJ/stb, J-1 = 1.5 cp,c

1 = 10-5Psi"1

, k = 80md, h=30ft, Pi= 2000psi

Well A: rw = 0.5ft, S=4

Well B: rw = 0.4ft, S=2

Pressure penetration into porous media: Let's use superposition to find pressure propagation in a porous medium by assuming flow with rate q is injected for a short period of time 8t into a well In an infinite


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reservoi~ 1be solution of drawdown pressure for a ell producing\at the constant rate q was, _ _. '·-..... . J .....----·--·-- ---cx:5"------

· .. ____ ~___.-/ t::..P == -70.6 quB J e-u du ---"J ('_~~\n.t 1'11.!_~(1) kh u . /1'1

~ (f;}jv~ 4~t /

I:luid injection is mathematically equivalent to fluid production except for the sign. Injection at a rate q for the elapse time 8t is equivalent to inject at rate q at ti~ 0 ~!1-d then produce at rate q at time 8t.

q ' c~ :- - - - - - - - - production

0~~--------------~ 0 &,t t

__ (,~ - · ...... - - - - - - - - injection

Using superposition principal the solution to the problem will be, t::..Protaz == f::..Pwell (1) + f::..Pwezz (2)

For injection well, 00

M = -70.6 ~: J e~u du

For production well,

qJ1B 11P == + 70.6 kh

rZ

00

4~(t-8t)

e -u -du u

Department of Chemical & Petroleum Engineering, SharifUniversity of Technology, Tehran, IRAN 93

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Substituting these in total pressure gives,

qJlB l:1Protal = 70.6 kh

co co

e-u J e-u -du- -du u u

r2 417 (t-ot) 41]t

41](t-Ot)

For small elapse time Jt this integral can be approxim_~!ed

by,

qJlB l:1Protal = 70.6 kh

41](t-Ot)

e-u -du u

r2

~ 706--X ------qJlB e -4rJt ( rz rz ) .

- . kh r 2 4T}t 4ry(t- ot) 41]t

which then by simplification gives, qJl.B ( _ _c_) (1 1 )

llProtal = 70.6 kh X te 41]t t- t- ot

qfiB ( -~) ( -at ) = 70.6 kh X e 41]t (t-ot) r2

qJiB ( -~) (1 1 ) qJl.B e-41Jt llProtal = 70.6 kh X te 4

1Jt t ~ t _ ot = -70.6ot kh X t

Pressure drawdown is a function of two time related terms: -~_pone_ri!laJ increase with t~and lit decayjHenc~-the drawdown increase and then decrease and face a maximum


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value in between.(The time for pressure pulse to reach ~ @:axtmum value at posthon r= /1.1 can he found bY'i)

( r

2 e -:;t) t- e-::t · a 11Protal 4ry t 2

--------- = 0 at r=A t2 Az yields Az 1 '4=-t -- 1 = o t = _ or /1, = '\/ '+l]t 4~t 4~

0.006328k 7]=

J1Ct¢

where k: md,Jl: cp,ct: psi-t .

Example: We wish to run a flow test on an exploratory well for a sufficient time to. ensure that the well will drain a cylinder of n1ore than I 000 ft radius. Well and fluid data

analysis St1ggests that k:100md,J1:0.5cp,c, :2x10-5 psr

1,¢=0.2.

What time flow do you recommend?

Solution: A.= J4iii

0.006328(100) mD 11 = (2 x 10-5 )(0.2)(0.5) =

316400 cp. Psi-1

yields 1000 = vf4(316400 )t t = 18.96 hours

4-4 Effect of nearby boundaries Reservoirs are often transected by nearly vertical faults, mostly impermeable to flow due to various reasons including geological process or deposition. These faults act


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•

Rese~voir Engineering I, Mohsen Masihi

as a no flow boundary condition on the nearby wells. The superposition theorem can also· be extended to predict the pressure of a well in a bounded reservoir.

A well close to a sealing fault Consider figure below which shows a well that is located a distanced from the non-flow boundary, e.g., sealing fault.

Sealing fault

~athematically th~ barrier is equivalent to the pressure cJisturbance produce<!J?y a secondJ___mage well pro_~ing ~~the same rate and liaving the same production history as

'-._ .. -··-- - ~-------- . . - -

the real well with both of the wells in aiili1fiilite acting reservoir. Assume there is no fault_then due to the --- ----- · ·~----~--~--··-· ----- ---symmetry no flow will cross tl}~ __ Qlane of the fault. ~- --------~--- '-----. ·--~~ .. -· · ·· · ·· · -~------· ·---

/ ·~ \ / () op ) · l ' ! - =o -~ /-(//r ':.r, 'Cir ,

"f-q , I ll I

Actual Well ... 1 . \ i . _j Image Well I ' u ! ,, . ~ I v'

;u'r "J 0 "'/ (J 'ttr Department of Chemical & Petroleum Engineering, SharifUniversity ofTechnology, Tehran, I~ . 96

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Hence using the superposition, the drawdown is the sum of the drawdown due to the actual well and the image well. The solution for the pressure in well is,

b..Protal = b..PActual well + b..Pnute to image well

qJ1B - . ( r2

) ll.Protal = -70.6 kh El - 4 t

7] r=rw

-70.6-Ei --q11B ( r2

)

kh 41}t r=Zd

This solution can be clq_s.sified based on the /!me re 1-Early time regime: the full expansion exponential integral function should be used.

2-Intermediate time: loga~ithmic approximation can be used for the actual well but there is no response yet from if!!_age well to the actual well (practically for the image well Ei<O .01 ). The solution for this regime is the drawdown for the well in the absence of the fault because the pressure pulse has not reach the J~ul_!~<)the slope of

. . . . : ·· . . _..)!q · drawdown curve on a semi log plot IS, m 7 4.1{kl;z.

3~Late ti!!ii] wh~~~ approximation is valis!J9r b.oth wells. The slope of drawdown curve on a semi log

plot is, . Here the system is semi-infinite rather

than infinite (expect to produce half so to fix rate drawdown will be doubled). Hence the occurrence of


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0


!!_ol!bling of __ the~ is an indica~_i~~----of a nea~~y impermeable fgult. __ -- ----... -----·- -------------------,

~u~)~ h~--- -k~~ )>9'1/

' ···-.. ··'

j -ft; /

\o z_ -t vv(zy, -\--\.yQ '( e -~ i .._,

Moreover, the distance "d' can be estimated from the

.fr£tl!~ition time het~tWOTeglffi~ = ±e·!~~~J' l

c-z.J.) :::.. 1-. ~-

1[-tt

Example: Well (A) is located 1500 ft apart from a sealing fault and is completed in a huge reservoir. Well (A) is opened to flow 340stb/d. What pressure drop theoretically·· will occur at observation well B (700 ft apart) from. the fault after 216 hrs of flow? Other data are: P; = 4000 psi,¢= 10%, Bo = 1.07 bbl/ stb, Jl = 0.8 cp,

c = 5 X I0-:-5Psi-l' k = 75md, h=100ft, rw = 0.5ft,SA = 0.2

........... ..... ············ ······· I 5001\

;e. -. 700 ft

' • Aciual wdl B Irnagcwdl


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' I r\L ' \

,/~ o()b'7", CQ

~/' \). eX

Reservoir En:in.<ecl•g I, ' .. 1 o .. h. s.e.n Mas. ih. i. ..··· /// ./7 \(/ r ;· Solution: /-- · · Ei approx-i~-~~i~rilnythod can be used, check it! /

7 0.006328(75) ) mD

~ = (0.8)(5 X 10-5Y(if.l) = 118650 cp. Psi-1

t>~A~tu~~~~;i~tn/ -70.6 (34~;~~(io~)07) Ei (- 4(1~~~~):)(9)) = 2.7396£i(-0.1498)

=4.014 Psi (340)(0.8) (1.07) . ( (2200) 2

)

b.Plmage well at B . -70.6 (75)(100) El - 4(118650) (9)

= 2.7396£i(-1.3311) =0.483 Psi l:lPTotal = 0.483 + 4.014 Psi

J ·• '- -- . -·

~gain sk~rt fa_ctor of well !A} did nQj considered in the calculation, why?

Exercise# 19 For the following two producing wells close to a sealing fault derive an expression for the wellbore pressure of well (2) versus time. Solve above problem with typical values of the reservoir parameters


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A well close to two perpendicular sealing faults The method of using image wells can also be used under other well configurations. Consider a well located equidistant from two perpendicular sealing faults. Again we try to have a no flow boundary for location of sealing faults. We should use three image wells (not two?) as below in order to get the proper symmetry in its flow field.

#3~. . ,® , ' #2 J I ,,'' l

~t ~ ,'~2,d . ' c . .,fL.,

') . \. }f .-;;;.;;;~;:;;..;;;;.;:;;..;;;;.~~~~ . .\_,

) "/ f; .) d : ,/ Jl.._b '?J.( : /

~ :)f.> . '. J j I .-:: ___ [ __ _

,'( ::r well v ' i

The solution for draw down is,

------ -0' #I

llProtal = llPactual well + llPwell (1) + llPwell (2) + llPwell (3)

+~ . . qJlB . ( r

2)

llPactual well = -70.6 kh Ez. - 41]t -. T-Tw

q11B . (. r2 )j

llPwell (l) = -70.6 kh Ez -4

t 1J r=2d

q11B . ( r2

) llPwell (Z) == -70.6 kh Ez -

4 t

1J r=2Vid

qJlB . ( r2

) 11Pwell (3) == -70.6 kh El - 4 t

1J r=2d

. Department of Chemical & Petroleum Engineering, Sharif University of Technology, Tehran, IRAN 100

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llProtal = -70.6 q:: Ei (- ;:t) _ r-rw

- 70.6 q:: Ei (- ;;t) . r=2d

-70.6 qJ1B Ei (-~) - 70.6 qJ1B Ei (- r2

)

kh 41]t r=Z..fid kh 41}t r=Zd

This solution shows that using logarithmic approximation the slo_pe of the well bore pressure versus time curve on the se111i_:!og ~ill be four times greater in magnituae than the corresponding slope for the well when there is not any fault! .

. E~ample: with the following data check the Ei applicability, ln approximation for the real and image wells and determine the wellbore pressure after 32 days. ¢ = 22%, Bo = 1.5res m 3 I ST m3

, h =36m, JL = 10-3 Pa.s, ~ = 240bar

c=9x 10-9 Pa-1, k = 89mD, · rw = 0.15m, re = 6000m,q = 100stm3 I day

Solution:

70n1, I I I

Gt------------------· 120m

well

For real ·well Ei applicability:

Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN !Ol

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100 ;upc1r!;, ;upc1re2

_.;:,._,;__...:._;;_ < t < -'----'---'-=--

k . 4k~~\ \ 50 sec < t < 2317 ~ OK \

. · . 25J1¢c,r!;, In approximation: t > k ~ t > 13 sec OK

• 25J1¢c (2d)2

For Image well# I : t > ~ · ~ t > 126 days not valid

P;- Pwf =!¥?well+ /¥Jimage#l + /¥Jimage#2 + /¥Jimage#3 => Pwf = 230.5bar

Exercise#20 How many image-wells should be considered in the case of a well close to two sealing fault with the angle of e = 60 and e = 45 as below. Using logarithmic approximation compare the slope of the well bore pressure versus time curve on the semi-log with the corresponding

. slope for the well when there is not any fault. /. 0 ·-/. 0 ~'(· l

(

/

lj) ii I

A well close to two parallel sealing faults '· , Consider a well located within a. channel. Again we try to have a no flow boundary for

1Iocation °~f sealing faults.

How many image wells should we use in order to get the proper symmetry in the flow field? Since Ei function has

. little influence above certain distance from actual well then


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we can consider a certain number of image wells with no error in the approximation.

Fault#! Fault#2

<-······································ ................ ............. ><·································· ·······> -<········································· ················> -<········································> 2~ 2~ 2~ 2~

Exercise#21 Assume a well located in the center of rectangular shape reservoir. Consider sufficient number of image wells and find pressure at the well for typical values of t>O. 02day.

~'--- 2oort

¢ = 22%, B0

= 1.78res m 3/ ST m3

, J.1 = 1.5x 10-3 Pa.s,~ = 3000psi

c=5x10-6 Psr', k = 80mD, rw = 0.25ft, q = 300stb/ day

A well close to constant pressure boundary A -similar ~PP~~~~h as before can be used for the case where an active well is close to a constant pressure


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boundary (e.g. fanned by a gas cap t~at borders a slightlydipping oil reservoir). In that case the Image well should be an injection well (-q) rather than producer.

Exercise#22 show that for a well close to a constant pressure boupdary for the late time solution using semi-log -------------- ··· --- . ----=-----------· approximation the slope of draw down curve o~a _§~mi log

-----------------··---- --------------------------~- - ... --- - -.. ... .. ------ -------~

plot is zero. · 1

,

~----- -----..... _.....- - Y 1 .,.,._ z) /'t2--:)l~-t r4! be v. ;;..~ c.> - ~~- ~~; L.O Vi )t~ vf.··' 1

_ .. --~- (! . '{;; J6 P,a Cj -~;.. ... --~ -- ----------. -----·

- · -:~,- : :f~ jJ'~ )_

Department of Chemical & Petrolewn Engineering, Sharif University ofTe<;hnology, Tehran, IRAN .. C <>' 9 < ___.. 104 ' , ___ __ ___ __.

---· l -~- ~--· - - - ... ···-- -·-- -~----

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5-Pseudo steady state single phase flow (pseudo steady state or sen1i steady state flow represents the condition that exists when the pressure throughout the r~ir changes linearly with time.) As said before this does not occur in the reservoirs with incotnpressible fluid therefore applications of the semi-steady state flo\v to describe the flow behavior include slightly compressible fluids (either linear, radial or spherical geometries) and compressible fluids (linear, radial or spherical geometries).

Here we only study the problem of slightly cotnpressible fluid under radial geometry. For a well located in a large reservoir and producing at a constant rate under unsteadystate regime, there is a pressure disturbance in the reservoir that travels throughout this infinite-size reservoir. During this transient flow period, reservoir boundaries have no effect on the pressure behavior of the well. As soon as the pressure. disturbance reaches all drainage boundaries, this ends a different flow regime begins that is called pseudo steady (semi-steady) state flow (see figure below).

No-flow Boundry

__ .., ...... ---. ·~--~ ------ ---~ -·_{/'/

.'~--'l_.!/

'.l;'-1

1//

I'I_,

.~,_,

/.,:,

•'/, j ........_ _____ .........___ _____ ____.,.--

. rw r r,

A)Pvs.r


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Reservoir Engineering I, Mohscn Masihi

At radius r --

(OP\ ----:- ) . =Constant OL ' '

tl t2 . t3 Time B)P vs.Tinie

5-1 Slightly compressible fluid and redial geometry Consider radial flow regime with semi-steady state condition into a well which leads to an equation for pressure drawdown in a well.

q =constant

Pressure t r--...,;..._ .

f h

I..____._ _ ____, r., r~

For a well producing at a constant flow rate under semisteady state the drainage volume is related to the pressure decline through material balance. . Consider material balance on reservoir (tank model) with an average pressurep where the total amount of oil is M

dM = pV r/Jc1

dp =-pq =mass flow rate out of reservoir dt dt

-ldV · dp dV c, = V dp => c,Vdp=-dV => ctV dt =- dt =-q

dp- -q - -5.615q dt - V ¢c

1 = trre2 h¢cr

A = tr re2 drinage area

Department of Chemical & Petroleuq~ Engineering, Sharif University of Technology, Tehran, IRAN 106

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Because the pressure at every point in the reservoir is changing at the same rate so, it leads to the conclusion that the average reservoir pressure is changing at the same rate. Substitution in diffusivity equation gives:

!~(r.·opJ = p.¢.ct op = -5.615qp r ar or 0.00633k at rcre2 kh

Integrating gives· (r. opJ = - 8873 q; r

2

+c. . or TC~ kh 2

Using outer boundary condition (no flow):

C = 141.2 qB,U l kh

Substitute C1 into the equation and integrating once again,-

op = 141.2 qBJl (~---;-J or kh r re

q J1 ( r r2

) p- Pw =141.2- ln---2 kh rw 2re

2

where r~ negligible re

for r=re

. / / • I I . / :-'--- -

• J • I . ./' 0 ...,:f.-e.-If . .__-o/ .;.. /' (:) ,, : £- r~/ / .-- \:,,.- • c\c••"J r / ~~-

-,..., ,._,____-~ 1 l--.. k p ___. u - r: ~.,.-e.·\\. 1--e,\<~, ; I c. ~- 1

The outer boundary pressure cannot be measure directly so

it used to write the dr~~~9.~Q_jn~~-=-&h~r_~_.1he~verage p~ssure obtained ~wel_L~t~s._tiil_gj A$--_- Qe~?re, the av~rage press~r~_£~!?:- ~e -~e~~E~!_l).ed as: -~ -f-~" v~~.\.1.. ~-Department of Chemical & Petroleum Engineering, Sharif University of Technology, Tehran, IRAN 107.

/

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ReseiVoir Engineering I, Mohsen Masihi

Note that P-Pw and P-Pw do not changing but P,Pw,P do change.

Note that the skin factor, if exists, can be accounted for by having an artificially changing well radius (called effective wellbore radius due to skin

The alternative solution for semi steady state flow regime can be obtained by solving diffusivity equation provided the appropriate boundary and. initial conditions for a well in a circular drainage area. ~onsider a well located at t ~enter of a circular servoir with no flow at the outer boundary and___Q_onstant flow rate into the wellbore; then, tile governing PDE with initial and boundary conditions are,

1 a ( aP) ct¢11 aP r ar r ar == 0.006328k at


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BC# 1: constant flow rate at the well bore khrw apl yields ap

q = o.oo7os a J1B r r=rw ar r=rw

qJ1B

0.00708khrw

BC#2: no flow boundary aP

=0 ar r=re

IC: initial pressure P ( r, t = 0) = Pi

(ijgrst an§ van Everdngell) ( 1949 found the solution of

~~d~ow~n~·~~th~e~w~e~ll~b~or~e~u~n~d~e~r~p~s~eu=:o~~s=e~a~y~~st~a=te c~ditiqns _(i.e. late time when the boun pressure disturbance) as,

·J u:' '~~ !.,'

\ / 'Ai I'' .j\ \· J'' ':'/' 0' ~ •/ c 11

@ -~ Y- /. 141.2qj1B ( re 3) qB

Pwf =Pi- kh ln~_-- -1.79 h 2 t rw 4 Ct¢ re ---- -- ---- ··- -- ·------·-----------

\)' .:i ", \f Sn"" ~·r(l' .. c:1

{J ~

This shows that w~llQ.gre pr-essi!re-declipe~ lineatJ.y_ with (~~ /' -- \ ... '-'

. .. dpw _ qB ) . \,;): tune -~-Q.~ -~lope ~s dt - -1. ~-~ re2 h¢sr '-which shows that

the slope of the derivative curve atlate time can gives the drainage a~ea. We also remember frotn material balance th_~t the time t is,

CtAh¢(Pi- Pr) Ct(nrl)h¢(Pi- Pr) t= =-------

5.615qB 5.615qB


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Hence, by substitution we get the same equation as before, kh P- Pwf

q == 0.00708 [ r. ] JiB ln _g_- 0.75

. rw

5-2 Effect of non -circular drainage area Often we see the cases where the drainage area is not circular and/ or the well is not located at the center of the drainage area.l A simple case rriay be a well bounded by~~ set of linear faults which may have a polygon drainage area./ As before, for this case we can use the superposing solution of image wells.

d well d --- -~-- ·-----------~--------

~~

For a well in a rectangular reservoir, we can use an infinite array of image wells arranged Of1 a squ~re lattice to find

. the solution by us~ng superpos-ition~B-~fore -thepressure pu_lse r~aches the h<llindary~j!!!_e_i_~_!!lfl~j~~--~~.~-~ry9ir behavior as,

P. = P· + 70 6 qJ1B Ei (- r~ ) · w l • kh 41]t

Let's(@'~ the long-!ime sellll_steaAY_ state solution as below, -


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p = p. _ 141.2qpB (ln re _ ~) _ 1. 79 qB t wf l kh rw 4 Ctcf>hri

With re = -JA/rr And except that the constant term may be different · so, incorporating the constant in the logarithm term yields,

2

1141.2qJ1B ( re ) qB Pwr =Pi-

2 kh ln - 3- - 1.79 2 t

-4 Ct¢hre e rw 1141.2qJ1B 4A qB

Pwf =Pi- 2 kh ln 3 - 5.615 Ctcf>hA t 4rre2r~

3

Defining t = 1.781 (Euler's number2 and cA = 4:ez (well

centered in the circular reservoir) and replacing them into · · the above equation gives,

70.6qJ1B 4A qB Pwf =pi- kh ln yCArJ- 5.615 Ctcf>hA t

Pi: Psia, Pw1: Psia, q: STB' 11= cp, B: bbl, k: mD, h: ft, A: te, rw: ft, Day STB

tt: Psi-1, t: Day ;} z ~ . / r/ _ , r/

V:JJ(__/J/~':-?' t y) 7t:_P~ r:-Wi~h the same units except for the tlme which expresses in hours we have, - --P 162.6qJ1B l 4A 5 5 qB If). 5 .S

wf = pi - kh og yCArJ - '61 Ct¢hA t

Hence, rather than developing a separate equation for each geometry; a correction factor that is called the Dietz shape factor, CA, is used. ~onsidering t to be the time for whic_h t~e well 4 has been at a reasonably steady state rate of production then using average reservoir pres·sure this is, --------------~~--~~------~------------

· Department of Chemical & Petroleum Engineering, Sharif University of Technology, Tehran, IRAN Ill

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70.6qi1B 4A Pwf == Pr- kh In 1.781CAr~

Using the material balance and rei?lacing the average reser~ with initial reservoir pressure gives,

!~P.~P 5.615qBt ------ ;1"~1~ dti,~~{;:_ 1/14R.-/

(__ r - ;(- C~5qBt) 70:;JlB if~f', :if Pwf == pi - CtAh¢ - kh In 1.781CArJ

This shows that for a well at the center of a circular reservoir,

3 4rre2

CA == == 31.6 y

The values of shape factors for other simple geometry are,

CA = 12.9

EB CA =22.6

[- --------f _________ ,,

• ' __________ j __________ _


' ' ' ' '

112

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Magnitude of CA depends on shape of area being drained andJ_he po.s-itien ofthe _w_e_ll~w!Jh respect to the boundary. Shape factors decreases as:

-drainage area become more non-circular -location of the well become off center

CA Hence, by determining J?ieJz----s-h -- -reasonably estimate th drainage vo u e.

one can

Example: a well has been on production in an oil reservoir for 100 hrs. Available data are: ¢ = 21%, Bo = 1.4 rm3 I stm3

, h = 15m, )1 = 2 x 10-3 Pa.s,

C=2x10-9 Pa"1,-P;=240bar, q=200stm3/day, rw =0.l5m

thr 0 1 . 3 5 8 15 30 50 . 70 100

Pw(bar) 240 201 199 198 196 193 185 176 170 159

(a) Check the applicability of_Jine source solution and log~rith~i~_~proxjll}ation validity.

lL s VU--

(b) Estimate the permeability, skin and drainage area using ~•; provided data. ---- ... --· - ·----~------- 'v()-.~ t'-. ?¥'

. 6' )'l 1 \ ( 1\ \o C.."

b.-~¢s \) . 0 ~ --? Solution:


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Th 1 · m- -jlq e s ope Is - 4trkh = -1.857 => k = 185mD

210

y = ·1.8571x + 201.01

200

190

3: a..

180

170

160

150

0 2 3 4 5

Ln t

To determine the skin:

2S =Pi - _Pw (t = lhr) kt 2-0.8091 => S = 3.6 m · . J.-U)c1r iP

To d~termine the drainage area we look at the late time behavior of semi steady state regime

- pq ( 2trk 11 ( 4A J sJ p=p.- t+-n + w

1

2trkh p¢c1A 2 yCAr; .

From the following graph the slope is,

\~ ·m' = -q = -0.36 ___ .. · h¢c

1A

:::> A= 51440m2

Department of Chemical & Petroleum Engineering, S]larifUniversity ofTechnology, Tehran, IRAN 114

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210 .----------------.

200

190

180

170

y = -0.3645x + 195.28

160

150 -1-----r---.---------r------r--~----~

0 20 40 60 80 100 120

time

To determine CAwe extrapolate this line to t=O which

gives Pw =Po= 195bar. Hence the equation becomes:

Po= P;- Jlq (In 4~ -InCA+ 2SJ

47rkh rrw

which gives CA=2.08

Exercise#23 For a well situated at the centre of a regular sh~ped drainage area find an approximate transition time from transient regime to semi steady state regime in terms

f d. . l . d fi d 0.006328k o a new Imension ess time e 1ne as tvA == r~. t J1Ct't'A

(hint: equate the transient and semi steady state solutions). Then with the following available data, estimate this transition time for a well situated at the centre of a square drainage area of side size (1OOft or 200ft). ¢=30%, k = 50mD, JL = lcp,c=l5xl0-6 psi-1,CA =30.9, rw = 0.3ft

Depanment of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN 115

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Example: Compare the time to reach pseudo steady state for a well producing in an oil reservoir (with f1 = 1.5cp, c1 = 15 x 1 o-6

psr•) and a well producing in a gas reservoir (withJL = 0.015cp,cr = 400xl0-6 psr1

).

5-3 Compressible fluid and redial geometry Again for compressible fluids ·such as gases, the definition of semi steady state regime is the same except instead

of~: the value of ~~ is conStant. Hence the same methodology as described for steady state flow regime can be potentially applicable here.

Exercise#24 Derive an expression_for radial flow of highly compressible fluids under semi steady state condition. Show all steps clearly.

5-4 Productivity"index and IPR curves ~~1ponly used m~_9sure_o£_the_ability of the well to produ<?~ i~-- th_e P~~~llct~Y..i..!.r __ !_~~ex( J) This is defined as tlie ratio of the total liquid flow rate-to._Uiepressure drawdown,


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0

ReseiVoir Engineering I, Mohsen Masihi

J =- q_ (P- Pwt )

It_ is obvious t~~~-- durin_g_!E~--!~~!~~!~g!_ _fl_ow period, the calculated values of the productivity index will vary depending upon the time at which the measurements of Pwf are made. Hence, practically this productivity index shows the well productivity potential tneasured at pseudosteady state conditions. f.gr_this_flo_w_r_e_gime in presence__.9f datnage \Y_~--h~ve, . . .. -------~ ---o:-t)670-8kh P - Pwr --,,"\ q- I

' · - J1B ln re - 0. 7 5 + S ) rw

The above· equation is combined with the productivity /"' index to give, . _____.-'" <'

I F' . 5 .1 r'~// :.~ ~ ;(-

' lr' / . --

0.00708kh STB .--- r:(J·t/Jcs_Y· ·,(Q 1'1 c:~· '· -'- _ PI- ~. /~ ~ . ~/. · .-

- J1B (In re - 0.75 + s) Day. Psi /0t ---u;- /a>?~;/ f)t~-a: ~I (r:v ~ rw '· ..-·e:_

Since most of the well life is spent' in a flow regime that is approximating the pseudo steady-state, the productivity il)dex is_a valuable tnethodology for predicting the future well per_formance. Ft_!E~h~~~'--~y l!.!QQ.!_!g_t!_QgJhe . pro_du~tivity index during the life of a well, it is possible to see if the well has-become __ .. damaged due to completion, workover, production, injection operations, or mechanical problems.

A comparison of productivity indices of different wells in the same reservoir should also indicate some of the wells


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Reservoir Engineering I, Moi1Sen Masihi

might have experienced unusual difficulties or damage during completion. Productivity index can be increases by: a) removal of skin (acid treatment) b) increase of permeabthty (~draulic ~~~ c) viscosity reduction (thermal stimulatio~) d) reduction ofBo (choice of separator)

Since the productivity indices may vary fro1n well to well ------- ~-because of the variation in thickness of the reservoir, it is helpful to Pofmalize t~indi~-es by dividing each by the ..._____ - -thickness of the wei . 1s Is defined as the specific

-~---·~·-........

product_i~i~y_1J.~<Ie:x]5 , f>. 2) -- ~ · ·- ... _ - -~ r _ _ q o.oo7oak ils-------

1 ·"---~---- ---~-~--=--~~~!-·--·-- -~-~-(~-~-~:--~ ?.~? 5 -~ -~}-Assuining that the well's productivity index ts __ const'!nt and rewriting, _:.r

fD · ) - (1) /I Gi!(L;_,,.L ;,'L q = ]\..r - Pwt or Pwt = P- ] q · 7/.

The above expression shows that the plot Pwt against q is a straight line with a slope of -1/J as shown in below .

. . · ~(~Jp. Gt~~-$ -~.~.Cu;: ; ,,-'-~ ; &te. S . .... .. . .. 7-

bt>l/d.~y


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· This graphical representation of the relationship that exists between the oil flow rate and botto1n-hole flowing pressure is called the Inflow Performance Relationship and referred to as IPR. ..----------·~ '

Lv£t ) • f(.,L "j1 CL{ ~ /', A or: J7Q l-en ll Ctt ----" r I .,.> L/

J'iote tha. t~J} __ ~ . __ .f · nd !!.' = AOF absolute open flow which is ~el~ ___ pgtenti~! or formation potential. Although in practice this may not be a condition at which th_e well can -produce, it is a useful definition that has widespread applications in the petroleu1n industry (e.g., companng flow potential of different wells in the field).

Example: Assume pseudo-steady state and drainage radius is 2980 ft in previous example. What portion· of the pressure drawdown is lost in the skin zone?

Solution:

3.7 0 3 0 --::-----:---- - 1 or 31 Yo

In ( 2980

J' -I+ 3. 7 - . 0.328 4

Example: Assume that the well has been stimulated ( acidized) and the original permeability restored in the skin zone. By 'Nhat percent PI increases?

Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN 1!9

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Solution:

ln(-re )-l_+s In( 2980 )-l_+ 3.7

rw 4 _ 0.328 4 = 1.44 In(!§_)_~ ln( 2980) _l_

4 0.328 4 rw

or 44%

Exercise#25 A well-reservoir system is given with the

following data: re = 11001 ,pi= 4200psia,k = 75md,J.1o = 30cp

h = 481

, rw = 3.511

, Bo = 1.25bbl I stb, f1p = 900 psia What is the rate

of production at ideal condition (no damage)? What is the &ate if the well is damaged so that the average rock

I / fse~e;~i~i;~~~s r:~~t~~;;:!~~ ~~~: ~~!~~~?v~~:~ :~:~ ~ damaged radius if damaged zone permeability is reduced ' to 12% of its original value? What is the flow efficiency

~\ (FE=PiactuaiiPiideat) of the well?

·Example: A productivity test has been conducted on a well. The test results indicate that the well is capable of

Producing at a stabilized flow rate of 110 STB and a .. Dey

bottom-hole flowing pressure of 900 Psia. After shutting the well for 24 hours, the botton1-hole pressure reached a static value of I 300 Psia. Calculate the productivity index, AOF, oil flow rate at a bottom-hole flowing pressure of 600 Psia and wellbore flowing pressure required to produce 250 ST B 1 d.

Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN . 120

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Solution: 110 STB

J = = 0.275---1300 - 900 Day. Psi

STB AOF = 0.275(1300- 0) == 357.5 -

Day STB

q = 0.275(1300- 600) = 192.5 _. -Day

. yields 250 == 0.275( 1300- Pwr) Pwr == 390.9 Psia

The effect of skin can also b~eell_Q_n pro4!:tctivityjndex as seen in figure below:

IPR for saturated oil/gas wells may not be a straight line and it may exhibit some curvature. Then J can be defined as the ratio of the rate to the drawdown. A decreasing in~ productivity in_d~x ___ (or_i~_creasing in rate may hap_Qen. (; Several __ _ ~!:l_'!!i~Q_Lpaye __ q~~n suggested to _ repLesent) n~PR in presence of gas and two phase flo~ (e.g.0 Vogel, Standing, Wiggin and Fetkovich etc.). For example, Vogel equation (1969) is:

Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN 12!

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Knowledge of IPR behavior need in the regulating allowable production in a well and correct forecasting of a well and field producing potentiality. In a radial flow most of pressure drop occurs close to wellbore. Suppose a reservoir with Pres> Ph but with P w <Psat then as oil approaches the wellbore gas come out of solution so GOR is increased and J decreased as the drawdown pressure

. increases. As long as P w remain about the psa\ J will remain fixed so this portion .of IPR curve is straight line.

In the case of stratification (layer reservoir) the IPR curve can also be obtained. Consider 3 different zones with k1=IO, k2=100 and k3=1 md. Figure below shows the individual IPR's of these zones . The composite IPR will be some of three curves.

Department of Chemical & Petroleum Engineering, SharifU~iversity of Technology, Tehran, IRAN 122

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ISOO

1200

~1000

.,j 1\ I \ I \ t- \ I'~ I \"',

I I ', I \ ' I I ', 1 I ' I 1 ' I \ ' I I ', I I \.

: ' ', '1-md l1o :m~ '\ IOO·rtld ' IM< I :ot~e '\ :ot~e : I '\ I I \ l ~ \ \ I \ I \ \

o~~---~~~~~~~~~----~LProd~ction rat~. t>~/d1y

Product;on rate


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6-Two phase fluid flow in porous media So far we have studied a single phase flow; however, in reality there is multiphase flow in any reservoir such as oil/water or gas/oil etc. N_Qte that we have represented the conductallce of a porous media by a paratpeter~ill!~d

~~~Now we define f!!ective perm~abdit~ kef!

as a measure of the conguctance of the porous medium to one_f!uid _g_hase in the presence of the others]Aiso relative

permeability kr is defined as the ratio of the effective permeability of one fluid and the base permeability (e.g. at 100% saturation) that depends on the phase saturation.

With the concept of the effective permeability to each phase and the associated physical properties _Ft;; CilD extel)g the applicability Q-f-W Darcy's equation to rnu1ti PQ.~~ ~For a radial system, the generalized form of the Darcy's equation is,

Water phase Flow:·

Oil phase Flow:

Gas phase Flow:

Vw = 0.001121_kkrw o:w Jl.w vX

V = 0.001121 kkro oPo o Jlo ox

Vg ~ 0.001121 kkrg oPg Jl.g ox

N~~ that pressures in ~e at each point in reservoir are related by capillary pressure ~ (defined as the pressure difference between non-wet and wet fluid) where kro, krw and Pcap =Po- Pw are functions of Sw and reflect the surface and wettability forces of the fluid-rock system. Department of Chemical & Petrolewn Engineering, Sharif University ofTechnology, Tehran, IRAN 124

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0

k';o I .

' I I

lrre.:ludble vVater

Saturation

· End point relative

k'rw parrnabihty . • ~---'---f--1 ..

Residual OH

Saturation

Also the definition of Ct is different from single phase as it . IS,

Ct = SwCw + S0 C0 + S9 C9 + Cr

Water flooding is one of the most important with which we can describe multiphase flow. For _this case and also for the gas injection process, analytic methods are used for analyzing displacement-type reservoir. Water flooding were done under fixed reservoir pressure (above the bubble point), at a fixed oil rate or total rate (oil and water) until decline. But availability of injection fluid makes the water injection as the main intervention method used in reservoir development; however disposal of produced oily water to the sea is one of the challenging and need a recycle process where rather than a once through process the water is re-injected into the formation. In this chapter


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we develop analytical methods for estimating the oil recovery in such situations.

B_cl_Qif3 start let's rev_!~_\:Y __ some_ b_ffi_~fits of water injection: Zone i~-olatzan:- -·Even with an activ~---aquifer with energy support faulting can result in isolated zones.

Sealing Faull

. . kmu

Oil viscosity: the endpoint water/oil mobility ratioM = k~:xJ.Lo <...... _____ .. -----····· ········· ·----··· ···· ···-··· .--···-· ·····---- .... ... --- -- · .. . . . ·· - · ro f.Lw

affe9t flo_w__hehavior. __ at_microscopic scale as shown below but at reservo~c_sQale_Qther_p~r.~m_e_te_rs_j._e_!_ _h_C!Jerogeneities aricfgrav1tY-~ffec_t.' -- ------------·-

lnrection PrOduction Injection ProrJuction

Oil

M<1 M>l Stable Dispfacernenl Unstable .Displacement

The residual oil at__mi.cro.sG.Q.P_i_~ __ )~y-~j __ js_ ____ b_o_ld _ _l2y the competition between Jbe)qt~rf£!~jal tensio_n_fQrQ~_s__Jin9 the

- ·-·--- - __ _______ ... _ ......... ~------- . . . -- --- . ·-· - ·· ··-··· ·-------


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viscous flow forces as_~.Q_gia.t~~:L_ with fluid flow (Sor~ 10-40% of the pore space).

S•.vept Zonr::

At the field scale formation and well locations causes some of the rock to be unswept by the water leads to two residual oil saturations in the swept portions oil at Sor and in the unswept portions oil at Swc

\Natt2r · Oil at Flosidu<::.l

We ~rite the governing PDE flow equation at larger scale b~t th~~a1r-scafep~y_$I~~-- ~r_e ___ f~pf.eserit~([.iri~iirms: of](r and Pc curves.

Department of Chemica(& Petroleum Engineering, SharifUni~ersity ofTechnology, Tehran, IRAN 127

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6-1 Buckley Leverett analysis 1?1!Pkl~y_:_LeYer.:ett.the.QrY-(19.42)js _has.~d __ Q.Q._!h~_Eo!!_~_~Qtof relative permeability that is _ used to d~scribe the diffuse flow--dls-iJi~ce~ent.-\- c·c;·n.sraer:··--t~0-=})i1as_e __ -- Iilcompressib Ie flow with no capillary pressure ( field-scg.l.e=waterflooding) in lD. This also called compl~te1~y![~persed--rn~ From M.B on an element in lD can be written and solved to find the saturation profile as a function of distance and time.

,'"1 A

I I I I

' ' x' X

1

+dx

PwUwAix dt- PwUwA/x+dx dt = Pw·SwA.dx.¢/t+dt ~ Pw·SwA.dx.¢/,

- apwUw = do apwSw _rP 0~ ax 'jf at _ 1 (\ . v-'.~

For constant density, \~) ~!177 !l

¢ OSw + Ouw =O ~;;// at ax .:\ >~

Similarly for oil: -< 15~· /:v

¢ as 0 + auo = 0 >./

at ax Adding these equations gives:

a( Sw +Sa) a( Uw + Ua) ¢ + =0 ar ax


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Defining total velocity as~ gives, ,_ aut ;ax ;!6 ~ Ut = cte

Means that 1 velocity in every cross section is constant in lD.Using Darcy's law or water an oil g1ve: ~---·-"-·· ~~ . _ Kkrw (apw )

Uw- --+pwgx Jlw ax

Kkro (8P0 J uo = -8

+pogx J.lo X


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Defining fw as the water fractional flow, uw = fw.ut

Rewriting the above equation,

Uw = Aw + AwA0 K[ 8Pcap ( _ )] 1 1 a gx Pw Po

Ut '"1 '"tUt ~

fw=~(l+~;K[a;P gx(Pw-Po)JJ For a reservoir with constant dip this fractional flow equation become, ........ c· -- .,,.:_ .

fw = _Aw (1 + Ao K[OPc- /£;~sin~ JJ .. \field units) At Ut ax 1.0133 X 106 ·

1

'•, ·_.

It shol!ld. beJJoted th~t allthe .~ariables have Darcy's unit . .····-· - . - -----2--- --------------------

system in above equation) (i.e., g in em and u in~). The ---- s s --

gravity term is positiv~ _ for oil _displ~cement in updip direction 8>0 and negative fo(4owndip 8~111

-------- --- ---- ·-·-····· -·· ··.

Department of Chemical & Petroleum Engineering, Sharif University ofTec~ology, Tehran, IRAN 130

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pefine a gravity number:

N _ K6.pg~ { edge water drive (North sea)

G- e.g. put bottom water drive (piston like) N G = 10.3

No =0.22

This term not only considers gravity effects but also i~hlcl~-~-~--~---yelocity term. Rewnte water mobility to -total mobility and inserting it in the fractional flow equation,

a) If water injected downdip then the gravity number IS

po~ve i.e. reducing the fractional flow of water. . · · ··~

( ape - ape as~ b) The capillary pressure term_.- ax - asw ax is positive i.e.

increasing the fractional flow (b~t - -is .. very ~mall).

I I I ··jO 1 ·L C I I I

' r l I I I I I


X

131

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c)IAtypic~ract~al flo~~(8 = 0) as a function of ~ km~ .

t~_e_ endpoint mobility ratio M = k~~;: . 1 r-~~~--~~==~~

fw

0 L-~~~~--------~-0 Swc

Viscous fingering can take place if the mobility of displacing phase is much greater than the mobility of the displaced phase. ~ typical fractional flow curve as __ a function of n = NGk;;ax sinB fo!__M 1.

0 ~~~------~-----r------0


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For sufficiently strong gravity (fw >·1 and fo <0) ther~_js counter-current flow of oil and water i.e. water moves downhill while oil moves uphill. {lfc;-t~ -- that the relative permeabi · tio rna be expressed as a funcfloi1o-f-~~-ter

------ . !k -bS saturation b ' krw ro = ae - w)

Buckley-Leverett solution _ T~~_$olution for the fractional flow equation (where the

total- flow is constant) under the diffuse flow conditions ~ich means that the saturations at any point in the 4irection of linear displacement are uniformly distribut~d over the thickness can be easily obtained.l Thi_s __ ~D?:~J~~-~ lD analysis for the flow dispj_~Q~ment modeling . ... _.____ -----------------·-·-----·-· ·· · ··· - - - -------~------- - ---- · ·

This is where the injection rates are high preventing the establishing of vertical equilibrium and for low injection rates where the thickness of the reservoir is small compared to thickness of transition zone.

\ \


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Re-writing the saturation equation,

¢ 8Sw + OUw = 0 => asw +!i_ atw = 0 at ox at ¢ ax

8Sw+v8fw=O at ax

where v = uw I¢ is an intersHti,~l-:elo;i~~~his is, ------- -----------·-··-·---------------~~

asw +v ofw 8Sw = O at 8Sw ax

I.C & B.C are:

{

Sw(t = 0) = Swi = Swc

Sw(x = 0) = Swo = 1- S0 r &fw(x = 0) = 1 at well

Using dimension less variable xn =xI L

; JytD = ~ 1 vdt= 1~!= 1 ¢~dt=* 1 qdt

11 ~ ' s~-is the eore volumes of flui<!_ inje~). Dimension less form of the equation will be,

8Sw +v 8fw asw · O => 8Sw + Bfw 8Sw =O at asw ax - 8tv asw 8xn

Define the new variable as, vn = xn I tn

asw 8vn + Bfw (asw BvnJ=O Ovn 8tn asw Ovn &n

dSw(- vn + 1 dfwJ=O => dvn tn tn dSw _

dSw ( dfw J-o - - -vn+ -dvn dSw


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Non-trivial solution is,·

This is B.l,_ equation which is velocity of a plane of ~~turatioii- Sw moving through the linear system. Hence, for a constant rate of Water injection the velocity of a plane of constant water saturation is directly proportional to the dei1Y.ative of the fractional flow equation evaluated for that saturation.

a)Very viscous oil

f.::: 1

I

I I

. I I I

s _ __.... 1 - s.,

~ ·\ !\

.. .____ __ .)

'

T~ ___ maximum veloqi~y is at Sw just greater than Swc and decreases to a minitnum at Sw==l-Sor.

Breakthough in very viscous oil occurs with low Sw (just greater than Swc) and then gradually increases.

Department of Chemical & Petroleu·m Engineering, Sharif University ofT~clmology, Tehran, IRAN 135

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VVater lr~jector Producer

X

b)Very light oil and l~rge gravitationa! force: f.,= 1 -------------------.7:

I :

/

I I I

L

s... s. -- 1 . s.. S~, S .• --+- 1 · S,

The max velocity is at highest saturation Sw = 1-S0 r related to a _guick build up of a shock so until the shock front arrives only water free oil is produced.

Water Injector

X


Producer

136

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c) Typical medium viscosity oil I density. (S shape): f .. = 1

------ -------~

I I I . \b ..

Water saturation profile in the displacement path indicates that multiple water saturations can co-exist at a given point in the reservoir which is physically impossible

What actually occurs is that the intermediate yalues of the water saturation which have the maximum velocity, will imtrnlly tend to overtake the lower saturations resulting in

. th~ fo~~~ion of a saturation discontinuity or shock front. To draw the correct water saturation profile using the BL requires the determination of the vertical dashed line such that shaded areas A and B are. equal.

Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN !37

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v\latcr Injector . . ProducEH

l X

Capillary and gravity may modify the sham front as shown -~

in figure below. It is clear that these cause a smearing effect on the interface .

. 1_- S~-

~\-Gravity

. Capi. l~.!Y

... ,...-Ct. '""-v .,,; ..,._ _____ __,_...;;...__ __ _

6-~eige ~ns~ Because of this discontinuity BL approach which assumes that saturation Sw is continuous and differentiable; wilf f?e 1nappro priate to describe .the.._situatiQll_Jl_!_!he __ _fr_2_~!_j tse I f. ~magine a shock moving at sj)eed Vsh-:j

Department of Chemical & Petroleum Engineering, Sharif University ofTechnolqgy, Tehran, IRAN 138

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X

Considering the change in mass as the shock moves in a time dt and writing M.B,

ut (Pwlt- Pwl~) = Vsh¢( S~ -S~ )Pw u1 (1t- ~~) u

1 !lfw

vsh = = ¢ ( s~ -s~) ¢ l'hlw

(Rankine-Hugoniot condition)

R{Sw =Swc r -0 Jw-

Swr is the saturation of the shock front. This can be found from the following graph.

J):f fwls _ Yw _ wf 1 or=-~----

Sw lfw =1 - Swc

Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Teluan, IRAN 139

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·. ------~-------------------; sw1, f,.. •. Is 4' II . . c •

. ~ I : I l I : I

I I

1 - s~ s ... Breakthrough occurs when x=L or xn= 1 so,

1 Swf -Swc vn=- ~ 'n= I

tn . fw s wf

Vfelg~ _ _(1952) construction c~sists of integrating the satl1I<!tiPlldi~tribution over the distance from the iQjection point to the~ front thus obtaining the 9.:_yerage water s~turati . ront used to calculat~recovery .... Water saturation at the front (Swr) and average saturation behind the fluid front can be determined graphically.

I · S.,t----

s ..

s...:t--------+o--,--.,..---

Department ofChem.ical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN !40

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Consider the saturation path at a fixed time The average saturation behind x1D is:

· xn dfw but we had vD = tn = dSw

One case at front shock,

_ 1- fwls Sw(tD) = Swf + wf

dfw

dSw S wf

dfw

dSw S wf


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Previously we have shown that,

dfw 1 VshD =-d'-(ll - S I S

uw s s w ~'w=l- we w= wf ;,

dfw 1- fwls wf 1

dSw S - Sw-Swf - Swlr =1-Swc wf Jw

Graphically the extension of the tangent to fw == 1 gives the average saturation behind the front

1- S"'

Before breakthrough oil recovered is simply equal to the volume of water injected no water production during this

-phase so Sw = Swc + t D . At breakthrough time the flood front

saturation reaches the producing well and the reservoir

water cut increases suddenly · from zero to fwlsw1 which

confirms the existence of a shock front.


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Exercise#26 Neglecting gravity, find the Buckley-Leverett shock speed, the water saturation at the shock, and time to breakthrough if Sor = 0.3, Swc = 0.2, kr:ax = 0.8, kr;:vax = 0.5,

f.iw =10-3 kgm-1s-1, f.io =10-3 kgm-1s-1, k=l0-13 m2,

= 10-2 ms-1, L = 0.4 m

(1-S -S )2 (s S )

2

k = kmax or w k = kmax w we

ro ro ( 1 - S or - S we ) 2 ' rw rw ( 1 - S or - S we ) 2

----------------- --- ----------....... .

Pet~I_f!!J~-~!!QJJ of oil - ~~~o~e-ry--

¢ = 0.3 '

Welge's work enables us to e ermine oil recovery). Plot of oil recovery versus time (Npn vs. tn). We know that,

Sw = Swc + N pD . --Choosing different saturation points __ Swt

(e.g. every 5% above the saturation at breakthrough) then from t D = l I f~t find Sw and hence Npn - . -

_ _ 1- fwls N pD = Sw- Swc = ( Swl -Swc )+( Sw -Swl )=(Swl -Swc )+ r

- . . dfwl dSw swl

t f;,

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Recovery can be plotted as a function of time. Before tbr . recovery is linear (oil recovered=water injected) after that recovery depend on fractional flow curve above the breakthrough saturation.

Brcaktlrrough

Time ---.-

Exercise#2 7 For each of the cases below provide a graph of: (i) Sw as a function of v0 =x0 /t0 ; and (ii) pore volumes of oil produced Npn as a function of t0 . In all cases Swi=Swc=0.2. There is no gravity and the relative permeabilities are given by:

k = kmax (Sw -Swcr k = kmax (1-Sor -Sw t M =.&_ e;~ax rw IW ( 1- sor - SWC r , ro ro ( 1- sor - swc t ' Jlw k,:ax

a) A strongly water-wet rock with a= 3, b = 1, k~ax = 0.18, k':oax =0.9, Swc = 0.2, Sar = 0.4 and Pw =f-Lo (M = 0.2). b) An oil-wet rock with a= 1, b = 3, k::Vax= 0.9, k::ax=0.18,

Swc = 0.2, Sar = 0.1 and l1w = Jlo (M = 5). c) Repeat part (a) forM= 5 and M =50. d) Repeat part (b) forM= 0.2 and M =50. e) Comment on the results. Is it better to waterf1ood a water-wet or an oil-wet reservoir? What is the effect of mobility ratio on recovery? Department of Chemical & Petroleum Engineering, Sharif University ofTechnology, Tehran, IRAN 144

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Example: For a constant rate water injection 1,200 bbl/d/well in a direct line drive in a reservoir assuming that diffuse flow conditions prevail and that the. injection project starts simultaneously with oil production from the reservoir, determine: 1) Time when breakthrough occurs 2) Cumulative oil production as a function of both the cumulative water injected and the time.

Data Sumnmr) Sw Krw •/ "ro

Watl'r visl:osit) u :::: 0.5 cp 0.20 0.000 0.880 ~ \\' 0.25 0.002 0.671 ()j! \'JSCOSilY ll =: 4.5 Cj) 0.30 0.010 0.517 '6

luitial H atu s.al urati on $\\·(::::: 0.20 035 0.021 0 .401

l~csidual oil saw ration Sor= -o ?(' 0..40 0.0:35 0.314

.... J 0.45 0.054 0.242 I \:r;~si t\ <.p:::: 0.22 0.50 0.079 0.179

Dip angle· 0:::: oo 0..55 0 .105 0.132

0.60 0.139 0.089 Rr~tT\ 'Oir rhickm·ss h :::: 50 fl 0.65 0.1/9 0.055

Di\lilllC\.' bri\H'CJI ir1jrctiun wdl ~ \\ 800 ft 0.70 0.218 0.030 - 0 "" 0.264 0.011 _/0

Di~taJII..\' - irl_jrctors it ltd producrrs f.;;:.: 2.000 ft O.BO 0.315 0.000

Solution:

Viscosity ratio is Jiw I Jlo = 0.11 and fractional flow equation

/, -11(1 + krof-lw J (horizontal flow) w_- krwf.lo can be calculated and

plotted versus Sw. From this graph:

Water saturation at breakthrough is: Swbt = 0.45 Fractional flow at breakthrough is: fwbt = 0.665 Extrapolated Savg at fw = 1 is: ··Savg = 0.572


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Fractional flow versus water satmat.ion 1 . . : ... . ..

0.9 I I I I I ------,-----·-•T·-·----·r·------, -------·r•• -I I I I •

0.8 , Water breiaktrougb ::-• • t I 1 I I I

............. ~--------- · ~ .................... ~ ....... · ..... ~................. •••••• ................. L .................. I ................... ~ ............. .. I 't I I . I I I

: : ' Ext~rapGiatkd Sw : : 1 , 1 I I t I I I

0.7 -- - - - ~- - -- - - - - t - - ... - -- - -:- - - - - - - - • --- - - -~ .. - - - -- - ~-- -- - -- - ~ - .. - - - - - -:- -- - ---- ~ - - - - -- -1 1 1 I I I I I

1 I I I : ; : : : 1 I I I I I I I I

~ 0.6 ------~--------f-------~--------~-1 ----~-------~--------~--------:--------f------ -0 ~

~ 0.5 -~ ... {;

0.4 Lt

1 I I I. I I I I I

0 0 0 ~· I I I I I 1 I I I I I I I

1 1 I • I 1 I I I 1 I I .. I I I I I

........ - ..... .., ....................... ,. ... ..................... , ................ -~ .. ... ... ... ........ r- ......... - - - ... , ......... - ......... - r ..................... , ....................... " ............... ... 1 I I # I I I f I

.6 I I I "t I I ,.- I I I I I I

• I I I I I I

-- .. -- .. .! .. ------ ... :- .. ---- ... :..: .. -- _J_ ... ~--.. -- ..... :.. ... -- ... -.-- ..! .. --- ... --- ~------- .. : .. - ..... ----!-- ... -- ..... 1 I I .. I I I I I I 1 1 I • 1 1 I I I 1 I I • · I I I I I t I I .. . I I I I I

0.3 I .I I • I I I I I I

-- ... - ..... ~·--:---- -.t- ..... ... --- ~:,_'!- -.. -.. ~ ... -.. -.. --:-- .. -.-...... --: .. --- ..... -.. ~ ...... ---- ·!- ............ --- t------ -: ··: . f : : . : : : : 1 I ..... I I t I I I I t. • I I I I I I I

0.2 -- ....... --\- ------·T- .......... ..;. .. : ................. ; --- ........... :.. ................ ..: .. :. ............ -~ --- ......... -:-- .. ----- ~ ....... ----1 I • I I I 1 I I I I I .6 I I I I I I

0.1

I I ,.- I I I I I I

: : ! : : : : : : : - .. - .. --., .. - ............. ·"" .. .. -.~..... .., ..................... , .. .. .. .. ...... ,.. ............... ., .................. r- .. - ........... , .................... .., ............... ..

: : : . : : 0 45 : : ' : : : .~~· : I •' : : :

0 0 0.1 0.2 .. 0.3 0.4 0.5 0.6 0.7 1

Water saturation

Oil recovery is: N pD =. Sw- Swc = 0.372(PV)

PV = hLW ¢ = 1.76 X 107 ft 3 q = 1200 bbl/d

tbr = Recovered oil 0.372PV = 971 days Flow rate q

G_~mul~tLY~~"'Qil_recovery calculation after breakthrough, Allow Swe water saturation at producing end of the block to riseillii1cfeffien~~--Qf5% and _g!aphically obtain Save then USe rt.--;: r ~ I ; · -~ ' ~! vJ.;.J--:-, t ~ ~JJ 0 ~ JvU.)

../ . rl -.. ......... _ V\0 '-\

Npv = Sw- Swc .... .....__ _______ __./

(Sw- Swe) + (Swe- Sw3Cswe- SwJ + ~;:, f'we

dSw S we

Department of Chemical & Petroleum Engineering, Sharif University of Technology, Tehran, IRAN . 146

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AtSw = 0.6

Since we must approximate dfwf with two known dSw S we

. . 1 1 t d t 0'6+0.

55 0 575 saturation so, It may ca cu a e . a Swe == 2

= . as,

dfw fw,at sw=0.6- fw,at sw=0.55 0.934- 0.878 dSw Sw=0.575 ""' 0.6 - 0.55 = 0.6 - 0.55

= 1.12

But according to above table fw,at sw=o.575 = 0.905 therfore,

1-0.905 Npv = (0.575- 0.2) + = 0.4598

·. 1.12

But as mentioned after breakthrough we have, 1 1 1

tv = Vv == dfw I = 1.12 = 0.8928

. . dSw Swe

Water injected yields R . . = 0.8928 water injected

eservozr pore volume = 0.8928 X 3134461.2 = 2798446.9 bbl

2798446.9 t = ---- = 6.389 Years

1200

One can do above procedure for other saturation and construct below table.

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. f J J _J J .J

Sw:c ....... .., .:.\S\'Ie ~w., .:.\1\v</ .:lS~.va Swe:·

0.45 0.665 <" At . brcak!h:ougtl 0.05 O.t34 2.68 0.475

0.5 0.799 0.05 0016 1.52 0.525

0.55 0.875 o:os 0 .051 1.14 0.575

0;6 0.932 0 .05 0.033 0.65 0.625

0.65 O.£<t>5 0.05 0016 0.32 0.675

0.7 0 .981 0.05 0.013 0.25 0.725

0.75 0 .994

0 0.05 (t0Q6 0.12 0]75

0.8 l.OOO

Cumulative tv Cumulative

fw dfwl Oil 1 Water t

Sw dSw Swe

Npv Production -dfw Production (Years)

(MSTB) dSw (MSTB) 0.475 0.735 2.62 0.376 6617.6 0.382 6723.2 2.734 0.525 0.841 1.58 0.426 7497.6 0.633 11140.8 4.53 0.575 0.905 1.12 0.46 8096 0.893 15716.8 6.391 0.625 0.950 0.66 0.501 8817.6 1.515 26664 " 10.842. 0.675 0.975 0.36 0.544 9574.4 2.778 48892.8 19.88 0.725 0.988 0.2 0.585 10296 5 88000 35.782 0.775 0.9975 0.1 0.6 10560 10 176000 71.563

12000 Curnml.uivc <dl pr~d'''ed Vef$tl~ culnmul.uive Wdltl Injecte-d

12000

10000 ------. ~-------- .: . --.-. - • .!. -.

0 0 ' ' : j ···, ··· ····.· ·······~····!·· · ····· ~ 10000 ________ ;__·---- - --~-------- ----···--·-·-· -··r··--··-r--------(-----·

! V\

~ 8000

~ ~ , 000 ~ ., >

--:···-·--·-:---------:·· ·----·-i· ---- - --:- -·· - -- - -~--- · -· - ·r · ·------~------- ·

-- ·- ..... --- ----. ~ . - ----- .... .... ... ; ...... .. .: ......... : ... . ... .

-_-_;==-· lOOO ---·-·j·····-··f··-·-··-j- -- ----f· -·· -·· y--- ----1···----·y- -· --·--j---· -- --

V i Btt.lktl,jough ~ 1 i ; : ; . . . : : : :

2000 ~-----i- --- -·-·i···-· ·- -f - - -- --·-+-- --- --+-------+ - ----- - +·------

0 0

: . ! . : : ~ f

L __ -L---L--~--~~ 20 JO GO 80 100 1l0 UO 1GO lBO

J : ···•·••••.:_c_c .r :.c_c ._i

i •ooo --------:-·-- · -----1 · - - -·- - --r·---···r·-- ·- --~-- - ---· · ·-r··-- ·- -· -;------·--~ . d - r"::~:~ , , ::··· ; : ....... .

Cuuuuul.uive water iuj•C14HI (ldMSTBt Time (ye.liSI

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Reservoir Engineering I Course Notes - Shariff University.pdf

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