Fluid Flow in Porous Media

Objectives:

•Understand forces responsible for driving fluid through reservoir

•Be aware of models available to represent reservoir and wells

•Assess flow properties of reservoir

•Introduce concepts used in welltesting

Summary

Fluid flow depends on:

• reservoir geometry• reservoir fluids• reservoir properties

Two methods to represent fluid flow:

• analytical solutions to diffusivity equation

• approximation methods to diffusivity equation

finite difference/ finite element simulations

Summary

Solutions to be examined:

Introduction

•Objective to understand mechanism of fluid migration in order to understand and improve recovery from the reservoir.

•Similar in concept to flow in pipes

•General energy equation not applicable because of geometry, interconnections etc.

•Dimensions give rise to scaling problem – capillary forces become relatively important (over viscous forces)

Introduction

•Surface chemistry effects between minerals and fluids

•Main effect is time taken for fluid to move from high to low pressure regions

•If similar to large body of water, then pressure same at every point at all times

•If rapid equilibration, then Darcys Law would be applicable

Introduction

•Illustrated with model:

–10 tubes representing section of reservoir from wellbore into reservoir

–Initially no flow and all tubes at same height

–Flow initiated and fluid expands

–Pressure profile develops as fluid expansion migrates along model

Introduction

•Illustrated with model:

–Unusual situation where although hydraulically connected, pressure varies along tubes

–Time taken for pressure wave to move along tubes

–Restrictions at base of tubes limits flow and mimics flow through pore structure

Initial steady state, no flow

Flow initiated at constant rate

Pressure profile develops

Pressure disturbance reaches outer boundary

Sealed therefore the pressure in system drops

Well shut in and pressure builds in well

Steady state, no flow

Initial pressure profile is transient – note back end of model still at initial pressure condition even though pressure has fallen in tubes nearest fluid exit

Pressure profile develops through model until the back tube is hit, then the entire system depressurises. Transition to semi-steady state.

If fluid entered the system at the back end of the model at the same rate as it outflowed, there would be a second steady state flow regime, i.e. the pressure in the back end tube would not change.

In this case the pressure in the model declines once the fluid at the back end has started to flow since there is no inflow, this is a special case of transient flow called semi steady state. The gradient remains constant; the absolute pressure falls.

The model can be seen to continue to flow after the well has been shut in. The fluid in the reservoir is still trying to reach equilibrium.

A major feature of this system is the appreciable time taken to change from one steady state to a second steady (or semi-steady) state.

This time dependence may be significant in real reservoirs, where the areal extent of the reservoir is such that it may take decades to equilibrate, or indeed there may never be equilibration.

Characterisation and Modelling of Flow Patterns

Complex patterns:shapes of oil bearing formations irregularheterogeneous formation properties:

porosity, permeability, saturation(saturation of hydrocarbons may

vary throughout the formation leading to variations in rel perm)

deviated wellbores through reservoirvarying production rates from different

wells – in general high rate wells drain larger areas

many wells do not fully penetrate the reservoir

Essentially two possibilities to cope with complexities:

• drainage area of well subdivided into small blocks which represent the variations in

propertiesseries of complex equations describing

fluid flow solved by numerical or semi-numerical methods

• single block which preserves global features and heterogenities – fluid properties averaged or substituted by simple relationship or pattern of features (like fracture pattern)- which allows analytical solution

Idealised Flow Patterns

Linear, radial, spherical, hemispherical

Linear and radial of most use

Assume oil system with cp<<1, i.e. small and constant compressibility

If gas reservoir, compressibility must be accounted for (by gas pseudo-functions for example)

General Case

Flow velocity, UResolved into x, y, z directions

The components of the flow velocity vector, U are: Ux = -(kx/)(P/x)

Uy = -(ky/)(P/y)

Uz = -(kz/)(P/z+g)

k = permeability (m2) in the direction of X, Y, Z. The Z direction has an elevation term, g, included to account for the change in head. P = pressure (Pa) = viscosity (Pas) = density (kg/m3)g = acceleration due to gravity (m/s2)U = flow velocity (m/s) = (m3/s/m2)

Linear Horizontal Model of a Single Phase Fluid

dx

x=0 x x+dx x=L

flowrate, qin x=0

x

x+dx

x=L

dx

area, A

porosity,

X axis

X axis

flowrate, qout

flowrate, qout

flowrate, qin

isometric view

plan view

Flow along x direction, no flow in y or z directions

Flow into cuboid at left, out of cuboid at right

Total length, L

Rock 100% saturated with one fluid

Flow equations:

x

PkU x

U x

t

0 ≤ x ≤ L

k = permeability (in the X direction), (mD)

= density, (kg/m3)

U = flow velocity (m/s)

t = time (s)

= porosity

= viscosity, Pas

P = pressure, Pa

x = distance, (m)

Fluid flows in at position x=0, out at x=L.

Element from x to position x+dx is examined.

The bulk volume of the element is the product of the area, A and the length, dx, i.e. bulk volume = A*dx.

The pore volume of element is product of the bulk volume and the porosity, , i.e. pore volume = A*dx*

If flow steady state then the flowrates into and out of the volume (qin and qout) would be identical and Darcy’s Law would apply.

Fluid Flow in Porous Media

If the flow rates vary from the inlet of the volume to the outlet, i.e. qin ≠ qout then either:

fluid is accumulating in the element and qin > qout

or:

fluid is depleting from the element qout > qin

(which is possible in a pressurised system since the pressure of the fluid in the element may reduce causing it to expand and produce a higher flow rate out of the element)

Fluid Flow in Porous Media

Therefore, there is a relationship between the change in mass, m, along the cuboid and the change in density, , over time as the mass accumulates or depletes from any element. In terms of mass flowrate, Mass flow rate through the area, A = q ((m3/s)*(kg/m3) = kg/s)Mass flow rate through the area, A at position x = (q)x

Mass flow rate through the area, A at position x+dx = (q)x+dx

Mass flowrate into a volume element at x minus mass flowrate out of element at x + dx =(q)x- (q)x+ dx

The mass flow rate out of the element is also equal to the rate of change of mass flow in the element, i.e.

q x

q xdx q

x * dx

Change in mass flow rate =

q x

* dx

(if change is +ve, element accumulating mass, if –ve depleting mass)

This must equal rate of change of mass in element with volume A*dx*

Rate of change of mass equal to t

A dx

hence q x t

=A

flow velocity, U = q/A, therefore

U x t

or

U x t

Substitution of parameters gives

tx

Pk

x

Equation shows areal change in pressure linked to temporal change in density. Measure pressure easier than density, therefore use isothermal compressibility to convert to pressure

c = - 1V

(VP

) T

The density equals mass per unit volume (mV),hence:

c = -

m (m/)

P =

1

P ; (Quotient Rule, constant mass system)

Since

t

= P

P t

= c P t (from above)

then

x

k

P

x

= c

Pt

Partial differential equation for linear flow of any single phase fluid in porous medium – relates spatial and temporal variations in pressure

•In core relates pressure distribution along core during flooding, during all time, i.e. from start of flood to staedy state conditions

•In linear reservoir where aquifer flows into reservoir as production proceeds

But, non-linear because of pressure dependence of density, compressibility and viscosity.Simple linearisation follows

x

k

P

x

= c

Pt

Linearisation of Fluid Flow Equation

Assume permeability and viscosity are independent of location

x

(Px

) = (c/k)Pt

The left hand side can be expanded to: x

P

x + P/x2)

Using equation 2.4 and since x =

P Px

the above becomes

c(P/x)2 + (2P/x2).

2P

x2 = (

c

k )P

t

c(P/x)2 is neglected compared to 2P/x2 since pressure gradient small, substituting gives

assumption that compressibility small and constantcoefficients c/k are constant and equation linearised(k/c) termed diffusivity constantassume cp<<1 for oil systemssaturation weighted compressibilityc=coSo+cwSwc+cf

c= saturation weighted compressibilityco = compressibility of oil

cw= compressibility of connate water

cf =compressibility of formation

So= oil saturation

Swc= connate water saturation

Conditions of Solution

Initial conditionsat time t=o, intial pressure Pi specified for every value of x•Boundary conditions•at end faces x-0, x=L flow rate or pressure specified for every value of x•solutions of linear diffusivity equation for linear flow from aquifers

Radial model for cylindrical reservoir , constant thickness, h

Radial Model

Distance, r from x-axis, flow velocity, U now radius dependent:

U = q/2rhFrom Darcys Law,

U = k Pr

The mass balance gives:

(q)r = 2rh

t

Eliminating U and q through equations gives the non linear equation:

1r r r

k Pr = c

Pt

Making assumptions as for linear flow, linearises the equation to:

1r r(rPr ) =

ck Pt

Range of Application

applied to water influx and wellbore production

•water encroachment - inner boundary corresponds to mean radius of reservoir, outer boundary mean radius of aquifer•wellbore pressure regime - inner boundary is wellbore radius, rw, outer boundary is the boundary of the drainage area.

•values of rw

open hole drilled close to gauge: 1/2 bit diameterwell cased cemented, perforated: 1/2 bit diameterslotted liner with gravel pack: 1/2 OD of linerout of gauge hole: average radius from caliper log

Condition of Solution

Initial: t=o, Pi specified at all locationsOuter boundary:

a) no flow: p/r = 0, flow velocity =ob) flow: p/r not equal to zero,

pressure maintained at boundaryInner boundary:

constant terminal rateproduction rate constant at wellaquifer influx constant

constant terminal pressureBHP constantaquifer pressure constant, influx

varies

Characterisation of flow regimes based on time

transientsemi-steady statesteady state

Steady State: pressure and rate distribution in reservoir constant with timeUnsteady State (transient): pressure and/or rate vary with timeSemi-Steady State: special case of unsteady state which resembles steady stateWorking solutions need to refer to the appropriate flow regime

CTR for radial models

flow rate constantoil flowing to fully perforated wellaquifer encroachment

radial flow of single phase fluid from outer radius b to inner radius a

assuming right hollow cylinder of homogeneous medium

for a well, a is rw, b is re (external boundary radius)flow rate, q is constant at wellborefor aquifer, a is mean reservoir radius, b mean aquifer radius, q volume flow rate of water across initial WOC

The radial constant terminal rate case is determined by the following system of equations:

1r r(r

Pr ) =

ck Pt

; a≤ r ≤ b (3.1)

r

prkh2q

with the initial condition that the pressure at all points is constanta≤ r ≤ b, t = o; P=Pi = constant (3.3)

; r =a (3.2)

and the boundary conditions that at the wellbore the flowrate is constant after the production starts r=a, t ≥ 0 : q = constant (3.4) and at the outer boundary, the pressure is either a constant (and equal to the initial pressure) in the case of pressure maintenance r=b, t ≥ 0 : P = Pi = constant (3.5a)

or there is a sealing boundary with no flow across it in which case the pressure gradient at the boundary is zero

r=b, t ≥ 0 : Pr = 0 (3.5b)

Solution to equations well known: Mathews and Russell, SPE monograph - very complex solutions - asymptotic solutions fair approximations of general solutionProblem to identify flow regime

Steady state is simplestNon-steady state involves time element

Steady State Solution

pressure at outer boundary, re, constant

flow rate, q, constantp/t = 0 for all values of radius, r, and time, t.

Pr dP

dr

and the flow equation becomes qdrr 2kh

dP

integrating between the limits rw and r gives:

P Pw q

2kh

ln

rrw

(3.6)

Integrating between the limits rw and re gives:

Pe P

w

q2kh

ln

re

rw

(3.7)

same as Darcy’s LawDefinition of pressure at external radius, re :

difficult to determine, use average reservoir pressure,

Defined by area drained by each well in a reservoir.Found by well test analysis and routine bottom hole pressure measurements

P

re

rw

PdVV

1P

(3.8a)

where dV = 2rhdr

(3.8b)

The volume of the well’s drainage zone, V, = (re2-rw2)h

and considering rw<<re, V≈ re2

h

re

rwe2 Prdr

r

2P

from equation 3.6,

ww

r

rln

kh2

qPP

4r

4r

r

rln

2r

r

rln

2r

kh2

q

r

2P-P

dr2

r

r

1

r

rlnr

2

1

kh2

q

r

2P-P

rdrr

rln

kh2

q

r

2P-P

rdrr

rln

kh2

qP

r

2P

2w

2e

w

w2w

w

e2e

2e

w

re

rw

2re

rww

2

2e

w

re

rw w2e

w

re

rw ww2

e

4r

2wassuming is negligible

4r

r

rln

2r

kh2

q

r

2P-P

2e

w

e2e

2e

w

2

1

r

rln

kh2

qP-P

w

ew

(3.10)

Example 1. A well produces oil at a constant flowrate of 15 stock tank cubic metres per day (stm3/d). Use the following data to calculate the permeability in milliDarcys (mD). Data porosity, 19%formation volume factor for oil, Bo 1.3rm3/stm3 (reservoir cubic meters per stock

tank cubic meter)net thickness of formation, h, 40mviscosity of reservoir oil, 22x10-3 Paswellbore radius, rw 0.15m

external radius, re 350m

initial reservoir pressure, Pi 98.0bar

bottomhole flowing pressure, Pwf 93.5bar

qreservoir = qstock tank x Bo

1bar = 105 Pa

Solution the steady state inflow equation (accounting for fluid flowrate at reservoir conditions in m3/s and pressure in Pa) is

w

eowfe

r

rln

kh2

BqPP

w

e

wfe

o

r

rln

)hP(P2

Bqk

341mD

m341x10

0.15

350.00ln

x4093.5)x10x(98.024x3600x2

x1.315x22x10k

215

5

3

Unsteady State Flow Regimes

Dimensionless variables

normalised parametersdefine solution to diffusivity equation for dimensionless variablesdetermine solutioncalculate specific reservoir values from dimensionless solutiondimensionless radius, rD :

wD

r

rr

dimensionless time, tD : 2w

Dcr

ktt

dimensionless pressure, PD :

)P)(Pq

kh2()t,(rP tr,iDDD

(at a dimensionless radius and dimensionless time)

where r = radius in question rw = wellbore radius

k = permeability t = time in question = porosity

= viscosity c = compressibility h = thickness of the reservoir

Pi = initial reservoir pressure Pr,t = pressure at the specified radius and time

then the radial diffusivity equation becomes

D

D

D

DD

DD t

P

r

Pr

rr

1

(3.11)

There are other definitions of dimensionless variables, such as dimensionless external radius

Unsteady State Solution

CTR solution obtained in several forms with different assumptions and mathematical analysesGeneral considerations

Wellbore pressure and flow rate responsePressure decline normally divided into 3 sections depending on the value of flowing time and reservoir geometry.

Initially, transient solution – infinite acting reservoir case – reservoir appears infinite in extent

Late transient – boundaries start to affect the response

Semi-steady state or pseudo-steady state – pressure perturbation affecting all parts of the reservoir – no influx from aquifer

Hurst and van Everdingen Solution

CTR solution in 1949

Solved radial diffusivity using Laplace transform for both CTR and CTP

Solution describes pressure drop as function of time and radius for fixed values of re and rw rock and fluid properties.

Dimensionless variables and parameters:

PD = f(tD,rD,reD)

where tD = dimensionless time

rD = dimensionless radius

reD = re/rw = dimesionless external radius.

If the reservoir is fixed in size, i.e. reD is a particular value,

then the dimensionless pressure drop, PD, is a function of the dimensionless time, tD and dimensionless radius, rD.

The pressure in a particular reservoir case can then be calculated at any time and/or radius.

One of the most significant cases is at the wellbore since the pressure can be measured routinely during production operations and compared to the theoretical solutions.

The determination of a reservoir pressure at a location remote from a well may be required for reasons of technical interest, but unless a well is drilled at that location, the actual value cannot be measured.

At the wellbore radius, r=rw (or rD=1.0) PD = f(tD, reD) (3.13)

i.e.

1m m21eDm

21

2m

eDm21

t

eD2eD

DDD

))()r((

)r(e2

4

3lnr

r

2t)(tP

JJJ

D

2

m

(3.14)

where m are the roots of 0)r()Y(J)()Yr(J eDm1m1m1eDm1

J1 and Y1 are Bessel functions of the first and second kind

This series has been evaluated for several values of dimensionless external radius, reD, over a wide range of values of dimensionless time, tD. The results are presented in the form of tables (from Chatas, AT, “A Practical Treatment of non-steady state Flow Problems in Reservoir Systems,” Pet. Eng. August 1953) in “Well Testing” by J Lee, SPE Textbook series, Vol 1. A summary of the use of the tables for constant terminal rate problems is as follows in Table 1.

Table Presents Valid for

2 i PD as a function of t D <1000 (from table) infinite acting reservoirs

ii P

D 2

tD

for t D <0.01 (an extension of the table)

infinite acting reservoirs

iii

PD 0.5(lnt

D 0.80907) for 100< t D <0.25 r eD

2

(an extension of the table)

infinite acting reservoirs

iv

PD as a function of t D <0.25 r eD2 (from table) finite reservoirs

3 i PD as a function of t D for 1.5< r eD2 <10 (from table) finite reservoirs, but if

the value of t D is smaller

than that listed for a

given value of r eD then

the reservoir is infinite

acting and therefore table

2 is used.

ii P

D

2 tD 0.25

reD

2 1

3reD

4 4reD

4 lnreD 2r

eD

2 1 4 r

eD

2 1 2

for 25 tD

and 0.25reD

2 tD

finite reservoirs

iiiP

D

2tD

reD

2 lnr

eD

3

4 for r

eD

2 > 1 finite reservoirs

Table 1 Hurst and Van Everdingen solutions to the Constant Terminal Rate Case

These equations are applicable to a well flowing at a constant rate or to a reservoir and aquifer with a constant flowrate across the oil water contact.

Most problems involving flow at a well involve relationship 2(iii) and 3(iii);

most problems involving aquifer influx involve Tables 8 and 9.

It can be seen that in using these solutions, the pressure can be calculated anywhere in the reservoir as long as the flow rate is known.

If the pressure in the reservoir at a location where the flow rate is unknown is required then an alternative solution is needed (the Line Source solution).

Example 2. A reservoir at an initial pressure, Pi of 83.0bar produces to a well 15cm

in diameter. The reservoir external radius is 150m. Use the following data to calculate the pressure at the wellbore after 0.01 hour, 0.1 hour, 1 hour, 10 hours and 100hours of production at 23stm3/d Data porosity, 21%formation volume factor for oil, Bo 1.13rm3/stm3

net thickness of formation, h 53mviscosity of reservoir oil, 10x10-3 Paswellbore radius, rw 0.15m

external radius, re 150m

initial reservoir pressure, Pi 83.0bar

permeability, k 140mDcompressibility, c 0.2x10-7Pa-1

 

Solution Using Hurst and Van Everdingen’s solution for CTR, the dimensionless external radius and the dimensionless time are calculated and used with the appropriate solution to determine the dimensionless pressure drop. The dimensionless pressure drop is then turned into the real pressure drop from which the bottomhole flowing pressure is calculated.

10000.15

150.00

r

rr

w

eeD

0.148tx0.15x0.2x100.21x10x10

xt140x10

cr

ktt 273-

-15

2w

D

time time tD PD expression (hour) (second) (0.148t)

0.01 36 5.3 1.3846 table 2 0.10 360 53.3 2.4146 table 2 1.00 3600 532.8 3.5473 table 2

10.00 36000 5328.0 4.6949 0.5(lntD+0.80907) 100.00 360000 53280.0 5.8462 0.5(lntD+0.80907)

the bottomhole flowing pressure, Pwf is

Do

iwf Pkh2

BqPP

Pa82.1x10= x1.3846x53140x1024x3600x2

x1.1323x10x1083.0x10P 5

15

35

0.01hourat wf

i.e. Pwf at 0.01 hour =82.1bar similarly for the rest of the times time PD Pwf (hour) (bar)

0.00 0 83.0 0.01 1.3846 82.1 0.10 2.4146 81.4 1.00 3.5473 80.7

10.00 4.6949 80.0 100.00 5.8462 79.2

Line Source Solution

Assumes radius at wellbore is vanishingly smallallows calculationof the pressure in the reservoir using the flowrate at the wellDisadvantage is that only works in Transient Flow RegimeBarriers alter applicability of Line Source SolutionHowever, principle of superposition allows combination of different wells and use of imaginary wells to compensate for the effect of barriers

In constant terminal rate problems, the flowrate at the well was given by

q 2rhk

Pr

rrw

(3.15)

and for a line source, the following boundary condition must hold:

lim

r 0rpr

q2kh

for time, t > 0.

Using the Boltzman Transformation

y cr 2

4kt and substituting into the diffusivity equation (

1r r(rPr ) =

ck Pt )

gives

yd2p

dy2 dp

dy(1 y)0

with the boundary conditions

p pi as y lim

y 02ypy

q2kh

If p'dp

dy then

ydp'

dy (1 y)p' 0

S epara tin g th e variab les an d in tegra tin g g ives ln p ’ = -ln y - y + C

i.e . p ' d p

d y

C1

ye y (3 .1 6 )

w here C an d C 1 a re co ns tan ts o f in tegratio n . S in ce lim

y 02yp

y

q2 kh

lim

y 02C 1 e y

then C1 q

4kh and equation 3.16 becomes

dp

dy

q4kh

e y

y which is integrated to give

p q

4kh

e y

y

y

dy C2 or

p q

4kh

e y

yy

dy C2

which can be rewritten as

p q

4khEi(-y) C 2

Applying the boundary condition that p pi as y then C2 = pi and the line source

solution is obtained:

pi p(r, t) q

4khEi(-

cr2

4kt)

(3.17)

The term Ei(-y) is the exponential integral of y (the Ei function) which is expressed as

Ei( y)e y

ydy

y

.

It can be calculated from the series

Ei( y) lny yn

n!n

where = 0.5772157 (Euler’s Constant). On inspection of the similarities in the Ei

function and the ln function, it can be seen that when y <0.01, Ei( y) lny and the

power terms can be neglected. Therefore,

Ei( y)ln(1.781y) = ln(y)

(1.781 = e e0.5772157)

Solutions to the exponential integral can be coded into a spreadsheet and used with the

line source solution. Practically, the exponential integral can be replaced by a simpler

logarithm function as long as it is representative of the pressure decline. The limitation

that y<0.01 corresponds to time, t, from the start of production t 25cr 2

k.

The equation can be applied anywhere in the reservoir, but is of significance at thewellbore (i.e. for well test analysis) where typical values of wellbore radius, rw, andreservoir fluid and rock parameters usually means that y<0.01 very shortly afterproduction starts. Therefore the line source solution can be approximated by

P Pi q

4kh(lncr2

4kt)

or, since -ln(y) = ln(y-1)

P Pi q

4kh(ln

4kt

cr2 ) (3.18)

and if the pressure in the wellbore is of interest,

Pwf Pi q

4kh(ln

4kt

crw2 ) (3.19)

The values of exponential integral have been calculated and presented in Matthews and

Russel’s Monograph and are produced in Table 4. The table presents negative values, i.e.

-Ei(-y). For values of y0.01, the ln approximation can be used. For values >10.9, the

decline in pressure calculated is negligible.

Range of Application and Limitations of Use

Ei function has limitations on applicationcannot represent the initial flow into wellbore

(line source)reservoir must be infinite acting

Analysis of real reservoirs has shown that Ei function valid for i) flowing time> 100crw2/k

rw is wellbore radius. constant 100 derived from reservoir responseii) time< cre2/4k

re is external radius, after this time, infinite acting period has ended

E x a m p l e 3 . A w e l l a n d r e s e r v o i r a r e d e s c r i b e d b y th e f o l lo w i n g d a t a : D a t a p o r o s i t y , 1 9 % f o r m a t i o n v o l u m e f a c t o r f o r o i l , B

o 1 .4 r m 3 / s t m 3

n e t t h i c k n e s s o f f o r m a t i o n , h 1 0 0 m v i s c o s i t y o f r e s e r v o i r o i l , 1 .4 x 1 0 - 3 P a s c o m p r e s s i b i l i t y , c 2 .2 x 1 0 - 9 P a - 1 p e r m e a b i l i t y , k 1 0 0 m D w e l l b o r e r a d i u s , r

w 0 .1 5 m

e x t e rn a l r a d i u s , re 9 0 0 m

i n i t i a l r e s e rv o i r p r e s s u r e , Pi 4 0 0 b a r

w e l l f l o w r a t e ( c o n s t a n t ) 1 5 9 s t m 3 / d a y = 1 5 9

2 4 x 3 6 0 0s t m 3 / s e c o n d

s k i n f a c t o r 0 D e t e r m i n e t h e f o l l o w i n g : 1 ) t h e w e l l b o re f l o w i n g p r e s s u r e a f t e r 4 h o u r s p r o d u c t i o n 2 ) t h e p r e s s u r e i n t h e r e s e rv o i r a t a r a d i u s o f 9 m a f t e r 4 h o u r s p ro d u c t io n 3 ) t h e p r e s s u r e i n t h e r e s e rv o i r a t a r a d i u s o f 5 0 m a f t e r 4 h o u r s p r o d u c t i o n 4 ) t h e p r e s s u r e i n t h e r e s e rv o i r a t a r a d i u s o f 5 0 m a f t e r 5 0 h o u r s p ro d u c t i o n

Solution The line source solution is used to determine the pressures required at the specified radii and at the specified times (i.e. using the flowrate measured at the wellbore, the pressures at the other radii and times are calculated by the line source solution). SI units will be used so time will be converted to seconds. Checks are made to ensure that: i) there has been adequate time since the start of production to allow the line source solution to be accurate ii) the reservoir is infinite acting. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made.

A C h e c k E i a p p l i c a b i l i t y l i n e s o u r c e n o t a c c u r a t e u n t i l

t 1 0 0 c r w

2

k

t 1 0 0 x 0 . 1 9 x 1 . 4 x 1 0 - 3 x 2 . 2 x 1 0 9 x 0 . 1 5 2

1 0 0 x 1 0 - 1 5

t > 1 3 . 2 s t i m e i s 4 h o u r s , t h e r e f o r e l i n e s o u r c e i s a p p l i c a b l e .

B C h e c k r e s e r v o i r i s i n f i n i t e a c t i n g

t h e r e s e r v o i r i s i n f i n i t e a c t i n g i f t h e t i m e , t c r e

2

4 k

i . e . t 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 9 0 0 2

4 x 1 0 0 x 1 0 - 1 5

t < 1 1 8 5 0 3 0 s t < 3 2 9 h o u r s t h e r e f o r e l i n e s o u r c e s o l u t i o n i s a p p l i c a b l e .

1 ) t h e b o t t o m h o l e f l o w i n g p r e s s u r e a f t e r 4 h o u r s p r o d u c t i o n , P w f a t 4 h o u r s i ) c h e c k l n a p p r o x i m a t i o n t o E i f u n c t i o n

t h e l n a p p r o x i m a t i o n i s v a l i d i f t h e t i m e , t 2 5 c r w

2

k

t 2 5 x 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 0 . 1 5 2

1 0 0 x 1 0 - 1 5

t > 3 . 3 s t h e r e f o r e l n a p p r o x i m a t i o n i s v a l i d .

i i ) P w f P i q B o

4 khl n

c r w2

4k t

( t a k i n g a c c o u n t o f t h e c o n v e r s i o n f r o m s t o c k t a n k t o

r e s e r v o i r c o n d i t i o n s v i a t h e f o r m a t i o n v o l u m e f a c t o r f o r o i l , B o , f l o w r a t e s i n r e s e r v o i r m 3 / s a n d p r e s s u r e s i n P a s c a l ) .

q B o

4 k h

1 5 9 x 1 . 4 x 1 0 3 x 1 . 4

2 4 x 3 6 0 0 x 4 x 1 0 0 x 1 0 1 5 x 1 0 0 = 2 8 7 0 3

c r 2

4 k t

0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 r 2

4 x 1 0 0 x 1 0 - 1 5 x 4 x 3 6 0 0 = 1 0 1 5 9 7 x 1 0 - 9 r 2

P w f = 4 0 0 x 1 0 5 + 2 8 7 0 3 x l n ( 1 . 7 8 1 x 1 0 1 5 9 7 x 1 0 - 9 x 0 . 1 5 2 ) = 4 0 0 x 1 0 5 - 3 5 6 2 4 9 = 3 9 6 4 3 7 5 1 P a = 3 9 6 . 4 b a r

2 ) t h e p r e s s u r e a f t e r 4 h o u r s p r o d u c t i o n a t a r a d i u s o f 9 m f r o m t h e w e l l b o r e i ) c h e c k l n a p p r o x i m a t i o n t o E i f u n c t i o n

t h e l n a p p r o x i m a t i o n i s v a l i d i f t h e t i m e , t 2 5 c r 2

k

t 2 5 x 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 9 2

1 0 0 x 1 0 - 1 5

t > 1 1 8 5 0 s t > 3 . 3 h o u r s t h e r e f o r e l n a p p r o x i m a t i o n i s v a l i d .

i i ) P P i q B o

4 k hl n

c r 2

4 k t

( t a k i n g a c c o u n t o f t h e c o n v e r s i o n f r o m s t o c k t a n k t o

r e s e r v o i r c o n d i t i o n s v i a t h e f o r m a t i o n v o l u m e f a c t o r f o r o i l , B o

a n d a l s o t h e f a c t t h a t t h e r a d i u s , r , i s n o w a t 9 m f r o m t h e w e l l b o r e ) .

q B o

4 k h

1 5 9 x 1 . 4 x 1 0 3 x 1 . 4

2 4 x 3 6 0 0 x 4 x 1 0 0 x 1 0 1 5 x 1 0 0 = 2 8 7 0 3

c r 2

4 k t

0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 r 2

4 x 1 0 0 x 1 0 - 1 5 x 4 x 3 6 0 0 = 1 0 1 5 9 7 x 1 0 - 9 r 2

P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x l n ( 1 . 7 8 1 x 1 0 1 5 9 7 x 1 0 - 9 x 9 2 ) = 4 0 0 x 1 0 5 - 1 2 1 2 0 9 = 3 9 8 7 8 7 9 1 P a = 3 9 8 . 8 b a r

3 ) t h e p r e s s u r e a f t e r 4 h o u r s p r o d u c t i o n a t a r a d i u s o f 5 0 m f r o m t h e w e l l b o r e i ) c h e c k l n a p p r o x i m a t i o n t o E i f u n c t i o n

t h e l n a p p r o x i m a t i o n i s v a l i d i f t h e t i m e , t 2 5 c r 2

k

t 2 5 x 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 5 0 2

1 0 0 x 1 0 - 1 5

t > 3 6 5 7 5 0 s t > 1 0 1 . 6 h o u r s t h e r e f o r e l n a p p r o x i m a t i o n i s n o t v a l i d a n d t h e E i f u n c t i o n i s u s e d .

i i ) P P i q B o

4 k hE i

cr 2

4 k t

( t a k i n g a c c o u n t o f t h e c o n v e r s i o n f r o m s t o c k t a n k t o

r e s e r v o i r c o n d i t i o n s v i a t h e f o r m a t i o n v o l u m e f a c t o r f o r o i l , B o

a n d a l s o t h e f a c t t h a t t h e r a d i u s , r , i s n o w a t 5 0 m f r o m t h e w e l l b o r e ) .

q B o

4 k h

1 5 9 x 1 . 4 x 1 0 3 x 1 . 4

2 4 x 3 6 0 0 x 4 x 1 0 0 x 1 0 1 5 x 1 0 0 = 2 8 7 0 3

c r 2

4 k t

0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 5 0 2

4 x 1 0 0 x 1 0 - 1 5 x 4 x 3 6 0 0 = 0 . 2 5 4

P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x E i ( - 0 . 2 5 4 ) E i ( - 0 . 2 5 4 ) = - 1 . 0 3 2 ( b y l i n e a r i n t e r p o l a t i o n o f t h e v a l u e s i n T a b l e 4 ) P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x - 1 . 0 3 2 = 4 0 0 x 1 0 5 - 2 9 6 2 2 = 3 9 9 7 0 3 7 8 P a = 3 9 9 . 7 b a r

4 ) t h e p r e s s u r e a f t e r 5 0 h o u r s p r o d u c t i o n a t a r a d i u s o f 5 0 m f r o m t h e w e l l b o r e i ) c h e c k l n a p p r o x i m a t i o n t o E i f u n c t i o n

t h e l n a p p r o x i m a t i o n i s v a l i d i f t h e t i m e , t 2 5 c r 2

k

t 2 5 x 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 5 0 2

1 0 0 x 1 0 - 1 5

t > 3 6 5 7 5 0 s t > 1 0 1 . 6 h o u r s t h e r e f o r e l n a p p r o x i m a t i o n i s n o t v a l i d a n d t h e E i f u n c t i o n i s u s e d .

i i ) P P i q B o

4 k hE i

cr 2

4 k t

( t a k i n g a c c o u n t o f t h e c o n v e r s i o n f r o m s t o c k t a n k t o

r e s e r v o i r c o n d i t i o n s v i a t h e f o r m a t i o n v o l u m e f a c t o r f o r o i l , B o

a n d a l s o t h e f a c t t h a t t h e r a d i u s , r , i s n o w a t 5 0 m f r o m t h e w e l l b o r e a n d t h e t i m e i s n o w 5 0 h o u r s a f t e r s t a r t o f p r o d u c t i o n ) .

q B o

4 k h

1 5 9 x 1 . 4 x 1 0 3 x 1 . 4

2 4 x 3 6 0 0 x 4 x 1 0 0 x 1 0 1 5 x 1 0 0 = 2 8 7 0 3

c r 2

4 k t

0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 5 0 2

4 x 1 0 0 x 1 0 - 1 5 x 5 0 x 3 6 0 0 = 0 . 0 2 0

P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x E i ( - 0 . 0 2 0 ) E i ( - 0 . 0 2 0 ) = - 3 . 3 5 5 P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x - 3 . 3 5 5 = 4 0 0 x 1 0 5 - 9 6 3 0 0 = 3 9 9 0 3 7 0 0 P a = 3 9 9 . 0 b a r

time radius pressure (hours) (m) (bar)

0 all 400.0 4 0.15 396.4 4 9.00 398.8 4 50.00 399.7

50 50.00 399.0

Summary

Skin Factor

Assumption of constant permeability around wellboreFormation damage during drilling and completion and during production causes alteration of permeability around wellbore.Extends up to a few feet from wellbore into reservoirIf reservoir fractured (naturally or by workover) permeability may be increasedEi function fails to account for these conditionsSkin zone defined as zone around wellbore with altered permeability

bo

ttom

ho

le fl

ow

ing

pre

ssu

re, P

wf

radius, r

rsrw

Pskinskin zone

permeability, kpermeability, ks

pressure profile if no skin zone was present

actual pressure profile through skin zonePwf(no skin)

Pwf(skin)

Pskin = Pwf(skin) - Pwf(no skin)

Assume skin zone equal to altered zone of uniform permeability, ks with an outer radius rs

Additional pressure drop across zone Ps can be modelled by steady state inflow

Assumed that after initial time, flow regime around wellbore close to steady state

Ps=q

2kshln

rs

rw

q2kh

lnrs

rw

q2kh

(k

ks

1)lnrs

rw

- =

Pi Pwf q

4khEi( y) Ps

q4kh

Ei( y) 2k

k s

1

ln

rs

rw

Pi Pwf q

4khln(crw

2

4kt) 2

k

ks

1

ln

rs

rw

If at the wellbore the logarithm approximation can be substituted for the Ei function,

then:

A skin factor, s, can then be defined as:

sk

k s

1

ln

rs

rw

Pi Pwf q

4khln(crw

2

4kt) 2s

and the drawdown defined as

Equation shows that positive skin means decrease in permeability around well

Measured as part of the objectives of well testingExtent of altered zone unknown

Altered zone around a particular well affects only the pressure near that well

Semi-Steady State Solution

Once boundaries felt, transience stops

where no flow across external boundary,re

P

r

P

t

dP

dtconstant

Similar to steady state, pressure drop between

wellbore and external radiuswellbore and average pressure

may be calculated.

Average pressure usually known in reservoirs - this is used to determine pressure drop

Under semi-steady state conditions, pressure profile averaged over reservoir drainage cell

If several wells in reservoir, cells stabilise drainage areas based on flow rates

cV(Pi P ) qt

where V = pore volume of the radial cell; q = constant production rate; t = total flowing

time, c = isothermal compressibility.

q dVdt

dV

dP

qdt

dPq

dt

dP

since c 1

V

dV

dP

T

q cVdPdt

dP

dt

q

cV

Substitution of equation 3.29 in the radial diffusivity equation

1r r(rPr ) =

ck Pt

gives

1

r

r

(rPr

) = c

k

q

cre2 h

which is

1

r

r

(rPr

) = qre

2 hk

which, for the drainage of a radial cell, can be expressed asdP

dt

q

cre2h

(3.29)

Integration gives

rdP

dr

qr2

2re2 kh

C1 (3.30)

at the outer boundary the pressure gradient is zero, i.e. dP

dr0 therefore

C1 q

2kh and substitution into equation 3.30 gives

dP

dr

q2kh

1

r

r

re2

(3.31)

When integrated, this gives

P Pwf

Pr q

2khlnr

r 2

2re2

rw

re

or

Pr Pwf q

2khln

r

rw

r2

2re2

(3.32)

The term rw

2

2re2 is considered negligible, and in the case where the pressure at the external

radius, re is considered (including the skin factor, s, around the well),

Pe Pwf q

2khln

re

rw

12 s

(3.33)

If the average pressure is used, then the volume weighted average pressure of the

drainage cell is calculated as previously in the steady state flow regime, i.e.

P 2

r2e

Prdrrw

re

where rw and re are the wellbore and external radii as before, and P is the pressure in eachradial element, dr at a distance r from the centre of the wellbore. In this case,

P Pwf 2re

2

q2kh

rrw

re

lnrrw

r 2

2re2

dr

and integrating gives

i) rrw

re

lnrrw

dr =

r2

2ln

rrw

rw

re

1rrw

re

r 2

2dr

= r 2

2ln

rrw

rw

re

r 2

4

rw

re

re

2

2ln

re

rw

re

2

4

ii) r3

2re2 dr

rw

re

= r 4

8re2

rw

re

re

2

8

and substitution into equation 3.32 with inclusion of the skin factor gives

P Pwf q

2khln

re

rw

34

+s

(3.34)

The pressure differences (Pr - Pwf), (Pe- P

wf), ( P -Pwf

) do not change with time, whereas Pr,

Pe, Pw and P do change.

If the pressure drop from initial pressure conditions is required then equation 3.27 may be

written as:

P Po q

cV

to t (3.35)

P Pi qtcV

(3.36)

where q is the volume flow rate, c is the isothermal compressibility, V is the originalvolume to is a reference time after which flow starts, t is the flowing time, Po is the

pressure at the reference time and P is the pressure at time t after the flow starts. P is the

average reservoir pressure after time, t. Subtracting equation 3.36 from equation 3.34

gives

Using Initial Pressure, Pi

Pi - Pwf = q

2kh (ln rerw

- 34 +

2ktcre2 ) (3.37)

Generalised Reservoir Geometry:Flow Equation under SSS

Key factor: SSS feels boundaries

Finite amount of fluid in reservoir

Equation built for radial geometries, but non-radial shapes can be accommodated by Dietz shape factor

Using the average reservoir pressure and assuming no skin factor, the pressure drop is

described by equation 3.34 as

P Pwf q

2khln

re

rw

34

(3.34)

Expressing the terms lnre

rw

34

as

12

2lnre

rw

32

12

lnre

rw

2

32

12

lnre

rw

2

ln e32

=12

ln

re

rw

2

e3

2

=12

lnre

2 rw

2e3

2

The area drained (for a radial geometry) is re2 therefore the logarithm term becomes

4re2

4rw2e

3

2

4A

1.781 x 31.6 x rw2

where A is the area drained, 1.781 and Dietz shape factor, CA (for a well in a radial

drainage area) =31.6.

The final form of the generalised semi steady state inflow equation for an averagereservoir pressure is

P Pwf q

2kh12

ln4ACArw

2 s

(3.38)

For the pressure drop between initial reservoir pressure conditions and some bottom hole

flowing pressure during semi steady state flow, equation 3.37 can be expressed as

Pi - Pwf = q

2kh(1

2ln

4A

CArw2

2kt

cA) (3.39)

or

Pwf = Pi -q

2kh(1

2ln

4A

CArw2

2kt

cA) (3.40)

In a convenient dimensionless form, this can be expressed as

2khq

(P- Pwf )12

ln4ACArw

2 2 ktcrw

2

rw2

Aor

PD tD 1

2ln

4A

CArw2 2tD

rw2

A (3.41)

The term involving the wellbore radius can be accommodated by using the following

modified dimensionless time

t DA tD

rw2

Ain which case

PDtD 1

2ln

4A

CArw2 2tDA

Series of common shape factors with wells located at particular positions

Values of dimensionless tDA

i) infinite solution with error < 1% for tDA <X

X is value of maximum elapsed time during which reservoir infinite acting and Ei function can be used

ii) solution with less than 1% error for tDA >X

SSS solution used with error < 1% for elapsed time t

iii) solution exact for tDA >X

SSS solution for exact results for an elapsed time t

Real reservoir:

volume drained by well related to its flow rate

volume correlated to structural map to determine shape

shape factor values then used to locate position of well near boundaries

not an exact procedure and heterogeneity can alter pressure distribution

Application of CTR Solution in Well Testing

So far, pressure drops in a reservoir have been considered as a result of a flow rate in a well for a period of time

Therefore for given values of porosity, permeability, reservoir geometry and flow regime, the pressure at a particular distance can be calculated

In reality, only flow rates and pressure can be measured at the wells, and the most significant parameter to be verified is the permeability

This is part of well test analysis where the pressure in a well is measured continuously over time, the flow regime is identified and the appropriate flow equation applied to the data to determine the permeability

It is important to note that this section considers an initially undisturbed reservoir in which a well is brought on production.

Pressure decline at well measured through time

It is reasonable to expect the reservoir to go through a flow regime that starts in transience than changes to steady state or semi steady state.

In the following example, the reservoir is considered to be an isolated block, therefore SSS flow regime is expected after

Initial transient solution used to calculate permeability and skin factor

SSS to determine reservoir limits

Example 5. A well is tested by producing it at a constant flow rate of 238stm3/day (stock tank) for a period of 100 hours. The reservoir data and flowing bottomhole pressures recorded during the test are as follows: Data porosity, 18% formation volume factor for oil, Bo 1.2rm3/stm3 net thickness of formation, h 6.1m viscosity of reservoir oil, 1x10-3 Pas compressibility, c 2.18 x10-9Pa-1 wellbore radius, rw 0.1m initial reservoir pressure, Pi 241.3bar well flowrate (constant) 238stm3/day

Time (hours) Bottomhole flowing pressure

(bar) 0.0 241.3 1.0 201.1 2.0 199.8 3.0 199.1 4.0 198.5 5.0 197.8 7.5 196.5

10.0 195.3 15.0 192.8 30.0 185.2 40.0 180.2 50.0 176.7 60.0 173.2 70.0 169.7 80.0 166.2 90.0 162.7 100.0 159.2

1. Calculate the effective permeability and skin factor of the well. 2. Make an estimate of the area being drained by the well and the Dietz shape factor. Solution The description of the test is such that this is the first time the well has been put on production and the reservoir pressure will decline at a rate dictated by the solutions of the diffusivity equation. The pressure decline has been recorded at the wellbore (as in the table of data) and it is expected that there will be an unsteady state (transient) period initially followed by a semi steady state or steady state flow period. It is thought to be an isolated block therefore there would be a depletion of the reservoir pressure under semi steady state conditions expected. The initial unsteady state or transient flow period can be used to determine the permeability and skin factor of the well, and the subsequent semi steady state flow period can be used to detect the reservoir limits. SI units will be used at reservoir conditions, therefore flowrates are in m3/s and the formation volume factor for oil is used to convert from stock tank to reservoir volumes. The pressure related items are in Pascal.

1. The permeability and skin factor can be determined from the initial transient period using the line source solution:

Pwf Pi q

4khln

4ktcrw

2

2s

or

Pwf m lntc

(3.19)

Examining the data, the following are constant: initial pressure, Pi, permeability, k, , porosity, , viscosity, , compressibility, c, wellbore radius, rw, and skin factor, s. Both permeability and skin factor are unknown (but they are known to be constant). Therefore in equation 3.19, there is a linear relationship between the bottom hole flowing pressure, Pwf and the logarithm of time, lnt, the slope of the relationship, m, equal to

mq

4kh

F r o m t h i s , t h e u n k n o w n v a l u e , i . e . t h e p e r m e a b i l i t y , k , c a n b e c a l c u l a t e d . O n c e t h e p e r m e a b i l i t y i s k n o w n , t h e e q u a t i o n c a n b e r e a r r a n g e d t o d e t e r m i n e t h e o t h e r u n k n o w n , t h e s k i n f a c t o r , a s :

2s P i P wf

m ln

4kt

cr w2

A n y c o h e r e n t s e t o f d a t a p o i n t s c a n b e u s e d t o d e t e r m i n e t h e p e r m e a b i l i t y a n d s k i n , h o w e v e r , i t i s n o t c l e a r w h e n t h e d a t a r e p r e s e n t t h e l i n e s o u r c e s o l u t i o n . T h e r e f o r e a l l o f t h e p r e s s u r e d a t a a r e p l o t t e d a n d a l i n e a r f i t a t t a c h e d t o t h o s e d a t a w h i c h s h o w t h e l i n e a r r e l a t i o n s h i p b e t w e e n t h e b o t t o m h o l e f l o w i n g p r e s s u r e , P w f a n d t h e l o g a r i t h m o f t i m e , l n t .

time (hours)

Bottomhole flowing pressure

(bar)

ln time

0.0 241.3 1.0 201.1 0.0 2.0 199.8 0.7 3.0 199.1 1.1 4.0 198.5 1.4 5.0 197.8 1.6 7.5 196.5 2.0

10.0 195.3 2.3 15.0 192.8 2.7 30.0 185.2 3.4 40.0 180.2 3.7 50.0 176.7 3.9 60.0 173.2 4.1 70.0 169.7 4.2 80.0 166.2 4.4 90.0 162.7 4.5 100.0 159.2 4.6

543210150

160

170

180

190

200

210

P

Pressure- time data (log to base e)

ln flowing time, t (hours)

Bott

om

hole

flow

ing

pre

ssu

re,

Pw

f (b

ar)

slope = 1.98 bar/unit

T h e p l o t s o f b o t t o m h o l e f l o w i n g p r e s s u r e s h o w t h a t t h e t r a n s i e n t p e r i o d ( f o r w h i c h t h e l o g a r i t h m a p p r o x i m a t i o n i s v a l i d ) l a s t s f o r a p p r o x i m a t e l y 4 h o u r s a n d f r o m t h e p l o t , t h e s l o p e , m , c a n b e d e t e r m i n e d t o b e 1 . 9 8 b a r / l o g c y c l e . S u b s t i t u t i n g t h i s i n t o t h e e q u a t i o n g i v e s :

k q B o

4 m h

2 3 8 x 1 . 2 x 1 x 1 0 3

2 4 x 3 6 0 0 x 4 x 1 . 9 8 x 1 0 5 x 6 . 1 2 1 8 x 1 0 1 5 m 2 2 1 8 m D

( c o n v e r t i n g f r o m s t o c k t a n k c u b i c m e t r e s / d a y t o r e s e r v o i r c u b i c m e t r e s / s e c o n d a n d f r o m b a r t o P a s c a l p r o d u c i n g a p e r m e a b i l i t y i n t e r m s o f m 2 w h i c h i s t h e n c o n v e r t e d t o m D ) .

To determine the skin factor, the slope, m, of the line is theoretically extrapolated to a convenient time. This is usually a time of 1 hour. The bottomhole pressure associated with this time is calculated and this is used to determine a pressure drop (Pi - Pwf ) during the time (t1 hour - t 0). This is then equal to the pressure drop calculated from the ln function plus an excess caused by the skin. In this case, a real pressure measurement was recorded at time 1 hour, but this is not necessarily the same number as calculated from the extrapolation of the linear section of the relationship since the real pressure recorded at time 1 hour may not be valid for use with the Ei function, i.e. although it was recorded, it may have been too early for the Ei function to accurately approximate the reservoir flow regime.

In this case P1 hour =201.2bar and therefore

2sPi P1 hour

m ln

4kt

crw2

241.3 201.2

1.98 ln

4x218x10 -15x3600

1.781x0.18x1x10 3 x2.18x10 9 x0.12

2s=20.25-13.02 = 7.23 s=3.6

2. To determine the area drained and the shape factor, the data from the semi steady state flow regime are required. From equation 3.29, there will be a linear relationship between bottomhole flowing pressure and time. This is related to the area of the drained volume and the shape factor. To determine the gradient of the pressure decline, the bottomhole flowing pressure and time are plotted using Cartesian co-ordinates

120100806040200150

160

170

180

190

200

210

Pressure- time data

Flowing time, t (hours)

Bott

om

hole

flow

ing

pre

ssu

re,

Pw

f (b

ar)

slope = 0.35 bar/hour

F r o m t h e p l o t , t h e g r a d i e n t i s d e t e r m i n e d t o b e - 0 . 3 5 b a r / h o u r o r - 9 . 7 2 P a / s . T h i s i s r e l a t e d t o t h e v o l u m e t r i c c o m p r e s s i b i l i t y o f t h e r e s e r v o i r , i . e .

d P

d t

q

c A h

w h e r e q i s t h e f l o w r a t e , c i s t h e c o m p r e s s i b i l i t y , A i s t h e a r e a o f t h e r e s e r v o i r , h i s t h e t h i c k n e s s a n d i s t h e p o r o s i t y . T a k i n g a c c o u n t o f t h e f o r m a t i o n v o l u m e f a c t o r , B

o,

A q B o

c h d P

d t

A 2 3 8 x 1 . 2

2 4 x 3 6 0 0 x 2 . 1 8 x 1 0 - 9 x 6 . 1 x 0 . 1 8 x - 9 . 7 2

A = 1 4 2 0 7 6 m 2

T h e s e m i s t e a d y s t a t e i n f l o w e q u a t i o n i s

P w f = P i -q

2 k h(

1

2l n

4 A

C A r w2

2 k t

c A+ s )

T h e l i n e a r e x t r a p o l a t i o n o f t h i s l i n e t o s m a l l v a l u e s o f t g i v e s t h e s p e c i f i c v a l u e o f P w f o f 1 9 4 . 2 b a r a t t = 0 . I n r e a l i t y , a t t = 0 , t h e f l o w r a t e h a s n o t s t a r t e d , s o t h i s w i l l b e n a m e d P 0 . I n s e r t i n g t h i s v a l u e i n e q u a t i o n 3 . 3 9 a t t = 0 , c o n v e r t i n g b a r t o P a s c a l a n d i n c l u d i n g t h e s k i n f a c t o r g i v e s :

P i P 0 q

4 k hln

4 A r w

2 ln C A 2 s

i . e .

(2 4 1 . 3 1 9 4 . 2 ) x 1 0 5 1 . 9 8 x 1 0 5 ln4 x 1 4 2 0 7 61 . 7 8 1 x 0 . 1 2 ln C A 2 x 3 . 6 2

1 7 . 2 8 + 7 . 2 4 - 2 3 . 7 9 = 0 . 7 3 = l n C A C A = 2 . 0 8

From Table 5, this is close to the configuration in Figure 11.

2

1

Figure 11 Well configuration for Dietz shape factor of 2.0769

In the constant terminal rate solution of the diffusivity equation, the rate is known to be constant at some part of the reservoir and the pressures are calculated throughout the reservoir.

Conversely, in the constant terminal pressure solution, the pressure is known to be constant at some point in the reservoir, and the cumulative flow at any particular radius can be calculated.

The Constant Terminal Pressure Solution

The constant terminal pressure solution is not as confusing as the constant terminal rate solution simply because less is known about it. Only one constant terminal pressure solution is available, so there is no decision to be made over which to use as in the case of the constant terminal rate solutions.

Hurst and Van Everdingen produced the solutions for cases of an infinite radial system with a constant pressure at the inner boundary and for constant pressure at the inner boundary and no flow across the outer boundary.

These can model, for example, a wellbore whose bottomhole flowing pressure is held constant whilst flow occurs in the reservoir, or they can model a reservoir surrounded by an aquifer.

The same geometrical and property conditions apply as for the constant terminal rate solutions: a radial geometry of constant thickness with a well in the centre, and with fixed rock and fluid properties throughout, however, in this case there is a pressure drop from an initial pressure to some constant value.

In the case of aquifer encroachment, the radius of the “well” is the radius of the initial oil water contact.

The constant terminal pressure solution is most widely used for calculating the water-encroachment (natural water influx) into the original oil and gas zone due to water drive in a reservoir.

Hurst and van Everdingen Solution

As mentioned previously, the radial diffusivity equation in dimensionless form is: 1

rD

rD

rD

PD

rD

PD

tD

(4.1)

where the dimensionless terms are:

dimensionless radius: rDr

ro

dimensionless external radius: reDre

ro

dimensionless time: t Dktcro

2

dimensionless pressure drop: PDPi P

Pi Pro

where ro = outer radius of the oil reservoir (i.e. the oil water contact)

PD = 0 at tD = 0 for all rD

PD = 1 at rD = 1 for all tD>0

PD

rD rDreD

=0 for all tD>0

Since the instantaneous production rate, q

q2rohk

Pr

then the cumulative produced water, We

We qdt0

t

Hurst and Van Everdingen developed solutions for the constant terminal pressure condition of the form:

qD

(tD

)q

2khP (4.2)

where qD(tD) = dimensionless influx rate evaluated at rD

= 1.0 and which describes the change in rate from zero to q due to pressure drop P applied at the outer

reservoir boundary ro at time t = 0.

The dimensionless cumulative produced water volume, QD(tD)

QD

(tD

) qDdt

D

0

t D

2khP

k

cro2

qdt0

t

(4.3)

from which the cumulative produced water is

We = 2hcr

o2PQ

D(t

D) (4.4)

where We is the cumulative water influx due to pressure drop P imposed at the radius ro at the initial time, t=0. QDtD is the dimensionless cumulative water influx function giving the

dimensionless influx per unit pressure drop, P imposed at the reservoir/ aquifer boundary at time t=0.

Equation 4.4 is often expressed as We = UPQD(tD) (4.5) where: U = 2fhcro2 (4.6)

f = “aquifer constant” for radial geometry describing the proportion of the aquifer in contact with the oil rim as shown in figure 12.

WATER

OIL

re

ro

fraction, f = (encroachment angle)

360

360

Figure 12 Illustration of nomenclature for water influx problems

The solutions, QD(tD) are prepared in Tables 8 and 9 as functions of tD. These tables are described in Lee and are also available as equations for direct use in

spreadsheets. The use of the tables depends on whetherthe reservoir is infinite or bounded. (a) Bounded Aquifer (Table 9). Irrespective of the geometry, there is a value of tD for which the dimensionless water influx reaches a constant value:

QDMax

1

2reD2 1 (4.7)

reD = re/ro If QD in equation 4.7 is used in equation 4.4, for a full aquifer (f = 1.0), the result is

We = 2hcro2P (re2 - ro2

2ro2 ) = ( re2 - ro2 )hcP (4.8)

(b) Infinite aquifer (Table 8). There is no maximum value of QD(tD) reached in this case since the water influx is always governed by transient flow conditions.

This is also equal to the total influx occurring, assuming that the P is instantaneously transmitted throughout the aquifer. Therefore, once the plateau level of QD(tD) has been reached, it means that the minimum value of tD at

which this occurs has been sufficiently long for the instantaneous drop P to be felt throughout the aquifer. The plateau level of Q D(tD) is then the maximum

dimensionless water influx resulting from such a pressure drop.

Example 6. Water influx: Hurst and Van Everdingen’s Constant Terminal Pressure solution A reservoir is surrounded by an aquifer with an external boundary as shown in figure 13.

radius of the oil reservoir, r o

external radius, re

1525m

4575m

oil

water

Figure 13 Plan of the reservoir/aquifer Data porosity, 23% net thickness of formation, h 50m viscosity of reservoir oil, 0.7x10-3 Pas compressibility, c 1.7 x10-9Pa-1 permeability, k 170mD oil reservoir radius, ro 1525m external radius, re 4575m instantaneous pressure change, P 10bar

1) Calculate the water influx at times of 0.1 year, 0.5 year, 1.0 year, 1.5 years 2.0 years and 2.2 years after the instantaneous pressure drop at the oil water contact. 2) Calculate the water influx if it is assumed that the same pressure drop is transmitted simultaneously through the aquifer.

Solution The constant terminal pressure solutions shown in Tables 8 and 9 are used to find the dimensionless cumulative water influx (at a particular dimensionless external radius) at a particular dimensionless time from which the cumulative water influx is calculated. The first step is to calculate the dimensionless time, then look up the table for the corresponding dimensionless cumulative water influx.

t Dktcro

2

t D170x10-15t

0.23x0.7x103x1.7x109x15252

tD = 2.7x10-7t

aquifer constant, f80º360º

0.22

cumulative water influx, We = 2fchro

2P QD(tD) We = 2 x 0.22x0.23x1.7x10-9x50x15252x10x105x QD(tD) We = 62847.6 QD(tD) dimensionless external radius,

reD

re

ro

4575

1525 3

time time tD QD (ReD=3)

We

(year) (s) (table 9)

(m3)

0.1 0.1x365x24x3600 = 3153600

2.7x10 -7x 3153600 = 0.85

1.41 88615

0.5 0.5x365x24x3600 = 15768000

2.7x10 -7x 15768000 = 4.26

3.32 208654

1.0 1.0x365x24x3600 = 31536000

2.7x10 -7x 31536000 = 8.52

3.87 243220

1.5 1.5x365x24x3600 = 47304000

2.7x10 -7x 47304000 = 12.77

3.97 249505

2.0 2.0x365x24x3600 = 63072000

2.7x10 -7x 63072000 = 17.03

4.00 251390

2.2 2.2x365x24x3600 = 69379200

2.7x10 -7x 69379200 = 18.73

4.00 251390

From the table the increase in cumulative water influx through time can be seen. After tD of 17.03, QD(tD) becomes constant at 4.00 indicating that the maximum water influx for this reservoir under 10 bar pressure drop is 251390m3. 2) If the pressure drop is instantaneously transmitted through the reservoir, the expansion and therefore encroachment of the water is We = (re

2 - ro2)fchP

We = x (45752 - 15252)x0.22x0.23x1.7x10-9x50x10x105 We = 251390m3 This is the same as calculated by the constant terminal pressure solution but without the variation in water influx through time. Note in the example the declining rate of water influx wit h time and also that unlike a steady-state system, the values of influx do not double for doubling of time. If the permeability of the aquifer rock is very low, for instance, the aquifer may provide only a small volume of water to the reservoir during it’s producing life, which may essentially produce as a depletion type reservoir.

Superposition

In the analyses so far, the well flow rate has been instantly altered from zero to some constant value.

In reality, the well flowrates may vary widely during normal production operations and of course the wells may be shut in for testing or some other operational reason.

The reservoir may also have more than a single well draining it and consideration must be taken of this fact.

There may be some combination of several wells in a reservoir and/or several flowrates at which each produce. The calculation of reservoir pressures can still be done using the previous simple analytical techniques if the solutions for each rate change, for example, are superposed on each other.

In other words, the total pressure drop at a wellbore can be calculated as the sum of the effects of several flowrate changes within the well, or it may be the sum of the effects caused by production from nearby wells.

There is also the possibility of using infinite acting solutions to mimic the effects of barriers in the reservoir by using imaginary or image wells to produce a pressure response similar to that caused by the barrier. Mathematically, all linear differential equations fulfill the following conditions:  (i) if P is a solution, then C x P is also a solution, where C is a constant.

(ii) if both P1 and P2 are solutions, then P1 + P2 is also a solution. These two properties form the basis for generating the constant terminal rate and constant terminal pressure cases.

The solutions may be added together to determine the total effect on pressure, for example, from several applications of the equation.

This is illustrated if a typical problem is considered: that of multiple wells in a reservoir.

Effects of Multiple Wells

In a reservoir where more than one well is producing, the effect of each well’s pressure perturbation on the reservoir is evaluated independently

(i.e. as though the other wells and their flow rate/ pressure history did not exist),

then the pressure drop calculated at a particular well at a particular time is the simple addition of all of the individual effects superimposed one effect upon the other.

Consider 3 wells, X, Y and Z, which start to produce at the same time from an infinite acting reservoir (figure 14).

Superposition shows that: (Pi-Pwf)Total at Well Y 

= (Pi -P)Due to well X + (Pi-P)Due to well Y + (Pi-P)Due to well Z

Assuming unsteady state flow conditions, the line source solution can be used to determine the pressure in well Y.

It is assumed here that the logarithm function can be used for well Y itself and that there will be a skin around the well.

The effects of wells X and Z can be described by the Ei function.

There is no skin factor associated with the calculation of pressure drop caused by these wells, since the pressure drop of interest is at well Y (i.e. even if wells X and Z have non-zero skin factors, their skin factors affect the pressure drop only around wells X and Z).

The total pressure drop is then:

Y

2wYY

Y at well totalwfi 2S4kt

crln

kh4

q)P(P

4kt

crEi

kh4

q 2XYX

4kt

crEi

kh4

q 2ZYZ

Where

qY is the flowrate from well Y

qX is the flowrate from well X

qZ is the flowrate from well Z

rwY is the radius of well Y

rXY is the distance of well Y from the X well

rZY is the distance of well Z from the X well

the rest of the symbols have their usual meaning

This technique can be used to examine the effects of any number of wells in an infinite acting reservoir.

This could be to predict possible flowing well pressures amongst a group of wells, or to deliberately use the interaction between wells to check reservoir continuity.

These interference tests and other extended well tests are designed to characterise the reservoir areally rather than to determine only the permeability and skin factor around individual wells.

Example 7. Two wells, well 1 and well 2, are drilled in an undeveloped reservoir.

Well 1 is completed and brought on production at 500stm3/day and produces for 40 days at which time Well 2 is completed and brought on production at 150stm3/day.

Using the data provided, calculate the pressure in Well 2 after it has produced for 10 days (and assuming Well 1 continues to produce at its flowrate).

Therefore, Well 1 produces for 50days when its pressure influence is calculated; Well 2 produces for 10 days when its pressure influence is calculated. The wells are 400m apart and the nearest boundary is 4000m from each well.  Data porosity, , 21%formation volume factor for oil, Bo 1.4rm3/stm3

net thickness of formation, h, 36mviscosity of reservoir oil, 0.7x10-3 Pascompressibility, c 8.7 x10-9Pa-1

permeability, k 80mD wellbore radius, rw (both wells) 0.15m

initial reservoir pressure, Pi 180.0bar

Well 1 flowrate (constant) 500stm3/dayWell 2 flowrate (constant) 150stm3/dayskin factor around both wells 0

Solution The line source solution is used to determine the bottomhole flowing pressure at Well 2 after 10 days prod uction, accounting for the effect of 50days production from Well 1. Checks are made to ensure that: i) there has been adequate time since the start of production to allow the line source solution to be accurate ii) the reservoir is infinite acting. A Check Ei applicability line source not accurate until

k

cr100t

2w

15-

29-3

80x10

x0.15x8.7x10.7x10100x0.21x0t

t >36s time is 50 days, therefore line source is applicable.

B Check reservoir is infinite acting

the reservoir is infinite acting if the time, 4k

crt

2e

i.e. 15-

293

4x80x10

x4000x8.7x1000.21x0.7x1t

t < 63945000 t <740 days therefore line source solution is applicable. The bottomhole flowing pressure at Well 2 is the sum of the pressure drops caused by its production and by the pressure drop generated by the production of Well 1. Pwf at Well 2 = Pi -Pwell2 flowing for 10 days -Pwell1 flowing for 40+10 days 400m away

A) At 10 days, contribution to pressure drop from production from Well 2 check ln approximation to Ei function

the ln approximation i s valid if the time, k

cr25t

2

15-

293

80x10

x0.15x8.7x107x1025x0.21x0.t

t > 9s therefore ln approximation is valid.

4kt

crln

kh4

BqPP

2wo

iwf

4kt

crln

kh4

BqP-P

2wo

wfi

x36x80x1024x3600x4

x1.4150x0.7x10

kh4

Bq15

3o

= -47011

0x10x24x3604x80x10

15.x0x8.7x10x0.7x101.781x0.21

4kt

cr15-

2932w

= 185x10 -9

Pi - Pwf = -47011x ln(1 85x10 -9) Pi - Pwf = -47011x -15.5 Pi - Pwf =728671Pa

B) At 10 days production from well 2, well 1 has been producing for 50 days and its contribution to pressure drop at Well 2 is calculated as follows. check ln approximation to Ei function

the ln approximation is valid if the time, k

cr25t

2

15-

293

80x10

x400x8.7x107x1025x0.21x0.t

t > 63945000s t > 740 days therefore ln approximation is not valid and the Ei function is used.

4kt

crEi

kh4

BqP-P

22-1o

1 by Well caused at Well2 wfi

x36x80x1024x3600x4

x1.4500x0.7x10

kh4

Bq15

3o

= 156704

0x50x24x3604x80x10

400xx8.7x1000.21x0.7x1

4kt

cr15-

29322-1

= 0.148

Ei(-0.148) = -1.476 Pi - Pwf at Well 2 caused by Well 1 = -156704x-1.476 Pi - Pwf at Well 2 caused by Well 1 = 231295Pa Pwf Well2 = 180.0 - 7.3 - 2.3 Pwf Well2 = 170.4bar

Principle of Superposition and Approximation of Variable - Rate Pressure

Histories The previous section illustrated the effect of the production from several wells in a reservoir on the bottomhole flowing pressure of a particular well.

Of equal interest is the effect of several rate changes on the bottomhole pressure within a particular well.

This is a more realistic situation compared to those illustrated previously where a well is simply brought on production at a constant flowrate for a specific period of time.

For instance, a newly completed well may have several rate changes during initial cleanup after completion, then during production testing then finally during production as rates are altered to match reservoir management requirements (for example limiting the producing gas oil ratio during production).

A simple pressure and flowrate plot versus time would resemble figure 15

The well has been brought onto production at an initial flowrate, q1.

The bottomhole flowing pressure has dropped through time (as described by the appropriate boundary conditions and the flow regime) until at time t1, the flowrate has been increased to q2 and this change from q1 to q2 has altered the bottomhole flowing pressure (again as described by the boundary conditions and the flow regime).

The total (i.e. the real bottomhole flowing pressure) is calculated by summing the pressure drops caused by the flowrate q1 bringing the well on production, plus the pressure drop created by the flowrate change q2 - q1 for any time after t1.

During the first period (q1) the pressure drop at a time, t, is described by

P(t) = Pi - Pwf = q1

2kh PD(t)

where PD(t) is the dimensionless pressure drop at the well for the applicable boundary condition.

For times greater than t1, the pressure drop is described by

P(t) = q1

2kh PD(t) + (q2 - q1)

2kh PD(t - t1) (5.3)

In this case, the pressure drop is that caused by the rate q1 over the duration t, plus the pressure drop caused by the flowrate change q2 - q1 over the duration t - t1.

In fact, the pressure perturbation caused by q1 still exists in the reservoir and is still causing an effect at the wellbore.

On top of that, the next perturbation caused by flowrate change q2 - q1 is added or superposed to give the total pressure drop (at the wellbore in this case).

In mathematical terms:

0 ≤ t ≤ t 1 : P(t) = q1

2kh PD(t) (5.4)

t > t1 P(t) = q1

2kh PD(t) + (q2 - q1)

2kh PD(t - t1) (5.5)

In the 2nd equation, the first term is P from flow at q1 :

2nd term is the incremental term P caused by increasing rate by an increment (q2-q1).

These expressions are valid regardless of whether q2 is larger or smaller than q1 so that even if the well is shut in, the effects of the previous flowrate history are still valid.

The dimensionless pressure drop function depends as mentioned on the flow regime and boundaries. If unsteady state is assumed and the line source solution applied, then

PD = Pi - Pwf

q/2kh = -

12 Ei (

-crw2

4kt ) (5.6)

and the equation for time, t less than or equal to t1 would be as expected

P(t) = - q1

4kh Ei (

-crw2

4kt ). (5.7)

For times greater than t1 the additional pressure drop is added to give

P(t) = - q1

4kh Ei (

-crw2

4kt ) - (q2 - q1)

4kh Ei (

-crw2

4k(t-t1) ) (5.8)

This approach can be extended to many flowrate changes as illustrated in figure 16.

This leads to a general equation

P(t)q

1

2khP

D(t)

(q2 q

1)

2khP

D(t t

1)

(q3 q

2)

2khP

D(t t

2) ...

(q

n q

n 1)2kh

PD

(t tn 1) (5.9)

or

P(t)q

1

2khP

D(t)

qi q

i 1

q1i2

n

PD

(t ti 1 )

(5.10)

This is the general form of the principle of superposition for

multi rate history wells. For the specific case where the well

is shut in and the pressure builds up, an additional term is

added to reflect this.

Assuming that the well was shut in during the nth flowrate

period, the pressure builds during the shut in timet (i.e.

t starts from the instant the well is shut in) back up

towards the initial reservoir pressure according to

( 5.11) where

Pws is the shut in bottomhole pressure

tn-1 is the total producing time before shut in

t is the closed in time from the instant of shut in.

Pi Pws q1

2khPD(t)

q i q i 1

q112

n

PD(t n-1 t i 1 + t)

qn-12kh

PD(t)

Effects of Rate Changes

Illustration of problem as follows:

Well brought onto production at q1 until time t1 then flowrate increased to q2

q2 continues till t2, flowrate increased to q3.

Assume reservoir in unsteady state and line source applicable

Seeking Pwf, skin may be present

Each flowrate change produces a pressure perturbation that moves into formation

Figure illustrates:

The pressure drop produced by bringing the well onto production is calculated by the logarithmic approximation of the Ei function (it is assumed that the checks have been made to the applicability of the Ei function and its logarithmic approximation).

P1 Pi Pwf

1 q14kh

lncrw

2

4kt

2s

The next pressure drop is that produced by the flowrate change q2 - q1 at time, t1.

It is still the bottomhole flowing pressure that is to be determined, therefore any skin zone will still exist and still need to be accounted for. The second pressure drop is:

P2 P

i P

wf

2(q

2- q

1)

4khlncr

w2

4k(t - t1 )

2s

And finally the third pressure drop is:

P3 P

i P

wf

3 (q

3- q

2)

4khlncr

w2

4k(t - t2)

2s

The total pressure drop at the wellbore caused by all of the flowrate changes is (Pi - Pwf )= P1 + P2 + P3

Example 8. A well is completed in an undeveloped reservoir described by the data below. The well flows for 6 days at 60 stm3/day and is then shut in for a day. Calculate the pressure in an observation well 100m from the flowing well. Data porosity, , 19%formation volume factor for oil, Bo 1.3rm3/stm3

net thickness of formation, h, 23mviscosity of reservoir oil, 0.4x10-3 Pascompressibility, c 3 x10-9Pa-1

permeability, k 50mD wellbore radius, rw (both wells) 0.15m

external radius, re 6000m

initial reservoir pressure, Pi 180.0bar

flowrate (constant) 60stm3/dayskin factor around well 0 The observation well is 100m from the flowing well.

Solution The line source solution is used to determine the pressure in the observation well after 6 days production from the flowing well then 1 day shut in at the flowing well. Checks are made to ensure that:  i) there has been adequate time since the start of production to allow the line source solution to be accurate ii) the reservoir is infinite acting. A Check Ei applicability line source not accurate until

t >10.3s time is 6 days, therefore line source is applicable.

t 100crw

2

k

t 100x0.19x0.4x10 -3 x3x10 9 x0.152

50x10-15

B Check reservoir is infinite acting the reservoir is infinite acting if the time,

i.e.

t < 41040000st <475 days therefore line source solution is applicable.

t cre

2

4k

t 0.19x0.4x10 3 x3x10 9 x6000 2

4x50x10-15

The pressure drop at the observation well is described by 

 

Checking for the validity of the ln approximation, the ln approximation is valid if the time,

t > 1140000st > 13 days therefore ln approximation is not valid.

)t4k(t

cr)Eiq(q

4kt

crEiq

kh4

BPP

1

2

12

2

1o

wellobsi

t 25cr2

k

t 25x0.19x0.4x10 3 x3x10 9 x1002

50x10-15

B

o

4kh

0.4x10 3 x1.3

4x50x1015 x23 = -35982857

cr1-22

4kt

0.19x0.4x103 x3x109 x1002

4x50x10-15 x7x24x3600 = 0.019

cr1-22

4k(t - t1)

0.19x0.4x10 3 x3x10 9x1002

4x50x10-15 x(7 - 6)x24x3600 = 0.132

Ei(-0.019) = -3.405 Ei(-0.132) = -1.576

Pi Pobs well 3598285760

24x3600x 3.405

0 - 60

24x3600x 1.576

Pi Pobs well 35982857 2.36x10 3 1.09x103 Pi - Pobs well = 45698Pa = 0.5bar Pobs well = 180.0 - 0.5 = 179.5bar

Simulating Boundary Effects

One of the intriguing possibilities of the application of the principle of superposition to reservoir flow is in simulating reservoir boundaries. It is clear that when a well in a reservoir starts production, there will be a period where the flow regime is unsteady while the reservoir fluid reacts to the pressure perturbation as if the volume of the reservoir was infinite (i.e. an infinite acting reservoir). Once the boundaries are detected, there is a definite limit to the volume of fluid available and the pressure response changes to match that of, for example, semi steady state or steady state flow. This assumes that the pressure perturbation reaches the areal boundary at the same time, i.e. if the well was in the centre of a circular reservoir, the pressure perturbation would reach the external radius at all points around the circumference at the same time (assuming homogeneous conditions).

If the well was not at the centre then some parts of the boundary would be detected before all of the boundary was detected. This means that some of the reservoir fluid is still in unsteady flow whilst other parts are changing to a different flow regime. This would appear to render the use of the line source solution invalid, however, the effect of the nearest boundary in an otherwise infinite acting reservoir has the same effect as the interaction of the pressure perturbations of two wells next to each other in an infinite acting reservoir. So if an imaginary well is placed at a distance from the real well equal to twice the distance to the boundary, and the flowrate histories are identical, then the principle of superposition can be used to couple the effect of the imaginary well to the real well in order to calculate the real well’s bottomhole flowing pressure. Figure 18 illustrates the problem and the effect of superposition. Figure 19 shows a simplification of the model.

This shows a plane-fault boundary in an otherwise infinite acting reservoir, as in the top figure . To determine the pressure response in the well, the line source solution can be used until the pressure perturbation hits the fault. Thereafter there are no solutions for this complex geometry. However, the reservoir can be modelled with an infinite acting solution if a combination of wells in an infinite-acting system that limit the drainage or flow around the boundary is found. The bottom of figure 18 indicates 1 image well with the same production rate as the actual well is positioned such that the distance between it and the actual well is twice the distance to the fault of the actual well. No flow occurs across the plane midway between the two wells in the infinite-acting system, and the flow configuration in the drainage area of each well is the same as the flow configuration for the actual well. Pressure communication crosses the drainage boundary, but there is no fluid movement across it and the problem of the flow regime has been resolved: the real well can be thought of as reacting to the flowrate in it and to the pressure drop produced by the imaginary well on the opposite side of the fault.

The pressure drop is therefore:

Pi Pwf q

4khln(crw

2

4kt) 2s

q4kh

Ei c(2L)2

4kt

where the symbols have their usual meaning, and L is the distance from the real well to the fault. The skin factor is used in the actual well, but not in the other (image) well since it is the influence of this image well at a distance 2L from it that is of interest.

Example 9. A well in a reservoir is produced at 120 stm3 /day for 50 days. It is 300m from a fault. Using the data given, calculate the bottomhole flowing pressure in the well and determine the effect of the fault on the bottomhole flowing pressure. Data porosity, , 19% formation volume factor for oil, Bo 1.4rm3/stm3 net thickness of formation, h, 20m viscosity of reservoir oil, 1x10-3 Pas compressibility, c 9 x10-9Pa-1 permeability, k 120mD wellbore radius, rw 0.15m external radius, re 4000m initial reservoir pressure, Pi 300.0bar flowrate (constant) 120stm3/day flowrate period, t 50days distance to fault, L 300m skin factor around well 0

S o l u t i o n T h e l i n e s o u r c e s o l u t i o n w i l l b e u s e d t o a s s e s s t h e e f f e c t s o f t h e r a t e a n d t h e b o u n d a r y o n t h e b o t t o m h o l e f l o w i n g p r e s s u r e . U s i n g a n i m a g e w e l l 6 0 0 m f r o m t h e r e a l w e l l ( i . e . 2 x d i s t a n c e t o t h e f a u l t ) w i t h i d e n t i c a l p r e s s u r e a n d r a t e h i s t o r y a s t h e r e a l w e l l , t h e e f f e c t o f t h e b o u n d a r y o n t h e i n f i n i t e a c t i n g r e s e r v o i r c a n b e o v e r c o m e . T h e b o t t o m h o l e f l o w i n g p r e s s u r e i n t h e r e a l w e l l w i l l b e t h e p r e s s u r e d r o p c a u s e d b y t h e p r o d u c t i o n f r o m t h e r e a l w e l l p l u s a p r e s s u r e d r o p f r o m t h e i m a g e w e l l 6 0 0 m a w a y . T h e l i n e s o u r c e s o l u t i o n w i l l b e u s e d . C h e c k s a r e m a d e t o e n s u r e t h a t : i ) t h e r e h a s b e e n a d e q u a t e t i m e s i n c e t h e s t a r t o f p r o d u c t i o n t o a l l o w t h e l i n e s o u r c e s o l u t i o n t o b e a c c u r a t e i i ) t h e r e s e r v o i r i s i n f i n i t e a c t i n g . A C h e c k E i a p p l i c a b i l i t y l i n e s o u r c e n o t a c c u r a t e u n t i l

t 1 0 0 c r w

2

k

t 1 0 0 x 0 . 1 9 x 1 x 1 0 - 3 x 9 x 1 0 9 x 0 . 1 5 2

1 2 0 x 1 0 - 1 5

t > 3 2 s t i m e i s 5 0 d a y s , t h e r e f o r e l i n e s o u r c e i s a p p l i c a b l e .

B Check reservoir is infinite acting the reservoir is infinite acting if the time, t

cre2

4k

t 0.19x1x10 3 x9x10 9 x4000 2

4x120x10 -15

t < 57000000s t <660 days therefore line source solution is applicable. Checking for the validity of the ln approximation, for the real well the ln approximation is valid if the time,

t 25cr 2

k

t 25x0.19x1x10 3 x9x10 9 x0.152

120x10 -15

t > 8s therefore ln approximation is valid. Checking for the validity of the ln approximation, for the image well

the ln approximation is valid if the time, t 25c(2L)2

k

t 25x0.19x1x10 3 x9x10 9 x6002

120x10 -15

t > 128250000s t> 1484 days therefore ln approximation is not valid. For this case, then, the ln approximation will predict the bottomhole flowing pressure around the real well, but the effect of the image well 600m away will need to be predicted by the Ei function.

Pi Pwf qBo

4khlncrw

2

4kt

qBo

4khEi

c(2L)2

4kt

qBo

4kh

120x1x10 3 x1.4

24x3600x4x120x10 15x20= -64473

crw2

4kt

1.781x0.19x1x10 3 x9x10 9 x0.152

4x120x10 -15 x50x24x3600= 33.1x10-9

c(2L)2

4kt

0.19x1x10 3 x9x10 9 x6002

4x120x10 -15x50x24x3600= 0.297

Ei(-0.297) = -0.914

Pi Pwf 64473x ln(33.1x10 -9 ) 64473x 0.914

Pi - Pwf = 1110466 + 58928 =1169394Pa = 11.7bar Pwf = 300.0 - 11.7 = 288.3bar The fault 300m away pulled the bottomhole flowing pressure down by an extra 58928Pa or 0.6bar.

There are other examples of the use of image wells to mimic the effect of boundaries on flow. The larger networks require computer solution to relieve the tedium. To complicate the simple fault boundary described earlier, consider the effect of a well near the corner of a rectangular boundary. In this case, there are more image wells required to balance the flow from the real well. Figure 20 shows the boundary and the image wells.

Four pressure drop terms are required to determine the pressure at the actual well. The total pressure drop then is the sum of the pressure drops caused by all of the wells at the actual well. Pi - Pwf = (P)rw + (P)2L1 + (P)2L2 + (P)r3 (Pi-Pwf)Total at the actual well = (Pi -P)at the actual wellbore radius, rw + (Pi-P)Due to image well 1 at distance 2L1

+ (Pi-P)Due to image well 2 at distance 2L2

+ (Pi-P)Due to image well 3 at distance R3

The number and position of image wells can become complex.

In the apparently simple geometry of an actual well surrounded by two equidistant barriers, such as illustrated in figure 21, the flow can be balanced as before by defining image well, i1 on the right. On the left side, the barrier is balanced by image wells i2 and i3 (because seen from i2, there is a barrier with 2 wells on the other side - a real well and an image well). Now there is an imbalance in production across the right barrier, so image wells i4 and i5 are added. This unbalances the left barrier and image wells i6 and i7 are added. This should continue to infinity, however, since the line source solution is known to have little influence above a certain distance from the actual well, the number of image wells used can be fixed with no error in the approximation.

Even more complex patterns can be devised. Mathews, Brons and Hazebroek (Matthews, CS, Brons, F and Hazebroek, P, A Method for the Determination of Average Pressure in a Bounded reservoir. Trans. AIME.201) studied the pressure behaviour of wells completely surrounded by boundaries in rectangular shaped reservoirs. Figure 22 shows the network of wells set up to mimic the effect of the boundaries.

Summary The basic partial differential equation expressing the nature of fluid flow in a porous rock has been illustrated in the context of petroleum reservoirs. Only oil and water have been used as the simplifications for solving the diffusivity equation have required the compressibility of the fluid to be small and constant. This is the reason that the compressibility of the fluid in the examples has not changed with pressure as would be expected. So, for instance, the same value of compressibility is used for the fluid at the wellbore which may be under a lower pressure than the same fluid at, for example, the external radius of the reservoir. In gasses, the same diffusion process occurs, but the pressure dependence of the gas is accommodated by various mathematical devices which again lead to simple working solutions.

The assumptions made concerning the geological structure and the petrophysical properties of the rock may appear radical: to assume a reservoir is circular, horizontal and has identical permeability in all directions is a great simplification of the problem. Yet these simple analytical solutions allow an appreciation of the role of the fluids and the rock in a producing reservoir. For more realistic treatments of real reservoirs, approximations to the diffusivity equation are made from which simple algebraic relationships can be formed. This process is encapsulated in reservoir simulation where the reservoir (with its properties) is subdivided into small blocks within which the flow equations have been approximated by simple relationships. These can then be solved by a process of iteration to achieve an acceptable result. The great potential of this process is the ability to represent the shape of the reservoir and the changing properties, vertically and horizontally, throughout the reservoir. Figure 23 summarises the route taken through the analytical solutions for radial flow regimes examined in this chapter. The number of solutions is mathematically infinite; only a few are suitable for real reservoir problems.

PD versus tD - infinite radial system, constant rate at inner

boundary

PD versus tD - finite radial system with

closed exterior

boundary, constant rate

at inner boundary

PD versus tD - finite radial system with

closed exterior

boundary, constant rate

at inner boundary

Values of Exponential Integral, -Ei(-x)

Dietz shape factors for single well drainage areas