resengch10.pdf

1010Fluid Flow In Porous Media

CONTENTS

1 INTRODUCTION

2 CHARACTERISATION AND MODELLING OFFLOW PATTERNS2.1 Idealised Flow Patterns2.2 General Case2.2.1 Linear Horizontal Model of a Single Phase

Fluid2.2.1.1 Linearisation Of Partial Differential Flow

Equation For Linear Flow2.2.1.2 Conditions of Solution2.2.2 The Radial Model2.2.2.1 Range Of Application And Conditions Of

Solution2.3 Characterisation of the Flow Regimes by

their Dependence on Time

3 BASIC SOLUTIONS OF THE CONSTANTTERMINAL RATE CASE FOR RADIAL MODELS3.1 The Steady State Solution3.2 Non-Steady State Flow Regimes and

Dimensionless Variables3.3 Unsteady State Solution3.3.1 General Considerations3.3.2 Hurst and Van Everdingen Solution3.3.3 The Line Source Solution3.3.3.1 Range of Application and Limitations to

Use3.3.4 The Skin Factor3.4 Semi-Steady-State Solution3.4.1 Using The Initial Reservoir Pressure, Pi3.4.2 Generalised Reservoir Geometry: Flowing

Equation under Semi-Steady State Conditions3.5 The Application of the CTR Solution in

Well Testing

4. THE CONSTANT TERMINAL PRESSURESOLUTION

5. SUPERPOSITION5.1 Effects of Multiple Wells5.2 Principle of Superposition and Approximation

of Variable - Rate Pressure Histories5.3 Effects of Rate Changes5.4 Simulating Boundary Effects (Image Wells)

6. SUMMARY

SOLUTIONS TO EXERCISES

2LEARNING OBJECTIVES:

Having worked through this chapter the student will be able to:

Understand the nature of fluid flow in a porous medium and the relation betweentime, position and saturation

Understand the assumptions used in the derivation of the diffusivity equation

Understand the characterisation of the reservoir flow regime on the basis of time

Understand the application of the solutions of the diffusivity equation to steadystate flow, semi-steady state flow and transient flow

Understand the use of the line source solution in radial systems to determine thepressure at any point in a reservoir under transient flow conditions

Understand the application of line source solution to multiple well/ multiple ratehistories in a transient flow reservoir

Understand the basis of well test analysis, and use of the line source solution todetermine the reservoir permeability and skin factor

Understand the application of semi-steady state solutions to determine reservoirboundaries and their influence on flow rates.

Department of Petroleum Engineering, Heriot-Watt University 3


1 INTRODUCTION

The ability to determine the productivity of a reservoir and the optimum strategy tomaximise the recovery relies on an understanding of the flow characteristics of thereservoir and the fluid it contains. The physical means by which fluid diffuses througha rock (or any other porous medium) depends on the interaction between the fluid (andits properties) and the rock (and its properties). In terms of energy, the process mayat first sight appear to be similar in concept to the application of the general energyequation to flow through pipes, although in this case the container through which thefluid flows is made of very small tubes. It is precisely because of the geometry anddimensions of the tubes that the application of the general energy equation would beimpossible: the description of a real pore network in a whole reservoir would be toocomplex. Coupled with this is the interaction between the material of the tubes (orpores) and the fluids. Surface chemistry effects start to dominate the flow when verysmall tubes are considered and when multiphase flow occurs in them. Thus, complexforce fields are produced from not only the viscous pressure drop but also the effectsof surface tension and capillary pressure.

The combination of these factors dictates the nature of the fluid flow and one of theinitially unusual aspects is the time taken for pressure to change in the reservoir or forfluid to migrate from one location to another. For instance, if a large body of water,such as a swimming pool were drained, for all intents and purposes, the level of waterin the swimming pool would be the same as the water drained out. It would take anappreciable amount of time for the water to drain (i.e. it would not be instantaneous),but the pressure or level of the water in the pool would be the same at all locations ofthe pool. The pressure in the pool would equilibrate almost immediately. Contrast thiswith, for example, a water saturated reservoir rock in which the water could flow, butwhere the permeability of the reservoir and the compressibility and viscosity of thewater dictated that the transfer of the water through the reservoir was not instantane-ous (as in a swimming pool), but took an appreciable time. In this case pressurechanges in one part of the reservoir may take days, even years to manifest themselvesin other parts of the reservoir. In this case, the flow regime would not be steady statewhile the pressure was finding its equilibrium and a major problem, therefore, wouldbe that Darcys Law could not be applied until the flow regime became steady state.In some way, the diffusion through the reservoir needs to be examined: Darcys Lawis one expression of that diffusion process, but time dependent scenarios must also beexamined.

To illustrate this, consider the following model of a linear reservoir with a well at theleft side (figure 1).

4

1 2 3 4 5 6 7 8 9 10

Outlet

(constant flowrate)

Initialwaterprofile

Profileaftertime t

Tube number

Interconnecting, small diameter pipes

10 vertical tubes, 100mm diameter, arranged linearly

time, t after start of flow

top of tubes 0

-50

-100

-150

-200

-2501 2 3 4 5 6 7 8 9 10

tube number

bottom of tubes

heig

ht o

f wate

r in

tube

s

t=0t=1t=2t=3t=4t=5t=6t=7

Each tube contains water, the height of which represents the pressure at that part of thereservoir. The tubes are connected to each other at the base by a small diameter tubewhich restricts the flow. Under initial conditions, the height of the fluid is identical ineach of the tubes (assuming the model is level). The outlet at one end is at a lower levelthan the model and when it is opened the fluid immediately drains from the model andthe level of the water in the tubes decreases. The energy to drive this system is thepotential energy stored in the height of the water columns: there is no high pressureinlet to the model. As is shown in figure 1, to reduce the pressure in the model, the fluidneeds to be expelled, but because of the permeability of the rock (the restrictions inthe bottoms of the tubes) it takes time for the fluid in the tubes nearest the outlet tomove (or expand in the case of pressurised fluid in a reservoir) and therefore it takestime for the pressure to change. When the flow is started from the outlet, there is animmediate reduction in the pressure in tube 1 and this pressure perturbation movesthrough the rest of the fluid at a rate dictated by the rock permeability and fluidproperties. This produces a variation in the pressure along the model. The pressureprofile takes time to develop from the outlet (at tube 1) to the tube farthest from theoutlet (tube 10) and at time, t=1, the pressure in tube 10 is still equal to the pressureat the initial time, t=0. This is termed a transient flow condition as the fluid is tryingto reach pressure equilibrium. When the fluid in tube 10 starts to expand and flow, allof the fluid in the whole model is now expanding and flowing to the outlet. Tube 10represents the limit of the fluid volume: there are no more tubes behind to supply fluidat the initial pressure. Therefore, as the pressure perturbation moves through the model

Figure 1Model of a linear reservoirand the pressure responsemeasured after differenttimes



from tube 1 to tube 10, the rate of pressure change in the fluid is not limited by thevolume of the fluid: it is as if the volume of fluid was infinite in extent. During thetransient period, the reservoir is often referred to as infinite acting.

On inspection, a profile has been developing across the tubes during the transientperiod. At the end of the transient period, the fluid in all of tubes is expandingproducing a decline in the pressure in all of the tubes. The shape of the pressure profileacross all of the tubes remains essentially constant and as time continues, the profilesinks through the model until the water in the tube nearest the outlet empties. Duringthis time, the water in the model has not been replaced so steady state conditions havenot been achieved, however, since the gradient between the pressures in each adjacenttube is not changing, the system can be considered to be in pseudo-steady state orsemi-steady state: the pressure gradient is constant but the absolute pressure isdeclining. This mimics the situation in a real reservoir where the pressure is perturbedaround a well and the pressure disturbance moves out into the rest of the reservoir untilit reaches the outer boundary. If this is sealing and no flow occurs across the boundary,then the reservoir pressure will decline (neglecting any injection into the reservoir) ina pseudo-steady state manner. If the boundary is nonsealing (i.e. it is the water oilcontact and the aquifer water is mobile) then the aquifer water will flow into thereservoir and a steady state will be achieved if the flowrates match.

The flow described in this model is trivial, but it illustrates the problem of applyingDarcys Law to real reservoirs: the effect of time on flow may be considerable and ifonly steady state flow relationships were available then either permeability of thereservoir would remain unknown or unrealistic flow periods would be required tomeasure an essentially simple rock property.

2 CHARACTERISATION AND MODELLING OF FLOW PATTERNS

The actual flow patterns in producing reservoirs are usually complex due mainly tothe following factors:

(i) The shapes of oil bearing formations and aquifers are quite irregular

(ii) Most oil-bearing and water bearing formations are highly hetereogenous withrespect to permeability, porosity and connate water saturation. The saturationsof the hydrocarbon phases can vary throughout the reservoir leading to differentrelative permeabilities and therefore flow patterns

(iii) The wellbore usually deviates resulting in an irregular well pattern through thepay zone

(iv) The production rates usually differ from well to well. In general, a high rate welldrains a larger radius than a lower rate well

(v) Many wells do not fully penetrate the pay zone or are not fully perforated

There are essentially two possibilities available to cope with complexities of actualflow properties.

6(i) The drainage area of the well, reservoir or aquifer is modelled fairly closely bysubdividing the formation into small blocks. This results in a complex series ofequations describing the fluid flow which are solved by numerical or semi-numerical methods.

(ii) The drained area is modelled by a single block to preserve the global featuresand inhomogeneities in the rock and fluid properties are averaged out orsubstituted by a simple relationship or pattern of features (such as a fracture set,for example). The simplifications allow the equations of flow to be solvedanalytically.

The analytical solutions will be examined in this chapter.

2.1 Idealised Flow PatternsThere are a number of idealised flow patterns representing fluid flow in a reservoir:linear, radial, hemispherical, spherical. The most important cases are the linear andradial models since both of them can be used to describe the water encroachment froman aquifer into a reservoir, and the radial model can be used to describe the flow of fluidaround the wellbore.

In the following sections, dealing mainly with oil, the compressibility of the flowingfluid may depend on the pressure. It will always be assumed that the product ofcompressibility and pressure, cP, is smaller than one, i.e. cP



The components of the flow velocity vector, U are:

Ux = -(k

x/)(P/x)

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

z = -(k

z/)(P/z+g) (2.1)

wherek = 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)

These components are similar to Darcys law in each of the three directions.

2.2.1 Linear Horizontal Model of a Single Phase FluidIn this geometry, the flow is considered to be along the axis (in the x direction) of acuboid of porous rock. The total length of the cuboid is L and fluid flows into the rockat the left end (x=0) and exits at the right end (x=L). There is no flow in the otherdirections at any time i.e. Uy = Uz =0 for all values of x, y, z and time, t (in a realreservoir, there may be flows in different directions in different parts of the reservoirand there may be cross flows from different layers within the reservoir). The rock is100% saturated with the fluid.

The flow equations are:

U k Px

x =

(2.2a)

Ux t

( )=

; 0 x L (2.2b)

wherek = permeability (in the X direction), (mD) = density, (kg/m3)U = flow velocity (m/s)t = time (s) = porosity = viscosity, PasP = pressure, Pax = distance, (m)

The latter equation is obtained from a mass balance as follows (figure 3):

8

flowrate, qinx=0

x

x+dx

x=L

dx

area, A

porosity, X axis

flowrate, qout

isometric view

dx

x=0 x x+dx x=L

X axis

flowrate, qoutflowrate, qin

plan view

In figure 3, fluid flows into the end of the cuboid at position x=0, through the rock onlyin the X direction and out of the cuboid at x=L. In the middle of the cuboid, an elementfrom position x to position x+dx is examined. The bulk volume of the element is theproduct of the area, A and the length, dx, i.e. the bulk volume = A*dx. The pore volumeof the element is therefore the product of the bulk volume and the porosity, , i.e. thepore volume = A*dx*. If the flow was steady state then the flowrates into and out ofthe volume (qin and qout) would be identical and Darcys Law would apply. If the flowrates vary from the inlet of the volume to the outlet, i.e. qin qout then either the fluidis accumulating in the element and qin > qout or the fluid is depleting from the elementq

out > qin (which is possible in a pressurised system since the pressure of the fluid inthe element may reduce causing it to expand and produce a higher flow rate out of theelement). Therefore, there is a relationship between the change in mass, m, along thecuboid and the change in density, , over time as the mass accumulates or depletesfrom any element. In terms of mass flowrate,

Figure 3Flow into and out of acuboid of porous rock



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 q qx

dxx dx x

( )

= ( ) + ( )+ *

Therefore the change in mass flow rate = ( ) qx

dx*

i.e. if the change in mass flowrate is positive it means the element is accumulatingmass; if the change is negative it is depleting mass.

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

The rate of change of mass is equal to

t

A dx

hence ( ) =

qx t

1A

since the flow velocity, U = q/A, this becomes

( ) =

Ux t

or

( )=

Ux t (2.2b)

Substituting the parameters of equation 2.2a in 2.2b gives

=

x x t

k P (2.3)

Equation 2.3 shows the areal change of pressure is linked to the change in density overtime. Realistically, it is pressure and time that can be measured successfully in alaboratory or a reservoir, therefore a more useful relationship would be between thechange in pressure areally with the change in pressure through time. The density canbe related to the pressure by the isothermal compressibility, c, defined as:

10

=

P

cV

VT

1

where V is the volume (m3) and P is the pressure (Pa).The density equals mass per unit volume ( = mV ), hence:

= =

P P

cm

m 1( / ) (Quotient Rule, constant mass system)(2.4)

Since

= =

t P t t

Pc

P(from above)

then

x

k

Px

= c

Pt (2.5)

This is the partial differential equation for the linear flow of any single phase fluid ina porous medium which relates the spatial variation in pressure to the temporalvariation in pressure. If it were applied to a laboratory core flood, it could describe thepressure variation throughout the core from the initial start of the flood when theflowrate was increased from zero to a steady rate (the transient period) as well as thesteady state condition when the flow into the core was balanced by the flow out of thecore. Inspection of the equation shows that it is non-linear because of the pressuredependence of the density, compressibility and viscosity appearing in the coefficients

k

and c. The pressure dependence of the coefficients must be removed before

simple solutions can be found, i.e. the equation must be linearised. A simple form oflinearisation applicable to the flow of liquids such as undersaturated oil is to assumetheir compressibility is small and constant. More complex solutions are required formore compressible fluids and gasses.

2.2.1.1 Linearisation Of Partial Differential Flow Equation For LinearFlowAssuming that the permeability and viscosity terms do not depend on location (i.e.distance along the cuboid), then



x

Px

Pt

=

c

k (2.6)

The left hand side can be expanded to:

x

P Px2

+

x

2

Using equation 2.4 and since

x P x

=P

the above becomesc(P/x)2 + (2P/x2).

Usually c(P/x)2 is neglected compared to 2P/x2 since the pressure gradient issmall, and substituting gives

2Px2

=c

k

Pt

(2.7)

This is termed the linear diffusivity equation

The assumption is made that the compressibility is small and constant, therefore the

coefficients c

k are constant and the equation is linearised. In equation (2.7) k

c

is termed the diffusivity constant. For liquid flow, the above assumptions arereasonable and have been applied frequently, but can be applied only when the productof the compressibility and pressure is much less than 1, i.e. cP

12

2.2.1.2 Conditions of SolutionThe solution of the equation requires initial conditions and the boundary conditions.

(i) Initial Solution Condition.At time t = 0, the initial pressure, Pi, must be specified for every value of x.

(ii) Boundary Conditions.At the end faces x = 0 and x = L, the flow rate or pressure must be specified for everyvalue of time, t.

Solutions of the linear diffusivity equation are needed when dealing with linear flowfrom aquifers. For solutions dealing with well problems a radial model is required.

2.2.2 The Radial ModelFigure 4 illustrates the geometry of this model in which the flow occurs in horizontalplanes perpendicular to the Z axis (i.e. in planes parallel to the XY plane) within a layerof constant height, h. The flow is radial and is either towards the Z axis or away from it.

Z

Y

X

h

re

rw

h

qr qr+dr

r

dr

radial element

section in the XZ plane

wellbore

Z

At a distance r from x-axis, the flow velocity, U is now radius dependent:

U = q/2rh (2.9)

Figure 4Radial horizontal flowgeometry geometry



From Darcys Law (taking account of the flow direction and the co-ordinate direction):

Pr

=U k (2.10)

The mass balance gives:

(( )qrh

r t= 2 (2.11)

Eliminating U and q through equations 2.9 to 2.11 gives the non-linear equation:

rr

kc

r

Pr

Pt

=1 (2.12)

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

1r

r

rPr

= c

kPt (2.13)

This is termed the radial diffusivity equation

2.2.2.1 Range Of Application And Conditions Of SolutionThe two main systems to which the radial diffusivity equation can be applied are waterinflux and wellbore production although there are others.

(a) In the case of water encroachment from an aquifer into a reservoir, the innerboundary corresponds to the mean radius of the reservoir, the outer boundary to themean radius of the aquifer.

(b) In the case of the pressure regime around a wellbore, the inner boundarycorresponds to the wellbore radius, r

w, the outer boundary to the boundary of the

drainage area. In general the wellbore radius, rw is a mathematical concept, however,

the following are widely treated as valid:

Open hole, drilled close to gauge : rw = 1/2 drill bit diameter

Well cased, cemented and perforated : rw = 1/2 drill bit diameter

Slotted liner with gravel pack : rw

= 1/2 outer diameter (OD) of the linerOut-of-gauge hole : r

w = average radius from caliper log

The solution of the equation requires the initial conditions and the boundary conditions.

(i) Initial Solution Condition.At time t=0, the initial pressure, Pi, must be specified for every point of the rangeof equation 2.13, i.e. in the reservoir or in the aquifer.

14

(ii) Boundary ConditionsThe boundaries consist of the outer and inner boundaries. The number ofsolutions depend on the number of boundary conditions, but in the main thereare a few sensible conditions representing the majority of reservoir performance.

Outer Boundary(a) If there is no flow across the outer boundary it is a closed system and the flow

velocity, U will equal zero. The pressure gradient, P/r will also be zero

(b) If there is flow across the outer boundary, the reservoir pressure will bemaintained at a constant value equal to the initial reservoir pressure, Pi.

Inner BoundaryThere are two main cases for the inner boundary which represent either maintaininga constant pressure or a constant flow rate. These are representative of possible flowregimes in the reservoir during either water flooding or production from a well.

(a) Constant Terminal Rate Case (C.T.R.)This can be applied to a wellbore in which the production rate of the well is heldconstant and the pressure varies through time. It can also be applied to waterencroachment in which the influx rate of water from the aquifer into thereservoir across the initial oil-water contact is constant.

(b) Constant Terminal Pressure Case (C.T.P.)Applied to a wellbore, the flowrate is varied to maintain a constant bottom holepressure in the producing well. In the case of water influx, the pressure at theinitial oil water contact of the reservoir remains constant and the flow rate varies.

2.3 Characterisation of the Flow Regimes by their Dependence on TimeTo apply the diffusivity equation to real reservoirs requires careful consideration ofthe boundary conditions. It will be shown that for most practical purposes, thesolutions to the diffusivity equation can be grouped according to the flow regime thatthey represent: steady-state, semi-steady-state (pseudo steady state) or unsteady state(transient).

Steady-state refers to the situation in which the pressure and the rate distribution in thereservoir remain constant with time. Unsteady state is the situation in which thepressure and/or the flow rate vary with time. Semi-steady is a special case of unsteadystate that resembles steady-state flow. These differences in the flow regimes haveramifications in practical reservoir engineering since working solutions to thediffusivity equation are usually limited to a particular flow regime. For instance, in apressure build up test in a well, the determination of an accurate average reservoirpressure will depend strongly on the flow regime the well is in and therefore whichworking solution is used.



3 BASIC SOLUTIONS OF THE CONSTANT TERMINAL RATE CASEFOR RADIAL MODELS

In this flow regime, one of the conditions for solution of the diffusivity equation is thatthe flow rate is constant. This can be applied to the flow of oil towards a full lengthperforated well, and to the flow of water to a producing reservoir from an aquifer. Theflow can be described approximately as the radial flow of a single phase from the outerradius b of a right hollow cylinder towards its inner radius, a. It is assumed thatthe cylinder consists of a homogeneous porous medium.

In the case of drainage by a well, a is the radius of the well, rw and b is the radius

of the external boundary, re. The flow rate, q at radius, r = r

w is the production rate of

the well. In the case of natural water influx into a reservoir, a is the mean reservoirradius, b is the mean aquifer radius, and q is the volume flow rate of water across theinitial oil-water contact.

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

1r

rc

ka r b

r

Pr

Pt

( ) ;= (3.1)

r a

q 2 rkh Pr

;=

=

(3.2)

with the initial condition that the pressure at all points is constant

a r b, t 0; P P constanti = = = (3.3)

and the boundary conditions that at the wellbore the flowrate is constant after theproduction starts

r=a, t 0 : q = constant (3.4)

and at the outer boundary, the pressure is either a constant (and equal to the initialpressure) 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 gradientat the boundary is zero

Pr

r = b, t 0 : = 0 (3.5b)

16

The solution of the equations 3.1 to 3.4 and 3.5a & equations 3.1 to 3.4 and 3.5b arewell known and can be referenced in Pressure buildup and flow tests in wells by CSMatthews and DG Russell, SPE Monograph Volume 1. These are too complex formost practical applications and asymptotic solutions which are fair approximations ofthe general solution are used, i.e. simple solutions which approximate certain flowregimes can be used. The problem is to identify accurately which flow regime andtherefore which asymptotic solution should be used. The steady state solution is thesimplest and is the same as Darcys Law. The non-steady state solutions involve a timeelement and are conveniently expressed in dimensionless form.

3.1 The Steady State SolutionIf a well is produced at a constant flow rate, q, and if the pressure at the external radius,r

e is maintained constant, flow will finally stabilise to steady state conditions.

i.e. flowrate, q = constant and the pressure gradient, Pt

= 0 for all values of radius,

r and time, t

therefore, Pr

dPdr

= and the flow equation becomes

q drr

2 kh dP=

integrating between the limits rw and r gives:

P P q2 kh

ln rr

w

w

=

(3.6)

Integrating between the limits rw and r

e gives:

P Pe w ==

q2 kh

ln rr

e

w

(3.7)

which is identical to the relationship described for a radial system by Darcys Law. Inthis case, the pressure at the external radius of the reservoir is required and the onlyway to measure it in the reservoir would be to drill a well at the external radius. Thisis uneconomic, therefore a mean reservoir pressure,P , is used. It is found fromroutine bottom hole pressure measurements and well tests conducted on the wells ina reservoir, it includes the effect of the area of influence of each well. In simple terms,the volume drained by each well is used to weight the bottom hole pressuremeasurements made in the well; all of the weighted pressures of all of the wells in thereservoir are then averaged. Figure 5 shows a well in a reservoir and its area ofinfluence. Volumetrically, this volume is drained by the well and the mean reservoirpressure,P , is related to the pressure, P of elements of volume, dV being drained. Thetotal volume is V.



h

wellbore

element of volume, dV, at radius, r and at pressure, P

initial pressure profile

pressure profile due to production rate, q

Pi

Pwf

P

rw

re

= P1V

PdVrw

re (3.8a)

where dV = 2rhdr (3.8b)The volume of the wells drainage zone, V = (r

e2-r

w2)h

and considering rw

18

assuming w2r

4 is negligible

=

=

we2

e2

e

w

e2

we

w

P - P 2r

q2 kh

r

2ln r

r

r

4

P - P q2 kh

ln rr

12

(3.10)

EXERCISE 1A 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).

Dataporosity, 19%formation volume factor for oil, B

o1.3rm3/stm3 (reservoir cubic metres per stock

tank cubic metre)net thickness of formation, h, 40mviscosity of reservoir oil, 22x10-3 Paswellbore radius, r

w0.15m

external radius, re

350minitial reservoir pressure, Pi 98.0barbottomhole flowing pressure, P

wf 93.5barq

reservoir = qstock tank x Bo1bar = 105 Pa

EXERCISE 2A well produces oil from a reservoir with an average reservoir pressure of 132.6bar.The flowrate is 13stm3/day. Use the following data to calculate the permeability.

Dataporosity, , 23%formation volume factor for oil, B

o1.36rm3/stm3

net thickness of formation, h 23mviscosity of reservoir oil, 14x10-3 Paswellbore radius, r

w0.15m

external radius, re

210maverage reservoir pressure, P 132.6barbottomhole flowing pressure, P

wf 125.0bar



EXERCISE 3A reservoir is expected to produce at a stabilised bottomhole flowing pressure of 75.0bar. Use the following reservoir data to calculate the flowrate in stock tank m3/day.


o1.41rm3/stm3

net thickness of formation, h 15mviscosity of reservoir oil, 21x10-3 Paswellbore radius, r

w0.15m

external radius, re

250maverage reservoir pressure, P 83.0barbottomhole flowing pressure, P

wf 75.0barpermeability, k 125mD

3.2 Non-Steady State Flow Regimes and Dimensionless VariablesAs mentioned previously, dimensionless forms of the diffusivity equation have foundwide application in the description of flow through porous media. They normalisethe equation for use with many different reservoirs and allow general solutions to befound which can be applied to specific data to determine the specific solution for aparticular reservoir. In such a way, general plots of, for example, the difference inpressure from the reservoir to the wellbore through time can be constructed which canthen be used to determine the actual pressure difference for a specific reservoir. Itshould be noted that solutions for a radial flow reservoir can only be sensible if thedimensionless variables and diffusivity equation have been developed for a radialflow reservoir.

If the dimensionless variables are defined as:

dimensionless time, rD : rr

rD

w

=

dimensionless time, tD : tktcr

Dw

2=

dimensionless pressure, PD : P (r , t ) 2 khq )(P P )D D D i r,t=

(

(at a dimensionless radiusand at a dimensionless time)

wherer = radius in questionr

w = wellbore radius

k = permeabilityt = time in question

20

= porosity = viscosityc = compressibilityh = thickness of the reservoirPi = initial reservoir pressureP

r,t = pressure at the specified radius and time

then the radial diffusivity equation becomes

11r r

rPr

PtD D

DD

D

D

D

= (3.11)

There are other definitions of dimensionless variables, such as a dimensionlessexternal radius, which may be used in particular instances.

3.3 Unsteady State SolutionThe constant terminal rate (CTR) solution can be obtained in several forms, usingdifferent assumptions and methods of mathematical analysis. The various solutionsoverlap, and all of them have particular uses and limitations.

3.3.1 General Considerations

zero flowrate

time

transient

late transient

semi - steady state

time

botto

mho

le fl

owin

g pr

essu

re, Pw

fflo

wra

te, q

Pi

(a)

(b)

Figures 6a and 6b show the response of a reservoir at a wellbore when a flow rate, q,

Figure 6Wellbore pressure responseto a change in flowrate



is suddenly applied. The pressure of the flowing fluid in the wellbore, Pwf falls from

the initially constant value, Pi (static equilibrium) through time and the constantterminal rate (CTR) solution of the diffusivity equation describes this change as afunction of time. The CTR solution is therefore the equation of P

wf versus t for aconstant production rate for any value of the flowing time. The pressure decline,Figure 6(b), can normally be divided into three sections depending on the value of theflowing time and the geometry of the reservoir or part of the reservoir being drainedby the well. This figure represents the pressure change at the wellbore through timewhich is equivalent to the pressure change (or change in the height of water) in thecylinder nearest the outlet in the model represented in Figure 1.

Initially, the pressure response can be described using a transient solution whichassumes that the pressure response at the wellbore during this period is not affectedby the drainage boundary of the well and vice versa. This is referred to as the infinitereservoir case, since during the transient flow period, the reservoir appears to beinfinite in extent with no limits to the fluid available to expand and drive the system.

The transient period is followed by the late-transient when the boundaries start toaffect the pressure response. This is analogous to the pressure disturbance havingmoved along the line of tubes in the model in figure 1. The nature of the boundariesaffects the type of solution used to describe the pressure change since a well may drainan irregularly shaped area where the boundaries are not symmetrical or equidistantfrom the well.

The next phase in the pressure decline is termed semi-steady state or pseudo steadystate where the shape of the pressure profile in the reservoir is not changing throughtime and the wellbore pressure is declining at a constant rate. It is analogous to themodel depicted in figure 1 where the level of water in all of the tubes is falling and noadditional water is being added to tube 10 to maintain absolute pressure profile. If thepressure profile developed in the reservoir around the well had remained constant, truesteady state conditions would have occurred and the steady state solutions asmentioned in the previous section would have applied.

3.3.2 Hurst and Van Everdingen SolutionThe constant terminal rate solution for all values of the flowing time was presentedby Hurst and van Everdingen in 19492. They solved the radial diffusivity equationusing the Laplace transform for both the constant terminal rate and constant terminalpressure cases. The full equation contains, as one of its components, an infinitesummation of Bessel functions which are required to describe the complex wellborepressure response during the late transient period. Simple solutions can be obtainedfor the transient and semi-steady state flow.

The solution describes pressure drop as a function of time and radius for fixed valuesof external radius, r

e, and wellbore radius, r

w, rock and fluid properties. It is expressed

in terms of dimensionless variables and parameters as:

PD = f(tD,rD,reD) (3.12)

22

wheretD = dimensionless timerD = dimensionless radiusr

eD = 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 canbe measured routinely during production operations and compared to the theoreticalsolutions. The determination of a reservoir pressure at a location remote from a wellmay be required for reasons of technical interest, but unless a well is drilled at thatlocation, the actual value cannot be measured.

At the wellbore radius, r=rw (or rD=1.0)

PD = f(tD, reD) (3.13)

i.e. P (t ) 2tr

lnr 34

2e J ( r )

(J ( r ) J ( ))D D DeD2 eD

t12

m eD

m2

12

m eD 12

mm 1

m

2D

= + +

=

(3.14)

where

m are the roots of J ( r )Y ( ) J ( )Y ( r ) 01 m eD 1 m 1 m 1 m eD =

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 theform of tables (from Chatas, AT3, A Practical Treatment of non-steady state FlowProblems 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 terminalrate problems is as follows in Table 1. It reports the dimensionless pressure at somedimensionless time for various configurations of reservoir. It is the solution toequation 3.14.



Table Presents Valid for

2 i PD as a function of tD

24

EXERCISE 5An experiment on a cylindrical sand pack is conducted to examine the wellborepressure decline. The sand pack is filled with pressurised fluid which is withdrawn fromthe wellbore at a constant flowrate of 0.1m3/d. There is no flow at the externalboundary. Calculate the wellbore pressure at times 0.001 hour, 0.005 hour and 0.1hour after the start of production. The figure below indicates the sand pack.

closed top, bottomand side

fluid production

sand pack withfluid filled pore space

flow to the wellbore

Dataporosity, 25%net thickness of formation, h 0.2mviscosity of fluid, 2x10-3 Paswellbore radius, r

w0.2m

external radius, re

2minitial reservoir pressure, Pi 2barpermeability, k 1200mDcompressibility, c 0.15x10-7Pa-1

EXERCISE 6A discovery well is put on test and flows at 2.9stm3/d. Using the following data.calculate the bottomhole flowing pressure after 5 minutes production.

Dataporosity, 17%net thickness of formation, h 40mviscosity of reservoir oil, 14x10-3 Pasformation volume factor of oil, B

o1.27rm3/stm3

wellbore radius, rw

0.15mexternal radius, r

e900m

initial reservoir pressure, Pi 200barpermeability, k 150mDcompressibility, c 0.9x10-9Pa-1



3.3.3 The Line Source SolutionThis solution assumes that the radius of the wellbore is vanishingly small relative tothe mean radius of the reservoir. It allows the calculation of the pressure at any pointin an unbounded reservoir using the flowrate at the well. The benefits are clear in thatno flow rates other than those measured in the producing well are required and fromwhich the pressure at any location can be calculated. The disadvantage is that thesolution works for infinite acting reservoirs only and if barriers are met, then thesolution fails to represent the true flow regime. The technique of superposition can beused to combine the effect of more than one well in an infinite acting reservoir and thistechnique can represent the effect of a barrier. The barrier is equivalent to the pressuredisturbance produced by a second, imaginary well producing at the same rate andhaving the same production history as the real well with both these wells in an infiniteacting reservoir. This solution has found a lot of use in well test analysis.

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

r r

q 2 rhk Pr

w=

=

(3.15)

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

limr 0

rpr

q2 kh

=

for time, t > 0.

Using the Boltzman Transformation

y cr4kt

2

=

and substituting into the diffusivity equation =

1r r

rPr

c

kPt

( )

gives

y d pdy

dpdy

(1 y) 02

2 + + =

with the boundary conditions

p pi as y

limy 0

2y py

q2 kh

=

If p'dpdy

= then

26

y dp'dy

(1 y)p' 0+ + =

Separating the variables and integrating giveslnp = -lny - y +C

i.e. p' dpdy

Cy

e1 y= = (3.16)

where C and C1 are constants of integration. Sincelim

y 02y p

yq

2 khlim

y 02C e1

y

= =

then C q1 =4 kh

and equation 3.16 becomes

dpdy

q4 kh

e

y

y

=

which is integrated to give

p q4 kh

e

ydy C

yy

2= +

or

p q4 kh

e

ydy C

y

2= +

y

which can be rewritten as

p q4 kh

Ei(-y) C2= +

Applying the boundary condition that p pi as y then C2 = pi and the line sourcesolution is obtained:

p p q4 kh

Ei(- cr4kt

)i (r,t)2

=

(3.17)

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

Ei( y) ey

dyy

y

=

It can be calculated from the series

Ei( y) lny yn!n

n

= +

where = 0.5772157 (Eulers Constant). On inspection of the similarities in the Eifunction and the ln function, it can be seen that when y



( = =1.781 = e e )0.5772157

Solutions to the exponential integral can be coded into a spreadsheet and used with theline source solution. Practically, the exponential integral can be replaced by a simplerlogarithm function as long as it is representative of the pressure decline. The limitation

that y

. The

equation can be applied anywhere in the reservoir, but is of significance at the wellbore(i.e. for well test analysis) where typical values of wellbore radius, r

w, and reservoir

fluid and rock parameters usually means that y100 crw

2/k (3.20)

where rw is the wellbore radius. The value of 100 has been derived form the analysis

of the responses of real reservoirs; it can be varied according to the nature of a specificwell and reservoir. The time involved here is not the same as the dimensionless time,tD calculated for other models of fluid flow in a reservoir (e.g. the input parameters forthe Hurst and van Everdingen solutions require the dimensionless time at the radiuswhere the dimensionless pressure drop is required - this may be the wellbore and r

w

would be used or it may be some other radius).

28

(ii) t < cre2/4k (3.21)

where re is the external radius. The reservoir boundaries begin to effect the pressure

distribution in the reservoir after this time, the infinite acting period ends and the linesource solution does not represent the fluid flow.

EXERCISE 7A well and reservoir are described by the following data:


o1.4rm3/stm3

net thickness of formation, h 100mviscosity of reservoir oil, 1.4x10-3 Pascompressibility, c 2.2 x10-9Pa-1permeability, k 100mDwellbore radius, r

w0.15m

external radius, re

900minitial reservoir pressure, Pi 400barwell flowrate (constant) 159stm3/day = 159

24x3600stm3/second

skin factor 0

Determine the following:

(1) the wellbore flowing pressure after 4 hours production

(2) the pressure in the reservoir at a radius of 9m after 4 hours production



EXERCISE 8A well flows at a constant rate of 20stm3/day. Calculate the bottomhole flowingpressure at 8 hours after the start of production.


o1.32rm3/stm3

net thickness of formation, h 33mviscosity of reservoir oil, 22.0x10-3 Pascompressibility, c 0.6x10-9Pa-1permeability, k 340mDwellbore radius, r

w0.15m

external radius, re

650minitial reservoir pressure, Pi 270barwell flowrate (constant) 20stm3/dayskin factor 0



EXERCISE 9Two wells are drilled into a reservoir. Well 1 is put on production at 20stm3 /day. Well2 is kept shut in. Using the data given, calculate how long it will take for the pressurein well 2 to drop by 0.5bar caused by the production in well 1. Well 2 is 50m from well 1.


o1.21rm3/stm3

net thickness of formation, h 20mviscosity of reservoir oil, 0.8x10-3 Pascompressibility, c 43x10-9Pa-1permeability, k 85mDwellbore radius, r

w0.15m

external radius, re

1950minitial reservoir pressure, Pi 210barwell flowrate (constant) 20stm3/dayskin factor 0Distance well 1 to well 2 50m

EXERCISE 10A well in a reservoir has a very low production rate of 2stm3/day. Calculate the flowingbottomhole pressure after 2 years production.


o1.13rm3/stm3

net thickness of formation, h 10mviscosity of reservoir oil, 5x10-3 Pascompressibility, c 14x10-9Pa-1permeability, k 10mDwellbore radius, r

w0.15m

external radius, re


30

EXERCISE 11A well is put on production at 15stm3/day. The following well and reservoir data arerelevant.


o1.2rm3/stm3

net thickness of formation, h 23mviscosity of reservoir oil, 5x10-3 Pascompressibility, c 22 x10-9Pa-1permeability, k 130mDwellbore radius, r

w0.15m

external radius, re


Determine the following:

(1) the wellbore flowing pressure after 2 hours production(2) the pressure in the reservoir at a radius of 10m after 2 hours production(3) the pressure in the reservoir at a radius of 20m after 2 hours production(4) the pressure in the reservoir at a radius of 50m after 2 hours production

3.3.4 The Skin FactorThe analysis of fluid flow encountered thus far has assumed that a constant permeabil-ity exists within the reservoir from the wellbore to the external boundary. In reality,the rock around the wellbore can have higher or lower permeability than the rest of thereservoir. This results from formation damage which may occur during drilling andcompletion (where the wellbore fluids alter the wettability of the near wellboreformation as fluid leaks off into it, or solids suspended in the drilling fluids aredeposited in the pore spaces and become trapped thereby physically hindering the flowof fluid and reducing the permeability) or during production (where sand or precipi-tates from the hydrocarbon fluids or from formation brines can alter wettability andplug pore spaces). Alternatively, wellbores intersecting fractures may exhibit en-hanced permeabilities as the fractures offer much greater conductive paths to the fluidsaround the wellbore, thus enhancing the permeability. This situation may also berequired as part of the reservoir management: hydraulic fractures or acidisingworkovers are performed on wells to bypass zones of reduced permeability whichhave developed during production.

In these cases, the Ei equation fails to model the pressure drop in these wells properlysince it uses the assumption of uniform permeability throughout the drainage areaof the well up to the wellbore. Figure 7 shows the effect of a reduction in permeabilityaround a wellbore. The skin zone does not affect the pressures in the rest of theformation remote from the wellbore, i.e. it is a local effect on the pressure drop at thewellbore.



pressure profile if no skin zone was present

actual pressure profile through skin zone

skin zone

permeability, Ks permeability, K

botto

mho

le fo

llowi

ng p

ress

ure,

Pwf

radius, r

rw rs

Pwf(no skin)P skinPwf(skin)

P skin = Pwf(skin) - Pwf(no skin)

It can be shown that if the skin zone is considered equivalent to an altered zone ofuniform permeability, k

s, with an outer radius, r

s, the additional drop across this zone

(Ps) can be modelled by the steady-state radial flow equation. It is assumed that after

the pressure perturbation caused by the start of production has moved off into the restof the formation, the skin zone can be thought of as being in a steady state flow regime.The pressure drop associated with the presence of a skin is therefore the difference inthe bottomhole flowing pressures at the well when skin is present and when skin is notpresent, i.e.

Ps ==

=

q2 k h

ln rr

q2 kh

ln rr

q2 kh

( kk

1)ln rrs

s

w

s

w s

s

w

(3.22)

Equation 3.22 simply states that the pressure drop in the altered zone is inverselyproportional to the permeability, k

s rather than to the permeability, k of the rest of the

reservoir and that a correction to the pressure drop in this region must be made.

When this is included in the line source solution it gives the total pressure drop at thewellbore:

= + =

P P

q4 kh

Ei( y) P q4 kh

Ei( y) 2 kk

1 ln rr

i wf ss

s

w

(3.23)

Figure 7Variation of thepermeability around thewellbore changes the localpressure profile

32

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

P P q4 kh

ln( cr4kt

) 2 kk

1 ln rr

i wfw

2

s

s

w

=

(3.24)

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

skk

1 ln rrs

s

w

=

(3.25)

and the drawdown defined as:

P P q4 kh

ln( cr4kt

) 2si wf w2

=

(3.26)

Equation 3.26 shows that a positive value of skin factor will indicate that thepermeability around the well has been reduced (by some form of formation damage).The absolute value reflects the contrast between the skin zone permeability and theunaltered zone permeability and the depth to which the damage extends into theformation. Part of the essential information from a well test is the degree of formationdamage (skin factor) around a well caused by the drilling and completion activities.Alternatively, a well may have a negative skin factor, i.e. the permeability of the skinzone may be higher than that of the unaltered zone, caused by the creation of highlyconductive fractures or channels in the rock. The extent of the damage zone cannot bepredicted accurately and there may be variations vertically in the extent of the damagezone therefore this simple model may not characterise the near wellbore permeabilityexactly.

An altered zone near a particular well affects only the pressure near that well, i.e. thepressure in the unaltered formation away from the well is not affected by the existenceof the altered zone around the well.

EXERCISE 12.A discovery well is put on well test and flows at 286stm3/day. After 6 minutesproduction, the well pressure has declined from an initial value of 227bar to 192bar.Given the following data, calculate the pressure drop due to the skin, P

skin , and themechanical skin factor.


o1.39rm3/stm3

net thickness of formation, h, 8.5mviscosity of reservoir oil, 0.8x10-3 Pascompressibility, c 2.3 x10-9Pa-1permeability, k 100mDwellbore radius, r

w0.15m

external radius, re

6100m



initial reservoir pressure, Pi 227barbottomhole flowing pressureafter 6 minutes 192barwell flowrate (constant) 286stm3/day

EXERCISE 13A reservoir and well are detailed in the following data. Use this data to calculate theskin factor around the well after producing for 1.5 hours.

Data

porosity, 23%formation volume factor for oil, B

o1.36rm3/stm3

net thickness of formation, h 63mviscosity of reservoir oil, 1.6x10-3 Pascompressibility, c 17 x10-9Pa-1permeability, k 243mDwellbore radius, r

w0.15m

external radius, re

4000minitial reservoir pressure, Pi 263.0barbottomhole flowing pressureafter 6 minutes 260.5barwell flowrate (constant) 120stm3/day

3.4 Semi-Steady-State SolutionOnce the initial pressure perturbation produced by bringing a well onto production hasmoved through the reservoir and met the boundaries, the infinite-acting nature of thefluid changes to become finite acting. As stated previously, this is termed pseudosteady state or semi steady state because the pressure drop with time is the same at allpoints around the flowing well, i.e.

Pt

dPdt

constant= =

and where there is no flow across the outer boundary at r = re of the drainage zone, i.e.

Pr

0 at r re= =

In a similar manner to the steady state flow regime, the pressure difference betweenthe wellbore and, say, the external radius, or the pressure difference between thewellbore pressure and the initial pressure, or the pressure difference between thewellbore pressure and the average reservoir pressure can be calculated depending onthe physical measurements which have been taken. Usually, an average pressure isknown in a reservoir and this is used to determine the pressure drop. Figure 8 showsthe pressure profile in the reservoir and the values which may be relevant.

34

well with constant flow rate, q

flowi

ng p

ress

ure,

P

Pe

rw re

Pwf

initial pressure

calculated average pressurePi

pressure profile in reservoir

radius, r

heig

ht o

f for

mat

ion

Under semi steady state conditions, the pressure profile can be averaged over thevolume of the reservoir. This gives the average reservoir pressure at a particular timein the stage of depletion of the reservoir. If there are several wells in a reservoir, eachwell drains a portion of the total volume. For stabilised conditions, the volume drainedby each well is stable and in effect the whole reservoir can be subdivided into severalportions or cells. The average pressure in each cell can also be calculated from thestabilised pressure profile. The calculation of the average pressure is determined fromthe material balance of the initial pressure and volume of fluid and its isothermalcompressibility. The expansion of the fluid in each cell manifests itself as a volume,or flow rate, at the well, i.e.

cV(P P) qti = (3.27)

where V = pore volume of the radial cell; q = constant production rate; t = total flowingtime, c = isothermal compressibility.

q dVdt

dVdP

qdtdP

q dtdP

since c 1V

dVdP

q cV dPdt

dPdt

qcV

T

=

= =

=

=

=

(3.28)

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

Figure 8Pressure profile in areservoir under semi steadystate flow conditions



dPdt

qc r he

2=

(3.29)

Substitution of equation 3.29 in the radial diffusivity equation

= 1r r

rPr

c

kPt

( )gives

1r r

(r Pr

) = ck

qc r he

2

which is

1r

r

(r Pr

) = qr hke

2

Integration gives

rdPdr

q r2 r kh

C2

e

2 1= +(3.30)

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

= 0 therefore

C1= q

2 kh and substitution into equation 3.30 gives

dPdr

q2 kh

1r

r

re2=

(3.31)

When integrated, this gives

P q2 kh

lnr r2rP

P2

e

2r

r

wf

r

w

[ ] =

or

P P q2 kh

lnr r2r

lnr r2rr wf

2

e

2 ww

2

e

2 =

36

P P q2 kh

ln rr

r

2rr wf w

2

e

2 =

(3.32)

The term r

2rw

2

e

2 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 PPq

2 khln r

r

12

swfe

w

= +

(3.33)

If the average pressure is used, then the volume weighted average pressure of thedrainage cell is calculated as previously in the steady state flow regime, i.e.

P 2r

Prdr2e rw

re

= (3.9)

where rw and r

e are the wellbore and external radii as before, and P is the pressure in

each radial element, dr at a distance r from the centre of the wellbore. In this case,

P P 2r

q2 kh

r ln rr

r

2rdrwf

e

2r

r

w

2

e

2w

e

=

and integrating gives

(i) r lnr

rdr = r

2ln r

r

1r

r

2dr

= r

2ln r

r

r

4

r

r

w

2

w r

r

r

r 2

2

w r

r 2

r

w

e

w

e

w

e

w

e

ww

er

e

2e

w

e

2

r

2ln r

r

r

4

(ii)w

e

w

e3

e

2r

r 4

e

2r

r

e

2r

2rdr = r

8rr

8

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

wfe

w

P P q2 kh

ln rr

34

+ s =

(3.34)



The pressure differences (Pr - P

wf), (Pe- Pwf), ( P -Pwf) do not change with time, whereasP

r, P

e, P

w and P

do change.

EXERCISE 14A well has been on production in a reservoir which is in a semi-steady state flowregime. For the following data, calculate the bottomhole flowing pressure, P

wf

Dataformation volume factor for oil, B

o1.62rm3/stm3

net thickness of formation, h 72mviscosity of reservoir oil, 1.2x10-3 Paspermeability, k 123mDwellbore radius, r

w0.15m

external radius, re

560maverage reservoir pressure, P 263.0barwell flowrate (constant) 216stm3/dayskin factor 0

3.4.1 Using The Initial Reservoir Pressure, PiIf the pressure drop from initial pressure conditions is required then equation 3.27 maybe written as:

o oP Pq

cVt t= +

( ) (3.35)

iP PqtcV

= (3.36)

where q is the volume flow rate, c is the isothermal compressibility, V is the originalvolume t

o is a reference time after which flow starts, t is the flowing time, P

o is the

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

isthe average reservoir pressure after time, t. Subtracting equation 3.36 from equation3.34 gives

i wfP - P == q

khr

r

ktcr

e

w e

2

34

22

ln +

(3.37)

3.4.2 Generalised Reservoir Geometry: Flowing Equation under Semi-Steady State ConditionsThe key aspect of the radial flow equation under semi-steady state conditions is thatthe boundary of the reservoir has an effect on the flow regime. The pressure declineis influenced by the fact that there is a finite limit to the amount of fluid present in thereservoir. The equations developed have been for radial geometries. However, thesemi-steady state flow regime in non-radial reservoirs can be examined by the radialequation if the shape of the reservoir can be attributed to a factor which encapsulates

38

the relative position of a producing well in a volume of reservoir fluid. This non-symmetrical geometry can be described by the Dietz shape factor (given the symbolCA ) as follows.

Using the average reservoir pressure and assuming no skin factor, the pressure dropis described by equation 3.34 as

P P q2 kh

ln rr

34wf

e

w

=

(3.34)

Expressing the terms ln rr

34

e

w

as

12

12

12

12

12

2

2

2

2ln rr

32

ln rr

32

ln rr

e

r

r

e

e

w

e

w

e

w

32

e

w

32

=

=

=

=

ln

ln

ln

r

r e

e

2

w

232

( )

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

( )4 r

4

e

2

r e

4A1.781 x 31.6 x r

w

232 w

2

=( )

( )

where A is the area drained, = 1.781 and Dietz shape factor, CA (for a well in a radialdrainage area) =31.6.



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

P P q2 kh

12

ln 4AC r

swfA w

2 = +

(3.38)

For the pressure drop between initial reservoir pressure conditions and some bottomhole flowing pressure during semi steady state flow, equation 3.37 can be expressedas

P P q2 kh

(12

ln 4AC r

2 ktcA

)i wfA w

2 = +

(3.39)

or

P P q2 kh

(12

ln 4AC r

2 ktcA

)wf iA w

2= +

(3.40)

In a convenient dimensionless form, this can be expressed as

2 khq

(P - P 12

ln 4AC r

2 ktcr

wfA w

2w

2= +

)

r

Aw

2

or

= +P t 12

ln 4AC r

2 t rAD D A w

2 Dw

2

(3.41)

The term involving the wellbore radius can be accommodated by using the followingmodified dimensionless time

=t tr

ADA Dw

2

in which case

= +P t 12

ln 4AC r

2 tD DA w

2 DA

The calculation of the Dietz shape factors and their limitations in use is presented inLee and reproduced in Table 5. There are a series of common simple shapes with wellslocated close to certain barriers and the shape factors associated with them. There arealso values of tDA which indicate the use of the shape factors.

(i) The infinite system solution with less than 1% error for tDA < X in this case, Xis the value of the maximum elapsed time during which a reservoir is infiniteacting and the Ei function can be used. The time, t is calculated by

40

X in this case, the semi steady statesolution can be used with the results having an error less than 1% for an elapsedtime, t

>t tcAkDA

(iii) The solution which is exact for tDA > X in this case, the semi steady state solutioncan be used with the results being exact for an elapsed time, t

>t tcAkDA

For a real reservoir under semi steady state conditions, the volume of reservoir drainedby a well can be determined from its flow rate, and this volume correlated to thestructural map of the reservoir to determine the shape. The values of shape factor canthen be used to locate the position of the well relative to the boundaries of the areabeing drained. This is not an exact procedure and variations in the heterogeneity of thereservoir can alter the pressure responses, however, it is an analytical step in thecharacterisation of the reservoir.

EXERCISE 15For each of the following geometries, calculate the time in hours for which the reservoiris infinite acting

Geometry1. Circle2. Square3. Quadrant of a square

DataArea of reservoir, A 1618370m2viscosity of reservoir oil, 1.0x10-3 Paspermeability, k 100mDporosity, , 20%compressibility, c 1.45 x10-9Pa-1

The times are calculated by the dimensionless time, diffusivity of the reservoir and thearea of the reservoir. The dimensionless time accounting for the reservoir drainagearea is found for the conditions in Table 5.



3.5 The Application of the CTR Solution in Well TestingThe study of fluid flow so far has related the pressure drop expected as a result of aflow rate from a well in a reservoir. If the appropriate parameters, such as porosity,permeability and fluid viscosity are known, then for a particular flow regime, such asunsteady state, the pressure drop at a certain distance from the well at a certain timeafter production starts can be calculated.

In reality, only flow rates and pressures at wells can be measured directly, and the mostimportant unknown factor in the diffusivity equation is the permeability. Therefore,rather than calculate a pressure drop for a given set of conditions, the pressure dropcan be continuously measured and the permeability calculated.

This is part of the objectives of well testing and for illustration, the following examplecalculates the permeability and skin factor for a well in a reservoir. It is important tonote that these examples all assume that an initially undisturbed reservoir is broughton production, i.e. that there has been no previous production in the reservoir thereforethe pressure is at its initial value. In well test analysis, the previous history of a wellmust be accounted for. The section on superposition will introduce the concepts of amulti-rate history for a well.

EXERCISE 16A well is tested by producing it at a constant flow rate of 238stm3/day (stock tank) fora period of 100 hours. The reservoir data and flowing bottomhole pressures recordedduring the test are as follows:


o1.2rm3/stm3

net thickness of formation, h 6.1mviscosity of reservoir oil, 1x10-3 Pascompressibility, c 2.18 x10-9Pa-1wellbore radius, r

w0.1m

initial reservoir pressure, Pi 241.3barwell flowrate (constant) 238stm3/day

42

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.

(Refer to solution to exercise 16 on page 93)

EXERCISE 17An appraisal well is tested by producing at a constant rate of 200stm3/day for 107hours. The following table of flowing bottomhole pressures and time were recordedduring the test. Using the data,

1. calculate the permeability and skin factor of the well2. estimate the shape of the drainage area

Data

porosity, 22%formation volume factor for oil, B

o1.3rm3/stm3

net thickness of formation, h 21mviscosity of reservoir oil, 1.9x10-3 Pascompressibility, c 4.3 x10-9Pa-1wellbore radius, r

w0.15m

initial reservoir pressure, Pi 378.7barwell flowrate (constant) 200stm3/day



Time (hours) Bottomhole flowing pressure

(bar)

0.0 378.7 1.1 326.41 2.1 324.7 3.2 323.8 4.3 323.1 5.4 322.1 8.0 320.5 10.7 318.8 16.1 315.5 21.4 312.2 32.1 305.6 42.8 300.8 53.5 296.0 64.2 291.2 74.9 286.3 85.6 281.5 96.3 276.7

107.0 271.9

4. THE CONSTANT TERMINAL PRESSURE SOLUTION

In the constant terminal rate solution of the diffusivity equation, the rate is known tobe constant at some part of the reservoir and the pressures are calculated throughoutthe reservoir. Conversely, in the constant terminal pressure solution, the pressure isknown to be constant at some point in the reservoir, and the cumulative flow at anyparticular radius can be calculated. The constant terminal pressure solution is not asconfusing as the constant terminal rate solution simply because less is known aboutit. Only one constant terminal pressure solution is available, so there is no decision tobe made over which to use as in the case of the constant terminal rate solutions. Hurstand Van Everdingen produced the solutions for cases of an infinite radial system witha constant pressure at the inner boundary and for constant pressure at the innerboundary and no flow across the outer boundary. These can model, for example, awellbore whose bottomhole flowing pressure is held constant whilst flow occurs in thereservoir, or they can model a reservoir surrounded by an aquifer. The samegeometrical 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 rockand fluid properties throughout, however, in this case there is a pressure drop from aninitial pressure to some constant value. In the case of aquifer encroachment, the radiusof the well is the radius of the initial oil water contact. The constant terminalpressure solution is most widely used for calculating the water-encroachment (naturalwater influx) into the original oil and gas zone due to water drive in a reservoir. Thistopic is covered in the chapter on water influx.

44

5. SUPERPOSITION

In the analyses so far, the well flow rate has been instantly altered from zero to someconstant value. In reality, the well flowrates may vary widely during normal produc-tion operations and of course the wells may be shut in for testing or some otheroperational reason. The reservoir may also have more than a single well draining it andconsideration must be taken of this fact. In short, there may be some combination ofseveral wells in a reservoir and/or several flowrates at which each produce. Thecalculation of reservoir pressures can still be done using the previous simple analyticaltechniques if the solutions for each rate change, for example, are superposed on eachother. In other words, the total pressure drop at a wellbore can be calculated as the sumof the effects of several flowrate changes within the well, or it may be the sum of theeffects caused by production from nearby wells.

There is also the possibility of using infinite acting solutions to mimic the effects ofbarriers in the reservoir by using imaginary or image wells to produce a pressureresponse 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 andconstant terminal pressure cases. The solutions may be added together to determinethe 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 areservoir.

5.1 Effects of Multiple WellsIn a reservoir where more than one well is producing, the effect of each wells pressureperturbation on the reservoir is evaluated independently (i.e. as though the other wellsand their flow rate/ pressure history did not exist), then the pressure drop calculatedat a particular well at a particular time is the simple addition of all of the individualeffects superimposed one effect upon the other. Consider 3 wells, X, Y and Z, whichstart to produce at the same time from an infinite acting reservoir (figure 9).



Well X

Flowrate, qx

rxy rzy

Flowrate, qy Flowrate, qz

Well Y Well Z

Initial Pressure, Pi

No Barrier Detected No Barrier Detected

Pressure in well Y after flowing time, t

P caused by well Xindependent of well Y

or well Z

P caused by well Zindependent of well Y

or well Z

Actual well pressure,Pwf

P caused by well Yindependent of well X

or well Z

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 ZAssuming unsteady state flow conditions, the line source solution can be used todetermine the pressure in well Y. It is assumed here that the logarithm function canbe used for well Y itself and that there will be a skin around the well. The effects ofwells X and Z can be described by the Ei function. There is no skin factor associatedwith the calculation of pressure drop caused by these wells, since the pressure drop ofinterest is at well Y (i.e. even if wells X and Z have non-zero skin factors, their skinfactors affect the pressure drop only around wells X and Z). The total pressure dropis then:

(P P ) q4 kh

ln cr4kt

2Si wf total at well Y Y wY2

Y =

q4 kh

Ei cr4kt

X XY2

+

+

q4 kh

Ei cr4kt

Z ZY2

(5.1)

whereqY is the flowrate from well YqX is the flowrate from well X

Figure 9The superposition ofpressure changes fromseveral wells

46

qZ is the flowrate from well Zr

wY is the radius of well YrXY is the distance of well Y from the X wellrZY is the distance of well Z from the X wellthe rest of the symbols have their usual meaning

This technique can be used to examine the effects of any number of wells in an infiniteacting reservoir. This could be to predict possible flowing well pressures amongst agroup of wells, or to deliberately use the interaction between wells to check reservoircontinuity. These interference tests and other extended well tests are designed tocharacterise the reservoir areally rather than to determine only the permeability andskin factor around individual wells.

EXERCISE 18Two wells, well 1 and well 2, are drilled in an undeveloped reservoir. Well 1 iscompleted and brought on production at 500stm3/day and produces for 40 days atwhich time Well 2 is completed and brought on production at 150stm3/day. Using thedata provided, calculate the pressure in Well 2 after it has produced for 10 days (andassuming Well 1 continues to produce at its flowrate). Therefore, Well 1 produces for50days when its pressure influence is calculated; Well 2 produces for 10 days whenits 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, B

o1.4rm3/stm3

net thickness of formation, h, 36mviscosity of reservoir oil, 0.7x10-3 Pascompressibility, c 8.7 x10-9Pa-1permeability, k 80mDwellbore radius, r

w (both wells) 0.15m

initial reservoir pressure, Pi 180.0barWell 1 flowrate (constant) 500stm3/dayWell 2 flowrate (constant) 150stm3/dayskin factor around both wells 0

5.2 Principle of Superposition and Approximation of Variable - RatePressure HistoriesThe previous section illustrated the effect of the production from several wells in areservoir on the bottomhole flowing pressure of a particular well. Of equal interest isthe 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 wellis 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 initialcleanup after completion, then during production testing then finally during produc-tion as rates are altered to match reservoir management requirements (for examplelimiting the producing gas oil ratio during production). A simple pressure and flowrateplot versus time would resemble figure 10.



flowr

ate,

q

time, t

(q2 - q1)q1q2

t1

t1

botto

mho

le fl

owin

g pr

essu

re, P

wf

P associated with (q1 - 0)

P associated with (q2 - q1)

continuation of the effect of q1 in the reservoir

time, t

initial reservoir pressure

The well has been brought onto production at an initial flowrate, q1. The bottomholeflowing pressure has dropped through time (as described by the appropriate boundaryconditions and the flow regime) until at time t1, the flowrate has been increased to q2and this change from q1 to q2 has altered the bottomhole flowing pressure (again asdescribed by the boundary conditions and the flow regime). The total (i.e. the realbottomhole flowing pressure) is calculated by summing the pressure drops caused bythe flowrate q1 bringing the well on production, plus the pressure drop created by theflowrate change q2 - q1 for any time after t1. During the first period (q1) the pressuredrop at a time, t, is described by

P(t) = P - P = P (t) qi wf D2

kh (5.2)

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

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

P(t) = q2 kh

P (t) + (q2 kh

P (t - t )D 2 D 1

q) (5.3)

In this case, the pressure drop is that caused by the rate q1 over the duration t, plus thepressure drop caused by the flowrate change q2 - q1 over the duration t - t1. In fact, thepressure perturbation caused by q1 still exists in the reservoir and is still causing aneffect at the wellbore. On top of that, the next perturbation caused by flowrate changeq2 - q1 is added or superposed to give the total pressure drop ( at the wellbore in thiscase).

In mathematical terms:

0 t t : P(t) = P (t) q2 kh1 D

(5.4)

Figure 10Effect of flowrate changeson the bottomhole flowingpressure

48

t > t P(t) = q2 kh

P (t) + q P (t - t )1 D 2 D 1

:

qkh

1

2

(5.5)

In this 2nd equation, the first term is P from flow at q1 : 2nd term is the incrementalterm P caused by increasing rate by an increment (q2-q1). These expressions are validregardless 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 regimeand boundaries. If unsteady state is assumed and the line source solution applied, then

P = Ei ( )D

=

/

P Pq kh

cr

kti wf w

212 4

2

(5.6)

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

P(t) = - Ei ( )

qkh

cr

ktw1

4 4

2(5.7)

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

P(t) = - qqkh

cr

ktq q

khcr

k t tw w1 2 1

14 4 4 4

2 2

Ei ( ) - Ei ( )

( )( )

(5.8)

This approach can be extended to many flowrate changes as illustrated in figure 11.

flowr

ate,

q

q1q2

q3q4

time, t

Botto

mho

le fl

owin

g pr

essu

re, P

wf

time, t

different flow rates

pressure responses cused by rate changes

Figure 11Multi rate pressureresponse in a wellbore



This leads to a general equation

P(t) q2 kh

P (t) (q q )2 kh

P (t t ) (q q )2 kh

P (t t ) ...

(q q )2 kh

1D

2 1D 1

3 2D 2

n n 1

= + + +

+

P (t t )D n 1

(5.9)

or

P(t) q2 kh

P (t) q qq

P (t t )1 D i i 11i 2

n

D i 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 additionalterm is added to reflect this. Assuming that the well was shut in during the nth flowrateperiod, the pressure builds during the shut in time, t (i.e. t starts from the instant thewell is shut in) back up towards the initial reservoir pressure according to

P P q2 kh

P (t) q qq

P (t t + t) q2 khi ws

1D

i i 1

11 2

n

D n-1 i 1n-1 = +

=

P ( t)D

(5.11)whereP

ws is the shut in bottomhole pressure

tn-1 is the total producing time before shut int is the closed in time from the instant of shut in.

5.3 Effects of Rate ChangesThe application of superposition to a well with several rate changes is illustrated asfollows. A well is known to have the flowrate history as presented in figure 12. It isseen that the well is brought onto production at a flowrate, q1 and this is maintainedconstant until time, t1 at which the flowrate is increased to q2. This second flowratecontinues until time t2 when the flowrate is decreased to q3. In terms of the reservoir,it is assumed that the reservoir is in unsteady state flow regime and the line source canbe used to describe the pressure drop caused by the flowrate changes. In this case, thefirst flow rate change is when the well is brought on production, so the change fromzero to q1 causes the first pressure perturbation to move into the reservoir.

It is the bottomhole flowing pressure, Pwf, that is of interest, and it can be calculated

using the line source solution. There is the possibility of a skin zone around the well,so this must be accounted for. If no other flowrate change occurred, then eventuallyunsteady state would give way to either semi steady state or steady state conditionsand the bottomhole flowing pressure would either decline at a steady rate or (if steadystate) would remain constant at some level. Assuming that this did not occur and thatunsteady state conditions still existed when the flowrate was changed to q2 then thechange q2 - q1 would cause a second pressure perturbation that would move out intothe reservoir, following the first one created when the well was put on production. Thereservoir is still in unsteady state conditions i.e. the first pressure perturbation has notmet any barriers so the reservoir fluid still reacts as if it were an infinite volume and

50

this behaviour is still causing a decline in the pressure at the wellbore even though asecond pressure perturbation has been created and is moving out into the reservoir.The pressure drop due to this flowrate change can be calculated by the line sourcesolution and added to that produced by bringing the well onto production.

Eventually at time t2, the flowrate is changed again. This time, the pressure perturba-tion caused by q3 -q2 follows the first and second perturbations into the reservoir, andagain, as long as the reservoir fluid still behaves as if it were infinite in volume, thepressure drop created by this flowrate change can be added to the changes producedby the others to give the total pressure drop.

flowr

ate,

q q1q2

q3

t1 t2time, t

real well flowrate history

time, t

q1

q2 - q1

q3 - q2

t1

t2

flowr

ate,

q

time, t

time, t

equivalent flowrate effects in the reservoir

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

P P P q4 kh

ln cr4kt

2s1 i wf 11 w

2

= ( ) =

The next pressure drop is that produced by the flowrate change q2 - q1 at time, t1. It isstill the bottomhole flowing pressure that is to be determined, therefore any skin zonewill still exist and still need to be accounted for. The second pressure drop is:

Figure 12The equivalence of flowratechanges in a reservoir



P P P (q - q )4 kh

ln cr4k(t - t 2s2 i wf 2

2 1 w2

1

= ( ) =

)

And finally the third pressure drop is:

P P P (q - q3 i wf 3 3 2= ( ) = ))

4 khln cr

4k(t - t 2sw

2

2

)

The total pressure drop at the wellbore caused by all of the flowrate changes is

(Pi - Pwf )= P1 + P2 + P3

EXERCISE 19Two wells are brought on production in an undeveloped reservoir. Using the databelow, calculate the bottomhole flowing pressure in each well. Well 1 produces at110stm3/day for 27 days at which time Well 2 starts production at 180stm3/day andboth produce at their respective rates for a further 13 days when the bottomholeflowing pressures are calculated. Therefore Well 1 produces for 40 days when itspressure influence is calculated; Well 2 produces for 13 days when its pressureinfluence is calculated.


o1.2rm3/stm3

net thickness of formation, h, 36mviscosity of reservoir oil, 1x10-3 Pascompressibility, c 10 x10-9Pa-1permeability, k 110mDwellbore radius, r

w (both wells) 0.15m

external radius, re

7000minitial reservoir pressure, Pi 250.0barWell 1 flowrate (constant) 110stm3/dayWell 2 flowrate (constant) 180stm3/da

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