Course Reader Petr3511 2013

8/12/2019 Course Reader Petr3511 2013

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Wednesday, February 27, 2013

Reservoir Engineering

PETR3511 PETR8503: Course Reader2013

JISHAN LIU

SCHOOL OF MECHANICAL ENGINEERINGTHE UNIVERSITY OF WESTERN AUSTRALIA

Email: [email protected]: 6488 7205

F, Production

E,SystemE

xpansion

EFFull Balance

EFPartial Balance

F, Production

E,SystemE

xpansion

EFFull Balance

EFPartial Balance


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CHAPTER 1 RESERVOIR & RESERVOIR ENGINEERING

Petroleum Reservoir: A petroleum reservoir is a continuous body of oil and/or gas that

occurs within a single geological trap. A reservoir may be small (a few acres) or it may extend

over many square miles. A field consists of one or more reservoirs.

Tp,

Overburden Weight

e p

0 pe

Grain Stress orEffective Stress

LiquidPressure

Figure 1.1 Schematic diagram of a petroleum reservoir and its surroundings

External Geometry: Defined by seals or flow barriers that inhibit the migration of

hydrocarbons, forming a hydrocarbon trap.

Internal Architecture: Vertical stacking defined by stratigraphy.

Reservoir A box with its bottom removed.

Reservoir pressure, p, can be calculated as

ep (1)


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ep

e ep MPastress,EffectiveMPastress,Overburden

MPapressure,pore

e

p

Fig.1.2 Relation between overburden stress, pore pressure, and grain stress (effective stress).

Terzaghi equation


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QUESTION 1: How to calculate pore pressures at points B, C and D? Assuming the pore

pressure at Point A is equal to zero

.

Overlying rocks

Gas

Oil

Water

50 m

100 m

50 m

2,000 m

B

A

C

D

Rock density=2,200 kg/m3

Pressure gradient= 0.022 MPa/m

Density of Gas= 200 kg/m3


Density of Oil= 600 kg/m3


Density of Water= 1,000 kg/m3


e ep0 pe


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QUESTION 2: How to calculate pore pressures at points B, C and D? Assuming the pore

pressure at Point A is equal to the effective stress: ep

.

Overlying rocks

Gas

Oil

Water

50 m

100 m

50 m

2,000 m

B

A

C

D

Rock density=2,200 kg/m3


Density of Gas= 200 kg/m3


Density of Oil= 600 kg/m3


Density of Water= 1,000 kg/m3


e ep ep


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QUESTION 3: The distribution of pore pressure is shown in the following figure. Please

determine the heights for oil zone, gas zone and water zone.


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RESERVOIR ENGINEERING

In this unit, Reservoir Engineering is defined as the underground science of evaluating the

original hydrocarbons in place, the performance of the full range of reservoirs from dry gases

to heavy oils, including gas condensates and volatile oils, and the ultimate recovery. These

evaluations will be conducted based on a single variable, i.e., pore pressure, only.

QUESTION 4: The distribution of pore pressure within a reservoir is shown in the following

figure. How do you determine what pressure to use for these evaluations?


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DEVELOPMENT HISTOTRY OF RESERVOIR ENGINEERING

FLUID & ROCK PROPERTIES

Shithuis, 1935; Fancher, 1933

FIRST MATERIAL BALANCE

EQUATION

Shithuis, 1936

FLOW PRINCIPLES

Muskat 1937

DISPLACEMENT THEORY

Buckley & Leverret 1942

FLUID & ROCK PROPERTIES

Shithuis, 1935; Fancher, 1933

FIRST MATERIAL BALANCE

EQUATION

Shithuis, 1936

FLOW PRINCIPLES

Muskat 1937

DISPLACEMENT THEORY

Buckley & Leverret 1942

AQUIFER INFLUX

Everdingen & Hurst 1949

REVIEW OF RESERVOIR

ENGINEERING

Moore 1955

MBE GRAPHICAL SOLUTION

Havlena &Oldel, 1963

MBE - VOLATILE RESERVOIRS

Walsh 1994

AQUIFER INFLUX

Everdingen & Hurst 1949

REVIEW OF RESERVOIR

ENGINEERING

Moore 1955

MBE GRAPHICAL SOLUTION

Havlena &Oldel, 1963

MBE - VOLATILE RESERVOIRS

Walsh 1994

Engineering Subject

Theory for Secondary

Recovery

TankModel


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CHAPTER 2 RESERVOIR CLASSIFICATIONS

pRP In this unit, we evaluate the performance of a reservoir based on its response to pore pressure

only. Therefore, we can design the classification schemes accordingly.

Pore pressure is related to the number of phases. We can classify reservoirs as single-phase

reservoir; two-phase reservoir; multi-phase reservoir.

Overlying rocks

Gas

Oil

Water

50 m

100 m

50 m

2,000 m

B

A

C

D

Reservoir pressure (MPa)

22 MPa

22.1 MPa

22.7 MPa

0.01MPa50

wwCD hp

MPa23.2 CDCD ppp23.2 MPa

Pore pressure is also related to the reservoir internal structure. Based on the internal structure,

we can classify reservoirs as un-fractured reservoir and fractured reservoir.

IN THIS UNIT, WE CLASSIFY RESERVOIRS BASED ON THE INITIAL

RESERVOIR PRESSURE.

From a technical point of view, the various types of reservoirs can be defined by the location

of the initial temperature and pressure with respect to the two-phase (gas and oil) region as

commonly shown on pressure temperature (PT) phase diagrams. This figure is the PT diagram

of a particular reservoir fluid. The area enclosed by the bubble-point and dew-point lines to

the lower left is the region of pressure-temperature combinations in which both gas and liquid

phases will exist.

Point A:initially at 300oF and 3700psi. Since this point lies outside the two-phase region, it

is originally in a one-phase state.


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Reservoir Fluid: If the reservoir pressure declines along path (1), it will remain in the

single-phase or gaseous state.

Production Fluid: Although the fluid left in the reservoir remains in one phase, the fluid

produced through the wellbore and into surface separators, may enter the two-phase region

owning to temperature decline, as along line (2).

Figure 1.2: P-T Diagram for a Typical Reservoir Fluid

Point B:initially at 180oF and 3300psi. Since this point lies outside the two-phase region, it

is originally in a one-phase state.

Reservoir Fluid:As pressure declines along path (1) because of production, the composition

of the reservoir fluid will remain constant until the dew-point pressure is reached at 2700psi.

Production Fluid:Because the condensed liquid adheres to the walls of the pore walls of the

rock, the gas produced at the surface will have a lower liquid content. This process of

retrograde continues until B2 is reached.

Point C:initially at 75oF and 2900psi. Since this point lies outside the two-phase region, it is

0 50 100 150 200 250 300350

Reservoir Temperature, oF

4000

3500

3000

2500

2000

1500

1000

ReservoirPressure,psi

A

A2

B2

B1

B

B3

A1

C1

C

D

0%

20%

10%

5%

80%40%

Bubb

lePo

int

Critical

Point

Dew

Point

Bubble Point or

Dissolved Gas

Reservoir

Dew Point or

Retrograde Gas

CondensateReservoir

Single-phase Gas

Reservoir

Path

ofReservoirFluid

Cricon

dentherm

Path

ofP

rod

ucti

on

0 50 100 150 200 250 300350


4000

3500

3000

2500

2000

1500

1000

ReservoirPressure,psi

A

A2

B2

B1

B

B3

A1

C1

C

D

0%

20%

10%

5%

80%40%

Bubb

lePo

int

Critical

Point

Dew

Point

Bubble Point or

Dissolved Gas

Reservoir

Dew Point or

Retrograde Gas

CondensateReservoir

Single-phase Gas

Reservoir

Path

ofReservoirFluid

Cricon

dentherm

Path

ofP

rod

ucti

on


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originally in a one-phase state.

Reservoir Fluid:As pressure declines along path (1) because of production, the composition

of the reservoir fluid will remain constant until the bubble-point pressure is reached at

2550psi.

Production Fluid:Below this point, bubbles or a free gas will appear. Eventually, the free

gas evolved begins to flow to the wellbore, and in ever increasing quantities.

According to the initial conditions, reservoirs can be classified as

Single Phase Gas Reservoir Gas-Condensate Reservoir Undersaturated Oil Reservoir Saturated Oil Reservoir


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QUESTION 1: Assuming an ideal gas reservoir, the original gas pressure is 8MPa. During

production, the total pore volume, 0V , remains unchanged. When the reservoir pressure

becomes 4MPa, calculate the total gas volume in the reservoir. Comparing this volume with

the total pore volume, what can you conclude based on this comparison?

Ground Surface

Gas

Reservoir


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CHAPTER 3 PRODUCING MECHANISMS

pRP In this unit, we evaluate the performance of a reservoir based on its response to pore pressure

only. Therefore, all of the producing mechanisms can be related to the reservoir pressure.

1p 2p

21 pp

Balloon: The balloon size is proportional to the pressure. When air flows out from inside, the

balloon pressure becomes smaller.


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QUESTION 1: Use the balloon phenomenon to analyze the gas production process. When

the reservoir pressure deceases, what would happen to both the pore volume and the gas

volume?

PORE


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Types of Reservoir Energy

The major types of energy available for primary petroleum production are

The energy of compression of the water and rock within a reservoir; The energy of compression of oil within a reservoir; The energy of compression of gas within a reservoir; The energy of waters that are contiguous to and in communication with the petroleum

reservoir;

The gravitational energy that causes oil and gas to segregate within the reservoir.In the thermodynamic sense, the energy in the first four items of the list refers to the

potential energy stored in the compressed fluids. It is equivalent to the potential energy stored

in a compressed spring. Through the first law of thermodynamics, such energy can be

converted to the pressure-volume work (force through a distance) needed for fluids to be

produced. The energy in the fifth term is the potential energy caused by the differences

between different distances (elevations) from the Earths center of gravity.


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Producing Mechanisms

We need to know and understand by what mechanism the oil is being produced. These are the

fundamental drive mechanisms. The main drive mechanisms include water drive, oil and gas

expansion drive, compaction drive.

The performance characteristics of hydrocarbon producing reservoirs depend largely on thetypes of energy available to move the hydrocarbon fluids to the wellbore.

A water drive reservoir is an unsealed petroleum reservoir that is in communication with

water-bearing reservoirs (aquifers). There is appreciable movement of water from the aquifer

into the petroleum reservoir.

If the volumetric rate of water intrusion into the reservoir approaches the volumetric rate of

fluid withdrawal from the reservoir, then the reservoir is a complete-water drive reservoir. A

complete water drive reservoir experiences a very little pressure decline; however, some

pressure decline must exist else no potential (pressure) difference between the reservoir and

aquifer would exist and no water influx would occur. If the volumetric rate of water intrusioninto the reservoir is substantial but substantially less than the volumetric rate of fluid

withdrawal from the reservoir, then the reservoir is a partial water drive reservoir. When a

water drive exists, the reservoir pressure will always be sensitive to the producing rate. If the

producing rate is too large relative to the water influx rate, the water drive will lose its

effectiveness and the reservoir pressure will decline.

Oil Recovery Processes

PRODUCING MECHANISMS

EXPANSION

DRIVE

WATER

DRIVE

COMPACTION

DRIVE

OIL DRIVE GAS DRIVE

SOLUTION GAS

DRIVE

GAS

DRIVE

GAS CAP

DRIVE

PRODUCING MECHANISMS

EXPANSION

DRIVE

WATER

DRIVE

COMPACTION

DRIVE

OIL DRIVE GAS DRIVE

SOLUTION GAS

DRIVE

GAS

DRIVE

GAS CAP

DRIVE


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The simplest case is that of a single-phase oil field. Such fields can be found among fields in

their very early stage of development, when the gas or oil is produced by simple natural

decompression. This stage ends rapidly, when the pressure equilibrium between the oil field

and the atmosphere is attained: the natural production of oil or gas stops, though only a small

percentage of the total amount of oil or gas has been produced. This first stage is called

primary recovery.

In order to recover part of the remaining oil, we could pump off at the wells, creating a

pressure drop which would draw the oil to these wells. This would have two draw backs: first,

the pressure around the wells could fall below the bubble pressure of the oil so that the wells

would produce almost only gas, and the heavier components would mainly remain trapped in

the field. Second, diminishing the pressure in the fluid phase could lead the rock to collapse,

resulting in a field with lower permeability and hence more difficult to produce. This is why

we use an alternative method called secondary recovery: we divide the available wells into

two sets: one set of injection wells, and one set of production wells. The injection wells are

used to inject an inexpensive fluid (usually water) into the porous medium, in order to push

the oil toward the production wells. During this process, the pressure inside the field ismaintained above its initial level, so that the two above mentioned draw backs may be more

easily avoided.

For secondary recovery process, two cases are to be considered: Either the pressure can be

maintained always above the bubble pressure of the oil. The flow in the reservoir is then of

the two-phase immiscible type, one phase being water and the other being oil, with no mass

transfer between the phases. Or the pressure may drop, at some points, below the bubble

pressure of the oil: then the oil may split into one liquid phase and one gaseous phase atthermodynamical equilibrium. This is the so-called black-oil reservoir, with one water phase,

which does not exchange mass with the other phases, and two hydrocarbon phases (one liquid

Control Ps & Vs & T & Cs

Why the Recovery Fraction is Small?

Control Ps & Vs & T & Cs

Why the Recovery Fraction is Small?


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phase and one gaseous phase), which exchange mass when the pressure and the temperature

change.

The above waterflooding technique makes it possible to recover a certain percent (up to 40%)

of the oil. Three reasons for low recovery:

There exists regions which are never flooded by the water, and hence whose oil is notgoing to be produced; Even in the flooded regions, a significant part of oil (up to 20% to 30%) remains

trapped in the pores by the action of the capillary forces;

When the oil is heavy and viscous, the water is extremely mobile in comparison to theoil. Then instead of pushing the oil towards the production well, the water finds very

quickly its own way to the production well, getting the oil moving very slowly toward

the production well.

The oil industry developed a set of different techniques known under the generic name of

tertiary recovery techniques or enhanced oil recovery (EOR) techniques or improved oil

recovery (IOR) techniques. One of the main goals of those techniques is to achievemiscibility of fluids, thus eliminating the residual oil saturation, which was one cause of low

recovery with the water flooding technique. This miscibility is sought using temperature

increase or the introduction of other components such as polymers.

20%

35%

50%

Primary Recovery

Water Flooding

EOR

Recovery Processes

Without Stimulation

faster oil

Incremental

Oil

Same Process with

Well Stimulation

TIME

CumulativeOil

Recovery

Oil Recovery Summary


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QUESTION 2: Can you thinks five ways to get the oil droplets out?


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CHAPTER 4 ROCK PROPERTIES

pRP

In this unit, we evaluate the performance of a reservoir based on its response to pore pressureonly. Therefore, we can relate rock properties changes to the reservoir pressure.

A Four-Component Reservoir Model:To understand and predict the volumetric behavior of

oil and gas reservoirs as a function of pressure, knowledge of the physical properties of

reservoir rocks and fluids must be obtained. These properties are usually determined by lab

experiments on samples of actual reservoir rocks and fluids. In the absence of experimentally

measured properties, it is necessary for the petroleum engineer to determine the properties

from empirically derived correlations. In this unit, we assume that a reservoir consists of four

components:

Gas Oil Water RockWith the following assumptions:

At most, there are four components, gas, oil, water and rock; At most, there are four phases: gas, oil, water, and solid; The gas component is defined by the composition of the reservoir gas at the reservoir

conditions;

The oil component is defined by the composition of the reservoir oil at the reservoirconditions;

Thermodynamic equilibrium exists.In the following chapters, we will define the fundamental properties of each component and

interactions between components. These properties are absolutely essential for the

determination ofthe volumetric behavior of oil and gas reservoirs as a function of pressure.

In this chapter, we will cover five major properties of reservoir rocks: Porosity; Saturation; Permeability; Relative permeability; Compressibility.Two things are required to remember:

Definitions of all concepts; Evolution of each property during the production process.POROSITY


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We will start with what is porosity and then how does the porosity change as a function of

pore pressure.

Definition: Although a reservoir rock looks a solid to the naked eye, a microscopic

examination reveals the existence of voids in the rock. These pores are the ones where

petroleum reservoir fluids are present. This particular storage capacity is called porosity. The

more porous a reservoir rock material is, the greater the amount of voids it contains, hencegreater the capacity to store petroleum reservoir fluids. From a reservoir engineering

perspective, porosity is probably one of the most important reservoir rock properties.

Porosity, is a volumetric fraction defined as the ratio of the pore volume,poreV in a

reservoir rock to the total volume (bulk volume),bulkV :

bulk

pore

V

V (2.1)

The porosity of a rock is a measure of the storage capacity (pore volume) that is capable of

holding fluids. It may by occupied by a single-phase fluid or mixtures. As the sediments were

deposited and the rocks were being formed during past geological times, some void spaces

that developed became isolated from the other void spaces by excessive cementation. Thus,

many of the void spaces are interconnected while some of the pore spaces are completely

isolated. This leads to two distinct types of porosity, namely:

Absolute porosity Effective porosity.

bulk

totalpore

aV

V (2.2)

bulk

dporeerconnecte

eV

Vint (2.3)

The effective porosity is the value that is used in all reservoir engineering calculations

because it represents the interconnected pore space that contains the recoverable hydrocarbon

fluids.

QUESTION 1: Assuming the rock porosity is, use the mass conservation law to work out

the relation between the bulk density and rock & pore densities.


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Figure 1: Relation between porosity and density.

QUESTION 2: Assuming the pores in the rock as shown in Figure 1 are occupied by water,what is the total mass in the rock?

bulk

rock

pore

V

V

V

bulk

rock

pore


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Original Porosity & Induced Porosity: Porosity may be classified according to the mode of

origin as original induced. The original porosity is that developed in the deposition of the

material, while induced porosity is that developed by some geologic process subsequent to

deposition of the rock. The intergranular porosity of sandstones and the intercrystalline and

porosity of some limestones typify original porosity. Induced porosity is typified by fracture

development as found in shales and limestones and by the slugs or solution cavities

commonly found in limestones. Rocks having original porosity are more uniform in theircharacteristics than those rocks in which a large part of the porosity is included. For direct

quantitative measurement of porosity, reliance must be placed on formation samples obtained

by coring. Since effective porosity is the porosity value of interest to the petroleum engineer,

particular attention should be paid to the methods used to determine porosity. For example, if

the porosity of a rock sample was determined by saturating the rock sample 100 percent with

a fluid of known density and then determining, by weighing, the increased weight due to the

saturating fluid, this would yield an effective porosity measurement because the saturating

fluid could enter only the interconnected pore spaces. On the other hand, if the rock sample

were crushed with a mortar and pestle to determine the actual volume of the solids in the core

sample, then an absolute porosity measurement would result because the identity of any

isolated pores would be lost in the crushing process.

One important application of the porosity is its use in determining the original hydrocarbon

volume in place.


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Question 3As shown in Figure 2, think about how to use the porosity for the determination

of the original hydrocarbon volume in place (OHIP) and answer the following questions:

If the pore is occupied by oil only, the calculated PV is the original oil in place (OOIP); If the pore is occupied by gas only, the calculated PV is the original gas in place (OGIP). If the pore is occupied by oil and gas, what would happen? If the pore is occupied by oil, gas and water, what would happen?

Figure 2: Illustration of a reservoir volume determination

Question 4As shown in Figure 2, assuming the pores are occupied by water only, calculatethe original water in place (OWIP) and discuss what factors will change OWIP?

Question 5If pores are occupied by a mixture of oil, gas and water, discuss how to calculate

the original water, oil, or gas in place (OWIP, OOIP and OGIP)?

RESERVOIR

bulk

rock

pore

V

V

V

bulk

rock

pore

bulkVPV

RESERVOIR

bulk

rock

pore

V

V

V

bulk

rock

pore

bulkVPV


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COAL POROSITY

s

b

Coal Matrix

Fracture

Fig.3 Fractured Coal

Fractured coals are made up of two porosity systems; one matrix formed by void spaces

between the grains of the coal, and a second formed by void spaces of fractures as shown in

Fig.1. In a fractured coal reservoir the total porosity ( t ) is the result of the simple addition of

the matrix and fracture porosities:

fracturematrixt (1)

Where

VolumeBulkCoal

VolumeVoidMatrixCoalmatrix

VolumeBulkCoal

VolumeVoidFractureCoalfracture


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QUESTION 5: Assuming cmsmmb 10.0 , calculate the fracture porosity and discuss

the result.

s

b

Coal Matrix

Fracture


28/125


FLUID SATURATION

While porosity represents the maximum capacity of a reservoir rock to store fluids, how can

we know how much of this available capacity, pore volume, or pore space distributed or

partitioned among the typical reservoir fluid phases: gas, oil and water? In order to achieve

this, we need to introduce the concept of fluid saturation.

Saturation is defined as that fraction, or percent, of the pore volume occupied by a particular

fluid (oil, gas, or water). This property is expressed mathematically by the following

relationship:

VolumePoreTotal

FluidtheofVolumeTotalSaturationFluid

Assumingwgo SSS ,, represent the oil saturation, gas saturation and water saturation,

respectively, andwgo VVV ,, for the oil volume, gas volume, and water volume, respectively,

and pV the total pore volume, applying the above mathematical concept to each to each

reservoir fluid gives

p

w

w

p

g

g

p

o

o

V

VS

V

VS

V

VS

(2.7)

Thus, all saturation values are based on pore volume and not on the gross reservoir volume.

The saturation of each individual phase ranges between zero to 100 percent. By definition, the

sum of the saturations is 100%, therefore

1 wgo

SSS (2.8)

Equation (2.8) is probably the simplest, yet the most fundamental equation in reservoir

engineering, and is used everywhere in reservoir engineering calculations. Moreover, many

important reservoir rock properties, such as capillary pressure and relative permeability, areactually related or linked with individual fluid-phase saturations.

The fluids in most reservoirs are believed to have reached a state of equilibrium and, therefore,

will have become separated according to their density, i.e., oil overlain by gas and underlain

by water. In addition to the bottom (or edge) water, there will be connate water distributed

throughout the oil and gas zones. The water in these zones will have been reduced to some

irreducible minimum. The forces retaining the water in the oil and gas zones are referred to as

capillary forces because they are important only in pore spaces of capillary size. Connate

(interstitial) water saturation Swc is important primarily because it reduces the amount of

space available between oil and gas. It is generally not uniformly distributed throughout the

reservoir but varies with permeability, lithology, and height above the free water table.

Another particular phase saturation of interest is called the critical saturation and it is


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associated with each reservoir fluid. The definition and the significance of the critical

saturation for each phase is described below.

Critical oil saturation,ocS : For the oil phase to flow, the saturation of the oil must exceed a

certain value which is termed critical oil saturation. At this particular saturation, the oil

remains in the pores and, for all practical purposes, will not flow.

Residual oil saturation,or

S : During the displacing process of the crude oil system from the

porous media by water or gas injection (or encroachment) there will be some remaining oil

left that is quantitatively characterized by a saturation value that is larger than the critical oil

saturation. This saturation value is called the residual oil saturation,orS . The term residual

saturation is usually associated with the nonwetting phase when it is being displaced by a

wetting phase.

Movable oil saturation,omS : Movable oil saturation is another saturation of interest and is

defined as the fraction of pore volume occupied by movable oil as expressed by the followingequation:

ocwcom SSS 1 (2.9)

wherewcS and

ocS are connate water saturation and critical oil saturation, respectively.

Critical gas saturation,gc

S : As the reservoir pressure declines below the bubble-point

pressure, gas evolves from the oil phase and consequently the saturation of the gas increases

as the reservoir pressure declines. The gas phase remains immobile until its saturation exceeds

a certain saturation, called critical gas saturation, above which gas begins to move.

Critical water saturation,gc

S : The critical water saturation, connate water saturation, and

irreducible water saturation are extensively used interchangeably to define the maximum

water saturation at which the water phase will remain immobile.


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QUESTION 4: The following is a summary of saturation definitions. Discuss their changes

during production processes.

1poreV

wiS

orS

wmS

omS1

omwmorwr SSSS

Irreduciblewatersaturation

Residualoilwatersaturation

Movablewatersaturation

Movableoilsaturation

1

1

1

1

**

*

*

ow

orwr

orw

o

orwr

wrw

w

SS

SS

SSS

SS

SSS

1

wo

oromo

wrwmw

SS

SSSSSS


31/125


ABSOLUTE PERMEABILITY

In above sections, we address the porosity and saturation both of which are used to define the

storage capacity of reservoir rock for reservoir fluids. However, merely having a large enough

porosity of reservoir rock is not sufficient because the petroleum reservoir fluids contained inthe pore spaces of reservoir rock have to flow, so that they can be produced or brought to the

surface from the reservoir. This particular property of a reservoir rock is called permeability.

Permeability is one of the most influential parameters in determining the production

capabilities of a producing formation.

Unlike porosity, permeability is a flow property (dynamic) and therefore can be characterized

only by flow experiments in a reservoir rock. Permeability is a property of the porous medium

that measures the capacity and ability of the formation to transmit fluids. The rock

permeability, k, is a very important rock property because it controls the directional

movement and the flow rate of the reservoir fluids in the formation. Absolute permeability is

the rock permeability when a reservoir rock is 100% saturated with a given fluid. It should be

noted that the absolute permeability is a property of the rock alone and not the fluid that flowsthrough it, provided no chemical reaction takes place between the rock and the flowing fluid.

All the equations used to describe fluid flow in reservoirs are based on Darcys law. Darcy

(1856), investigated the flow of water through sand filters. He observed the following

relationship between velocity and pressure gradient as

x

pk

A

Qux

(2.10)

Where

direction-xtheinCoordinate

Pressure

flowtoopenareasectional-Cross

ViscosityFluid

tyPermeabili

RateFlow

direction-xtheinvelocityDarcy

x

p

A

k

Q

ux

Basic assumptions:

It is assumed that the porous medium is saturated with a single fluid. The flowing fluid is incompressible. The linear dependence of flow velocity on the pressure gradient implies laminar. The flow takes place under the viscous regime (i.e., the rate of flow is sufficiently low

so that it is directly proportional to the pressure differential or the hydraulic gradient).

The flowing fluid does not react with the porous medium.

The negative sign in the above equations indicates that pressure decreases in the direction of

flow. The sign convention is therefore that distance is measured positive in the direction of

flow.


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33/125


QUESTION: Assuming 1-D flow, determine the pressure distribution.

Figure 2.4 One-dimensional horizontal flow.

Flow-InFlow-Out

A

1p

2p?

Flow-InFlow-Out

A

Flow-InFlow-Out

A

1p

2p?


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Dimensional Analysis for Permeability Unit:

2

2

23

222

2

3

1

Lk

LM

LT

LT

MkL

T

L

LT

Mtime

area

force

LTM

LTML

areaforcep

LA

T

LQ

21 m-12101Darcy

One Darcy is a relatively high permeability as permeabilities of most reservoir rocks are

below one Darcy. In order to avoid the use fractions in describing permeabilities, the term

millidarcy is used. As the term indicates, one millidarcy, i.e., 1 md, is equal to one-thousand

of one Darcy or, 1 Darcy = 1000 md.


35/125


QUESTION: A fluid of viscosity of 1.2cp flows through a cylindrical core at a rate of

scm /25.0 3 with a pressure drop of 2.5 atm. Core dimensions are length of 12cm and a 5-cm2

flow area. Determine the core permeability.

2

3

5

5.2

/25.0

2.1

???

cmA

atmp

scmq

cp

k


36/125


2.4 Relative Permeability

Darcys law was considered to apply when the porous medium was 100% saturated with a

homogeneous single-phase fluid. However, petroleum reservoirs having much simple

single-phase fluid systems seldom exist; usually reservoir rock systems are saturated with atleast two or more fluids such as water, oil and gas. These multiphase fluid systems play a very

important role in the reservoir flow processes when petroleum reservoirs are produced by

primary recovery mechanism or immiscible displacement methods involving the injection of

water or gas. It is under these circumstances that more than one fluid phase is flowing or is

mobile through a porous medium; thus the flow of one fluid phase interferes with the other.

This interference is a competition for the flow paths.

Numerous laboratory studies have concluded that the effective permeability of any reservoir

fluid is a function of the reservoir fluid saturation and the wetting characteristics of the

formation. It becomes necessary, therefore, to specify the fluid saturation when stating the

effective permeability of any particular fluid in a given porous medium. Just as k is theaccepted universal symbol for the absolute permeability, ko, kg, and kw are the accepted

symbols for the effective permeability to oil, gas, and water, respectively. The saturations, i.e.,

So, Sg, and Sw, must be specified to completely define the conditions at which a given

effective permeability exists.

Effective permeabilities are normally measured directly in the laboratory on small core plugs.

Owing to many possible combinations of saturation for a single medium, however, laboratory

data are usually summarized and reported as relative permeability. The absolute permeability

is a property of the porous medium and is a

measure of the capacity of the medium to transmit fluids. When two or more fluids flow at the

same time, the relative permeability of each phase at a specific saturation is the ratio of theeffective permeability of the phase to the absolute permeability, or:

k

kk

k

kk

k

kk

wrw

g

rg

oro

(2.11)

For example, if the absolute permeability k of a rock is 200 md and the effective permeability

ko of the rock at an oil saturation of 80 percent is 60 md, the relative permeability kro is 0.30

at So = 0.80. Since the effective permeabilities may range from zero to k, the relative

permeabilities may have any value between zero and one, or:


37/125


0.1,,0 rorgrw kkk (2.12)

It should be pointed out that when three phases are present the sum of the relative

permeabilitiesrwrgro

kkk is both variable and always less than or equal to unity.

Initially, it might appear that the sum of the phase permeabilities equals the total or absolute

permeability, which would mean that the relative permeabilities should sum to unity. However,

this is not true. When two or more phases are present, capillary forces exist that reduce the

flow rate of each individual phase in a non-linear fashion. This means that the sum of the

phase permeabilities is always less than the total or absolute permeability and the sum of

relative permeabilities is always less than one. Indeed, at irreducible water saturation, the

relative permeability to water becomes zero while the relative permeability to oil or gas is less

than one because the immobile water is occupying some of the flow volume. Similarly, at

residual oil saturation, the relative permeability to water or gas is less than one.

Question 2.7Relative Permeability Calculations from Steady-State TestsA set of steady state oil/water relative flow rates at different saturations is given below. Core

dimensions are a length of 12 cm and a 5-cm2 flow area (i.e. the area perpendicular to the

direction of flow). Oil viscosity is 5 cp and water viscosity is 1.2 cp.

Determine the oil and water relative permeabilities; Determine the oil- and water-phase permeabilities.Solution:

1. Relative Permeabilities2.

Sw qo(cm3/s) qw(cm3/s)

0 0.06 0

0.2 0.042 0

0.3 0.03 0.01

0.4 0.02 0.02

0.5 0.013 0.035

0.6 0.0075 0.051

0.7 0.004 0.068

0.8 0.001 0.085

0.85 0 0.0961 0 0.25

00

///

///

w

w

w

w

So

So

Soo

Sooo

roq

q

dpdxAq

dpdxAq

k

kk

Similarly,

11

///

///

w

w

w

w

So

So

Sww

Swww

rwq

q

dpdxAq

dpdxAq

k

kk


38/125


kro krw

1 0

0.7 0

0.5 0.04

0.333333 0.08

0.216667 0.14

0.125 0.2040.066667 0.272

0.016667 0.34

0 0.384

0 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Sw

Kro

,

krw

2. Phase Permeability

mddarcy

dx

dp

A

qk ooo

4.62624.0

5.2

12

5

5013.0/

ko kw

0.288 0

0.2016 0

0.144 0.01152

0.096 0.02304

0.0624 0.04032

0.036 0.058752

0.0192 0.078336

0.0048 0.097920 0.110592

0 0.288

2.5 Compressibility

A reservoir thousands of feet underground is subjected to an overburden pressure caused by

the weight of the overlying formations. Overburden pressures vary from area to area

depending on factors such as depth, nature of the structure, consolidation of the formation,

and possibly the geologic age and history of the rocks. Depth of the formation is the most

important consideration, and a typical value of overburden pressure is approximately one psiper foot of depth. The weight of the overburden simply applies a compressive force to the

reservoir. The pressure in the rock pore spaces does not normally approach the overburden


39/125


pressure. A typical pore pressure, commonly referred to as the reservoir pressure, is

approximately 0.5 psi per foot of depth, assuming that the reservoir is sufficiently

consolidated so the overburden pressure is not transmitted to the fluids in the pore spaces.

The pressure difference between overburden and internal pore pressure is referred to as the

effective overburdenpressure. During pressure depletion operations, the internal pore pressure

decreases and, therefore, the effective overburden pressure increases. This increase causes thefollowing effects:

The bulk volume of the reservoir rock is reduced.

Sand grains within the pore spaces expand.

These two volume changes tend to reduce the pore space and, therefore, the porosity of the

rock. Often these data exhibit relationships with both porosity and the effective overburden

pressure.

Compressibility typically decreases with increasing porosity and effective overburden

pressure. Geertsma (1957) points out that there are three different types of compressibilitythat must be distinguished in rocks:

Rock-matrix compressibility,r

c : The rock matrix compressibility is defined as the

fractional change in volume of the solid rock material (grains) with a unit change in pressure.

Mathematically, the rock compressibility coefficient is given by

dp

dV

Vc rock

rock

r

1 (2.13)

Rock-bulk compressibility,Bc : The rock-bulk compressibility is defined as the fractional

change in volume of the bulk volume of the rock with a unit change in pressure. The

rock-bulk compressibility is defined mathematically by:

eff

B

B

Bd

dV

Vc

1 (2.14)

WhereB

V is the bulk volume and eff is effective pressure.

Pore compressibility,pc : The pore compressibility coefficient is defined as the fractional

change in pore volume of the rock with a unit change in pressure and given by the following

relationship:

dp

dV

Vc

p

p

p

1 (2.15)

Wherep

V is the pore volume.

Equation 2-15 can be expressed in terms of the porosity by noting that porosity increases


40/125


with the increase in the pore pressure; or:

dp

dcp

1 (2.16)

For most petroleum reservoirs, the rock and bulk compressibility are considered small in

comparison with the pore compressibilityp

c . Theformation compressibilityf

c is the term

commonly used to describe the total compressibility of the formation and is set equal top

c ,

i.e.:

dp

d

dp

dV

Vcc

p

p

pf

11 (2.17)

Typical values for the formation compressibility range from16103 psi to 161025 psi .

It should be pointed out that the total reservoir compressibility ct is extensively used in the

transient flow equation and the material balance equation as defined by the following

expression:

fggwwoot ccScScSc (2.18)

Question 2.8: Measured volumes and pressures are listed below:

p(psi) Vp(cm3) Vb(cm3)

9800 3.42 20.53

9000 3.379 20.498

8000 3.337 20.447

7000 3.303 20.413

6000 3.276 20.3825000 3.257 20.367

4000 3.243 20.353

3000 3.23 20.34

2000 3.213 20.332

1000 3.177 20.329

500 3.144 20.254

Calculate porosity, rock compressibility, bulk compressibility and pore compressibility.


41/125


porosity cp cb

16.65855 1.49854E-05 1.94837E-06

16.48454 1.24297E-05 2.48805E-06

16.32024 1.01888E-05 1.66284E-06

16.18087 8.17439E-06 1.51864E-06

16.07301 5.79976E-06 7.35943E-07

15.99155 4.29843E-06 6.87386E-0715.93377 4.00863E-06 6.38726E-07

15.88004 5.26316E-06 3.93314E-07

15.80268 1.12045E-05 1.47551E-07

15.62792 2.07743E-05 7.37862E-06

15.52286

0.00E+00

5.00E-06

1.00E-05

1.50E-05

2.00E-05

2.50E-05

0 2000 4000 6000 8000 10000

Pressure, psi

Compress

ibility,1

/ps

i


42/125


DecreasingPermeability

Figure 2.5 Resource Triangle.

QUESTIONHow to relate the resource triangle to the rock properties?


43/125


CHAPTER 3 FLUID PROPERTIES

To understand and predict the volumetric behavior of oil and gas reservoirs as a function of

pressure, knowledge of the physical properties of reservoir fluids must be obtained. These

fluid properties are usually determined by lab experiments on samples of actual reservoir

fluids. In the absence of experimentally measured properties, it is necessary for the petroleum

engineer to determine the properties from empirically derived correlations. The objective of

this chapter is to define fundamental PVT properties for the following reservoir fluids:

Natural Gases Crude Oil Water

3.1 Gas Properties

3.1.1 Gas Law

A gas is defined as a homogeneous fluid of low viscosity and density that has no definite

volume but expands to completely fill the vessel in which it is placed. Generally, the natural

gas is a mixture of hydrocarbon and nonhydrocarbon gases. The hydrocarbon gases that are

normally found in a natural gas are methanes, ethanes, propanes, butanes, pentanes, and small

amounts of hexanes and heavier. The nonhydrocarbon gases (i.e., impurities) include carbon

dioxide, hydrogen sulfide, and nitrogen.

Behavior of Ideal Gases:The kinetic theory of gases postulates that gases are composed of a

very large number of particles called molecules. For an ideal gas, the volume of these

molecules is insignificant compared with the total volume occupied by the gas. It is also

assumed that these molecules have no attractive or repulsive forces between them, and that all

collisions of molecules are perfectly elastic. Based on the above kinetic theory of gases, a

mathematical equation called equation-of-state can be derived to express the relationship

existing between pressure p, volume V, and temperature T for a given quantity of moles of

gas n. This relationship for perfect gases is called the ideal gas law and is expressedmathematically by the following equation:

nRTPV (3.1)

Where is the absolute pressure, psi; V is the volume, cft; T is the temperature, oR; n is the

number of moles of gas, lb-mole; R is the universal gas constant which, for the above units,

has the value of 10.730psi cft/lb-mole oR.

Behavior of Real Gases: Basically, the magnitude of deviations of real gases from theconditions of the ideal gas law increases with increasing pressure and temperature and varies

widely with the composition of the gas. Real gases behave differently than ideal gases. The


44/125


reason for this is that the perfect gas law was derived under the assumption that the volume of

molecules is insignificant and that no molecular attraction or repulsion exists between them.

This is not the case for real gases. Numerous equations-of-state have been developed in the

attempt to correlate the pressure-volume-temperature variables for real gases with

experimental data. In order to express a more exact relationship between the variables p, V,

and T, a correction factor called thegas compressibility factor, gas deviation factor, or simply

thez-factor, must be introduced into Equation 3-1 to account for the departure of gases fromideality. The equation has the following form:

znRTPV (3.2)

where the gas deviation factor z is a dimensionless quantity and is defined as the ratio of the

actual volume of n-moles of gas at T and p to the ideal volume of the same number of moles

at the same T and p:

pnRT

V

V

Vz

ideal

actual

/

(3.3)

Figure 3.1: Effect of Pressure on the Gas Deviation Factor.

Many gases near atmospheric PT conditions approach ideal behavior (Z=1). All molecules

have two tendencies: 1. to fly apart from each other because of their constant kinetic motion,

and 2. to come together because of electrical attractive forces between molecules. Because the

molecules are quite far apart, the attractive forces are negligible, and the gas behaves close to

ideal. Also at high temperatures the kinetic motion, being greater, makes the attractive forces

comparatively negligible, and, again, the gas approaches ideal behavior. When the gas is

highly compressed, the gas appears to be more difficult to compress.

Question 3.1 For an ideal gas under the isothermal condition, prove the following relation

PRESSURE, PSI

1000 2000 6000

Z

1.0

PRESSURE, PSI

1000 2000 6000

Z

1.0


45/125


between two P-V states:

1

2

2

1

V

V

p

p (3.4)

Assuming psipcfVpsip 5001010002

6

11

, calculate ?2

V

Question 3.2 For an ideal gas reservoir under the isothermal condition as shown in Figure

3.2, use equation (3.40 to calculate the surface volume if we move all of the gas in the

reservoir to the surface.

Figure 3.2 Volumetric Relations

3.1.2 Gas Formation Volume Factor

The gas formation volume factor is used to relate the volume of gas, as measured at reservoir

conditions, to the volume of the gas as measured at standard conditions, i.e., 60F and 14.7

psia. This gas property is then defined as the actual volume occupied by a certain amount of

gas at a specified pressure and temperature, divided by the volume occupied by the same

amount of gas at standard conditions. In an equation form, the relationship is expressed as

Ground Surface

cfV

psip

610

1000

?

7.14

V

psip

Ground Surface

cfV

psip

610

1000

?

7.14

V

psip


46/125


SCg

Tpg

gV

VB

, (3.5)

Applying the real gas equation-of-state gives

p

zT

T

p

p

nRTz

p

nzRT

BSC

SC

SC

SCSC

g (3.6)

Assuming that the standard conditions are represented by psc =14.7psia and Tsc = 520, the

above expression can be reduced to the following relationship:

p

zTBg

02827.0 (3.7)

For an ideal gas, equation (3.7) becomes

p

TBg 02827.0 (3.8)

Under the isothermal condition, the gas formation volume factor is inversely proportional to

pressure as shown in Figure 3.3.

gBgB


47/125


Figure 3.3 Relation between gas formation volume factor and pressure.

3.1.3 Gas Compressibility

Knowledge of the variability of fluid compressibility with pressure and temperature is

essential in performing many reservoir engineering calculations. For a liquid phase, the

compressibility is small and usually assumed to be constant. For a gas phase, the

compressibility is neither small nor constant.

By definition, the isothermal gas compressibility is the change in volume per unit volume for

a unit change in pressure or, in equation form:

p

V

Vc

g

g

g

1 (3.9)

From the real gas equation-of-state:

p

nRTzVp

(3.10)

Differentiating the above equation with respect to pressure gives

2

1

p

z

p

z

pnRT

p

Vp

(3.11)

Substituting (3.6) into (3.4) gives

p

z

zpcg

11 (3.12)

For an ideal gas, z=1 and 0/ pz , therefore,


48/125


pcg

1 (3.13)

3.1.4 Dissolved Gas-Oil Ratio

Dissolved gas-oil ratio is defined as the ratio of surface gas to stock-tank oil in a reservoir

liquid phase at reservoir conditions. The volumes of surface gas and stock-tank oil are those

measured at standard conditions. The dissolved gas-oil ratio, vR is defined as

SCo

SCg

sV

VR (3.14)

For a particular gas and crude oil to exist at a constant temperature, the solubility increases

with pressure until the saturation pressure is reached. At the saturation pressure (bubble-point

pressure) all the available gases are dissolved in the oil and the gas solubility reaches its

maximum value. Rather than measuring the amount of gas that will dissolve in a given

stock-tank crude oil as the pressure is increased, it is customary to determine the amount of

gas that will come out of a sample of reservoir crude oil as pressure decreases.

A typical gas solubility curve, as a function of pressure for an undersaturated crude oil, is

shown in Figure 3.4. As the pressure is reduced from the initial reservoir pressure pi, to the

bubble-point pressure pb, no gas evolves from the oil and consequently the gas solubility

remains constant at its maximum value of Rsb. Below the bubble-point pressure, the solutiongas is liberated and the value of Rs decreases with pressure.

Figure 3.4 Dissolved Gas-Oil Ratio and Pressure Diagram

3.1.5 Volatilized Oil-Gas Ratio


49/125


50/125


51/125


At pressures below the bubble-point pressure, the oil compressibility is defined as

p

R

B

B

p

B

Bc s

o

go

o

o

1 (3.20)

Note that the first term in Eq (3.20) is negative. The second term is necessary because FVFs

contain the effect of solution gas on the change in liquid volume caused by gas going into

solution as the pressure is increased.

3.3 Water

Similarly, we can define the water formation volume factor as

SCw

Tpw

wV

VB

, (3.21)

3.4 Summary of Single-Phase PVT Properties

Two-Component Model

1. At most, there are two hydrocarbon pseudo-components: surface gas and stock-tank-oil;2. At most, there are two hydrocarbon phases: gas and oil;3. The surface gas pseudo-component is defined by the composition of the gas at standard

conditions;

4. The stock-tank-oil pseudo-component is defined by the composition of the stock-tank oilat standard conditions;

5. The surface gas can partition between the oil and gas phases;6. The stock-tank oil can partition between the oil and the gas phases;7. Thermodynamic equilibrium exists.Assumption 5: The partitioning of surface gas into the oil phase allows for dissolved gas;

Assumption 6: The partitioning of stock-tank oil into gas phase allows for volatilized oil;

Partitioning implicitly exists at all conditions except standard conditions. Standard conditionsto measure a standard cubic food are defined as 14.7psi and 60oF.

The summary of PVT properties definitions is shown in Figure 3.5.

2 5

3

1

gB

3

4vR

3

Gas

Oil

Expanded

gas

5

6

4

Gas

Oil

Expanded

oil

OIL

MoveablePiston

BPpp

1

2

Gas

Oil

BPpp

Reservoir temperature

Surface conditions

22 55

3

1

gB

3

1

gB

3

4vR

3

4vR

3

Gas

Oil

Expanded

gas

3

Gas

Oil

Expanded

gas

5

6

4

Gas

Oil

Expanded

oil

5

6

4

Gas

Oil

Expanded

oil

Expanded

oil

OIL

MoveablePiston

BPpp

OIL

MoveablePiston

BPpp

1

2

Gas

Oil

BPpp

1

2

Gas

Oil

BPpp

Reservoir temperature

Surface conditions


52/125


Figure 3.5: Summary of Gas PVT Properties

3.5 Two-Phase Formation Volume Factors

Two-phase or total formation volume factors are secondary PVT properties. They are strictly a

function of the standard PVT relations ( vsgo RandRBB ,,, ); consequently, they can always be

calculated from the standard PVT.

Two-Phase Oil Formation Volume Factor ( toB ): the total (liquid-phase plus gas-phase)

volume at reservoir conditions divided by its resulting oil-phase volume at standard

conditions.

SCo

Tpgo

toV

VVB

, (3.22)

Figure 3.6: Schematic interpretation of two phase oil formation volume factor.


53/125


Two-Phase Gas Formation Volume Factor ( tgB ): the total (liquid-phase plus gas-phase)

volume at reservoir conditions divided by its resulting gas-phase volume at standard

conditions.

SCo

Tpgo

tg

V

VVB

,

(3.23)

Figure 3.6: Schematic interpretation of two phase gas formation volume factor.

3.6 Relations between Two-Phase and Single-Phase PVT Properties

Consider a reservoir oil sample composed of N surface volumes (e.g. STB) of oil and G

surface volumes (scf) of gas. The two-phase oil FVF is defined as

N

VB totalto (3.24)

Where Vtotal is the total volume (liquid+gas) of the hydrocarbon systems, and is defined as

fgfototal VVV (3.25)

Where Vfo and Vfg are defined as the total volume of the free-oil phase and the total volume

of the free-gas phase, respectively. They are given by

gfgfg

ofofo

BGV

BNV

(3.26)

Where Nfo and Gfg are the surface volume of oil in the free-oil phase and the surface volume

of gas in the free-gas phase.

The total surface volumes of the stock-tank oil and surface gas are


54/125


NNN fgfo (3.27)

GGG fgfo (3.28)

Where

phaseoil-freein thegasof(SCF)volumesurface

phasegas-freein theoilof(STB)volumesurface

fo

fg

G

N

fgN and foG are given by

vfgfg RGN (3.29)

sfofo RNG (3.30)

Solving equations (3.8) through (3.11) gives

GGRRGNR fgsvfgs (3.31)

Solving for Gfg gives

vs

s

fgRR

NRGG

1 (3.32)

Similarly, we can obtain

vs

v

foRR

GRNN

1 (3.33)

We know

fgfo

gfgofofgfototal

toNN

BGBN

N

VV

N

VB

(3.34)

Substituting above equations into (3.34) gives

sv

ssigvsio

toRR

RRBRRBB

1

1 (3.35)

Similarly, we can obtain

G

VB Ttg (3.36)

sv

vviosvig

tg

RR

RRBRRBB

1

1 (3.37)

Equations (3.35) and (3.37) apply to any fluid whose overall composition is defined by a


55/125


gas-oil ratio, siR , or by a oil-gas ratio, viR .

For the special case of a black oil, 0vR , they become

gtg

ssigoto

BB

RRBBB

(3.38)

3.5 Expansivities

The oil-phase expansivity is defined as the total expansion of a unit mass of oil phase between

two pressures at the reservoir temperature, expressed in units of reservoir volume per unit

volume of oil at standard conditions. The two subject pressures are a reference pressure and a

terminal pressure. The terminal pressure is invariably less than the reference pressure, hence

the term expansivity. The reference pressure is usually the initial reservoir pressure. The

expansion volume includes a gas-phase volume if one should emerge from the expanding oil

phase.

The gas expansivity is defined analogously: the total expansion of a unit mass of gas phase

between two pressures at the reservoir temperature, expressed in units of reservoir volume per

unit volume of gas at standard conditions. The expansion volume includes liquid dropout if

such condensation occurs.

itgitgig

itoitoio

pBpBppE

pBpBppE

,

, (3.39)

Example:How to calculate the two-phase formation volume factors.

p (psi) Bg (RB/Mscf) Bo (RB/STB) Rs (scf/STB) Rv (STB/MMscf)

4320 0.83 1.3647 710 23.9

4225 0.845 1.3578 693.5 23

4130 0.86 1.3509 677 22

4030 0.875 1.344 660 21

3930 0.89 1.337 643 20

3830 0.91 1.3301 626 19.1

3730 0.93 1.3232 609 18.2

3630 0.95 1.3164 592 17.3

3530 0.97 1.3095 576 16.4

3800.1

10235.6931

5.69371010845.0102371013578.1

)4225(

3647.1)4320()4320(

1

1

6

36

pB

pBpB

RR

RRBRRBB

to

oto

sv

ssigvsio

to


56/125


3.6 Immaterial Concept

As illustrated in Figure 3.7, we force the two-phase fluid PVT properties to be equal to the

single-phase PVT properties for the single-phase (oil or gas) reservoirs. Assuming an oilreservoir, we define

o

SCo

RCo

SCo

RCtotal

to BV

V

V

VB (3.40)

vgogv

oso

SCg

SCo

SCo

RCo

SCg

RCtotal

tg RBBorBR

BRB

V

V

V

V

V

VB (3.41)

v

sR

R1

(3.42)

Assuming a gas reservoir, we define

g

SCg

RCg

SCg

RCtotal

tg BV

V

V

VB (3.43)

ovgSCoSCg

SCg

RCg

SCo

RCtotal

to BRBV

V

V

V

V

V

B

(3.44)

Figure 3.7 Illustration of immaterial concept

v

sgso

g

SCg

RCtotal

tg

o

SCo

RCtotal

to

RRBRB

BV

VB

BV

VB

1

Above the Bubblepoint:

GAS.asorOILaseithertreatedbecouldRCtotal

VReservoir Oil

Surface Oil

Surface

Gas

v

sgso

g

SCg

RCtotal

tg

o

SCo

RCtotal

to

RRBRB

BV

VB

BV

VB

1

Above the Bubblepoint:

GAS.asorOILaseithertreatedbecouldRCtotal

VReservoir Oil

Surface Oil

Surface

Gas


57/125



58/125


CHAPTER 4 GENERALIZED MATERIAL BALANCE EQUATION

LIST OF SYMBOLS

4.1 Generalized Material Balance Equation

Injected

MassOverall

Initially

MassOverall

Produced

MassOverall

Finally

MassOverall

Ground Surface

Total Mass

PressureInitial

productionbefore0

0pp

t

Produced

Mass

Residual

Mass

PressureReservoir

productionDuring

p

t

PressureStandard7.14

productionDuring

psip

t

a

+=

Figure 4.1 Illustration of Mass Balance for a Petroleum Liquid

ratiooilgasSolution

factorvolumeformationOil

ratioproductionoil

-productiongasveAccumulati

factorcompactionformationandWater

factorvolumeformationgasInitial

factorvolumeformationoilInitial

influxNet water

factorexpansionGas

factorexpansionOil

PlaceInGasOriginal

placeingasphase-freeOriginal

PlaceInOilOriginal

placeinphase-freeOriginal

expansionfluidsNet

withdrawndundergrounfluidsNet

so

o

p

fw

gi

oi

g

o

fgi

foi

R

B

N

E

B

B

W

E

E

G

G

N

N

E

F

factorvolumeformationWater

productionWater

InfluxWater

intervalstimeobetween twdifferencepressure

saturationwaterconnateInitialilitycompressibwater

ilitycompressibrockFormation

factorvolumeformationgasphasetwoInital

factorvolumeformationgasphasetwo

factorvolumeformationoilphasetwoInital

factorvolumeformationoilpahseTwo

ratiogasoilvolatileInitial

ratiogasoilVolatile

factorvolumeformationGas

ratiooilgassolutioninitial

w

p

e

wi

w

f

tgi

tg

toi

to

vi

v

g

so i

B

W

W

p

Sc

c

B

B

B

B

R

R

B

R


59/125


Apply this principle to oil, gas and water respectively:

NNRB

SV

B

SVpv

g

gb

o

ob

(4.1)

Ips

o

ob

g

gbGGGR

B

SV

B

SV

(4.2)

Ip

w

wb WWWB

SV

(4.3)

Constraint:

1 wgo SSS (4.4)

Solving equations (4.1) through (4.4) gives

gogo

sv

b

gogo

sv

pI

go

s

p

og

v

pI

BBBB

RRV

BBBB

RRWWW

BB

RNN

BB

RGGG

1

111

(4.5)

We note that the initial pore volume is

oifoigifgiwifoifgifwiib BNBGWBVVVV (4.6)

Where Vfwi, Vfgi and Vfoi are free water- gas- and oil-phase volumes. The pore volume at

any time is defined as

pi

p

oifoigifgiwibV

VBNBGWBV (4.7)

The original surface gas in place is the sum of the resident gas in the free gas and oil phases

sifoifgifoifgi RNGGGG (4.8)

Analogously, the original stock-tank oil in place is

vifgifoifgifoi RGNNNN (4.9)


60/125


wIp

sv

sgo

p

sv

vog

Ip

pi

p

wiw

sv

vog

si

pi

p

oi

sv

sgo

foi

sv

sgo

vi

pi

p

gi

sv

vog

fgi

BWWRR

RBBN

RR

RBBGG

V

VBBW

RR

RBBR

V

VB

RR

RBBN

RR

RBBR

V

VB

RR

RBBG

11

11

11

(4.10)

The next step is to eliminate the pore volume and replace certain terms with their

expansivities. The rock expansivity is defined as

f

pi

p

pi

ppif E

V

V

V

VVE

1 (4.11)

The gas, oil and water expansivities are defined by

tgitgg BBE (4.12)

toitoo BBE (4.13)

wiww BBE (4.14)

Solving equations (10) to (14) gives

WIpsv

sgo

p

sv

vog

Ip

efpiwofoigfgi

BWWRR

RBBN

RR

RBBGG

WEVWEENEG

11

(4.15)


61/125


Figure 4.2: Volumetric Interpretation of General Material Balance Equation

Equation (4.15) applies to the full range of reservoir fluids. Substituting the following

relations

wwipi BSVW / (4.16)

oifoigifgiwipi BNBGWBV (4.17)

Into equation (4.15) gives

WIpsv

sgo

p

sv

vog

Ipeowffoigwffgi BWWRR

RBBN

RR

RBBGGWENEG

11 (4.18)

Where

f

wi

wwi

wi

oi

oowf EB

ES

S

BEE

1 (4.19)

f

wi

wwi

wi

gi

ggwf EB

ES

S

BEE

1 (4.20)

The water and rock expansivities can be replaced by compressibilities. Water and rock

compressibilities are defined as

pB

BpVV

VVc w

w

scww

scww

w

1//1 (4.21)

INITIAL PRESSURE LOWER PRESSURE

InitialVolume

Produced

Volume

Expanded

Volume


InitialVolume

Produced

Volume

Expanded

Volume


InitialVolume

Produced

Volume

Expanded

Volume


62/125


p

E

pV

VV

p

V

Vc

f

f

ffif

f

f

11 (4.22)

pcE ff (4.23)

pcBE wwiw (4.24)

Substituting equations (4.23) and (4.24) into (4.19) and (4.20) gives

fwoioToioowf EBEpcBEE (4.25)

fwgigTgiggwf EBEpcBEE (4.26)

Where

pS

Scc

pcEwi

wiwf

Tfw

1 (4.27)

The generalized MBE can be further simplified as

EF (4.28)

Where

eowffoigwffgi WENEGE (4.29)

WIp

sv

sgo

p

sv

vog

ps BWWRR

RBBN

RR

RBBGF

11

(4.30)

And

Ipps GGG (4.31)

4.2 MBE APPLICATIONS

The MBE can be used to

Estimate the original oil and gas in place (OOIP and OGIP); Estimate the gas-cap size; Estimate water influx; Estimate water influx model parameters; Estimate geophysical parameters;

Confirm producing mechanisms.

The application forms may be different for different reservoir types. In the following, we


63/125


present some general principles.

Definitive Tool or Diagnostic Tool: The simplest form of MBE can be written as

EF (4.32)

where, F and E are the hydrocarbon production and the reservoir expansion in volume. If thematerial is fully balanced, F vs. E is a straight line; if the material is NOT fully balanced, F vs.

E is a curve, as illustrated in Figure 4.2.

Figure 4.3: Illustration of Material Balance and Imbalance.

We write the material balance equation as

eowffoigwffgiWIp

sv

sgo

p

sv

vog

ps WENEGBWWRR

RBBN

RR

RBBGEF

11(4.33)

When the material is balanced,

0EF (4.33)

Equation (4.34) can be used to solve only for one unknown. If eW or fwE are not known

and are dropped off from the equation, equation (4.33) becomes

0EF (4.34)

If the water production data are not available and are dropped off from the MBE, equation

(4.34) becomes

0EF (4.35)

F

E

BALANCE

IMBALANCE

F

E

BALANCE

IMBALANCE


64/125


Figure 4.4: MBE can be used either as a definitive tool or as a diagnostic tool.

Recovery Ratios:Oil and gas recoveries as a fraction of the original amount in place can be

solved from the material balance equation. Dividing equation (18) byfoiN gives

sv

spsgpsvo

foi

gwffgi

owf

foi

ewpI

foi

p

RR

RRBRRB

N

EGE

N

WBWW

N

N

1

1 (4.36)

Where, psR , is defined as the cumulative produced sales gas-oil ratio,

p

Ip

p

ps

ps

N

GG

N

GR

(4.37)

When 0IG , equation (4.37) becomes

p

p

p

p

pN

G

N

GR (4.38)

In practice, we calculate the fraction of the total OOIP recovered. The oil recovery fraction

and gas recovery fraction are defined as

N

NF

p

o (4.49)

F

E

0

11

eowffoigwffgi

WIp

sv

sgo

p

sv

vog

ps

WENEG

BWWRR

RBB

NRR

RBB

GEF

0

11

eowffoigwffgi

sv

sgo

p

sv

vog

ps

WENEG

RR

RBBN

RR

RBBGEF

0

11

ofoigfgi

WIp

sv

sgo

p

sv

vog

ps

ENEG

BWWRR

RBBN

RR

RBBGEF

F

E

F

E

0

11

eowffoigwffgi

WIp

sv

sgo

p

sv

vog

ps

WENEG

BWWRR

RBB

NRR

RBB

GEF

0

11

eowffoigwffgi

sv

sgo

p

sv

vog

ps

WENEG

RR

RBBN

RR

RBBGEF

0

11

ofoigfgi

WIp

sv

sgo

p

sv

vog

ps

ENEG

BWWRR

RBBN

RR

RBBGEF


65/125


G

GF

p

g (4.50)

respectively.

Drive Indices: We write the general material balance equation as

sv

sgo

p

sv

vog

Ip

WIpefwoifoigifgiofoigfgi

RR

RBBN

RR

RBBGG

BWWWEBNBGENEG

11

(4.51)

We define

WIpefwoifoigifgiofoigfgit BWWWEBNBGENEGE

Dividing each term in Equation (4.51) by tE gives

1 wdcdodgd IIII (4.52)

IndexDriveWater

IndexDriveCompaction

IndexDriveExpansionOil

IndexDriveExpansionGas

t

wIpe

wd

t

fwoifoigifgi

cd

t

ofoi

od

t

gfgi

gd

E

BWWWI

E

EBNBGI

E

ENI

E

EGI

(4.53)

These drive indices provide a convenient means to rank the different producing mechanisms.

They change with time. For example, the gas drive may dominate early in the life of a

reservoir while the water drive dominates later after water influx commences.

Calculation of Saturations:The oil saturation at any time can be obtained

sgo

pp

vog

vpp

ow

o

RBBN

N

G

GBBB

N

G

RPG

G

G

N

NBS

S

11

111

(4.54)

At the initial condition: 0 pp GN , equation (4.54) becomes


66/125


sigioivioigi

voiwi

oi

RBBBBBN

G

RP

GBS

S

11

(4.55)


67/125


CHAPTER 5 DRY-GAS RESERVOIRS

A reservoir that produces only gas and no appreciable hydrocarbon liquids is called a dry-gasreservoir. Dry-gas reservoir is the simplest type to evaluate. Gas reservoirs often have high

recovery factors irrespective of the drive mechanism. Recovery factors of more than 80% are

not uncommon, and can even be found in volumetric reservoirs. Natural gas is very

compressible and has a very low viscosity, and both factors contribute to the high recovery

factor.

The recovery factor is lower in waterdrive gas reservoirs primarily because waterdrive leaves

behind a residual gas saturation that is unrecoverable. This residual gas saturation is usually at

high pressure. The residual gas saturation at high pressure has a large volume of gas

compressed into a small space. In heterogeneous waterdrive reservoirs, water tends to flow

along the high permeability streaks, bypassing gas that gets trapped in low permeability areas.This becomes a severe problem if the reservoir is highly fractured and has a strong waterdrive.

The waterdrive flushes the gas out of the fractures and bypasses the great majority of the gas

contained in the matrix.

Figure 5.1: P-T Diagram of A Typical Reservoir Fluid

5.1 Volumetrics and Recovery Factors

0 50 100 150 200 250 300 350


4000

3500

3000

2500

2000

1500

1000

ReservoirPressure,

psi

A

A2

B2

B1

B

B3

A1

C1

C

D

0%

20%

10%5%

80%40%

Bubb

lePo

int

Critical

PointDewPoint

Bubble Point or

Dissolved Gas

Reservoir

Dew Point or Retrograde

GasCondensate

Reservoir

Single-phase Gas

Reservoir

PathofReservoirFluid

Cricondentherm

Path

ofPr

od

ucti

on

Wet-andDry-GasReservoirs

0 50 100 150 200 250 300 350


4000

3500

3000

2500

2000

1500

1000

ReservoirPressure,

psi

A

A2

B2

B1

B

B3

A1

C1

C

D

0%

20%

10%5%

80%40%

Bubb

lePo

int

Critical

PointDewPoint

Bubble Point or

Dissolved Gas

Reservoir


GasCondensate

Reservoir

Single-phase Gas

Reservoir


Cricondentherm

Path

ofPr

od

ucti

on

0 50 100 150 200 250 300 350


4000

3500

3000

2500

2000

1500

1000

ReservoirPressure,

psi

A

A2

B2

B1

B

B3

A1

C1

C

D

0%

20%

10%5%

80%40%

Bubb

lePo

int

Critical

PointDewPoint

Bubble Point or

Dissolved Gas

Reservoir


GasCondensate

Reservoir

Single-phase Gas

Reservoir


Cricondentherm

Path

ofPr

od

ucti

on

Wet-andDry-GasReservoirs


68/125


The OGIP (original gas in place) can be determined by

gi

wi

B

SAhG

1 (5.1)

Where

SaturationWaterInitial

FactorVolumeFormationGasInitial

HeightReservoir

AreaReservoir

PlaceInGasOriginal

wi

gi

S

B

h

A

G

At the abandonment pressure, the remaining gas is calculated as

ga

gr

aB

SAhG

(5.2)

PressuretAbandonmenat theSaturationGas

PressuretAbandonmenat theFactorVolumeFormationGas

PressuretAbandonmenat thePlaceInGas

gr

ga

a

S

B

G

Recovery factor is defined as

wiga

grgia

RSB

SB

G

GGF

11 (5.3)

If the reservoir is homogeneous and volumetric (no waterdrive), 0wiS and 1grS ,

equation (5.3) can be simplified as

ga

gia

RB

B

G

GGF

1 (5.4)

5.2 MBE Analysis

Two conditions:

0

0

vvi

foi

RR

N (5.5)

General Material Balance Equation

eowffoigwffgi WENEGE (5.6)


69/125


WIp

sv

sgo

p

sv

vog

ps BWWRR

RBBN

RR

RBBGF

11

(5.7)

pS

SccpcE

NW

wi

wiwf

Tfw

foi

e

1

00

fwoioToioowf EBEpcBEE (5.8)

fwgigTgiggwf EBEpcBEE

Substituting equation (5.8) into (5.6) and (5.7) gives

fwgifgigfgi EBGEGE (5.9)

gpBGF (5.10)

If 0fwE , substituting gigg BBE into (5.9) gives

gigfgigp BBGBG (5.11)

Equation (5.11) is the equation most commonly used to analyze the normal volumetric gas

reservoirs.

5.3 P-Z Plot

We note

sc

sc

gpT

zTpB (5.12)

Substituting (5.12) into (5.11) gives

sci

sci

sc

sc

fgi

sc

scp

TpTpz

pTzTpG

pTzTpG (5.13)

fgi

p

i

i

i

i

G

G

z

p

z

p

z

p (5.14)

Use MBE as a Diagnostic Tool

efwgigfgi WEBEGF (5.15)

fwgig

e

fgi

fwgig EBEWG

EBEF

(5.16)


70/125


Overpressured Gas Reservoir: 0fwE

Example 1:A volumetric dry-gas reservoir with an initial pressure of 3000 psi (zi=0.912) and

formation temperature 190F has produced 384MMscf of gas, and the pressure has dropped to

2,876 psi (z=0.907). Determine the original gas in place (OGIP) by material balance.

Solution:The material balance equation for this problem is

p

fgii

i

i

i GGz

p

z

p

z

p

Wherep

G is the accumulative gas production. Rearranging Equation above gives

Bscf

scfzppz

GG

ii

p

fgi

7.10

10652.10907.03000/912.028761

10384

/19

6


71/125


Example 2

A volumetric dry-gas reservoir has the measured production and pressures given below

P(psi) Gp(MMscf) z

3000 0 0.912

2876 384 0.9072824 550 0.905

2755 788 0.903

2688 1002 0.902

2570 1445 0.901

2435 1899 0.900

2226 2670 0.901

2122 3113 0.903

1866 3982 0.905

Determine the OGIP from appropriate material balance plots.

Solution:

The material balance equation can be written as

gigfgigp BBGBG (E2.1)

Alternatively,

p

fgii

i

i

i GGz

p

z

p

z

p (E2.2)

We can use both equations to solve for Gfgi.

Method 1: p/z plot

y = -0.3063x + 3290.1

R2= 0.9998

0

500

1000

1500

2000

2500

3000

3500

0 1000 2000 3000 4000

Accumulative Gas Production, Gp

p/z,psi

y = -0.3063x + 3290.1

R2= 0.9998

0

500

1000

1500

2000

2500

3000

3500

0 1000 2000 3000 4000

Accumulative Gas Production, Gp

p/z,psi


72/125


From the plot, Gfgi=10.74MMscf.

Method 2: GpBg versus Eg plot.

005586.0460603000

7.14460190912.0

sc

sc

gpT

zTpB

Example 3:

If we do not know if there is a waterdrive, we can use the material balance equation as a

diagnostic tool. The material balance equation for a waterdrive gas reservoir can be written as

g

fgi

g

gp

g

gfgigp

E

WGE

BG

E

F

WEGBGF

The graph verifies the volumetric nature of the gas reservoir.

Bg-Bgi=Eg

GpBg

y = 10740x

R2= 0.9998

0

5

10

15

20

25

30

35

40

0 0.001 0.002 0.003 0.004

9000

9500

10000

10500

11000

11500

12000

0 1000 2000 3000 4000 5000

gE

F

pG


73/125


CHAPTER 6 GAS/CONDENSATE RESERVOIRS

When a gas reservoir produces significant quantities of liquids along with the gas, it is called

a wet-gas or retrograde condensate reservoir. A wet-gas reservoir can be produced

volumetrically, like a dry-gas reservoir, but the liquids are an added valuable product. In

analyzing and managing such reservoirs, h

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