dispersed flow -...

Dispersed Flow

Reservoir Fluid Displacement Core

Learning Objectives

By the end of this lesson, you will be able to:

Recognize immiscible fluid displacement and sources of dataused for calculations

Describe the assumptions and limitations of the Buckley-Leverett theory

Describe the derivation of the dispersed fractional flow equation

Recognize the impact of mobility, gravity, and capillary pressureon the dispersed fractional flow equation

Identify how the Welge solution can be applied before, during,and after breakthrough

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═════════════════════════════════════════════════════════════════════════Reservoir Fluid Displacement Core

© PetroSkills, LLC., 2017. All rights reserved._____________________________________________________________________________________________

1

In This Section

Immiscible Fluid Displacement

Where does immiscible fluid displacement occur?

Buckley-Leverett Theory

Fractional Flow

Capillary Pressure

Fluid Mobility

Gravity Term

Frontal Advance and Welge Solution

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Reservoir Fluid Displacement Core═════════════════════════════════════════════════════════════════════════

2_____________________________________________________________________________________________

© PetroSkills, LLC., 2017. All rights reserved.

What is the Buckley-Leverett Theory?

Calculates pore level displacement efficiency

Some moveable oil is bypassed, due to: Viscosity contrast Relative permeability Capillary pressure

Assumes “piston-like” displacement and linear flow

Immiscible displacement characterized as: Uniformly dispersed Segregated

This module uses an oil-water system but analogous equations apply for a gas-oil system and a water displacing gas system, these will be covered in the fundamental module of this series.


Buckley-Leverett only applies to linear systems• Most injection patterns and real formation displacement

systems are not linear

Problem B, however, will cover a linear displacementsystem

• Sweep efficiency

• Horizontal and non-horizontal reservoirs

Linear Systems

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3


Front Boundary

Rapid saturation change

Ahead of the Front

Displaced oil bank

Assume that a stabilized flood front develops, including:

Behind the Front

Two-phase flow of oil and water

Buckley-Leverett Theory Assumptions

Front

Horizontal Reservoir

Variable Saturations Initial Conditions

Assumptions

1. Incompressiblein-situ and injectedfluids

2. Constant averagesystem pressure

3. Steady state flow

• Same reservoirflow out as in

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4_____________________________________________________________________________________________


Fractional Flow at Reservoir Conditions

Where:

qo = oil flow rate per unit cross-sectional area, at any point along the core

qw = displacing phase (here water) flow rate per unit cross-sectional area, at the same point in the core as qo

qt = total flow rate per unit cross-sectional area, at any point in the system

Assuming water as the displacing phase:

qin qout= = qo + qw = qt

Fractional Flow at Reservoir Converted to Surface Conditions

fw = qw / qt = qw / (qw + qo)

Where:

fw = fraction of the total flow rate at a given point that is the displacing phase at reservoir conditions of temperature and pressure

Assuming water as the displacing phase:COPYRGHT



5

Fractional Flow at Reservoir Converted to Surface Conditions

Where:

Wc = the amount of water produced at surface, as a fraction of the total production of oil and water

Bo = reservoir volume of oil (at reservoir temperature and pressure)/surface volume of oil at standard conditions

Bw = Reservoir volume of water/surface volume of water

))1)1((*/( owc BfBBW wo

Fractional Flow – Reservoir Conditions

Where:u = the linear direction of flow (measured from the inlet end),∂po / ∂u = the pressure gradient in the oil phase∂pw / ∂u = the pressure gradient in the displacing phaseα = the angle of the fluid flow with respect to the horizontal (updip flow is assigned to be "+"; downdip is "-")qo = displaced fluid (oil) flow rate per unit of cross-sectional area normal to uqw = displacing fluid flow rate per unit of cross-sectional area normal to uρ = fluid densityµ = fluid viscosityo = subscript indicating displaced or oil phase, and

w = subscript indicating displacing phase

sino

o

o

oo gu

p

k

q

sinw

w

w

ww gu

p

k

q

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6_____________________________________________________________________________________________



o

w

w

o

c

to

o

w

kkup

qk

f

1

sin)(434.0001127.0

1

Thanks to:

Capillary pressure

(Pressure difference across the interface of two fluids in a confined space)

and Algebra, you get:

Units: md, cp, Bbl/D/ft2, feet, psi; (=w‐o)

Pc = Po- Pw

Absolute permeability, grain size, rock type

Saturation, wettability, interfacial tension, pore structure


1

2

3

4

Flow rate and angle of dip

Fluid viscosity and density

Capillary pressure

Effective fluid permeabilities

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7

Define Fluid Mobility

w = kw / w = k krw / w

o = ko / o = k kro / o

Where:

λW = mobility of the displacing phase (water), darcy/cp

kW = effective permeability of the displacing phase, darcies

µW = viscosity of the displacing phase, cp

λo= mobility of the displaced phase (oil), darcy/cp

ko = effective permeability of the displaced phase, darcies,

µo = viscosity of the displaced phase, cp

Mobility Ratio

Where:

krw (Sw) = the relative permeability of the displacing phase evaluated at the average displacing-phase saturation behind the front

kro (Swi) = relative permeability to oil evaluated at the initial water saturation in the oil bank

wwiro

owrw

wwio

oww

aheado

behindw

Sk

Sk

Sk

SkM

)(

)(

)(

)(

)(

)(

Of greater importance is the mobility ratio – how the displacing or behind fluid moves with respect to the displaced or ahead fluid:

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Reservoir Fluid Displacement Core ═════════════════════════════════════════════════════════════════════════

8_____________________________________________________________________________________________


Mobility Ratio

Mobility Ratio

Compares fluid mobilities at two points in the system

Favorable displacement

M < 1

Oil has greater mobility

Unfavorable displacement

M > 1Displacing fluid moves

more easily

Fractional Flow Influences

Sw

fw

0

0.5

1

0.2 0.5 0.8

-30º

30º

1 - SorSwirr

Dip angle,

Displacementupwards

Displacementdownwards

In a waterflood:

Displacement efficiency is increased when angle of dip is increased

• Water injected low on structure

Water injection rate is decreased, slower displacement is better

Density difference is greatest

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9

Fractional Flow Influence on Wettability

o

w

w

o

c

to

o

w

kkup

qk

f

1

sin)(434.0001127.0

1

Kw will be lower for a water-wet system. Thus, the mobility term (denominator) will be lower.

Water-wet reservoirs yield a higher displacement efficiency than comparable oil-wet reservoirs.

Fractional Flow Viscosity Influences on Mobility Term

Displacement efficiency is improved by• Increasing water viscosity with polymers

• Decreasing oil viscosity using hot water orsteam

Viscosity Mobility RatioCOPYRGHT


10_____________________________________________________________________________________________


Gravity Term

sin)(434.0

001127.0

u

p

q

k c

ot

o

Maximum displacement of oil will resultwhen fw is kept to a minimum.

Thus, you would like the Gravity term,

to be as large as possible.

Fractional Flow Influence on Gravity Term

o

w

w

o

c

to

o

w

kkup

qk

f

1

sin)(434.0001127.0

1

o

w

rw

ro

to

roa

w

k

k1

sinq

Akk000488.01

f

Neglect ∂pc /∂u since it is only significant at the front Front is region of rapid

saturation change

Neglecting derivativeresults in shock front

Field Units: k (md)μ (cp)qt (RB/D)A(ft2)α (degrees) = (water - oil)

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11

Fractional Flow – Capillary Pressure Pc and Gravity Ignored

(1-Sorw)Swirr

1.0

0

0

fw

Sw

Neglecting capillary pressure and the effects of gravity:

o

w

w

ow

kk

f

1

1

This is the most common form of the fractional equation

The resulting curve:

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12_____________________________________________________________________________________________


Why Calculate Fractional Flow Curves?

Used to compute the saturation distribution in a linear waterflood system as a function of time

Saturation distribution is a function of the slope of thefractional flow curve

Used to get the oil recovery behind the front at breakthrough

Can be used to predict oil recovery and required water injection vs time

Shock Front Travels through Reservoir

Time 1 32 4

Each saturation plane travels at its own constant speed andis proportional to gradient of the fw curve;

The displacing fluid saturation moves at the same velocity atthe leading edge of the front.

0

1-Sor

Swc

0% 100%Distance through reservoir

Capillary trapped (residual) oil

Capillary trapped (connate) water

Time 1 2 3 4COPYRGHT



13

Frontal Advance and the Welge Solution

1.0

0

1.0

0

Capillary Trapped (Residual) Oil

Capillary Trapped (Irreducible) Water

Injection Production

Wat

er S

atu

rati

on

Distance

1- Sor

Swirr

1- Sor

Swirr

WaterAccumulation

Flow ofWater In

Flow ofWater Out

(∆Αφx) dSw/dt = Qt∆fw

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Efficiency of Displacement

1.0

0

1.0

0

Capillary Trapped (Residual) Oil

Capillary Trapped (Irreducible) Water

Injection Production

Wat

er S

atu

rati

on

Distance

1- Sor

Swirr

1- Sor

Swirr

Piston Versus Non-piston Displacement


14_____________________________________________________________________________________________


Mobility Ratio

Where:

x (ft) = distance traveled by a fixed saturation. For a linear system, the front is traveling perpendicular to the cross-sectional area A (ft2), refer to the lavender Time bars on our displacement diagram.

Wi = cumulative water injected, bbls

t = time, days

dfw / dSw = derivative of the fractional flow curve at the saturation of interest, and must be equal for all saturations in the stabilized zone

Via substitution and integration:

x = (5.615 Wi / A) * dfw / dSw

Welge Graphical Solution

Sw

fw

0

0.2

0.4

0.6

0.8

1

0.2 0.3 0.4 0.5 0.6 0.7 0.8

fw,bt

SwfSwirr

wS

wcwSw

w

SSdS

df

fw

1

Simple graphical solution atwater breakthrough

Shock front saturation

Average saturation behindfront

Water saturation at flood frontis determined from point oftangency on fractional flowcurve (greatest slope) change(Part IV of Problem)

Average water saturationbehind the flood front isdetermined from where thetangent line intersects fw = 1.0

Problem A Parts II & III – Calculate the Sw at breakthrough and the average Sw

behind the front for each Reservoir in Part I

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15

Recovery Efficiency at Water Breakthrough

wi

wiwBTD

S

SSE

1 oi

oBToiD

S

SSE

or

Where:

ED = displacement efficiency or fractional recovery of the total original oil in place, at pore level (fraction)

Sw = mean displacing-phase saturation behind the front at breakthrough (found by extending tangent of fw curve to fw = 1.0), fraction

Swi = initial (irreducible) displacing-phase saturation, fraction

Problem A Part IV – Using Part III, the average Sw, calculate the ED at breakthrough for each of the three reservoirs

Surface Water Cut

Where:

Wc = Surface fraction of water of total oil and water (qw/(qo + qw)

Bw = Reservoir volume of water/surface volume of water

Bo = Reservoir volume of oil (at reservoir temperature and pressure)/surface volume of oil at standard conditions

Convert to a surface water cut by converting reservoir volumes to surface volumes using:

Problem A Part V – Using Part III and reading the Fw at breakthrough, calculate the corresponding Wc for each oil that has a different Bo

)1)/1((*(

ww

oc

FB

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Welge Solution Results

Swf provides the fractional flow at the front, fwf, which is used to calculate the surface water cut at breakthrough

Sw provides the average oil saturation behind the front which is used to calculate the pore displacement efficiency, ED

Breakthrough time (days) can be estimated from ED, OOIP and the producing oil rate o

D

q

EOOIPt

Problem A Parts VI and VII – Calculate mobility ratio and, using Part IV as well as the data on OOIP and qo, calculate the t (days) for each reservoir to breakthrough.

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17

Welge Solution after Breakthrough

1

0

1 - Sor

Swc

Capillary trapped (residual) oil

Capillary trapped (connate) water0% 100%Distance through reservoir

Swe

Swf

0.8

0.85

0.9

0.95

1

0.55 0.6 0.65 0.7Sw

fw

SwBT

fwe

Swe

Sw

fw

0

0.2

0.4

0.6

0.8

1

0.2 0.3 0.4 0.5 0.6 0.7 0.8

BTwS Sw

Welge Procedure after Breakthrough

1

2

3

4

Construct tangents to fractional flow curve at increasing values of water saturation.

Average water saturation behind the front is determined from intersection of tangents with fw = 1.0

Calculate recovery efficiency

Calculate cumulative water injected

wi

wit

S

SSER

1

..

w

w

i

S

fQ

1 Where:

Qi = cumulative water injected, pore volumes

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Welge Solution Formulas

After breakthrough, Sw = Swe

Water saturation at the production well

Where:

Wid = the Cumulative water injected in the displacement process in bblor m3

PV = pore volumes in reservoir bblor reservoir m3

qi = the rate of injection of water in bblsper day or m3 per day

t = time in days

Npd = Cumulative oil produced in thedisplacement process in bbl or m3

Welge Solution

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 5 10 15

Water injection displacing oil, Wid

Cu

mu

lati

ve o

il re

cove

red

fr

om

dis

pla

cem

ent,

Np

d

After breakthrough, an extended period of high water cut production takes place to recover remaining mobile oil. If a tangent line cannot be constructed, then a stabilized front or oil bank will not form.COPYR

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19

Map of Reservoir

The reservoir being studied is a thin, shallow, under-saturated oil accumulation, it is under-going line drive dispersed flow water flood, as shown below.

3000

3100

3200

3300

3400

3500

OWC 3530 3600

0 1000' 2000'

DEPTH: ft ss

Cross Section (Part 1)

The LD-A reservoir consists of four patterns with dimensions as follows:

1870'

1000'

15.5°

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Understanding the Fractional Flow Equation (Part 2)

Set up a table in Problem B.xls as shown below. Fill it in using Table 1, Figure 2, and your knowledge of the Welge solution to the fractional flow equation.

Plot of Fractional Flow (Part 3)

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0

Fra

ctio

nal

Flo

w o

f W

ater

Water Saturation (fraction)

At Breakthrough

Calculate the frontal saturation and the average saturation of water behind the front.

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21

Oil Recovery at Breakthrough (Part 4a)

Use either of the equations below to calculate Part 4. You should also prove to yourself that they are the same.

wi

wiwBTD

S

SSE

1 oi

oBToiD

S

SSE

Using your units of choice, calculate oil recovery at breakthrough and time to breakthrough.

Water Injection and Front Maintenance (Part 4b)

Review the reservoir rate of water to maintain reservoir pressure and surface rate from the total field of 2000 bopd. How much is it?

2400 bwpd (381 m3wpd)

The fractional flow equation has to calculated at reservoir conditions and the difference between oil reservoir volume and water reservoir volume is 1.2/1 (or 20%).

Refer to the problem statement Bo and Bw

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22_____________________________________________________________________________________________


Water Drive Recovery Efficiency, in Real Reservoir

Where:

RE = water drive recovery efficiency, percent

ED = pore level recovery from displacement

EV = recovery vertically

EA = recovery areally

Boi = initial oil formation volume factor, rb/STB

Bo = reservoir volume of oil (at reservoir temperature and pressure)/surface volume of oil at standard conditions

Water drive oil recovery efficiency (above pb)

Typical range ofrecovery efficiency:under 40% to over60%

Statistical estimate forwater drive yields anaverage recoveryefficiency of 51%

Water Injection and Front Maintenance (Part 4c)

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0

Fra

ctio

nal

Flo

w o

f W

ater

Water Saturation (fraction)

At Breakthrough

Calculate the fractional flow of water at the time of breakthrough from the curve on the left.

What is it?

After breakthrough, do you think the required water rate will go up or down?

77%

Down, because of the 20% volume difference between oil and water.

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23

Oil Recovery at Breakthrough (Part 4d, Part 5)

Finish the problem using the equation below for water injection rate after breakthrough to maintain reservoir pressure.

Water Component = (0.77*2000*1.0) + Oil Component(0.23*2000*1.20) =

To understand the Welge solution after breakthrough, review the calculations in Part 5, as well as the solution.

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24_____________________________________________________________________________________________


Learning Objectives

You are now able to:

Recognize immiscible fluid displacement and sources of dataused for calculations

Describe the assumptions and limitations of the Buckley-Leverett theory

Describe the derivation of the dispersed fractional flow equation

Recognize the impact of mobility, gravity, and capillary pressureon the dispersed fractional flow equation

Identify how the Welge solution can be applied before, during,and after breakthrough

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25

Aquifers


Learning Objectives


Determine the rock, fluid, operational, and reservoir geometricfactors affecting water influx (We) into your reservoir

Calculate We using different models

Understand which descriptive parameters you should change soyou can match the calculated We to the observed reservoirpressureCOPYR

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26_____________________________________________________________________________________________


In This Section

Aquifer Models

Van Everdingen and Hurst

Pot Aquifers

Fetkovich Finite Aquifers

Carter-Tracy Aquifers

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27

Artesian Aquifers

Continuity from point of recharge

The fact that they are recharged Often, aquifers under the

sea are recharged better

Connection they have at point of contact with hydrocarbon reservoir

Head it can exert on hydrocarbon reservoir

Critical aspects of an aquifer:

Water Influx from Aquifers

Account for additional drive energy

Match historical reservoir pressure (OOIP)

Predict future pressure-production performance (rate vs. time)

Estimate oil-water contact movement

Why is it important to account for aquifer water encroachment?

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28_____________________________________________________________________________________________


Water Influx from Aquifers

Aquifer influx depends more on aquifer propertiesthan hydrocarbon reservoir properties

• Location of wells relative to aquifer and watermovement

• Aquifer size

• Aquifer porosity and permeability

Why is it difficult to account for aquifer water encroachment?

Initial until transient reaches boundary Declines with continuing water influx Pressure may be maintained if aquifer outcrops

Water Influx Mechanism

1 Pressure drop in the reservoir

2 Water flows from aquifer to reservoir

3 Pressure gradient grows in the reservoir andaquifer pressure drops from depletion

4 Rock and water expansion in the aquifer balancesaquifer voidage

5 Aquifer pressure at outer boundary

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29

Aquifer Expansion

How long water encroachment takes is a function of:

Maximum water encroachment, We,aq

aqiaqaqwfaq,e ppPVccW

Aquifer size (PVaq = aquifer pore volume)

Aquifer permeability

Connection of reservoir and aquifer

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30_____________________________________________________________________________________________


Water Influx Models

Pot Aquifer Small well

connected aquifer,limited use

Fetkovich Semi-Steady State

Good for lateperiods andlimited aquifers,but doesn’tdefine thetransient period

Van Everdingen and Hurst Most complete

and useful

Schilthuis Steady-State Almost never

applicable

Carter-Tracy Analytical model

good forsimulation


Transient and bounded aquifer flow

Solution to hydraulic diffusivity equation

Most comprehensive approach to water influx

Requires superposition

Defines aquifer characteristics

Implemented in material balance programs,including MBAL

Characteristics

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31

Behavior of a Large Aquifer

RESERVOIR

AQUIFER

RESERVOIR

1 2 3 4

1 2 3 4

Transient moves out until boundaries reached,then PSS (pseudo-steady state) behavior

Reservoir/Aquifer Boundaries

ra

ro

Aquifer

ReservoirCOPYR

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32_____________________________________________________________________________________________


Aquifer Pressure Distribution

t=0

t=1

t=10t=100

t=1000

pi

p

Radius

Pressure

raro

t=infinite

van Everdingen Hurst Solution (Field Units)

doee QpfrchW )119.1( 2

Where:We = cumulative water influx, reservoir bbls

= porosity, fraction

h = effective aquifer thickness, ft

ce = total compressibility of the aquifer, cw + cf, psi -1

ro = radius of the hydrocarbon reservoir, ft

f = fraction of the reservoir boundary exposed to the aquifer (0 < f < 1)

p = pressure drop across the original reservoir/aquifer boundary, psi

Qd = dimensionless cumulative influx term

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33

Dimensionless Influx, Qd

o

ad r

rr 2

* 0.00633

oewd rc

tkt

Where:ra = radius of the aquifer, ft or m

ro = radius of the original reservoir/aquifer interface, ft or m

Where:

k = aquifer permeability, mdt = time, days

Aquifer influx depends on the aquifer size, aquifer rock properties, water compressibility, the pressure drop the aquifer sees at the reservoir/aquifer interface, and how long that pressure drop has occurred.

Function of Two Dimensionless Terms

Dimensionless Influx, Qd 2 ≤ rd≤ 4

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34_____________________________________________________________________________________________


Dimensionless Influx, Qd 5 ≤ rd ≤ 10

Calculation Procedure

1

2

3

4

Estimate a dimensionless radius rd from regional geology

Calculate a dimensionless time td based uponaquifer properties and real time

Determine dimensionless water influx, Qd as a function of rd and td

Using the observed reservoir p, calculate water influx for the time period

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35

Application of Superposition

If you want to calculate the We that has occurred after 24 months, you need to take into account everything that has occurred before then

At 24 months:p1 has been occurring since time = 0 mo., or for 24 months,

+ p2 has been occurring since time = 6 mo., or for 18 months,

+ p3 has been occurring since time = 12 mo., or for 12 months,

+ p4 has been occurring since time = 18 mo., or for 6 months

So the calculated We at any time must take into account all of the

p’s that have happened before then, and for how long they have happened

Application of Superposition

Etrc

ktt

oewd

2

00633.0

We = (1.119hcero2f)Σ(pQd )=C Σ(pQd )

rd = ra/ro

Water Influx Calculations Let:

(E-constant)

(C-constant)

Time, yrs p, psi p p, psi

0 4500 -- --

1 4250 4375 125

2 3900 4075 300

3 3500 3700 375

For example: rd = 10, E = 0.01 days-1, C = 150 rb/psi

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36_____________________________________________________________________________________________


Application of Superposition Example

After 1 Year:

After 2 Years:

After 3 Years:

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37

1

Predicting Aquifer Pressure Support

We We We

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38_____________________________________________________________________________________________


Fetkovich Finite Aquifer

Used for finite aquifers• Neglects transient effects

Does not require superposition

Closely approximates van Everdingen and Hurstunder pseudo-steady state conditions

Based upon productivity index and pressuredepletion concepts

Overview


Where:qw = water influx rate from aquifer; rbbl/d

J = productivity index for the aquifer, rbbl/d/psi

pa = average aquifer pressure, psi

pr = pressure at the aquifer-reservoir boundary, psi

Where:pi = initial aquifer pressure, psi

Wei = encroachable water-in-place at pi, rbbls

We = cumulative water influx, rbbls

Based on two equations:

)( rawe ppJq

dt

dW ie

ei

ia pW

W

pp

Productivity index for the aquifer Aquifer material balance

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39


ei

ninrna

i

eine W

tJppp

p

WW exp1)( ,1,,

Solved over n repeated time increments:

Where:

ei

eina W

Wpp 11,

21,,

,

nrnrnr

ppp nee WW ,

Fetkovich Method

Where:k = absolute permeability, md

w = width, ft

L= length of linear aquifer, ft

f = fraction of reservoir boundary exposed to aquifer (0<f<1)

ra = radius of aquifer, ft

ro = radius of the original reservoir/aquifer interface, ft

h = aquifer thickness, ft

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40_____________________________________________________________________________________________


Aquifers in Numerical Simulation

Analytical Carter-Tracy (numerical approximation to

van Everdingen and Hurst), Fetkovich, etc.

Coupled to edge or bottom of grid

Numerical Aquifer represented with discrete grid cells

Each cell can have its own uniqueproperties

Aquifers Summary

Constant terminal pressure solution

Most comprehensive approach

Time-based (superposition principle)

Applicable to transient, late transient and PSS flow

Tedious calculation procedure

Assumes PSS relationship

Not applicable for transient response

Avoids superposition calculations

Assumes instantaneous response (no time dimension)

Quick and simple

Calculates maximum influx


Fetkovich

Pot Aquifer

Carter-Tracy

de QpCW

)( rae ppJ

dt

dW

pPVcW aqte

Constant terminal rate solution

Applicable to transient, late transient and PSS flow

Avoids extensive superposition calculations

Often preferred to analytical

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41

This description can then be used to project future performance,but the reliability of this forecast is limited if the aquifer has notyet reached pseudo-steady state

Key Learnings

1 Aquifer performance is predominantly determined by the aquifer properties: size, degree/manner of connection to reservoir and permeability

2 The aquifer goes through the same transient flow into pseudo-steady state behavior that we see in oil or gas wells

3 Aquifer behavior can be modeled by various means, with the van Everdingenand Hurst and Carter-Tracy models being the most technically correct

4Historical reservoir production and pressure data is used to allow the engineer to “identify” an aquifer description that makes the material balance calculations converge on a reservoir and aquifer description that honors field performance

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42_____________________________________________________________________________________________


Learning Objectives


Determine the rock, fluid, operational, and reservoir geometricfactors affecting water influx (We) into your reservoir

Calculate We using different models

Understand which descriptive parameters you should change soyou can match the calculated We to the observed reservoirpressure

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43

Segregated Flow


Learning Objectives


Identify the theoretical differences between segregated anddispersed flow

Describe how segregated flow is characterized

Identify what causes segregated flow instability in oildisplacement by water

Identify the critical rate calculation for unstable displacement

Identify the pore level displacement efficiency calculation forsegregated flow

Explain why coning occurs and how it can be avoided

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44_____________________________________________________________________________________________


In This Section

Gravity

MobilityRatio

Contact Tilt

Stability of Displacement Processes

Water Under-Running

Maximum Pore Level DisplacementEfficiency

Special Segregated Flow Case - Coning

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45

Gravity

Very thin transition zone

Water-oil-contact

Saturations are verticallysegregated based on fluiddensity

Residual oil saturationbehind the front

Low rates of flooding

Large fluid densitydifference

Good vertical permeability

Small thickness with smallcapillary transition zone

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46_____________________________________________________________________________________________


Mobility Ratio

wSro

oSrw

wi

orw

k

kM

@

@

Water

Sw=1 - Sor

So=Sor

Oil

Sw = Swi

So = Soi

Thin transition zone

Mobility ratio based on endpoint saturations

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47

Contact Tilt

Water-oil Contact

Contact tilt occurs for dipping reservoirs as a function of reservoirprocessing rate (qt )

wt

Srwa

w

o

ro

rw

q

AkkG

k

kM

G

TanMGTan

orw

sin000488.0

1

@

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Stability of Displacement Processes

M <= 1

Unconditionally Stable

UnstableViscous forces are strongerthan gravity (density) forces

G < M-1M > 1

Oil Zone(light oil)

Oil Zone(viscous oil)

Water Zone

Water Zone

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Maximum Pore Level Displacement Efficiency

• This equation is referring to only the lab-derived fraction of pore levelrecoverable oil from kr curves of total oil in pore space of thereservoir, since behind the front in segregated flow, the oil saturationis at the end point, Sor

• Two phases are present, but only one phase is flowing

oi

rooiD S

SSE

• Dispersed flow ED atbreakthrough depends on theaverage saturation behindthe front at that time

• Segregated flow ED atbreakthrough depends on theresidual saturation of oilbehind the front at that timeResidual Oil Saturation

Sw@ which Kro 0) (1

Sor

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Wow! Nice Oil Well!

Oil

Kh

Kv

Qc Here is a well that is capable ofproducing 1347 bopd

It has no geologic barriers through theinterval and it has a relatively highmacro kv/kh

Qo = 0.00708 * K * Kro * H * (Preservoir – Pbhf)

(μo * Bo * (ln(re/rw) + s))

Where: K (md) = 15Kro = 0.9H (ft) = 25Preservoir (psia) = 3500Pbhf = 500

Oil Viscosity = 0.5Bo (RB/STB) = 1.4Skin = 0Drainage Radius (ft) = 500Wellbore Radius (ft) = 0.25

Qo (STBOPD) = 1347 per well

Whoa, Not So Fast!

∆P > 0.433 * ( γw - γo ) * hc

Qc

Oil

Water

Kh

KvD

hc

h

Water coning into a vertical well is due to exceeding a critical drawdown toovercome gravity (see cone of water in the figure below)

When the ∆P exceeds the valueshown below under the geometricconditions indicated, water willcone into a well of relativelyhomogeneous reservoir character

By staying below the critical rate(see animation on water_drive), it ispossible to avoid coning

However, this rate is, in the majorityof cases, sub-economical

Criteria for Coning

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Bottom Water Drive Vertical Well - Coning

Oil

Water

Kh

KvD

hc

h

Critical Rate qDC - maximumoil flow rate that will avoid acone breakthrough into thewell

qDC is a function of:• Oil zone thickness, h

• Perforation interval, D

• FWL distance away

For a vertical well, you cannow write the qDC

qDC = 0.0246 * k * (ρw - ρo) * (h2 - D2)

μo * Bo * [ln(re/rb)]

How Much Can You Produce to Avoid Water?

Where: K (md) = 15Kro = 0.9H (ft) = 25D (perfs) = 0.5Density Oil (psi/ft) = 0.35Density Water (psi/ft) 0.45Oil Viscosity = 0.5Bo (RB/STB) = 1.4Skin = 0Drainage Radius (ft) = 500Wellbore Radius (ft) = 0.25

Qo (STBOPD) = 4 per well

Qo = 0.0246 * Ka * Kro * H * (ρw - ρo) * (h2 - D2)

μo * Bo * (ln(re/rw) + S)

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If You Exceed the Critical Drawdown an Upside Down Cone Forms Perpendicular to the Iso-Potential Lines

Vertical Well

FWL

xe

With a vertical well the Xe is the distance you can perforate the well, and since you want to get away from water, you perforate as little as possible

xe = effective length

If you had a horizontal well, you could place the well parallel to the FWL, and the drawdown would be spread over the longer Xe

FWL

xe = length of horizontal

xe

Horizontal Well

If You Exceed the Critical Drawdown an Upside Down Cone Forms Perpendicular to the Iso-Potential Lines

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When is Vertical Well Coning Impact Greatest?

The oil is viscous or heavy (low API or high SG)

The effective distance to the water is low

Production rate is higher than the critical rate

Reservoir is oil wet

The macro kv/kh is high, i.e., there is good communication to water, and there are no barriers to perforations

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Learning Objectives


Identify the theoretical differences between segregated anddispersed flow

Describe how segregated flow is characterized

Identify what causes segregated flow instability in oildisplacement by water

Identify the critical rate calculation for unstable displacement

Identify the pore level displacement efficiency calculation forsegregated flow

Explain why coning occurs and how it can be avoided

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Vertical Layering of Reservoir Units


Learning Objectives


Recognize the techniques available to separate a reservoir intovertical layers

Explain how Lorenz techniques can divide flow units by the porestructure that occursCOPYR

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In This Section

Dykstra-Parsons

CoefficientLorenz

Techniques

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k vs. % greater than

Permeability Variation (VDP Steps)

Dykstra Parsons Coefficient of Variability VDP

1 Arrange core samples in order of decreasing kvalue

2 Prepare cumulative frequency plot on log-normalpaper

3 Draw a straight line through data points puttingmore weight on central points

4 Calculate V from k50 , k84.1

Permeability Variation VDP

Percent Greater Than

K84.1

K50

K,

md

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Where:V = coefficient of variation

= standard deviation

x = mean value

log-normal distribution: x = geometric mean 68.3% of values within mean +/-1 standard deviation

k84.1 = mean + 1 standard deviation


Advantages• Ability to compare

across reservoirsworldwide for useas an analog

• Allows for easyelimination ofoutliers

Disadvantages• Eliminating outliers

may result in data1 – 2 standarddeviations awayfrom the mean

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Sample Lorenz Plot – SPE Monograph 3

Lorenz Coefficient

Coefficient values range from 0 to 1 in order of increasing heterogeneity

It is important to retain the hin both sides of the axis

• kh is in Darcy’s Law, indicating cumulative flow capacity at pores

Most often coupled with porethroat size

Alternatively, Φ multiplied byh in the X axis as in thevolumetric formula andindicates storage capacity, orpore body size

Therefore this is a plot of thevariation in pore throatgeometry


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Increasing Heterogeneity

Lorenz Coefficient

Coefficient values range from 0 to 1 in order of increasing heterogeneity

Lorenz Plot Honoring Stratigraphy

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Learning Objectives


Recognize the techniques available to separate a reservoir intovertical layers

Explain how Lorenz techniques can divide flow units by the porestructure that occurs

While the fractional flow curve is usually calculated ignoring the effect of gravity, the gravity term (for non horizontal flow) can be utilized to calculate the potential benefit of processing the reservoir at a slower rate.

A mobility ratio of less than 1.0 is favorable and leads to pistonlike displacement and better performance in both technical and economic terms.

Key Learnings

A fractional flow curve is calculated from the relative permeability curve, coupled with the reservoir oil and water viscosities.

The shape of the fractional flow curve can be graphically analyzed to identify the efficiency of the displacement process before water breakthrough and after

The mobility ratio is the mobility of the displacing fluid (behind the front) / the mobility of the displaced fluid.

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PetroAcademyTM Applied Reservoir Engineering Skill Modules

This is Reservoir Engineering Core

Reservoir Rock Properties Core

Reservoir Rock PropertiesFundamentals

Reservoir Fluid Core

Reservoir Fluid Fundamentals

Reservoir Flow Properties Core

Reservoir Flow PropertiesFundamentals


Reservoir Fluid Displacement Fundamentals

Properties Analysis Management

Reservoir Material Balance Core

Reservoir Material Balance Fundamentals

Decline Curve Analysis and Empirical Approaches Core

Decline Curve Analysis and Empirical Approaches Fundamentals

Pressure Transient Analysis Core

Rate Transient Analysis Core

Enhanced Oil Recovery Core

Enhanced Oil Recovery Fundamentals

Reservoir Simulation Core

Reserves and Resources Core

Reservoir Surveillance Core

Reservoir Surveillance Fundamentals

Reservoir Management Core

Reservoir Management Fundamentals


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dispersed flow -...

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