Darcy’s Law for Linear Flow...

Introduction

Reservoir Flow Properties Fundamentals

Why This Module is Important

Fundamental understanding of the flow through rocks is extremely important to understand the behavior of the reservoir

Permeability

The property which defines the conductivity of the fluid in the

rocks

Darcy’s Law

Describes the behavior of flow, and depending on the nature of

flow, can take many forms

It is important to understand how to extend Darcy’s law equation to account for the nature of flow

Equations to calculate the rate changes based on whether it is a single phase oil, single phase gas, or a multi-phase fluids

Reservoir Flow Properties Fundamentals ═════════════════════════════════════════════════════════════════════════

© PetroSkills, LLC. All rights reserved._____________________________________________________________________________________________

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Reservoirs can encounter:


Pe

Pwf

re

rw

Linear Flow Radial Flow

It is common to see linear flow in horizontal

wells

It is unusual to see radial flow in vertical

wells


During the drilling process:• The nearby well region is often damaged

• The permeability near the well bore is altered

How to handle the changes in the flow behavior due to damagenear the wellbore is important to correctly predict the ratebehavior of the well

Skin factor is normally considered to properly account for thisbehavior

When a well is stimulated or the near wellbore properties arealtered, the productivity equation can be manipulated todetermine how that alteration is going to impact the productivityof the well

Reservoir Flow Properties Fundamentals═════════════════════════════════════════════════════════════════════════


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Reservoirs are rarely homogeneous and we need to account for the permeability heterogeneity in properly calculating the productivity of the well

Depending on the distribution of the permeability heterogeneity, different sets of equations need to be used to calculate the effective flow behavior

For gas wells, because of low viscosity of the gas and relatively high velocities near the wellbore, the traditional Darcy’s law equation may not be applicable

We need to account for additional pressure drop due to high velocity of gas by supplementing Darcy’s law


Darcy’s law and all its deviations and extensions are crucial to understand the behavior of oil and gas wells

Without understanding these extensions, no reservoir engineer would be able to correctly predict the rate at which the well is capable of producing



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Darcy’s Law for Linear Flow


This section will cover the following learning objectives:

Learning Objectives

Solve linear flow problems using Darcy’s law equation

Differentiate between gas and oil flows and why the equations are different for two phases

Summarize the importance of gravity and pressure gradients and their influence on flow in linear systems



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1.127 10 0.4335

8.5245 10 9.81

Recall Darcy’s Law for Linear Flow

Symbol Field Unit SI Unit

q Barrels per day m3/per day

k Permeability in md Permeability in md

A ft2 m2

cp Pa.s

p psia kPa

I ft m

ϒ Specific gravity of fluid

Specific gravity of fluid

θ Dip angle of inclination measured counter-clockwise from horizontal flow

* For upwardvertical flow θ is90 degrees

Dip angle of inclination measured counter-clockwise from horizontal flow

* For verticallyupward flow, θ is90 degrees

For linear flow, assuming constant density:

SI Units

Field Units



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Example 1

What is the permeability of the sand pack?

If the tube was turned vertical and water flows from the top to bottom, at

what rate water will flow if the pressures at the inlet and the outlet were

maintained the same as in the previous experiment?

What would the flow rate be if the tube was turned at 45° and the liquid

flowed in the upward direction?

A cylindrical tube 4 ft [1.22 m] long with a 3 in. [7.62 cm] diameter is packed with sand. The tube was laid horizontally and water, at a rate of 1 gallon/hour [3788 cc/hr], was flowed through the tube. The pressure at the inlet was observed to be 25 psig [172.4 kPag] and at the exit, it was 10 psig[68.9 kPag].

Assume the viscosity of water equal to 0.8 cp

[0.0008 Pa.s] and the specific gravity equal to 1

Solution: Example 1

Flow Rate

Field Units 1 1 24 0.5714

SI Units 3788 3788 24 0.091

AreaField Units 0.0491ft

SI Units.

0.00456

Horizontal Flow =0

Substituting in Darcy’s law equation

Field Units 0.5714 1.127 10.

.

SI Units 0.091 8.5245 10.

.

. .

.

Solving for kField Units k 2,203md

SI Units k 2,203md



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Solution: Example 1

1.127 102,203 0.0491

0.825 10

40.4335 1

0.6374 1.115 /

8.5245 102,203 0.00456

0.0008172.4 68.9

1.229.81 1

0.101 4227 /

Downward flow: gravity assists the pressuredrop, hence higher rate is observed.

Field Units

SI Units

Vertical Flow θ= -90o (downward flow)

1.127 102,203 0.0491

0.825 10

40.4335 45

0.5247 0.918

8.5245 102,203 0.00456

0.0008172.4 68.9

1.229.81 45

0.0835 3479

Field Units

SI Units

Solution: Example 1

Upward flow: gravity opposes the pressure gradient, hence the rate is lower.

Inclined Flowθ= 45o (upward flow)



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Field Units

Gas Flow Equations for Linear Flow

Gas flow equation for linear flow can be written as (in terms of pressure squared method)

112 10

You can write a similar equation in the form of pseudo-real pressure as follows:

112 10

SI Units

Gas flow equation for linear flow can be written as (in terms of pressure squared method)

1.215 10

You can write a similar equation in the form of pseudo-real pressure as follows:

1.215 10


q MSCFD 3DM

k md md

A 2ft m2

Centipoise Pa.s

T Rankineankine

L ft m

p psi kPa

m(p) psi2/cp kPa2/Pa.s

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A horizontal well – which is 5000 ft [1525 m] long – is subjected to multi-stage fractures. The total number of transverse fractures is 20 and the fracture length is 200 ft [61 m].

Example 2

A horizontal well – which is 5000 ft [1525 m] long – is subjected to multi-stage fractures. The total number of transverse fractures is 20 and the fracture length is 200 ft [61 m].

Example 2

Assume the following:

Average distance between the fractures is 250 ft [76.2 m] and the flow is linear from reservoir into the fracture.

Each fracture is draining about half the distance between two fractures.

Pressure in the middle of two fractures is 4,000 psia [27,579 kPa] and the pressure inside the fracture is 500 psia [3,447 kPa].

The well is producing dry gas. Assume the permeability to be 0.01 md.

Height of each fracture is 100 ft [30.5 m].

The temperature of the reservoir is 180° F [82.6° C].

Average values of and z to be 0.017 cp [0.000017 Pa.s ] and 0.88 respectively.

What is the rate at which the horizontal

well is producing?



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Solution: Example 2

Area of each fracture

Field Units 2 x 100 x 200 = 40,000 ft2

SI Units 2 x 61 x 30.5 = 3,718 m2

Distance gas travels for linear flow

Field Units 125 ft (half the distance)

SI Units 38.1 m (half the distance)

Gas rate for linear flow

Field Units 112 10

SI Units 1.215 10

SubstitutingField Units 112 10

. ,

. .4000 500 589.6

SI Units 1.215 10. ,

. . .27579 3447 16,685 /

Multiplying by 20, total rate at which the well will produce:

Field Units 11.8 MMSCFD

SI Units 333,703 /

This section has covered the following learning objectives:

Learning Objectives

Solve linear flow problems using Darcy’s law equation

Differentiate between gas and oil flows and why the equations are different for two phases

Summarize the importance of gravity and pressure gradients and their influence on flow in linear systems



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Darcy’s Law for Radial Flow



Learning Objectives

Solve simple problems for radial flow across porous medium using Darcy’s law for radial flows

Differentiate between oil and gas flows

Define and calculate productivity index

Predict the inflow performance relationship for oil and gas wells



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Recall Radial Flow

For petroleum reservoirs,radial flow is much morecommon as the flowconverges into the wellbore

The pressure at the boundaryof radius re is pe

The pressure at the wellboreradius, rw, is pwf

Pe > pwf and re > rw

Pe

Pwf

re

rw

.

5.3562 10

SI Units

Field Units

Recall Radial Flow


q Barrels per day m3/day

k Permeability in md Permeability in md

h ft m

p psia kPa

cp Pa.s

Radii ft m

You can write the previous equation for radial flow in field units as:



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Productivity Index

Productivity index is defined as units of production per day per unit changein the bottom hole pressure.

Productivity index is an important indicator of how productive the well is.

For oil wells, typically, in field units, the value of J ≥ 1 (bbl/d/psi) [0.023m3/d/kPa in SI units] is considered a highly productive well.

If the value of productivity index, J ≤ 0.001 (bbl/d/psi) [0.000023 m3/d/kPa],it is considered a marginal well.

The rigorous definition of productivity index is . This definition

tells us about the incremental oil (or gas) that can be produced per unitchange in BHP.

For single phase oil, J is constant. Taking the derivative of rate equationwith respect to BHP, the value will be constant. For radial flow it has the

form where C depends on the units.

Ideally it does not change with reservoir pressure. So, if indeed there is areduction over time, it is an indication of reservoir damage.

Why?



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Example 3

If the permeability of the reservoir is 20 md, the thickness is 30 ft[9.15 m] and the drainage radius is 1,000 ft [304.9 m], at whatrate will the well produce? The well bore radius is 6 in. [0.152 m].

If, by applying an artificial lift method, the bottom hole pressure isreduced to 3,000 psi [20,684 kPa], at what rate will the wellproduce?

What is the value of productivity index? Plot rate vs. bottom holepressure for this well.

A well is producing in a radial reservoir at a bottom hole pressure of 5,500 psi [37,921 kPa]. The reservoir pressure is 6,000 psi [41,369 kPa]. Oil viscosity is 0.25 cp [0.00025 Pa.s].

Field Units SI Units

Solution: Example 3

7.08 10 20 30 6,000 5,500

0.2510000.5

1,118 /

7.08 10 20 30 6,000 3,000

0.2510000.5

6,707 /

7.08 10 20 30

0.2510000.5

2.24/

Use the radial flow equation and substitute all the numbers:

5.362 10 20 9.146 41,368 37,921

0.00025304.90.152

177.9 3/

Choosing 3,000 psia allows you to calculate the rate as:

5.362 10 20 9.146 41,368 20,684

0.00025304.90.152

1067.8 3/

Productivity Index, J:

5.362 10 20 9.146

0.00025304.90.152

0.0523/



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Solution: Example 3

Once the productivity indexis calculated, calculate therate at any bottom holepressure using the equation:

The plot of rate vs. BHP(bottom hole pressure)indicates a straight line

This graph is also called IPRor Inflow Performance Curve

The slope of the line is J orproductivity index

SI Units

0

10000

20000

30000

40000

50000

0 500 1000 1500 2000 2500

pwf, kPa

q, m3/day

Pwf, kPa

q, m3/day

Field Units

0

1000

2000

3000

4000

5000

6000

7000

0 5000 10000 15000

pwf, psia

q, bbl/day

Pwf, psia

q, bbl/day

q J pe – pwf



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As can be seen from this equation, the rigorous definition of productivity index, provides the following value of productivity index:


703 10

7.633 10

Gas Wells

For gas wells, the productivity index decreases as the bottom hole pressure (BHP) decreases. Don’t expect to see the same incremental rate increase as BHP decreases further. This is an important observation and has a practical significance.

For gas wells, write the equation for the rate as:

703 10 2

7.633 10 2



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Example 4

Generate the inflow performance curve. Calculate the productivityindex as a function of BHP.

A main reason we install compressor on a gas well is to reducethe BHP. Based on your understanding of IPR curves, is it betterto install compressor when the BHP is higher or lower?

A gas well is producing from a reservoir with permeability of 3 md and thickness of 50 ft [15.2 m]. The reservoir pressure is 5,000 psia [34,474 kPa]. The average viscosity is 0.02 cp [0.00002 Pa.s] and the z factor is 0.87. The reservoir temperature is 200 F [93.7 C]. The drainage radius is 1,500 ft [457.3 m] and the wellbore radius is 4 in. [0.102 m].

Field Units

SI Units

Solution: Example 4

703 10 3 50 5000 2500

0.02 0.87 66015000.33

20,465 20.5

7.633 10 3 15.2 34,474 17,237

0.00002 0.87 366.7457.30.102

579.7 3

Calculation of Flow Rate



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Field Units

SI Units

Solution: Example 4

703 10 2

703 10 3 50 2 2500

0.02 0.87 66015000.33

5.46 /

7.633 10 2

7.633 10 3 15.2 2 17,237

0.00002 0.87 366.7457.30.102

22.42 3/ /

By substituting different values of BHP, the IPR curve can be generated:

To calculate the productivity index at 2500 psia [17,237 kPa], use this equation

Solution: Example 4

SI Units

Field Units

0

1000

2000

3000

4000

5000

6000

0 10 20 30

pwf, psia

q, MMSCF/day

0

5000

10000

15000

20000

25000

30000

35000

40000

0 200 400 600 800 1000

pwf, kPa

q, Mm3/day

Pw

f, p

sia

Pw

f, k

Pa

q, Mm3/day

q, MMSCF/day

The graph shows the inflowperformance curve for thegas well

Unlike single phase oil well,it is not a straight line;instead, it is concaveupwards

The IPR curve is gentle atthe top and becomes steeperat the bottom



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Solution: Example 4

SI Units

Field Units

0

1000

2000

3000

4000

5000

6000

‐ 5.00 10.00 15.00

pwf, psia

J, MSCF/day/psi

0

5000

10000

15000

20000

25000

30000

35000

40000

‐ 10.00 20.00 30.00 40.00 50.00

pwf, kPa

J, m3/day/kPa

Pw

f, p

sia

Pw

f, k

Pa

J, m3/day/kPA

J, MSCF/day/psi

The graph shows theproductivity index as a functionof bottom hole pressure (BHP)

Notice that productivity indexincreases with an increase inBHP

For the same change of BHP,the improvement in rate is muchbetter when BHP is higher thanlower

It is better to install acompressor if the BHP is higherso that improvement will belarger


Learning Objectives

Solve simple problems for radial flow across porous mediumusing Darcy’s law for radial flows

Differentiate between oil and gas flows

Define and calculate productivity index

Predict the inflow performance relationship for oil and gas wells



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Flow Regimes and Their Impact on the Performance of the Wells



Learning Objectives

Outline the differences between transient state, steady state and pseudo-steady state flow regimes

Calculate the productivity of both oil and gas wells under different flow regimes

Describe the importance of the shape of the reservoirs and how they impact the behavior of oil and gas wells



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Equations for Other Flow Regimes

If the reservoir has different shapes than a perfect circle, different boundaries are felt at different times, resulting in a late transient period where the flow is between end of true transient and beginning of pseudo-steady state period

Transient State Pseudo-State

Transient state occurs when the production is initiated and the well is producing without seeing the influence of the boundary

Pseudo-steady state occurs when the boundary effect is felt and the pressure across reservoir starts changing uniformly as a function of time due to depletion

Transient to Late Transient to PSS

Time

Transient Late Transient Pseudo-steady State

dP/dt = c

Boundaries

p fn Log time

Key requirement of pseudo-steady state when well is producing:Change in the pressure with respect to time in any part of the reservoir is constant



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k md md

h ft m

q bbI/d m3/d

μ cp md

P psi kPa

t hrs hrs

ct psi-1 kPa-1

rw ft m

pi represents initial pressure which will also be equal to the pressure at the boundary or pe.

pwf represents the pressure at the wellbore.

Transient Flow Equation for Radial Flow

Equation for Transient Flow:

Field Units

7.08 1012 1688∅

OR

. ∅

Recall that time to end transient state is:

Field Units: ∅

SI Units: . ∅

where teTF is in hours

SI Units

5.3562 1012 1.25 10 ∅

OR

5.3562 10

12 1.25 10 ∅



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Example 5

Calculate the time to end transient state.

Calculate the rate at which the well will flow at 1, 5, 10, 20, 30, 50and 70 days.

A well is producing from a reservoir with a permeability of 1 md, thickness of 50 ft [15.2 m], rw of 0.36 ft [0.11 m] and re of 1,500 ft [457.3 m]. Porosity is 0.2, viscosity is 0.4 cp [0.0004 Pa.s], ct is 10 x 10-6 psi-1[1.45 x 10-6 kPa-1], the pi is 5,000 psia [34,474 kPa] and pwf is 1,000 psia [6,895 kPa].

Field Units

SI Units

Solution: Example 5

948 0.2 0.4 10 10 15001

= 1,706hours 71days

7.034 10 0.2 0.0004 1.45 10 457.31

= 1,706hours 71days

Calculate the time to end the transient state:

At all the times for which you are calculating the rate, you are in transient state.



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Field Units

SI Units

Solution: Example 5

0.47.08 10 1 50

12

1 5 241688 0.2 0.4 10 10 0.36

=7.59

5,000 1,0007.59

= 527barrels/day

0.00045.356 10 1 15.2

12

1 5 241.25 108 0.2 0.0004 1.45 10 0.11

=330

34,474 6,895330

= 83.6m3/day

By substituting different values of BHP, the IPR curve can be generated:

To calculate the rate at 10 days,Replace this value with “10”

Solution: Example 5

SI Units

40.0

50.0

60.0

70.0

80.0

90.0

100.0

0 20 40 60 80

q, m

3/day

t, days

q,m

3/day

t, days

Field Units

300

350

400

450

500

550

600

650

0 20 40 60 80

q, b

bl/day

t, days

q,bbl/day

t, days

Rate profile as a function of time indicates rapid decline in the rate as time progresses

For radial flow, as the time reaches 71 days, the flow regime will change to pseudo-steady state period



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Average vs. Boundary Pressure

The difference between and pwf is smaller thanthe difference between pe and pwf.

When the transition from transient to pseudo-steady state flow begins, the rate equations can be written in terms of either boundary pressure or average pressure.

Boundary Pressure Average Pressure

The average pressure represents volumetric average of the pressure. It will fall between pwf and pe but it is closer to pe

because most of the pressure drop in radial flow occurs near the well bore and the volume in that region is very small.

Denote the average pressure by .

The boundary pressure represents the pressure at the boundary.

Terminology

rw re

pwf

pi , pe

r

p

p

p - pwf

pi - pwf

The average pressure is a lot closer to the drainage

pressure or initial pressure than the bottom hole

pressure



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Field Units

SI Units

Pseudo-steady State

7.08 10

0.5or

7.08 10

0.75

5.3562 10

0.5or

5.3562 10

0.75

For pseudo-steady state conditions, define the rate equation for radial flow as:

It is more common to use the average pressure instead of boundary pressure

Field Units

SI Units

Pseudo-steady State

7.08 10

0.5or

7.08 10

0.75

5.3562 10

0.5or

5.3562 10

0.75

Considering oil properties (from), write the equation in terms of STB/day as:

In this equation B is formation volume factor



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SI Units

5.3562 10 12

4

Field Units

7.08 10 12

4

Non-radial Reservoirs

Variable Definition

A Area of the reservoir in ft2

CA The shape factor which depends on the shape of the reservoir

Euler’s constant and has a value of 1.781

Wellbore radius squared

For circular reservoir, CA is 31.6. substituting A as re

2, you obtain the equation on the previous slide.

If the reservoir has a shape which is not circular, we can still write an equation for pseudo-steady conditions.

Shape Factors

The figure shows the values of shape factors for some shapes. It includes the beginning of pseudo-steady state when dimensionless time tDA reaches a certain value, defined as:


k md md

t hrs hrs

∅ Fraction Fraction

cp Pa.s

ct psi-1 kPa-1

A ft2 m2

2.814 10 ∅

3792∅

SI Units

Field UnitsStart of PSS End of Infinite Acting

<1% errorfor tDA >

Transient solution has <1% error

for tDA <

Exactfor tDA>

In CACA

For Radial Flow:The end of transient

period and the start of pseudo-steady state is the

same



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Example 6

Calculate the time to reach pseudo-steadystate.

Calculate the rate at which the well willproduce under pseudo-steady state conditions.

Calculate the value of the productivity index.

A well is located as shown in the figure here. The area of the reservoir is 4,000,000 ft2 [3.718x105 m2], permeability is 1 md, thickness is 70 ft [21.3 m], viscosity is 0.5 cp [0.0005 Pa.s], ct is 13 x10-6 psi-1 [1.89 x10-6 kPa-1] porosity is 0.15, rw is 0.45 ft [0.137 m], is 4,000 psia [27,579 kPa] and pwf is 1,500 psia [10,342 kPa]. The value of B is 1.2 bbl/STB [1.2 m3/Sm3].

Field Units

SI Units

Solution: Example 6

0.6 3792 0.15 0.5 13 10 4 101

8,873 369.7

0.6 2.814 108 0.15 0.0005 1.89 10 3.718 101

8,873

369.7

The time to reach pseudo-steady state is calculated by knowing that the value of tDA = 0.6

It will take approximately one year to reach pseudo‐steady state



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Field Units

SI Units

Solution: Example 6

7.08 10 1 70 4,000 1,500

1.2 0.512

4 4 101.781 4.5 0.45

256STB/day

5.356 10 1 21.3 27,579 10,342

1.2 0.000512

4 3.718 101.781 4.5 0.137

= 40.8Sm3/day

Calculate the rate as:Productivity Index, J = 256/(4,000 – 1,500) = 0.103 STB/d/psi

Productivity Index, J = 40.8/(27,579 – 10,342) = 0.0024 Sm3/d/kPa



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For transient state:

Equations for Gas for Other Flow Regimes

Field Units

703 1012 1688∅

SI Units

7.6326 1012 1.25 10 ∅

where


k md md

h ft m

t hrs

cp Pa.s

ct psi-1 kPa-1

rw ft m

Fraction Fraction

P psia kPa


Pressure Squared Method

Equations for Gas for Other Flow Regimes

Field Units

703 1012 1688∅

where

SI Units

7.6326 1012 1.25 10 ∅

where


k md md

h ft m

t hrs

cp Pa.s

ct psi-1 kPa-1

rw ft m

Fraction Fraction

P psia kPa




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Pseudo-Steady State Rate Equation

Gas Flow: Pseudo-steady State

Field Units

703 10

3/4

SI Units

7.6326 10

3/4


k md md

h ft m

m(p) psi2/cp. kPa2/Pa.s

T R K

Pressure Squared

Gas Flow: Pseudo-steady State

Field Units

703 10

3/4

SI Units

7.6326 10

3/4


k md md

h ft m

t hrs

cp Pa.s

ct psi-1 kPa-1

rw ft m

Fraction Fraction

P psia kPa




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Example 7

Calculate the time required to reach pseudo-steady state.

Calculate the rate at which the well will produce at different times until it reaches pseudo-steady state.

Assume that the average reservoir pressure is reduced to 4,970 psia[34,267 kPa] at the beginning of pseudo-steady state period. Calculate the rate at the beginning of pseudo-steady state period and compare it to the end of transient period. What do you observe?

A gas well is producing from a reservoir with initial reservoir pressure of 5,000 psia [34,474 kPa] and bottom hole pressure of 1,500 psia [10,342 kPa]. The permeability is 0.2 md, h is 50 ft [15.24 m], re is 1,500 ft [457.3 m], and rw is 0.4 ft [0.122 m]. The reservoir temperature is 200 F [93.6 C]. The gas gravity is 0.65. The porosity is 0.12.

Solution: Example 7

Field Units

948 0.12 0.0262 131 10 1,5000.2

= 4,392hours= 183days

Calculate time to reach pseudo-steady state:

SI Units

7.034 107 0.12 0.0000262 1.9 10 457.30.2

4,392hours 183days



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Solution: Example 7

Field Units

660703 10 0.2 50

12

0.2 5 241688 0.12 0.0262 131 10 0.4

= 576,533

SI Units

Calculate ag for 5 days as follows:

366.77.6326 10 0.2 15.24

12

0.2 5 241.25 10 0.12 0.0000262 1.9 10 0.122

= 9.67x108

x

Solution: Example 7

Field Units

660 0.0262 1.007703 10 0.2 50

12

0.2 5 241688 0.12 0.0262 131 10 0.4

=1.52x104

SI Units

366.7 0.0000262 1.0077.6326 10 0.2 15.24

12

0.2 5 241.25 10 0.12 0.0000262 1.9 10 0.122

=2.55x104

Calculate the value of “a” for pressure squared method:



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Solution: Example 7

For using m(p) method, you will need to calculate m(p) as a function of pressure by numerically integrating p/(z) as a function of pressure.

Field Units

0.E+00

2.E+08

4.E+08

6.E+08

8.E+08

1.E+09

0 1000 2000 3000 4000 5000 6000

m(p)

p, psia

Pwf, psia

q, bbl/day

SI Units

0.E+00

1.E+13

2.E+13

3.E+13

4.E+13

5.E+13

0 10000 20000 30000 40000

m(p)

p, kPa

Pwf, kPa

q, m3/day

Solution: Example 7

Field Units

=9.45 10 1.60 10

576,533=1,362 MSCFD

SI Units

7.6326 10

3/4

Calculate rate using m(p) method as:

SI Units

20,000

25,000

30,000

35,000

40,000

45,000

50,000

55,000

0 50 100 150 200

q, Sm

3/D

t, days

m(p) p squared

q, Sm

3/D

t, days

Field Units

600

800

1,000

1,200

1,400

1,600

1,800

0 50 100 150 200

q, M

SCFD

t, days

m(p) p squared

q, M

SCFD

t, days



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Solution: Example 7

Field Units

/= 1,496MSCFD

SI Units

7.6326 10

3/4

Calculate rate using pressure squared method as:

SI Units

20,000

25,000

30,000

35,000

40,000

45,000

50,000

55,000

0 50 100 150 200

q, Sm

3/D

t, days

m(p) p squared

q, Sm

3/D

t, days

Field Units

600

800

1,000

1,200

1,400

1,600

1,800

0 50 100 150 200

q, M

SCFD

t, days

m(p) p squared

q, M

SCFD

t, days

The match between m(p) and p2 method is not exact since the initial pressure is significantly greater than 3,000 psia [21,000 kPa]

Solution: Example 7

Field Units

703 10

3/4=703 10 0.2 50 9.39 10 1.60 10

66015000.4 3/4

= 1,109MSCFD

SI Units

7.6326 10

3/4=7.6326 10 0.2 15.24 4.46 10 7.61 10

366.7457.30.122 3/4

=31,422Sm3/D

Calculate the rate under pseudo‐steady state conditions as:



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Solution: Example 7

Field Units

703 10

3/4=703 10 0.2 50 4970 1500

0.0262 1.007 66015000.4 3/4

1,203MSCFD

SI Units

7.6326 10

3/4=7.6326 10 0.2 15.24 34,267 10,342

0.0000262 1.007 366.7457.30.122 3/4

= 34,061Sm3/D

For pressure squared method, the rate is:

There is some difference between the two answers because the reservoir pressure higher than 3,000 psia [21,000 kPa].


Learning Objectives

Outline the differences between transient state, steady state and pseudo-steady state flow regimes

Calculate the productivity of both oil and gas wells under different flow regimes

Describe the importance of the shape of the reservoirs and how they impact the behavior of oil and gas wells



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Vertically Fractured and Horizontal Wells



Learning Objectives

Outline the difference between vertically fractured and horizontal wells and their advantages and disadvantages

Calculate the rates and productivity indices for vertically fractured and horizontal wells using the concept of effective well bore radius

Describe different flow regimes encountered by vertically fractured and horizontal wells

Evaluate efficacy of horizontal wells and compare the performance to vertically fractured wells



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Other Types of Wells

Hydraulically Fractured Vertical Well: A vertical well drilled in tight formations but is hydraulically fractured to improve the productivity

Horizontal Well: Instead of vertical, a horizontally drilled well to improve the connectivity with reservoir

Fractured Vertical Well

Assume that the fracture is vertical and contained within formation

Half fracture length is also denoted as xf

Fracture has significantly higher conductivity (permeability) than the formationfL

er

Vertical Well

2fL



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Flow Regimes

Similar to vertical wells,

vertically fractured as

well as horizontal wells

also exhibit different flow

regimes during the production

Unlike vertical wells, the flow regimes are

more complex and many

variations in transient flow regimes exist

There will be a demonstration on the physical reasons for the

existence of such flow

regimes but will also

concentrate on pseudo-radial transient flow

Unless the permeability is extremely low, the existence of other flow regimes is

short lived, and pseudo-radial

flow starts within days

Flow Regimes (Transient Flow)

Fracture Linear Flow

Bi‐Linear Flow

Formation Linear

Flow

Pseudo‐radial Linear Flow

When the well is open, the first thing which drains into the well bore is the fluid inside the fracture. And because we assume that the fracture is linear, the flow will be linear inside the fracture.



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k, mdXf Linear

Flow Startt, Days Linear

Flow EndsRadial Flowft m

0.0001 1,000 305 1,316 6,782 175,958

0.001 500 152 354 1,825 47,350

0.1 250 76 1 5 118

Importance of Permeability on Flow Regimes

The lower the permeability, the more likely is the possibility that linear or bi-linear flow can

extend over a long time.



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Vertically Fractured vs. Horizontal Well

When comparing a vertically fractured with a horizontal well, it is observed that the vertically fractured well with the same length as the horizontal well will be superior compared to the horizontal well.

Because the horizontal well only contacts a limited portion of the reservoir in a horizontal direction; whereas, the vertically fractured well contacts the entire thickness in a vertical direction.

Horizontal well performance is controlled by vertical permeability but this is not the case for vertically fractured wells.

We still drill horizontal wells because we have much better geological control over horizontal wells compared to vertical fractures.

The assumption is that once the horizontal well reaches a pseudo-radial flow the performance of the horizontal well can be estimated.

Horizontal Well

hL

er

wzh

Z

X

hL

We need to know:

The distance from the bottom of the formation where the center of the horizontal well is drilled

The length of the horizontal well



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What About Horizontal Wells that are Vertically Fractured?

These wells overcome the biggest problem of horizontal wells; i.e., vertical communication

To determine the performance of these wells analytically is difficult

However, as an approximation, consider each vertical fracture as less efficient vertically fractured well

The efficiency can be as low as 10% to as much as 75%



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Example 8

Calculate the rate with and without hydraulic fracture underpseudo-steady state conditions.

• Use pressure squared method.

• Generate the IPR curve for both the conditions.

A vertically fractured well is producing from a gas reservoir. The average pressure is 2,800 psia [19,305 kPa] and the BHP is 700 psia [4,826 kPa]. The reservoir has a permeability of 0.1 md, thickness of 60 ft [18.3 m], re

is 1,700 ft [518.3 m], and rw is 0.38 ft [0.116 m]. The half fracture length is 100 ft [30.5 m]. Assume the fracture to be infinite conductivity. The viscosity of gas is 0.0186 cp [0.0000186 Pa.s], z is 0.853 and the reservoir temperature is 130 F [54.8 C].

Field Units

SI Units

Solution: Example 8

/=

.

. ..

/= 433MSCFD

.

/=

. . . , ,

. . ..

/= 12,253Sm3/D

Rate without Fracture



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Field Units

SI Units

Solution: Example 8

703 10

3/4=703 10 0.1 60 2800 700

0.0186 0.853 590170050 3/4

1,193MSCFD

7.6326 10

3/4=7.6326 10 0.1 18.3 19,305 4,826

0.0000186 0.853 328518.315.25 3/4

= 33,788Sm3/D

Rate with FractureThe effective wellbore radius is 100/2=50 ft [30.5/2 = 15.25 m]

Solution: Example 8

The IPR curve can be generated by calculating the rate at other BHP’s

Examining the IPR curve, one can clearly see the difference between the productivity of the well with and without a hydraulic fracture

SI Units

600

5600

10600

15600

20600

25600

‐ 10,000 20,000 30,000 40,000 50,000

pwf, kPa

q, Sm3/D

w/o Fracture w/ Fracture

Pwf, kPa

q, Mm3/day

Field Units

600

1100

1600

2100

2600

3100

‐ 500 1,000 1,500

pwf, psia

q, MSCFD

w/o Fracture w/ Fracture

Pwf, psia

q, MSCFD



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Example 9

Calculate the inflow performance of the horizontal well

Compare the performance with the inflow performance of a vertical welldrilled in the same reservoir

A horizontal well is planned to be drilled in an oil reservoir. The permeability of the formation is 1 md, vertical to horizontal permeability ratio is 0.1, thickness is 60 ft [18.3 m], the oil viscosity is 0.6 cp [0.0006 Pa.s], the compressibility is 20 x 10-6 psi-1 [2.81 x 10-5 kPa-1], the porosity is 0.12, the rw is 0.4 ft [0.122 m], and re is 2000 ft [609.8 m]. The length of the well is proposed to be 1,000 ft [304.9 m]. The initial reservoir pressure is 2,800 psia [19,305 kPa] and the initial formation volume factor is 1.15 bbl/STB [1.15 m3/Sm3]. Assuming that permeability in x and y directions are identical and are equal to 1 md. The distance from bottom of the formation to the well is 30 ft [9.15 m].

Field Units

SI Units

Solution: Example 9

42

22

42

22

To determine the performance of the horizontal well:First determine the value of F:



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Field Units

Solution: Example 9

601000

10 42 60

2 30 0.4 0.12 60

0.4 0.1 0.8204

2 ln 21000

2 ln 2 0.8204110ft

SI Units

18.3304.9

10 42 18.3

2 9.15 0.122 0.12 18.3

0.122 0.1 0.8204

=.

.= 34m

Field Units

SI Units

Solution: Example 9

7.08 10

0.75=7.08 10 1 60 2,800 1,000

0.6 1.152000110 0.75

= 515.3STB/day

5.362 10

0.75=5.362 10 1 18.3 19,305 6,895

0.0006 1.15609.834 0.75

= 82Sm3/day

The rate at any well flowing pressure using standard rate equation for radial flow can be calculated.

Rate at 1,000 psia

Rate at 6,895 kPa



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Field Units

SI Units

Solution: Example 9

7.08 10

0.75

7.08 10 1 60 2,800 1,000

0.6 1.1520000.4 0.75

142.7STB/day

5.362 10

0.75=5.362 10 1 18.3 19,305 6,895

0.0006 1.15609.80.122 0.75

= 22.7Sm3/day

Contrast this to standard vertical well, where the rate is equal to:

Solution: Example 9

SI Units

600

5600

10600

15600

20600

0.0 50.0 100.0 150.0

pwf, kPa

q, Sm3/D

Vertical Horizontal

Pw

f, k

Pa

q, Sm3/D

Field Units

600

1100

1600

2100

2600

3100

0.0 200.0 400.0 600.0 800.0 1000.0

pwf, psia

q, STB/D

Vertical Horizontal

Pw

f, p

sia

q, STB/D

The figure on the right compares the IPR curves for vertical vs. horizontal wells

Horizontal wells are more than three times prolific than vertical wells

If the cost of drilling horizontal wells are significantly smaller than vertical wells, horizontal wells are clearly a more effective method of producing this formation



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Learning Objectives

Outline the difference between vertically fractured and horizontal wells and their advantages and disadvantages

Calculate the rates and productivity indices for vertically fractured and horizontal wells using the concept of effective well bore radius

Describe different flow regimes encountered by vertically fractured and horizontal wells

Evaluate efficacy of horizontal wells and compare the performance to vertically fractured wells



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Heterogeneous Systems and Skin Factor



Learning Objectives

Calculate the effective permeability of heterogeneous systems when the layers are either in parallel or in series

Distinguish between effective permeability calculations for linear and radial flows

Describe the concept of skin factor and its influence on the performance of the well

Evaluate the well performance with limited production data



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Application of Darcy’s Law for Heterogeneous Systems

k1,h1

k2,h2

kn‐1,hn‐1

kn,hn

Layers in Parallel

Layers in Series

Flow direction is perpendicular to the layersFlow direction is parallel to the layers

k1,L1 k2,L2 kn,Lnk1,L1 k2,L2 kn,Lnk1,h1k2,h2

Kn‐1,hn‐1

Kn,hn

k1,L1 k2,L2 kn,Ln

Layers in Parallel

k1,h1

k2,h2

kn‐1,hn‐1

kn,hn

Assume that we have n layers with different permeabilities and different thicknesses. The flow is parallel to the layers as shown by the direction of the arrows.



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Layers in Parallel – Basic Derivation

If we define the total thickness of all the layers as:

The effective permeability is calculated as:

∑

It is easy to show that the same equation can also be used for radial flow when the flow is through parallel layers



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Example 10

Assume that we have n layers with different permeabilities and different thicknesses. The flow is parallel to the layers as shown by the direction of the arrows.

k = 100 md, h = 5

k = 10 md, h = 10

k = 1 md, h = 60

Solution: Example 10

. Calculate the effective permeability as:

.Assuming that the highest permeability is 1000 instead of 100 (an order of magnitude higher), the effective permeability will be:

..

Assuming that the lowest permeability is 0.1 instead of 1 (an order of magnitude smaller), the effective permeability will be:

Notice that the effective permeability is much more sensitive to the largest value but not as sensitive to the smallest value. This is an important characteristics of arithmetic average.



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Layers in Series

k1,L1 k2,L2 kn,Ln

Linear Flow Radial Flow

kn

k1

re

By applying Darcy’s law and making similar assumptions, we can also derive an effective permeability value for layers in series.

The flow direction is shown for both linear and radial flow below.

Layers in Series – Linear vs. Radial Flow

If we define the total length of all the layers as:

∑We can write the expression for effective permeability as:

∑

Radial FlowWe can write the expression for effective permeability as:



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Example 11

Following system consists of three layers with different permeabilities and thicknesses. Calculate the effective permeability of the system if the flow is perpendicular to the layers.

k = 100 md, L = 5

k = 10 md, L = 10

k = 1 md, L = 60


.

.

.

.

Calculate the effective permeability

If we assume that the highest permeability is 1000 instead of 100 (an order of magnitude higher), the effective permeability will be

On the other hand, if we assume that the lowest permeability is 0.1 instead of 1 (an order of magnitude smaller), the effective permeability will be

Notice that the effective permeability is much more sensitive to the smallest value but not as sensitive to the largest value. This is an important characteristic of harmonic average



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Near Wellbore Damage

Earlier there was a discussion on how a small region near the wellbore can alter the productivity of a well

• This damage can also be explained in an alternate equation using a concept of skin factor. The equation can be written as:

0.75

In the equation, C is a constant depending on the units, and S is called a skin factor – which is a dimensionless quantity and represents the near wellbore alteration in permeability.

Physical Meaning of Skin Factor

1 1

0.75

0.75

If we assume that up to a radius of rd, the original permeability k is altered to kd, we can write an equation for effective permeability as:

It is known that the rate calculated using effective permeability or using original permeability with skin factor has to be the same. The

equation can be written as:



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Physical Meaning of Skin Factor

Substituting the value of keff in the previous equation, and with some algebraic manipulation, you can write:

Where we assume that very close to the wellbore, there is an instantaneous pressure drop due to damage.

Thin Skin

The assumption that the damaged zone has a finite thickness.

Thick Skin

This expression allows us to calculate the value of skin factor in terms of damaged zone.



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Example 12

Calculate the skin factor under the current conditions.

If the permeability in the damaged zone is restored to original value, what would be the new rate?

If by using stimulation, the permeability up to a distance of 3 ft [0.91 m] is increased to 100 md, what would be the skin factor? How much would be the rate under these conditions?

A well is currently producing at a rate of 130 bbl/day [20.7 m3/day]. As an operator you suspect that there is a near wellbore damage where the original permeability of 20 md has been altered to 5 md up to a distance of 3 ft [0.91 m]. The drainage radius is 900 ft [274.4 m], and the wellbore radius is 0.35 ft [0.107 m].

A well is currently producing at a rate of 130 bbl/day [20.7 m3/day]. As an operator you suspect that there is a near wellbore damage where the original permeability of 20 md has been altered to 5 md up to a distance of 3 ft [0.91 m]. The drainage radius is 900 ft [274.4 m], and the wellbore radius is 0.35 ft [0.107 m].


Using the equation for skin factor, you can calculate the value as the following:

..

Field Units

..

.

SI Units



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Altering the near wellbore permeability to 20, the skin factor would be zero. Examining the rate equation for the rate in terms of skin factor, write the following:

.

.

Field Units and SI Units


Substituting

. . .

. . /

Field Units

. . .

. .. . /

SI Units



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By restoring the permeability to the original value, there is a significant improvement in the rate.

The negative skin factor implies that the near wellbore alteration has resulted in improved permeability which exceeds the reservoir permeability.

30.35

20100

1 1.71

Field Units

0.910.107

20100

1 1.71

SI Units


Using the same logic, calculate the new rate for negative skin factor as, substituting:

Big improvement in the rate is obtained by changing the permeability over a very small interval.

9000.35 0.75 6.44

9000.35 0.75 1.71

130 327 /

Field Units

9000.35 0.75 6.44

9000.35 0.75 1.71

20.7 52.1 3/

SI Units



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77

Rate Equation Approximation

For a typical vertical well, the value of ⁄ -0.75 can be approximated by a constant 7. Although both the drainage radius and wellbore radius can vary, the ratio is such that the log term is close to 7.

• Assuming this value, write the rate equation for different skin factors as:

The advantage of using this equation is that without knowing reservoir parameters, you can still approximate the rate improvement by altering the near wellbore permeability.


Learning Objectives

Calculate the effective permeability of heterogeneous systems when the layers are either in parallel or in series

Distinguish between effective permeability calculations for linear and radial flows

Describe the concept of skin factor and its influence on the performance of the well

Evaluate the well performance with limited production data



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Deviations from Single Phase Darcy’s Law Equations



Learning Objectives

Evaluate the well performance in the presence of non-Darcy flow using both pressure squared and pseudo-real pressure methods

Evaluate the multi-rate test and generate the inflow performance curve for gas wells

Use the single rate test to predict inflow performance of oil wells producing below bubble point using both Fetkovich and Vogel methods



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Non-Darcy Flow

Darcy law is applicable for

wells when the velocity at which

the fluid is moving is very slow.

If the velocity is higher, the

pressure drop cannot be

calculated using Darcy’s law

equation. Instead we will need to

use Forchheimer’s

equation.

The non-Darcy flow is important

near the well bore and mostly for

gas wells.

For oil wells, non-Darcy flow is rare and need not be

considered.

Gas expands as it approaches the

wellbore.

At lower pressures the

gas will expand, and hence the

velocity will increase.

Non-Darcy Flow

Darcy law is applicable for

wells when the velocity at which

the fluid is moving is very slow.

If the velocity is higher, the

pressure drop cannot be

calculated using Darcy’s law

equation. Instead we will need to

use Forchheimer’s

equation.

The non-Darcy flow is important

near the well bore and mostly for

gas wells.

For oil wells, non-Darcy flow is rare and need not be

considered.

Gas expands as it approaches the

wellbore.

At lower pressures the

gas will expand, and hence the

velocity will increase.

For GAS wells, non-Darcy flow is more of a RULE, whereas

for OIL wells non-Darcy flow is more of an EXCEPTION.



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Forchheimer’s Equation

Forchheimer’s equation is written as:

This equation requires the knowledge of which is called high velocity coefficient.

Firoozabadi and Katz empirical equation provides us with a value of for different reservoirs

Rewrite Forchheimer’s equation

Gas Flow

For gas flow rate

and , for radial flow, 2

Substituting these two equations into the first equation

21

2

In the above equation, you can simplify further by assuming that viscosity can be defined at near wellbore conditions on the right hand side



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or or

Non-Darcy Flow

That equation is written as:

If non-Darcy term is negligible, the equation reduces to:

When non-Darcy flow is important, a quadratic equation is governing the flow. For given pressures, solve the quadratic equation to calculate the rate.

Note that the values of a and b are different for pressure squared and m(p) methods.

This is the same equation we have seen before.

or =

or =



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Example 13

Calculate the inflow performance of the well with and withoutnon-Darcy flow. Use pressure squared method.

A gas well is producing from a reservoir with a reservoir pressure of 3,500 psia [24,132 kPa]. The drainage radius is 1,500 ft [457.3 m], the wellbore radius is 0.4 ft [0.122 m], the permeability is 16 md, and the thickness is 45 ft [13.7 m]. The average viscosity is 0.02 cp[0.00002 Pas], z factor is 0.87, and the reservoir temperature is 180 F [82.6 C]. The perforated interval is 30 ft [9.15 m]. The specific gravity of gas is 0.65. Assume skin factor to be zero. Assume consolidated formation.


2.33 1016 . 8.34 10 ft−1

7.64 1016 . 2.74 10 m−1


2.226 10. .

. .

1.21 10

. . ,

.0.75

1.65 10

. .

1.21 10 2.66

10

First calculate the value of

2.226 10. . .

. . .

4.26 10

. . .

. .

.

.0.75

2.76 10

. . .

. .4.26 10

1.57 10



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Field Units

SI Units

For Calculating Rate without Non-Darcy Flow

1.65 10 1.65 10 4 2.66 10 3500 20002 2.66 10

32.8

2 1.57 102.76 10 2.76 10 4 1.57 10 24132 13789

2 1.57 10929 3/

BHP of 2000 psia [13,789 kPa], the rate is

Field Units

SI Units

For Calculating Rate without Non-Darcy Flow

50.13500 20001.65 10

50.1

24132 137891,420 /

24132 137892.76 10

1,420 3/

For calculating rate without non-Darcy flow, use:

The rate at other BHP’s can be calculated using similar equation.



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SI Units

Field Units

Solution: Example 13 (continued)

The IPR relationship shows large differences with and without non-Darcy flow

If we ignore non-Darcy flow for this well, the difference in the rate prediction can be off by 40%

At low rates, the two curves converge, indicating that non-Darcy flow becomes important at high rates

0

5000

10000

15000

20000

25000

30000

‐ 500 1,000 1,500 2,000 2,500

pwf,kPa

q, Mm3/day

non‐Darcy w/o non‐Darcy

Pwf, kPa

q, Mm3/day

0

1000

2000

3000

4000

‐ 20.0 40.0 60.0 80.0

pwf, psia

q, MMSCF/day

non‐Darcy w/o non‐Darcy

Pwf, psia

q, MMSCF/day



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Well Productivity for Multiphase Cases

If oil produces below bubble point, both free gas and

oil flow in the reservoir

The well is producing under multiphase flow

conditions

As the bottom hole pressure changes,

so does the saturation

surrounding the wellbore; hence,

the productivity of both oil and gas will

be impacted

So far, we have considered well productivity for single phase oil and single phase gas.

What happens if we have oil

producing below bubble point?



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Based on numerical studies and field observations, we can reasonably assume that

where a and b are constants

Substituting in the above equation, integrate to obtain:

7.08 10

ln 0.75 2

Equation for Below Bubble Point

7.08 10

ln 0.75

The relative permeability of oil is changing as the saturation changes. Write the equation for rate under pseudo-steady state as the following:

If single phase oil is present throughout the reservoir, kro

is 1 and the equation will reduce to standard single phase equation.

Equation for Below Bubble Point

Both Fetkovich and Vogel’s equations require that to generate IPR, we will need a single rate and the corresponding BHP.

2

7.08 10

ln 0.75 2

We can also write the equation for maximum flow rate which corresponds to a case when BHP is zero

11

If we assume that b is zero and take ratios of the two rates, we obtain Fetkovich’sequation

1 0.2 0.81 0.2

0.8

There is also another equation called Vogel’s equation which has a slightly different form



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Example 14

What would be the incremental production from this well? • Use both Fetkovich and Vogel methods.

A well is producing at a rate of 72 STB/day [11.5 Sm3/day] at BHP of 1,000 psia [6,895 kPa]. The reservoir pressure is 2,000 psia[13,790 kPa.]. If by installing rod pump, the BHP can be reduced to 100 psia [690 kPa].

Field Units

SI Units

Solution: Example 14 – Fetkovich Method

10001

=72

110002000

= 96STB/day

68951

=11.5

1689513790

= 15.3Sm3/day

Calculate qmax as:



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Field Units

SI Units

Solution: Example 14 – Fetkovich Method

1 1 96STB/day1

=11002000

Therefore,q=96STB/day

457.37.633 10 2

7.633 10 3 15.2 2 17,237

0.00002 0.87 366.7457.30.102

22.42 3/ /

Calculate the rate at 100 psia (690 kPa) as:

The improvement would be 24 STB/day (3.7 Sm3/day)

Field Units

SI Units

Solution: Example 14 – Vogel Method

1000 10001 0.2 0.8

72

1 0.210002000

0.810002000

103STB/day

/day1 0.2 0.8

11.5

1 0.2689513790 0.8

689513790

16.4Sm3/day

Calculate qmax as:



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Field Units

SI Units

Solution: Example 14 – Vogel Method

1 0.2

0.8

1 0.21002000

0.81002000

Therefore,q 102STB/day

1 0.2

0.8

1 0.269013790

0.869013790

Therefore,q 16.2Sm3/day

Calculate the rate at 100 psia (690 kPa) as:

The improvement would be 30 STB/day (4.7 Sm3/day)

Vogel's equation is always going to be a little bit more optimistic than Fetkovich method.



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Back to Work Suggestions

Leverage the skills you’ve learned by discussing the skill module objectives with your supervisor to develop a personalized plan to implement on the job. Some suggestions are provided.

How can you better predict the performance of an existing well?

Identify if the well is vertical, horizontal, or fractured.

What type of flow regime is it producing from?

Is the well producing under single phase or multi‐phase conditions?

If the well is a gaswell, consider non‐Darcy effect in predicting the performance of the well.

Compare the performance of the well with the predicted performance.

Have you made adjustments in the reservoir parameters to match the performance?

Reservoir Flow Properties Fundamentals How can you better predict the

performance of an existing well?

Identify if the well is vertical, horizontal, or fractured.

What type of flow regime is it producing from?

Is the well producing under single phase or multi‐phase conditions?

If the well is a gaswell, consider non‐Darcy effect in predicting the performance of the well.

Compare the performance of the well with the predicted performance.

Have you made adjustments in the reservoir parameters to match the performance?

Back to Work Suggestions

Leverage the skills you’ve learned by discussing the skill module objectives with your supervisor to develop a personalized plan to implement on the job. Some suggestions are provided.

Reservoir Flow Properties Fundamentals What does the history of

well performance tell us?

Do you have sufficient information about the reservoir pressure as a function of time?

Is the well performance consistent with what is expected from that well?

Does the past performance indicate potential skin damage?

What possible solutions you can implement to improve the performance of the well in the future?

What does the history of well performance tell us?

Do you have sufficient information about the reservoir pressure as a function of time?

Is the well performance consistent with what is expected from that well?

Does the past performance indicate potential skin damage?

What possible solutions you can implement to improve the performance of the well in the future?



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Learning Objectives

Evaluate the well performance in the presence of non-Darcy flow using both pressure squared and pseudo-real pressure methods

Evaluate the multi-rate test and generate the inflow performance curve for gas wells

Use the single rate test to predict inflow performance of oil wells producing below bubble point using both Fetkovich and Vogel methods

Applied Reservoir Engineering

This is Reservoir Engineering Core

Reservoir Rock Properties Core

Reservoir Rock Properties Fundamentals

Reservoir Fluid Core

Reservoir Fluid Fundamentals

Reservoir Flow Properties Core


Reservoir Fluid Displacement Core

Reservoir Fluid Displacement Fundamentals

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

Properties Analysis Management




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Darcy’s Law for Linear Flow...

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