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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
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Reservoirs can encounter:
Why This Module is Important
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
Why This Module is Important
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
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Why This Module is Important
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
Why This Module is Important
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
Reservoir Flow Properties Fundamentals
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
Symbol Field Unit SI Unit
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
Reservoir Flow Properties Fundamentals
This section will cover the following learning objectives:
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
Symbol Field Unit SI Unit
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:
Field Units SI Units
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
This section has covered the following learning objectives:
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
Reservoir Flow Properties Fundamentals
This section will cover the following learning objectives:
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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Symbol Field Unit SI Unit
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:
Symbol Field Unit SI Unit
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
Symbol Field Unit SI Unit
k md md
h ft m
t hrs
cp Pa.s
ct psi-1 kPa-1
rw ft m
Fraction Fraction
P psia kPa
m(p) psi2/cp kPa2/Pa.s
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
Symbol Field Unit SI Unit
k md md
h ft m
t hrs
cp Pa.s
ct psi-1 kPa-1
rw ft m
Fraction Fraction
P psia kPa
m(p) psi2/cp kPa2/Pa.s
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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
Symbol Field Unit SI Unit
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
Symbol Field Unit SI Unit
k md md
h ft m
t hrs
cp Pa.s
ct psi-1 kPa-1
rw ft m
Fraction Fraction
P psia kPa
m(p) psi2/cp kPa2/Pa.s
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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].
This section has covered the following learning objectives:
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
Reservoir Flow Properties Fundamentals
This section will cover the following learning objectives:
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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This section has covered the following learning objectives:
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
Reservoir Flow Properties Fundamentals
This section will cover the following learning objectives:
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
Solution: Example 11
.
.
.
.
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].
Solution: Example 12
Using the equation for skin factor, you can calculate the value as the following:
..
Field Units
..
.
SI Units
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Solution: Example 12
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
Solution: Example 12
Substituting
. . .
. . /
Field Units
. . .
. .. . /
SI Units
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Solution: Example 12
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
Solution: Example 12
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.
This section has covered the following learning objectives:
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
Reservoir Flow Properties Fundamentals
This section will cover the following learning objectives:
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.
Field Units SI Units
2.33 1016 . 8.34 10 ft−1
7.64 1016 . 2.74 10 m−1
Solution: Example 13
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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This section has covered the following learning objectives:
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 Flow Properties Fundamentals
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
Reservoir Flow Properties Fundamentals
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