Feketes Course Notes
Feketes Course Notes
Theory/Software Course
Introduction to Well Performance
Analysis
Traditional
- Based on analogy
- Deliverables:
- Production forecast
- Recoverable Reserves under current
conditions
Modern
- Deliverables:
- OGIP / OOIP and Reserves
- Permeability and skin
- Drainage area and shape
- Production optimization screening
- Infill potential
Recommended Approach
4.50
4.00
q = qie − Dit
3.50
fd
3.00
,M
te s
Mc
Slope
2.50
a
sR
Di =
a
G
2.00
1.50
1.00
q
0.50
0.00
2001 2002 2003 2004 2005 2006
Dit
log q = log qi −
5 4.00
q = qi − DiQ
4
3.50
3
2
2.302
Di = Slope
3.00
Gas Rate, MMscfd
Di = 2.302* Slope
G as Rate, MMscfd
2.50
1.0
2.00
7
6
5
1.50
4
3
1.00
2
0.50
0.00
10-1
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50
2001 2002 2003 2004 2005 2006
Gas Cum. Prod., Bscf
The Hyperbolic Decline Curve
Unnamed Well Rate vs. Cumulative Prod.
4.50
4.00
3.50
qi
q=
(1 + bDit )1/ b
3.00
Gas Rate, MMscfd
2.50
Di b
2.00 D= b q
1.50
qi
1.00
0.50
D = f (t )
0.00
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60
Gas Cum. Prod., Bscf
Hyperbolic Exponent “b”
Unnamed Well Rate vs. Cumulative Prod.
4.50
4.00
Mild Hyperbolic – b ~ 0
3.50
3.00
M
,Msfd
c
2.50
sR
a te
a
2.00
G
1.50
1.00
0.50
0.00
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60
Gas Cum. Prod., Bscf
NBU 921-22G R a t e v s . C u m u la t iv e P r o d .
3 .2 0
3 .0 0
2 .8 0
2 .6 0
2 .4 0
Strong Hyperbolic – b ~ 1
2 .2 0
2 .0 0
Gas Rate, MMscfd
1 .8 0
1 .6 0
1 .4 0
1 .2 0
1 .0 0
0 .8 0
0 .6 0
0 .4 0
0 .2 0
0 .0 0
0 .0 0 0 .0 5 0 .1 0 0 .1 5 0 .2 0 0 .2 5 0 .3 0 0 .3 5 0 .4 0 0 .4 5 0 .5 0 0 .5 5 0 .6 0 0 .6 5 0 .7 0 0 .7 5 0 .8 0 0 .8 5 0 .9 0 0 .9 5 1 .0 0 1 .0 5
G a s C u m u la t iv e , B s c f
Analytical Solutions
Definition of Compressibility
pi pi-dp
dV
V V
1 ∂V
c=−
V ∂p
Compressibility Defines Material Balance of
a Closed Oil Reservoir (above bubble point)
Δp = pi - p ΔV = Np
V=N
1 Np
c=
N pi − p
Np
p = pi −
ctN
p = pi − mpssNp
Assumptions: 1. c is constant
2. Bo is constant
Illustration of Pseudo-Steady-State
p1
1
p2
2
p3
pressure
3
pwf1
pwf2 time
Constant Rate q
pwf3
rw Distance re
Steady-State Inflow Equation
pi
p
p − pwf = qbpss
pressure
141.2Bμ q ⎛ re 3 ⎞
bpss = ⎜ ln − ⎟
kh ⎝ rwa 4 ⎠
pwf
rw Distance re
The Two Most Important Equations in
Modern Production Analysis
p = pi − mpssNp
p = pwf + qbpss
Operating Conditions - Simplified
q q
pwf pwf
Constant Flowing Pressure Solution
pi − pwf
mpss
− t
q(t ) = e bpss
bpss
pi − pwf
Np max = = ( pi − pwf ) ctN
mpss
Constant Flowing Pressure Solution – Relate
back to Arps Exponential, Determine N
pi − pwf
qi =
bpss
mpss
Di =
bpss
qi
Np max =
Di
ct ( pi − pwf ) ct ( pi − pwf ) Di
N= =
Np max qi
Constant Rate Solution
y = mx + b
Np 141.2Bμ q ⎛ re 3 ⎞
pi − pwf = + ⎜ ln − ⎟
ctN kh ⎝ rwa 4 ⎠
pi − pwf
1
m= = mpss
cN
Np
Constant Rate Solution – Relate back to
Arps Harmonic
q 1 1
= =
pi − pwf (t ) mpssNp + bpss mpsst + bpss
q
1
q bpss
=
pi − pwf (t ) mpss t + 1
bpss
Plot Constant p and Constant q together
0. 9
Constant rate q/Δp (Harmonic)
0. 8
1
q bpss
=
pi − pwf (t ) mpss t + 1
0. 7
0. 6 bpss
0. 5
0. 4
mpss
0. 2
q(t ) 1 − bpss t
= e
0. 1
pi − pwf bpss
0 5 10 15 20 25 30 35 40 45
Transient Flow
3600
3400
3200
3000
Cross Section 2800
2600
Transient Well Performance = Boundary Dominated
2400 f(k, skin, time) Well Performance =
2200
f(Volume, PI)
ps
2000
Plan View
Radius (Region) of Investigation
3600
3400
3200
3000
Cross Section 2800
2600
2400
kt
2200 rinv =
948φμ c
ps
2000
π kt
Ainv =
948φμ c
Plan View
Transient Equation
q kh 1
=
( pi − pwf ) 141.2 μ B 1 ⎛ 0.0063kt ⎞
ln ⎜ ⎟ + 0.4045 + s
2 ⎝ φμ ct ⎠
q(t)’s compared
1. 6
1. 4
1. 2
0. 6
0. 4
0. 2
0
0 5 10 15 20 25 30 35 40 45
Blending of Transient into
Boundary Dominated Flow
3
2. 5
Complete q(t) consists of:
Transient q(t) from t=0 to tpss
2 Depletion equation from t = tpss and higher
1. 5
0. 5
0 5 10 15 20 25 30 35 40 45
Log-Log plot: Adds a new visual dynamic
Comparison of qD with 1/pD
Cylindrical Reservoir with Vertical Well in Center
1000
0.1
0.01
0.001
0.0001
Constant Pressure Solution Exponential
0.00001
0.000001
0.000001 0.0001 0.01 1 100 10000 1000000 100000000 1E+10 1E+12 1E+14
tD
Type Curves
Type Curve
- Log-log plot
7
6
5
4
Harmonic
q (t ) 1
3
qDd = qDd =
1 + tDd
2 qi
tDd = Dit
qDd
Rate,
10-1
9
7
6
5 Exponential Hyperbolic
4 qDd = e − tDd qDd =
1
3 (1 + btDd )1/ b
2
10-2
2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 2 3 4 5 6 7 8
10-1 1.0 101
tDd Time
Plotting Fetkovich Type Curves- Example
Time (years) Rate (MMscfd) tDd qDd
Well 1 (exponential) Well 1 Well 2 Well 1 Well 2 Well 1 Well 2
0 2.50 10.00 0.00 0.00 1.00 1.00
qi = 2.5 MMscfd 1 2.26 8.19 0.10 0.20 0.90 0.82
Di = 10 % per year 2 2.05 6.70 0.20 0.40 0.82 0.67
3 1.85 5.49 0.30 0.60 0.74 0.55
4 1.68 4.49 0.40 0.80 0.67 0.45
Well 2 (exponential) 5 1.52 3.68 0.50 1.00 0.61 0.37
6 1.37 3.01 0.60 1.20 0.55 0.30
qi = 10 MMscfd
7 1.24 2.47 0.70 1.40 0.50 0.25
Di = 20 % per year 8 1.12 2.02 0.80 1.60 0.45 0.20
9 1.02 1.65 0.90 1.80 0.41 0.17
10 0.92 1.35 1.00 2.00 0.37 0.14
12.00 1.00
10.00
Rate (MMscfd)
8.00
Well 1 Well 1
qDd
6.00
Well 2 Well 2
4.00
2.00
0.00 0.10
0 5 10 15 0.01 0.10 1.00 10.00
Time (years) tDd
Fetkovich Typecurve Matching
In most cases, we don’t know what “qi” and “Di” are ahead of time. Thus, qi and Di
are calculated based on the typecurve match (ie. The typecurve is superimposed on
the data set
NBU 921-22G Fetkovich Typecurve Analysis
q (t )
qi = 1.0
qDd 8
7
q6
5
tDd
Di =
4
3
qDd
Rate,
t 2
10-1
9
8
7
6
5
3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 2
1.0 101
tDd
Time
6
4
3
2
1.09
6
Transient Flow
4
3
2
Rate,
10-29
6
4
3
2 Boundary Dominated Flow
Exponential
2 3 4 5 6 7 9 -3 2 3 4 5 678 -2 2 3 4 5 678 -1 2 3 4 5 678 2 3 4 5 678 1 2 3 4 5 67
10-4 10 10
tDd
10
Time
1.0 10
Dimensionless Variable Definitions (Fetkovich)
141.2q μ B ⎡ ⎛ re ⎞ 1 ⎤
qDd = ⎢ ln ⎜ ⎟− ⎥
kh( pi − pwf ) ⎣ ⎝ rwa ⎠ 2 ⎦
0.00634kt
φμ ctrwa 2
tDd =
⎡
1 ⎡ ⎛ re ⎞ 1 ⎤ ⎛ re ⎞
2
⎤
⎢ ln ⎜ ⎟ − ⎥ ⎢⎜ ⎟ − 1⎥
2 ⎣ ⎝ rwa ⎠ 2 ⎦ ⎢⎣⎝ rwa ⎠ ⎥⎦
Type Curve Matching (Fetkovich)
141.2μ B ⎡ ⎛ re ⎞ 1 ⎤ q
k= ⎢ ln ⎜ ⎟− ⎥
h( pi − pwf ) ⎣ ⎝ rwa ⎠ 2 ⎦ qDd match
0.00634k 1 t ⎛ rw ⎞
rwa = s = ln ⎜ ⎟
φμ ct 1 ⎡ ⎛ re ⎞ 1 ⎤ ⎡⎛ re ⎞ 2 ⎤ tDd ⎝ rwa ⎠
⎢ ln ⎜ ⎟ − ⎥ ⎢⎜ ⎟ − 1⎥
2 ⎣ ⎝ rwa ⎠ 2 ⎦ ⎣⎢⎝ rwa ⎠ ⎦⎥ match
141.2 B 0.00634 q t
re = 2
h( pi − pwf ) φ ct qDd match tDd match
Type Curve Matching - Example
10 Fetkovich Typecurve Analysis
101
8
6
k = f(q/qDd)
4
3 reD = 50 s = f(q/qDd * t/tDd, reD)
2 re = f(q/qDd * t/tDd)
q
1.0
8
6
4 Transient Flow
3
2
Rate,
qDd 10-1
8
6
4 t
3
2
10-2
8
6
4
3
2 Boundary Dominated Flow
Exponential
10-3
2 3 4 5 678 -3 2 3 4 5 678 -2 2 3 4 5 678 -1 2 3 4 5 678 2 3 4 5 678 1 2 3 4 5 6 78
10-4 10 10
tDd
10
Time
1.0 10
What about Variable Rate / Variable Pressure Production?
The Principle of Superposition
Superposition in Time:
q2
q q1
pi − pwf = q1 f (t ) + (q 2 − q1) f (t − t1)
pwf
Effect of (q2-q1)
t1
The Principle of Superposition - Continued
N - Rate History
N
pi − pwf = ∑ (qj − qj − 1) f (t − tj − 1)
j =1
f(t) is the Unit Step Response
Superposition Time
Convert multiple rate history into an equivalent single rate history by re-plotting
data points at their “superposed” times
pi − pwf N
(qj − qj − 1)
=∑ f (t − tj − 1)
qN j =1 qN
The Principle of Superposition – PSS Case
pi − pwf N
(qj − qj − 1)
=∑ f (t − tj − 1)
qN j =1 qN
pi − pwf t 141.2Bμ ⎛ re 3 ⎞
f (t ) = = + ⎜ ln − ⎟
q ctN kh ⎝ rwa 4 ⎠
pi − pwf 1 N
(qj − qj − 1) 141.2Bμ ⎛ re 3 ⎞
qN
=
ctN
∑
j =1 qN
(t − tj − 1) + ⎜ ln − ⎟
kh ⎝ rwa 4 ⎠
pi − pwf 1 Np 141.2Bμ ⎛ re 3 ⎞
= + ⎜ ln − ⎟
qN ctN qN kh ⎝ rwa 4 ⎠
Q
Q
actual material
time (t) balance = Q/q
time (tc)
Features of Material Balance Time
1000 1.2
Very early time radial flow
Ratio (qD to 1/pD) ~ 90%
100
1
0.97
10
Ratio 1/pD to q
qD and 1/p
0.1
Beginning of "semi-log" radial flow (tD=25) 0.6
Ratio (qD to 1/pD) ~ 97%
0.01
0.001 0.4
0.0001
Constant Pressure Solution qD
0.2
Corrected to Harmonic
0.00001
0.000001 0
0.000001 0.0001 0.01 1 100 10000 1000000 100000000 1E+10 1E+12 1E+14
tD
Corrections for Gas Reservoirs
Corrections Required for Gas Reservoirs
qt 141.2qBoμ ⎛ re 3 ⎞
pi − pwf = + ⎜ ln − ⎟
coN kh ⎝ rwa 4 ⎠
Darcy’s Law Correction for Gas Reservoirs
Solution: Pseudo-Pressure
p
pdp
pp = 2 ∫
0
μZ
Depletion Correction for Gas Reservoirs
0.012
0.01
0.008
Compressibility (1/psi)
0.006
1
cg ≈
0.004
p
0.002
0
0 1000 2000 3000 4000 5000 6000
Pressure (psi)
Depletion Correction for Gas Reservoirs:
Pseudo-Time
Solution: Pseudo-Time
ta = (μcg )i ∫
dt
t
0 μc g
μ , c g → Evaluated
pressure
at average reservoir
Pseudo-pressure Pseudo-time
2 pi 1.417e6 * Tq ⎛ re 3 ⎞
Δpp = ppi − ppwf = qta + ⎜ ln − ⎟
( μcgZ )iGi kh ⎝ rwa 4 ⎠
1 t
tc = ∫ qdt
q 0
1 ta
tca = ∫ qdta =
(μcg )i t qdt
q 0 q ∫0 μ c g
Corrected Material Balance Pseudo-time
( μct )i t q (t )
tca =
q ∫0 μc t [1 − cf ( pi − p)] dt
Where, Evaluated at average
reservoir pressure
c t = cf + s oco + s wcw + s gcg
Practice
- Traditional
- Blasingame
- Agarwal – Gardner and NPI
- Flowing p/z analysis
- Transient
- Models and History Matcning
Notes About Drive Mechanism and b Value (from
Arps and Fetkovich)
4
EUR = 9.5 bcf
Gas Rate, MMscfd
0
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00 6.50 7.00 7.50 8.00 8.50 9.00 9.50 10.00 10.50
Gas Cum. Prod., Bscf
Example 1 (cont’d)
Flowing Pressure and Rate vs Cumulative Production
Rates
5 1200
4.5
1000
4
1
200
0.5
0 0
0 1 2 3 4 5 6 7 8 9 10
Cumulative Production (bcf)
Example 2: Decline Underpredicts Reserves
Unnamed Well Rate vs. Cumulative Prod.
8.50
8.00
7.50
7.00
6.50
6.00
5.50
EUR = 3.0 bcf
5.00
Gas Rate, MMscfd
4.50
4.00
3.50
3.00
2.50
2.00
1.50
1.00
0.50
0.00
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00 3.10 3.20
Gas Cum. Prod., Bscf
Example 2 (cont’d)
Unnamed Well Flowing Material Balance
0.085
Legend
Decline FMB
0.080
0.075
0.070
0.065
0.060
OGIP = 24 bcf
Normalized Rate, MMscfd/(106psi2/cP)
0.055
0.050
0.045
0.040
0.035
0.030
0.025
0.020
0.015
0.010
0.000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Normalized Cumulative Production, Bscf
Example 2 (cont’d)
Unnamed Well Data Chart
18 Legend 1300
Pressure
17 Actual Gas Data
1200
16
15 1100
14
1000
13
900
12
11
Operating conditions: Low drawdown 800
Pressure, psi
Gas, MMscfd
10
700
9
600
8
7 500
6
400
5
4 300
3
200
2
100
1
0 0
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380 400 420 440 460 480 500 520 540 560 580 600 620 640 660 680 700 720
Time, days
Example 3 – Illustration of Non-Uniqueness
10
1
Gas Rate (MMscfd)
0.1
0.01
Dec-00 May-06 Nov-11 May-17 Oct-22 Apr-28 Oct-33
Time
Blasingame Typecurve Analysis
Fetkovich Blasingame
log(q) log(q/Δp)
log(qDd) log(qDd)
log(t) log(tca)
log(tDd) log(tDd)
t DA
⎛ q ⎞ 1 q
t
c
⎛ q ⎞ t ca
∫ qDd (t )dt
1
⎟ = ∫ ⎜ ⎟ = 1 q
Rate Integral q Ddi =
t DA
⎜
⎝ ΔP ⎠i tc 0 ΔP
dt
⎜ ΔP ⎟ t ∫ ΔP dt
0 ⎝ p ⎠ i ca 0 p
Q
Q
actual actual
time time
Rate Integral: Like a Cumulative Average
t1 t2
Effective way to remove noise
Rate Integral: Definition
⎛ q ⎞ 1 q
tc
⎜⎜ ⎟⎟ = ∫ dt
⎝ Δp ⎠i tc 0 Δp
Typecurve Interpretation Aids: Integrals, Derivatives
Rate Integral
Rate (Normalized)
Oil:
q ⎛ 141 .2 βμ ⎞⎛⎜ ⎛⎜ re ⎞ 1⎞
q Dd = ⎜ ⎟⎜ ln ⎜ ⎟⎟ − ⎟
Δ p ⎝ kh ⎠⎝ ⎝ rwa ⎟
⎠ match 2 ⎠
⎛ q ⎞
⎜ Δp ⎟ ⎛ 141 .2 βμ ⎞⎛⎜ ⎛⎜ re ⎞ 1⎞
k =⎜ ⎟ ⎜ ⎟⎜ ln ⎜ ⎟⎟ − ⎟
⎜ q Dd ⎟ ⎝ ⎠⎝ ⎝ rwa ⎟
h ⎠ match 2 ⎠
⎝ ⎠ match
0 .006328 kt c
t Dd =
⎛
2 ⎜ ⎛ re ⎞
2
⎞⎛ ⎛ r ⎞ 1⎞
− 1 ⎟⎜ ln ⎜⎜ e
1
φμ c t rwa ⎜⎜ ⎟⎟ ⎟⎟ − ⎟
2 ⎜ ⎝ rwa ⎠ ⎟⎜ ⎝ rwa ⎟
⎠ match 2 ⎠
⎝ match ⎠⎝
⎛ t ⎞
= ⎜ c ⎟ 0 .006328 k
r
wa ⎜t ⎟ ⎛ 2 ⎞⎛ ⎛ ⎞
⎝ Dd ⎠ match 1 ⎜ ⎛⎜ re ⎞⎟ ⎟⎜ r ⎞ 1⎟
φμ c ⎜ − 1 ⎟⎜ ln ⎜ e ⎟ − ⎟
t 2 ⎜⎜ r ⎟ ⎟⎜ ⎜⎝ rwa ⎟ 2⎟
⎝ ⎝ wa ⎠ match
⎠⎝ ⎠ match ⎠
⎛ r ⎞
s = ln ⎜⎜ w ⎟⎟
⎝ rwa ⎠
Blasingame Typecurve Analysis-
Boundary Dominated Calculations- Oil
qDd =
(q / Δp ) and tDd = Ditc
(q / Δp )i
Recall the Fetkovich definition for the harmonic typecurve and the PSS equation for oil in
harmonic form:
Definition of Harmonic PSS equation for oil in
typecurve 1 harmonic form, using
material balance time
1 q b
q Dd = and =
1 + tDd Δp 1
tc + 1
ctNb
From the above equations:
⎛
⎜
⎜
q ⎞⎟⎟
⎜⎜
Δp ⎟⎟⎠ i ⎛ q ⎞ 1
q
= ⎝
where ⎜ ⎟ = , and Di =
1
Δp 1 + Ditc ⎜ Δp ⎟ b ctNb
⎝ ⎠i
Blasingame Typecurve Analysis- Boundary
Dominated Calculations- Oil (cont’d)
1
N=
ctDib
Now, substitute the definitions of qDd and tDd back into the above equation:
1 1 ⎡ tc ⎤ ⎡ (q / Δp ) ⎤
N= = ⎢ ⎥⎢ ⎥
⎡ tDd ⎤ ⎡ qDd ⎤ ct ⎣ tDd ⎦ ⎣ qDd ⎦
ct ⎢ ⎥ ⎢ ⎥
⎣ tc ⎦ ⎣ (q / Δp ) ⎦
q Dd =
(q / Δpp ) and tDd = Ditca
(q / Δpp )i
Recall the Fetkovich definition for the harmonic typecurve and the PSS equation for gas
in harmonic form:
Definition of Harmonic PSS equation for gas in
typecurve harmonic form, using
1 material balance pseudo-
time
1 q b
qDd = and =
1 + tDd Δpp 2 pi
tca + 1
(Zμct )iGib
From the above equations:
⎛
⎜
⎜
q ⎞⎟⎟
⎜⎜
Δp ⎟⎟⎠ i ⎛ q ⎞ 1
q
= ⎝
where ⎜ ⎟ = , and Di =
2 pi
Δp 1 + Ditc ⎜ Δpp ⎟ b
⎝ ⎠i
(Zμct )iGib
Blasingame Typecurve Analysis- Boundary
Dominated Calculations- Gas
2 pi
Gi =
Di (Zμct )ib
Now, substitute the definitions of qDd and tDd back into the above equation:
2 pi 2 pi ⎡ tca ⎤ ⎡ (q / Δpp ) ⎤
Gi = =
⎛ tDd ⎞ ⎛ ⎞
(Zμct )i ⎢⎣ tDd ⎥⎦ ⎢⎣ qDd ⎥⎦
⎜ ⎟(Zμct )i ⎜⎜
⎜ q Dd ⎟
⎟
⎝ ( q / Δpp ) ⎠
⎟⎟
⎝ tca ⎠ ⎜
Agarwal and Gardner Rate vs. Time typecurves are the same as
conventional drawdown typecurves, but are inverted and plotted in
tDA (time based on area) format.
qD vs tDA
1/pD(der) = ( t(dpD/dt) ) -1
Agarwal-Gardner - Rate vs. Time typecurves
Rate Integral-
Derivative
Inv. Pressure
Integral-
Derivative
141.2 μB q (t )
qD =
kh pi − pwf (t )
1 Q 1 pi − p
QDA = qD * tDA = or alternatively
2π ctN ( pi − pwf ) 2π pi − pwf
Agarwal-Gardner - Rate vs. Cumulative typecurves
1.417e6 * T q (t )
qD =
kh ψ i −ψ wf (t )
1 2qtca 1 ψ i −ψ
QDA = qD * tDA = or alternatively
2π (ctμZ )iGi (ψ i −ψ wf ) 2π ψ i − ψ wf
Agarwal-Gardner - Rate vs. Cumulative typecurves
⎛ ΔP ⎞ ΔP ⎛ ΔPp ⎞ 1 ca ΔPp
t DA tc t
∫ Pp (t )dt
1 1
Pressure Integral PDi =
t DA 0
⎜⎜ ⎟⎟ =
⎝ q ⎠i tc
∫0 q dt ⎜⎜
⎝ q
⎟⎟ =
⎠i t ca
∫0
q
dt
dPDi ⎛ ΔP ⎞ ⎛ ΔPp ⎞
PDid = t DA d ⎜⎜ ⎟⎟ t ca d ⎜⎜ ⎟⎟
Pressure Integral - dt DA ⎛ ΔP ⎞ ⎝ q ⎠i ⎛ ΔPp ⎞ ⎝ q ⎠i
Derivative ⎜⎜ ⎟⎟ = tc ⎜⎜ ⎟⎟ =
⎝ q ⎠ id dt c ⎝ q ⎠ id dt ca
NPI (Normalized Pressure
Integral): Diagnostics
Transient
Normalized Pressure
Typecruve
Integral - Derivative
Typecurve
Boundary
Dominated
NPI (Normalized Pressure Integral): Calculation
of Parameters- Oil
Oil - Radial
khΔP 0.00634kt c
PD = t DA =
141.2qβμ πφμC t re2
⎛ ⎞
⎜ ⎟
141.2 βμ ⎜ PD ⎟
k=
h ⎜ ΔP ⎟
⎜ ⎟
⎝ q ⎠ match
0.00634k ⎛ tc ⎞
re = ⎜ ⎟
πφμCt ⎜⎝ t DA ⎟⎠ match
re ⎛r ⎞
rwq = S = ln⎜⎜ w ⎟⎟
⎛ re ⎞ ⎝ rwa ⎠
⎜⎜ ⎟⎟
⎝ rwa ⎠ match
⎛ ⎞
⎜ ⎟
0.00634 ⎛ 141.2 S 0 ⎞⎜ PD ⎟ ⎛ tc ⎞
N= ⎜ ⎟ ⎜⎜ ⎟⎟ (MBBIS)
⎜
C t ⎝ 5.615 * 1000 ⎠ ΔP ⎟ ⎝ DA ⎠ match
t
⎜ ⎟
⎝ q ⎠ match
NPI (Normalized Pressure Integral):
Calculation of Parameters- Gas
Gas – Radial
khΔPp 0.00634kt ca
PD = t DA =
1.417Ε6Tq πφμ i C ti re2
⎛ ⎞
⎜ ⎟
1.417Ε6T ⎜ PD ⎟
k= ⎜ ΔPp ⎟
h
⎜⎜ ⎟⎟
⎝ q ⎠ match
0.00634k ⎛ tca ⎞
re = ⎜ ⎟
πφμ i Cti ⎜⎝ t DA ⎟⎠ match
re ⎛r ⎞
rwa = S = ln⎜⎜ w ⎟⎟
⎛ re ⎞ ⎝ rwa ⎠
⎜⎜ ⎟⎟
⎝ rwa ⎠ match
⎛ ⎞
⎜ ⎟
(0.00634)(1.417Ε6)S g PiTsc ⎛ t ca ⎞ ⎜ PD ⎟
G= ⎜⎜ ⎟⎟ ⎜ ⎟ * 10 9 (bcf)
μ i cti z i Psc ⎝ t DA ⎠ match ⎜ ΔPp ⎟⎟
⎜ q
⎝ ⎠ match
Flowing p/z Method for Gas – Constant Rate
Measured at well
Gi
during flow
Gp
Graphical Flowing p/z Method for Gas –
Variable Rate
pi
zi
Graphical Method
Doesn’t Work!
pwf
zwf
Gi ?
Measured at well
during flow
Gp
Flowing p/z Method for Gas – Variable Rate
pi
zi Pressure loss due to flow
in reservoir is NOT
constant
pwf
zwf p ⎛ p⎞
= ⎜ ⎟ + qbpss
z ⎝ z ⎠ wf
Unknown
Gi
Measured at well
during flow
Gp
Variable Rate p/z – Procedure (1)
Unnamed Well Flowing Material Balance
Legend 550
Static P/Z*
P/Z Line
Flowing Pressure
500
450
Step 1: Estimate OGIP and plot
a straight line from pi/zi to OGIP. 400
250
200
150
100
0
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70
Cumulative Production, Bscf
Variable Rate p/z – Procedure (2)
Unnamed Well Flowing Material Balance
Legend
550
Static P/Z*
4.40 P/Z Line
Flowing Pressure
500
Productivity Index
4.00
⎛ p⎞ ⎛ p⎞
1.20 150
0.80 100
0.00 0
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70
Cumulative Production, Bscf
Variable Rate p/z – Procedure (3)
Unnamed Well Flowing Material Balance
Legend
550
Static P/Z*
4.40 P/Z Line
Flowing Pressure
500
Productivity Index
4.00
Step 3: 1/bpss should tend 450
3.60
towards a flat line. Iterate on
OGIP estimates until this
Productivity Index, MMscfd/(106psi2/cP)
400
3.20
happens
300
2.40
250
2.00
200
1.60
1.20 150
0.80 100
0.40 50
Original Gas In Place
0.00 0
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70
Cumulative Production, Bscf
Variable Rate p/z – Procedure (4)
Unnamed Well Flowing Material Balance
Legend
550
Static P/Z*
4.40 P/Z Line
Flowing P/Z*
500
Flowing Pressure
4.00 Productivity Index
Step 4: Plot p/z points on the p/z 450
3.60
line using the following formula:
Productivity Index, MMscfd/(106psi2/cP)
400
3.20 ⎛ p⎞ ⎛ p⎞
⎜ ⎟ = ⎜ ⎟ + qbpss
2.40
“Fine tune” the OGIP estimate 300
250
2.00
200
1.60
1.20 150
0.80 100
1/bpss
0.40 50
Original Gas In Place
0.00 0
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70
Cumulative Production, Bscf
Transient (tD format) Typecurves
log(qD) log(qDd)
log(tD) log(tDd)
Transient (tD format) Typecurves: Definitions
−1 −1 −1
⎡ dP ⎤ ⎡ ⎛ ΔP ⎞ ⎤ ⎡ ⎛ ΔPp ⎞ ⎤
Inverse Presssure 1 / PDid = ⎢t DA Di ⎥ ⎢ d ⎜⎜ ⎟⎟ ⎥ ⎢ tca d ⎜⎜ ⎟⎟ ⎥
⎣ dt DA ⎦ ⎛ ΔP ⎞ ⎝ q ⎠i ⎥ ⎛ ΔPp ⎞ ⎝ q ⎠i ⎥
Integral - Derivative
Inv⎜⎜ ⎟⎟ = ⎢tc Inv⎜⎜ ⎟⎟ =⎢
⎝ q ⎠id ⎢ dtc ⎥ ⎝ q ⎠id ⎢ dtca ⎥
⎢ ⎥ ⎢ ⎥
⎣⎢ ⎦⎥ ⎢⎣ ⎥⎦
Transient (tD format) Typecurves:
Diagnostics (Radial Model)
Normalized Rate
Typecurve
Transient (tD format) Typecurves:
Finite Conductivity Fracture Model
Increasing Fracture
Conductivity (FCD stems)
Increasing Reservoir
Size (xe/xf stems)
Transient (tD format) Typecurves:
Calculations (Radial Model)
O il W e lls : Gas Wells:
U s in g th e d e fin itio n o f q D ,
For gas wells, qD is defined as follows:
141 . 2 qB μ
qD =
kh ( p i − p wf )
1.417 E 6TR q
qD =
kh Δpp
p e rm e a b ility is c a lc u la te d a s fo llo w s :
141 . 2 B μ ⎛ q / Δ p ⎞
k = ⎜⎜ q D ⎟⎟ The permeability is calculated from above, as follows:
h ⎝ ⎠ match
r wa =
0 . 00634 ⎛ 141 . 2 B ⎞ ⎛ q / Δ p
⎟ ⎜⎜
⎞
⎟⎟
⎛ tc ⎞ 0.00634 ⎛ 1.417 E 6TR ⎞⎛ tca ⎞ ⎛ q/Δpp ⎞
φct
⎜
⎝ ⎠ ⎝ qD ⎠
⎜ ⎟
⎝ t D ⎠ match rwa = ⎜ ⎟⎜ ⎟ ⎜⎜ ⎟⎟
h match
φμicti ⎝ h ⎠⎝ tD ⎠ match ⎝ qD ⎠ match
S k in is c a lc u la te d a s fo llo w s :
Skin is calculated as follows:
⎛ rw ⎞
s = ln ⎜ ⎟
⎝ r wa ⎠
⎛ rw ⎞
s = ln⎜ ⎟
⎝ rwa ⎠
Modeling and History Matching
1. Pressure Constrained System:
Constraint Signal
(Input) (Output)
Constraint Signal
(Input) (Output)
Models - Horizontal
Models - Radial
Rectangular reservoir with a horizontal well located anywhere inside.
Rectangular reservoir with a vertical well located anywhere inside.
Models - Fracture
Rectangular reservoir with a vertical infinite conductivity fracture located anywhere inside.
A Systematic and Comprehensive
Method for Analysis
Modern Production Analysis Methodology
24 5.00 1400
1300
22
4.50
1200
20
4.00 1100
18
1000
Liquid Rates , bbl/d
3.50
Pressure , psi
16
G as , MMcfd
900
14 800
3.00
12 700
2.50 600
10
500
8 2.00
400
6
1.50 300
4
200
1.00
2 100
0 0.50 0
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380 400 420 440 460 480 500 520 540
Tim e, days
“Face Value” Analysis of Data
OGIP = 90 bcf
Go Back: Diagnostics
Unnam ed Well Data Chart
28
Legend 1600
5.50
26 Pressure 1500
Actual Gas Data
24 5.00 1400
1300
22
4.50
1200
20
4.00 1100
18
1000
Liquid Rates , bbl/d
3.50
Pressure , psi
16
G as , MMcfd
900
Unnam e d We ll Data Chart
14 Legend
800
3.00 Pre s s ure
Actual Gas Data
12 700
2.50 600
10
500
8 2.00
Pressures are not 400
6
1.50 representative of bh 300
4
deliverability 200
1.00
2 100
0 0.50 0
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380 400 420 440 460 480 500 520 540
Tim e, days
Correct Data Used
UnnamedWell DataChart
7400
Legend
6.00 5.50
Pressure 7200
Actual Gas Data
5.50 Oil Production
5.00 7000
Water Production
5.00 6800
4.50
4.50 6600
4.00
4.00 6400
Liquid Rates , bbl/d
3.50 6200
Pressure , psi
Gas , MMcfd
3.50
6000
3.00 3.00
5800
2.50
2.50
5600
2.00
2.00 5400
OGIP = 19 bcf
1.50
5200
1.50
1.00
5000
1.00
0.50 4800
Radial Model
Blasingam e Typecurve Match
10-7
8
5
3
2
10-8
Transient
8
5
Base Model:
3
2
10-10
5
8
- Vertical Well in Center of Circle
3
2
- Homogeneous, Single Layer
10-11
8
5
3
2
4 56 8 -1 2 3 45 79 2 3 45 7 9 1 2 3 4 56 8 2 2 3 4 56 8 3 2 3 4 56 8 4 2 3 4 56 8 5 2 3 4 56 8 6 2 3 45 7 2 3 45 7
10 1.0 10 10
tDd
10 10 10 10 107
Diagnostics using Typecurves
Radial Model
Blasingam e Typecurve Match
10-7
8
5
3
2 Reservoir With
10-8
8
Pressure Support
5
qDd 32
Dual
10-9 Depl
e
Syst tion
8
5 em
3
2
Infi
n
10-10 Pre ite A
8 ssu ctin
re S g
Vo
5 upp
ort
lu
3
m
et
2
ric
10-11
5
8
Leaky Reservoir
3 (interference)
2
4 56 8 -1 2 3 45 79 2 3 45 7 9 1 2 3 4 56 8 2 2 3 4 56 8 3 2 3 4 56 8 4 2 3 4 56 8 5 2 3 4 56 8 6 2 3 45 7 2 3 45 7
10 1.0 10 10
tDd
10 10 10 10 107
Diagnostics using Typecurves
Productivity Diagnostics
Radial Model
Blasingam e Typecurve Match
10-7
8
Increasing Damage (difficult to identify)
5
3
2
10-8
8
5
Productivity
qDd 3
2 Shifts (workover,
10-9
8 unreported
5
tubing change)
3 Well Cleaning Up
2
10-10
8 Liquid Loading
5
3
2
10-11
8
5
3
2
4 56 8 -1 2 3 45 79 2 3 45 7 9 1 2 3 4 56 8 2 2 3 4 56 8 3 2 3 4 56 8 4 2 3 4 56 8 5 2 3 4 56 8 6 2 3 45 7 2 3 45 7
10 1.0 10 10
tDd
10 10 10 10 107
Diagnostics using Typecurves
10-9
8
Radial Flow
5
3
2
10-10
8
5
3
2
10-11
8
5
3
2
4 56 8 -1 2 3 45 79 2 3 45 7 9 1 2 3 4 56 8 2 2 3 4 56 8 3 2 3 4 56 8 4 2 3 4 56 8 5 2 3 4 56 8 6 2 3 45 7 2 3 45 7
tDd
10 1.0 10 10 10 10 10 10 107
Diagnostics using Typecurves
10-7
8
Δp in reservoir is too low
5
3
-Tubing size too small ?
2 - Initial pressure too low ?
10-8 - Wellbore correlations
8
5
overestimate pressure loss ?
qDd 3
2
10-9
8
5
Δp in reservoir is too high
3
2 -Tubing size too large ?
10-10
- Initial pressure too high ?
8
- Wellbore correlations
5
3
underestimate pressure loss ?
2
10-11
8
5
3
2
4 56 8 -1 2 3 45 79 2 3 45 7 9 1 2 3 4 56 8 2 2 3 4 56 8 3 2 3 4 56 8 4 2 3 4 56 8 5 2 3 4 56 8 6 2 3 45 7 2 3 45 7
10 1.0 10 10
tDd
10 10 10 10 107
Selected Topics and Examples
Tight Gas
Industry Migration to Tight Gas Reservoirs
Production Analysis – Tight Gas versus Conventional Gas
1.00E-05
1.00E-07 1/2
qDd
1.00E-08
1.00E-09 1
Linear flow
1.00E-10
dominated Limited, bounded
drainage area
1.00E-11
1.00E-05 1.00E-04 1.00E-03 1.00E-02 1.00E-01 1.00E+00 1.00E+01 1.00E+02 1.00E+03 1.00E+04
tDd
Tight Gas Model 1
1800 psi
Pi = 2000 psi
Pi = 1500 psi
Infinite Acting System
Tight Gas Type Curves
1.00E-05
1.00E-06
1.00E-07 1/2
qDd
1.00E-08
1.00E-09
1.00E-10
1.00E-11
1.00E-05 1.00E-04 1.00E-03 1.00E-02 1.00E-01 1.00E+00 1.00E+01 1.00E+02 1.00E+03 1.00E+04
tDd
Example#1 – Infinite Acting System
10 Agarwal Gardner Rate vs Time Typecurve Analysis
10 Agarwal Gardner Rate vs Time Typecurve Analysis 2
2
102
102
6
6 4
4 3
3 2
2
101
101 7
7 5
5
3
3 2
2
Normalized Rate
Normalized Rate
1.09
1.09
6
6 4
4 3
3 2
2
10-1
10-1 7
7 5
5
3
3
2
2
10-2
10-2 7
7 5
5
3
3
2
2 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 -2 2 3 4 5 6 78 -1 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 -2 2 3 4 5 6 78 -1 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 10-5 10-4 10-3 10 10 1.0 101 102
10-5 10-4 10-3 10 10 1.0 101 102
Material Balance Pseudo Time
Material Balance Pseudo Time
k = 0.08 md k = 0.08 md
xf = 53 ft xf = 53 ft
OGIP = 10 bcf Minimum OGIP = 2.6 bcf
Tight Gas Model 2
1.00E-05
1.00E-06
1.00E-07 1/2
qDd
1.00E-08
1.00E-11
1.00E-05 1.00E-04 1.00E-03 1.00E-02 1.00E-01 1.00E+00 1.00E+01 1.00E+02 1.00E+03 1.00E+04
tDd
Normalized Rate
1.0
7
8 5
3
2
10-1
7 7
5
3
2
.
6 10-2
2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
10-3 10-2 10-1 1.0 101 102
Material Balance Pseudo Time
O G IP (b c f)
5 35 120%
30
100%
4 25
80%
Frequency
20
3 60%
15
40%
10
2
20%
5
0 0%
1 10 20 30 40 50 60 70 80 90 100 More
Frequency Cumulative %
0
0 100 200 300 400 500 600
xf (feet)
Tight Gas Model 3
kx
ky
Infinite Systems versus Linear Flow Systems
Establish
permeability and xf
independently
1.00E-05
1.00E-06
1.00E-07 1/2
qDd
1.00E-08
1.00E-11
1.00E-05 1.00E-04 1.00E-03 1.00E-02 1.00E-01 1.00E+00 1.00E+01 1.00E+02 1.00E+03 1.00E+04
tDd
5
4
3
k = 1.1 md
xf = 511 ft
2 ye = 5,500 ft
yw = 2,900 ft
10-7
ye
7
yw
4
2
2xf
10-8
9
7
2 3 4 5 6 7 89 1 2 3 4 5 6 7 89 2 2 3 4 5 6 7 8
10 10 103
More Examples
Example #3- Multiple Layers
Blasingame Typecurve Analysis
3
Multi Layer Model
Well Blasingame Typecurve Match
10-8
2
7
5
4
1.0 3
8
2
Normalized Rate
5
10-9
4 8
6
3
4
3
2
10-1
10-10
9
2 3 4 5 6 7 89 1 2 3 4 5 6 7 89 2 2 3 4 5 6 7 89 3 2 3 4 5 6 78
1.0 10 10 10 104
3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9
10-1 1.0
Material Balance Pseudo Time
- Blasingame typecurve match, using Fracture - Three-Layer Model (one layer with very low
Model permeability) used, late-time match improved
- Pressure support indicated
Example #4- Shale Gas
Well Agarwal Gardner Rate vs Time Typecurve Analysis
5
4
- Multi-stage fractures, horizontal well
3
- Analyzed as a vertical well in a circle
2
1.0
7
6
Normalized Rate
10-1
9
7
6 k = 0.02 md
5
4
s = -4
3
OGIP = 4.5 bcf
6 7 8 9 -3 2 3 4 5 6 7 8 9 -2 2 3 4 5 6 7 8 9 -1 2 3 4 5 6 7 8 9
10 10 10 1.0
Material Balance Pseudo Time
Tight Gas: Assessing Reserve
Potential – Recovery Plots
¾ Objectives
10
9 1 md reservoir, unfractured
8 (~10 bcf / section)
7 100% Recovery
6
EUR (bcf)
0
0 1 2 3 4 5 6 7 8 9 10
Original Gas in Place (bcf)
Typical Recovery Profile
Recovery Curves for k = 1 md
10
9
1 md reservoir, unfractured
8 (~10 bcf / section)
7 100% Recovery
6
EUR (bcf)
3
Actual EUR (qab = 0.05 MMscfd)
2
0
0 1 2 3 4 5 6 7 8 9 10
Original Gas in Place (bcf)
10
9
1 md reservoir, unfractured
8 (~10 bcf / section)
7 100% Recovery
6
30 Year Limited
EUR (bcf)
3
Actual EUR (qab = 0.05 MMscfd)
2
0
0 1 2 3 4 5 6 7 8 9 10
Original Gas in Place (bcf)
10
9
1 md reservoir, unfractured
8 (~10 bcf / section)
7 100% Recovery
6
30 Year Limited
EUR (bcf)
5
20 Year Limited
4
3
Actual EUR (qab = 0.05 MMscfd)
2
0
0 1 2 3 4 5 6 7 8 9 10
Original Gas in Place (bcf)
10
0
0 1 2 3 4 5 6 7 8 9 10
Original Gas in Place (bcf)
10
4
30 Year
3
0
0 1 2 3 4 5 6 7 8 9 10
Original Gas in Place (bcf)
10
4
30 Year
3 20 Year
2
0
0 1 2 3 4 5 6 7 8 9 10
Original Gas in Place (bcf)
10
4
30 Year
3 20 Year
2
0
0 1 2 3 4 5 6 7 8 9 10
Original Gas in Place (bcf)
10-8
5
4
Sqrt k X xf = 155
3
Min OGIP = 4.2 bcf
2
10-9
9
7
2 3 4 5 6 7 89 2 3 4 5 6 7 89 1 2 3 4 5 6 7 89 2 2 3 4 5 6 78
1.0 10 10 103
Example – South Texas, Deep Gas Well
Recovery Plot - Linear System
0
0 100 200 300 400 500 600
ROI (acres)
Water Drive Models
Water Drive (Aquifer) Models:
Models for reservoirs under the influence of active water encroachment can
be categorized as follows:
Advantages:
Disadvantages:
- Does not provide a full time solution (transient effects are ignored)
- Does not work well for infinite acting or very low mobility aquifers
Water Drive (Aquifer) Models:
Pseudo Steady-State Model- Equations
We =
Wei
( pi-p ) ⎛⎜1 − e − Jpit /Wei ⎞⎟ Initial encroachable water
pi ⎝ ⎠
Reservoir boundary pressure
The above equation applies to the water influx due to a constant pressure difference
between aquifer and reservoir. In practice, the reservoir pressure “p” will be declining
with time. Thus, the equation must be discretized as follows:
ΔW e n =
Wei
pi
(
pa n−1- p n )⎛⎜⎝1 − e Jp t /W
− i ei ⎞⎟
⎠
(1)
The average aquifer pressure at the previous timestep (n-1) is evaluated explicitly, as follows:
⎛ n −1
⎞
⎜ ∑ ΔWej ⎟
= pi ⎜⎜1 − ⎟
j =1
pa n−1
Wei ⎟
⎜⎜ ⎟⎟
⎝ ⎠
Water Drive (Aquifer) Models:
Pseudo Steady-State Model- Equations
Now, we have one equation with two unknowns (water influx “We” and reservoir boundary
pressure “p”)
But there is another equation that relates the average reservoir pressure to the amount of
water influx: the material balance equation for a gas reservoir under water drive.
Cumulative Production
-1
p pi ⎛ Gp ⎞ ⎛ WeBi ⎞
= ⎜1 − ⎟ ⎜1 − ⎟ FVF at initial conditions
z zi ⎝ Gi ⎠ ⎝ Gi ⎠ Gas-in-place
As with the water influx equation, the material balance equation can be discretized in time:
-1
⎛ p⎞ pi ⎛ Gp n ⎞ ⎛ W e n Bi ⎞
⎜ ⎟ = ⎜1 − ⎟ ⎜1 − ⎟ (2)
⎝ ⎠n
z z i ⎝ Gi ⎠ ⎝ Gi ⎠
Equations 1 and 2 are now solved simultaneously at each timestep, to obtain a discretized
reservoir pressure and water influx profile through time.
Water Drive (Aquifer) Models:
Transient Models
Transient models use the full solution to the hydraulic DIFFUSIVITY
EQUATION to model rates and pressures.
Advantages:
Disadvantages:
1. Reservoir fluid-in-place
2. Aquifer mobility
Reservoir Aquifer
6
20 12000 5
4
18 11000
3
10000 2
16
9000
14 1.0
8000 8
Pressure, psi
Gas, MMscfd
12 6
Normalized Rate
7000 5
4
10 6000
3
5000 2
8
4000
6
condensate reservoir
6
2000 5
4
2
1000
3
0 0 2
Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct
2002 2003
10-2
2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
10-3 10-2 10-1 1.0 101 102
Material Balance Pseudo Time
Example F Agarwal Gardner Rate vs Time Typecurve Analysis Example F Blasingame Typecurve Analysis
101
8
1.0
8
Transient Water Drive 6
5 PSS Water Drive Model
6
5
Model 4
3
4
2
3
2
1.0
8
Normalized Rate, Derivative
6
10-1
Normalized Rate
5
8
4
6 3
5
4
3 k = 8.5 md 2
k = 3.1 md
2
s=0 10-1 s = -4
8
M = 0.001
8
6 3
10-2
2 3 4 5 6 7 8 9 -1 2 3 4 5 6 7 89 2 3 4 5 6 7 89 1 2 3 4 5 6 7 8 2 3 4
10 1.0 10 102 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
10-3 10-2 10-1 1.0 101 102
Material Balance Pseudo Time
Material Balance Pseudo Time
Multiple Well Analysis
Multi-well / Reservoir-based Analysis-
Available Methods
3. Reservoir Simulation
Well A Well B
Q Q
Correcting Interference Using
Blasingame et al Method
Q tot QA + QB
tce = ⇒ (for analyzing Well A)
q qA
log(q/Δp)
28000
4.50 2.20
26000
2.00
4.00
Aggregate production of well group 24000
1.80
Oil / Water Rates, bbl/d
22000
3.50
Pressure, psi
Gas, MMscfd
1.60 20000
3.00 18000
1.40
16000
2.50 1.20
14000
1.00 12000
2.00
0.80 10000
1.50
8000
0.60
1.00 6000
0.40
4000
0.50 0.20 Production history of well to be analyzed
2000
0.00 0.00 0
1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002
Multi-Well Analysis- Example
1.0
1.0
7
7
Normalized Rate
Normalized Rate
5
5
4
4
3
3
2
2
10-1
10-1
7
“Leaky reservoir” diagnostic 7
5
4
5 Corrected using multi-well model
Total OGIP = 7 bcf
4
3
3
2
2
10-2
10-2
2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
10-3 10-2 10-1 1.0 101 102 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
10-3 10-2 10-1 1.0 101 102
Material Balance Pseudo Time
Material Balance Pseudo Time
Multi-Well Analysis- Example
1600
1600
1500
Total OGIP = 7.0 bcf
OGIP for subject well = 3.5 bcf
1400
1400
1300
1200
1200
1100
P/Z*, psi
P/Z*, psi
1000
1000
900
800
800
700
600 600
500
400 400
300
200 200
Original Gas In Place Original Gas In Place
100
0 0
0.00 0.40 0.80 1.20 1.60 2.00 2.40 2.80 3.20 3.60 4.00 4.40 4.80 5.20 5.60 6.00 6.40 6.80 7.20 7.60 0.00 0.40 0.80 1.20 1.60 2.00 2.40 2.80 3.20 3.60 4.00 4.40 4.80 5.20 5.60 6.00 6.40 6.80 7.20 7.60 8.00
Cumulative Production, Bscf Cumulative Production, Bscf
Overpressured Reservoirs
Overpressured Reservoirs
3.00E-04
gas
Formation Formation
2.50E-04 energy is energy may Formation energy is critical in this region
negligible in be influencial
Compressibility (1/psi)
1.50E-04
1.00E-04
5.00E-05 formation
0.00E+00
0 2000 4000 6000 8000 10000 12000
Reservoir Pressure (psi)
p/z* Model – Corrects Material Balance
⎛ p⎞ ⎡
p 1 Gp ⎤⎥
= ⎜⎜ ⎟⎟ ⎢1−
⎜ ⎟ ⎢
⎥
z ⎢⎣1 − cf ( pi − p )⎥⎦ ⎝ z ⎠ i ⎢⎣ OGIP ⎥⎦
⎡ ⎤
*
Flowing MB
p ⎛⎜ p ⎞⎟ ⎡⎢ Gp ⎤⎥
= ⎜⎜ ⎟⎟ ⎢1− ⎥
z ⎝ z ⎠ i ⎢⎣ OGIP ⎥⎦
( μ ct ) i t q (t ) Typecurves
tca =
q ∫0 μ ct ⎡⎣⎢1 − cf ( pi − p)⎤⎦⎥ dt
Geomechanical Model – Corrects Well Productivity
One way to account for a variable permeability over time is to modify the definition of
pseudo-pressure and pseudo-time.
2qpi 1.417e6 * Tq ⎛ re 3 ⎞
Δp p * = ta * + ⎜ ln − ⎟
( μctZ )iGi kih ⎝ rwa 4 ⎠
where
6
5 Gulf Coast, deep gas condensate reservoir
4
1.0
8
6
Normalized Rate
5
4
10-1
8
10-2
2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
10-3 10-2 10-1 1.0 101 102
Material Balance Pseudo Time
Overpressured Reservoirs - Example
Radial Model
218 Prod and Pressure Data History Match
80 18000
70 16000
60 14000
Pressure, psi
OGIP = 17 bcf
40 10000
30 8000
6000
20
4000
10
2000
0
June July August September October
2003
Overpressured Reservoirs - Example
Radial Model
218 Prod and Pressure Data History Match
80 18000
70 16000
60 14000
Pressure, psi
OGIP = 29 bcf
40 10000
30 8000
6000
20
4000
10
2000
0
June July August September October
2003
Overpressured Reservoirs - Example
k (p) Permeability
218 Prod and Pressure Data k (p)
1.05
Legend
1.00 Default
Custom
0.95 Interpolation
0.90
0.85
0.80
0.75
0.70
Assumed permeability profile
0.65
0.60
0.55
k / ki
0.50
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0 500 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 11000 12000 13000 14000 15000 16000 17000 18000
Pressure, psi(a)
Horizontal Wells
Horizontal Wells
1. As a vertical well,
• if lateral length is small compared to drainage area
2. As a fractured well,
• if the formation is very thin
• if the vertical permeability is high
• if the lateral is cased hole with single or multiple stage
fractures
• to get an idea about the contributing lateral length
The procedure for matching horizontal wells is similar to that of vertical wells. However, for
horizontal wells, there is more than one choice of model. Each model presents a suite of
typecurves representing a different penetration ratio (L/2xe) and dimensionless wellbore radius
(rwD). The definition of the penetration ratio is illustrated in the following diagram:
Plan
Cross Section
L
h rwa
L
2 rwa
rwD =
L
2xe
The characteristic dimensionless parameter for each suite of horizontal typecurves is defined as
follows:
L
LD =
2βh
Where is the square root of the anisotropic ratio:
kh
β=
kv
For an input value of “L”,
Horizontal Wells – Example
Unnamed Well Blasingame Typecurve Analysis
102
8
6
4 L/2xe = 1
3
2
rwD = 2e-3
Ld = 5
101
8
Le = 1,968 ft
6
4 k (hz) = 0.18 md
3
2
k (v) = 0.011 md
Normalized Rate
4
3
10-1
8
6
4
3
10-2
2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
10-3 10-2 10-1 1.0 101 102
Material Balance Pseudo Time
Oil Wells
Oil Wells
110
Pressure, psi
Gas, MMscfd
2200
100 0.06
2000
90 1800
0.05
80 1600
70 1400
0.04
60 1200
50 0.03 1000
40 800
0.02
30 600
20 400
0.01
10 200
0 0.00 0
Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct
2001 2002
Oil Wells – Example
6
5
4
3 k = 1.4 md
s = -3
2
OOIP = 2.4 million bbls
1.0
8
6
Normalized Rate
5
4
10-1
8
6
5
4
10-2
2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78 2 3 4 5 6 78
10-3 10-2 10-1 1.0 101 102
Material Balance Time
Oil Wells – Example
example7 Numerical Radial Model - Production Forecast
300 4000
Legend 3800
280 History Oil Rate
240 month forecast Flow Press 3600
260 EUR = 265 Mbbls Syn Rate
History Reservoir Press 3400
Forecasted Press
240 Forecasted Reservoir Press 3200
Forecasted Rate
3000
220
2800
200
2600
180 2400
Oil Rate, bbl/d
Pressure, psi
2200
160
2000
140
1800
120 1600
1400
100
1200
80
1000
60 800
600
40
400
20
200
0 0
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