Diagnostic Tools: Production Data Analysis and Well Testing Prepared for: ENPE 533 Mehran Pooladi-Darvish
A few questions to answer • What will be rate of production in future? • Can we do anything to improve production rate? By how much? • Which well is worth more?
Two methods of finding the answers • Well testing (PTA):
• Production data (Decline curve) analysis (RTA):
• Estimate parameters • Use the historical production data and that affect production predict (extrapolate) the rate and model future future response response
• Modern RTA/PTA analysis
Well-Testing: What is a test? • Impose a change in rate A drawdown test
• Measure change in pressure with time
Which reservoir has …? 9000
Higher permeability? Lower Skin? Larger reserve?
8000 7000
Pressure, kPa
6000 5000
Reservoir 1 Reservoir 2
4000 3000 2000 1000 0 0
100
200
300 Time, hour
400
500
600
What is well-test interpretation? • Analysis of p, q vs. t To obtain • Permeability • Skin (Damage) • Reservoir size • Reservoir pressure • Reservoir description
What is the basis of PTA? • Determine the relation between change in pressure (because of change in rate) and reservoir properties, k, S, etc. • Combine mass balance, Darcy’s law and EOS to obtain the diffusivity equation 1 ∂ ∂p φµct ∂p r = r ∂r ∂r k ∂t
• Solve diffusivity equation with the following two BC’s • q = const. r = rw • q=0 r = re
Determination of K and S • Transient flow regime
pi − pwf (t ) =
k 162.6qµ B log ( t ) + log − 3 . 23 + 0 . 87 S φµc r 2 kh t w
• Interpretation method?
Semi-log plot 8000 7000 6000
Pressure, kPa
slope = 5000
162.6 qµB kh Reservoir 1 Reservoir 2
4000 3000 2000 1000 0 0.001
0.01
0.1
1 Time, hour
10
100
1000
SEMI-LOG PERM=1 PERM=10 PERM=100 Slope gives permeability
SEMI-LOG ∆P k s' = 1.151 − log (∆ t ) − log 2 φµ c r m t w
+ 3.23
Skin = 10 Skin = 0 Skin = -5
Intercept gives skin
Detection of transient flow regime is necessary Semi-log plot Analysis 1 4500 4000
k kh s'
20.974 m D 20.97 mD.m 0.291
∆p/q, kPa/(m 3 /d)
3500 3000 2500 2000 1500 1000 500
∆p/qdata ∆p/qmodel
0 10-4
2 3 4 56 8 10-3
2 3 4 56 8 10-2
2 3 4 56 8 10-1
2 3 4 56 8 1.0
∆t , h
2 3 456 8 101
2 3 4 56 8 102
2 3 4 56 8 103
Methods of Analysis • Traditional (e.g semi-log/Horner) • Derivative: Slope of the semi-log plot • Computer-aided integrated (History matching)
DERIVATIVE • Derivative is defined as: Der = dP/d(ln(t)) • Derivative is obtained by plotting the data on a semi-log scale of “Pressure versus Log time”. The slope of this semi-log curve gives the derivative. • Derivative is a misnomer. It should have been called “semi-log Derivative”. It should not be confused with the mathematical derivative of pressure with time. This latter term dP/dt has been given the name PPD (Primary Pressure Derivative), and is used to differentiate between wellbore dynamics and reservoir transients.
Diagnostic Analysis Method: Derivative Analysis Radial 0 k s'
19.392 mD -0.147
Semi-log plot
Log-Log Derivative Plot 104
Analysis 1
7
4500
4 2
4000
7
3500
4
∆ p/q, kPa/(m3/d)
∆p/q , kPa/(m 3/d)
103
2 102 7 4 2
k kh s'
20.974 m D 20.97 mD.m 0.291
3000 2500 2000 1500
101 7
∆p/q data ∆p/q model
4
Derivativedata Derivativemodel
2 1.0 10-4
2 3 4 56 8 10-3
2 3 4 56 8 10-2
2 3 4 56 8 10-1
2 3 456 8 1.0
∆t , h
2 3 456 8 101
2 3 4 56 8 102
2 3 4 56 8 103
1000 500
∆p/q data ∆p/q model
0 2 10-4
3 4 56 8 10- 3
2
3 4 56 8 10 -2
2 3 4 56 8 10 -1
2 3 4 56 8 1.0
∆t , h
• Flat portion: Transient radial flow, gives K • The magnitude of the hump, gives S
2 3 4 56 8 101
2 3 4 56 8 102
2
3 4 56 8 103
Effect of skin on Derivative plot
Radial 0 k s'
Radial 0
19.392 mD 9.482
k s'
Log-Log Derivative Plot
Log-Log Derivative Plot
104
104 7
7
4
4
2
2
103
103 7
7
4
4
∆p/q , kPa/(m 3/d)
∆p/q , kPa/(m 3/d)
19.392 mD -0.147
2 102 7 4
2 102 7 4
2
2
101
101 7
Derivativedata Derivativemodel
2 1.0 10-4
7
∆p/q data ∆p/q model
4
2 3 4 56 8 10-3
2 3 4 56 8 10-2
2 3 4 56 8 10-1
2 3 456 8 1.0
∆t , h
2 3 456 8 101
2 3 4 56 8 102
2 3 4 56 8 103
∆p/q data ∆p/q model
4
Derivativedata Derivativemodel
2 1.0 10-4
2 3 4 56 8 10-3
2 3 4 56 8 10-2
2 3 4 56 8 10-1
2 3 456 8 1.0
∆t , h
2 3 456 8 101
2 3 4 56 8 102
2 3 4 56 8 103
How to determine reservoir size?
Compressibility – Cause of production 9000
• Expansion vs. displacement
7000 6000
Pressure, kPa
1 ∂V c=− V ∂p
8000
5000
Reservoir 1 Reservoir 2
4000 3000 2000 1000 0 0
100
200
300 Time, hour
dp wf 0.234qB =− = constant dt ct Ahφ
400
500
600
Determination of Reservoir Area – Derivative Analysis Radial 0
Radial 0 k s'
k s'
19.392 m D 9.482
Log-Log Derivative Plot
Log-Log Derivative Plot 104
104 7
7
4
4 2
2 103
103 7
7
4
4
∆p/q , kPa/(m3/d)
∆p/q , kPa/(m3/d)
19.392 m D 9.482
2 2
10
7 4
2 102 7 4 PSS 1
2
2
Vp 1.92e+05 m3 V 9.59e+05 m3 OIPanalysis 1.36e+02 103m 3
101
101
7
7
∆ p/qdata ∆ p/qmodel
4
1.0 2 3 4 56 8 10-3
2 3 4 56 8 10- 2
2 3 4 56 8 10- 1
2 3 4 56 8 1.0
∆t, h
2 3 456 8 101
2 3 4 56 8 102
2 3 4 56 8 -4 103 10
∆ p/qdata ∆ p/qmodel Derivative data Derivative model
2 Derivative data Derivative model 1.0
2
10-4
4
2 3 4 56 8 10-3
2 3 4 56 8 10- 2
2 3 4 56 8 10- 1
2 3 4 56 8 1.0
∆t, h
2 3 456 8 101
2 3 4 56 8 102
2 3 4 56 8 103
Different Plotting Techniques
Reservoir Description through Diagnostic Analysis • • • • • •
Hydraulically fractured wells Naturally fractured reservoirs Channel-shaped reservoirs Composite and layered reservoirs Horizontal and vertical wells Different boundary types
Fractured well 104 7 4 2
Linear Fracture 1/2 k mD xf(sqrt(k)) 124.21 mD1/2m Xf m s Xf
103
∆p/q , kPa/(m 3/d)
7 4 2 102 7 4 2 101 7
∆ p/qdata ∆ p/qmodel
4
Derivative data Derivative model
2 1.0 10-4
2 3 4 56 8 10-3
2 3 4 56 8 10-2
2 3 4 56 8 10-1
2 3 4 56 8 1.0
∆t , h
• Objectives • What other flow regime?
2 3 4 56 8 101
2 3 4 56 8 102
2 3 4 56 8 103
Naturally Fractured Reservoirs (NFR)
Derivative Plot - NFR TD Vertical 4 Typecurve 103 7 5 3
∆p/q , kPa/(m3/d)
2 102 7 5 3 2 101 7 5
∆p/qdata ∆p/qmodel
3 2
Derivative data Derivative model
1.0 10-4
2 3 4 56 8 10-3
2 3 4 56 8 10-2
2 3 4 56 8 10-1
2 3 4 56 8 1.0
∆t , h
2 3 4 56 8 101
2 3 4 56 8 102
2 3 4 56 8 103
• Slope of Derivative:
Flow Regime:
• Early Time slope = 1 • =½
Wellbore storage Fracture
• Middle Time • • Late Time
slope = 0
slope = ½ slope = 1
Radial flow; (semilog straight line) Channel Reservoir Volume
Methods of Analysis TD Vertical 1 Total Test 4000
pi (syn) 4000.0 psi p* 3979.3 psi
kh h k sd
3995 3990
5000.00 md.ft 50.000 ft 100.000 m d -2.000
Xe Ye Xw Yw
1500.0ft 0.0010 1500.0ft 100.0 ft 750.0 ft 0.0008 0.0006
3985 Pressure , psi
• Traditional (e.g semi-log/Horner) • Derivative: Slope of the semi-log plot • Computer-aided integrated (History matching)
0.0004
3980
0.0002
3975
-0.0000
3970
-0.0002
3965
-0.0004
3960
p data -0.0006 p model % Error -0.0008 100 110 120 130 140 150 160 170 180 190 200 210
3955 0
10
20
30
40
50
60
70
80
90
Time , h
Build-up vs. Drawdown 500 450
Change in Pressure, kPa
400
Higher permeability? Lower Skin?
350 300
Reservoir 1 Reservoir 2
250 200 150 100 50 0 0
50
100
150 Time, hour
200
250
Well-Testing (PTA) vs. Production Data Analysis (RTA) • Well-Testing:
• Study of wellbore pressure when produced at a constant rate
• Production Data Analysis:
• Study of wellbore rate when produced at a constant pressure
• Many wells are produced at (near) constant bottomhole pressure
Arps Decline Curve Analysis Exponential, b=0 hyperbolic Harmonic, b=1 log(q)
t
q(t ) 1 = 1 qi (1 + bDi t )b
log(t)
q(t ) = exp(− Dt ) qi
Traditional - Arps • Empirical – Single-phase and two-phase (0 < b < 1)
• Boundary-dominated regime • Data q vs. t • Constant operating conditions
Arps Type Curves by Fetkovich
Example Example 1-Single Phase Depletion
Flow Rate, STB/day
1000
100
10 0
200
400
600
800 Time, Days
1000
1200
1400
1600
Example Example 1-Single-Phase Depletion 10000
Flow Rate, STB/day
1000
100
10
1 0.1
1
10
100
1000
Time, Days
10000
100000
1000000
Meaning of different stems • The empirical exponent b is a measure of effectiveness of the recovery process • Depletion above bubblepoint, b = 0 • Solution-Gas Drive, b ≅ 0.3 • Active water drive, b ≅ 0.5 • Gas Cap drive with gravity drainage b ≅ 1 • Decline may start with b = 0 and change to b>0
Theoretical meaning of the decline curves Solve diffusivity equation with the following two boundary conditions • p = pwf r = rw :Constant BHP Solution • q=0 r = re • Has an analytical solution
Theoretical Meaning of the decline curves
Boundary-Dominated Flow Start of Boundary-Dominated Flow pi Transient Flow Pressure ( p )
Boundary-Dominated
re Distance ( r )
Depletion above the bubble-point pressure • Dimensionless solutions are presented in terms of q
141.3qµB qD = kh( pi − pwf )
0.00634kt tD = φµct rw2
1 − 4πtD qD (tD ) = exp 4A 4A A ln µ ln 2 2 2 γC Arw r γ C r A w w
t
Exponential Decline
Transient Flow is a single curve; Boundary-Dominated Flow is a family of curves
Same Transient for all re/rw 's
qD
Different re/rw 's
tD
Merging the boundary-dominated portion tDd =
tD 2 1 re 1 re ln − − 1 2 rw 2 rw
re 1 qDd = qD ln − rw 2
Fetkovich Type Curves Rate Decline Curves for "Constant Wellbore Flowing Pressure"
10
Transient Flow Transient rFlow Different e/rw
Boundary Boundary Dominated Flow becomes Flow Dominated Exponential Decline is Exponential Decline
qDd
1
Analytical solution for constant flowing pressure
0.1
Matching will give reservoir parameters 0.01 0.0001
0.001 re/rw=10 re/rw=200
0.01 re/rw=20 re/rw=1000
tDd
0.1 re/rw=50 re/rw=10000
1
10 re/rw=100 exponential
Questions • Do transient stems depend on reservoir size? • What parameters can be determined by matching the transient data?
Non-Uniqueness Of Matching When Only Early-time Data Are Available 10
re/rw = 10
1
q, q Dd
re/rw = 50
Two equally plausible matches!
0.1
0.01 0.0001
0.001
0.01
0.1
1
t, tDd re/rw=10
re/rw=50
Exponential (b=0)
10
Type Curve Matching • The rate and transient stem matches are used for kh calculations qDd
141.2 Bo µ re 1 ln − = q kh ( p − p ) r 2 i wf wa
log q Dd
141.2 B µ r 1 o ln e − = log q + log kh( p − p ) r 2 i wf wa
• The time and rate matches are used for Pore Volume calculations
Fetkovich Theory – Boundary Dominated and Transient Fetkovich Decline Type Curves 10
Boundary DominatedEmpirical Stems
qDd
1
0.1
TransientAnalytical Stems
0.01
0.001 0.0001
0.001
0.01
0.1
1
10
100
tDd re/rw=10 re/rw=1000
re/rw=20 re/rw=10000
re/rw=50 b=0
b=0.6
b=0.8
b=1.0
re/rw=100 b=0.2
re/rw=200 b=0.4
Fetkovich • Empirical and theoretical – Single-phase and two-phase (0 < b < 1)
• Analytical solution for single-phase flow – Exponential decline (b = 0) – Introduction of transient stems (k and S)
• Boundary-dominated regime and transient • Constant bottomhole pressure • Data q vs. t
Merging Well-Testing (PTA) and Production Data Analysis (RTA) • Well-Testing:
pD =
kh ( pi − pwf (t )) 141.3qµB
• Production Data Analysis:
141.3q (t )µB qD = kh( pi − pwf )
• Study of wellbore pressure when produced at a constant rate • Study of wellbore rate when produced at a constant pressure
0.00634kt tD = 2 φµct rw
How to account for variable BHP?
Search for a Common Solution • Can one relate the PTA and RTA solutions and therefore account for both variable BHP and variable rate? PTA:
RTA:
kh[pi − pwf (t )] pD = 141.2 µBq 141.2µBq(t ) qD = kh( pi − pwf ) qD =
?
1 pD
0.00634kt tD = φµcrw2 0.00634kt tD = φµcrw2
The Two Solutions – Transient
The Two Solutions – Boundary Dominated Circles: constant P Lines: constant q
Material Balance Time – Blasingame t
tc =
∫ q(t )dt 0
q(t )
Constant Rate Case
=
N p (t ) q(t )
Constant BHP Case
t
tc =
∫ qdt 0
q
t
=t
tc =
∫ q(t )dt 0
q (t )
q t
>t
The Two Solutions – Boundary Dominated
Importance of Material Balance Time To analyze field data that exhibit variable rate and variable BHP
Condition for use • Application of Material Balance Time allows use of constant BHP (or rate) solutions for the analysis of variable rate and/or variable BHP data • Both rate and BHP need to be smooth functions of time – Sharp transient will deviate from the analytical solutions
Use of PTA Techniques
Blasingame • Theoretical – Analytical solution – Single-phase
• Accounts for variable BHP – Introduction of MB Time – Exponential decline turned to Harmonic (b = 1)
• Boundary-dominated regime and transient • Data q/∆p vs. tc (makes use of pressure data)
Exponential stem turned harmonic
Use of RTA Techniques
Agarwal-Gardner • The flowing material balance plot allows an alternative representation of data – Very advantageous for determination of OGIP 0.0014
0.0012
q/Dm(p)
0.001
0.0008
0.0006
0.0004
0.0002
0 0
0.2
0.4
0.6 Dm(p-bar)/Dm(p)
0.8
1
1.2
Application to Gas Reservoirs Oil Reservoirs Gas Reservoirs Pseudo-pressure
1 ∂ ∂p φµct ∂p r = r ∂r ∂r k ∂t 1 ∂ p ∂p φµ g ct p ∂p r = r ∂r µZ ∂r k µZ ∂t p p ψ ( p ) = m( p ) = 2 ∫ dp 0 µz 1 ∂ ∂ψ ( p ) φµ g c g ∂ψ ( p ) r = r ∂r ∂r k ∂t
When average reservoir pressure does not change much, the RHS coefficient may be assumed constant, i.e. liquid solutions apply. Application: Transient flow, PTA
Gas Reservoirs (cont.) When analyzing long-term data (RTA), average reservoir pressure changes with time. A pseudo-time function is Defined to account for these changes.
1 ∂ ∂ψ ( p ) φµ g cg ∂ψ ( p ) r = r ∂r ∂r k ∂t dt ta = (µ g cg )i ∫ µ g ( p )cg ( p ) 0 t
tac =
(µ c )
g g i
q(t )
q(t )dt ∫0 µg ( p )cg ( p )
Pseudo-time
t
Material-Balance pseudo-time
Typecurves as a Diagnostic Tool
Transient (infinite acting)
Transition (influenced by boundaries / heterogeneities)
Boundary Dominated
log(q/∆p)
Energizing reservoir (aquifer, partially sealing fault)
Decreasing skin
Volumetric depletion
Wellbore effects (liquid loading) or Interference
log(material balance time)
Typecurves as a Diagnostic Tool – Example 1
Stimulated well, volumetric depletion
log(q/∆p)
log(material balance time)
Typecurves as a Diagnostic Tool – Example 2
Damaged well, liquid loading problems
log(q/∆p)
log(material balance time)
Typecurves as a Diagnostic Tool – Example 3
Boundaries not yet reached ?
log(q/∆p)
log(material balance time)
Typecurves as a Diagnostic Tool – Example 3
Or aquifer supported reservoir?
log(q/∆p)
log(material balance time)
Advanced Decline Curve Analysis • Uses production data – No need to shut-in • Accounts for change in bottomhole pressure • Type curve matching and history matching • Focus is on reserve determination
