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Introduction Model Example Equalisation Conclusions References Proof
Application of Inflow Control Devices toHeterogeneous Reservoirs
Andrei Bejan(joint work with Vasily Birchenko, Alexander Usnich, and David Davies)
ECMOR XII, Oxford, September 2010
Introduction Model Example Equalisation Conclusions References Proof
Advanced well completions: ICDs and ICVsIncreasing well-reservoir contact
advantages: drainage area, sweep efficiency and delayed wateror gas breakthrough, well productivity
challenges: drilling, completion and production (uneven inflowdistribution)
Inflow Control Devices and Interval Control Valves: methods of study
using well modelling software and simulators (from basicfunctionality to capturing the annular flow effect)
analytical models (still can play a role in (i) quick feasibilitystudies, e.g. screening ICD installation candidates; (ii)verification of numerical/simulation results; (iii) communicatingbest practices without referring to specific products)
We present an analytical model for quantifying the inflowequalisation effect of ICD applications in heterogeneous reservoirs.
Introduction Model Example Equalisation Conclusions References Proof
Advanced well completions: ICDs and ICVsIncreasing well-reservoir contact
advantages: drainage area, sweep efficiency and delayed wateror gas breakthrough, well productivity
challenges: drilling, completion and production (uneven inflowdistribution)
Inflow Control Devices and Interval Control Valves: methods of study
using well modelling software and simulators (from basicfunctionality to capturing the annular flow effect)
analytical models (still can play a role in (i) quick feasibilitystudies, e.g. screening ICD installation candidates; (ii)verification of numerical/simulation results; (iii) communicatingbest practices without referring to specific products)
We present an analytical model for quantifying the inflowequalisation effect of ICD applications in heterogeneous reservoirs.
Introduction Model Example Equalisation Conclusions References Proof
Advanced well completions: ICDs and ICVsIncreasing well-reservoir contact
advantages: drainage area, sweep efficiency and delayed wateror gas breakthrough, well productivity
challenges: drilling, completion and production (uneven inflowdistribution)
Inflow Control Devices and Interval Control Valves: methods of study
using well modelling software and simulators (from basicfunctionality to capturing the annular flow effect)
analytical models (still can play a role in (i) quick feasibilitystudies, e.g. screening ICD installation candidates; (ii)verification of numerical/simulation results; (iii) communicatingbest practices without referring to specific products)
We present an analytical model for quantifying the inflowequalisation effect of ICD applications in heterogeneous reservoirs.
Introduction Model Example Equalisation Conclusions References Proof
17
Courtesy Hulliburton
Courtesy Hulliburton
Introduction Model Example Equalisation Conclusions References Proof
Model assumptionsflow through the reservoir can be described by Darcy’s law andthe inflow into the well is in steady or pseudo-steady state
distance between the well and reservoir boundary is muchlonger than the well length (or parallel to the well)
perpendicular-to-the-well components of the reservoir pressuregradients are much greater than the along-hole ones.
friction and acceleration pressure losses between the toe andthe heel are small compared to the drawdown (Birchenko et al (2010))
the fluid is incompressible
the completion interval does not need to be perfectly horizontal(hence TVD may vary along the completion)
no flow in the annulus parallel to the base pipe
ICDs installed are of the same strength
the flow distribution along the wellbore’s internal flow conduitq(ℓ) is smooth
Introduction Model Example Equalisation Conclusions References Proof
Model assumptionsflow through the reservoir can be described by Darcy’s law andthe inflow into the well is in steady or pseudo-steady state
distance between the well and reservoir boundary is muchlonger than the well length (or parallel to the well)
perpendicular-to-the-well components of the reservoir pressuregradients are much greater than the along-hole ones.
friction and acceleration pressure losses between the toe andthe heel are small compared to the drawdown (Birchenko et al (2010))
the fluid is incompressible
the completion interval does not need to be perfectly horizontal(hence TVD may vary along the completion)
no flow in the annulus parallel to the base pipe
ICDs installed are of the same strength
the flow distribution along the wellbore’s internal flow conduitq(ℓ) is smooth
Introduction Model Example Equalisation Conclusions References Proof
Model assumptionsflow through the reservoir can be described by Darcy’s law andthe inflow into the well is in steady or pseudo-steady state
distance between the well and reservoir boundary is muchlonger than the well length (or parallel to the well)
perpendicular-to-the-well components of the reservoir pressuregradients are much greater than the along-hole ones.
friction and acceleration pressure losses between the toe andthe heel are small compared to the drawdown (Birchenko et al (2010))
the fluid is incompressible
the completion interval does not need to be perfectly horizontal(hence TVD may vary along the completion)
no flow in the annulus parallel to the base pipe
ICDs installed are of the same strength
the flow distribution along the wellbore’s internal flow conduitq(ℓ) is smooth
Introduction Model Example Equalisation Conclusions References Proof
Specific inflow rate and specific productivity index:
U(ℓ) =dqdℓ
= j(ℓ)∆Pr = j(ℓ)(Pe(ℓ)− Pa(ℓ))
Model
∆P(ℓ) = ∆Pr (ℓ) + ∆PICD(ℓ) = U(ℓ)/j(ℓ) + aU2(ℓ), where
a =
(
ρcal µρµcal
)1/4ρ
ρcalℓ2
ICDB2aICD for channel ICDs,Cuρℓ
2ICDB2
C2d d4 for nozzle or orifice ICDs
Solution
qw =
∫ L
0U(ℓ)dℓ =
∫ L
0
−1 +√
1 + 4a∆P(ℓ)j2(ℓ)2aj(ℓ)
dℓ
Introduction Model Example Equalisation Conclusions References Proof
Specific inflow rate and specific productivity index:
U(ℓ) =dqdℓ
= j(ℓ)∆Pr = j(ℓ)(Pe(ℓ)− Pa(ℓ))
Model
∆P(ℓ) = ∆Pr (ℓ) + ∆PICD(ℓ) = U(ℓ)/j(ℓ) + aU2(ℓ), where
a =
(
ρcal µρµcal
)1/4ρ
ρcalℓ2
ICDB2aICD for channel ICDs,Cuρℓ
2ICDB2
C2d d4 for nozzle or orifice ICDs
Solution
qw =
∫ L
0U(ℓ)dℓ =
∫ L
0
−1 +√
1 + 4a∆P(ℓ)j2(ℓ)2aj(ℓ)
dℓ
Introduction Model Example Equalisation Conclusions References Proof
Specific inflow rate and specific productivity index:
U(ℓ) =dqdℓ
= j(ℓ)∆Pr = j(ℓ)(Pe(ℓ)− Pa(ℓ))
Model
∆P(ℓ) = ∆Pr (ℓ) + ∆PICD(ℓ) = U(ℓ)/j(ℓ) + aU2(ℓ), where
a =
(
ρcal µρµcal
)1/4ρ
ρcalℓ2
ICDB2aICD for channel ICDs,Cuρℓ
2ICDB2
C2d d4 for nozzle or orifice ICDs
Solution
qw =
∫ L
0U(ℓ)dℓ =
∫ L
0
−1 +√
1 + 4a∆P(ℓ)j2(ℓ)2aj(ℓ)
dℓ
Introduction Model Example Equalisation Conclusions References Proof
Inflow equalisation using ICD: an example
Introduction Model Example Equalisation Conclusions References Proof
Conventional completion (no ICD)
U = j∆Pw , where ∆Pw = const in ℓ
CV U = CV j
ICD application
aU2 +1jU −∆P = 0, where a,∆P > 0, sup j ⊆ R+
CV U?
≤ CV j
Introduction Model Example Equalisation Conclusions References Proof
Conventional completion (no ICD)
U = j∆Pw , where ∆Pw = const in ℓ
CV U = CV j
ICD application
aU2 +1jU −∆P = 0, where a,∆P > 0, sup j ⊆ R+
CV U?
≤ CV j
Introduction Model Example Equalisation Conclusions References Proof
0
0.2
0.4
0.6
0.8
1
1.2
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007
Channel ICD strength, bar/(Rm3/day)
2
CV
ratio
0
500
1000
1500
2000
2500
Well
flow
rate
, S
m3/d
ay
CoV ratio Well flow rate
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12 14 16
Effective nozzle diameter per 40 ft joint, mm
CV
ratio
0
500
1000
1500
2000
2500
Well
flow
rate
, S
m3/d
ay
CoV ratio Well flow rate
Dependence of inflow equalisation and well productivity on ICD strength for channel (left) and nozzle/oriffice (right) ICDs in a prolific reservoir
0
0.2
0.4
0.6
0.8
1
1.2
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007
Channel ICD strength, bar/(Rm3/day)
2
CV
ratio
0
500
1000
1500
2000
2500
Well
flow
rate
, S
m3/d
ay
CoV ratio Well flow rate
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12 14 16
Effective nozzle diameter per 40 ft joint, mm
CV
ratio
0
500
1000
1500
2000
2500
Well
flow
rate
, S
m3/d
ay
CoV ratio Well flow rate
Dependence of inflow equalisation and well productivity on ICD strength for channel (left) and nozzle/oriffice (right) ICDs in a medium reservoir
Introduction Model Example Equalisation Conclusions References Proof
Describe the distribution of j , η(j), rather than j(l)
qw =
∫ L
0U dℓ = L
∫ j2
j1
−1 +√
1 + 4a∆Pw j2
2ajη(j)dj
Calculation of CV ratio
〈U〉 = qw/L, 〈U2〉 =∫ j2
j1
(
−1 +√
1 + 4a∆Pw j2
2aj
)2
η(j) dj
CV U =
√
〈U2〉 − 〈U〉2
〈U〉 , CV j =
√
〈j2〉 − 〈j〉2
〈j〉
Results1 CV U ≤ CV j2 explicit analytical formulae for CV U/CV j when j has
uniform or triangular distribution (+ numerical estimation isalways possible)
Introduction Model Example Equalisation Conclusions References Proof
Conclusions
1 Negative effect of frictional pressure losses and reservoirheterogeneity on oil sweep efficiency and ultimate oil recovery
2 ICDs: an equalisation effect on the inflow profile
3 Derived a simple mathematical model for ICDs’ effectivereduction of the inflow imbalance caused by reservoirheterogeneity:
quantify the equalisation effect using the ratio CV U/CV jrigorous proof that CV U/CV j ≤ 1give explicit form analytic expressions for the CV ratio foruniform and triangular distribution of jthe model can be used in quantifying the trade-off betweenthe well productivity and inflow equalisation
4 Generalisation of the model (accounting for both frictionalpressure losses and reservoir heterogeneity)?
Introduction Model Example Equalisation Conclusions References Proof
Birchenko, V.M., Usnich, A.V., Davies, D.R. (2010) Impact offrictional pressure losses along the completion on wellperformance. J. Pet. Sci. Eng., doi: 10.1016/j.petrol.2010.05.019.
Birchenko, V.M., Muradov, K.M., Davies, D.R. (2010) Reductionof the Horizontal Well’s Heel-Toe Effect with Inflow ControlDevices. Under review in J. Pet. Sci. Eng.
Birchenko, V.M., Bejan, A.Iu., Usnich, A., Davies, D.R. (2010)Application of Inflow Control Devices to HeterogeneousReservoirs. Under review in J. Pet. Sci. Eng.
Birchenko, V.M., Al-Khelaiwi, F., Konopczynski, M., Davies, D.R.(2008) Advanced wells: how to make a choice between passiveand active inflow-control completions. In SPE Annual Tech.Conf.&Exhib., dx.doi.org/10.2118/115742-MS
More references in the proceedings paper.
Introduction Model Example Equalisation Conclusions References Proof
aU2 + U/j −∆P = 0, a,∆P > 0, supp j ⊆ R+
task: to prove that CV U/CV j ≤ 1
Y = U/√∆P, X = j
√∆P ⇒ aY 2 + Y/X − 1 = 0
CV Y = CV U, CV X = CV j, so will prove for X and Y
Define a cont. bijection x = y/(1 − ay2) from (0, a−1/2) to (0,∞)
F (a) =
∫
∞
0 x(y ; a)2 dh(y)(∫
∞
0 x(y ; a)dh(y))2 = 1 + (CV X)2 ; F (0) = 1 + (CV Y )2
Theorem
The function F satisfies the inequality F (0) ≤ F (a), where, except forthe case when a = 0, equality holds iff h is a point mass distribution.
Introduction Model Example Equalisation Conclusions References Proof
aU2 + U/j −∆P = 0, a,∆P > 0, supp j ⊆ R+
task: to prove that CV U/CV j ≤ 1
Y = U/√∆P, X = j
√∆P ⇒ aY 2 + Y/X − 1 = 0
CV Y = CV U, CV X = CV j, so will prove for X and Y
Define a cont. bijection x = y/(1 − ay2) from (0, a−1/2) to (0,∞)
F (a) =
∫
∞
0 x(y ; a)2 dh(y)(∫
∞
0 x(y ; a)dh(y))2 = 1 + (CV X)2 ; F (0) = 1 + (CV Y )2
Theorem
The function F satisfies the inequality F (0) ≤ F (a), where, except forthe case when a = 0, equality holds iff h is a point mass distribution.
Introduction Model Example Equalisation Conclusions References Proof
aU2 + U/j −∆P = 0, a,∆P > 0, supp j ⊆ R+
task: to prove that CV U/CV j ≤ 1
Y = U/√∆P, X = j
√∆P ⇒ aY 2 + Y/X − 1 = 0
CV Y = CV U, CV X = CV j, so will prove for X and Y
Define a cont. bijection x = y/(1 − ay2) from (0, a−1/2) to (0,∞)
F (a) =
∫
∞
0 x(y ; a)2 dh(y)(∫
∞
0 x(y ; a)dh(y))2 = 1 + (CV X)2 ; F (0) = 1 + (CV Y )2
Theorem
The function F satisfies the inequality F (0) ≤ F (a), where, except forthe case when a = 0, equality holds iff h is a point mass distribution.
Introduction Model Example Equalisation Conclusions References Proof
aU2 + U/j −∆P = 0, a,∆P > 0, supp j ⊆ R+
task: to prove that CV U/CV j ≤ 1
Y = U/√∆P, X = j
√∆P ⇒ aY 2 + Y/X − 1 = 0
CV Y = CV U, CV X = CV j, so will prove for X and Y
Define a cont. bijection x = y/(1 − ay2) from (0, a−1/2) to (0,∞)
F (a) =
∫
∞
0 x(y ; a)2 dh(y)(∫
∞
0 x(y ; a)dh(y))2 = 1 + (CV X)2 ; F (0) = 1 + (CV Y )2
Theorem
The function F satisfies the inequality F (0) ≤ F (a), where, except forthe case when a = 0, equality holds iff h is a point mass distribution.
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