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8/13/2019 Part 12 Superposition Single and Multi Well
1/22
Well Test Analysis
Superposition
Superposition
It is a fundamental tool of pressuretransient test analysis and reservoirengineering.
ma es us o cons ruc reservo rresponse functions in complex situations(boundaries, variable rate, variablepressure producti on) using only simplebasic model solutions , namely constant-rate (or constant-pressure) solutions.
Superposition
It can be used to represent the response dueto several wells by adding up the individualwell responses (multi-well applications).
By appropriate choice of flow rate and welllocation, it is also possible to representvarious types of r eservoir boundaries (no-flow and cons tant-pressure).
It should be noted that principle ofsuperposition holds only for linear systems(in the mathematical sense).
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Well Test Analysis
Superposition However, these include most o f the standard response
functions used i n pressure transient analysis, makingthe assumption of slightly compressible fluid of
constant viscosity and compressibility.
Infinite acting radial flow for homo geneousreservoirs
Double-porosity and double-permeability models
Fractured, horizontal wells
Bounded systems, etc.
It can be used for gas wells with t he appropriatetransformations (e.g., real-gas pseudo-pressure) andcorrections (e.g., material balance correction) as to bediscussed later in gas pressure transient testing.
Principle of Superposition
It states that the response of the systemto a number of perturbations is exactlyequa o e sum o e responses oeach of the perturbations as if they werepresent by themselves.
Principle of Superposition
Well 2q2
Well 4q4
Well 1
q1 Well 3q3
p(M,t) = ?
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Well Test Analysis
Principle of Superposition
= r1
+
r2
q1
Mq2
q3
q4
q1
M Mq2
p(M,t) = p1(r1,q1,t) + p2(r2,q2,t) + p3(r3,q3,t) + p4(r4,q4,t)
++ r3 r4M
q3
Mq4
Superposition
Using the principle of superposition and theimage well concept (to be discussed), it isrelatively straightforw ard to account for theeffects of complex boundary shapes,
constant-pressure boundaries.
As said previous ly , can be used to combinea series of different constant-rate solutions(or response functions) to describe thepressure response in a variable-rate
pressure transient t est.
Homogeneous Infinite System:
Principle of Superposition
= r1
+r2
q1
Mq2
q3
q4
q1
M Mq2
p(M,t) = p1(r1,q1,t) + p2(r2,q2,t) + p3(r3,q3,t) + p4(r4,q4,t)
++ r3 r4M
q3
Mq4
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Well Test Analysis
Homogeneous Infinite System:
Principle of Superposition
Suppose that that four wells in the the previous
slide start to production at different rates, butconstant, at the same time.
How could we describe each w ells pressure
change pj(rj,qj,t) that will created at the point M
at time t in the reservoir i f each well were
producing alone with a constant-rate of qj?
==
kt
rc948.05Ei
kh
B70.6q)r,p(qpt),r,(qp
2
tj
jjijjj
j
for j = 1,2,3,4.
Homogeneous Infinite System:
Principle of Superposition
=
rc948.05B70.6q
kt
rc948.05Ei
kh
B70.6qt)p(M,p
2
2t2
2
t1i
1
kt
rc948.05Ei
kh
B70.6q
kt
rc948.05Ei
kh
B70.6q
ktkh
2
4t4
2
3t3
Homogeneous Infinite System:
Principle of Superposition
Now, suppose that we wish to compute the
pressure drop at Well 1 in our previous 4- well
example, i.e., the point M is at w ell.
Well2
q2
Well1Well3
Well4
M
q3
q4
q1
r2
r3
r4
How to take r1? We take as the wellbore radius o f Well 1, r1 = rw1
r1
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Well Test Analysis
Homogeneous Infinite System:
Principle of Superposition
Then, the pressure drop at Well 1, with the
presence of other three production wells, canbe written as:
rc948.05B70.6q 2
wt1 1
=
kt
rc948.05Ei
kh
B70.6q
kt
rc948.05Ei
kh
B70.6q
kt
rc948.05Ei
kh
B70.6q
ktkh,
2
4t4
2
3t3
2
2t2
1w1i
Skin factor
at Well 1
Homogeneous Infinite System:
Principle of Superposition: Example
500 ft350 ft
Well 3
(a) Compute pr essure and pressur e drop at Well 3 at t = 60 hr.
(b) Compute pressure and press ure drop at Well 1 at t = 100 hr.
650 ftWell 2Well 1
Homogeneous Infinite System:
Principle of Superposition: Example
(a) Compute pressu re and pressure drop at Well 3at t = 60 hr.
= rc948.05EiB70.6qrc948.05EiB70.6qt),(r
2
2t2
2
1t1w3i
thth
=
605
3501.5x100.70.2.05948Ei
1005
0.71.27512070.6
605
5001.5x100.70.2948.05Ei
1005
0.71.27545070.6t),p(rp
25
25
w3i
[ ] [ ] psi9.725.244.480.74Ei15.11.66Ei56.7t),p(rp w3i =+==
p(rw3,t = 60 hr)=5000-9.72= 4990.3 psi
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Well Test Analysis
Homogeneous Infinite System-
Classical Two Well Problem
For t t* at point M,p i-p(x,y,t)=p1(x,y,t)
For t t* at point M,p i-p(x,y,t)=p1(x,y,t)+ p2(x,y,t-t*)
Homogeneous Infinite System-
Classical Two Well Problem
For t t* at point M,p i-p(x,y,t)=p1(x,y,t) =
kt
rc948.05Ei
kh
B70.6q 2
1t1
For t t* at point M,p i-p(x,y,t)=p1(x,y,t)+ p2(x,y,t-t*)
=
)tk(t
rc948.05Ei
kh
B70.6q
kt
rc948.05Ei
kh
B70.6q
*
22t2
2
1t1
Homogeneous Infinite System-
Classical Two Well Problem
Now suppose that both wells start
production wit h the same rate q at t = 0,
then the pressure drop at point M for t 0is iven b
p i-p(x,y,t)=p1(x,y,t)+ p2(x,y,t)
=
kt
rc948.05Ei
kh
70.6qB
kt
rc948.05Ei
kh
70.6qB
2
2t
2
1t
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Well Test Analysis
Homogeneous Infinite System-
Classical Two Well Problem
Then, we can rewrite by using:
2221 yx)(dr ++=
222
2 yx)(dr +=
[ ]
[ ]
+
++=
kt
yx)(dc948.05Ei
kh
70.6qB
kt
yx)(dc948.05Ei
kh
70.6qBt)y,p(x,p
22
t
22
ti
0x
p
y)0;(x
=
=From this equation, we obtain:
Homogeneous Infinite System-
Classical Two Well Problem
y
d d
xWell 2, q
No-flow boundary
Well 1,q
Homogeneous Infinite System-
Single No-Flow (Fault ) Problem
dd d
Sealing Fault
Well, q
Well, q Image
Well, q
No-flow boundary
+
=
t4
(2d)Ei
kh
70.6qB2s
t4
rEi-
kh
70.6qB(t)pp
22
wwfi c
k2.637x10
t
4
=
(prod.)
Actual System
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Well Test Analysis
Homogeneous Infinite System-
Single No-Flow (Fault ) Problem
2D Pressure
Profile, hereZ-axis represents
pressure
3D Pressure
profile
No-flow boundary
Homogeneous Infinite System-
Single No-Flow (Fault ) Problem
3D Pressure profiles
No-flow boundary
Image well
Numerical Simulation
with a single no-flow boundaryCondition in a semi-infini te sys.
Numerical Simulation
With two wells producing at thesame constant rate q , separated
by a distance 2d, in an infinite sys.
Homogeneous Infinite System-
Classical Two Well Problem
Now suppose that Well 1 start produc tion
with rate q at t = 0, but Well 2 start injection
with rate q at t = 0, the pressure drop at
oint M for t 0 is iven bp i-p(x,y,t)=p1(x,y,t)+ p2(x,y,t)
( )
=
kt
rc948.05Ei
kh
Bq-70.6
kt
rc948.05Ei
kh
70.6qB
2
2t
2
1t
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Well Test Analysis
Homogeneous Infinite System-
Classical Two Well Problem
Then, we can rewrite by using:
2221 yx)(dr ++=
222
2 yx)(dr +=
[ ]
( ) [ ]
+
++=
kt
yx)(dc948.05Ei
kh
Bq-70.6
kt
yx)(dc948.05Ei
kh
70.6qBt)y,p(x,p
22
t
22
ti
tallforpt)y,0,p(x0t)y,0,p(x-p ii ====
From this equation, we obtain:
Homogeneous Infinite System-
Classical Two Well Problem
y
d d
Well 1,q
x
Well 2, -q
constant-pressureboundary (C-P Boundary)
Homogeneous Infinite System-
Single C-P Boundary Problem
dd d
C-P boundary
Well, q
Well, q Image
Well, -q
C-P boundary
+
=
t4
(2d)Ei
kh
70.6(-q)B2s
t4
rEi-
kh
70.6qB(t)pp
22
wwfi c
k2.637x10
t
4
=
(Inj.)
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Well Test Analysis
Examples: Imaging Based on
Superposition
dary
N-F boundary
d2
Image
Well, q(prod.) Image
Well, q
(prod.)
d2
N-F
bou
Well, q
1
Well, q
d1d2
Image
Well, q(prod.)
Actual System
d1
Write down an expression for computin g pressure drop at the well.
Examples: Imaging Based on
Superposition
dary
N-F boundary
d2
Image
Well, q(prod.) Image
Well, -q
(Inj.)
d2
C-P
bou
Well, q
1
Well, q
d1d2
Image
Well, -q(Inj.)
Actual System
d1
Write down an expression for computin g pressure drop at the well.
Examples: Imaging Based on
Superposition
d
ary
nd
ary
N-F
bou
Well, q
1
Well, qN-F
bou 2
Image
Well, q
Actual System
2
Image
Well, q
1 2 1 1 2.. ..
Infinitely many image wells (all prod .)
Write down an expression for computin g pressure drop at the well.
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Well Test Analysis
Examples: Imaging Based on
Superposition
dary
ndary
C-P
bou
Well, q
1
Well, qN-F
bou 2
Image
Well, q
Actual System
2
Image
Well, -q
1 2 1 1 2.. ..
Infinitely many image wells (prod + inj .)
Write down an expression for computing pressure drop at the well.
Examples: Imaging Based on
Superposition
Infinitely many array of image wells (all prod .)
Examples: Imaging Based on
Superposition
Infinitely many array of image wells (both prod. and inj .)
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Well Test Analysis
Superposition in Time
Homogeneous Infinite System-
Classical Two Well Problem
r2 = 2d
y-axis2
w
2
w1
2
1 rrr ==22
2 (2d)r =
Well 1, q Well 2, -q
x,y)r1 = rw1=rw
d d
0,0x-axis
Suppose Well 1 starts production w ith q from t = 0
Suppose Well 2 start producti on with -q from t = t*, t* >0
How can we compute pressure drop at Well 1 for agiven time t>0?
Homogeneous Infinite System-
Classical Two Well Problem
For t t* at point M, i.e., at Well 1p i-p(rw,t)=p1(rw,t)
For t t* at Well 1,p i-p(rw,t)=p1(rw,t)+ p2(r2,t-t*)
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Well Test Analysis
Homogeneous Infinite System-
Classical Two Well Problem
For t t* at Well 1,p i-p(rw,t)=p1(rw,t) =
+
12s
kt
rc948.05Ei-
kh
B70.6q 2
1t1
For t t* at Well 1,p i-p(rw,t)=p1(rw,t)+ p2(r2,t-t*)
( ) ( )
+
=
)tk(t
2dc948.05Ei
kh
Bq-70.6
2skt
rc948.05Ei
kh
70.6qB
*
t
1
21t
2
Rate Superposition
Now, we bring Well 2 to the location of Well1 so that two well s are operating at thesame location of Well 1:
One production well started productionwith q at t = 0, and
an injection well started injection with qat t = t *, t* > 0.
Rate Superposition
Rate histories of the wells are:
Rate
q > 0 Well 1
Well 1
Rate
0
Time
Time- q < 0
t*Well 2
0
Rate
0Time
q = 0
t*
Buildup Test
+
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Well Test Analysis
Rate Superposi tion Equation
For t t* at Well 1,p i-p(rw,t)=p1(rw,t) =
+
12s
kt
rc948.05Ei-
kh
70.6qB2
wt
For t t* at Well 1,p i-p(rw,t)=p1(rw,t)+ p2(rw,t-t*)
( )
+
+
+
=
1*
2
wt
1
2
1t
2s)tk(t
rc948.05Ei-
kh
Bq-70.6
2skt
rc948.05Ei-
kh
70.6qB
Buildup Test Pressure/Rate History
ressure
P
Time Time
Examples: Two Step-Rate Changes:
Single WellRate Histories
+
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Well Test Analysis
Example: Two Step-Rate Changes:
Single Well
+
p
Timet10
p
pi
pi
+
p
Timet10
pp
pi
pi
Pressure Histories
p
0Time
t2Timet1
0t2
pip
0Time
t2Timet1
0t2
Timet10
t2
pi
Write down the pressure equation usin g superposition:
Assume infi nit e-acting homogeneous reservoi r,
fully penetrating line-source well.
Examples: Two Step-Rate Changes:
Single WellRate Histories
q1 > 0qq
q1
q1 > 0qq
q1
q
0
Time
Time
q2-q1
t1
t2
Timet10
t2
2
q
0
Time
Time
q2-q1> 0
t1
t2
Timet10
t2
2
Examples: Two Step-Rate Changes:
Single Well
+
p
Timet10
p
p i
p i
+
p
Timet10
p
p i
p i
Pressure Histories
p
0Time
t2Timet1
0t2
pip
0Time
t2Timet1
0t2
pi
Write down the pressure equation usin g superposition:
Assume infi nit e-acting homogeneous reservoi r,
fully penetrating line-source well.
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Well Test Analysis
Multirate Superposition:Illustration
Multirate: illustration
q1
q2
q3 q5q6
q8
q1q3 q5
q6
q8
t1 t2 t3 t4 t5 t6 t7q4 q7
t1 t2 t3 t4 t5 t6 t7
q2
q4 q7
Multirate Superposition Equation
Although , so far, we have consideredinfinite acting homogeneous reservoirsolution, superposition is quite general canbe used for all models assuming slightl y
compressible fluid of constant viscosityan compress y, e.g., ou e poros y,layered systems, etc.
All we need is to have the cons tant -rateresponse of the system considered togenerate the solution for the multirate case.
Next, I develop the general equation ofmultirate for a single-well case.
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Well Test Analysis
Multirate Superposition Equation
Let pu
(r,t) be the rate-normalized pressurechange in psi /(B/D) created at location rinthe reservoir at a time t from a well
- .
It includes the skin factor if it is evaluatedat the production/injection w ell.
In well testing literature, it is also referredto as the constant uni t-rate pressurechange.
Multirate Superposition Equation
For example, for a fully-penetrating lin e-sourcewell with no w ellbore storage effects in aninfinite-homogeneous reservoir, pu(r,t) would begiven by:
pu(r,t) =
=
+
>
=
w
2
wt
w
2
t
i
rr2skt
rc948.05Ei-
kh
70.6B
rr,kt
rc948.05Ei
kh
70.6B
q
t)p(r,p
,
Multirate Superposition Equation
Now, lets consid er the following rate historyat the producing well:
Time
q1q2
q3
qi
qn
t0=0 t1 tn-1tit2 ti-1t3
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Well Test Analysis
Multirate Superposition Equation
Now, we apply superposition to ob tain thesolution:
( ) ( ) )t(tpqq)t(tpqq(t)pqt)p(r,p 2u231u12u1i ++=( ) ( ) )t(tpqq)t(tpqq 1nu1nn1iu1ii ++++ LL
( ) ( )=
+ =n
0j
juj1ji ttr,pqqt)p(r,p
where q0 = 0, and t0 = 0.
( ) ( )=
=n
1j
j-1uj-1ji ttr,pqqt)p(r,p
OR:
Multirate Superposition Equation
For a line-source fully penetrating well in aninfinite acting reservoir, we can express thesuperposition equation as:
( ) w
n
1j
tj-1ji rr,
kt
rc948.05Eiqq
kh
70.6Bt)p(r,p >
= =
( ) w
n
1j
2
wtj-1ji rr,2s
kt
rc948.05Eiqq
kh
70.6Bt)p(r,p =
+
=
=
Single-Well Convolution Equation
The superposition equation can also bewritten in an m ore general or compactintegral form as:
( ) ( )
ddt
tr,dpqt)p(r,p u
t
0
i
=
For a line-source well:
( )
=
kt
rc948.05exp
kh
70.6B
dt
tr,dp 2tu
convolution equation
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Well Test Analysis
Multiwell/MultirateSu er osition
Multiwell/Multirate Superposition Now, we extend to a more general case
where we can have more than one wellproducing wi th a variable history i n thereservoir.
Lets consider a two-well system, Well Aand Wel l B with the fol lowin rateh istor ies ,at each w ell.
100
700
qB, stb/D
t, hour5
How do we calculate pressure at Well A at t = 25 hr?
0 (buildup)
250
100
qA, stb/D
10 20t, hour
Multiwell/Multirate Superposition
Combining superpositi on in space andsuperposition in time (or ratesuperposition), we can wri te the
case:
( ) ( )= =
=w
kN
1k
n
1j
k
j-1
k
u
k
j-1
k
ji tt,rpqqt)p(r,p
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Well Test Analysis
Multiwell/Multirate Superposition
Suppose, we wish to compute pressure at
Well A at t = 25 hr for the exampleconsiderd, then w e will apply thesuperposition equation as:
( ) ( )
( ) ( ) ( ) ( )
= =
= =
+==
=
3
1j
2
1j
B
j-1
BB
u
B
j-1
B
j
A
j-1
A
w
A
u
A
j-1
A
j
A
wi
2
1k
n
1j
k
j-1
k
u
k
j-1
k
j
k
wi
tt,rpqqtt,rpqq25)t,p(rp
tt,rpqqt),p(rp
k
Pressure change caused
at Well A by Well A itself.
Pressure change caused
at Well A by Well B
Multiwell Convolution Equation
The multiwell/multirate superpositionequation can also be written in an moregeneral or co mpact integral form as:
( ) ( )
=
=
wN
1j
kj
u
t
0
jkk
i ddt
tdpqt),p(rp
Exercise 1
700
qB, stb/D250
100
qA, stb/D
100
t, hour5
Calculate pressure at Well A at t = 25 hr.
0 (buildup)
10 20t, hour
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Well Test Analysis
Exercise 2
qA, stb/D 450250
500
0 shut-in
qB, stb/D
Calculate pressure at Well A at t = 72 hr.
24 48 t, hour 80 t, hour
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