Post on 03-Apr-2018
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Inflow PerformanceRelationship
for Multiphase Flow
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Multiphase Flow
Bubblepoint pressure (pb)
Pressure at which first bubble of gas is released
from reservoir oils
Darcys law is no longer valid below the
bubble point.
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IPR Below the Bubblepoint
qmax
qO
O
qb
Rate
pwf
pb
Pressure
p
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Vogels IPR
Vogels Behavior
IPR Curve - Vogel plotted the data using the
following dimensionless variables
maxq
qand
P
P
r
wf
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Vogel Curve
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
q/qmax
pwf/p
r
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Vogels Equation
Mathematical model for Vogels curve
2
8.02.01p
p
p
p
q
q wfwf
max
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Vogels Equation
qma
x
q2
q
qb
Pwf
Pb
Pressure
Pr
q
x
bx qqq max
bqqq 2
2
2
2 8.02.01
b
wf
b
wf
x P
P
P
P
q
q
For a test @ point 2 (below bubble point):
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Vogels Equation
2
26.12.0
b
wf
b
x
wf P
P
Pq
P
q1. Differentiate the previous equation with respect toPwf
2. PI is equal to the slope atPwf=Pb , so:
br
b
x
bbrb PPP
1.8qqsoPPPIq
b
xbrb
P1.8qPIthen,PPPIqsince,
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Vogels Equation
b
x
P
1.8qPI,thatknowing
and, qmax = qb + qx
Then,
So: qmax =
8.1
PPIq bx
8.1PPIq bb
Note that Vogel does no account for damaged or stimulated wells
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Flow Efficiency
Standing came up with the concept of Flow
Efficiency. Ifpwf is defined as the BH flowing
pressure for an undamaged well, then:
0'
'
1
0''
2
1
sforpp
pp
sforppppFE
wfr
wfr
wfr
wfrDamaged well
Undamaged well
Stimulated well
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Flow Efficiency
Standing extended the effect of skin on
Vogels IPR equation and
ppFEpp wfrrwf )(' 1
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Flow Efficiencygraphical representation
Pr
Pwf
Pwf
rw
rd
Positive skin ~ Damaged wellbore or
Reduced wellbore radius
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Standings Extesion of Vogels
IPR So Vogels IPR can be re-written as:
2'
8.0'
2.01r
wf
r
wf
max p
p
p
p
q
q
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Multiphase Flow:Combination Darcy/Vogel
qmax
PI pb
1.8
qO
O
qb
Rate
pwf
pb
Pressure
p
8.1
max
bb
pPIqq
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Composite IPR
2
max
max
8.02.01)(
8.1
b
wf
b
wf
bb
wf
b
b
p
p
p
pqqqq
:pointbubblethebelowPanyfor
pPIqq
A
Aq-qqq bmaxb
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Exercise 2
qo= 200 bpd pb= 3,000 psi
pwf= 2,000 psi pr= 4,000 psi
Find: qmax and qo for pwf= 1,000 psi
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Answers to Exercise 2
5111.0
2
000,3
000,28.0
000,3
000,22.01
A
1080.0
85.851,1
200
5111.08.1
000,3000,3000,4
200
PI
1. Solve for A:
2. Solve for PI:
3. Solve for qb:
bpdPPPIq brb 10830004000108.0
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Answers to Exercise 2, cont..
4. Solve for qmax:
bpd
PPIqq bb 288180108
8.1
3000108.0108
8.1max
5. Find: qo for Pwf= 1000
8444.0
2
000,3
000,18.0
000,3
000,12.01
A
bpdAq-qqq bmaxbo 2608444.0108288108
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Transient Flow Equation
K
rCt ethours
2
948
tSrC
K
PPKhq
wt
oo
wfr
o
log87.023.3log6.1622
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Jones Gas IPR
bqaqPP wfr 222
qKH
Srr0.472lnTZ10x
qrh
ZTxP
w
e3
w
2
p
g
424.1
1016.32
12
2
a
PabbAOFP
r
2
422