2_Introduction to Nodal Analysis

7/28/2019 2_Introduction to Nodal Analysis

1/20

Inflow PerformanceRelationship

for Multiphase Flow


2/20

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Darcys law is no longer valid below the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.


3/20

IPR Below the Bubblepoint

qmax

qO

O

qb

Rate

pwf

pb

Pressure

p


4/20

Vogels IPR

Vogels Behavior

IPR Curve - Vogel plotted the data using the

following dimensionless variables

maxq

qand

P

P

r

wf


5/20

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


6/20

Vogels Equation

Mathematical model for Vogels curve

2

8.02.01p

p

p

p

q

q wfwf

max


7/20

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):


8/20

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,


9/20

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


10/20

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


11/20

Flow Efficiency

Standing extended the effect of skin on

Vogels IPR equation and

ppFEpp wfrrwf )(' 1


12/20

Flow Efficiencygraphical representation

Pr

Pwf

Pwf

rw

rd

Positive skin ~ Damaged wellbore or

Reduced wellbore radius


13/20

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


14/20

Multiphase Flow:Combination Darcy/Vogel

qmax

PI pb

1.8

qO

O

qb

Rate

pwf

pb

Pressure

p

8.1

max

bb

pPIqq


15/20

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


16/20

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


17/20

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


18/20

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


19/20

Transient Flow Equation

K

rCt ethours

2

948

tSrC

K

PPKhq

wt

oo

wfr

o

log87.023.3log6.1622


20/20

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

2_Introduction to Nodal Analysis

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