IPR - Leslie Thompson

8/17/2019 IPR - Leslie Thompson

1/57

Inflow Performance

Relationship

IPR

by

Dr. Leslie Thompson


2/57

Production System


3/57

Fluid Production

• Path of produced fluids

– Reseroir

– Perforations! "rael pac#! etc.

– Downhole e$uipment! casin"! tubin".

• %i&in" with "as'lift "as

– (ellhead! production cho#es – Flow lines! manifolds

– Separator


4/57

Fluid Production

• In each flow se"ment! the fluids interact

with the production components

– Pressure! temperature and flow elocityare altered.

– Fluid properties constantly chan"in".


5/57

Pressure Losses


6/57

Driin" Force

• The driin" force that moes fluids alon" the

reseroir and production system is the ener"y

stored in the form of compressed fluids in thereseroir.

– )s the fluids moe alon" the system components!

pressure drop occurs. The pressure in the

direction of flow continuously decreases from thereseroir pressure to the final downstream

pressure alue at the separator.


7/57

IPR

• The flow of oil! "as and water from the

reseroir is characteri*ed by the Inflow

Performance relationship. – IPR is a measure of +Pressure losses, in the

formation.

• The functional relationship between flow rate

and bottomhole pressure is called IPR. – Indicator of well performance.

( ) ( ) p f p p f q wf ∆=−=


8/57

Productiity Inde&

• %easure of the well-s capacity to

produce fluids from reseroir to

wellbore.• Definition Fluid production rate for / psi

pressure drop from reseroir to

wellbore.


9/57

Productiity Inde&

( )

( ) ( )( )t pt pt q

PI wf −

=

PI Productiity Inde&! ST01d1psi

( )t p )era"e pressure in the well-s draina"e area! psi.

( )t pwf 0ottomhole flowin" pressure! psi

q(t ) Production rate! ST01d


10/57

Parameters 2ontrollin" IPR or PI

• Roc# Properties – Permeability! Relatie permeability

• Fluid Properties – 0o! µo! 0w! µw! co! cw – 0"! µ"! *! c"

• Flow Re"ime in the Reseroir – Darcy Flow

– 3on'Darcy 4inertial! Forchheimer5 flow

• Phases Flowin" – Sin"le Phase 46il! 7as5

– Two Phases 4687! 68(! 78(5

– Three Phases 46878(5


11/57

Parameters 2ontrollin" IPR or PI

• Fluid Saturation Distribution

• Reseroir Drie %echanisms

• Formation Dama"e of Stimulation

• Relationship between IPR1PI and other

ariables 4#! fluid properties! relatie

permeability! etc.5 can be ery comple&.


12/57

Determination of PI• Testin" the (ell

– %easure q, pwf ,

• Deelop PI models 4e$uations5

– PI 9 f (k, k ro , k rg , k rw , Bo , B g , Bw , etc.5

– e.".! steady'state sin"le phase flow

p

( )

( ) ( )( )t pt pt q

PI wf −=

+−

µ

=−

=

s

r

r B

kh

p p

q PI

w

ewf

2

1ln2.141


13/57

Steady'State Radial Flow

( e l l ! r w D a m a " e d : o n e , ( r s , k s )

R e s e r o i r , ( r e , k )


14/57

Radial Flow

• 2onsider sin"le'phase radial flow in a cylindrical

reseroir of radius r e with a well in the center

produced at a constant rate q sc ST01D.

• If the outer boundary is held at a constant pressure

pe, after some time we will hae steady'state flow!

q(r )/ B = q sc for all r and t where q(r ) is the flow rate in

R01D throu"h a cylinder of radius r . %oreoer! ∂ p/∂t =0.

• Darcy-s Law becomes

( ) ( )

r

pr

B

kh

r

p

B

k rhr q sc

∂

∂

µ

=

∂

∂

µ

π×= −

2.141

210127.1 3


15/57

Radial Flow

• In the outer *one! r e > r > r s

• Separatin" ariables and inte"ratin"

( ) r p

r B

khr q sc ∂

∂µ= 2.141

( )

( )

µ=−

µ= ∫ ∫

r

r

kh

Bqr p p

r

dr

kh

Bqdp

e sce

r

r

sc

p

r p

ee

ln2.141

2.141


16/57

Radial Flow

• )t r = r s! the pressure is "ien by

• For the inner 4dama"ed *one5 Darcy-s

law

( )

µ−= s

e sc

e s r

r

kh

Bq

pr p ln

2.141

r

pr

B

hk q s sc ∂

∂µ

=2.141


17/57

Radial Flow

• Separatin" ariables and inte"ratin"

• ;liminatin" p4r s5

( )

( )

( ) ( )

µ=−

µ=

∫ ∫ r

r

hk

Bqr pr p

r

dr

hk

Bqdp

s

s

sc s

r

r s

sc

r p

r p

s s

ln2.141

2.141

( )

+

µ=

µ+

µ=−

r

r

k

k

r

r

kh

Bq

r

r

hk

Bq

r

r

kh

Bqr p p

s

s s

e sc

s

s

sc

s

e sce

lnln2.141

ln2.141ln2.141


18/57

Radial Flow

• ;&pandin" and simplifyin"

• ;aluatin" at the wellbore

( )

−+

µ=

+

−

+

µ=−

r

r

k

k

r

r

kh

Bq

r

r

k

k

r

r

r

r

r

r

kh

Bqr p p

s

s

e sc

s

s

s s

s

e sce

ln1ln2.141

lnlnlnln2.141

−+

µ=−

w

s

sw

e scwf e

r

r

k

k

r

r

kh

Bq p p ln1ln

2.141


19/57

S#in

• (e define

• The term s is #nown as the 4


20/57

3otes on S#in

• S#in is a dimensionless number

• If k = k s! the reseroir is not dama"ed!and s = 0.

• If k > k s! the reseroir is dama"ed! and s

> 0.• If k < k s! the reseroir is stimulated! and

s < 0.

−=

w

s

s r

r

k

k s ln1


21/57

Productiity Inde&

• For steady'state radial flow

• (ell-s productiity is increased if – s is reduced

– r w is increased

– µ is reduced – h is increased

+

µ

=

−

=

sr r B

kh

p p

q PI

w

ewf e ln2.141


22/57

;&ample

• ) reseroir with the followin" properties

is flowed at a bottomhole pressure of

=>?? psi. 2alculate the flow rate.Su""est two ways of increasin" the

wells production rate by a factor of @.


23/57

PropertiesProperty Value Source

Permeability, md 5.2 Pressre !ransient "nalysis

!#i$%ness, &t 53 'ell ls

*is$sity, $+ 1.7 lid "nalysis

rmatin *lme a$tr, -/! 1.1 lid "nalysis

'ellbre radis, &t 0.32

'ell drainae area, a$re 40 'ell s+a$in

%in &a$tr 10 Pressre !ransient "nalysis

"erae reserir +ressre, +si 535 Pressre !ransient "nalysis


24/57

Solution

• First! we conert the well-s draina"e area to an

e$uialent draina"e radius usin"

• The reseroir area is A=? acre 9 A=? acre×=B>A?ft@1acre 9 @CC=?? ft@. The e$uialent reseroir

draina"e area is r e 9 @ECE ft.

Ar e =π 2

+−

µ

=−

=

sr

r

B

kh

p p

q PI

w

ewf

2

1

ln2.141

( ) ( )

( ) ( ) ( )

!/+si 051.0

105.032.0

27ln7.11.12.141

532.5=

+−

= PI


25/57

Solution

• Solin" for Rate

• To double the rate! we could double the pressure

drop. The current pressure drop is //B> psi! so

double this alue is @@C? psi! which means that we

must reduce the bottomhole pressure to >AB> – @@C?

9 B?E> psi.

( ) ( ) !/d .34500535 051.0 S p p PI q wf =−=−=


26/57

Solution

• )nother method of doublin" the well-s rate would beto double the Productiity Inde& to a alue of ?.//@@ST01psi.

• (e would do this by decreasin" the s#in factor bystimulatin" the well.

• To determine the new s#in factor

• snew = 0.7.

( ) ( )

( )( )( )

+−

==

new s

PI

5.032.0

27ln7.11.12.141

532.51122.0


27/57

Pseudosteady State Flow

• Rate of chan"e of pressure with time at

each point in a closed reseroir is

constant. – ;ach +point, in the reseroir contributes

e$ually to the flow.

• Productiity Inde&

+−

µ

=−

=

sr

r B

kh

p p

q PI

w

ewf

4

3ln2.141


28/57


29/57

Diet* Shape Factors


30/57

3otes

• (ith all else e$ual! asymmetric well'reseroirconfi"urations hae lower PI and flow ratecompared with a symmetric well'reseroirconfi"uration

• For psuedosteady state flow with constant wellflowin" pressure! aera"e reseroir pressure andflow rate decline continuously due to depletion.

• For sin"le phase flow! PI does not chan"e with

chan"es in chan"es in flow rate and aera"ereseroir pressure due to depletion.


31/57

)bsolute 6pen Flow Potential

• )6F – For a "ien well'reseroir pair and aera"e

reseroir pressure! )6F is the ma&imum theoretical

flow rate that the well can proide.

• )6F is useful in analy*in" IPR in terms of

( ) p PI qq

p p PI q

AOF

wf

==−=

ma

ma

erssq

q

p

pwf


32/57

Sin"le Phase IPR

q

wf P

( )wf p p PI q −=r P

maq

pwf p

q PI

∂∂

−=

p PI q =ma


33/57

Future Linear IPR

q

wf P

( )wf r P P q −=

r P

maq

)s time t incresases! reseroir

pressure P r decreases and cumulatie

production N p increases.


34/57

IPR for 7as (ells

• 7as PT properties are a function of pressure 4µ)! ! , B g , c g 5

• If p G @>?? psi and steady'state "as flow 4$ %scf1D5

• Pseudosteady'state flow

+

µ=− s

r

r

kh

" # q p p

w

ewf e ln

142422

+

µ=− s

C r

A

kh

" # q p p

Aw

wf 2

22 245.2ln2

11424


35/57

PI for 7as (ells

• Pseudosteady'state flow

• PI for "as wells is a function of pressure.

– (hen aera"e reseroir pressure chan"es! "as propertieschan"e.

– PI is not constant durin" the well-s life

( )

+

µ

+=

−

==

sC r A" #

p pkh

p p

q PI PI

Aw

wf

wf 2245.2ln2

11424


36/57

Typical 7as IPR

Gas IPR

?

/???

@???

B???

=???

>???

A???

? /???? @???? B???? =???? >???? A???? C????

Rate, MSCF/d

W e l l P r e s s u r e , p s

i


37/57


38/57

Pseudosteady State 7as Flow

• For Pseudosteady state flow

• 0ac# Pressure ;$uation 4Rawlins – Schellhardt5

++

µ=− $q s

C r

A

kh

" # q p p

Aw

wf 2

22 245.2ln2

11424

222 %q&q p p wf +=−

+

µ= sC r

A

kh

" # & Aw

2

245.2ln2

11424

kh

"$ # %

µ=

1424


39/57

Limitin" 2ases

• (hen non'Darcy flow is ne"li"ible 4%


40/57

7enerali*ed 7as Flow

• Limitin" forms of the "as flow e$uation can be

"enerali*ed as

– 3e"li"ible non'Darcy flow! n 9 /

– Purely non'Darcy flow! n 9 ?.>

– 0oth components play a role! ?.> G n G /

• ;$uation can be rearran"ed as

( )

n

wf p pC q

22

−=

( ) ( ) ( )C n

qn

p p wf l1

l1

l 22 −=−


41/57

0ac# Pressure )nalysis

• )t constant reseroir pressure! flow at

four different flow rates.

– %easure stabili*ed bottomhole pressure ateach rate.

• Plot

• Strai"ht line with slope /1n.• (ith n! can construct the IPR.

( ) ( )q p p wf l()erssl() 22 −


42/57

%ultiphase Flow

• 3eed rates of oil! "as! water

– PI o , PI g , PI w

+

µ

=−= s

C r

A B

hk

p p

q PI

Aw

oo

o

wf

oo

2

245.2ln

2

12.141

+

µ

=−

=

sC r A B

hk

p p

q PI

Aw

ww

w

wf

ww

2245.2ln212.141

+

µ

=−

=

sC r

A B

hk

p p

q PI

Aw

g g

g

wf

g

g

2

245.2ln2

12.141

roo kk k =

rww kk k =

rg g kk k =


43/57

%ultiphase IPR

• IPR under multiphase flow conditions cannot beeasily calculated.

• The most accurate method is by solin" thee$uations "oernin" the flow in the porousmedia throu"h a reseroir simulator.

• The IPR is so important to Production;n"ineers that simplified or empirical methodsto estimate it are necessary.

• The most common correlations are o"el andFet#oich


44/57

o"el IPR

• o"el used a numerical reseroir simulator to

"enerate the IPR.


45/57

*el 6P-

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1mar f w

2

ma

.02.01

−

−=

p p

p p

qq wf wf


46/57

o"el IPR

• o"el IPR can be obtained from well tests.

• )lthou"h the method was deeloped for

solution "as drie reseroirs! the e$uation is"enerally accepted for other drie

mechanisms as well.

• It is found to "ie e&cellent results for any well

with a reseroir pressure below the oil bubblepoint! i.e.! saturated reseroirs.


47/57

Hndersaturated Reseroirs

P r

P %

P

q

( )wf r P P q −=

?

q% q'&(


48/57


P r

P %

P

q

q% q'&(

q) = q * q%

P r ) = P %


49/57


P r

P %

P

q

( )wf r P P q −=

q% q'&(

−

−=

−−

%

wf

P

P

qq

qq

P

P

%

wf

%

%

2

ma

.02.01


50/57


51/57

;&ample

ien data,

p = 2,400 +sia

qo = 100 !/d

wf = 100 +sia

enerate in&l8 +er&rman$e $re sin

bt# *el9s and et%i$#9s (n = 1)e:atins.


52/57

;&ample

Solution •*el9s ;:atin

etermine qo,ma

2

ma, 2400

100.0

2400

1002.01

100

−

−=

oq

250ma, =oq !/d

*els e:atin be$mes

2

2400.0

24002.01

250

−

−= wf wf o p pq


53/57

;&ample

•et%i$#9s ;:atin

etermine qo,ma

2

ma, 2400

1001

100

−=

oq

.22ma, =oq !/d


54/57

;&ample

*el et%i$#

pwf , +sia qo, !/d pwf , +sia

qo, !/d

0 250.0 0 22.

00 225.0 00 214.3

1200 175.0 1200 171.5

100 100.0 100 100.0

2400 0.0 2400 0.0


55/57


56/57

Multi-layer inflow Performance


57/57

rss?l8 bet8een 2 @ayers

Documents

IPR - Leslie Thompson