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8/8/2019 Ch-2 PVT-oil-2007 http://slidepdf.com/reader/full/ch-2-pvt-oil-2007 1/20 Chapter 2 Pressure-Volume-Temperature for Oil PVT analysis ± PVT relationship int) ( po bubble of because complex  simple condition undergroun Oil condition  scf Gas p p p

Ch-2 PVT-oil-2007

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Page 1: Ch-2 PVT-oil-2007

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Chapter 2

Pressure-Volume-Temperature for Oil

PVT analysis ± PVT relationship

int)( pobubbleof  becausecomplex

 simple

conditiond undergroun

Oil 

condition scf  

Gas

p

p

p

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Three main oil PVT parameters

The three main parameters required to relate

surface to reservoir volumes for an oil

reservoir:

- Rs: The solution (or dissolved) gas-oil ratio

- Bo

: The oil formation volume factor 

- Bg: The gas formation volume factor 

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Definition of the basic oil PVT parameters

- Rs: The solution (or dissolved) gas-oil ratioat reservoir P&T

- Bo: The oil formation volume factor 

at reservoir P&T

- Bg: The gas formation volume factor 

Note : Standard Condition

oil STB

 gasSCF ][!

oil S 

 gasdissolved oil  R )(

][

 gasSCF 

 gas free B][!

 psia p

 F T 

7.14

60

!

r!

),(,, T  P  f   B B  g o s ! )(,, P  f   B B  g o s !For T = const.

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Determination and Conversion of PVT

Data Determination of three main oil PVT parameters(RS, Bo, Bg)

-> PVT = f (P only) by routine laboratory analysis

Conversion of PVT data, as presented by the laboratory, to

the form required in the field,

Laboratory - an absolute set of measurements

Field - depend up as the manner of surface

separation of the gas and oil

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The complexity of Oil PVT

For gas, PVT relation

-- Simple relation

For oil, PVT relation > Complex;

PVT parameters must be measured by

laboratory analysis of crude oil samples.

>Relationship between surface and

reservoir hydrocarbon volumes.

rcf   scf  

 zT 

 p E 

nz T  pV 

][37.35 !!

!

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The complexity of Oil PVT The complexity in relating surface volumes of hydrocarbon

production to their equivalent volumes in the reservoir canbe appreciated by considering the following figures:

Undersaturated oil

gas saturated oil

+ free gas (or liberated solution gas)

They are traveling in reservoir at different velocity

saturated oil

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How to divide the observed surface gas

production into liberated and dissolved gas

volumes in the reservoir?

Control in relating surface volumes of production to underground withdrawal isgained by knowing the three oil PVTparameters which can be measured by

laboratory experiments performed onsamples of the reservoir oil, plus itsoriginally dissolved gas.

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Oil Reservoir & Surface Volume

 ± Above Bubble PointUndersaturated oil

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Oil Reservoir & Surface Volume

 ± Below Bubble Point

gas saturated oil

+ free gas (or liberated

solution gas)

They are traveling in

reservoir at different

velocity

Saturated oil

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The instantaneous gas-oil ratio or 

producing gas-oil ratio

the instantaneous gas-oil

ratio or producing

gas-oil ratio

)()()(

S  R R R

S withdrawal d Undergroun

 g  so v

)(S 

S  R

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Bo as Function of Pressure

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Rs as Function of Pressure

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Bg and E as Function of Pressure

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Producing Gas-oil Ratio (R) as Function of Pressure

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Exercise 2.1 - Underground withdrawal

Given: measured at t during the

producing life

Calculate:

(1) Underground withdrawal rate expressed in

terms of  x & y (2) Underground withdrawal rate

if  pres = 2400 psia , qo =2500 STB/D , qg =2.125 MMscf/D and

PVT data in fig. 2.5(a)~(c) (p.51) or table 2.4 (P.65)

(3) Pressure gradient of oil =?

if 

)(

)(

 DS  yq

 DS  xq

 g 

o

!

!

?)] _ ([ ! D

 Rin

?)] _ ([ ! D

 R Bin

 psia pair c sat  ft lb  g mo 2400),1(67.0),.(3/8.52 r  !!!! V

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(1) R = y/x = qg/qo [=] SCF

/STBp is known Bo ; Rs ; Bg

Underground withdrawal

)())(()(

)()()(

SCF  R B B

STBSCF  R

 x

 y

 DSTB xq

 D R B B x

STB R B B

 DSTB xq

 g  s g 

ooo

!

!!

)()( D

 R B B R x

 y x  g  s!

 g  so g ot  B R x

 y x B xqqq !! )(

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P = 2400 psia

from table 2.4 Bo =1.1822RB/STB

Rs = 352 SCF/STB

Bg = 0.0012 RB/SCF

x = qo = 2500 STB/D

;y = qg = 2.125 MMSCF/D

 D Rq o 5.29551822.12500 !v!

 D

 R Bq  g  14940012.0)352

2500

10125.2(2500

6

!v

vv!

 D Rqqq  g ot  5.444914945.2955 !!!

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(3)

To find , applying mass balance, such as

Mass of 1 STB of oil Mass of Bo RB of oil

+ = +

Rs scf dissolved gas at dissolved gas in the

standard condition reservoir condition

 g d  D

d o V!

o V

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3

3

3

/37.471822.1615.5

]076

3.067.0

[352

8.

5261

5.

5

615.5

]0763.0[615.5

615.5

615.5

615.5615.5

1

61458.5)()(1)(

)](1)[(]1

615.5)(1)([

 ft lb

r  R R

 R

 R

 ft 

 RS 

 ft 

lb

S S 

S  R

lb

S S 

lb

m

o

 g  sosc

o

 gsc sosc

or 

or o gsc sosc

om

or 

 sm

 gscm

osc

!v

vv

!

v!

!

!

vvv!

vv

 V V V V

 V V V

 V

 V V

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 ft  psi

 ft inlb

in ft 

 ft lb

 s

 ft 

 ft 

 slug 

 s ft 

lb

 slug 

 ft 

lb g 

d  D

 f   f  

m

mor 

/329.0

1

144

37.47

144

137.47

37.47)2.32()2.32

137.47(

22

2

3

2323

!

!!

vv!vv!! V