Correlations PVT SCAL

Technical DescriptionRock compressibility

325

Chapter 7Technical Description

PVT property correlations

Rock compressibility

Newman

Consolidated limestone

psi [EQ 7.1]

Consolidated sandstone

psi [EQ 7.2]

Unconsolidated sandstone

psi, [EQ 7.3]

where

is the porosity of the rock

Hall


psi [EQ 7.4]

Cr exp 4.026 23.07�– 44.28�2+( )

6–�10=

Cr exp 5.118 36.26�– 63.98�2+( )

6–�10=

Cr exp 34.012 � 0.2–� �( )6–�10= 0.2 � 0.5� ��

�

Cr3.63 5–�10

2�-------------------------PRa

0.58–=

326 Technical Description Water correlations


psi, [EQ 7.5]

psi,

where

Knaap


psi [EQ 7.6]


psi [EQ 7.7]

where

Water correlations

Compressibility

Meehan

[EQ 7.8]

Cr7.89792 4–�10

2----------------------------------PRa

0.687–= � 0.17�

Cr7.89792 4–�10

2----------------------------------PRa

0.687– �0.17----------� � 0.42818–

�= � 0.17�

is the porosity of the roc

is the rock reference pressure

is

�

Pa

PRa depth over burden gradient 14.7 Pa–+�� 2�

Cr 0.864 4–�10PRa

0.42 PRi0.42–

� Pi Pa–� �--------------------------------- 0.96 7–�10–=

Cr 0.292 2–�10PRa

0.30 PRi0.30–

Pi Pa–--------------------------------- 1.86 7–�10–=

is the rock initial pressure

is the rock reference pressure

is the porosity of the rock

is

is

Pi

Pa

�

PRi depth over burden gradient 14.7 Pi–+�� 2�

PRa depth over burden gradient 14.7 Pa–+�� 2�

cw Sc a bTF cTF2

+ +� � 6–�10=

Technical DescriptionWater correlations

327

where

[EQ 7.9]

where

Row and Chou

[EQ 7.10]

[EQ 7.11]

[EQ 7.12]

[EQ 7.13]

[EQ 7.14]

[EQ 7.15]

[EQ 7.16]

[EQ 7.17]

[EQ 7.18]

a 3.8546 0.000134p–=

b 0.01052– 4.77 7–�10 p+=

c 3.9267 5–�10 8.8 10–�10 p–=

Sc 1 NaCl0.7 0.052– 0.00027TF 1.14 6–�10 TF2

– 1.121 9–�10 TF3

+ +� �+=

is the fluid temperature in ºF

is the pressure of interest, in psi

is the salinity (1% = 10,000 ppm)

TF

p

NaCl

a 5.916365 100 TF 1.0357940– 10 2– TF 9.270048�+��

1TF------ 1.127522 103 1

TF------ 1.006741 105��+�–� �

�+

�+�=

b 5.204914 10 3– TF 1.0482101 10 5– TF 8.328532 10 9–��+�–� �

1TF------ 1.170293–

1TF------ 1.022783 102 ��+� �

�+

�+�=

c 1.18547 10 8– TF 6.599143 11–�10�–�=

d 2.51660 TF 1.11766 2–�10 TF 1.70552 5–�10�–� ��+–=

e 2.84851 TF 1.54305 2–�10 TF 2.23982 5–�10�+–� ��+=

f 1.4814–3–�10 TF 8.2969 6–�10 TF 1.2469 8–�10�–� ��+=

g 2.7141 3–�10 TF 1.5391–5–�10 TF 2.2655 8–�10�+� ��+=

h 6.2158 7–�10 TF 4.0075–9–�10 TF 6.5972 12–�10�+� ��+=

Vw a p14.22------------- b p

14.22------------- c�+� �

NaCl 1 6–�10

d NaCl 1 6–�10� e�+� �

NaCl 1 6–�10� p14.22------------- f NaCl 1 6–�10� g 0.5 p

14.22------------- h ��+�+� �

��–

�

�+�–=

328 Technical Description Water correlations

[EQ 7.19]

Formation volume factor

Meehan

[EQ 7.20]

• For gas-free water

[EQ 7.21]

• For gas-saturated water

[EQ 7.22]

[EQ 7.23]

where

cw

b 2.0 p14.22------------- c NaCl 1 6–�10� f NaCl 1 6–�10� g p

14.22------------- h�+�+� �

�+��+� �

Vw 14.22�------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------=




is the specific volume of Water

is compressibility of Water

TF

p

NaCl

Vw cm3 gram� �

cw 1 psi� �

Bw a bp cp2+ +� �Sc=

a 0.9947 5.8 6–�10 TF 1.02 6–�10 TF2

+ +=

b 4.228 6–�10– 1.8376 8–�10 TF 6.77 11–�10 TF2

–+=

c 1.3 10–�10 1.3855 12–�10 TF– 4.285 15–�10 TF2

+=

a 0.9911 6.35 6–�10 TF 8.5 7–�10 TF2

+ +=

b 1.093 6–�10– 3.497 9–�10 TF– 4.57 12–�10 TF2

+=

c 5 11–�10– 6.429 13–�10 TF 1.43 15–�10 TF2

–+=

Sc 1 NaCl 5.1 8–�10 p 5.47 6–�10 1.96 10–�10 p–� � TF 60–� �

3.23 8–�10– 8.5 13–�10 p+� � TF 60–� �2

+

+

�

+=




TF

p

NaCl

Technical DescriptionWater correlations

329

Viscosity

Meehan

[EQ 7.24]

[EQ 7.25]

Pressure correction:

[EQ 7.26]

where

Van Wingen

[EQ 7.27]

where

Density

[EQ 7.28]

where

Water Gradient:

[EQ 7.29]

�w Sc Sp 0.02414446.04 Tr 252–� ��

�10� �=

Sc 1 0.00187NaCl0.5– 0.000218NaCl2.5

TF0.5 0.0135TF–� � 0.00276NaCl 0.000344NaCl1.5

–� �

+

+

=

Sp 1 3.5 12–�10 p2 TF 40–� �+=




TF

p

NaCl

�w e 1.003 TF 1.479 2–�10– 1.982 5–�10 TF�+� ��+� �=

is the fluid temperature in ºFTF

�w62.303 0.438603NaCl 1.60074 3–�10 NaCl2+ +

Bw-------------------------------------------------------------------------------------------------------------------=


is the formation volume factor

is the Density of Water

NaCl

Bw

�w lb ft3� �

g�w

144.0------------- [psi/ft]=

330 Technical Description Gas correlations

Gas correlations

Z-factor

Dranchuk, Purvis et al.

[EQ 7.30]

[EQ 7.31]

[EQ 7.32]

[EQ 7.33]

[EQ 7.34]

[EQ 7.35]

[EQ 7.36]

where

z 1 a1a2TR�---------

a3

TR3�

---------+ +� ��

Pr a4a5TR�---------+

� ��

Pr2 a5a6Pr

5

TR�-------------------

a7Pr2

TR3�

------------ 1 a8Pr2

+� �exp a8Pr2

–� �

+ + +

+

=

TR�TRTc�--------=

Tc� Tc5E3

9---------� � –=

E3 120 YH2SYCO2

+� �0.9

YH2SYCO2

+� �1.6

–� � 15 YH2S

0.5 YH2S4

–� � +=

Pr0.27PprZTR�

-------------------=

PprPPc�---------=

Pc�PcTc�

Tc YH2S1 YH2S

–� �E3+-----------------------------------------------------------=

is the reservoir temperature, ºK

is the critical temperature, ºK

is the reduced temperature

is the adjusted pseudo critical temperature

is the mole fraction of Hydrogen Sulphide

TR

Tc

TR�

Tc�

YH2S

Technical DescriptionGas correlations

331

[EQ 7.37]

Hall Yarborough

[EQ 7.38]

where

Reduced density ( ) is the solution of the following equation:

[EQ 7.41]

This is solved using a Newon-Raphson iterative technique.

is the mole fraction of Carbon Dioxide

is the pressure of interest

is the critical pressure

is the adjusted pseudo critical Pressure

is the critical temperature, ºK

YCO2

P

Pc

Pc�

Tc

a1 0.31506237=

a2 1.04670990–=

a3 0.57832729–=

a4 0.53530771=

a5 0.61232032–=

a6 0.10488813–=

a7 0.68157001=

a8 0.68446549=

Z0.06125Pprt

Y------------------------------� � exp

1.2 1 t–� �2–� �=

is the pseudo reduced pressure

is

is the reduced density

(where is the pressure of interest and is the critical pressure)

[EQ 7.39]

(where is the critical temperature and is the

temperature in ºR) [EQ 7.40]

Ppr

t 1 pseudo reduced temperature�

Y

PprPPcrit-----------= P Pcrit

tTcritTR

----------=Tcrit TR

Y

0.06125Pprte1.2 1 t–� �2–

– Y Y2 Y3 Y4–+ +

1 Y–� �3----------------------------------------

14.76t 9.76t2– 4.58t3+� �Y2–

90.7t 242.2t2– 4.58t3+� �Y 2.18 2.82t+� �

+

+ 0=

332 Technical Description Gas correlations

Viscosity

Lee, Gonzalez, and Akin

[EQ 7.42]

where


[EQ 7.43]

where

Compressibility

[EQ 7.44]

where

Density

[EQ 7.45]

[EQ 7.46]

�g 10 4– K XpY� �exp=

� 1.4935 10 3–� �pMgzT--------=

BgZTRPscTscP

-------------------=

is the Z-factor at pressure

is the reservoir temperature

is the pressure at standard conditions

is the temperature at standard conditions


Z P

TR

Psc

Tsc

P

Cg1P---

1Z--- Z�

P�------� � –=


is the Z-factor at pressure

P

Z P

�g35.35�scP

ZT-------------------------=

�sc 0.0763�g=

Technical DescriptionOil correlations

333

where

Condensate correction

[EQ 7.47]

where

Oil correlations

Compressibility

Saturated oil

McCain, Rollins and Villena (1988)

[EQ 7.48]

where

is the gas gravity


is the Z-factor

is the temperature in ºR

�g

P

Z

T

�gcorr0.07636�g 350 �con cgr� �� +

0.002636350 �con cgr� �

6084 �conAPI 5.9–� �-------------------------------------------------� ��

+

------------------------------------------------------------------------------------=

is the gas gravity

is the condensate gravity

is the condensate gas ratio in stb/scf

is the condensate API

�g

�con

cgr

�conAPI

co 7.573– 1.450 p� �ln– 0.383 pb� �ln– 1.402 T� �ln 0.256 �API� �ln 0.449 Rsb� �ln+ + + �exp=

is isothermal compressibility, psi-1

is the solution gas-oil ratio at the bubble point pressure, scf/STB

is the weight average of separator gas and stock-tank gas specific gravities

is the temperature, oR

Co

Rsb

�g

T

334 Technical Description Oil correlations

Undersaturated oil

Vasquez and Beggs

[EQ 7.49]

where

• Example

Determine a value for where psia, scf /STB, ,

��API, �F.

• Solution

[EQ 7.50]

/psi [EQ 7.51]

Petrosky and Farshad (1993)

[EQ 7.52]

where


Saturated systems

Three correlations are available for saturated systems:

• Standing

co5Rsb 17.2T 1180�g– 12.61�API 1433–+ +� � 5–�10

p------------------------------------------------------------------------------------------------------------------------------=

is the oil compressibility 1/psi

is the solution GOR, scf/STB

is the gas gravity (air = 1.0)

is the stock tank oil gravity, �API

is the temperature in �F

is the pressure of interest, psi

co

Rsb

�g

�API

T

p

co p 3000= Rsb 500= �g 0.80=

�API 30= T 220=

co5 500� � 17.2 220� � 1180 0.8� �– 12.61 30� � 1433–+ +

3000 5�10--------------------------------------------------------------------------------------------------------------------------------=

co 1.43 5–�10=

Co 1.705 7–�10 Rs0.69357�� g

0.1885�API0.3272T0.6729p 0.5906–=


is the average gas specific gravity (air = 1)

is the oil API gravity, oAPI

is the temperature, oF

is the pressure, psia

Rs

�g

�API

T

p


335

• Vasquez and Beggs

• GlasO

• Petrosky

These are describe below.

Standing

[EQ 7.53]

where

• Example

Use Standing’s equation to estimate the oil FVF for the oil system described by the data �F, scf / STB, , .

• Solution

[EQ 7.55]

[EQ 7.56]

bbl / STB [EQ 7.57]

Vasquez and Beggs

[EQ 7.58]

where

Bo 0.972 0.000147F1.175+=

= Rs(�g/�o)0.5 + 1.25 T [EQ 7.54]

and

is the oil FVF, bbl/STB



is the oil specific gravity = 141.5/(131.5 + �API)


F

Bo

Rs

�g

�o

T

T 200= Rs 350= �g 0.75= �API 30=

�o141.5

131.5 30+------------------------- 0.876= =

F 350 0.750.876-------------� � 0.5

1.25 200� �+ 574= =

Bo 1.228=

Bo 1 C1Rs C2 C3Rs+� � T 60–� ��API�gc

-----------� ��

+ +=



is the stock tank oil gravity, �API

is the gas gravity

Rs

T

�API

�gc


, , are obtained from the following table:

Table 7.1 Values of C1, C2 and C3 as used in [EQ 7.58]

API � 30 API > 30

C1 4.677 10 -4 4.670 10-4

C2 1.751 10 -5 1.100 10-5

C3 -1.811 10 -8 1.337 10 -9

C1 C2 C3


337

• Example

Use the Vasquez and Beggs equation to determine the oil FVF at bubblepoint pressure for the oil system described by psia, scf / STB,

, and �F.

• Solution

bb /STB [EQ 7.59]

GlasO

[EQ 7.60]

[EQ 7.61]

[EQ 7.62]

where


[EQ 7.63]

where

Undersaturated systems

[EQ 7.64]

pb 2652= Rsb 500=

�gc 0.80= �API 30= T 220=

Bo 1.285=

Bo 1.0 10A+=

A 6.58511– 2.91329 Bob�log 0.27683 Bob�log� �2

–+=

Bob� Rs�g�o-----� �� 0.526

0.968T+=



is the oil specific gravity,


is a correlating number

Rs

�g

�o �o 141.5 131.5 �API+� ��=

T

Bob�

Bo 1.0113 7.2046 5–�10 Rs0.3738

�g0.2914

�o0.6265

------------------� ��

0.24626T0.5371+3.0936

+=

is the oil FVF, bbl/STB



Bo

Rs

T

Bo Bobexp co pb p–� �( )=


where

Viscosity

Saturated systems

There are 4 correlations available for saturated systems:

• Beggs and Robinson

• Standing

• GlasO

• Khan

• Ng and Egbogah

These are described below.

Beggs and Robinson

[EQ 7.65]

where

Taking into account any dissolved gas we get

[EQ 7.66]

where

• Example

Use the following data to calculate the viscosity of the saturated oil system. �F, , scf / STB.

• Solution

is the oil FVF at bubble point, psi.

is the oil isothermal compressibility, 1/psi


is the bubble point pressure, psi

Bob pb

co

p

pb

�od 10x 1–=

is the dead oil viscosity, cp

is the temperature of interest, �F

is the stock tank gravity

x T 1.168– exp 6.9824 0.04658�API–( )=

�od

T

�API

�o A�odB

=

A 10.715 Rs 100+� � 0.515–=

B 5.44 Rs 150+� � 0.338–=

T 137= �API 22= Rs 90=


339

�cp

�cp

Standing

[EQ 7.67]

[EQ 7.68]

where

[EQ 7.69]

[EQ 7.70]

[EQ 7.71]

where

Glas�

[EQ 7.72]

[EQ 7.73]

[EQ 7.74]

and

[EQ 7.75]

[EQ 7.76]

x 1.2658=

�od 17.44=

A 0.719=

B 0.853=

�o 8.24=

�od 0.32 1.8 7�10

�API4.53

-------------------+� �� 360

T 260–------------------� � a=

a 100.43 8.33

�API-----------+� �

=



T

�API

�o 10a� � �od� �b=

a Rs 2.2 7–�10 Rs 7.4 4–�10–� �=

b 0.68

108.62 5–�10 Rs

----------------------------------- 0.25

101.1 3–�10 Rs

-------------------------------- 0.062

103.74 3–�10 Rs

-----------------------------------+ +=

is the solution GOR, scf/STBRs

�o 10a �od� �b=

a Rs 2.2 7–�10 Rs 7.4 4–�10–� �=

b 0.68

108.62 5–�10 Rs

----------------------------------- 0.25

101.1 3–�10 Rs

-------------------------------- 0.062

103.74 3–�10 Rs

-----------------------------------+ +=

�od 3.141 10�10 T 460–� � 3.444– �APIlog� �a=

10.313 T 460–� �log� � 36.44–=


where

Khan

[EQ 7.77]

[EQ 7.78]

where

Ng and Egbogah (1983)

[EQ 7.79]

Solving for , the equation becomes,

[EQ 7.80]

where

uses the same formula as Beggs and Robinson to calculate Viscosity

Undersaturated systems

There are 5 correlations available for undersaturated systems:

• Vasquez and Beggs

• Standing



T

�API

�o �obppb-----� � 0.14–

e2.5 4–�10–� � p pb–� �

=

�ob0.09�g

0.5

Rs1 3� �r

4.5 1 �o–� �3---------------------------------------------=

is the viscosity at the bubble point

is

is the temperature, �R

is the specific gravity of oil

is the specific gravity of solution gas

is the bubble point pressure


�ob

�r T 460�

T

�o

�g

pb

p

�od 1+� �log �log 1.8653 0.025086�API– 0.5644 T� �log–=

�od

�od 10101.8653 0.025086�API– 0.5644 T� �log–� �

1–=

is the “dead oil” viscosity, cp



�od

�API

T


341

• GlasO

• Khan

• Ng and Egbogah

These are described below.

Vasquez and Beggs

[EQ 7.81]

where

where

Example

Calculate the viscosity of the oil system described at a pressure of 4750 psia, with �F, , , scf / SRB.

Solution

psia.

�cp

�cp

Standing

[EQ 7.82]

�o �obppb-----� � m=

= viscosity at

= viscosity at

= pressure of interest, psi

= bubble point pressure, psi

�o p pb�

�ob pb

p

pb

m C1pC2exp C3 C4p+( )=

C1 2.6=

C2 1.187=

C3 11.513–=

C4 8.98 5–�10–=

T 240= �API 31= �g 0.745= Rsb 532=

pb 3093=

�ob 0.53=

�o 0.63=

�o �ob 0.001 p pb–� � 0.024�ob1.6 0.038�ob

0.56+� �+=


where

GlasO

[EQ 7.83]

where

Khan

[EQ 7.84]

where

Ng and Egbogah (1983)

[EQ 7.85]

Solving for , the equation becomes,

[EQ 7.86]

where

uses the same formula as Beggs and Robinson to calculate Viscosity

is the viscosity at bubble point



�ob

p

pb

�o �ob 0.001 p pb–� � 0.024�ob1.6 0.038�ob

0.56+� �+=




�ob

p

pb

�o �ob e9.6 5–�10 p pb–� �

�=




�ob

p

pb

�od 1+� �log �log 1.8653 0.025086�API– 0.5644 T� �log–=

�od

�od 10101.8653 0.025086�API– 0.5644 T� �log–� �

1–=

is the “dead oil” viscosity, cp



�od

�API

T


343

Bubble point

Standing

[EQ 7.87]

where

Example:

Estimate where scf / STB, �F, , �API.

Solution

[EQ 7.88]

psia [EQ 7.89]

Lasater

For

[EQ 7.90]

For

[EQ 7.91]

[EQ 7.92]

For

[EQ 7.93]

For

Pb 18Rsb�g

---------� �� 0.83 yg�10=

= mole fraction gas =

= bubble point pressure, psia

= solution GOR at , scf / STB

= gas gravity (air = 1.0)

= reservoir temperature,�F

= stock-tank oil gravity, �API

yg 0.00091TR 0.0125�API–

Pb

Rsb P Pb�

�g

TR

�API

pb Rsb 350= TR 200= �g 0.75= �API 30=

�g 0.00091 200� � 0.0125 30� �– 0.193–= =

pb 18 3500.75----------� � 0.83 0.193–�10 1895= =

API 40�

Mo 630 10�API–=

API 40�

Mo73110

�API1.562

---------------=

yg1.0

1.0 1.32755�o MoRsb�� +-----------------------------------------------------------------=

yg 0.6�

Pb0.679exp 2.786yg( ) 0.323–� �TR

�g-----------------------------------------------------------------------------=

yg 0.6�


[EQ 7.94]

where

Example

Given the following data, use the Lasater method to estimate .

, scf / STB, , �F, . [EQ 7.95]

Solution

[EQ 7.96]

[EQ 7.97]

psia [EQ 7.98]

Vasquez and Beggs

[EQ 7.99]

where

Example

Calculate the bubblepoint pressure using the Vasquez and Beggs correlation and the following data.

, scf / STB, , �F, . [EQ 7.100]

Solution

Pb8.26yg

3.56 1.95+� �TR�g

----------------------------------------------------=

is the effective molecular weight of the stock-tank oil from API gravity

= oil specific gravity (relative to water)

Mo

�o


API < 30 API > 30

C1 0.0362 0.0178

C2 1.0937 1.1870

C3 25.7240 23.9310

pb

yg 0.876= Rsb 500= �o 0.876= TR 200= �API 30=

Mo 630 10 30� �– 330= =

yg550 379.3�

500 379.3� 350 0.876 330�� +------------------------------------------------------------------------- 0.587= =

pb3.161 660� �

0.876--------------------------- 2381.58= =

PbRsb

C1�gexpC3�APITR 460+----------------------� ��

--------------------------------------------------

1C2------

=

yg 0.80= Rsb 500= �g 0.876= TR 200= �API 30=


345

psia [EQ 7.101]

GlasO

[EQ 7.102]

[EQ 7.103]

where

for volatile oils is used.

Corrections to account for non-hydrocarbon components:

[EQ 7.104]

[EQ 7.105]

[EQ 7.106]

[EQ 7.107]

pb500

0.0362 0.80� �exp 25.724 30680---------� �

------------------------------------------------------------------------------

11.0937----------------

2562= =

Pb� �log 1.7669 1.7447 Pb�� log 0.30218 Pb�� log� �2

–+=

Pb�Rs�g-----� �� 0.816 Tp

0.172

�API0.989

---------------

� ��

=

is the solution GOR, scf / STB

is the gas gravity

is the reservoir temperature,�F

is the stock-tank oil gravity, �API

Rs

�g

TF

�API

TF0.130

PbcPbc

CorrCO2 CorrH2S CorrN2��=

CorrN2 1 a1�API a2+– TF a3�API a4–+ �YN2

a5�APIa6 TF a6�API

a7 a8–+ YN22

+

+

=

CorrCO2 1 693.8YCO2TF1.553–

–=

CorrH2S 1 0.9035 0.0015�API+� �YH2S– 0.019 45 �API–� �YH2S+=


where

Marhoun

[EQ 7.109]

where


[EQ 7.111]

[EQ 7.108]



is the mole fraction of Nitrogen



a1 2.65 4–�10–=

a2 5.5 3–�10=

a3 0.0391=

a4 0.8295=

a5 1.954 11–�10=

a6 4.699=

a7 0.027=

a8 2.366=

TF

�API

YN2

YCO2

YH2S

pb a· Rsb �g

c �od TR

e� � � �=


is the gas gravity

is the reservoir temperature,�R

[EQ 7.110]

Rs

�g

TR

a 5.38088 3–�10=

b 0.715082=

c 1.87784–=

d 3.1437=

e 1.32657=

pb 112.727Rs

0.5774

�g0.8439

-------------------X�10 12.340–=


347

where

GOR

Standing

[EQ 7.112]

where

Example

Estimate the solution GOR of the following oil system using the correlations of Standing, Lasater, and Vasquez and Beggs and the data:

psia, �F, , . [EQ 7.113]

Solution

scf / STB [EQ 7.114]

Lasater

[EQ 7.115]

For

[EQ 7.116]

For


is the average gas specific gravity (air=1)

is the oil specific gravity (air=1)


X 4.561 5–�10 T1.3911 7.916 4–�10 �API1.5410–=

Rs

�g

�o

T

Rs �gp

18yg�10

--------------------� �� 1.204

=

is the mole fraction gas =





yg 0.00091TR 0.0125�AP–

Rs

�g

TF

�API

p 765= T 137= �API 22= �g 0.65=

Rs 0.65 765

18 0.15–�10----------------------------� � 1.204

90= =

Rs132755�oygMo 1 yg–� �-----------------------------=

API 40�

Mo 630 10�API–=

API 40�


[EQ 7.117]

For

[EQ 7.118]

For

[EQ 7.119]

where is in �R.

Example


psia, �F, , . [EQ 7.120]

Solution

[EQ 7.121]

[EQ 7.122]


Vasquez and Beggs

[EQ 7.124]

where C1, C2, C3 are obtained from Table 7.3.

Example


psia, �F, , . [EQ 7.125]


API < 30 API > 30

C1 0.0362 0.0178

C2 1.0937 1.1870

C3 25.7240 23.9310

Mo73110

�API1.562

---------------=

p�g T� 3.29�

yg 0.359ln1.473p�g

T---------------------- 0.476+� � =

p�g T� 3.29�

yg0.121p�g

T---------------------- 0.236–� �

0.281=

T

p 765= T 137= �API 22= �g 0.65=

yg 0.359ln 1.473 0.833� � 0.476+ � 0.191= =

Mo 630 10 22� �– 410= =

Rs132755 0.922� � 0.191� �

410 1 0.191–� �------------------------------------------------------- 70= =

Rs C1�gpC2exp

C3�APITR 460+----------------------� ��

=

p 765= T 137= �API 22= �g 0.65=


349

Solution


GlasO

[EQ 7.127]

[EQ 7.128]

[EQ 7.129]

where

Marhoun

[EQ 7.130]

where

Rs 0.0362 0.65� � 765� �1.0937exp 25.724 22� �

137 460+--------------------------- 87= =

Rs �g�API0.989

TF0.172

---------------

� ��

Pb�1.2255

=

Pb� 102.8869 14.1811 3.3093 Pbc� �log–� �0.5

– �=

PbcPb

CorrN2 CorrCO2 CorrH2S+ +---------------------------------------------------------------------------=




is the mole fraction of Nitrogen



�g

TF

�API

YN2

YCO2

YH2S

Rs a �gb

�oc Td pb� � � ��

e=

is the temperature, �R

is the specific gravity of oil


is the bubble point pressure

[EQ 7.131]

T

�o

�g

pb

a 185.843208=

b 1.877840=

c 3.1437–=

d 1.32657–=

e 1.398441=



[EQ 7.132]

where

Separator gas gravity correction

[EQ 7.134]

where

Rspb

112.727------------------- 12.340+� � �g

0.8439 X�101.73184

=

[EQ 7.133]

is the bubble-point pressure, psia


X 7.916 4–�10 �g1.5410 4.561 5–�10 T1.3911–=

pb

T

�gcorr �g 1 5.912 5–�10 �API TFsepPsep114.7-------------� � log� � �+� �

=

is the gas gravity

is the oil API

is the separator temperature in �F

is the separator pressure in psia

�g

�API

TFsep

Psep


351

Tuning factors

Bubble point (Standing):

[EQ 7.135]

GOR (Standing):

[EQ 7.136]

Formation volume factor:

[EQ 7.137]

[EQ 7.138]

Compressibility:

[EQ 7.139]

Saturated viscosity (Beggs and Robinson):

[EQ 7.140]

[EQ 7.141]

[EQ 7.142]

Undersaturated viscosity (Standing):

[EQ 7.143]

Pb 18 FO1Rsb�g

---------� �� 0.83 �g�10�=

Rs �gP

18 FO1�g�10�

-----------------------------------� �� 1.204

=

Bo 0.972 FO2� 0.000147 FO3 F1.175� �+=

F Rs�g�o-----� �� 0.5

1.25TF+=

coFO4 5Rsb 17.2TF 1180�g– 12.61�API 1433–+ +� � 5–�10

P---------------------------------------------------------------------------------------------------------------------------------------------=

�o A�odB

=

A 10.715 FO5 Rs 100+� � 0.515–�=

B 5.44 FO6 Rs 150+� � 0.338–�=

�o �ob P Pb–� � FO7 0.024�ob1.6 0.038�ob

0.56+� � �+=

352 Technical Description Oil / water

SCAL correlations

Oil / waterFigure 7.1 Oil/water SCAL correlations

where

Corey functions

Water

(For values between and )

[EQ 7.144]

Kro

Krw

0 1

Swmin

Kro(Swmin)

Swmin Swcr 1-Sorw

Sorw’Krw(Sorw)

,Swmax,

Krw(Swmax)

is the minimum water saturation

is the critical water saturation (�� )

is the residual oil saturation to water ( )

is the water relative permeability at residual oil saturation

is the water relative permeability at maximum water saturation (that is 100%)

is the oil relative permeability at minimum water saturation

swmin

swcr swmin

sorw 1 sorw– swcr�

krw sorw( )

krw swmax( )

kro swmin( )

Swcr 1 Sorw–

krw krw sorw( )sw swcr–

swmax swcr– sorw–---------------------------------------------------

Cw=

Technical DescriptionGas / water

353

where is the Corey water exponent.

Oil


[EQ 7.145]

where

Gas / waterFigure 7.2 Gas/water SCAL correlations

Cw

swmin 1 sorw–

kro kro swmin( )swmax sw– sorw–

swmax swi– sorw–-----------------------------------------------

Co=

is the initial water saturation and

is the Corey oil exponent.

swi

Co

KrgKrw

0 1Swmin Swcr Sgrw

Swmin,Krg(Swmin)

Sgrw,Krw(Sgrw)

Swmax,Krw(Smax)

354 Technical Description Gas / water

where

Corey functions

Water


[EQ 7.146]

where is the Corey water exponent.

Gas


[EQ 7.147]

where


is the critical water saturation (�� )

is the residual gas saturation to water ( )

is the water relative permeability at residual gas saturation


is the gas relative permeability at minimum water saturation

swmin

swcr swmin

sgrw 1 sgrw– swcr�

krw sgrw( )

krw swmax( )

krg swmin( )

swcr 1 sgrw–

krw krw sgrw( )sw swcr–

swmax swcr– sgrw–---------------------------------------------------

Cw=

Cw

swmin 1 sgrw–

krg krg swmin( )swmax sw– sgrw–

swmax swi– sgrw–-----------------------------------------------

Cg=


is the Corey gas exponent.

swi

Cg

Technical DescriptionOil / gas

355

Oil / gasFigure 7.3 Oil/gas SCAL correlations

where

Corey functions

Oil


[EQ 7.148]

0

Sliquid

1-Sgcr 1-SgminSwmin Sorg+Swmin

Swmin,Krg(Swmin)

Sorg+Swmin,Krg(Sorg)

Swmax,Krw(Smax)


is the critical gas saturation (�� )

is the residual oil saturation to gas ( )

is the water relative permeability at residual oil saturation


is the oil relative permeability at minimum water saturation

swmin

sgcr sgmin

sorg 1 sorg– swcr�

krg sorg( )

krg swmin( )

kro swmin( )

swmin 1 sorg–

kro kro sgmin( )sw swi– sorg–

1 swi– sorg–------------------------------------

Co=

356 Technical Description Oil / gas

where:

Gas


[EQ 7.149]

where

Note In drawing the curves is assumed to be the connate water saturation.


is the Corey oil exponent.

swi

Co

swmin 1 sorg–

krg krg sorg( )1 sw– sgcr–

1 swi– sorg– sgcr–--------------------------------------------------

Cg=


is the Corey gas exponent.

swi

Cg

swi

Technical DescriptionPseudo Variables

357

Pseudo variables

Pseudo pressure transformationsThe pseudo pressure is defined as:

[EQ 7.150]

It can be normalized by choosing the variables at the initial reservoir condition.

Normalized pseudo pressure transformations

[EQ 7.151]

The advantage of this normalization is that the pseudo pressures and real pressures coincide at and have real pressure units.

Pseudo time transformationsThe pseudotime transform is

[EQ 7.152]

Normalized pseudo time transformationsNormalizing the equation gives

[EQ 7.153]

Again the advantage of this normalization is that the pseudo times and real times coincide at and have real time units.

m p� � 2 p� p� �z p� �---------------------- pd

pi

p

�=

mn p� � pi�izipi

--------- p� p� �z p( )--------------------- pd

pi

p

�+=

pi

m t� � 1� p� �ct p� �------------------------ td

0

t

�=

mn t� � �ici1

� p� �ct p� �------------------------ td

0

t

�=

pi

Documents

Correlations PVT SCAL