1 Fluids Characterization & Sampling

A special course by IFP Training for PDVSA

Gas Fields Reservoir Engineering & DevelopmentFluids Characterization and SamplingSalvatore Zammito

EP ‐ Fluids characterization & sampling

©2014 ‐IFP

Training

©2014 ‐IFP

Training

Fluids classification

EP ‐ Fluids characterization & sampling 3

©2014 ‐IFP

Training

Pressure

Temperature

Tres, Pres

Critical point

Tc

p2

p1

Separator

C

Dry gas


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Training

Dry gas

Dry gas is virtually puremethane

The two‐phase envelope issmall and lies below reservoirconditions and to the left ofsurface condition

The fluid is theoretically gasboth in the reservoir and atthe surface


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Training

Wet gas

The composition of a wet gas containsstill fewer heavy molecules. Since, thephase diagram covers a much smallertemperature range; the pressuredepletion path in the reservoir does notenter the two‐phase region

The composition of a wet gas containsstill fewer heavy molecules

Since, the phase diagram covers a muchsmaller temperature range; the pressuredepletion path in the reservoir does notenter the two‐phase region

The reservoir fluid is gas throughout thelife of the reservoir

However, separator conditions lie withinthe two‐phase envelope, indicating thatsome liquid will condense at the surface


©2014 ‐IFP

Training

Pressure

Temperature

Tres, PresCritical point

Tc

p2

p1

Separator

C

Wet gas


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Training

Wet gas

The composition of a wet gas containsstill fewer heavy molecules. Since, thephase diagram covers a much smallertemperature range; the pressuredepletion path in the reservoir does notenter the two‐phase region

The composition of a wet gas containsstill fewer heavy molecules

Since, the phase diagram covers a muchsmaller temperature range; the pressuredepletion path in the reservoir does notenter the two‐phase region

The reservoir fluid is gas throughout thelife of the reservoir

However, separator conditions lie withinthe two‐phase envelope, indicating thatsome liquid will condense at the surface


©2014 ‐IFP

Training

Dry gas

Dry gas is virtually puremethane

The two‐phase envelope issmall and lies below reservoirconditions and to the left ofsurface condition

The fluid is theoretically gasboth in the reservoir and atthe surface


©2014 ‐IFP

Training

Condensate gas

Retrograde gases have even fewerheavy molecules than volatile oils

The critical point shifts to left anddownward in the phase diagram and thecritical temperature is usually less thanreservoir temperature

Retrograde condensate appears in thereservoir pore spaces at pressure belowthe dew point pressure. Throughoutmost of the reservoir, since the amountof liquid in the pore space is usually lessthan critical oil saturation the effectivepermeability to this condensate is zeroand little is produced

Along line 2 to 3, the condensate buildsup at first and then revaporizes at thelower pressures

This behavior is typical for constantcomposition expansion type application

At reservoir conditions can we see re‐evaporation?


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Training

Condensate gas

Retrograde gases have even fewer heavymolecules than volatile oils

The critical point shifts to left anddownward in the phase diagram and thecritical temperature is usually less thanreservoir temperature

Retrograde condensate appears in thereservoir pore spaces at pressure belowthe dew point pressure. Throughoutmost of the reservoir, since the amountof liquid in the pore space is usually lessthan critical oil saturation the effectivepermeability to this condensate is zeroand little is produced

Along line 2 to 3, the condensate buildsup at first and then revaporizes at thelower pressures

This behavior is typical for constantcomposition expansion type application

At reservoir conditions can we see re‐evaporation?


©2014 ‐IFP

Training

Oil classification

API Definition

°API =

where d is the specific gravity of stock tank oil: relative to water at 60°F (15.6°C)

Condensate or very light oils: d < 0.80 (above 45°API)Light oil: 0.80 < d < 0.86 (33 to 45°API)Medium oils: 0.86 < d < 0.92 (22 to 33°API)Heavy oils: 0.92 < d < 1 (lower than 22°API)

131.5 d

141.5


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Training

Petroleum fluids classification

Basic data characterizing well liquid effluent

1. Production data

• API gravity

• Gas oil ratio− GOR < 500 m3/m3 Oil

− 500 < GOR < 1000 m3/m3 Oil or gas condensate

− GOR > 1000 m3/m3 Gas condensate

− GOR > 15000 m3/m3 Wet gas

2. Chemical composition


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Training

Fluids classification

Separator Density Composition

Well test GOR of STO C1 C2 C3 C4 C5 C6+

(Sm 3/m 3) (kg/m 3)

Heavy oil < 10 > 900 3 4 5 8 80

Standard oil < 500 800 ‐ 900 45 4 4 3 2 42

Critical fluid # 700 750 ‐ 850 55 10 8 5 6 16

Gas condensate 700 ‐ 800 700 ‐ 800 75 8 5 2 2 8

Wet gas > 15000 700 ‐ 800 90 4 3 1 1 1

Dry gas infinite 95 3 1 1

(mole %)


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Training

Gas properties


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Training

UNDERSATURATED OIL RESERVOIRS

GAS- CONDENSATE RESERVOIRS

SINGLE PHASE

GAS RESERVOIRS

Phase envelope


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Training

Phase envelopes for various fluids


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Training

Van der Waals EOS:

p =RT

(V‐b)

a

V2

Peng‐Robinson EOS:

p =RT

(V‐b)

aα

V(V+b)+b(V‐b)

Constants a and b are calculated based on the gas molar composition:

a = 0.45724 R2Tc2/pc and b = 0.07780 R Tc/pc α = α (TR, pR)

Equations of state


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Training

p = pressure

V = volume

n = number of moles

R = gas constant (R = 8.314 kJ kgmole ‐1 K‐1 or 10.73 psia ft3 lbmole ‐1 R‐1 )

T = absolute temperature

z = compressibility factor

• Negligible molecular volume

• Negligible molecular attraction/repelling forces

pV = nRT

Ideal gas:

Real gas:

pV = znRT

Gas equation of state


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Training

Function of:

gas composition

pressure

temperature

Ideal gas: z = 1

Real gas, S.C.: z = 1

Real gas, R.C.: z = 0.8 ‐ 0.9

IDEAL GAS

z

p

T

z = 1

p = 1 atm

Gas compressibility factor


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Training

z

p

T=Tr

rnnn RTnzVp

scscscsc RTnzVp

r

sc

sc

n

sc

nn T

T

V

V

p

pz

For a generic pressure condition (n)

and reservoir temperature Tr:

nzRTpV

EOS for real gas:

z Determination


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Training

For one molecular species:

• Reduced pressure

Critical pressurepr =

Absolute pressure

• Reduced temperature

Tr =Absolute temperature

Critical temperature

Natural gas, in the same pseudo‐reduced pressure and pseudo‐reducedtemperature conditions, present the same volumetric behavior and thus the same zvalue

Law of Corresponding States

z Factor


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Training

For n‐ component gas

• Pseudo‐critical pressure

• Pseudo‐critical temperature

Molar fraction yi =Mol ith‐ component

Moltot

ppc =

n

1ii,ci py

Tpc =

n

1ii,ci Ty

z Factor


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Training

Pseudo‐reduced pressure

Pseudo‐reduced temperature

ppr =Absolute pressure

Pseudo-critical pressure

p

ppc

=

Tpr =Absolute temperature

Pseudo-critical temperature

T

Tpc

=

z Factor


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Training

Gas: determination of Tpc and Ppc


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Training


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Training

Z Calculation

Exercise

Calculate the Z factor of a gacondensate

dg= 0.7 Pres= 450 b Tres= 150 °C


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Training

Volume of free gas at reservoir conditions in m3 (or bbl) occupied by 1 m3 (or 1 scf) of the same gas measured at standard conditions

Bg =V(p,T)

Vsc

Bg =zpsc

pTsc

T

Bg

p

Bg =nzRT

p

psc

nzscRTscBg ranges between:

0.002 m3/sm3 (0.0004 rb/scf) 0.05 m3/sm3 (0.009 rb/scf)

Gas Formation Volume Factor (FVF)


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Training

ρg =Mp

zRT

M

V=

ρair = 1.225 kg/m3

ρairG =

ρg

SC

Mair = 28.96

Mair

G =Mg

ρg(p,T) =ρg,sc

Bg

Gas density and gas gravity


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Training

cg 1

p

dpVcg =

dV1‐

cg

p

cg = 2÷4 10-4 psi-1

cg = 1.5÷3.5 10-2 MPa-1

Gas compressibility


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Training

cg =1

p

RT

pM

RT

M

dp

d

Substituting in

dp

dc

1

Ideal gas


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Training

dz

dp‐

1

pcg =

1

z

zRT

pM

For constant temperature:

dp

dz

zRT

pM

zRT

M

dp

d2

1

dp

dz

z

p

zRT

M

dp

d1

Substituting in

dp

dz

z

p

zRT

M

pM

zRTc 1

dp

dc

1

Real gas


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Training

In dimensionless terms:

prpcprpcg dpp

dz

zppc

11

prprgpcr dp

dz

zpcpc

11

Pseudo‐reduced pressurep

ppr ppc

=

Real gas


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Training


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Training

μg = 0.01 ÷ 0.02 cP (mPa s)

mg

p

T

Gas viscosity


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Training

Assuming

• Porous medium saturated by a single fluid

• No interactions between matrix and fluid

• Laminar flow

dpkμ

q

A=

dzdL

+( )dLg ρ

pp+dp

dL

XY

ZL

Generalized Darcy’s law


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Training

The permeability is an intrinsic property of the porous medium,independent from the flowing liquid. It indicates how easily fluidscan flow through the rock

k permeability [L2]

1 μm2 = 10‐12 m2

1 Darcy = 0.987 10‐12 m2 1 μm2

Absolute permeability


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Training

pm =p1+p2

2pmkg =

bkL 1 +( )

PER

MEA

BILITY

KL

1pm

Propane

Ethane

Methane

Klinkemberg effect

For increasing molecular weight, gases exhibit a behavior whichtends to that of liquids because – all other conditions being the same– the kinetic energy is smaller


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Training

ko

kwkeff

kg

water

oil

gas

0 < S 1

0 keff kabs

Effective permeability

When a second or third phase is introduced, the resultingpermeability to each phase is called “effective”

It represents the conductivity of each phase at a specificsaturation

It provides an extension of Darcy’s law to presence andmovement of more than a single fluid within the pore space

The fluids interfere with each other, and individual effectivepermeability of each phase as well as their sum is lower thanabsolute permeability


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Training

dpkg

μg

qg A= ( )dr

Oil

Water

Gas

mw

dpkwqw A= ( )dr

dpko

mo

qo A= ( )dr

Multiphase flow


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Training

WATER SATURATION

RE

LA

TIV

E P

ER

ME

AB

ILIT

Y

Relative permeability: gas‐water system


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Training

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Training

Oil and gas behavior between the

reservoir and the surface


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Training

Black‐Oil model

Definitions:

VGF

VHF

VGHS

VHS

VGS

BoVHFVHS

BgVGFVGS

RsVGHSVHS

T , PF F

T , Ps s


©2014 ‐IFP

Training

VGF

VHF

VGHS

VHS

BoV

VBg

V

V

RsV

VRv

V

V

HF

HS

GF

GS

GHS

HS

HGS

GS

'

T PF F,

T Ps s,

VGS

VHGS

Extended Black‐Oil model


Definitions:

©2014 ‐IFP

Training

VGF

VHF

BoV

VBg

V

V

RsV

VRv

V

V

HF

HS

GF

GS

GHS

HS

HGS

GS

'

T PF F,

T Ps s,

VGS

VHGS

Gas Condensate model


Definitions:

©2014 ‐IFP

Training

Vgres

Bg = Vgstd

Pres x Vres = Z x R x Tres

Pstd x Vstd = 1 x R x Tstd

Pstd

Bg = Tstd

x Z x Tres

Pres

Field Units

Pstd = 14.7 psia Tstd = 520 ° R Bg = 0.028269 xZ x T

Pvol/vol

Metric Units

Pstd = 1.01325 barsa Tstd = 288 ° K Bg = 0.00352 xZ x T

Pvol/vol

SI Units

Pstd = 101325 Pa (a) Tstd = 288 ° K Bg = 351.8 xZ x T

Pvol/vol

Vol of 1 mole of gas at std conditions {1 atm, 288 K (15°C)} : 23.63 dm3

° R

psia

° K

bars a

Pa

° K

Eg = 1 / Bg

Vol of 1 mole of gas at normal conditions {1 atm, 273 K (0°C)} : 22.414 dm3

460 +60°F(T st.)

273+15(Tst)

Main gas properties


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Training

Main gas properties

Definitions for gas

Gas Specific Gravity

(air =1) = gas/air = Mwgas/Mwair

= Mwgas /28.978

Gas density

gas = Mwgas/Vair(molar) = Mwgas/ 23.645

Vair(1 mole of air) = 23.645

gas = * 1.225 kg/m³

Air density

Mwair = 28.9784

air =28.9784/23.645 =1.225 kg/m³


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Training

Reservoir fluid sampling

Sampling objectives

Bottom hole sampling

Surface sampling


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Training

Sampling objectives

The objective is to obtain a sample of fluid that is identical(representative) to the reservoir fluid

Those samples are needed to

• Determine the fluid type

• Estimate Hydrocarbons in place and reserves

• Measure or estimate the fluid characteristics that will be used todesign the production facilities or used in numerical models topredict reservoir performance


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Training

Pwf

Separatorgas

oil

Production tubing

Vo Reservoir : Pr, Tr

Psep. ‐ Tsep.Pt


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Training

Bottom hole sampling

This type of sampling is preferred since it guarantees the bestfluid representativity

Disadvantage: high cost

Various tools are used

• MDT (modular dynamic formation tester)

• SRS (single phase reservoir sampler)

• MFE (multiple flow evaluator)

• PCT (pressure controlled test system)

• APR (annulus pressure responsive tool)


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Training

Surface sampling

Oil and gas samples are collected from separator, at surface

These samples are recombined in the laboratory, based on themeasured gas/oil ratio, in order to make up a reservoir fluid asrepresentative as possible


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Training

Surface sampling

Psep, Tsep

gas

GORsep = Qgas/Qoil

Psto, TstoGORsto

Tank oilambient conditions

Reservoir fluidPr, Tr

Gas sampling bottle

Oil sampling bottle


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Training

PVT studies


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Training

Constant Composition Expansion (Oil)

Differential Vaporization (Oil)

Separator test/Flash Liberation (Oil)

Constant Composition Expansion (Condensate Gas)

Constant Volume Depletion (Condensate Gas)


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Training

PVT experiment: constant volume depletion for a

condensate gas

Objective

• Simulation of reservoir production for condensate gas and volatile oil

Main properties derived from this experiment

• Liquid drop out curve

• Volumetric factor of the wellstream during depletion

• Wellstream composition during depletion

Remarks

• Be sure to have vapor and liquid at each step

• Problems imply negative calculated liquid compositions.

• Plot Ki(P) = Yi(P) / Xi(P)− All lines monotonic and smooth, No crossing, Ordered with volatilities

• Use the Hoffman‐Crump‐Hocott plot:− Log(KiP) is linear regarding B(1/Tbi‐1/T) with B=(log(Pci) –log(Pref)) / (1/Tbi‐1/Tci)


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Training

Constant volume gas depletion

Gas

Oil

P1 = Psat P3

V sat

P2 < Psat

V 2

V 3

P3 < P2P2P2

V sat

Steps 0 1 2 3 1 2

Steps

0: Measure reference Volume Vsat à P1=Psat

1: Decrease P to P2<P1 lowering the piston, condensation occurs

2: Reset the cell volume to Vsat, pushing up the piston

3: Measure Oil and gas volumes

1: Decrease P to P3<P2, …


©2014 ‐IFP

Training

PVT experiment: constant volume depletion for a gas

condensate Objective

• Simulation of reservoir production

Main properties derived from this experiment

• Liquid drop out curve

• Volumetric factor of the wellstream during depletion

• Wellstream composition during depletion

P1 = Psat P3

V sat

P2 < Psat

V 2

V 3

P3 < P2P2P2

V sat


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Training

PVT experiment: constant mass expansion for a gas

condensate

Temperature

Pressure

T res.

P1 > Psat

P2 = Psat (retrograde)

P3

P4

P5

P6

P7< P sat (normal)

C

P1 > Psat P6P2 = Psat P4P3 < Psat P5 P7


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Training

CCE – Gas condensate

Gas condensateReservoir Temperature : 105°CBubble point pressure : 286.72 abs bar

CCE

Bo = (Vol Liq at P,T)/(Vol Liq tank at 1b, 15)CGRres = (Vol Liq(PT)/Bo)/(VolVap(PT)/B'g) (CGR of liquid of CCE)CGRvap = Vol Liq(1,15)/Vol Gas Export Std (CGR of vapor of CCE)

Pressure Liquid CGR res CGR vap Gas Z factor Gas FVF Bg Gas viscosity Liq viscosity Liquid FVF Tank liquid STD gasdrop out B'g (P,T) (P,T) Bo density Sp Gravity Z'

(abs bar) (% DPPV) (Sm3/Sm3) (Sm3/Sm3) (m3/Sm3) (m3/Sm3) (cP) (cP) (m3/Sm3) (kg/m3)300 0,00000 0,000335 0,9371 0,00437 0,00410 0,0306 0,0000 736,5542 0,66771 0,99955

286,72 0,00000 0,000000 0,000335 0,9234 0,00451 0,00423 0,0296 0,3958 1,50266 736,5513 0,66771 0,98497280 0,18990 0,000006 0,000330 0,9169 0,00458 0,00430 0,0290 0,3950 1,49726 734,8984 0,66767 0,97764260 0,72054 0,000022 0,000313 0,8992 0,00483 0,00454 0,0273 0,3897 1,48252 730,2538 0,69179 0,95741240 1,21693 0,000038 0,000297 0,8839 0,00514 0,00483 0,0255 0,3804 1,46926 726,0515 0,69163 0,93979220 1,69928 0,000053 0,000281 0,8716 0,00552 0,00520 0,0238 0,3688 1,45612 722,2864 0,69147 0,92519200 2,18046 0,000069 0,000265 0,8626 0,00600 0,00566 0,0221 0,3582 1,44088 718,9011 0,69130 0,91407180 2,65607 0,000085 0,000247 0,8576 0,00661 0,00625 0,0205 0,3525 1,42099 715,8132 0,69112 0,90694160 3,10163 0,000101 0,000230 0,8568 0,00742 0,00703 0,0190 0,3549 1,39472 712,9628 0,69095 0,90420140 3,48253 0,000117 0,000212 0,8604 0,00849 0,00807 0,0177 0,3679 1,36180 710,3388 0,69081 0,90609120 3,76650 0,000130 0,000197 0,8683 0,00998 0,00950 0,0165 0,3937 1,32326 707,9747 0,69073 0,91270100 3,92883 0,000140 0,000184 0,8803 0,01213 0,01155 0,0156 0,4355 1,28072 705,9404 0,69073 0,9240180 3,94906 0,000145 0,000176 0,8963 0,01542 0,01470 0,0148 0,4989 1,23578 704,3532 0,69082 0,9400260 3,79928 0,000145 0,000175 0,9160 0,02102 0,02003 0,0142 0,5957 1,18974 703,4348 0,69102 0,9608540 3,41620 0,000135 0,000184 0,9394 0,03239 0,03082 0,0137 0,7569 1,14332 703,7007 0,69129 0,9869720 2,61173 0,000108 0,000215 0,9668 0,06694 0,06384 0,0133 1,1156 1,09628 706,7299 0,69162 1,019991 0,76071 0,000033 0,000300 0,9981 1,39472 1,33172 0,0130 3,4851 1,03850 720,5066 0,69192 1,06264


©2014 ‐IFP

Training

CVD gas condensate

CVD

Bo = (Vol Liq at P,T)/(Vol Liq tank at 1b, 15)CGRres = (Vol Liq(PT)/Bo)/(VolVap(PT)/B'g) (CGR of liquid of CCE)CGRvap = Vol Liq(1,15)/Vol Gas Export Std (CGR of vapor of CCE)

Pressure Liquid CGR res CGR vap Gas Z factor Gas FVF Bg Gas viscosity Liq viscosity Liquid FVF Tank liquid STD gasdrop out B'g (P,T) (P,T) Bo density Sp Gravity

(abs bar) (% DPPV) (Sm3/Sm3) (Sm3/Sm3) (m3/Sm3) (m3/Sm3) (cP) (cP) (m3/Sm3) (kg/m3) z'300 0,00000 0,000335 0,9371 0,00437 0,00410 0,0306 0,0000 736,5542 0,66771 0,99955

286,72 0,00000 0,00000 0,000335 0,9234 0,00451 0,00423 0,0296 0,3958 1,50266 736,5527 0,66771 0,98497280 0,18990 0,00001 0,000330 0,9169 0,00458 0,00430 0,0290 0,3950 1,49726 734,8984 0,66767 0,97764260 0,71165 0,00002 0,000313 0,8992 0,00483 0,00454 0,0273 0,3897 1,48251 730,2536 0,66752 0,95741240 1,17507 0,00004 0,000297 0,8839 0,00514 0,00483 0,0255 0,3807 1,46906 726,0471 0,66738 0,93979220 1,60059 0,00006 0,000281 0,8716 0,00552 0,00520 0,0238 0,3701 1,45531 722,2668 0,66722 0,92520200 1,99701 0,00008 0,000264 0,8628 0,00600 0,00566 0,0221 0,3617 1,43877 718,8485 0,66705 0,91411180 2,35475 0,00011 0,000246 0,8579 0,00661 0,00625 0,0205 0,3595 1,41694 715,7107 0,66687 0,90704160 2,65082 0,00015 0,000227 0,8572 0,00742 0,00703 0,0190 0,3669 1,38848 712,8026 0,66671 0,90436140 2,86316 0,00019 0,000209 0,8609 0,00850 0,00807 0,0177 0,3864 1,35365 710,1205 0,66659 0,90628120 2,98103 0,00023 0,000193 0,8689 0,00998 0,00950 0,0165 0,4201 1,31392 707,6961 0,66654 0,91288100 3,00630 0,00030 0,000180 0,8809 0,01213 0,01156 0,0156 0,4713 1,27121 705,5845 0,66662 0,9241380 2,94877 0,00038 0,000172 0,8966 0,01542 0,01471 0,0148 0,5448 1,22730 703,8681 0,66687 0,9400860 2,81934 0,00052 0,000171 0,9160 0,02102 0,02003 0,0142 0,6485 1,18359 702,6973 0,66736 0,9610140 2,62107 0,00076 0,000187 0,9387 0,03242 0,03080 0,0137 0,7978 1,14097 702,4529 0,66821 0,9880420 2,32134 0,00145 0,000250 0,9650 0,06735 0,03089 0,0133 1,0369 1,10000 704,6793 0,66965 1,026231 1,22196 0,01788 0,000944 0,9966 1,51858 0,03124 0,0122 2,3203 1,05053 733,4063 0,67168 1,157016


©2014 ‐IFP

Training

Liquid drop out

0,00000

0,50000

1,00000

1,50000

2,00000

2,50000

3,00000

3,50000

4,00000

4,50000

0 50 100 150 200 250 300 350

CCE

CVD

PRESSURE

Liquid drop out %


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Training

CVD – Comparison Z and Z’

0,6000

0,7000

0,8000

0,9000

1,0000

1,1000

1,2000

0 50 100 150 200 250 300 350

Z CVD

Z' CVD

PRESSURE

Z factor


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Training

PVT experiment: constant mass expansion

Fluide initialP = Pgis. (450 bar)


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Training

P = Dew point press – 100 mbar358 bar



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Training

P = Dew point pressure – 3 bar5% liquid drop out



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Training

During the depletion



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Training

Near the maximum liquid drop out



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Training

Maximum liquid drop out:41% P = 205 bar



Documents

1 Fluids Characterization & Sampling