api - behavoiur of hydrocarbons.pdf

8/14/2019 API - Behavoiur of Hydrocarbons.pdf

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BEH VIOR OFHYDROC RBONS


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PHYSICAL BEHAVIOR OF HYDROCARBONSINDEX

INTRODUCTION . . . . .Solid PhaseLi quid Phase . .Vapor Gas) PhaseEquations of State

BASIC PHASE BEHAVIOR . . . . . .The Pressure Temperature Diagram for a Pure SubstanceVapor Pressure of Liquids. .Vapor Pressure and Temperature . . . . . .Vapor Pressure of Mixtures .Reid Vapor Pressure . .Phase Diagram for a Multi-Component MixtureBl ack Oil Reservoi r . . .Volatile Oil Reservoir .as Condensate ReserYoirGas Reservoir . . .Effect of Depth . . . . .API Gravity . . . . . . . .Volume Shr inkage Upon MixingBEHAVIOR OF IDEAL GASES . . . . . . . . . . . .Change of Volume with Mass P and T constant)Change of Volume with Temperature P and mass constant)

Change of Volume with Pressure T and mass constant) .Genera l Ideal Gas Equation .Gas Mixtures .Relationship Between Wight and Mole (or VolumeApparen t Molecular Weightas Partial Pressure .Ideal Gas Density .

NON-IDEAL GASES . . . . . . . .The Compressibility FactorOPTIONAL SECTION . . . . . . Density (Compressibility) of Gas MixturesSUMM RY . . . . . . . . . . .V LIDATION - METRIC UNITS . .SOLUTIONS TO PROB LEM S - METRIC UNITSVALIDATION - ENGLISH UNITSSOLUTIONS TO PROBLEMS - ENGLISH UNITS

1234556911

11131617171818192021222324252729313233333439394347484950


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6T6Tt rItr/hrm or mtremkmm3mJ/hrm3/dkcalkcal/hr

kcal/m3keal/kggmkgkg/ cm 2 or barbars a

galgpmgphfefmefdMefdcfd

BTUBTU/hrMBTU hrMMBTU hrBTU/Cll i tBTU 1b1bpsipsiabb1BPDOPD

BWPDMBPDs i t

ABBREVIATIONS SYMBOLS USED IN THIS M NU LMEANINGTemperature difference

Pressure differenceEXAMPLETemp i f f is 10

Pres i f f i s 15 psi or barsMETRIC UNIT ABBREVIATIONS

l i t erl i ters per hourmetercentimeterkilometercubic meterscubic meters percubic meters perkilocalo rieskilocalories perkilocalories per

meter of gaskilocaloriesgramkilogram

per

hourdayho urcubickilogram

kilograms per square cen-timeter of pressure

kilograms per sq em ofabsolute pressure

square meter

10 1tr : 10 l i te rs20 Itr/hr : 20 l i ters per hour15 m: 15 meters10 em: 10 centimeters5 km: 5 kilometers10 m3 : 10 cubic meters10 m3/hr: 10 cu mtr per hr10 m3 /d 10 cu mtr per day20 kcal: 20 kilocalories20 kcal/hr: 20 kilocalories per hr9500 kcal/m3 : 9500 kilocalories

per cubic meter10 000 kcal/kg: 10 000 kilocaloriesper kilogram10 gm: 10 grams25 kg : 25 kilograms50 bars: 50 kg per sq cm

50 bars a : 50 kg per sq cm absolute100 m2: 100 square meters

ENGLISH UNIT ABBREVIATIONSgallongallons per minutegallons per hourcubic feetcubic feet per minutecubic feet per daythousand cu f t per daymillion cu f t per dayBritish Thermal UnitBritish Thermal Units/h rthousand BTU per hourmillion BTU per hourBTU per cubic foot of gasBTU per poundpoundpounds per square inch ofpressurepounds per square inch of

absolute pressurebarrelbarrels per daybarrels of o i l per daybarrels of water per daythousand barrels per daysquare foot

v

10 gal: 10 gallons25 gpm 25 gallons per minute25 gph: 25 gallons per hour20 cf: 20 cubic feet50 efm: 50 cubic feet per min50 cfd: 50 cubic feet per day50 Mcfd : 50,000 cu f t per day50 MMcfd: 50 , 000 ,000 ell t per day50 BTU hr: 50 BTU per hour30 MBTU/hr: 30,000 BTU per hr10 MMBTU hr: 30 000 000 BTU per hr1000 BTU/cu i t 1000 BTU per cu i t20,000 BTU/1b: 20,000 BTU per 1b10 lb: 10 pounds750 psi: 750 Ib per sq in750 psia : 750 Ib per sq in abs20 bb1:100 BPD:100 BOPD:100 BWPD:10 MBPD:25 sq i t

20 barrels100 barrels per day100 bbl o i l per day100 bbl water per day10,000 bbl per day25 square feet


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W keep t a l king about atoms and molecules because the behavior of thesubstance is controlled by the behavior of al l the molecules and atoms ofwhich that substance 1s composed. These atoms and molecules come in a l lsizes and shapes. but i t is convenient to show them 8S l ttle spheres ormarbles. We wil l do this to i l lus t ra te the principles involved .o l id hase

The properties of a s ubstance depend on the number of moleculespacked into a given volume and the energy of these molec ules We find tconvenient to divide properties into three general categories - solid.liquid and vapor gas) - which are called phases.

We can see a solid with our naked eye. I t retains i t s shape . We canhold i t in our hands. When placed in 8 container, solids do not changeshape to completely f i l l that container ; there is space between the solidpieces. A solid has a def ini te shape and a defini te volume.

A solid behaves this way because i t s molecules are very close togetherand do not move about freely . These molecules vibrate in place but do no tmove about as they do in a l iquid or gas.

The picture below is of a crystal of an iron compound. with the for-mula Fe52, takenof iron Fe) and

with a microscope. Each molecule cons ists of one atomtwo atoms of sulfur 5).

and molecules because their size has beenYou actually can see the atoms

magnified increased) 44 milliontimes.


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Notice tha t each large dark spot has two smaller spots on each s ideof it Thi s l arge spo t is an atom of iron combined with two a toms ofs ul f ur smaller spots) to form a molecule of FeS2 ,

o t ice a l so that these molecules form a r egula r pa t t e rn. Each ofthese molecules is vibra t ing in place but is not changing i ts position inthe crys t a l This is true of a l l crystal l ine solids. o t a l l solids arecrystall ine. In some amo rphous solids the molecules move from place toplace but do so in a very slow manner.

For a given s ub s tan ce , the solid phase is the most dense phase fo rthe molecules are packed closely together. Like a crowd of peop le thereis not much free space to move about.

The behavior of molecules in any phase i s a co mbina t ion of the kineticenergy veloci ty of the molecules and the at t rac t ive forces between them.In a solid mix tur e these molecules are c l ose .toge ther so that the attrac t i ve forces holding them together are greater t han the kinetic ene r gyforces which would le t them move about freely. Thus, the molecules simpl yvibrate in place in a crys t a l l in e solid l ike tha t shown i n the picture onpage 2.Liquid Phase

In the l iquid phase the molecules have greater kinetic energy than inthe solid phase. The molecules move more freely throughout t he l iq uid.Gr oup s or c lust ers of molecules tend to form because of the at t ract ivef orces, but these molecules t end to move from group to group because theypossess s uff icien t energy to do so. Th is is shown below.

-


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Since a l iquid is less dense - and less r ig id - than a sol id tassumes shape the same 8e the container in which t i s placed. I t iss t i l l dense enough so that we can see t with our eyes , even though i tmay be clear in color. A l iquid has a defini te volume but no defini teshape.

In a l iquid mixture of hydrocarbons each different size and shape ofmolecule present .poisesses a different kinetic energy. Those moleculeswith a low kinet ic energy l iquefy easier than those with a high kineticenergy. As a general rule the larger the hydtocarbon molecule the lessis the kinetic energy. Propane i s thus easier to liquefy than methane.

I t i s not convenient to measure kinetic energy. We measure relat ivekinetic energy indirec t ly by the measurement of vapor pressure. Thehigher the vapor pressure the higher i s the kinetic energy.Vapor Gas) Phase

In the gas phase the molecules are s t i l l farther apart . Thus theat t ract ive forces between these molecules is less than with a l iquid.Along with this . their kinetic energy is much greater .

Gases have neither defini te shape or def ini te volume. They willoccup y a l l volume open to them regardless of the shape or size of thatvolume. In a vessel part ia l ly f i l led with l iquid gas will occupy a l l ofth e remaining space.

Since gas and l iquid have no def ini te shape they are called f luids.A f luid can be a l l gas a l l l iquid or a mixture of the two. Solids pos-sess no tru e f lu id ity even though they can flow through a l ine or vesseli f ground up in to very small pieces .

We define two types of gases. An ideal gas is one where the a t t rac-t ive forces are negligible. Behavior is governed entirely by kinet icenergy. No gas is truly ideal but this idea is convenient to estimate gasbehavior. Around atmospheric pressure up to about 4 bars [60 psi ] theideal gas model is a suitab le approximation of t rue behavior.

This idea l gas model assumes tha t the gas molecules have no effect on


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each o the r S they mov e about in the gas in a random manner. There areno attractive forces between these molecules .

Real actual) gases depart from ideal behavior because of a tt r ac t iveforces. s the pressure increases the molecules are c loser together andthese at t ract ive forces become larger. Thus prediction of actual gasbehavio r depends on pressure. I t also depends on temperature, whichaffects energy.Equations of State

Prediction of phase behavior involves prediction of the relationshipbetween press ure, volume and temperature. The P-V-T equations used by theenginee r fo r th is purpose are cal led equations of s ta te .

Since so l ids and l iqu ids possess a defini t e volume. pressure and temperature onl y have a small effect on their volume and density. In mostproduct ion- pr ocessing ope ra t ions we can assume t hat the effec t is ze ro.n th e oi l reservoir , where volumes a r e very l arge, even smal l per centage

changes n volume can have an e f ~ e t on reservoir performance.Since gas possesses no defini te volume, this volume 1s affected by

both pressure and temperature. Much of this manual is devoted to theeffect of pressure and temperature on volume and density of gases - bothideal gases and act ual gases.

B SIC PH SE BEH VIORSince press ure and temperature affect behavior . i t is convenient to

prepare figures and tables of data to represent th is effect on phase behavior. These fig ures are called phase diagrams.

If al l of the molecules in the s ubs tance are the same i t is calleda pure s ubstance. Water is a pure substance because a l l molecules are thesame and are represented by th e eq uat ion H20.

A mixture containing more than one kind of molecule is cal led a mul t i -component mixtur e. Each t ype of molecule is a separa te component affectingbehavior .


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The behavior of pure substances 1s eas ier to represent so we wil ldiscuss th is f i r s t .

The Pressure Temperature Diagram for a Pure SubstanceThe phase behavior may be shown using pressure and temperature. The

f igure below is fo r a pure s ubstance. There are th ree regions - sol idl iquid and vapor. The l ines separating these regions are known as thesaturat ion or equilibrium l ines .

np -c

1SOLID

I - _IIII

SoUd Vapor VAPOR

F

TEMPERATURE

Vapor

Along l ine HD the s ubstance may be a l l sol id a l l l iquid or a mixtureof the two. Along l ine FH the substance may be a l l solid , a l l vapor or amdxture of the two . Point is known as the t r ip le point. At th is onepoin t a l l th ree phases can ex i s t toge the r . As s t rang e os t may Deem, tthe t r ip le point of water ice l iquid w t e r ~ and steam can exis t together.The ice would not melt and the steam would not cool. Of cour se we seldomobserve things a t their t r ip le point .


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Along l ine H no l iquid can form. One example is dry ice (solidcarbon dioxide). When t melts no l iquid is formed. I t melts fromsolid direct ly t o vapor. That is why t is often used to keep thingscold. There is no l iquid mess when t melts.

Line H is not of any practical importance with hydrocarbons becausethe solid us uall y melts to form liquid.

Line e is th e most important one for most systems . I t sepa ratesthe liquid and vapor regions. Along this l ine the substance may be al lliquid, al l vapor or a mixture of the two. One cannot te l l from th isfigure alone which is present. This depends on the energy of the s ubstance.Line He is also known by other names - vapor pressure l ine. bubble pointcurve and dew point curve . Each of t hese terms will be used in future discussions in th is and other manuals.

The upper l imit of the vapor pressure lin e is the point f . This isknown as the cr i t ica l point. The temperature and pressure represented byth is point are called the cr i t i ca l temperature Tc) and the cr i t ica l pres sure P) . At this point the proper t ies of the l iq uid phase and the vaporcphase become identical and they are no longer distinguishable. o r asingle- component system the cr i t i ca l temperature may also be defined asthe temperature above which a vapor cannot he l iquefied, regardless of theapplied pressure. Similarly , the cr i t i ca l pressure of a single-componentsystem may be defined as the minimum pressure necessary fo r l iquefactionof vapor a t the cr i t ica l temperature. I t ~ also the pressure above whichl iquid and vapor cannot coexist regardless of the temperature.

Each pure hYdrocarbon has a pressure - temperature diag ram similar tothe one shown in the figure on page 6. Please understand that the actualvapor pressures. cr i t i ca l values, etc. , are different for each s ubstance,but the general shape of the curves i s similar . If such a diagram isavailable for a given substance, i t may be used to predict the phase ofthat substance as t he temperature and pressure vary.

Let us suppose that the system in the phase diagram is in i t ia l ly a ta pressure temperature represented by the point I , and the system is


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heated at constant pressure unti l the point J is r eached. For thisconstant-pressure temperature increase the se phase changes occur: Thesystem is originally in the solid state and no phase change occurs untilthe temperature Tl is reached. At this temperature, which 1s the meltingpoint at this pressure liquid will begin to form and the temperaturewill remain constant at T unti l al l the solid has melted.

Liquid

Solid

At Point I At Temperature Tl

As heat is added at Tl some of the molecules obtain enough energy tobreak out of the solid s tructure and behave l ike liquid. This continuesuntil al l of the molecules possess enough energy to move around freely .At this point the pure substance is al l liquid. Once this occurs, anyfurther heat will begin to increase the temperature of the l iquid as themolecules absorb the heat energy.

As the temperature is further increased the system will be in theliquid state until the temperature T is r eached . T is the boiling

bubble) point a t this pressure. Vapor begins to form and again the temperature will remain constan t at T until al l of the liquid has vaporized.The temperature of this vapor system can now be increased until the pointJ is reached. t should be emphasized that, in the process just described.only the phase changes were cons idered.

Obviously, some physical pr operties of the liquid change as the tempera ture is increased . For instance, the increase in temperature causesan increase in volume with a resultant decrease in density. Simi lar ly,the other physical properties of the liquid are altered, but the properties


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of the system are those of a l iquid and no other phases appear during thispart of the isobaric (constant - pressure) temperature increase.

Suppose that the figure is for water and l ine 1 is at atmosphericpressur e. Temperature Tl would be at OO [100F], the freezing point ofwater. Temperature T would be 100C [212F], the boiling point of water .Liquid and/or vapor a t the boiling point would be called saturated. Thes team a t point would be called a superhea t ed vapor.

What happens at temperature T2? The molecules are packed less closelytogether and have more kinetic energy. They move around more wildly andsome begin t o ac t l ike molecules do in a gas . What we cal l gas begins toform. Firs t one bubble of gas forms . This i s why T is called the bubblepoint - the pressure and t empera ture at which th i s f i rs t bubble of gas forms.

As additional heat is added, more and more bubbles of gas form. Theprocess continues, a t constant temperature T2, unti l the substance is a l lvapor. The substance is ca l l ed saturated from the time the f i r s t bubbleof gas is formed unti l i t is a l l gas. As more heat is added the temperature of the gas increases and i t becomes superheat ed.

Suppose we reverse the process and s tar t with a s uperheated gas a tp o i n t ~ The gas will cool in t empe r ature as heat is removed unti l temperature T is reached. As energy is taken out by cooling, the moleculesenergy decreases. At T some of the molecules reach a level typical ofl iquid behavior. The f i r s t drop of l iquid forms . This drop of dew canbe detected in an instrument for that purpose.

As cooling continues, more and more l iquid forms unti l the substanceis a l l l iquid. Temperature T is thus both the dew point and bubble pointfor the pure substance, at the pressure of l ine IJ.

Vapor Pressure of LiquidsVapor pressure is defined as the pressure exerted by a vapor in equi

librium with i t s liquid. The term equilibrium means that the amount ofl iquid and vapor is constant with t ime. Consider a closed container which


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has been par t ia l ly f i l led with a l iquid. The molecules of the l iquid arein constant motioo, but not a l l the molecules move with the same velocity.Ther e will be some which possess more energy than others. I f one of thesefast-moving molecules reaches the l iquid surface t may possess suff i -cient energy to overcome the forces a t the surface of the l iquid and passin to the vapor space above.

As the number of molecules in the vapor phase increases, the ra te ofreturn to the l iquid phase also increases, and eventually a condition ofequilibrium is attained when the number of molecules leaving the l iquid i sequal to the number returning. The molecules in the vapor phase obviouslyexert a pressure on the containing vessel. This pressure i s known as vaporpressure, the pressure shown by l ine e a t a given temperature.

Gas

..o ALiquid

Molecules in the l iquid phase are moving about. Some ar e going fastenough to break through the boundary between the gas and l iquid. and endup in the gas. Some of the gas molecules do the same thing and end up inthe l iquid phase.

Remember pressure i s caused by molecules col l iding with the wall ofthe container. The greater the number and energy of these molecules, thegreater is the press ure. So vapor pressure is an indirect measure ofmolecular energy.


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desired temperature . The gauge then reads the vapor pressure. Thispressure plus the barometric pressur e give the absolute vapor pressure.

In order to read a true vapor pressure the instrument must containonly the substance being tested. I f any other substance. l ike ai r i spresent the instrument will not give a true reading.

I t is inconvenient to remove a i r from the instrument. So, the easies t thing to do 1s to run a simpler tes t . The result is called Reid VaporPressure, abbreviated RVP. I t uses a standard instrument of the same name.

Many l iquid specifications are stated in Reid Vapor Pressure. Gasol ine for an automobile may be specified as 9 psi RVP. I t may be specifiedfor tanker shipment that crude oi l may not exceed 11-12 psi RVP. The l iq uid formed from natural gas in processing plants natural gasoline) oftenvaries from 14 -34 psi RVP. Note: The Reid Vapor Pressure i s defined interms of psi. Approximate pressur e equivalents are: 9 psi - 0.62 bars;

to 12 psi - 0.76 to 0.83 bars; 14 to 34 psi - 0.97 to 3.3 bars.The term IIpsi RVP refers to the absolute pressure measured with the

Reid apparatus. I t may be converted to true vapor pressure using a standard correlation.

Problem 1In the phase diagram for a pur e substance:1. Poin t H is the point.2. Along l ine H the solid melts t o f orm _ _ _3. At any pOint along line He the substance may consist of what

phases?4. At paint J the substance is a _ _ _ _ _ gas .

Fi l l in the bl anks.)


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Phase Diagram for a Multi-Component MixtureThe phase diagram for a mixture con ta ining mo re than one type of mole

cule is more complicated than fo r a pure substance. Line He in t he previous diagram for a pure substance is replaced by an envelope. The dewpoint and bubble point no longer occur a t the same pressure and temperature .

The diagram t ha t f ollows i s a pressure- t emperatu re diagram for amil t i- component sys t em wi th a fixed overall composition . The bubble pointcurve i s r epresented by the l in e AG , and the dew point curve is representedby the l ine BG. Th e point where these curves meet i s known as the c r i t i ca lpoint . Symbols for the c r i t i ca l pressure and t emperature are Pc and Te respect ively. Points within the envelope ACB represent s ystems consist ingof two phases. Points to t he r ight of the dew point curve represent vaporand points to the l e f t of t he bubble point curve represent l iquid .

The press ure- temperature d iagr am indicates the phase changes that mustoccur when the pressure and temper a t ure of a sys t em are varied. Suppose asystem original ly a t point is cooled a t cons t an t pressure a t a temperature below Tc) along the path 1M. These phase changes occur: The systemis or ig ina l ly in the vapor s tate . At the dew point l iquid begin s to form.In passing from to more l iquid condenses. At the bubble p o i n t ~ thesystem is a saturated l iquid . At point the system is in the sub-cooledliquid s ta te . A l iquid a t po i n t M is also cal led undersa tu rated.

Cricondenbar e"C .cw, ~\ " , 11l/ \ ,LIQUID I \ 0/ I 0/ I - "" M L / I K u" - - - - _- -_J I I < , , ," ...,'" , , 0 TWO r PHASf.., ,'

/A , B1007. 80 60 40 20 07. LiguidTemperatu re


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The horizontal l ine drawn tangent to the phase curve at the point ofhighest pressure i s called the cricondenhar l ine. t r ep r esents the maxi-mum pressure a t which l iquid can occur.

The ver t ica l l ine drawn tangent to the r ight side of the curve i scalled the cricondentherm l ine. t is the maximum temperature a t whichl iquid can occur.

No l iquid c ~ occ ur above or to the r ight of these two l ines . Reser -voirs in either of these areas contain only gas.

The dashed l ines inside the phase curve are sometimes called qualityl ines . They represen t a constant percentage of l iquid. The dew pointc urve i s 0 l i quid; the bubble point curve represents 100 l iquid. Atpoint K the system contains SOX l iquid and 50 vapor.

Any t ime you have both l iquid and vapor present a t any point in yoursystem. you know that t is a t a pressure and temperatur e somewher e insidethe phase curve. The amount of l iquid depends on where you are inside thecurve.

An oi l and gas separator must operate inside the phase curve. Thepressure is normally fixed by contract or some system re quirement. There-fore the lower th e temperature the more the l iquid that wil l be produced.This is why effective cooling or refr igerat ion i s used to produce more l iq uid.

The gas leaving the separator is a t i t s dew point . This means that i fthis gas cools . some l iquid wil l be formed. This applies t o both water andhydrocarbons.

Cooling is natural in a pipeline . On land the l ine is normally buriedbelow the frost l ine. At this dep t h t he temperature normally is about 20 Ct o 4C [3SoF to 40F] . For offshore l ines the temperature will vary from4C to 320 C [400 F to 90F] depending on dep t h and t he location of t he waterbody. These t emperatur es normally a r e less than normal separation tempe ra-tures. Thus . f ormation of l iquids fr om separated gas is normal unless thestream is processed to pr event i t .


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Problem 2

Match the foll owing l etters with the correct description for theportion of the two component sys tem as given.

G

Critical Pre ss ureVapor Reg onCritical Point

BO

Vapor Liquid Region

H

oC

F

Critical mperatureLiqui d Regionubble Point Curveew Po nt Curve

Another form of phase curve s useful for those who work with reser-vo irs or producing wells

p0 ic /1N 2:/ X: /10 I~ ( U / Il /1 I /11 r1wj jr0 /10 I I' I II wI Jr/ Jwi

Temperature


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Points ~ . ~ and represent four dif ferent types of petroleumreservoirs. The vertical l ines from these points represent any reservoi rpressure decline that may occur . This decline i s a t constant reservoirtemperature.

The curved l ines from these points represent the drop in both pressure and temperature that occurs as the fluid flows up the wellbore. Point

represents the pressure and temperature at the wellhead. I f Point ~ isinside the phase curve, the wellstream is two-phase, l iquid and vapor . Thera t io of the two is often expressed as a gas-oil rat io. In metric units ,this ra t io may be m3 /1 t r of l iquid; in English units , t i s sci/API bbl.

I f Point W i s outside the phase cur ve, the wellstream i s single-phase,gas l iquid.

In most cases, both gas and l iquid are present a t the wellhead. Therelat ive amount of gas is a measure of how close Point W is to the phasecurve.

lack Oil ReservoirA reservoir whose in i t ia l pressure is Point is called a black oi l

reservoir . The name comes from the fact that the oi l is usually black incolor . But, the fact the o i l is black does not mean the reservoir behavesl ike a black oi l one .

I f the in i t ia l pressure is as s o ~ Point M , the reservoir is calledundersaturated . I f the reservoir is on, or below, the bubble point curvethe reservoir is saturated.

At Point M the reservoir is a l l l iquid. No gas will form in the reservoir unt i l the pressure declines to the bubble point. Until that time,any gas present a t the wellhead will have formed in the wellbore.

Once gas begins to form in the rese r voir , the behavior of the reservoir might change noticeably. The gas formed may flow to the wellbore;a l l or a part of i t may form a gas cap in the reservoir .


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lati le Oil ReservoirA reservoir whose in i t i a l pressure is Point i s cal led a volat i le

oi l reservoir. I ts general behavior is very much l ike that of a black oi lreservoir .

At Point the reservoir is undersaturated. I t will be a l l l iqu idunti l the pressure declines to the bubble point curve, where gas will begin to form.

Since Point W for this reservoir is inside the phase curve, the wellstream will be two-phase a t the surface. The gas-al l r a t i o normally willbe higher than for a black a l l reservoir; the l iquid also will be lessdense and may vary in color from l ight brown to black. One cannot t e l lfrom the color alone whether the reservoir is black oi l or volat i le oi l

GaS Condensate ReservoirFor most hydrocarbon reservoirs the cr i t ical point ~ occurs to the

le f t of the maximum phase curve pressure cricondenbar) point as shown inthe figure on page 15.

A gas-condensate reservoir has a temperature between and the maximum phase curve cricondentherm) temperature. Point represents one suchreservoir .

I t will decline a t constant temperature in the normal manner. BetweenPoint and the phase curve i t wil l be a single- phase reservoir containingwhat is usually called a dense phase fluid. This f luid can be r ega r ded asa very dense gas e c ~ u s e gas equations a r e used to predict i t s per fo rmance.

Once the pressure declines to the dew point curve, l iquid forms inthe reservoir . At th is point both the flow rate and analysis of the wellstream wil l vary. The f i r s t l iquid that forms seldom flows to the wellbore.

Pressure maintenance is often used to keep the reservoir pressureabove the dew point curve. The usual system involves injection of gas orwater into the reservoir to replace the f l uids produced .


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The cross-hatched area in thephase curve is known as the retrograderegion. The dashed l ines in the sketcha t l e f t are the quali ty l ines. Oncel iquid begins to form a t the dew pointa fur the r r educ t ion i n p ressu re causes

l iqu id to form. This i s the re -verse of normal behavior - cal led re t ro -grade. s the pressure decl ines morel iquid will continue to form unt i l the

~ s r v o i r pressure is below the re t rograde region.

The gas-oil ra t io in this type of reservoir probably wil l be verylarge. Any l iqu id produced normally has a low density. Often t wil l bea ra ther clear straw- colored l iquid. Once again though one cannot ident i fy the kind of reservoir based on color.

Gas ReservoirPoint and the ver t ica l dashed l ine from t represent a true ga s

reservoir. At no pressure does l iquid form in the reservoir since t i sto the r ight of the phase curve.

Because of the drop in pressur e and temperature in the wellbore. somel iquid might form in the wellbore. as shown in the phase diagram. Thusthe presence of l iquid a t the surface does not mean that there is any l iq-uid in the reservoir . Moral: Do not assume that what you see a t the surface necessari ly is occurring in the reservoir .

Effect of Depths we dr i l l deeper into the earth s crust reservoir pressures and

temperatures increase . At shallow depths we tend to find black o i l reser voirs . As we dr i l l very deep wells the l ikelihood of finding gas-condensateor gas increases.


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The shape of the phase curve depends on reservoir fluid analysis .This means i t is possible to find gas a t shallow depths . When this happensthe gas is very lean and has migrated underground from where t was formed.There are very few la rge gas re servoir s a t shallow depths.

The only rel iable way to determine the phase cu r ve is to obtain ar e l iable re servoir f luid sample. This sample is very important and shouldbe taken with care. A laboratory can take this sample and produce th ephase data needed to predict properly r eservo i r behav i or .

These phase data are the f ingerpr in t s used to pos i t ive ly ident i fythe reservoir . Without such data one must guess a lo t . Gas-oil r a t i o.color and other surface data are important but they cannot t el l U5 enoughabout the re servoir .

Problem 3The following information is available on a re se rvo i r :

eservoir pressure SS3 bars [8 psig]Reservoir temperature IISoC [24 F]Rese rvoir fluid critical temperature 8SoC [18SoF]Circondentne rm temperature 143C [ 9 ]Gas -oil ratio 1.1 m3/l tr [6 cu ft/API bbl]

What kind of reservoir would one expect this to be? ___________________

API GravityThe density of produced l iquids is often measured in API. This

actually i s the measurement of spec i f i c gravity. The conversion from APIto specif ic gravity is made using the formula:

141. 5Sp Gr K 131.5 OAPIRemember--Specific gravi ty of a l iquid multiplied by the density of watergives the density of tha t l iquid.


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The API of a liquid may be found with ahydrometer. This is merely a device that floatsin oi l . I t has a scale on the sid e that meas-ures how deeply the f loat sinks into the l iq -uid. The API gravity of the l iquid i s measuredby th is scale . The process is similar to theprocess of measuring the specif ic gravity ofant i- f reeze for determining i t s freezing point.

Vol e Shrinkage Upon ixingWhen mixing l iquids, one volume plus one volume equals two volumes,

right? Not necessarily: When mixing two different hydrocarbons someshrinkage occurs . This shrinkage may be very small or quite noticeable.Why does t occur?

Each l iquid i s composed of molecules of different sizes. One canimagine what is going on y magnifying these molecules. Imagine onel iquid is a box ful l of fair ly large rocks. There wil l be many openspaces .

The other l iquid contains very small molecules equivalent to smallrocks. The two groups of rocks are mixed. Some of the small rocks willf i t into the open spaces between the bigger rocks. I f the mixture of rockscontains less open spaces than were present in i t ia l ly . the volume of themixture is no t the sum of the original volumes.


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As one would guess from this example the amount of shrinkage dependson the relative size of the molecules involved .

What happens to density on mixing when s hrinkage occurs? Density ismass divided by volume. The mass cannot change on mixing. If the finalvolume is le ss than the sum of the ini t i l volumes the density mustincrease.

Stated another way the fewer the open spaces be twe en molecules thegreater the density.

Prob l em 428 l t r [ cu ft] of liquid Aweighs 20 kg [44 lb ] 28 l tr [1 cu ft]

of liquid 8 weighs 15 kg [32 lb] The two liquids are mixed The resultingvolume is 54 l tr [1 90 cu ft] What is the density of the mixture in kg/ltr[lb/cu ft]?

MetricEnglish:

o 0 72 kg/ltr o 065 o 0 52o 44 lb/cu f t o 40 o 32

8EHAVIOR OF IDEAL GASES

0 0 62o 38

The pressure-voluMe-tempe rature of gases is most important to thepetroleum industry. The prediction of behavior is an important part ofthe design and operation of equipment. This is one reason why accuratemeasurement of t emperature and pressure is so important.

e wil l f irs t consider ideal gas principles for these serve as a basisfor predicting behavior of actua l gases.

e f irst need to define again some basic terms.Mass - weight of gas contained in a given volume

Units - Metric : kg gmor molesEnglish: lb or moles


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Moles - mass in kg , gm or I b multiplied by molecular weight ofthe gas

Molecular weight - found from the gas analysis or by theequation MW z Sp Gr)(29)

Density - weight ofUnits - Metric:

gas contained per unit3 3kg/m or gm cmvolume of gas

English : Ib /cu f tSpecific Volume - reciprocal of density ( l /density)3 3

Units - Metric m /kg, em IgmThe units of

Metric:English:

English : u f t / lbpressure to be used in this discussion are:2bars 1 bar = 1 kg/em 2 1 atmosphere)

psi absolute (psia)Absolute pressure always must be used in gas calculation s - gauge pressureplus atmospheric (barometric) pressure.

The units of temperature to be used in this discussion are :Metric: KEnglis h: OR

K = D 73OR OF + 460

Absolute temperature always must be used in gas calculations.

Change of VolumeThe volume of gas depends on pressure, temperature and mass. Let us

consider the effect of each one of these, with the other two held constant.Change of Volume with Mass (p and T constant) - A simple example of

th is is blowing up a child 's balloon.

State (1)

The balloon a t le f t has been part ia l lyblown up. The pressure and temperaturein the balloon are the same as the a t -mospheric air If we regard the a i r asan ideal gas, the density of the a i r inthe balloon is the same as that of the


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Z

surrounding a i r The mass of a i r in the balloon is the density of t h a ta i r times the volume of the balloon . I t i s convenient to express the massas n, the number of moles of gas present .

Sta te (2)

ow suppose we place more a i r inthe balloon. The pressure and temper -ature inside the balloon do not change;they are s t i l l the same as the atmos-pheric conditions. But. the volumehas increased because there are morem o ~ s of gas (n) in the balloon.

We can express this change from (1) to (2) by an equation.

This equation applies i f there is no change in pressure and temperature ingoing from 1) to 2) .

Change of Volume with Temperatur e P and mass constant)

HEAT

TZ

State (1) State (2)The above drawing i l lus t ra tes th is condition. I f there is no leak-

age the mass is the same in both s ta tes The pressure does not changebecause the weight on the piston stays constant. In order for the pres-sure to remain constant the col l ision forces of the molecules beating onthe walls must remain constant. These col l ision forces are the sum of twofactors, the energy of the molecules and their concentra t ion (density) .


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As we heat them up the i r energy increases. To compensate f or this thedensity must decrease to keep pressure constant. Since the to ta l mass i sfixed, the on ly way to decrease density is to increase volume. This isexactly what happens.

This pr inc ip le i s expressed by Charles t Law which s t a t e s t ha t Atconstant pressure, the volume of a constant weight of gas varies direc t lywith abso l u t e temperature .

This law can be written in equation form as:

or

where: k is a proportionali ty constant independent of V and T.Exampl e :

Gas i s held in a container a t constan t pressure . At 40C{10 40 Fj the volume i s 3m3 l06 u ft . What i s the volume ithe temperature i s increased to 60C [140F]?Met r i c Xl = 40 273 = 313K, K -2 60 + 273 K

3 V2 3) (333) 3313 = 333 r V2 313 3 .2 mEnglish : R1 104 460 564oR R 2 = 140 460 - 6000R

106 V2 V2106) 600 ) 113 u ft564 = 600 or 564

Change of Volume with Pressure T and mass constant) - The drawingbelow i l lus t ra tes th is condition.

Sta t e l ) Sta te (2)


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If there is no leakage the mass is the same n both states. If thetemperature is held constant the energy of the molecules remains constant .The only way the pressure can increase inside the cylinder to equal theincreased external pressure in s ta te (2) is for the density of moleculesto increase. Since mass Is constant the only way for density to incr easeis for the volume to decrease.

This princ ip le is known as Boyle's Law. t s tates t ha t the Volumeof a constant weight of gss varies inversely with the pressure exerted onit i f the temperature i s maintained constant .

This law may be expresse d in equation form as:

where: k' is a proportionality constant , different in value fromthe k in Charles I Law.

ExampleGas i s held in a container a t constant temperature. The in i t i l

volume is 4 m3 {142 cu f t} a t absolute pressure o f bars {29 psia} .The pressure on the gas i s increased to 4 bars {58 psia} . What i sthe f inal volume? The equation i s P1VlMetric:English: 29) 142) = 58)

2 , V

2= 71 cu f t

General Ideal Gas EquationBoyle's and Charles' Laws may be combined to make an equation that

r elates pressure, volume and temperature behavior of an ideal gas. I t is :PV IE: nRT

In this equation n is the number of moles of gas present in the con tainer holding the gas and R 1s a constant that is often called theuniversal gas constant. The value of constant R depends on the units usedfor P, V and T. The table on the next page summarizes common values of R.


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Common Values o f Constant R

- p V T n Rbars* l i t e r s K gm-moles 0.0821bars* cm3 K gm- moles 82 . 1bars* 3 K kg-moles 0 . 082pSia cu ft o 19-moles 10.73

2pressure m y also be expressed n kg/emo r atmospheres .

The value of m y be found from the gas analysis or by the equation:n = mass of gas) MW gas)

Example:

I f mass i s expressed ingramskilog ramspounds

n is ingm- moleskg-molesIb - moles

10 kg 22 Ib] DE a 0 .6 sp gr gas i s conta ined in a tank a t2 bars 29 ps i ) abso lu t e pressure a nd a t emperature o f 40C {104F}.What s the volume o f the tank?

Gas W = 0 .6) 29) = 17.4Metr ic En l i s h

n 10) 17.4) = 174 kg-moles 22) 17 .4 ) = 383 Ib-rrc>lesR 0.082 10.73T 40 + 273 = 313 K 460 104 - 5640 RP bars 29 ps iaV - nRT 174) 0 . 082) 313) 383) 10.73) 564)p - 2 29V - 2233 m3 79 900 cu t


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Problem 52 kg [4.4 lb of an ideal gas having a molecular weight of 16 is

placed in a tank at 15C [60F . If the pressure on the tank is 7 bars[1 psia . what is the volume of the tank?

Metric: o 0.44 m3 o 65 m3English: o 156 cu ft o 908 cu ft o 105 cu ft

Gas Mi xture sIn gas calculations we are usually concerned with the behavior of gas

mixtures. The compositions of gas mixtures are usually exp ressed asweight , or The weight of any component is definedas th e weight of that component divided by the total weight of the mixture(multiplied by 100 so the result will be on a percentage bas i s . Thus,for any component of weight Wtt:

Weight 100where Wrepresents the total weight of the system. Simi larly the volumeof any component is defined as:

Volume 100where Vi represen t s the volume of the component of interest and V is thetotal volume. Mole of a component is defined as:

Molen - x 100n

where n1 is the number of moles of the component of interest and n i s thetotal number of moles in the system

In expressing quantities l ike percentages i t is common to use a sub-script i to express the mass or volume of any component in the mixture;


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the mass or volume of the total mixture is written without a subscript .Often the term "mole fraction " is used instead of mole % The mole

fraction of a component is

where Y repre sents the mole f ract ion, n 1 represents the owr,ber of molesof the-Component of interes t . and ~ i s the t ota l number of moles in thesystem.

Mole fraction is used the most for calculation s on gas mixtures. I tis normally ob t ained directly from the analys is r eceived from the laboratory .

ComEonentMethaneEthanePropaneIso-ButaneNormal ButaneIso-PentaneNormal-PentaneHexaneHeptanes Plus

Mol76.409.504.612.4 13.700.771.450.75

--2.:.i .100.00

This analysis usually will be in theform shown a t the l e f t Mole percent i soften shown . In order to obtain molefraction simply divide each number by100 (move the decimal point to the le f ttwo places . For example, 76.40 mole %for methane is 0.7640 mole fraction.

~ o r r e s p o n l i n g equations may bewritten for weight fraction or volumefraction. Notice that the only difference between percent and fraction iswhether or not the rat io is multipliedby 1 .

The concepts of weight %and volume % are self-explanato r y . However,the concep t of mole % or mole fraction should be described more fully sothat i t s meaning is clearly understood.

BaSically, t he mole fraction represents the fraction of molecules inthe system that are of a given kind. This follows from the fact that onemole of any gas con t ains the same number of molecules. Fo r example, s up pose a system contains one mole of methane and two moles of ethane. Inthis system the mole fraction of CH4 is 1/3 and that of C2H6 is 2/3. I t


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i s also true that 1/3 of the molecules are 4 molecules and 2/3 are C2H6molecules .f each gas i n a mixture obeys the perfect gas laws, the volume of a

component would be proportional to the number of moles of that component.So. for gases which obey the perfec t gas laws. volume and mole areequivalent .

Relationship Between Weight and Mole (or VolumeThe process of conversion from weight to mole can be explained

best y a definite example using a gas mixture whose components a re givenin Column 1 in the table below. The molecular weight MW) i s l i s ted inColumn 2 and the weight of each component i s l i s ted in Column 3 . Assume100 weight units either Ib or gr) of t he gas mixture are present.

In Column 4 the actual weigh t of each component i s sted . The number of moles n i of each component i s given in Column 5 and i s obtainedby dividing the weight of each component by i t s molecular weight. Thetotal number of moles in the system n) i s the sum of the figures in Column 5. The mole of each componen t i s l i s ted in Co lumn 6 and is obtainedby dividing ni by n and multiplying by 100. t is obvious tha t the resul t swould be the same no matter wha t weight of gas had been taken as the basis .One hundred weight units were chosen, since th i s simpl i f ies the calculation .

1) 2) 3) (4) I 5) 6)Wt Wti Moles n i ) MolesComponents MW per 100 wt units n /n ) x 100

H 4 16 60 60 60 3.750 77.8716-C2H6 30 20 20 20 0.667 13.8530 -C3H8 44 10 10 10 0.227 4. 7144 -C4HIO 58 10 10 10 0.172 3.5758

n 4.816 100.00Gr - moles. i f grams; Lb-moles. i f pounds


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What have we done n the exampl e on page 291 The molecular weight isexpressed as wt per mole in Column 2. The wt in Column 3 is wt of anycomponent per 100 wt uni ts of mixtur e (Column 4). Dividing Column 4 byColumn 2 :

wt component x mole component100 wt uni t s mixture wt component mo l es component100 wt un i t s mixtureEach of the entries in Column 5 is the r esul t of this calculat ion .

The conversion from mole or volume since they are ident ical forgas) t o weight wil l be expla ined y an example which is the reverse ofthe preced ing conve r sion . The componen t s are l i s ted in Column 1, nextt ab le .

The molecular weight is l i s ted in Column 2 and the mole of eachcomponent i s l i s t e d in Column 3 . One hundred moles o f the mixture 1schosen as a bas i s .

Th e we ight of each component is tabula ted in Co lumn 4 and is obtainedby mult i plying th e number of moles of each component by ts molecularweight. Th e sum of the f ig ure s in Column 4 wil l represent the to ta l weightof the s yst em. The weight of each component is gi ven in Column 5 . Hereagain, any quanti ty of gas could have been chosen as a s ta r t i ng basis andthe f inal r es uts would have been the same .

( 1) 2) ( 3) (4) (5)Mol e Weight (Wt i ) Weight -Componen t s W (Wti / Wt)(bas i s - one mo le ) x 100

CH4 16 77 . 86 0.7786 x 16 = 12 . 46 60.0 C2H6 30 13 . 85 0.1 385 x 30 - 4 .1 6 20 . 0C3H8 44 4 .71 0 . 0 4 71 x 44 = 2. 0 7 10 . 0C4H10 58 3. 5 7 0 .0 35 7 x 58 2 . 0 7 10 . 0- Wt = 20 . 76 100.0*I n wha t eve r weight un1t s are used (gr, l b, kg , e t c .)


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Problem 6A gas has the following composition:

Component Mole Fraction W t WtMethane 0.89Ethane 0.05Propane 0.02Iso Butane 0.01Butane 0. 03

1.00What i s the weight of each of the componen ts in this mixture?

Apparent Molecular Weightt 1s not proper theoretically) to speak of the molecular weight of

a gas mixture. However, a gas mixture behaves as though t had a defini temolecular weight. As a result , we speak of the apparent molecular weightof a gas mixture. We may calculate the apparen t molecular weight of a gasmixture y multiplying the mole fraction by the molecular weight fo r eachcomponent and then summing these for the mixture.

This is exact ly what we did in Column 4 of the previous example. Thus,for that example. we may say that the apparent molecular weight of thismixture is 20 . 76.

Problem 7Calculate the apparent molecular weight of the gas mixture in Problem 6

Components Mole Fraction


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Gas Par tial PressureIn gas mixtures containing more than one kind of molecule the tota l

pressure exerted is the sum of the pressures exerted by each type of mole~ These individual molecule pressures are called part ia l pressure.

Suppose we represent molecules by marbles in a box. I f we have 1marbles in a box and then add 1 more. we would expect, and indeed do obse rve, that the average number of wa l l col l i s ion doubles and, consequently,the pressure doub le s.

Pressure .. 2 bars[29 psi] Pressure s 3 bars[44 psi] Pressure - 5 bars[73 psi]

The above f igure i l lus tra tes this principle. If the two gases aremixed, as shown, the part ial pressure in the mixture is equal to the pres sure exerted by each gas i f separated into containers of the same volume.

The equation for this principle is

where: PP)i - par t ia l pressure exerted by each component type ofgas molecule) in the gas mixtureP - absolute total pres sure of the gas mixture

Yi - mole fraction of each component i n the gas mixtureExample : What is the partial pressure o ethane in a gas mixture con-

taining 1 mole ethane i the to tal pressure absolute) is3 b rs [44 psia]?


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Metr ic : pp) i = 3) 0.10) = 0 .3 barsEngl i sh: PP) i 44) 0.1) ' ' 4.4 ps ia

Ideal Gas DensityThe density p) of an ideal gas may be found from the equation

PV = nRT. This equation may be converted to the form:P) (MW)P RT

where: density may be in gm/cm3

, kg/m3

or Ib/cu f t . The proper units touse in this equation are shown in the table below.

Densi t l p) _ _ ...L.gm/cm3 bars 82.1 Kkg/m3 bars 0.082 KIb/cu t pSia 10.73 OR

Example:A 0.7 sp g r i dea l gas i s contained in a vesse l a t 3 bars

44 ps ia ] nd a temperature o f 40De [10 4F . What i s thedens i ty o f the natural gas in kg/m3 f ib /e l l f tJ?

Metr i c En 1 i shW = D.7) 29) 20.3 20.3T - 40 273 => 313K 104 460 = 56 40 RP 3) 2D.3) 44) 2D.3)D.DB2) 3 l 3 ) lD .3 ) {564)

3 0 .1 5 l b / cu t2 . 4 kg/m

IDEAL GASESAc tua l gases approach perfect gas behavior a t h igh temperatures and

low press ures . However. we usually do not have the se conditions whileproducing. transporting. and processing hydrocarbon gases . Hydrocarbongases are usually exposed t o conditions which do not allow the gas mixtureto follow the perfect ga s laws. Therefore. we must be able to predic tactual gas behavio r .


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A number of techniques are used to correct ou r P- V- T equations ofsta te for real non-ideal gas behavior. Many of these are rather complicated. Most engineers prefer using a correction factor Z, known as thecompressibil i ty factor.

The Compressibility FactorThe comp ress ib i l i ty fac t or 2) is easy to use because t is simpl y

added to the equations for ideal gases . This value of Z corrects for thefact that the molecules are not t ruly independent but are affected by theattract ive forces between them.

PV ' ZnRT or P) MW)P ' ZRTZ is an empirical factor, determined experimentally, Which makes the

above equations t rue a t a p r ~ i c u l r temperature and pressure.For an ideal gas, Z is equal to unity. For a non-ideal gas, Z is

greater or less than unity, depending on the temperature and pressure.Under most conditions i t is less than unity. At a gi ven temperature, theZ factor plotted as a function of pressure usually takes the form shownin Figure 1. There will be an individual curve for each value of temperature. Charts of th is kind are available giving Z as a function of temperature and pressure for various pure hydrocarbons of the type i l lustra tedfor methane, ethane, and propane in Figures 2 3 and 4.

TT10 _ _ ._ - - - J_ . L

FIGURE 1

z

PRESSURE -


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Pressur e bars0 50 100 150 21 250 300, , , I I 1 I 1 1' ' 1 1' 1 1 1, , 1-- j j I- .' J I-., 1 ~'. .. ~ = ~ ~L I,

1.00

1l, ,

I r ~ :~ . 1 ,1 < 1 Y ~K V

/ v.%jV oo

0.7PV ;00 .T 0f ~.'.6 .,


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o 51

1

1

J ~ ?1 I . ,, j

\. 1 .,1 I

, I0. ~ I, I

r ,

II,

I

I I,I

, ,

I i I, 1 . . / :

api - behavoiur of hydrocarbons.pdf

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