Reservoir Fluid

RESERVOIR FLUID NOTES

PREPARED BY: SHOAIB (PE-O22) AND BILAL (PE-004) Page 1

INTRODUCTION OF PHASE BEHAVIOR Phase behavior is a condition of temperature and pressure for which different phases can exist.

PHASE

Phase is any homogeneous and physically distinct part of a system which is separated from other parts

by definite boundary surfaces.

PHASE DIAGRAM

We understand phase behavior by using phase diagram and phase diagram is a graph of pressure

plotted against temperature showing the conditions under which various phases of a substance will be

present.

Q. How we recognize the phase?

A. If the reservoir temperature is greater than critical temperature then it is gas reservoir and if the

reservoir temperature is less than critical temperature then it is oil reservoir. No of wells, perforations

and reserves calculation all based on phase present in reservoir.

PURE SUBSTANCE

Pure substance is a single substance like C1, C2 etc. If we have combination of substances like

combination of propane and heptane then it does not remain pure substance.

PHASE DIAGRAM OF PURE SUBSTANCE

Vapor pressure line

Vapor pressure line is a line separates the pressure, temperature conditions for which the substance is a

liquid from the conditions for which substance is a gas. In simple words, it separates liquid and gas.



For example, water in Karachi at room temperature is in liquid state and if we boil it in beaker which is

open to atmosphere (i.e. pressure remains same) then temperature increases and at vapor pressure

line, it converts from liquid to gas.

At constant temperature

As we remove Hg, pressure declines but we let the temperature constant by providing heat by external

means. As we further remove Hg, pressure declines to vapor pressure and liquid starts converting into

gas but as we remove more Hg, pressure does not decline but convert remaining liquid into gas. When

liquid completely converts into gas and we remove Hg then pressure declines.

Bubble point and dew point of a pure substance occurs at same point which is on vapor pressure line.

Bubble point pressure is a pressure at which first bubble of gas escape. For pure substance, bubble point

pressure and vapor pressure are same because complete liquid becomes gas at same pressure. But

when two or more components are present then bubble point pressure and vapor pressure will be

different. Bubble point pressure will be the pressure at which first bubble of gas escapes but vapor

pressure of different components occur at different pressures.



When we go from 1 to 2 then point on vapor pressure line is called bubble point. If we go from 2 to 1

then point on vapor pressure line is called dew point.

At constant pressure

As we provide heat, temperature increases but we let the pressure constant by providing heat by

removing Hg. As temperature increases further, liquid starts converting into gas but as we provide more

heat, temperature does not increase but convert remaining liquid into gas. When liquid completely

converts into gas and we remove Hg then temperature increases.



Bubble point and dew point of a pure substance occurs at same point which is on vapor pressure line.

When we go from 1 to 2 then point on vapor pressure line is called bubble point. If we go from 2 to 1

then point on vapor pressure line is called dew point.

Variation of bubble point with temperature

Bubble point pressure increases with temperature. At higher temperature T2, bubble point pressure PB2

is greater as compared to bubble point pressure PB1 at low temperature T1.

How phase diagram is made?

Phase diagram can be made by getting points of different vapor pressures at different constant

temperatures.

At constant temperature 500C, we get point of vapor pressure (we decrease pressure and pressure at

which liquid converts into gas is vapor pressure) and similarly for 600C, 700C etc. and plot the curve from

these points to get the phase diagram.

Vapor pressure can occur at any temperature for which we should know corresponding pressure. That is

the reason at high altitude vapor pressure is less because temperature is less.



Critical point

The upper limit of vapor pressure line is the critical point and the temperature and pressure represented

by this point is called critical temperature and critical pressure respectively.

Critical temperature for pure substance

Critical temperature may be defined as the temperature above which gas cant be liquefied.

Above critical temperature, no liquid phase and only gas phase is present and it cant be liquefied as

well.

Gas at state 1 can be liquefied as state 2 is liquid and hence, temperature is less than critical

temperature that is why reservoir temperature less than critical temperature are oil reservoir. On the

other hand, gas at state 4 cant be liquefied and hence temperature is greater than critical temperature.

Triple point

The lower limit of vapor pressure line is triple point. The point represents the temperature and pressure

at which solid, liquid and gas co-exist.

Sublimation pressure line

Sublimation pressure line separates the condition for which substance in solid to a substance in a gas.



Melting point line

Melting point line separates the condition for which substance in solid to a substance in a liquid.

PRESSURE-VOLUME DIAGRAM

Considering a pressure starting at point with a substance in liquid phase, temperature is held constant

and volume is increased (slightly)by removal of mercury (because liquid is nearly incompressible). This

caused a reduction in pressure from initial pressure to vapor pressure. When the pressure is reduced to

vapor pressure, gas begins to form and further increases in volume caused vaporization of the liquid so

at constant pressure, all liquid converts to gas (It is similar to the case of constant pressure in which as

temperature reaches to boiling point, the heat we provide is utilized in changing the phase of liquid and

not in increase of temperature, same is here, the expansion after bubble point pressure does not

decrease the pressure but utilized in changing the phase of liquid). Further decrease in pressure only

causes expansion of gas.

Illustration

Initially liquid is incompressible and hence with decrease in pressure due to removal of mercury, volume

slightly increases. When first bubble of gas comes out then it is called bubble point and pressure does

not change until all liquid converts into gas. A point at which all liquid converts into gas or first liquid

drop is formed from gas (if we see from reverse direction) is called dew point. Pressure of both bubble

point and dew point remains same. After all liquid converts into gas, further decline of pressure by

removal of mercury only cause expansion of gas i.e. increase of volume.

If the temperature is above critical temperature then only gas expansion occurs because only gas phase

is present above critical temperature.

The bubble point and dew point of pure substance lie at same point or pressure because for pure

substance there will be a particular pressure at which the liquid transforms into gas (vapor pressure) and

this will be the same pressure (if we move from low pressure to high pressure) at which the gas will

convert into liquid (dew point).



This diagram is drawn at one temperature but if we draw at higher temperatures then vapor pressure

decreases.

As we go towards higher temperatures, the bubble point pressure increases and the region in which the

liquid and gas both exists decreases and as we move towards higher temperatures then a temperature

comes when there will be no distinction between liquid and gas phase such that the region of liquid +

gas phase is represented by a single point called critical point and the temperature and pressure at this

point are called critical temperatures and pressures, after this point as temperature is increased, only

gas exists and the graph will be curve as shown in the following graph:

The above figure shows more nearly completed pressure volume diagram. The dashed line shows the

locus of all bubble points and dew point. The area within the dashed line indicates condition for which

liquid and gas co-exist, often this area is called saturation envelope.



DENSITY-TEMPERATURE DIAGRAM FOR PURE SUBSTANCE

How this graph is drawn can be understood by the pressure volume diagram of pure substance, if we

draw the PV diagram at temperatures T1 and T2 such that T2>T1 then as we have already discussed that

at higher temperature, the region in which both liquid and gas exists shortens such that at higher

temperature the saturated liquid (liquid that is about to vaporize) volume increases while saturated

vapor (vapor that is about to condense) decreases, this is because at higher temperature the heat of

vaporization is less (heat that is required to completely convert the liquid phase into vapor phase) as

molecules are already energetic and need less heat to be able to break the bonding forces of liquid and

converts into gaseous phase. Since the mass of sample is constant in the cell so the density of saturated

liquid decreases as temperature increases while that of saturated vapor increases with increase in

temperature. The densities of saturated liquid and saturated vapor becomes identical at critical point (as

at critical point there will be no distinction between the saturated liquid and saturated gas phase and

both are represented by single point having same volume so same density at critical point. If we take

average of densities of liquid and gas phases and plot them against their relevant temperatures then

graph will be a straight line passing through the critical point and this property is known as Law of

Rectilinear Diameter. From this graph we can find the critical temperature of the substance and

moreover the densities of liquid and gas at any given temperature.



TWO COMPONENT MIXTURES A petroleum engineer never deals with any mixture containing two pure substances but 100s of

pure substances are present in petroleum obtained from well so first we will discuss the phase behavior

of two component mixture and then amplify the concept for multi component mixture. Moreover, the

petroleum engineer is concerned with only liquid and gaseous form of hydrocarbons so the phase

diagram we study will be for liquid and gas only and not for solids. The solid consideration is important

when we have to avoid the hydrate formation in pipe lines during transportation so we have to maintain

the temperature in transportation line so that it may not reach to hydrate forming temperature at which

the gas molecules form hydrate crystals with water molecules.

PHASE DIAGRAM OF TWO COMPONENT MIXTURES

The behavior of two component mixtures is not as simple as the behavior of pure substance. In

case of pure substance the two phase region is represented by a single line because for pure substance

the bubble point and dew point lie at the same pressure but in case of two component mixtures we

have a broad region in which the two phases co-exists and the bubble point and dew point also do not

lie at same pressure because in case of mixture at bubble point some part of mixture or some molecules

from liquid phase are able to leave the liquid phase but not the whole liquid converts into vapor phase

as occurs in case of pure substance which has a definite vapor pressure but in case of mixture the

fraction of the components of the mixture are vaporized at each step with pressure until dew point

when all become gas. The PT phase diagram for two component mixture is as follows:

Brief explanation:

The phase diagram is bounded by bubble point pressure line and dew point line. At pressure P1,

the mixture is liquid, the liquid expands and pressure drops until the pressure reaches the point at which

a few molecules are able to leave liquid and form a small bubble of gas this is the bubble point, as

pressure is further reduced below bubble point pressure additional gas appears, further expansion will



lead to a point where the two components in the mixture have completely transformed into vapor that

is the dew point (which can also be defined as the point where first drop of liquid forms), as pressure is

further reduced below due point only gas will appear and no liquid. The dotted line shows how much of

the mixture contains liquid contents such that if our state lies at line of 25% means 25% of the mixture is

in liquid state and 75% is gas or vapor, these dotted lines are called as iso-vols or Quality lines which

shows the temperature and pressure conditions of the same amount of liquid means along the 75% line,

75% liquid will exists at different temperature and pressure conditions. These can be plotted such that

for example if we are performing the experiment at constant temperature then during expansion, when

the liquid volume reaches to 75% of total cell volume (as cell is transparent and can be visualized), the

cell pressure at this point can be determined and can be plotted on the graph.

Illustration:

The diagram is obtained from the same experiment as that for pure substance, the experiment

is called the Flash Vaporization in which we find pressure and volume of the cell at constant

temperature and perform the experiment at different temperatures and form the PV diagram and then

form the phase envelope, the main purpose of the experiment is to find the bubble point pressure. If we

talk about the constant temperature process that takes place from pressure 1 to 2 as shown by the

vertical line in the diagram, then initially at pressure P1 the two components (say methane and ethane)

were liquid, as mercury is removed from the cell, expansion of liquid takes place but to negligible extent

and pressure drops. The pressure continue to drop until bubble point pressure reaches at which the first

bubble of gas will liberate now as further mercury is removed the pressure of the mixture still drops

because in case of mixture, the whole mixture cannot be vaporized at same pressure because of the

variation among the bonding forces of the components as pressure continue to drop up to dew point as

at every step the vapors are produced from liquid but this phase change process is not ordinary as it the

phase change process of mixture and not the pure substance.

So in short ultimately, we will reach finally to a point when both methane and ethane will be

completely transformed into gas or vapor phase this the dew point or we can say that dew point is the

point when the first drop of liquid will form (such that when we talk about from low pressure to high

pressure the heaviest component will first condense into liquid as it has more ability of being in liquid

phase as compared to lighter components so its first drop of liquid when forms, the pressure at this

point is the dew point).

Critical point:

For two component mixture, it is defined as the point where the bubble point pressure line and

dew point line joins. The definition of critical point that is the point above which only gas exists (such

that above temperature and pressure at critical point) is false for two component mixture because

seeing the phase diagram, the maximum pressure of the diagram is above the critical point and also

maximum temperature of the diagram is also not lie at critical point but at point ahead of it so it is not

necessary that always gas will exist at temperature and pressure above critical point but both phases

can co-exist.



Cricondentherm:

It is the maximum temperature of the mixture above which only gas will exist regardless of

pressure and saturation envelope ends there.

Cricondenbar:

The maximum pressure of phase envelope is cricondenbar and it is the pressure above which no

gas will form regardless of temperature

The phase diagram of two component mixture will always form between the vapor pressure

lines of the components in pure form as shown in the following figure:



The critical temperature of the mixture is in between the critical temperatures of the two

components in pure form while critical pressure of the mixture will always be at a greater value as

compared to critical pressures of both components in pure form.

The location of the phase diagram of the mixture between the vapor pressure lines of the

components in pure form depends upon the percentage composition of each component. In the

diagram as the % composition of CA is increased, the phase diagram tends to switch towards its (CAs)

vapor pressure line and same is the case when we increase the % composition of CB, the diagram will

switch towards its vapor pressure line, In short at 50% CA and 50% CB composition, the phase diagram

will be at center also the percentage composition affects the size of the envelope. The following figure

shows the phase diagrams of eight mixtures of methane and ethane with different percentage

compositions (shown in small box at left hand side of figure), the figure illustrates the above three

paragraphs including this. The dotted line joins the critical points of all phase diagrams.

Problem:

The densities of methane liquid and gas in equilibrium along vapor pressure line (liquid and

gas both coexist at vapor pressure line) are given below; estimate the density of methane at its critical

point of -116.7oF.

Temperature Density (lb/ft3) Mean Density

Saturated liquid Saturated Vapor -253 26.17 0.1443 13.15715

-235 25.25 0.2747 12.76235 -217 24.25 0.4766 12.3633 -199 23.18 0.7744 11.9772

-181 21.89 1.202 11.546



RESULT: the density of methane at its critical point of -116.7oF is 10.15 lb/ft3.

PV DIAGRAM FOR TWO COMPONENT MIXTURE

Explanation:

If the experiment is performed at constant temperature, the phase diagram will be like as

shown in above figure. If we take example of methane and ethane mixture then at high pressure P1 the

mixture will be liquid, as mixture is made to expand the pressure will reduce until point 2. The expansion

in liquid is negligible as compared to pressure drop as liquids are incompressible. At pressure P2, bubble

point has reached such that few bubbles of the gas have evolved. After bubble point further expansion

causes the pressure drop of the mixture as at each step the liquid fraction of the mixture converts into

vapor phase up to dew point where all mixture converts into vapor phase. After dew point only gas

exists which expands largely with small pressure drop as gas is highly compressible.

During the flash vaporization experiment, we can also find the composition of the fluid in the

cell. The composition is important in designing the surface plant facility for production and also in



economics. For example we have made the phase diagram of reservoir fluid now we have compressed

the fluid at initial reservoir pressure and reservoir temperature in the cell and find the composition of

the fluid (say which is in liquid phase) by flowing the fluid to chromatograph, the composition will tell us

that at this reservoir pressure we will have these components in the fluid, decrease pressure and

similarly at each pressure we find the composition which gives the idea about the reservoir fluid

composition that is being produced at each step because at each pressure drop the composition will be

varying as at each step the heavy fractions will continue to vaporize and start converting into gas.

RETROGRADE CONDENSATION:

We know that at constant temperature a decrease in pressure causes a liquid to convert into gas for

pure substance, this same phenomenon occurs for mixture of two or more components but the

restriction is that the temperature should be below critical temperature, which we have already studied.

We also know that for pure substance, it is the critical temperature that above which if temperature is

there then gas cannot be liquefied but for mixture it is the cricondentherm. The region between the

critical temperature and cricondentherm is called retrograde gas condensate as in this region with

expansion and pressure drop, gas is liquefied to liquid such that a reverse phenomenon occurs which

one expects. Retrograde means reverse. These reservoirs are initially gas reservoirs and only gas exists if

pressure is above the first dew point Pd1. If we have a gas at temperature between Tc and

cricondentherm and pressure P1, then as expansion is made at constant temperature, the pressure is

dropped and separation is occurred between the heavy and light components and the molecular

attraction among heavy components increases which causes them to condense to form liquid as

pressure reaches to Pd1 (first dew point pressure or retrograde dew point). As pressure is much

lowered, the concentration of liquid increases and up to pressure P2 25% of the mixture will exist as

liquid and only 75% will be gas. As pressure is much lowered from P2, then the heavy components that

have condensed again convert into vapors as molecular forces tend to be weaker, at this low pressure

and as pressure is dropped up to second dew point Pd2 all liquid become vapor and at pressure P3 only

gas exists. In case of mixture, the gas that exists between critical point and cricondentherm can be

liquefied but not the gas that exists above the cricondentherm.

Pd1

Pd2



Retrograde gas condensate reservoir can be of two types:

Lean gas reservoir

Rich gas reservoir

RICH RETROGRADE GAS CONDENSATE RESERVOIR

In these types of reservoirs, heavier component quantity is more (more liquid contents will be there

condensed). Reservoir temperature is near to critical temperature and liquid drop out will be more in

reservoir.

LEAN RETROGRADE GAS CONDENSATE RESERVOIR

In these types of reservoirs, heavier component quantity is less (less liquid contents will be there

condensed). Reservoir temperature is near to cricondentherm and liquid drop out will be less in

reservoir.

FIELD IDENTIFICATION

At surface, we can also distinguish between rich and lean retrograde gas condensate reservoir if we

dont have phase diagram because it is too expensive job. Field identification of rich retrograde reservoir

is that quantity of oil at surface is greater than 100bbl/MMscf and for lean retrograde reservoir; quantity

of oil is less than 100 bbl/MMscf.

CONDENSATE

The liquid which drops out after the dew point pressure is called condensate. It causes two main

problems:



It reduces the gas relative permeability so overall productivity of gas reservoir decreases.

A large amount of valuable condensate is left in the reservoir.

PRESSURE PROFILE

As we move towards perforation, pressure declines. We try to maintain well flowing pressure above

dew point pressure.

Initially say reservoir pressure is 5000psi and well flowing pressure is 3000psi then pressure difference is

2000psi. We have to maintain pressure difference because E&P company sign agreement at particular

flow rate. As the reservoir pressure decline, say to 4700 psi we have to equally drop well flowing

pressure i.e. to 2700 psi so that flow rate remains same. Dew point pressure is say 2500 psi and as we

decrease well flowing pressure due to decrease in reservoir pressure then point come when well flowing

pressure decreases below dew point pressure then liquid drop out occurs in well bore and region of

liquid drop out forms known as condensate banking which results in two main problems. Due to

formation of condensate banking, it reduces relative permeability of gas because it restricts flow of gas.

Condensate is very valuable for us to sell. Before dew point pressure, if condensate gas ratio (CGR) is

150bbl/MMscf then after dew point pressure we get less condensate gas ratio (CGR) say 100bbl/MMscf.

As the reservoir pressure declines more, condensate banking region increases and relative permeability

of gas decreases furthermore and more valuable condensate left in the reservoir.



We try to maintain well flowing pressure above dew point pressure so that liquid drop out occur either

in the well bore or in surface equipments.

Reservoir engineer main task is to increase net present value (NPV).

TREATMENT METHOD OF CONDENSATE BANKING

Pressure maintenance method

Production optimization method

PRESSURE MAINTENANCE METHOD

Water flooding

Gas recycling

Water flooding

Through injection well, we inject water so that reservoir pressure maintains and initial pressure curve

remains same and hence, well flowing pressure always remains higher than dew point pressure and

liquid drop out will not occur in reservoir.

This method is not feasible for gas reservoir as in case of water drive, water encroaches and the main

mechanism of gas i.e. expansion ends.



Gas recycling

To maintain condensate gas ratio (CGR), we inject same gas through injection well which we has

produced which maintain pressure fully or partially. We sell condensate to get profit and inject the

recovered gas.

Gas recycling can be done by;

Partial pressure maintenance

Full pressure maintenance

Partial pressure maintenance

If we inject only the gas produce then it partially maintain the pressure. It is because we have not

injected same amount of mass which we produced from reservoir as we separate the condensate and

sell it.

Full pressure maintenance

If we inject same mass which we produced from reservoir then it fully maintain the pressure. When we

separate the condensate and gas, gas recovered will be combined with bought gas equivalent to the

mass of condensate separated and injected through injection well so that pressure can be fully

maintained.

TYPES OF RESERVOIR FLUID

Black oil

Volatile oil

Retrograde reservoir

Dry gas

Wet gas

FIVE SPOT AND SEVEN SPOT



PRODUCTION OPTIMIZATION

Production optimization can be done in two ways:

Hydraulic fracturing

Solvent injection

Hydraulic fracturing

Hydraulic fracturing is a well stimulation technique in which a fluid is pumped down casing under high

pressure to artificially fracture a reservoir rock.

Condensate banking area is removed by hydraulic fracturing. Well flowing pressure now becomes

greater than dew point pressure but it is not a permanent solution because condensate banking region

starts forming again because reservoir pressure declines with production and we have to

correspondingly decrease well flowing pressure to maintain flow rate and point comes when the well

flowing pressure decreases below dew point pressure.

Solvent injection

Solvent is injected which mobiles the condensate so that it can be recovered.

CLAUSIUS-CLAPEYRON EQUATION

According to Clapeyron equation

( )

This equation expresses the relationship between the vapor pressure and temperature where

is the

rate of change of vapor pressure with temperature which is the slope of vapor pressure line.

Where;

LV = heat of vaporization

VMG VML = change in volume of 1 mole as it goes from liquid to gas



Since, VMG >>> VML

------ (1)

Since,

So, equation (1) becomes

(

)



EXAMPLE 2-1:

Plot the vapor pressure of n-hexane in such a way that it will result in a straight line:

SOLUTION:

For this we use the following formula:

(

)

Temperature (T) Temperature (T) Vapor Pressure

(Pv) 1/T ln(Pv)

(In Fahrenheit) (In Rankine) (In psia) (R-1)

155.7 615.7 14.7 0.001624168 2.687847494

199.4 659.4 29.4 0.00151653 3.380994674

269.1 729.1 73.5 0.001371554 4.297285406

331.9 791.9 147 0.001262786 4.990432587

408.9 868.9 293.9 0.00115088 5.683239573

454.6 914.6 435 0.001093374 6.075346031

0

1

2

3

4

5

6

7

0.001 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017

l

n(

P

v)

1/T



Problem:

Vapor pressure data of CO2 are given below, Calculate vapor pressure at 40F.

Temperature (Rankine) Vapor pressure

(Pv) ln (Pv) 1/T

420.9 147 4.990433 0.002376

458 293.9 5.68324 0.002183

482.5 440.9 6.088818 0.002073

At 40 degree fahrenheit or 500 degree rankine, we have 1/T = 0.002.

So at 1/T = 0.002, lnPv from graph is 6.4

The value of R in field units is 10.73 psia.ft3/lb-mole.oR.

For pure substance if temperature is above Tc then it is gas and if below Tc then it could either be totally

liquid, liquid and gas both or only gas depending on pressure such that for pressure above Pv, liquid; for

pressure equal to Pv, both; and for pressure below Pv, gas only.

NOTE: Exercise and example questions are in assignment

0

1

2

3

4

5

6

7

0.002 0.00205 0.0021 0.00215 0.0022 0.00225 0.0023 0.00235 0.0024



Problem:

Pure 2-methyl pentane is held in a closed container at 150oF. Both liquid and gas are present,

what is the pressure?

Solution:

The pressure will be equal to the vapor pressure of it at 150oF which is 17.5 psia.

Problem:

A sealed container with volume of 3 cu-ft holds 7 lb of n-butane at 300oF, what is the volume

of gas and liquid in the container?

Solution:

Since the temperature is below critical temperature and pressure is not given, we assume it to

have both liquid and gas as mentioned in the question.

The total mass in the container will be:

( )

From the n-butane density temperature graph see values of densities at given temperature.

Density of liquid = 0.31 gm/cc = ( )

(as 1kg=2.2lb and

1ft=30.48 cm)

Density of gas = 0.15 gm/cc = ( )

( )

COMPOSITION DIAGRAM FOR TWO COMPONENT MIXTURE:

The composition diagram shows the composition of mixture at any pressure and temperature

conditions. By composition we mean mole fractions of gas and liquid present in the mixture at that

pressure and temperature and moreover in the gaseous phase we can determine the mole fractions of

each component exists as gas and similarly mole fractions of each component exists as liquid in the

liquid phase of the mixture.



The composition diagram is shown below:

The diagram is made with the same experiment as the phase diagram was made up of. We will

be taking the example of mixture of methane and ethane. The phase diagram is made with reference to

the composition (in mole %) of one component in the mixture. In two phase mixture, knowing mole % of

one component, the mole % of the other component is simple to evaluate as their sum is 100%.

We are assuming that this diagram is made with reference to methane rather than ethane. If we

have prepared the mixture taking 70% lb-moles of methane (or we can say in mass of 1 lb-mole mixture,

methane will be 0.7 lb-mole and ethane will be 0.3 lb-mole). Now we will take this mixture to PVT cell

and apply pressure of P4 at which all mixture will be liquid. Since the pressure of the cell will remain

constant so the diagram is made for constant temperature and it changes with change in temperature.

As up to bubble point pressure, all mixture remain liquid so no problem is up to here, as bubble point

reaches, we can plot the point on graph at 70% methane composition (as during the whole process, the

composition of methane or mole % of methane will remain same in the mixture, how ever the total

mole % of methane will be divided between liquid and gas in the region between bubble point pressure

and dew point pressure) similarly at pressures below dew point pressure, the whole mixture remain gas

so no problem will be there too (when dew point comes, we can also plot the point), similarly repeating

same experiment for different composition of methane we can note bubble points and dew points and

can obtain the curve. The problem will arise in the region between bubble point and dew point in which

the two phases will exist.

Now if we want to find the composition of mixture (composition of liquid and gas in the mixture)

at a particular pressure and temperature value knowing the mole % of the components in the mixture

(binary mixture). Seeing the mole % of the component in the mixture for which composition diagram is

4

70%



given along x-axis and the pressure on y-axis and plot the point in the envelope (envelope at given

temperature) such that point 1. From this point, we draw horizontal line joining bubble point line and

dew point lines at points 2 and 3. Now we can evaluate the following parameters:

( )

( )

Further the composition of gas can be determined by flashing the gas to gas chromatograph.



IDEAL GAS

The assumptions for ideal gases are:

The volume occupied by the molecules is insignificant.

There are no attractive forces or repulsive forces.

The collisions are perfectly elastic.

For ideal gas we can use equation of state, the equation is called state equation as it includes all

variables such as temperature, pressure and volume, needed to define state of gas. The equation of

state is:

P=pressure in psia

V=volume in cu-ft

N=no. of moles in lb-mole=mass (lb)/molecular mass (lb-mole)

T=temperature in rankine

R= gas constant= 10.73 (psia.cu-ft/lb-mole.oR)

Problem:

Calculate mass of ethane gas at 100 psia and 68oF in a cylinder with volume of 3.2 cu-ft, assume

ethane is an ideal gas.

Solution:

Using equation of state:

( )

SPECIFIC GRAVITY OF GAS:



Problem:

Calculate density of ethane at the conditions given in previous example. Assume it as ideal gas.

Solution:

The molecular weight of mixture of gas (such that air or any other) is called apparent molecular

weight.

Problem:

Dry air is a gas mixture consisting of nitrogen, oxygen and small amount of other gases.

Calculate apparent molecular weight of air.

Components Composition (mole fraction)

Nitrogen 0.78

Oxygen 0.21

Argon 0.01

Solution:

Problem:

Calculate specific gravity of gas with the following data:

Components Composition (mole fraction)

C1 0.85

C2 0.09

C3 0.04

n-C4 0.02



Solution:

( )

( )

( )



For Water

Specific weight of gas is given by;

GAS DEVIATION FACTOR (Z)

Vactual and Videal are both measured at same temperature and pressure condition.

At low pressure, molecules are very far so, no attraction and repulsion is present between

molecules and Vactual = Videal

At high pressure, molecules are very near so, attraction and repulsion between molecules is

present and Videal > Vactual

At particular pressure, we calculate Videal and Vactual. Videal will also be different at different pressure

which can be understood by the ideal gas equation PV=nRT, V here is ideal volume which is different at

different pressure.

The curve is isotherm i.e. temperature is constant during experiment. At low pressure, molecules are

very far so, no attraction and repulsion is present between molecules and Vactual=Videal and Z is equal

to 1 as shown by point (1). With increase in pressure from point (1) to point (2), molecules come closer

to each other and force of attraction between the molecules starts dominating. Due to attraction,

Vactual decreases from Videal and value of Z decreases from 1 and if we remove attraction between

molecules at same pressure then we obtain Videal. At point (2), repulsion between the molecules starts

and with further increase in repulsion with increase in pressure, Vactual starts increasing and value of Z

increases. At point (3), force of attraction and repulsion counterbalance each other and Vactual



becomes equal to Videal and value of Z becomes 1. As pressure is further increased, Vactual increases

from Videal and value of Z increases from 1 due to further increase in repulsion between molecules.

PROBLEM

Calculate the mass of methane contained at 1000psia and 680F in a cylinder with volume of 3.2ft3,

assume methane is a real gas?

SOLUTION

680F > TC so methane will be gas regardless of pressure

From figure 3.2, Z=0.89

( )

FACTORS AFFECTING VALUE OF Z

Z depends upon pressure, temperature and composition.



Effect of pressure on value of Z is discussed in Z vs. P curve.

Effect of temperature

At same pressure, when we increase temperature value of Z increases because kinetic energy increases

and molecules go far i.e. Vactual increases.

Effect of composition

Critical temperature of methane is 343.30R and ethane is 5500R. Suppose, temperature is 5000R then

methane will be gas regardless of pressure but ethane can be gas or liquid depending on pressure

whether it is above or below vapor-pressure line and near vapor pressure line attraction is more due to

change of phase hence, behavior of methane and ethane is different so deviation of both will be

different as well. It means we have different graphs for different compositions so, to make it valid for all

we use law of corresponding state.

LAW OF CORRESPONDING STATE

Law of corresponding state says that all pure gases have the same Z factor at the same value of reduced

pressure and reduced temperature.

P and PC both have units of psi and T and TC both have units of 0R. TC and PC are of particular composition

by which it becomes independent of composition.

PROBLEM

Determine specific volume of ethane at 918psi and 1170F when Tc = 549.90R and Pc = 706.5 psi?

SOLUTION



From figure 3.6, Z=0.39

( )



FOR MIXTURES

Law of corresponding states has been extended to cover mixtures of gases which are closely related.

First pseudo critical temperature and pseudo critical pressure is calculated by Kays mixture rule;

Where, y is mole fraction and Tc is critical temperature of particular component.

After calculating Tpc and Ppc, pseudo-reduced temperature Tpr and pseudo-reduced pressure Ppr is

calculated from;

Using Tpr and Ppr, calculate Z from figure 3.7

PROBLEM

Calculate the pseudo-critical temperature and pseudo-critical pressure of the gas mixture given

below:

Component Mole fraction Tc (0R) Pc (psia) yjTcj yjPcj C1 0.850 343.3 666.4 291.81 566.44

C2 0.090 549.9 706.5 49.49 63.59

C3 0.040 666.2 616.0 26.65 24.64

n-C4 0.020 765.6 550.0 15.31 11.00

383.26 665.67

Tpc = 383.260R

Ppc = 665.67 psia

PSEUDO CRITICAL PROPERTIES OF GAS WHEN COMPOSITION IS UNKNOWN

Chromatograph gives us composition but if composition is unknown then we determine specific gravity

of gas. For finding specific gravity of gas mixture, we have device Schilling Effusionmeter in PVT lab.

From specific gravity, Ppc and Tpc can be obtained from figure 3.11

PROBLEM

Determine Z factor of a natural gas with specific gravity of 1.26 at 2560F and 6025psi?

SOLUTION



Ppc = 587 psia and Tpc = 4920R

Now using figure 3.7, Z=1.154



HEPTANE PLUS TERM

C7, C8, C9, C10, is written as C7+ because their composition is very less and known as heptane plus. Tc

and Pc of C7+ can be determined using figure 3.10 with the help of molecular weight of C7+ and specific

gravity.

PROBLEM

Determine pseudo critical temperature and pseudo critical pressure for the gas given below:

SOLUTION

Heptane plus



critical critical

mole temperature pressure

fraction deg R psia

component yj Tcj yj Tcj Pcj yj Pcj

H2S 0.0491 672 33.0 1300.0 63.8

CO2 0.1101 548 60.3 1071.0 117.9

N2 0.0051 235 1.2 493.1 2.5

C1 0.577 343 198.1 666.4 384.5

C2 0.0722 550 39.7 706.5 51.0

C3 0.0455 651 29.6 616.0 28.0

i-C4 0.0096 740 7.1 527.9 5.1

n-C5 0.0195 764 14.9 550.6 10.7

i-C6 0.0078 833 6.5 490.4 3.8

n-C5 0.0071 845 6.0 488.6 3.5

C6 0.0145 910 13.2 436.9 6.3

C7+ 0.0835 1157 96.6 367.0 30.6

506.2 707.9

Tpc = 506.20R

Ppc = 707.9 psia



PROBLEM

Specific gravity of C7+ is 0.807 and molecular weight of C7+ is 142 lb/lb-mol. Calculate Tc and Pc?

Using figure 3.10,

Tc = 11570R

Pc = 367 psia

CONCLUSION

PROBLEM

A cylinder has a volume of 0.5 ft3 and contains a gas ata pressure of 2000 psia and 1200F. The pressure

drops to 1000psi after 0.0923 lb-mol gas is removed, temperature is constant. Z factor was 0.90 at

2000psi. What is the Z factor at 1000 psi?

SOLUTION

( )

( )

PROBLEM

A 20ft3 tank at 1000F is pressured to 200 psi with a pure paraffin gas. 10 lbs of ethane are added and

specific gravity of the gas mixture is measured to be 1.68. Assume that gases exist as ideal gas, what

was the gas original in tank?



SOLUTION

( )

Mole fractions

WICHERT AZIZ CORRELATION (Effect of non-hydrocarbon components)

Natural gases commonly contain hydrogen sulfide, carbon dioxide and nitrogen. Gases are sour and

sweet as well. Gas which contains one grain of H2S per 100ft3 of gas is called sour gas. When sour gas is

present, value of Z is over-estimated. The remedy to this problem is to adjust the pseudo-critical

properties. If non-hydrocarbon contents are greater than 15% then error increases and correction is

required for which we use Wichert Aziz Correlation.

The equations used for the adjustment of pseudo-critical properties are:

( )



Where; = pseudo critical temperature adjustment factor and is calculated from figure 3.12

PARTIAL PRESSURE

Pressure exerted by individual gases in a total gas mixture is called partial pressure.

Where;

Pi = partial pressure of particular gas

yi = mole fraction of particular gas

P = total pressure

It is Daltons law of partial pressure.

PROBLEM

Calculate the volume occupied by 1 lb-mole of natural gas at standard conditions?

SOLUTION

Standard condition

T = 600F + 460 = 5200R

P = 14.7 psi

Molar volume: 1 lb-mole of any gas occupies 379.5 ft3 at surface condition.

FORMATION VOLUME FACTOR OF GAS

It is the amount of gas in reservoir requires to produce 1 SCF of gas at surface.

( )

In PVT cell, we simulate reservoir condition. Amount of gas at surface condition remains same which is

same if we reduce the pressure of cell to 14.7 psi. At 3000 psi, to produce 1 SCF of gas reservoir volume

requires less and at 2500 psi, to produce 1 SCF of gas reservoir volume requires more. With decreases in

pressure, volume of gas at reservoir condition increases with constant volume of gas at surface

condition so value of Bg increases with decrease in pressure.



At high pressure, molecules are very near so, they behave as liquid and with decrease in pressure

volume of gas at reservoir condition does not increase rapidly so, there is no significant change in the

value of Bg. At low pressure, molecules are far and with decrease in pressure volume of gas at reservoir

condition increases rapidly so, there is significant change in the value of Bg.

DERIVATION

BG= VRESERVOIR / VSURFACE. (1)

From gas law;

For 1 mole of gas

PV = ZRT

At reservoir condition

PRVR = ZRRTR

VR = ZRRTR / PR



At surface condition

PSVS = ZSRTS

At surface condition, gas behaves ideally

ZS = 1

PSVS = RTS

VS = RTS / PS

Put the values in (1)



Formation volume factor of gas

( )

PRESSURE VRESERVOIR VSURFACE 3000 1 ft3 100 SCF

2500 1 ft3 80 SCF

2000 1 ft3 60 SCF

As the reservoir pressure decreases, 1 ft3 of reservoir gas volume produces less SCF of gas at surface.

PROBLEM

Calculate the value of formation volume factor of a dry gas with specific gravity of 0.818 at reservoir

temperature of 2400F and reservoir pressure of 2100psi?

SOLUTION



Ppc = 647 psia & Tpc = 4050R

Now, from figure 3.7

Z = 0.86

Now,

( )

PROBLEM

A vessel contains 100 lb-mol of hydrocarbon gas. Calculate the SCF of gas in vessel?

SOLUTION

Molar volume = 379.5 SCF / lb-mol

1 lb-mol = 379.5 SCF

100 lb-mol = 37950 SCF



COEFFICIENT OF ISOTHERMAL COMPRESSIBILITY OF GAS

(

)

( )

UNIT: psi- or sip

The relation of gas compressibility and reservoir pressure (or simply the pressure) is given below:

The graph shows that at any particular value of pressure, how much the gas compressible is. As it is seen

from the graph that at a state of low pressure, the gas is highly compressible means with unit change in

pressure V/V (fractional change in volume) will be large as molecules are so far apart. At moderate

pressure, the compressibility is moderate and at high pressure, the compressibility is negligible (gas

behaves as incompressible) i.e. with unit change in pressure V/V (fractional change in volume) will be

small as molecules are very close to each other.

For graphs of Cg and Z, temperature is above critical temperature so, liquid cant be formed



PROBLEM

The following table gives volumetric data at 1500F for a natural gas. Determine the coefficient of

isothermal compressibility for this gas at 1500F and 1000psi?

PRESSURE (psia) MOLAR VOLUME (ft3/lb-mol)

700 8.5

800 7.4

900 6.5

1000 5.7

1100 5.0

1200 4.6

(

)

(

)



DERIVATIONS

For ideal gas

For real gas



Value of Cg can be greater or smaller than ideal. From 1 to 2, slope is negative so putting negative slope

gives value of Cg higher than ideal. On the contrary from 2 to 3, slope is positive so putting positive

slope gives value of Cg lower than ideal. It can be understand physically by that from 1 to 2, attraction is

dominant so molecules come close and compressibility will be more.

PROBLEM

Estimate the coefficient of isothermal compressibility of gas at 1700 psi; assume that gas behaves like

an ideal gas

SOLUTION

PSEUDO CRITICAL TEMPERATURE ADJUSTMENT FACTOR ( )

Both CO2 and H2S factor is present in fig 3.12 to calculate . For adjustment in pseudo critical pressure,

formula contains only mole fraction of H2S but in factor both CO2 and H2S factor is present. N2 is not

possible to remove because its plant is very expensive so if 5% N2 is present then error in Z is considered

1% and similarly, if 10% N2 is present then error in Z is considered 2%.



THE COEFFICIENT OF VISCOSITY OF GAS The coefficient of viscosity is the measure of the resistance to flow offered by the fluid. Its unit

is centipoises. The graph of viscosity of a gas with pressure is shown below:

The graph shows that viscosity of a gas decreases with

decrease in pressure at constant temperature because as

pressure is decreased molecules get far away from each

other and the collision among themselves are reduced

and viscosity is also reduced. The graph also shows the

relationship between gas viscosity and temperature. The

relation of viscosity with temperature can be studied in

two cases:

At higher pressure:

At constant higher pressure, molecules are very near to each other, as temperature is

increased at high pressure molecules tend to go away from each other and viscosity is reduced.

At low pressure:

At low pressure, molecules are already far away from each other, so as temperature is

increased, the collision among the molecules increases and viscosity increases.

Note from the graph, we can see that there is a point at which all curves are meeting, at this pressure

point the viscosity of gas will become independent of temperature means whatever the temperature is

the gas viscosity will be the same at this particular pressure.

METHODS OF FINDING GAS VISCOSITY

Now we will discuss how to find the viscosity of a gas. Basically, the viscosity of gas cannot be

determined by experimental methods in laboratory but we have to use various correlations and charts

to determine gas viscosity. Now the methods of finding gas viscosity will be discussed as per the

following cases:

Pure Gases

Mixture of gases (further divided as)

At atmospheric pressure

At high Pressure

High pressure

Low Pressure



FOR PURE HYDROCARBON GASES:

For pure hydrocarbon gases, we will be having the graphs between viscosity and

temperature with various isobars from which we can read the viscosity value at particular pressure,

temperature value. For example the graph of ethane is as follows:

FOR GAS MIXTURE

AT ATMOSPHERIC PRESSURE:

At atmospheric pressure, the viscosity of gas can be determined by the following formula:

Where;

viscosity of each component gas at atmospheric pressure determined by the following graph (6-7)

Mole fraction of each gas.

Molecular mass of each gas component.



Problem:

Calculate the viscosity of gas mixture given below at 200oF and pressure of 1 atmosphere.

Given Calculated

Component Composition(yj) gj Mj

C1 0.85 0.013 16.04 4.004997 3.404247 0.044255

C2 0.09 0.0112 30.07 5.483612 0.493525 0.005527

C3 0.04 0.0098 44.1 6.640783 0.265631 0.002603

n-C4 0.02 0.0091 58.12 7.623647 0.152473 0.001388

4.315877 0.053773



So, viscosity of the gas will be equal to:

If the gas composition is not known (as the composition of gas is expensive to determine using gas

chromatograph) and the specific gravity of the gas is known to us then we use the following graph to

determine gas viscosity at atmospheric pressure and at any temperature and specific gravity:



The above graph can also be used if we know the composition and specific gravity is unknown but still

we are given graph 6-8 (above graph) instead of 6-7 because from composition we can find apparent

molecular weight of the gas and knowing the molecular weight of air we can find the gas specific gravity.

In short, if we are given the composition of gas then first we look for the graph which is given to us, if it

is 6-7 then we will use the formula technique as discussed in the problem but if the given graph is 6-8

then we will use the specific gravity technique. But if the composition is unknown then surely sp.gr of

the gas will be known to us and graph 6-8 will be given which we will use to find gas viscosity.

There are three small graphs made above the graph 6-8, which shows the correction factor. Means if we

have the gas composition in which H2S is 10 mole % (0.1 in composition) and the overall sg.gr of the gas

is 1.5 then in the final gas viscosity value we determine, we will ad the factor 0.0005 to correct it. Same

is the case with N2 and CO2.

AT HIGHER PRESSURE:

Steps that are to be followed for calculating viscosity of gas at some higher pressure value above the

atmospheric pressure are as follows:

Calculate viscosity at atmospheric pressure

Find Tpc and Ppc from any of the following methods:

1-

2- If specific gravity of the gas is known then find it from the graphs discussed earlier in which one

was between sp.gr and Tpc and the other between sp.gr and Ppc.

Calculate Ppr and Tpr using Ppc and Tpc and the given temperature and pressure.

Calculate the ratio

( )from either of the following graphs such that use the graph

which contains such range of sp.gr values of the gas that contains your calculated value, in that

graph see the value of the ratio at calculated Tpr and Ppr.

From the ratio calculate the gas viscosity at high pressure by multiplying the ration with gas

viscosity calculated at atmospheric pressure in the first step.

HEATING VALUE:

The heating value of a gas is the quantity of heat produced when the gas is burnt completely

to CO2 and water.

The pricing of the gas is done on the basis calorific value or heating value whose unit is

BTU/SCF. When agreement is made, the company providing the gas has to maintain the heating

value of the gas which they have committed in the agreement. H2S, water nitrogen and CO2 are not

removed from the natural gas just because they are hazardous but because they reduce the heating

value of the gas that has to be maintained or improved.



GROSS HEATING VALUE:

It is the heat produced in complete combustion under constant pressure with combustion

products cooled to standard conditions and water in the combustion products condensed to the

liquid state. Gas is sold on the basis of gross heat value

NET HEATING VALUE:

It is defined similarly except that water of combustion remains vapor at standard condition.

The difference between net and gross heating values is the heat of vaporization of water.

PRIOR TO COMBUSTION:

Means we have discussed that as a result of gas combustion, water is produced that absorbs

some of the vital amount of heat but this is not the only water that cause problem. The water may

be associated with gas initially when it was produced (such that gas that we have before combustion

may also contain water) so on the basis of quantity of water present with the gas, we have two

types of gases:

o Wet gas: that contains greater than 1.75 volume % of water.

o Dry gas: that contains less than 1.75 volume % of water.

Water contents in the gas are to be reduced or minimized by dehydration of the gas. The

dehydration of gas is required because the water molecules in the gas combine with the gas

molecules at low pressure and temperature value to form hydrate crystal (ice like). The formation of

hydrate crystals mostly occur in pipelines particularly where nozzle action is taking place or at joints

because at these points due to sudden expansion of gas like at the choke point the pressure is

greatly reduced and according to Joules Thomson effect whenever expansion occurs cooling takes

place so when pressure and temperature is reduced, formation of hydrate may occur that may

possibly block the line, so at these crucial points we use heaters to maintain temperature.

Moreover, more the water contents are in gas less will be the heating value of the gas.

So, the water in the gas is removed at dehydrating plant using Glycol but when gas is passed

through this plant, not all water is removed some water still remain in gas, if we have threat that

this water may cause problem in the pipelines where the gas travel then we use Condensate Trap

that is operated at different low temperatures to remove further water from the gas, the actual

purpose of this is to remove the heavy contents in the gas by condensing them as liquid because the

sale price of oil is greater than gas. In this we can reduce the temperature to so much extent that

hydrates are formed in this plant, we do this because the hydrates that may form in the pipeline are

produced here because these hydrates can be removed from the plant during servicing but it will be

difficult to locate the place of hydrate formation in the pipeline and then remove those hydrate

crystals.



HEATING VALUE FOR DRY GAS MIXTURE:

Heating value of a gas mixture can be given as:

( )

Where, Lcj is the heating value of each component (that can be either gross or net heat value), if net

heat value is used then Lc (the total heat value of the mixture) will be the net value too. Yj is the

mole fraction of each component. In above relation Yj is the mole fraction of water.

In above equation, Z is given by:

( )

Problems:

Determine the gross calorific value of a separator gas of composition given below:

Component Composition(Yj) Lcj Z YjLcj (1-Zj) Yj(1-Zj)

CO2 0.0167 0 0.9943 0 0.075498 0.001261

N2 0.0032 0 0.9997 0 0.017321 5.54E-05

CH4 0.7102 1010 0.998 717.302 0.044721 0.031761

C2H6 0.1574 1769.6 0.919 278.535 0.284605 0.044797

C3H8 0.0751 2516.1 0.9805 188.9591 0.139642 0.010487

1-butane 0.0089 3251.9 0.9711 28.94191 0.17 0.001513

2-butane 0.0194 3262.3 0.9667 63.28862 0.182483 0.00354

i-butane 0.0034 4000.9 0.948 13.60306 0.228035 0.000775

n-pentane 0.0027 4008.9 0.942 10.82403 0.240832 0.00065

Hexane 0.0027 4755.9 0.91 12.84093 0.3 0.00081

Heptanes+ 0.0003 5502.5 0.852 1.65075 0.384708 0.000115

1315.945 0.095765

( )

Now for Lc(real):

First calculating value of Z for mixture:



( )

( )

Now ideal heating value is given by:

The value of calorific value for ideal and real gases are approximately the same because as per the

definition of calorific value, it is the heat liberated at standard temperature and pressure conditions

so the values of Z are taken at standard temperature and pressure conditions for each component

which is approximately equal to one.

HEATING VALUE FOR WET GAS MIXTURE:

For Lc(wet), we calculate first Lc(real) considering the gas to be a dry gas by adopting the same

procedure as discussed above but in the end we use the following relation to calculate calorific

value for wet gas depending whether we need to calculate the net or gross calorific value.

For Net calorific value:

The real calorific value for wet gases (containing water) can be given by:

( )

For Gross caloric value:

( )

In case of wet gas there are different formulae for net and gross however in case of dry gas both net

and gross are calculated by the same method or formula.



For Lcwet, we multiply the Lcdry by the factor (1-Ywater) because the amount of water present in

the gas will not burn and will not produce any heat. While calculating gross Lcwet, we add a factor

0.9 because in wet gas as heat is produced by burning of the fuel, the water molecules that are

formed as result of chemical reaction not only absorb heat but also the water prior to combustion

absorbs heat so when we condense it then this heat will also liberate that approximately

0.9BTU/SCF, this was not added in the net heat of wet gases because according to definition of net

heat, products are not condensed to STP.

The chart shows how we have to proceed in problems:

1) First whether the heat calculated is net or gross.

2) For example say we have to calculate gross heat then see whether gas is dry or wet.

3) If gas is wet then first we calculate Lcgross dry using

and then Lcwet using

( )

Some points:

If we are given mole fraction of each component is a mixture, then we can calculate weight

fraction of each by multiplying the mole fraction of each by its molecular weight. Volume

fraction is same as that mole fraction.

Heating value

Gross

Wet

Lc(dry)(1-ywater)+0.9

dry

(YjLcj)/Z

Net

wet

Lc(1-ywater)

dry

(YjLcj)/Z



Individual volume or partial volume of a component in mixture and partial pressure are

given by:

and

Formation volume factor of gas is the shrinkage factor and its reciprocal is the expansion

factor.

Formation volume factor of oil is the expansion factor.

The gas compressibility helps in understanding the drive mechanism for gas reservoirs.

We use STP conditions to calculate volume because all the reserves are reported at standard

temperature and pressure conditions.

The importance of this subject is:

o Fluid identification

o Drive mechanism

o Bubble point pressure (suggesting the type of oil reservoir) and dew point pressure

(important in retrograde reservoir).

Experiment to find specific gravity of gas:

We can determine the specific gravity of gas in the laboratory using Schilling Effusion meter.

Schilling Effusion meter:

Principle:

Effusion meter works on the principle of Grahams law of effusion, which states that

rate of effusion of gas is inversely proportional to the square root of its density.

Methodology:

In effusion meter, first liquid such as water is filled through the glass tube in to the

container such that water reaches up to the mark. Now air is injected by connecting the compressor

pipe to the inlet nozzle valve, the air with high pressure when enter through the orifice above the glass

tube, it displaces the water in the tube and we will continue injecting gas in the tube until the water

level reaches to the bottom mark. Now the inlet valve is closed and the outlet valve is opened and the

air will flow out as water is applying pressure on it, as air start to remove, the water level start rising. We

will continue to note the time such that how much time the gas or air in the tube will take to evacuate

fully. The full evacuation of gas can be identified when the water level reaches to its initial level. We take

three time readings. Similarly we repeat the process for the gas whose specific gravity is to be

determined and take three readings for it.



We can use Grahams law of effusion for two gases as:

( )

Putting the above values in equation (A);

Since Vgas=Vair so, above equation becomes:

Final water

level

Gas/air

filled



(

)

(

)



Properties of oil:

SOLUTION GAS OIL RATIO

It is defined as the quantity of gas dissolved in oil at reservoir condition or it can also be

defined as the amount of gas that liberates from the oil (not come with oil as it may include the free

gas too) as the oil is transferred from the reservoir to the surface condition. It is denoted by Rs and its

graph with pressure is as follows:

When the reservoir pressure is above bubble point pressure then no gas is liberated in the reservoir and

all the gas dissolved, as we remove the Hg from the cell to reduce pressure above bubble point pressure

then no gas will liberate and all gas will remain dissolved and so the Rs will remain same. As pressure is

reduced below bubble point pressure then the gas that was dissolved in oil now start to escape and the

quantity of gas in oil decreases which is equal to the previous volume dissolved minus the free gas cap

now forms. So as dissolved gas quantity decreases so Rs will also decrease below Pb.

The quantity of oil in the cell is constant say 2 STB because at any condition when we take the oil out of

the cell and confine it to STP then it will give the same volume, the only thing that changes is the

quantity of gas dissolved which remain same above bubble point pressure and when pressure is

dropped below Pb, the quantity of gas dissolved decreases as

.

Pb Pressure

Rs

oil oil

oil

gas

P>Pb P=Pb P



Difference between solution gas oil ration (Rs) and producing gas oil ration (GOR):

GOR is the amount of gas that comes with oil at surface and it has same unit of SCF per STB.

Above bubble point pressure the graph between pressure and GOR repeats the same path as that of Rs

because the gas coming at surface is the gas that is dissolved in oil. Now as pressure declines below Pb,

there will be two cases:

If the free gas cap that is formed is not producing the gas through perforation such that gas

coning has not started yet then again the GOR graph VS pressure repeats the same curve as that of Rs

means the value of GOR start decreasing, but when the free gas start to produce the GOR start to

increase as shown below:

Note: The change is Bo value is the change in reservoir volume as the surface volume is taken to be

1STB.

Method to calculate the solution gas oil ratio:

The laboratory method to calculate the solution gas oil ratio is the differential vaporization.

Before performing the differential vaporization, we first perform the flash vaporization in which confine

the reservoir fluid in the laboratory cell at high pressure and reservoir temperature, we increase the

volume of the cell and the pressure drops, we see when the gas bubbles start forming, that is the bubble

point pressure and similarly we find the dew point pressure. Finding the bubble point and dew point

pressure is main purpose of flash vaporization. In this process we do not take the free gas out of the cell

such that it is a constant mass expansion.

In differential vaporization, we set the initial pressure of the cell at bubble point pressure

(determined through flash vaporization) and only oil is present in the cell, now we drop the pressure to

some value of our interest, the gas from the oil will liberate out and can be seen as free gas, the cell is

agitated so that gas and liquid will come to equilibrium. For example if Pb of the fluid is 2500 psig and

the pressure is dropped to 2300 psig, from the cell we can determine the reservoir volume of the free

gas. Now we will remove the free gas from the cell keeping the cell pressure constant at 2300 psig by

injecting mercury in the cell as the gas is removed. We will analyze the properties of the gas that is taken

Rp

or G

OR

P



out such that Z, Bg, composition through gas chromatograph, etc. The purpose of the gas removal from

the cell is that we can determine the gas volume dissolved in oil by subtracting the free gas volume in

SCF from the initial volume of the gas dissolved in SCF. Similarly we drop the pressure to 2100 psig, the

gas that is liberated out is again removed by keeping the cell pressure constant and we will repeat the

same process by decreasing the cell pressure and in the end we will decrease the cell pressure to 0 psig

or 14.7 psia and atmospheric temperature. When the cell reaches to this condition it will give us the STB

of oil present in the cell that was constant throughout the experiment. Summation of gas liberated

volume in SCF will give SCF gas dissolved initially.

Reservoir temperature:________ Bubble point pressure:_________ Volume of oil in cell:___y___STB

Pressure of the cell (psig)

Oil volume (ft3)

Gas liberated volume (SCF)

Dissolved gas volume(SCF)

2500=Pb 0 x

2300 K x-k

2100 L (x-k)-l ( )

1700 M (x-k-l)-m ( )

1200 N (x-k-l-m)-n ( )

800 O (x-k-l-m-n)-o ( )

0 Z P (x-k-l-m-n-o)-p ( )

Z obtained is at atmospheric pressure 0 psig or 14.7 psia but not at atmospheric temperature so we

reduce the cell temperature to 600F due to which some gas will liberate and we remove that gas and

find volume of oil which will be the volume of oil at surface condition in SCF. Convert SCF into oil volume

in STB y.

Rs at each pressure is obtained by;

( )



PROBLEM

Data from a differential vaporization on a black oil at 2200F are given below. Prepare a table of

solution gas oil ratio.

PRESSURE (psig)

GAS REMOVED

(cc)

GAS REMOVED

(SCF)

OIL VOLUME

(cc)

CUMMULATIVE GAS REMOVED

(SCF)

CUMULATIVE GAS DISSOLVED

(SCF) Rso (SCF/STB)

2620 0 0 63.316 0 0.21256 854.00 (Rsoi)

2330 4.396 0.02265 61.496 0.02265 0.18991 763.00

2100 4.292 0.01966 59.952 0.04231 0.17025 684.01

1850 4.478 0.01792 58.522 0.06023 0.15233 612.01

1600 4.96 0.01693 57.182 0.07716 0.1354 543.99

1350 5.705 0.01618 55.876 0.09334 0.11922 478.99

1100 6.891 0.01568 54.689 0.10902 0.10354 415.99

850 8.925 0.01543 53.462 0.12445 0.08811 354.00

600 12.814 0.01543 52.236 0.13988 0.07268 292.00

350 24.646 0.01717 50.771 0.15705 0.05551 223.02

159 50.492 0.01643 49.228 0.17348 0.03908 157.01

0 _ 0.03908 42.54 0.21256 0 0.00

0.21256

Oil volume @ 14.7 psia and 600F = 39.572 standard cc



FLASH VAPORIZATION OR CONSTANT MASS EXPANSION (CME)

It is used to find

bubble point pressure or dew point pressure

Bo above bubble point pressure (above Pb, Bo is obtained from differential vaporization)

Bubble point pressure is calculated for black oil and volatile oil and dew point pressure is calculated for

retrograde gas condensate reservoir. For wet gas and dry gas, we dont perform flash vaporization and

differential vaporization. Mass remains constant during experiment that divides into liquid and gas.

A sample of reservoir is placed in a lab cell. Pressure is adjusted equal to or greater than Pi. Temperature

is set at reservoir temperature. At each step, pressure is reduced and volumes of all are measured.

Process is repeated till atmospheric pressure.

Some volatile or black oil reservoirs are present at their bubble point and retrograde reservoirs at their

dew point initially when reservoir is discovered. Initial pressure is equal to bubble point or dew point so

to identify this, we use cell pressure greater than initial pressure.

Oil reservoirs are classified as saturated and undersaturated reservoir. Saturated reservoirs can be at

bubble point pressure or below bubble point pressure.

If we take sample from oil reservoir below bubble point pressure then that oil sample will be saturated

and at every pressure decline, gas liberates out from that sample.



Similarly if we take sample from retrograde reservoir below dew point then gas will be saturated i.e. at

every pressure decline, oil drops out.

Method to find Pb

From PVT cell it is not easy to find Pb because gas is transparent and we cant find where the gas is

liberated from. It is not possible to measure gas and liquid volume separately initially so, we calculate

total volume and then

Plot pressure VS volume

At Pb, compressibility of the system changes

Small change in pressure below bubble point pressure would occupy higher volume

Incompressible fluids are those which follow following relation



PROBLEM

Data from a flash vaporization on black oil at 2200F are given above. Determine the bubble point

pressure?

PRESSURE (psig) TOTAL CELL VOLUME (cc)

5000 61.03

4500 61.435

4000 61.866

3500 62.341

3000 62.866

2800 63.088

2600 63.455

2400 65.532

2250 67.4

2090 69.901

1890 73.655

1700 78.677

1450 86.224

1300 95.05

1050 112.717

Pb = 63.455 psig



PROPERTIES OF OIL

Solution gas oil ratio (gas solubility)

Bubble point pressure

Formation volume factor of oil

Temperature Decrease in Reservoir

Since 1952, temperature of SUI decreases from 2330F to 2170F. Temperature of reservoir slightly

decreases due to expansion because expansion causes cooling. It is Joule Thomson Effect that when gas

molecules pass through a choke/valve, its temperature generally decreases. Scales are formed due to

temperature decrease when it passes through choke/ valve/ perforation/ SSSV. Helium, nitrogen etc.

has inverse phenomenon but we are concerned with petroleum fluids in which temperature decreases

due to expansion.

Effect of temperature on solubility of gas

With temperature increase, solubility of gas decreases because kinetic energy increases and bonding

forces between gas and oil decreases so gas separates out.

It is reverse phenomenon for solid solubility in liquid.

FORMATION VOLUME FACTOR OF OIL Bo

It is the volume of oil at reservoir condition requires to produce 1 SCF of gas at surface.

( )

Value of Bo is always greater than 1 because Bo is expansion factor (from surface to reservoir volume

expand). Value of Bg is always less than 1 because Bg is shrinkage factor (from surface to reservoir

volume shrink).

Bo is a strong function of dissolved gas. As dissolved gas decreases, volume of reservoir decreases with

constant surface volume hence Bo decrease. If no gas is dissolved in oil (Rso = 0 SCF/STB) then value of

Bo is 1.

With increase in temperature, Bo decreases as gas dissolved liberates out due to increase in kinetic

energy so, volume of oil at reservoir condition decreases.

With higher viscosity, dissolved gas decreases so Bo value decreases.

Three factors that affect Bo are;

Liberation of gas: Due to decrease in pressure, gas liberates out so volume of oil at reservoir

condition decreases so Bo decreases



Decrease in temperature: We do not consider gas liberation effect in that. If only oil is present

without any dissolved gas then due to decrease in temperature, kinetic energy decreases and oil

volume shrink.

Decrease in pressure: Not considering effect of above two, with decrease in pressure oil volume

expand.

All these effects are simultaneously present. Decrease in temperature (shrinkage) and decrease in

pressure (expansion) effect nearly cancel out each other and the main factor that effect Bo is liberation

of gas.

GRAPH OF Bo

Above bubble point pressure, value of Bo increases due to decrease in pressure because of

expansion of gas present in oil.

Below bubble point pressure, value of Bo decreases due to decrease in pressure because of

liberation of gas from oil.

Bo is maximum at bubble point pressure.

When pressure is decreased from initial reservoir pressure to bubble point pressure, the formation

volume factor increases because of the expansion of liquid in the reservoir i.e. dissolved gas which is

present in the oil expands. A reduction in reservoir pressure below bubble point pressure results in the

liberation of gas in the pore spaces of a reservoir to form gas cap. The liquid remaining in the reservoir

has less dissolved gas and smaller FVF as reservoir oil volume decreases. BO = Vreservoir / Vsurface



Initially, when the pressure is above bubble point pressure then dissolved gas is contracted and volume

of oil will be less but as the pressure declines, volume of oil increases up to bubble point pressure.

When the pressure increases from bubble point pressure, dissolve gas evolve and form gas cap. Due to

decrease in pressure volume (of oil + gas) increases but volume of oil decreases. V2 is greater than V1 but

volume of oil decreases as free gas is not considered part of oil while dissolved gas is a part of oil.

Bo for volatile oil

Curve indicates that with small pressure decline, large amount of gas liberates out. Bubble point

pressure of volatile oil also comes early which can be seen from phase diagram.

Bo for black oil



Bo IDENTIFICATION

Bo 2.0 bbl/STB (Black oil)

Bo > 2.0 bbl/STB (Volatile oil)

It is not field identification because we dont know reservoir volume at field.

Rs FIELD IDENTIFICATION

Rs < 2000 SCF/STB (Black oil)

2000 SCF/STB < Rs < 3300 SCF/STB (Volatile oil)

API GRAVITY IDENTIFICAITION

Condensate oil (45-60)

Volatile oil (40-45)

Black oil (



Above bubble point pressure, when pressure declines viscosity decreases because gas expands

and molecules can easily flow

Below bubble point pressure, when pressure declines viscosity increases because gas liberates

so lighter component decreases in oil and due to heavier components viscosity increases.

Maximum viscosity is present at bubble point pressure.

The curve above is for undersaturated reservoir and for saturated reservoir curve is

Curves for volatile oil and black oil are shown below. Since, less gas is liberated with decrease in

pressure in case of black oil so, viscosity does not increase so much as compared to volatile oil.

IMPORTANCE OF VISCOSITY

Viscosity is the main designing factor, whether to do secondary recovery or tertiary recovery.

Flow rate depends upon pressure difference and viscosity. With increase in viscosity, flow rate

decreases. Below bubble point pressure, with decrease in pressure viscosity increases in tens of times so

designing point should be bubble point pressure where we get minimum viscosity.



If viscosity is very high then it requires high pressure to move the oil. If pressure required exceeds

fracture pressure then it fractures the rock. In this case, only feasible technique is thermal injection. The

feasible point for injection is at bubble point pressure where viscosity is minimum.

PROPERTIES OF OIL

1. Solution gas oil ratio

2. Formation volume factor of oil

3. Coefficient of viscosity

4. Bubble point pressure

5. Specific gravity of oil

SPECIFIC GRAVITY OF OIL:

Specific gravity of oil can be given as:

The reference fluid in this case is water

Documents

Reservoir Fluid