Reservoir Fluid

RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES

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 can’t be liquefied. Above critical temperature, no liquid phase and only gas phase is present and it can’t 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 can’t 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.

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RESERVOIR FLUID NOTES 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).

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES

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.

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RESERVOIR FLUID NOTES 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

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES 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:

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RESERVOIR FLUID NOTES 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 (CA’s) 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 -253 -235 -217 -199 -181

Density (lb/ft3) Saturated liquid Saturated Vapor 26.17 0.1443 25.25 0.2747 24.25 0.4766 23.18 0.7744 21.89 1.202

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Mean Density 13.15715 12.76235 12.3633 11.9772 11.546

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RESERVOIR FLUID NOTES

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 P 1 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 P 2, 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

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES 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 don’t 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:

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RESERVOIR FLUID NOTES  

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.

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RESERVOIR FLUID NOTES

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.

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RESERVOIR FLUID NOTES 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

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RESERVOIR FLUID NOTES 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

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RESERVOIR FLUID NOTES

Since, VMG >>> VML ------ (1) Since,

So, equation (1) becomes

∫

∫ ( )

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RESERVOIR FLUID NOTES 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) (In Fahrenheit)

Temperature (T) (In Rankine)

Vapor Pressure (Pv) (In psia)

1/T (R-1)

ln(Pv)

155.7 199.4 269.1 331.9 408.9 454.6

615.7 659.4 729.1 791.9 868.9 914.6

14.7 29.4 73.5 147 293.9 435

0.001624168 0.00151653 0.001371554 0.001262786 0.00115088 0.001093374

2.687847494 3.380994674 4.297285406 4.990432587 5.683239573 6.075346031

7 6 l 5 n 4 ( )

P 3 v 2 1 0 0.001

0.0011

0.0012

0.0013

0.0014

0.0015

0.0016

0.0017

1/T

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RESERVOIR FLUID NOTES Problem: Vapor pressure data of CO2 are given below, Calculate vapor pressure at 40F.

Temperature (Rankine) 420.9 458 482.5

Vapor pressure (Pv) 147 293.9 440.9

ln (Pv) 1/T 4.990433 0.002376 5.68324 0.002183 6.088818 0.002073

7 6

5 4 3

2 1 0 0.002

 

0.00205 0.0021 0.00215 0.0022 0.00225 0.0023 0.00235 0.0024

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

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES The composition diagram is shown below:

4

70% 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

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES 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:

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RESERVOIR FLUID NOTES 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

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RESERVOIR FLUID NOTES Solution: ( ) (

)

(

)

∑

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Page 28

RESERVOIR FLUID NOTES 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

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES  

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

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RESERVOIR FLUID NOTES

From figure 3.6, Z=0.39 ̃ ̃ (

)

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RESERVOIR FLUID NOTES 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 Kay’s 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 C1 C2 C3 n-C4

Mole fraction 0.850 0.090 0.040 0.020

Tc (0R) 343.3 549.9 666.2 765.6

Pc (psia) 666.4 706.5 616.0 550.0

yjTcj 291.81 49.49 26.65 15.31 383.26

yjPcj 566.44 63.59 24.64 11.00 665.67

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

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RESERVOIR FLUID NOTES

Ppc = 587 psia and Tpc = 4920R

Now using figure 3.7, Z=1.154

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RESERVOIR FLUID NOTES 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

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RESERVOIR FLUID NOTES

component H2S CO2 N2 C1 C2 C3 i-C4 n-C5 i-C6 n-C5 C6 C7+

mole fraction yj 0.0491 0.1101 0.0051 0.577 0.0722 0.0455 0.0096 0.0195 0.0078 0.0071 0.0145 0.0835

critical temperature deg R Tcj 672 548 235 343 550 651 740 764 833 845 910 1157

yj Tcj 33.0 60.3 1.2 198.1 39.7 29.6 7.1 14.9 6.5 6.0 13.2 96.6 506.2

critical pressure psia Pcj 1300.0 1071.0 493.1 666.4 706.5 616.0 527.9 550.6 490.4 488.6 436.9 367.0

yj Pcj 63.8 117.9 2.5 384.5 51.0 28.0 5.1 10.7 3.8 3.5 6.3 30.6 707.9

Tpc = 506.20R Ppc = 707.9 psia

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RESERVOIR FLUID NOTES 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?

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RESERVOIR FLUID NOTES 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:

(

)

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Page 38

RESERVOIR FLUID NOTES 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 Dalton’s 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.

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Page 39

RESERVOIR FLUID NOTES

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

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Page 40

RESERVOIR FLUID NOTES At surface condition PSVS = ZSRTS At surface condition, gas behaves ideally ZS = 1 PSVS = RTS VS = RTS / PS Put the values in (1)

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RESERVOIR FLUID NOTES Formation volume factor of gas ( PRESSURE 3000 2500 2000

)

VRESERVOIR 1 ft3 1 ft3 1 ft3

VSURFACE 100 SCF 80 SCF 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

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RESERVOIR FLUID NOTES 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

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RESERVOIR FLUID NOTES 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 can’t be formed

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Page 44

RESERVOIR FLUID NOTES 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) 700 800 900 1000 1100 1200

(

)

(

)

MOLAR VOLUME (ft3/lb-mol) 8.5 7.4 6.5 5.7 5.0 4.6

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RESERVOIR FLUID NOTES DERIVATIONS For ideal gas

For real gas

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RESERVOIR FLUID NOTES

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%.

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Page 47

RESERVOIR FLUID NOTES 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: 

Low Pressure High pressure

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

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RESERVOIR FLUID NOTES 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.

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RESERVOIR FLUID NOTES

Problem: Calculate the viscosity of gas mixture given below at 200oF and pressure of 1 atmosphere. Given Component Composition(yj) µgj Mj C1 0.85 0.013 16.04 C2 0.09 0.0112 30.07 C3 0.04 0.0098 44.1 n-C4 0.02 0.0091 58.12

Calculated √ 4.004997 5.483612 6.640783 7.623647

√ 3.404247 0.493525 0.265631 0.152473

√ 0.044255 0.005527 0.002603 0.001388

4.315877

0.053773

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RESERVOIR FLUID NOTES 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:

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RESERVOIR FLUID NOTES 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: ∑ ∑ 12- 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.

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RESERVOIR FLUID NOTES

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 o

Wet gas: that contains greater than 1.75 volume % of water. 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 Joule’s 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.

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RESERVOIR FLUID NOTES

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. Y j 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) CO2 N2 CH4 C2H6 C3H8 1-butane 2-butane i-butane n-pentane Hexane Heptanes+

0.0167 0.0032 0.7102 0.1574 0.0751 0.0089 0.0194 0.0034 0.0027 0.0027 0.0003

Lcj

Z

0 0 1010 1769.6 2516.1 3251.9 3262.3 4000.9 4008.9 4755.9 5502.5

0.9943 0.9997 0.998 0.919 0.9805 0.9711 0.9667 0.948 0.942 0.91 0.852

YjLcj

√(1-Zj)

Yj√(1-Zj)

0 0 717.302 278.535 188.9591 28.94191 63.28862 13.60306 10.82403 12.84093 1.65075

0.075498 0.017321 0.044721 0.284605 0.139642 0.17 0.182483 0.228035 0.240832 0.3 0.384708

0.001261 5.54E-05 0.031761 0.044797 0.010487 0.001513 0.00354 0.000775 0.00065 0.00081 0.000115

1315.945

(

)

0.095765

∑

Now for Lc(real): First calculating value of Z for mixture:

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RESERVOIR FLUID NOTES (∑

√

)

(

)

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.

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RESERVOIR FLUID NOTES

Heating value

Gross

Net

Wet

dry

wet

dry

Lc(dry)(1-ywater)+0.9

∑(YjLcj)/Z

Lc(1-ywater)

∑(YjLcj)/Z

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.

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RESERVOIR FLUID NOTES 

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



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).

and

   

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 Graham’s 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.

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RESERVOIR FLUID NOTES

Gas/air filled

Final water level

We can use Graham’s law of effusion for two gases as: √ √ √

( )

√

Putting the above values in equation (A); √ √ Since Vgas=Vair so, above equation becomes: √ √

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RESERVOIR FLUID NOTES

(

) (

)

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Page 59

RESERVOIR FLUID NOTES 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:

Rs

Pb

Pressure

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. P>Pb

P=Pb

oil

P 2.0 bbl/STB (Volatile oil) It is not field identification because we don’t 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 ( 1 at below bubble point.

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RESERVOIR FLUID NOTES COMMERCIAL VALUE OF PETROLEUM OIL Commercial value means that how much the extracted crude oil of particular composition is valuable in the market and how much revenue can be generated by it, following are the factors governing the commercial value,      

Specific gravity Gasoline and kerosene content (octane number) Sulphur content Asphalt content Cloud point Pour point SPECIFIC GRAVITY: The more the specific gravity of the crude the lesser will be API gravity and heavier will be crude more will be the value of the crude. Diesel and jet fuels have the least API gravity and hence have more commercial value. GASOLINE AND KERSOENE CONTENT: The more the contents of gasoline and kerosene heavier will be the crude oil and greater will be the crude value because they burn effectively and gasoline (C7 to C8) have best anti knocking ability means the fuel burns more efficiently and engine makes no noise, that is why we often try to crack higher hydrocarbons into compounds of C8 to increase their octane number. SULPHUR CONTENT: Presence of sulphur content is not desirable in the crude it causes multiple problems. It will form H2SO4 after reaction which is corrosive for engine plus when it will be released to atmosphere then it will cause acid rain which is hazardous for humans and agriculture. More over the formation of H2S can cause sudden death to humans. Sulphur contents also decreases the heating value of the crude. However the small amount of H2S (4 to 10 ppm) is bearable at maximum. ASPHALT CONTENT: Asphalts are usually very heavy hydrocarbons in ranges of 20s to 40s in terms of carbon atom. They are solid contents when separated in refineries is used to form coke and graphite. If there quantity is less then we sale it out cheaply. But if they are found enormously then they are converted in to lighter hydrocarbons through thermal cracking. CLOUD POINT: Crude oil is mixture of heavier and lighter hydrocarbons which are present at high temperature in subsurface, but when the temperature decreases they solidify. Cloud point of a fluid is the temperature at which dissolved solids (paraffin wax) are no longer completely soluble in fluid and start precipitating as second (solid) phase. Significance of cloud point is that the cloud point of the fluid should be very less so that if doesn’t appear in operating conditions. At cloud point liquid starts solidifying so liquid phase is still there to flow because not all the liquid solidifies.

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RESERVOIR FLUID NOTES POUR POINT: The point at which liquid completely converts into solid and liquid flow ceases is called POUR POINT. So pour point is lesser than cloud point but for pure substance cloud point and pour point are equal because all the liquid converts into solid at melting point line.

NOTE      

CNG – C1, C2 LPG – C3, C4 Gasoline – C7, C8 Kerosene – range of C12 Diesel – range of C16 Jet fuel – higher than C16

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RESERVOIR FLUID NOTES CONSTANT VOLUME DEPLETION TEST: This test is performed for retrograde reservoirs in place of differential vaporization. This test is basically differential vaporization at constant volume. We perform flash vaporization and separator test for both retrograde and oil fluid but for retrograde we perform constant volume depletion rather than differential vaporization. Before test is performed, we perform flash vaporization to find Pd. The test is performed to see how much liquid is condensed in the cell at each pressure drop below Pd such that whether reservoir can give us valuable amount of liquid condensate or not. This is necessary to find because in condensate reservoirs we apply pressure maintenance techniques such that gas recycling or hydraulic fracturing to maintain the pressure to increase the oil and gas recovery (otherwise if condensate is formed in reservoir cannot be produced until hydraulic fracturing and decrease the relative permeability of the gas also so it should condense in the wellbore or at surface), so we have to see if we are applying pressure maintenance technique then whether this reservoir can give us such amount of condensate that will help to repay our revenue and we will be in profit. PROCEDURE:      

The test is used for retrograde reservoirs. The sample of reservoir gas is brought to dew point pressure and reservoir temperature in the lab cell. Volume of cell at Pd is noted. Pressure is reduced in steps by increasing the volume. Remove the excess gas in each step and note he composition and other properties. At the end of each pressure step, volume of the cell is same as the volume at Pd.

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RESERVOIR FLUID NOTES

We continue to perform the experiment until maximum amount of liquid is formed and then we flash that liquid and find its properties.

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RESERVOIR FLUID NOTES RESERVOIR FLUIDS WHY WE NEED TO IDENTIFY THE FLUID TYPE: Fluid identification is the decision making factor in many of the decisions. The fluid type affects the following: 

METHOD OF SAMPLING:

The main thing that is to be kept in mind while collecting the sample is that when we take the sample, the fluid we have collected should be in the original form as it was in the reservoir means while collecting the sample of oil, the collecting pressure should be above bubble point pressure and for retrograde, it should be above Pd. There are two methods of fluid sampling:  SURFACE SAMPLING It is done if the fluid type is dry gas or wet gas and is less expensive. In this the fluid comes to the surface and the sample is collected on the surface.  Sub-surface sampling It is done if the fluid type is oil or retrograde and is more expensive, in this the sample is collected at he sub-surface. It is more expensive. Now for example, if we have assumed that the sub-surface fluid is the wet gas and we prepared to take the surface sample and the reservoir fluid came to be retrograde then at surface the pressure will drop below Pd and the original sample cannot be achieved so fluid identification is necessary before collecting the sample.  SIZE OF SURFACE EQUIPMENT AND COMPLETION EQUIPMENT:  If the reservoir fluid is condensate, then we know that we place maximum separators such that three.  Completion equipments also depend on fluid type. For example, in case of oil reservoir, we know that its pressure will drop early and we will need artificial lift methods such that we place side pocket mandrel for gas lift and more over tubing dia is also suggested on the basis of fluid type.  If we have two wells such that gas well and oil well both volumetric then in both cases the design of perforation interval will be different. In case of oil well, we know that after some time gas cap will form which will come to the top so we do not perforate at the top of reservoir but either at bottom or middle of the reservoir. While in case of gas well, we perforate the whole zone.  Separator design is also affected by fluid type such that for high GOR we use horizontal, for low GOR we use vertical and for moderate GOR we use spherical separator.

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RESERVOIR FLUID NOTES  In case of gas wells, we use large diameter tubing, but in case of oil wells, we use small diameter tubing as liquid hold up case can be there.  Mud type is also affected by the fluid type, in gas wells more care should be taken as gas can readily mix with water so we keep hydrostatic pressure far greater than formation pressure. If gas mixes with mud, the mud density decreases and kick may come but in case of oil well, not too much care is required.  Gases are more compressible and exist at high pressure, so we use BOP of high pressure rating as kick may readily comes. 

The calculation procedure for determining oil and gas in place: We know there are different formulae and methods to determine the oil and gas initial in place

so in place hydrocarbon calculations are also dependent on the fluid type. We use material balance for black oil and gases but no material balance for condensate and volatile oil. 

The selection of enhanced recovery

In EOR methods, not all methods are for all for example;  If fluid type is light then we use water injection or gas cap.  If fluid type is heavy then we use chemical injection.  If fluid type is very heavy then we use thermal injection. 

The plan of depletion It means how you drill and complete a well. As shown in figure, we do not drill at point A as gas cap will form but we try to drill at B. Gas wells have large drainage radius as compared to oil wells so in two fields of same sizes, one of gas field and the other of oil then in gas field the number of wells drilled be less (large spacing among wells due to large drainage radius) and vice versa for oil field.

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RESERVOIR FLUID NOTES The different types of reservoir fluids are: 

Black oil



Volatile oil

  

Retrograde Dry gas Wet gas

IDENTIFICATION OF FLUID TYPE The fluid type can be confirmed only by observation of the sample in the laboratory yet available production information on the field can also help to identify the fluid type. The production or field data are:   

Initial production GOR Stock tank oil gravity Color of stock tank liquid

Lab data involves:  

Initial formation volume factor Weight fraction of C7+ component

DIFFERENTIATION BETWEEN BLACK OIL AND VOLATILE OIL:          

BLACK OIL Phase envelope is extremely large such that temperature range is large. Contains much heavier components Isovals are equally spaced Much larger amount of liquid arises at the stock tank. Reservoir temperature is far below Tc

    

VOLATILE OIL Phase envelope is small as temperature range is small. Contains fewer heavier components Isovals are not equally spaced Relatively smaller amount of liquid reaches the stock tank. Tc is much lower than Tc f black oil and moreover reservoir temperature is near Tc

LAB ANALYSIS Bo2 Molecular weight of C7+ >=20%  Molecular weight of C7+ is between 12.5%-20% FIELD IDENTIFICATION Initial producing GOR is 2000 SCF/STB  Initial producing GOR is between 2000 and 3300 SCF/STB API is below 45  API is greater than 40 Stock tank liquid is dark colored  Stock tank liquid is light colored

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RESERVOIR FLUID NOTES

Explanation of above points: In reservoir fluid, there are three types of components:    











Lights (C1, C2) Intermediate (C3-C6) Heavy (C7+) In black oil, heavy components are larger in amount so the two phase region is larger and the temperature range is larger therefore phase diagram is widely spread while in volatile oil, intermediate and light components are greater in quantity so phase envelope is small as temperature range is small therefore we use large number of separators in case of volatile oil to increase the liquid recovery by making the most intermediate components, the part of liquid. If we take production from the two reservoirs at same temperature and pressure, then due to large heavy components in black oil more liquid recovery occurs but incase of volatile oil less liquid recovery as heavy components are small in amount. Black oil is low shrinkage crude oil as it has less gas dissolved in it and reservoir temperature is far below Tc but volatile oil is high shrinkage crude oil as it has large gas dissolved in it and reservoir temperature is near to Tc. The API value of volatile oil is above 40 and decreases dominantly below Pb as lighter components are rapidly removed in reservoir on small pressure reduction below Pb so density of oil increases and API decreases The API of volatile oil also decreases below Pb but decrease is not dominant. The isovals are near to each other and near to bubble point line in case of volatile oil so small pressure dropped below Pb causes larger gas evolution from oil bt in case of black oil, the isovals are moderately spaced and heavy components are greater in amount (which are less energetic) so large pressure drop is required below Pb to evolve the same amount of gas as evolved from volatile oil on small pressure drop. At abandonment pressure, more liquid remained in black oil as compared to volatile so for EOR black oil is more suitable for black oil reservoir.

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RESERVOIR FLUID NOTES  

We should not rely on single property for differentiation but all properties should be checked. Liquid shrinkage curves:

The vertical line indicates the bubble point pressure where liquid is 100%. As pressure is decreased below Pb, gas evolves with equal amounts as isovals are equally spaced so almost a straight line is formed showing constant decrease in the liquid volume. In case of volatile oil as pressure is dropped below Pb then initially with small pressure dropped large gas is produced as isovals are not equally spaced and concentrated near Pb so a decreasing curve will form showing large decrease in liquid volume which will not be a constant decrease due to irregular amount of gas liberation. Then after some time, the decrease in liquid volume becomes small. Some points: Secondary recovery is the pressure maintenance recovery In tertiary recovery, we change the fluid properties (EOR is tertiary recovery) In primary recovery, operations are done on existing wells to produce oil, fluid comes from reservoir to well bore. Problem: The average GOR produced from a field is 275 SCF/STB. The gravity of produced oil is 26 API. The color of the stock tank liquid is black. What type of reservoir fluid is in formation? Answer: Black oil

Problem:

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RESERVOIR FLUID NOTES A field had an initial producing GOR of 2000 SCF/STB, the stock tank liquid was medium orange and had gravity 51.2 API. Classify this reservoir fluid? Answer: Volatile oil Problem: The field yielded an absolute open flow of 76 MMMSCF/day with 60 bbls of stock tank liquid per MMSCF of gas. Classify the fluid type, what other information you would like to have to confirm your result. Answer:

Since GOR is above 3300 so it is gas reservoir. Now gas reservoirs are dry gas, wet gas or retrograde. This is not a dry gas as GOR is given. Now for retrograde GOR is between 3300 and 50000 and wet gas it is greater than 50000. The fluid type may be a wet gas but other properties like color and API are also required to confirm the result.   

 

WET GAS Wet gas exists solely as a gas in reservoir throughout the reservoir life. Thus no liquid is formed in the reservoir. Separator condition lies within the phase envelope, causing some liquid to be formed at surface due to presence of intermediate components. Producing GOR is above 50000 SCF/STB and will remain constant. Liquid is above 60 API and is water white.

  

DRY GAS Dry gas is primarily methane. Hydrocarbon is solely gas in the reservoir and in the separator condition. Dry means gas does not contain enough of the heavy components to form liquid at surface.

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RESERVOIR FLUID NOTES

Some points: Flash calculation is a technique by which we can make phase diagram of a fluid such that wet gas or dry gas or any oil type mathematically. In dry gas we also use separator to remove water is produce. Dehydration plant separates the water that is part of gas (saturated water) but aquifer water is separated by separator. Three phase separator is used to separate all oil, water and gas. DIFFERENCE BETWEEN VOLATILE OIL AND RETROGRADE GAS CONDENSATE:    

VOLATILE OIL Initial oil in reservoir TR3300 Molecular wt of C7+ is less 12.5%

RETROGRADE GAS: 

 

The phase diagram of retrograde gas is some what like as shown in the figure (smaller than that of oils). It contains fewer of the heavy components. Reservoir temperature is less than cricondentherm but greater than critical temperature.

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RESERVOIR FLUID NOTES Field identification:  



Producing GOR is between 3300 and 150,000. Producing GOR will increase when pressure falls below dew point as liquid start to condense in the reservoir such that heavy components remained in the reservoir and lighter components come to the surface forming gas only with small liquid so GOR will increase. API gravity is between 40 and 60 and increases as pressure falls below Pd as specific gravity of the oil or condensate coming to the surface will decrease because the heavy components had already been condensed in the reservoir so comparatively lighter components will now be the part of liquid at surface.

Temperature range in the phase diagram is smaller as compared to volatile and black oils and critical temperature is also less than both of these (the larger the heavy components in the fluid the greater the temperature range and larger the critical temperature and more widely spread is the phase envelope. Molecular wt of C7+ is less than 12.5% for retrograde gas, Rich retrograde has near to 12.5% and lean has near to 1%. Therefore in lean, we do not appreciate any pressure maintenance technique as small liquid will drop out in reservoir that will cause no significant problem but in rich as pay back (means in how much time the investment is recovered) is good so we appreciate pressure maintenance technique. (

)

For retrograde Rs= 3300 to 50000 but generally we take range of 3300 to 150,000. If Rs is near to 150000 then it is lean more near to wet gas (in wet gas no liquid drop out in reservoir but in retrograde whether lean or rich, liquid drops out in reservoir during any period of reservoir life. In any field breakeven point may come when expenses are equal to reserves.

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RESERVOIR FLUID NOTES COMPONENTS OF NATURALLY OCCURRING PETROLEUM FLUIDS HYDROCARBONS: 1- Paraffins (Alkanes): Paraffins are straight chain saturated hydrocarbons. The chain may be straight or branched. Valency of carbon atom is filled by 4 atoms. 2- Olefins: Olefins are unsaturated hydrocarbons such as Alkenes and Alkynes. 3- Naphthenes (cyclo-alkanes): They are saturated hydrocarbons but present as ring structure. Aromatics: They are cyclic compounds that contain the benzene ring. 4- Resins and Asphaltenes: They are hydrocarbons with sulfur, oxygen or nitrogen atom. Note: Learn IUPAC names for exams. NON HYDROCARBONS: Nitrogen, Carbondioxide, Helium, Sulphur, etc.

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RESERVOIR FLUID NOTES In refinery, the crude is first heated in more than one furnace so that all components in crude are vaporized such that we heat the crude in furnaces to such temperature that all crude becomes vapor. The heating is not done in one furnace because we do heating in steps so that thermal cracking may not have occurred otherwise the coke being produced can block the passage towards CDU. If the crude contains heavy fractions then we heat them in vacuum so that boiling temperature lowers due to lowering of the surrounding pressures and all heavy components could be vaporized in furnace. When crude has been heated to desire temperature and it becomes in vapor phase then it is injected into the Crude Distillation Unit (CDU) from some distance above the bottom of the unit. In this unit the separation of different components in the crude oil occurs on the basis of difference in their boiling points. When crude in vapor phase enters into the unit then at first the heaviest components are condensed and settle at bottom which we call as residuum, the remaining mixture in vapor phase tends to flow up towards the low temperature region. The trays are set in unit at different temperatures where fractions of the crude oil after being condensed from vapor to liquid phase at their melting points are achieved at their respective trays. For o

example among the various trays in the unit, one tray is set at 450 oC and the other is set at 350 C. At tray of 450, the kerosene components will be condensing mostly that has boiling point range of 350-450 and at try of 350 the components of B.P. range of 100-350 will achieved, however impurities will be there as during separation from mixture other components beside the kerosene components may also condense at this tray, so for further purification we again heat the mixture and re-inject it again into the CDU, the process is continued for many times then the components from each tray or the two are carried to separate refinery units where temperature difference between the successive trays will be much smaller so that more pure substances can be achieved. If heavy components are in large amount in crude oil that are separated as residuum at bottom of CDU, then we will use VDU (vacuum distillation unit), which will be operating at vacuum so that boiling temperatures of the components could be lowered and maximum separation of relatively light components in the residuum can be achieved.

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RESERVOIR FLUID NOTES In order to decide, which components should be more is dependent on demand and heating value for example if demand of gasoline is more then we can use techniques to convert heavy hydrocarbons into gasoline.

CNG is C1 and C2 compressed under high pressure. LPG is C3 and C4 that remains liquid under ordinary conditions. LNG is liquefied natural gas such that C1 and C2 are compressed at high pressure and low temperature. This is because the transportation through pipelines could be easy. Homologous Series: A family of organic chemical is known as homologous series. Members of homologous series have similar molecular structure, same chemical properties, physical properties are different.

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RESERVOIR FLUID NOTES VAPOR LIQUID EQUILIBRIUM The area bounded by the Pb and Pd curves on the phase diagram of a multicomponent mixtures define the condition for the gas and liquid to exist in equilibrium. Quantities and composition of the two phases vary at different points. The purpose of flash calculation (vapor liquid equilibria) is: 1) Calculation of bubble point of the mixture 2) Calculation of dew point of the mixture 3) Calculation of quantities and composition of the gas and liquid at different points.

In order to make phase diagram experimentally, it takes months and years to be completed but we need a phase diagram at initial phases so this technique is used (vapor liquid equilibria or flash calculation). For flash calculation, the initial composition of liquid mixture should be known (by gas chromatograph or anything else). Some symbols to remember: Symbol

( (

( ) ) )

meaning Total number of moles in the mixture Total number of moles in liquid Total number of moles in gas Represents the mole fraction of jth component in total mixture Represents the mole fraction of jth component in liquid Represents the mole fraction of jth component in gas Total moles of the jth component in the total mixture Moles of the jth component of liquid Moles of the jth component of gas

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RESERVOIR FLUID NOTES

We knew initially Z3 and Z4 and ‘n’ only, now we have to find and pressures to draw the phase diagram.

at different temperatures

A material balance on the jth component results in: ( )

we have

To eliminate

, By comparing Dalton’s law and Roult’s law.

So, ( ) ( )

The calculation is simplified if n=1 therefore:

(

)

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RESERVOIR FLUID NOTES

(

)

Phase diagram depends on the composition of liquid initially as we had seen in previous chapters that phase diagram for C1 50% and C2 50% is different from C1 70% and C2 30% therefore we are taking in this constant composition of mixture means the liquid mixture that we had initially will have same composition and on this mixture we will see the effect of change in temperature and pressure. During production or injection, the fluid properties in the reservoir never remain same so the phase diagram changes at every period of production (question: if phase diagram changes at each step of production then the bubble point will also change then how we will predict that what will be the bubble point as bubble point calculated from sample will be at that particular composition which will change). In order to calculate ∑

∑

(

and

, we need to know

and

for which we have:

)

between 0 and 1 as ‘n’ is taken as 1, if answer of summation comes less than 1 then we will put another value for which sum should be greater than one, then we will interpolate among the two values to get required value that will give us sum as 1. For we have and then we can find and . In above equation, we will assume one value of

Problem: Calculate the composition and quantities of the gas and liquid when 1 lb-mole of the following mixture is brought to be in equilibrium at 150 oF and 200 psia. Assume ideal solution. Component C3 n-C4 n-C5

Composition 0.61 0.28 0.11

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350 105 37

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RESERVOIR FLUID NOTES Solution: First finding ∑

∑

assume it to be 0.2 so, (

) (

Now assume ∑

∑

)(

)

(

)(

)

(

)(

)

)(

)

(

)(

)

(

)(

)

to be 0.6 so (

) (

Interpolating between the two values we get:

(

)

(

)

(

)

(

)

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RESERVOIR FLUID NOTES

(

)

(

)

(

)

(

)

Problem: Perform flash calculation at 50 psia and 100oF Component C3 i-C4 n-C4 i-C5 n-C5 C6

First finding ∑

∑

at 100oF 190 72.3 51.6 20.44 15.57 4.956

Zj 0.2 0.1 0.1 0.2 0.2 0.2 assume it to be 0.2 so,

(

) ( (

)( )(

)

)(

( )

(

)(

)

)(

( )

(

) )(

)

∑

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RESERVOIR FLUID NOTES Now assume ∑

∑

to be 0.95 so

(

) ( (

)( )(

)

)(

( )

(

)(

)

)(

( )

(

) )(

)

∑ Interpolating between the two values we get:

(

)

(

)

(

)

(

)

(

)

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RESERVOIR FLUID NOTES

(

)

(

)

(

)

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