Upload
pundarik-kashyap
View
87
Download
8
Tags:
Embed Size (px)
DESCRIPTION
v
Citation preview
CONTENTS
1 IDEAL GASES
1.1 Boyle's Law
1.2 Charles' Law
1.3 Avogadro's Law
1.4 The Equation of State For an Ideal Gas
1.5 The Density of an Ideal Gas
1.6 Standard Conditions
1.7 Mixtures of Ideal Gases
1.7.1 Dalton's Law of Partial Pressures
1.7.2 Amagat's Law
1.8 Apparent Molecular Weight
1.9 Specific Gravity of a Gas
2 BEHAVIOUR OF REAL GASES
2.1 Compressibility Factor For Natural Gases
2.2 Law of Corresponding States
2.3 Pseudocritical Properties of Natural Gases
2.4 Impact of Nonhydrocarbon Components on
z Value
2.5 Standard Conditions For Real Reservoir
Gases
3 GAS FORMATION VOLUME FACTOR
4 COEFFICIENT OF ISOTHERMAL
COMPRESSIBILITY OF GASES
5 VISCOSITY OF GASES
5.1 Viscosity
5.2 Viscosity of Mixtures
6 EQUATIONS OF STATE
6.1 Other Equations-of-State
6.2 Van de Waals Equation
6.3 Benedict - Webb - Rubin Equation (BWR)
6.4 Redlich - Kwong Equation
6.5 Soave, Redlich Kwong Equation
6.6 Peng Robinson Equation of State
6.7 Application to Mixtures
Behaviour of Gases
2
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Present the ideal equation of state, PV=nRT.
• Calculate the mass of an ideal gas given PV 7T values.
• Derive an equation to calculate the density of an ideal gas.
• Convert a mixture composition between weight and mole fraction.
• Present an equation and calculate the apparent molecular weight of a mixture.
• Define and calculate the specific gravity of a gas.
• Present the equation of state, EOS, for a ʻreal gas ̓and explain what ʻZ ̓ is,
PV=ZnRT.
• Define the pseudocritical pressure and psuedocritical temperature and be able
to use them to determine the ʻZ ̓value for a gas mixture.
• Express and calculate reservoir gas volumes in terms of standard cubic
volumes.
• Define the gas formation volume factor and derive an equation fore it using the
EOS.
• Calculate the volume of gas in a reservoir in terms of standard cubic volumes
given prerequisite data.
• Calculate the viscosity of a gas of a specific composition given perquisite
equations and figures.
• Be aware of the development of EOSʼs to predict reservoir fluid properties.
Institute of Petroleum Engineering, Heriot-Watt University 3
INTRODUCTION
A gas is a homogenous fluid that has no definite volume but fills completely the vessel
in which it is placed. The system behaviour of gases is vital to petroleum engineers
and the laws governing their behaviour should be understood. For simple gases these
laws are straightforward but the behaviour of actual hydrocarbon gases particularly
at the conditions occurring in the reservoir are more complicated.
We will review the laws that relate to the pressure, volume and temperatures of gases
and the associated equations. These relationships were previously termed gas laws;
it is now more common to describe them as equations of state.
1 IDEAL GASES
The laws relating to gases are straightforward in that the relationships of pressure,
temperature and pressure are covered by one equation. First consider an ideal gas.
An ideal gas is one where the following assumptions hold:
• Volume of the molecules i.e. insignificant with respect to the total volume of
the gas.
• There are no attractive or repulsive forces between molecules or between
molecules and container walls.
• There is no internal energy loss when molecules collide.
Out of these assumptions come the following equations.
1.1 Boyleʼs LawAt constant temperature the pressure of a given weight of a gas is inversely proportional
to the volume of a gas.
i.e.
V
1
P or PV = constant, T is constant
(1)
P = pressure, V = volume, T = temperature.
1.2 Charles ̓LawAt constant pressure, the volume of a given weight of gas varies directly with the
temperature:
i.e.
V T or
V
T = constant, P is constant
(2)
The pressure and temperature in both laws are in absolute units.
Behaviour of Gases
4
1.3 Avogadroʼs LawAvogadroʼs Law can be stated as: under the same conditions of temperature and
pressure equal volumes of all ideal gases contain the same number of molecules. That
is, one molecular weight of any ideal gas occupies the same volume as the molecular
weight of another ideal gas at a given temperature and pressure.
Specifically, these are:
(i) 2.73 x 1026 molecules/lb mole of ideal gas.
(ii) One molecular weight (in lbs) of any ideal gas at 60˚F and 14.7 psia
occupies a volume of 379.4 cu ft.
One mole of a material is a quantity of that material whose mass in the unit system
selected is numerically equal to the molecular weight.
eg. one lb mole of methane CH4 = 16 lb
one kg mole of methane CH4 = 16kg
1.4 The Equation of State for an Ideal GasBy combining the above laws an equation of state relating pressure, temperature and
volume of a gas is obtained.
PV
T constant
(3)
R is the constant when the quantity of gas is equal to one mole.
It is termed the Universal Gas Constant and has different values depending on the
unit system used, so that;
R in oilfield units = 10 732.
cu ft psia
lb mole R
Table 1 gives the values for different unit systems.
P V T n R � �
�
psia �� cu ft �� R �� lb - mole � 10.73 ��
atm �� cu ft �� K �� lb - mole � 1.3145 �
atm �� cc �� K �� gm - mole � 82.06 �
atm �� litre �� K �� gm - mole � 0.08206 �
atm �� cu ft �� R �� lb - mole � 0.730 �
mm Hg � litre �� K �� gm - mole � 62.37 �
in.Hg �� cu ft �� R �� lb - mole � 21.85 �
�
Table 1 Values of R for different unit systems
Institute of Petroleum Engineering, Heriot-Watt University 5
For n moles the equation becomes:
PV = nRT (4)
T= absolute temperature oK or oR where
ºK=273 +oC and oR=460 +oF
To find the volume occupied by a quantity of gas when the conditions of temperature
and pressure are changed from state 1 to state 2 we note that:
n PV
RT is a constant so that
P V
T =
P V
T
1 1
1
2 2
2
EXERCISE 1.
A gas cylinder contains methane at 1000 psia and 70°F. If the cylinder has a vol-ume of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in
the cylinder.
1.5 The Density of an Ideal GasSince density is defined as the weight per unit volume, the ideal gas law can be used
to calculate densities.
g = weight / volume =
m
V
where g is the gas density
For 1 mole m = MW MW = Molecular weight
V RT
P
= MW.P
RTg
(5)
EXERCISE 2.
Calculate the density of the gas in the cylinder in exercise 1.
Behaviour of Gases
6
1.6 Standard ConditionsOil and gas at reservoir conditions clearly occur under a whole range of temperatures
and pressures.
It is common practice to relate volumes to conditions at surface, ie 14.7 psia and
60˚F.
ie
P V
T
P V
T
res res
res
sc sc
sc (6)
sc - standard conditions res - reservoir conditions
This relationship assumes that reservoir properties behave as ideal. This is NOT the
case as will be discussed later.
EXERCISE 3.
Assuming methane is at the conditions of exercise 1, calculate the volume the gas would occupy at standard conditions.
1.7 Mixtures of Ideal GasesPetroleum engineering is concerned not with single component gases but mixtures
of a number of gases.
Laws established over early years governing ideal gas mixtures include Daltonʼs
Law and Amagatʼs Law.
1.7.1 Daltonʼs Law of Partial PressuresThe total pressure exerted by a mixture of gases is equal to the sum of the pressures
exerted by its components. The partial pressure is the contribution to pressure of
the individual component.
Consider a gas made up of components A, B, C etc
The total pressure of the system is the sum of the partial pressures
ie
P = P + P + P + .....A B C (7)
where A, B and C are components.
therefore
Institute of Petroleum Engineering, Heriot-Watt University 7
P = n RT
V n
RT
V n
RT
V
i.e. P = RT
Vn
P
P =
n
n = y
A B C
j
j j
j
(8)
where yj = mole fraction of jth component.
The pressure contribution of a component, its partial pressure, is the total pressure
times the mole fraction.
1.7.2 Amagatʼs LawAmagatʼs Law states that the volume occupied by an ideal gas mixture is equal to the
sum of the volumes that the pure components would occupy at the same temperature
and pressure. Sometimes called the law of additive volumes.
i.e.
V = V + V + VA B C (9)
V = n RT
P + n
RT
P + n
RT
P
V = RT
Pn
V
V =
n
n = y
A B C
j
j j
ji e. . (10)
i.e, for an ideal gas the volume fraction is equal to the mole fraction.
It is conventional to describe the compositions of hydrocarbon fluids in mole terms.
This is because of the above laws. In some circumstances however weight compositions
might be used as the basis and it is straight forward to convert between the two.
EXERCISE 4.
A gas is made up of the following components; 25lb of methane, 3 lb of ethane and 1.5 lb of propane. Express the composition of the gas in weight and mole fractions.
Behaviour of Gases
8
1.8 Apparent Molecular WeightA mixture does not have a molecular weight although it behaves as though it had a
molecular weight. This is called the apparent molecular weight. AMW
If yj represents the mole fraction of the jth component:
AMW = y MWj j
AMW for air = 28.97, a value of 29.0 is usually sufficiently accurate.
EXERCISE 5.
What is the apparent molecular weight of the gas in exercise 4
1.9 Specific Gravity of a GasThe specific gravity of a gas,
g is the ratio of the density of the gas relative to that of
dry air at the same conditions.
g
g
air
=
(11)
Assuming that the gases and air are ideal.
g
g
air
g
air
g =
M P
RTM P
RT
= M
M =
M
29
Mg = AMW of mixture, Mair = AMW of air.
EXERCISE 6.
What is the gas gravity of the gas in exercise 4 ?
2 BEHAVIOUR OF REAL GASES
The equations so far listed apply basically to ideal systems. In reality, however,
particularly at high pressures and low temperatures the volume of the molecules are
no longer negligible and attractive forces on the molecules are significant.
Institute of Petroleum Engineering, Heriot-Watt University 9
The ideal gas law, therefore, is not too applicable to light hydrocarbons and their
associated fluids and it is necessary to use a more refined equation.
There are two general methods of correcting the ideal gas law equation:
(1) By using a correction factor in the equation PV = nRT
(2) By using another equation-of-state
2.1 Compressibility Factor for Natural GasesThe correction factor ʻz ̓which is a function of the gas composition, pressure and
temperature is used to modify the ideal gas law to:
PV = znRT (12)
where the factor ̒ z ̓is known as the compressibility factor and the equation is known
as the compressibility equation-of-state or the compressibility equation.
The compressibility factor is not a constant but varies with changes in gas composition,
temperature and pressure and must be determined experimentally (Figure 1).
To compare two states the law now takes the form:
P V
z T =
P V
z T
1 1
1 1
2 2
2 2 (13)
z is an expression of the actual volume to what the ideal volume would be.
i.e.
z
Vactual
Videal
(14)
Tem
pera
ture
= c
onst
ant
00
0.5
1.0
PRESSURE, P
Co
mp
ressib
ility
fa
cto
r, Z
Figure 1 Typical plot of the compressibility factor as a function of pressure at constant
temperature.
Behaviour of Gases
10
Although all gases have similar shapes with respect to z the actual values are component
specific. However through the law of corresponding states all pure gases are shown
to have common values.
2.2 Law of Corresponding StatesThe law of corresponding states shows that the properties of many pure liquids and
gases have the same value at the same reduced temperature (Tr) and pressure (P
r)
where:
T = T
T and P =
P
Pr
c
r
c (15)
Where, Tc and P
c are the pure component critical temperature and pressure.
The compressibility factor ʻz ̓follows this law. It is usually presented vs Tr and Pr.
Although in many cases pure gases follow the Law of Corresponding States, the gases
associated with hydrocarbon reservoirs do not. The Law has however been used to
apply to mixtures by defining parameters called pseudo critical temperature and
pseudocritical pressure .
For mixtures a pseudocritical temperature and pressure, Tpc
and Ppc
is used such
that:
T = y T and P = y Ppc j cj pc j cj (16)
where y is the mole fraction of component j and Tcj and P
cj are the critical temperature
and pressure of component j.
It should be emphasised that these pseudo critical temperature and pseudocritical
pressures are not the same as the real critical temperature and pressure. By definition
the pseudo values must lie between the extreme critical values of the pure components
whereas the actual critical values for mixtures can be outside these limits, as was
observed in the Phase Behaviour chapter.
EXERCISE 7.
Calculate the pseudo critical temperature and pseudocritical pressure of the mixture in exercise 4 .
For mixtures the compressibility factor (z) has been generated with respect to natural
gases 1, where ʻz ̓ is plotted as a function of pseudo reduced temperature, Tpr and
pseudo reduced pressure Ppr where
Institute of Petroleum Engineering, Heriot-Watt University 11
0
1.0
1.1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.25
1.1
1.0
0.9
1.01.05
1.05
1.1
1.2
1.3
1.4
1.5
1.6 1.7
1.8 1.9
2.0 2.2
2.4
2.6
3.0
3.02.8
1.21.3
1.1
1.10.95
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1.0
0.9
1 2 3 4 5 6 7 8
7 8 9 10 11 12 13 14 15
Compressibility of
Natural Gases
(Jan. 1, 1941)
Compressibility Factors for Natural Gases as a
Function of Pseudoreduced Pressure and Temperature.
Co
mp
ressib
ility
Fa
cto
r, z
Pseudo Reduced Temperature
Pseudo Reduced Pressure, Pr
3.02.82.62.42.2
2.01.91.8
1.7
1.6
1.5
1.45
1.35
1.4
1.3
1.25
1.2
1.15
1.1
2.6 2.4
2.22.0 1.9
1.71.6 1.4
1.3
1.2
1.1
1.05
1.05
1.8
1.4
1.5
Pseudo Reduced Pressure, Pr
Figure 2 Compressibility factors for natural gas1 (Standing & Katz, Trans AIME, 1942)
Behaviour of Gases
12
TT
Tand
P
Ppr
pc pc
Ppr
(17)
The use of this chart , figure 2 ,has become common practise to generate z values for
natural gases. Poettmann and Carpenter 2 have also converted the chart to a table.
Various equations have also been generated based on the tables.
EXERCISE 8.
For the gas of exercise 4 determine the compressibility factor at a temperature of 150°F and a pressure of 3500psia.
2.3 Pseudocritical Properties of Natural GasesThe pseudocritical properties of gases can be computed from the basic composition
but can also be estimated from the gas gravity using the correlation presented in
Figure 3.
0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2
Pseudocritical Properties of Natural Gases
Pseudocritical Tem
pera
ture
, R
Pseudocritical P
ressure
, psia
Gas Gravity (air = 1)
700
650
600
550
500
450
400
350
300
Condensate Well Fluids
Miscellaneous Gases
Miscellaneous G
ases
Condensate Well F
luids
Figure 3 Pseudocritical properties of natural gases 3
Institute of Petroleum Engineering, Heriot-Watt University 13
2.4 Impact of Nonhydrocarbon Components on z value.Components like hydrogen sulphide, and carbon dioxide have a significant impact
on the value of z. If the method previously applied is used large errors in z result.
Wichert and Aziz 4 have produced an equation which enables the impact of these
two gases to be calculated.
T'pc
= Tpc
- (18)
and
pp T
T y ypc
pc pc
pc H S H S2 21
(19)
T'pc
and p'pc
are used to calculate Tpr and p
pr. The value for is obtained from
the figure 4 from the Wichert and Aziz paper
0 10 20 30 40 50 60 70 800
10
20
30
40
50
60
70
80
PER CENT H2S
PE
R C
EN
T C
02
510
15
20
25
30
15
20
25
30
E
34.5
Figure 4 Adjustment factors for pseudocritiacl properties for non hydrocarbon
gases(Wichert & Aziz)
Behaviour of Gases
14
EXERCISE 9.
Calculate the pseudo critical properties of the gas in exercise 4 if it also contained 3
lb of hydrogen sulphide, 10lb of carbon dioxide and 2.5lb of nitrogen
1
2
3
Gas
Components
Mol
weight
Mole
fraction
Pc-psi Tc °R Ppcpsia
Methane 25 0.56 16.04 0.035 0.743 667.00 344 495.8 255.70
Ethane 3 0.07 30.07 0.002 0.048 708.00 550 33.7 26.17
Propane 1.5 0.03 44.09 0.001 0.016 616.00 666 10.0 10.81
Hydrogen 3 0.07 34.08 0.002 0.042 1306 673 54.8 28.25
sulphide
Carbon 10 0.22 44.01 0.005 0.108 1071 548 116.1 59.38
Dioxide
Nitrigen 2.5 0.06 28.02 0.002 0.043 493 227 21.0 9.66
Total 45 1.00 0.0466 1.000 731 390
TpcWeight Wgt
fraction
lb moles
4
5
6
From Wichert & Azis chart for compositions of H2S and CO
2 = 19
T = T - = 371 R
P = 694.3
pc pc
o
pc
pp T
T y ypc
pc pc
pc H S H S2 21
2.5 Standard Conditions for Real Reservoir GasesAs indicated in section 1.6 for ideal gases it is convenient to describe the quantity of
gas to a common basis and this is termed the standard conditions, giving rise to the
standard cubic foot and the standard cubic metre. The petroleum engineer is primarily
interested in volume calculations for gaseous mixtures. Throughout the industry gas
volumes are measured at a standard temperature of 60˚F (15.6˚C) and at a pressure of
14.7 psia (one atmosphere). These conditions are referred to as standard temperature
and pressure STP. Standard Cubic Feet, the unit of volume measured under these
conditions is sometimes abbreviated SCF or scf (SCM is Standard Cubic Metres). It
is helpful to consider these expressions not as volumes but as an alternate expression
of the quantity of material. For example a mass of gas can be expressed as so many
standard cubic feet or metres.
EXERCISE 10.
Express the quantity of 1 lb mole of a gas as standard cubic feet.
Institute of Petroleum Engineering, Heriot-Watt University 15
EXERCISE 11.
Express the mass of gas in exercise 4 as standard cubic feet.
3 GAS FORMATION VOLUME FACTOR
The petroleum industry expresses its reservoir quantities at a common basis of surface
conditions which for gases is standard cubic volumes. To convert reservoir volumes
to surface volumes the industry uses formation volume factors. For gases we have
Bg, the gas formation volume factor, which is the ratio of the volume occupied at
reservoir temperature and pressure by a certain weight of gas to the volume occupied
by the same weight of gas at standard conditions. The shape of Bg as a function of
pressure is shown in figure 5.
Bvolume occupied at reservoir temperature and pressure
volume occupied at STPg
The gas formation volume factor can be obtained from PVT measurements on a gas
sample or it may be calculated from the equations-of-state discussed previously.
One definition of the gas formation volume factor is: it is the volume in barrels
that one standard cubic foot of gas will occupy as free gas in the reservoir at the
prevailing reservoir pressure and temperature.
Depending on the definition the units will change and the units will be; rb free
gas/scf gas or rm3 free gas/scm gas
.008
.006
.004
.002
1000 2000 3000
Bg
rb/scf
PRESSURE (psig)
Figure 5 Gas Formation Volume Factor, Bg
Behaviour of Gases
16
For example Bg for a reservoir at condition 2 is;
BV
V
P T z
P T zg
2
sc
sc 2 2
2 sc sc (20)
ʻsc ̓refers to standard conditions. z at standard conditions is taken as 1.0
The reciprocal of Bg is often used to calculate volumes at surface so as to reduce the
possibility of misplacing the decimal point associated with the values of Bg being
less than 0.01, ie:
volume at surface
volume in formation Bg
1E
E is sometimes referred to as the expansion factor.
Usually the units of Bg are barrels of gas at reservoir conditions per standard cubic
foot of gas, ie bbl/SCF or cubic metres per standard cubic metre.
BV
Vg
R
sc (21)
R and sc are reservoir and standard conditions respectively.
V
znRT
PR
(22)
T and P at reservoir conditions:
Vz nRT
Psc
sc sc
sc (23)
z = 1 for standard conditions
B zT
T
P
P
cu. ft
SCFg
sc
sc. .
(24)
Since Tsc = 520˚Rm Psc = 14.7 psia for most cases
B 0
zT
P
cu. ft
SCFg .0283
B 0 zT
P
cu. ft
SCF
bbl
5.615 cu ftg .0283
or
Institute of Petroleum Engineering, Heriot-Watt University 17
B 0
zT
P
res bbl
SCFg .00504
(25)
EXERCISE 12.
Calculate the gas formation factor for a gas with the composition of exercise 4 existing at the reservoir conditions given in exercise 8.
EXERCISE 13.
A reservoir exists at a temperature of 150°F (as for exercise 8) suitable for storing gas. It has an areal size of 5 miles by 2 miles and is 200ft thick. The average porosity is 20% and there is no water present. How much gas of the composition of exercise 4 can be stored at a pressure the same as in exercise 8 i.e. 3500 psia ? (1 mile= 5280 ft.)
4 COEFFICIENT OF ISOTHERMAL COMPRESSIBILITY OF GAS-
ES
The compressibility factor, z, must not be confused with the compressibility which is
defined as the change in volume per unit volume for a unit change in pressure, or
cV
V
P or
V
V
Pg
m
m1 1
(26)
Vm is the specific volume or volume per mole.
cg is not the same as z, the compressibility factor.
For an ideal gas:
PV = nRT or:
dV
dP
nRT
P
c = 1
V
nRT
P=
1
P
2
g 2
(27)
For real gases:
V =
znRT
P
Behaviour of Gases
18
V
PnRT
Pdz
dP
P
cP
nRTz
nRT
PP
z
Pz
cP
1
z
z
P
T
2
g 2
g
z
1.
(28)
dz/dP can be obtained from the slope of the z vs P curve.
The Law of Corresponding states can be used to express the above equation in
another form
P = P P
z
P
P
P
z
P
z
P
z
P
pc pr
pr
pr
pr
P
P P
P
pr
pc
pc
1
1
Combining this equation with eqn 28 above yields
cP P
1
zP
z
P
c PP
1
z
z
P
g
pc pr pc pr
g pc
pr pr
1
1
T
T
pr
pr (29)
Units of cg = P-1, and c
gP
c is dimensionless
cpP
pc is called pseudo reduced compressibility, c
pr
Institute of Petroleum Engineering, Heriot-Watt University 19
Since the pseudo reduced compressibility is a function of ʻz ̓and pseudo reduced
pressure, the graph of Figure 2 can be used with Equation 29 to calculate values of
cpr.
5 VISCOSITY OF GASES
5.1 ViscosityViscosity is a measure of the resistance to flow. It is given in units of centipoise.
A centipoise is a gm/100 sec.cm. The viscosity term is called dynamic viscosity
whereas kinematic viscosity is the dynamic viscosity divided by the density.
kinematic vis itydensity
cos dynamic viscosity
Kinematic viscosity has units of cm2/100 sec and the term is called centistoke.
Gas viscosity reduces as the pressure is decreased. At low pressures an increase in
temperature increases gas viscosity whereas at high pressures gas viscosity decreases
as the temperature increases. Figure 6 gives the values for pure component ethane.
1000900800700600
500
400
300
200
100908070
50 100 150 200 250 300 350 400
Temperature, deg F
Vis
co
so
ty,
mic
rop
ois
es
Viscosity of ethane
Pressure, psia5000
40003000
200015000
1000750
600
14.7
Figure 6 Viscosity of ethane
The viscosity of gases at low pressures can be obtained from correlations presented
by different workers.
Behaviour of Gases
20
50 100 150 200 250 300 350 400
Vis
cosity,
cp
Temperature, ?ºF
0.020
0.022
0.024
0.018
0.016
0.014
0.012
0.010
0.008
0.006
0.004
Helium
Nitrogen
Carbon D
ioxide
Methane
Ethylene
Ethane
Propane
i-Butane
n-Butane n-Pentane
n-Hexane
n-Heptane n-Octane
n-Nonane
n-Decane
Hydrogen Sulfide
Air
Figure 7 Viscosity of paraffin hydrocarbon gases at one atmosphere
Figure 7 and Figure 8 give the viscosities of individual components and paraffin
hydrocarbons at one atmosphere. For systems greater than 1 atmos the viscosities
can be obtained from the literature. Another way is by calculating the reduced
temperature and reduced pressure and use the chart developed by Carr6 which gives
a ratio of µ at reservoir conditions. This is given in Figure 9 in terms of pseudo
reduced conditions.
Institute of Petroleum Engineering, Heriot-Watt University 21
400 º F
300 º F
200 º F
100 º F
0.5 1.0 1.5 2.0 2.5 3.0 3.5
10 20 30 40 50 60 70 80 90 1000.004
0.005
0.006
0.007
0.008
0.009
0.010
0.011
0.012
0.013
0.014
0.015
0.016
Molecular Weight
Vis
cosity,
at 1 a
tm,
1, centipois
e
Gas Gravity (Air = 1)
N2
Mole per cent N2
G = 20
G = 06
G = 20
G = 06
1.5
1.0
1.5
1.0
G = 20
G = 06
1.5
1.0
0
0.0015
0.0010
0.0005
05 10 15
Co
rre
cti
on
ad
de
d t
o
Vis
co
sit
y,
c.p
.
0.0015
0.0010
0.0005
Co
rre
cti
on
ad
de
d t
o
Vis
co
sit
y,
c.p
.
0.0015
0.0010
0.0005
Co
rre
cti
on
ad
de
d t
o
Vis
co
sit
y,
c.p
.
CO2
Mole per cent CO2
00
5 10 15
H2S
Mole per cent H2S0
05 10 15
Figure 8 Viscosity of gases at atmospheric pressure6
Pseudoreduced Pressure, PR
0.8 1.0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
5.0
6.0
1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4
= Viscosity at operating temperature
and pressure, centipoises
A = Viscosity at 14.7 psia (1atm) and
operating temperatures, centipoises
20
15
10
8
6
4
3
2
1
Pseudoreduced Temperature, TR
Vis
cosity,
/
A
Figure 9 Viscosity ratio vs pseudo reduced temperature and pseudo pressure.
Behaviour of Gases
22
5.2 Viscosity of MixturesAnother formula that is used for mixtures is:
mix
j j j
j j
y M
y M (30)
j = 1, n
where:
y = mole fraction of jth component
M = molecular weight of component
= the viscosity of jth component
n = number of components
j
j
j
The presence of other gases can also make a significant difference on the viscosity
(Figure 7).
EXERCISE 14.
Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of
one atmosphere.
EXERCISE 15.
Use the gas gravity method to calculate the viscosity of the gas in exercise 4
EXERCISE 16.
Determine the viscosity of the gas in exercise 4 at 150°F and 3500 psia (ref ex 4, 7, &8)
Institute of Petroleum Engineering, Heriot-Watt University 23
6 EQUATIONS OF STATE
6.1 Other Equations-of-StateAs indicated at the start of section 2 the compressibility factor evolved out of the
need to use an equation derived out of ideal gas behaviour and incorporating it
into it a correction factor to suit real gas behaviour. One of the difficulties of the
compressibility equation:
PV = ZnRT
to describe the behaviour of gases is that the compressibility factor is not constant and
therefore mathematical manipulations cannot be made directly but must be carried out
through graphical or numerical techniques. Rather than use this modified equation
of state many have developed equations specifically to represent the behaviour of
real gases. It is an irony however that because of the long use of the equation above
incorporating z many of the real gas equation of states have been worked to calculate
z for use in the above equation.
6.2 Van de Waals Equation 1873The well known van der Waalʼs equation was one of the earliest equations to represent
the behaviour of real gases. This most basic EOS, which corrects for the volume of the
molecules and attractive and collision forces using empirical constraints a and b.
(P + a/V2) (V-b) = RT (31)
The two corrective terms to overcome the limiting assumptions of the ideal gas
equation are:
(i) The internal pressure or cohesion term , which accounts for the cohesion forces,
is a/V2.
(ii) The co-volume b, which represents the volume occupied by one mole at infinite
pressure and results from the repulsion forces which occur when the molecules
move close together.
The equation can also be written as:
V3 - (+ b) V2 + (a/P)V - ab/P = 0
Such equations are therefore called cubic equations of state.
The equation written to solve for z, the compressibility factor , becomes:
Z3 - Z2 (1 + B) + Z A - AB = 0 (32)
where
AaP
RTand B
bP
RT( )2
(33)
Behaviour of Gases
24
Values of a and b are positive constants for a particular fluid and when they are
zero the ideal gas equation is recovered. One can calculate P as a function of V for
various values of T. Figure 10 is a figure of 3 isotherms. Also drawn is the curve
for saturated liquid and saturated vapour.
Isotherm T1 is the single phase isotherm, T
c is the critical isotherm and T
2 gives the
isotherm below the critical temperature.
Vsat (liq) Vsat (vap)
V
P
Psat
c
T1>T
c
Tc
T2<T
c
Figure 10 PV behaviours of pure components predicted by EOS.
At the critical point , for a pure substance , the equation of state should be such
that:
P
V
P
VT T T Tc c
2
20
That is the critical isotherm exhibits a horizontal inflection point at the critical
point.
Institute of Petroleum Engineering, Heriot-Watt University 25
The application of these conditions to the van de Waals equation yields:
aR T
Pb
RT
P
c
c c
27
64 8
2 2
and
(34)
EXERCISE 17.
Calculate the critical constants for n- heptane.
For the curve, T2<Tc, the pressure decreases rapidly in the liquid region with increasing
V; after crossing the liquid saturated line a minimum occurs, rises to a maximum
and then decreases at the saturated vapour line. Real behaviour does not follow this
behaviour. They contain a horizontal segment where saturated liquid and saturated
vapour coexist in varying proportions.
This equation is not able to represent gas properties over a wide rage of temperatures
and pressures and over subsequent years many equations have been developed. A
number are given including those which are finding favour in their application in
this industry.
6.3 Benedict-Webb - Rubin Equation (BWR) 1940This equation developed for pure light hydrocarbons found considerable application
in predicting thermodynamic properties of natural gases, since natural gases are
essentially mixtures of light hydrocarbons and it can be written in a form similar to
Van der Waals equation.
PPT
V
B RT A C T
V
bRT a
V
a
V
C
V T V V
o o o
o
/
exp
2
2 3
6 3 2 2 21
(35)
where a, b, c, Ao, Bo and Co are constants for a given gas.
These equations are derived for pure components for which the empirical parameters
need to be obtained. For mixtures mixing rules are required to obtain these
constants.
6.4 Redlich-Kwong Equation 1949Numerous equations were developed with increasing numbers of constants specific to
pure components. More recently there has been a move back to the cubic equations
like van der Waals. We will describe briefly those which have found favour in the
oil and gas sector.
This modern development of cubic equations of state started in 1949 with the Redlich
and Kwong equation which involves only two empirical constants.
Behaviour of Gases
26
P = RT
V - b
a T
V V b (36)
where a and b are functions of temperature.
The term a(T) depends on the temperature and Redlich Kwong expressed this as a
function of the reduced temperature Tr using
a Ta
T
c
R
By applying the limiting condition at the critical points yields values of ac and b
related to critical constants. Such that ;
aR T
Pb
RT
Pc
c
c
c
c
0 42748 0 086642 2
. . and
(37)
6.5 Soave, Redlich Kwong equationSoave, in 1972, modified the Redlick-Kwong (RK) equation and replaced the a/T0.5
term with a temperature dependent term aT where a
T = a
c. .
The Soave, Redlich-Kwong (SRK) equation is therefore:
PRT
V b
a
V V b
c
(38)
where
is a non dimensionless temperature dependent term which has a value of 1.0 at the
critical temperature.
is obtained from
1 12
2
m Tr
where m = 0.480 + 1.574 - 0.176
where is the Pitzer accentric factor .8
6.6 Peng Robinson Equation of State 1975Peng and Robinson modified previous equations in relation to the attractive term.
They introduced it to improve the predictions of the Soave modification in particular
for the calculation of liquid densities.
Institute of Petroleum Engineering, Heriot-Watt University 27
PRT
V b
a
V V b b V b
c
(39)
aR T
Pb
RT
Pc
c
c
c
c
0 457235 0 07782 2
. . and
(40)
and
is the same function as for the Soave equation except the function is
different;
where m = 0.37464 + 1.54226w - 0.26992w2
These equations, in particular the SRK and PR equation are widely used in simulation
software used to predict behaviour in reservoirs, wells and processing. There are
other equations of state which are as competent at predicting physical properties
which have been developed mainly focusing on the need to improve the accuracy of
liquid volumes predictions. There is, however, great reluctance to change from those
presently used because of the investment in their associated parameters. An excellent
review of these equations and application is given by Danesh 9.
6.7 Application to MixturesWhen properties of mixtures are required mixing rules are required to combine the
data from pure components.
For both the SRK and PR equation
b y b y y a aj kj j
j
i j i
ji
ij and a 1
(41)
where the term kij is termed the binary interaction coefficients which are independent
of pressure and temperature. Values of binary interaction coefficients are obtained
by fitting equation of state (EOS) predictions to gas-liquid data for binary mixtures.
They have NO physical property significance. Each equation has its own binary
interaction coefficient.
Effort is underway and methods exist to not use binary interaction parameters but to
use physical property related parameters to enable good quality predictions.
Behaviour of Gases
28
EXERCISE 18.
A PVT cell contains 0.01 cu ft ( 300cc) of gas with at composition of ; methane 0.67 mol.frac, ethane 0.235 and n-butane 0.05. The temperature is increased to 300°C. Use the SRK equation to calculate the pressure at this increased temperature. Use
binary interaction coefficients of C1-nC4 0.02, C2-nC4 0.01 and C1-C2 0.0
Solutions to Exercises
EXERCISE 1.
A gas cylinder contains methane at 1000 psia and 70 oF. If the cylinder has a volume
of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in the
cylinder.
SOLUTION
PV = nRT
n = m/M
where n = number of moles
m = mass
M = molecular weight
m = PMV/RT
m
psialb
lbmolecuft
psia cuft
lbmole RR
o
o
1000 16 04 3
10 73 530
.
..
.
Mass of methane, m = 8.46 lb
Institute of Petroleum Engineering, Heriot-Watt University 29
EXERCISE 2.
Calculate the density of the gas in the cylinder in exercise 1.
SOLUTION
g = MW.P
RT
g
g
psialb
lbmolepsia cuft
lbmole oRR
Density of gaslb
cu ft
1000 16 04
10 73 530
2 82
0
.
..
.
, .. .
EXERCISE 3.
Assuming methane is at the conditions of exercise 1, calculate the volume the gas
would occupy at standard conditions.
SOLUTION
P V
T =
P V
T
P V
T
= P
P T
T V
= 1000
1 520 R
530 R x3ft
= 200.23 scf
1 1
1
2 2
2
sc sc
sc
1
sc
sc
1
psia
psia
o
o
3
V
V
V
sc
sc
sc
4 7.
EXERCISE 4.
A gas is made up of the following components; 25lb of methane, 3 lb of ethane and 1.5
lb of propane. Express the composition of the gas in weight and mole fractions.
Behaviour of Gases
30
SOLUTION
Gas Components
AWeight
B Mol weight
Clb moles
DMole fraction
Methane 25 16.04 1.559 0.921
Ethane 3 30.07 0.100 0.059
Propane 1.5 44.09 0.034 0.020
Totals 29.05 1
1
2
3
EXERCISE 5.
What is the apparent molecular weight of the gas in exercise 4
SOLUTION
Apparent Molecular weight= 17.43
Gas Components
AMol weight
BMol fraction
C
Methane 16.04 0.921 14.77
Ethane 30.07 0.059 1.77
Propane 44.09 0.020 0.89
1.000 17.43
1
2
3
mw yi A*B
EXERCISE 6.
What is the gas gravity of the gas in exercise 4 ?
SOLUTION
g
g
air
gM
M
M
29
g = AMW = 17.43
Gas gravity = 0.6
EXERCISE 7.
Calculate the pseudo critical temperature and pseudocritical pressure of the mixture
in exercise 4 .
Institute of Petroleum Engineering, Heriot-Watt University 31
SOLUTION
1
2
3
Gas Components
AMol weight mw
BMole fraction yi
C Pc-psi
D.
Tc °R Ppc
Methane 16.04 0.921 667.00 344 614.3 316.81
Ethane 30.07 0.059 708.00 550 41.7 32.42
Propane 44.09 0.020 616.00 666 12.4 13.39
Total 1.0 668.4 362.6
Tpc
Pseudocritical pressure = 668.4 psia
Pseudocritical temperature = 362 oR
EXERCISE 8.
For the gas of exercise 4 determine the compressibility factor at a temperature of 150 oF and a pressure of 3500psia.
SOLUTION
Ppr = P/P
pc, T
pr = T/T
pc
From exercise 6 Ppc
= 668 psia, Tpc
= 362.6°R
P = 3500 psia, and T = 150°C ie. 610°R
Ppr = 5.24, and Tpr = 1.68
From standing Katz chart, figure 2
Compressibility factor, z = 0.88
EXERCISE 9.
Calculate the pseudo critical properties of the gas in exercise 4 if it also contained 3
lb of hydrogen sulphide, 10lb of carbon dioxide and 2.5lb of nitrogen
1
2
3
Gas
Components
Mol
weight
Mole
fraction
Pc-psi Tc °R Ppcpsia
Methane 25 0.56 16.04 0.035 0.743 667.00 344 495.8 255.70
Ethane 3 0.07 30.07 0.002 0.048 708.00 550 33.7 26.17
Propane 1.5 0.03 44.09 0.001 0.016 616.00 666 10.0 10.81
Hydrogen 3 0.07 34.08 0.002 0.042 1306 673 54.8 28.25
sulphide
Carbon 10 0.22 44.01 0.005 0.108 1071 548 116.1 59.38
Dioxide
Nitrigen 2.5 0.06 28.02 0.002 0.043 493 227 21.0 9.66
Total 45 1.00 0.0466 1.000 731 390
TpcWeight Wgt
fraction
lb moles
4
5
6
From Wichert & Azis chart for compositions of H2S and CO
2 = 19
Behaviour of Gases
32
T = T - = 371 R
P = 694.3
pc pc
o
pc
pp T
T y ypc
pc pc
pc H S H S2 21
EXERCISE 10.
Express the quantity of 1 lb mole of a gas as standard cubic feet.
SOLUTION
Equation of state PV = RT for 1 mole
R = 10.732 psia. cu.ft/lb.mole °R T = 60+460 = 520 °R, P = 14.65 psia
or V for 1 lb.mole = RT/P = 380.9 scf/lb.mole.
EXERCISE 11.
Express the mass of gas in exercise 4 as standard cubic feet.
SOLUTION
Total mass of gas = 29.5 lb.
Apparent mol.wgt of gas exercise 5 = 17.43 lb./lb.mole
lb.moles of gas = 1.6924
Standard cubic feet of gas = 380.9 x 1.6924
= 644.68 scf
EXERCISE 12.
Calculate the gas formation factor for a gas with the composition of exercise 4 existing
at the reservoir conditions given in exercise 8.
SOLUTION
T = 150 oF ie 610 oR and P = 3500 psia
Compressibility factor at these conditions from exercise 8 = 0.88
Bg using equation above = 0.0008 res bbl/scf
Institute of Petroleum Engineering, Heriot-Watt University 33
EXERCISE 13.
A reservoir exists at a temperature of 150oF (as for exercise 8) suitable for storing
gas. It has an areal size of 5 miles by 2 miles and is 200ft thick. The average porosity
is 20% and there is no water present. How much gas of the composition of exercise
4 can be stored at a pressure the same as in exercise 8 i.e. 3500 psia. ? (1 mile=
5280 ft.)
SOLUTION
Volume of reservoir pore space = 5x2 x (5280)2 x 200 x 0.2
= 11,151,360,000 cu. ft.
=1,985,994,657 bbls
Bg , exercise 11 =0.00077299 res. bbls/SCF
Volume of gas =2.56923E+12 scf
EXERCISE 14.
Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of
one atmosphere.
SOLUTION
Gas
Components
Mol Weight Mole fraction
yj
Viscosity
from fig 7
j
Mj yj Mj
Methane
Ethane
Propane
jyj Mj
16.04
30.07
44.09
0.921
0.059
0.020
1.000
0.013
0.0112
0.0098
4.0050
5.4836
6.6400
SUM
3.6884
0.3233
0.1335
4.1451
0.0470
0.0036
0.0013
0.529
mix
j j j
j j
y M
y M
mix = 0.0529/4.1451
mix =0.01275 cp
EXERCISE 15.
Use the gas gravity method to calculate the viscosity of the gas in exercise 4
SOLUTION
Gas Components
Mol Weightmw
Mole fractionyj
Methane
Ethane
Propane
16.040
30.070
44.090
0.000
0.921
0.059
0.020
1.000
14.7720
1.773
0.886
17.431
Behaviour of Gases
34
g=AMW/M
air
g=AMW/29 Temperature = 150°F
Mol weight air = 29.000
AMW of gas = 17.431
Gas Gravity = 0.601
g = 0.01265 from fig 8
EXERCISE 16.
Determine the viscosity of the gas in exercise 4 at 150oF and 3500 psia (ref ex 4, 7,
&8)
SOLUTION
From exercise 7
P
T
P
TT
pc
pc
r
rc
668 4
362 6
.
.
PP =
3500
668.4 = 5.24
T = 610
362.6 = 1.68
c
From Lee correlation
/ atmos = 1.75
Viscosity at atmospheric pressure
From exercise 13 and 14 = 0.01275 cp
Viscosity at conditions = 0.0223 cp
EXERCISE 17.
Calculate the critical constants for n- heptane.
SOLUTION
R = 10.732. Tc for heptane = 973 oR and P
c = 397 psia
Using equations above a = 115,872 cu ft 2/lb mole
and b = 3.2878 cu ft./lb mole
Institute of Petroleum Engineering, Heriot-Watt University 35
EXERCISE 18.
A PVT cell of volume 0.01 cu ft ( 300cc) contains 0.008 lb mole. of gas with
a composition of; methane 0.67 mol.frac, ethane 0.235 and n-butane 0.05. The
temperature is increased to 300°C. Use the SRK equation to calculate the pressure
at this increased temperature. Use binary interaction coefficients of C1-nC4 0.02,
C2-nC4 0.01 and C1-C2 0.0
SOLUTION
Calculate the constants a and b for each component
aR T
Pb
RT
P
m T
cc
c
c
c
r
0 42748 0 08664
1 1
2 2
2
. . and
where m = 0.480 + 1.574 - 0.176 2
Components Pc ac m a a=a*Y Tc°R) bj
Methane
ethane
n-butane
0.67 344 667 0.4759 8735 0.0104 0.49635 0.57546 5027
0.235
0.05
550
766
708
551
0.7223
1.2926
21036
52429
0.0979
0.1995
0.63241
0.78701
0.79033
1.00619
16625
52753
Now calculate the mixture values.
b y b y y a aj kj j
j
i j i
ji
ij and a
where a = (1- k )(a a )ij ij i j
0.5
1
Components kijMethane
kijn-butane
aijMethane
aijethane
aijn-butane
sumYi b kijethane
Methane
ethane
n-butane
0.67 0.312 0.00 0.00 0.02 2123.7 1485.5 1037.29 4646.52
0.235
0.05
0.181
0.129
0.00 0.01
0.00
1485.5
1037.3
1039.1
732.97
732.969
527.535
3257.56
2297.8
1 0.622 sum 10201.9
Now use SRK to calculate pressure.
Behaviour of Gases
36
PRT
V b
a
V V b
c
V = 1.25 cu ft / lb mole
b = 0.622 a = 10201.9
P = 8617.6 psia
m
c
REFERENCES
1. Standing MB and Katz DL Density of Natural Gases. Trans AIME, 146(1942).
p140
2. Poettmann FH and Carpenter PG The Multiphase Flow of Gas and Water through
Vertical Flow Strings with Application to the Design of Gas Lift Installations.
API Drilling and Production Practise. 1952, pp 279-91
3. Brown GG et al. Natural Gasoline and Volatile Hydrocarbons” National Gasoline
Assoc. of America, Tulsa, Okl. 1948
4. Wichert, E and Aziz,K “ Calculate Zʼs for sour gases” Hyd Proc.(May 1972)
51, 119-122
5. Katz, D.L., Handbook of Natural Gas Engineering, McGraw Hill, NY, 1959
6. Carr N et al. Viscosity of natural gases under pressure. Trans AIME 201, 264,
(1954)
7. Lee et al “The viscosity of natural gases.” Trans AIME 1966 237, 997-1000
8. Pitzer K S et al The Volumetric and Thermodynamic Properties of Fluids II.
Compressibility Factor, Vapour Pressure and Entropy of Vaporisation. J .Am.
Chem. Soc. (1955) 77, No 13,3433-3440
9. Danesh, A PVT and Phase Behaviour of Petroleum Reservoir Fluids. 1998
Elsevier ISBN:0 444 82196 1 p129-162
CONTENTS
1 COMPOSITION BLACK OIL MODELS
2 GAS SOLUBILITY, Rs
3 OIL FORMATION VOLUME FACTOR, Bo
4 TOTAL FORMATION VOLUME FACTOR, BT
5 BELOW THE BUBBLE POINT
6 OIL COMPRESSIBILITY
7 BLACK OIL CORRELATIONS
8 FLUID DENSITY
8.1 Specific Gravity of a Liquid
8.2 Density Based on Ideal Solution Principles
9 FORMATION VOLUME FACTOR OF GAS
CONDENSATE, Bgc
10 VISCOSITY OF OIL
11 INTERFACIAL TENSION
12 COMPARISON OF RESERVOIR FLUID
MODELS
Properties of Reservoir Liquids
2
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Define gas solubility, Rs and plot vs. P for a reservoir fluid.
• Define undersaturated and saturated oil.
• Explain briefly flash and differential liberation
• Define the oil formation volume factor Bo, and plot B
o vs. P for a reservoir
fluid.
• Define the Total Formation Volume factor Bt, and plot B
t vs. P alongside a Bo
vs. P plot.
• Present an equation to express Bt in terms of B
o, R
s and B
g.
• Express oil compressibility in terms of oil formation volume factor.
• Use black oil correlations and their graphical form to calculate fluid
properties.
• Calculate the density of a reservoir fluid mixture, using ideal solution principles,
at reservoir pressure and temperature, using density correction chart for C1 &
C2 and other prerequisite data.
• Define the formation volume factor of a gas condensate
• Calculate the reserves and production of gas and condensate operating above
the dewpoint, given prerequisite data.
• Use viscosity equations and correlations to calculate viscosity of fluid at reservoir
conditions.
• Calculate the interfacial tension of equilibrium gas-oil systems given prerequisite
equations and data.
• List the comparisons of the black oil and compositional model in predicting
liquid properties
Institute of Petroleum Engineering, Heriot-Watt University 3
1 COMPOSITION - BLACK OIL MODEL
As introduced in the chapter on Composition, petroleum engineers are requiring
a compositional description tool to use as a basis for predicting reservoir and well
fluid behaviour. The two approaches that are commonly used are the multicomponent
compositional model described in the earlier chapter and the two component black oil
model. The latter simplistic approach has been used for many years to describe the
composition and behaviour of reservoir fluids. It is called the “Black Oil Model”.
The black oil model considers the fluid being made up of two components - gas dissolved
in oil and stock tank oil. The compositional changes in the gas when changing pressure
and temperature are ignored. To those appreciating thermodynamics this simplistic
two component model is difficult to cope with. The Black Oil Model, illustrated in
Figure 1, is at the core of many petroleum engineering calculations, and associated
procedures and reports.
Associated with the black oil model are Black Oil model definitions in relation to
Gas Solubility and Formation Volume Factors.
Reservoir Fluid
Solution Gas
Stock Tank Oil
/ = Rs
/ = Bo
Bo = Oil Formation Volume Factor
Rs = Solution Gas to Oil Ratio
Figure 1 "Black Oil Model"
Properties of Reservoir Liquids
4
2 GAS SOLUBILITY
Although the gas associated with oil and the oil itself are multicomponent mixtures
it is convenient to refer to the solubility of gas in crude oil as if we were dealing with
a two-component system.
The amount of gas forming molecules in the liquid phase is limited only by the
reservoir conditions of temperature and pressure and the quantity of light components
present.
The solubility is referred to some basis and it is customary to use the stock tank
barrel.
Solubility = f (pressure, temperature, composition of gas
composition of crude oil)
For a fixed gas and crude, at constant T, the quantity of solution gas increases with
p, and at constant p, the quantity of solution gas decreases with T
Rather than determine the amount of gas which will dissolve in a certain amount of
oil it is customary to measure the amount of gas which will come out of solution as
the pressure decreases. Figure 2 illustrates the behaviour of an oil operating outside
the PT phase diagram in its single phase state when the reservoir pressure is above
its reservoir bubble point at 1. Fluid behaviour in the reservoir is single phase and
the oil is said to be undersaturated . In this case a slight reduction of pressure causes
the fluid to remain single phase. If the oil was on the boundary bubble point pressure
line at 2 then a further reduction in pressure would cause two phases to be produced,
gas and liquid. This saturated fluid is one that upon a slight reduction of pressure
some gas is released. The concept of gas being produced or coming out of solution
gives rise to this gas solubility perspective. Clearly when the fluids are produced to
the surface as shown by the undersaturated oil in figure 2 the surface conditions lie
within the two phase area and gas and oil are produced. The gas produced is termed
solution gas and the oil at surface conditions stock tank oil. These are the two com-
ponents making up the reservoir fluid, clearly a very simplistic concept.
The gas solubility Rs is defined as the number of cubic feet (cubic metre) of gas
measured at standard conditions, which will dissolve in one barrel (cubic metre)
of stock tank oil when subjected to reservoir pressure and temperature.
In metric units the volumes are expressed as cubic metre of gas at standard conditions
which will dissolve in one cubic metre of stock tank oil.
Institute of Petroleum Engineering, Heriot-Watt University 5
Solution Gas
Stock Oil Tank
Oil Reservoir
Oil and Dissolved Gas
Rsi scf/stb
1 st b. oil
Bo rb.oil
Pre
ssure
Temperature
Pi 1
2
P
+
Surface
Phase Diagram
Figure 2 Production of reservoir hydrocarbons above bubble point
Figure 3 gives a typical shape of gas solubility as a function of pressure for a reser-
voir fluid at reservoir temperature. When the reservoir pressure is above the bubble
point pressure then the oil is undersaturated, i.e. capable of containing more gas. As
the reservoir pressure drops gas does not come out of solution until the bubble point
is reached, over this pressure range therefore the gas in solution is constant. At the
bubble point pressure, corresponding to the reservoir temperature, two phases are
produced, gas and oil. The gas remaining in solution therefore decreases.
The nature of the liberation of the gas is not straight forward. Within the reservoir
when gas is released then its transport and that of the liquid is influenced by the relative
permeability of the rock ( discussed in Chapter 10). The gas does not remain with its
associated oil i.e. the system changes. In the production tubing and in the separator
it is considered that the gas and associated liquid remain together i.e. the system is
constant. The amount of gas liberated from a sample of reservoir oil depends on the
conditions of the liberation. There are two basic liberation mechanisms:
Properties of Reservoir Liquids
6
1000 2000 3000
200
600
400
Pressure (psig)
Pb
Rsi
Rs scf/stb
Figure 3 Solution Gas - Oil Ratio as a Function of Pressure.
Flash liberation - the gas is evolved during a definite reduction in
pressure and the gas is kept in contact with the liquid
until equilibrium has been established.
Differential liberation - the gas being evolved is being continuously
removed from contact with the liquid and the liquid is in
equilibrium with the gas being evolved over a finite
pressure range.
The two methods of liberation give different results for Rs. This topic is covered in
more detail in the PVT analysis chapter.
Production of a crude oil at reservoir pressures below the bubble point pressure occurs
by a process which is neither flash or differential vaporisation. Once enough gas is
present for the gas to move toward the wellbore the gas tends to move faster than the
oil. The gas formed in a particular pore tends to leave the liquid from which it was
formed thus approximating differential vaporisation, however, the gas is in contact with
liquid throughout the path through the reservoir. The gas will also migrate vertically
as a result of its lower density than the oil and could form a secondary gas cap.
Fluid produced from reservoir to the surface is considered to undergo a flash process
where the system remains constant.
3 OIL FORMATION VOLUME FACTOR, B o
The volume occupied by the oil between surface conditions and reservoir or other
operating changes is that of the total system; the ʻstock tank oil ̓ plus its associated
or dissolved ʻsolution gasʼ. The effect of pressure on the complex stock tank liquid
and the solution gas is to induce solution of the gas in the liquid until equilibrium is
reached. A unit volume of stock tank oil brought to equilibrium with its associated
Institute of Petroleum Engineering, Heriot-Watt University 7
gas at reservoir pressure and temperature will occupy a volume greater than unity
(unless the oil has very little dissolved gas at very high pressure).
The relationship between the volume of the oil and its dissolved gas at reservoir
condition to the volume at stock tank conditions is called the Oil Formation Volume
Factor Bo. The shape of the B
o vs. pressure curve is shown in Figure 4. It shows
that above the bubble point pressure the reduction in pressure from the initial pres-
sure causes the fluid to expand as a result of its compressibility. This relates to the
chapter on Phase Behaviour where for an oil the PV diagram shows a large decline
in pressure for a small increase in volume, being again an indication of the com-
pressibility of the liquid. Below the bubble point pressure this expansion due to
compressibility of the liquid is small compared to the ʻshrinkage ̓of the oil as gas is
released from solution.
The oil formation volume factor, is the volume in barrels (cubic metres) occupied in
the reservoir, at the prevailing pressure and temperature, by one stock tank barrel
(one stock tank cubic metre) of oil plus its dissolved gas.
1000 2000 30001.0
1.2
1.1
Pressure (psig)
Pb
Bo
rb
./stb
Units - rb (oil and dissolved gas)
Figure 4 Oil formation volume factor
These black oil parameters, Bo and Rs are illustrated in Figure 5 a,b,&c from Craft
and Hawkins 1 reservoir engineering text., where they present the Rs and B
o curve for
the Big Sandy field in the USA. The visual concept of the changes during pressure
and temperature decrease is also presented.
Properties of Reservoir Liquids
8
P01
P01
= 3500 PSIA
T01
= 160º F
A
PB
= 2500 PSIA
T01
= 160º F
B
P = 1200 PSIA
T01
= 160º F
C
PA
= 14.7 PSIA
T01
= 160º F
D
PA
= 14.7 PSIA
T01
= 60º F
E
PB
P
PA PA
Free Gas
676 Cu. Ft.Free Gas
2.990 Cu. Ft.
Free Gas
567 Cu. Ft.
1,000 BBL1,040 BBL1,210 BBL1,333 BBL1,310 BBL
567SCF/STB
AT 1200 PSIA
RS = 337
BU
BB
LE
PO
INT
PR
ES
SU
RE
INIT
IAL P
RE
SS
UR
E
Solu
tion G
as, S
CF
/ST
B
600
500
400
300
200
100
00 500 1000 1500 2000
Pressure, PSIA
2500 3000 3500
(a)
(b)
Figure 5 Gas to oil ratio and oil formation volume factor for Big Sandy Field reservoir
oil 1.
Form
ation V
olu
me F
acto
r, B
BL/S
TB
0 500
1.40
1.30
1.20
1.10
1.001000 1500 2000
Pressure, PSIA
2500 3000 3500
BU
BB
LE
PO
INT
PR
ES
SU
RE
INIT
IAL P
RE
SS
UR
E1200 PSIA
BO
= 1.210
14.7 PSIA & 160º F
BO
= 1.040
2500 PSIA
BOB
= 1.333
3500 PSIA
BOI
= 1.310
14.7 PSIA & 60º F
BO
= 1.000
(b)
Figure 5b
Institute of Petroleum Engineering, Heriot-Watt University 9
The reciprocal of the oil formation volume factor is called the ʻshrinkage factor bo
bB
o
o
1
The formation factor Bo may be multiplied by the volume of stock tank oil to find
the volume of reservoir required to produce that volume of stock tank oil. The
shrinkage factor can be multiplied by the volume of reservoir oil to find the stock
tank volume.
It is important to note that the method of processing the fluids will have an effect
on the amount of gas released and therefore both the values of the solution gas-oil
ratio and the formation volume factor. A reservoir fluid does not have single Bo or R
s
values. Bo & R
s are dependant on the surface processing conditions. This simplistic
reservoir model (Figure 6) demonstrates that the black oil model description of the
reservoir fluids is an after the event, processing, description in terms of the produced
fluids. This simplistic approach to modelling reservoir fluids becomes more difficult
to consider when one is involved in reservoirs which become part of a total reservoir
system (Figure 7).
Rs
BO
Figure 6 Black oil description of reservoir fluid
Properties of Reservoir Liquids
10
Rs 3
Bo 3
Rs 2
Bo 2
Rs 4
Bo 4
Rs 1
Rs
Bo
Bo 1
?
Multi Reservoir System
Figure 7 Integrated system of reservoir common pipeline and final collection system.
4 TOTAL FORMATION VOLUME FACTOR, Bt
In reservoir engineering it is sometimes convenient to know the volume occupied
in the reservoir by one stock tank barrel of oil plus the free gas that was originally
dissolved in it. A factor is used called the total formation-volume factor Bt, or
the two-phase volume-factor and is defined as the volume in barrels that 1.0 STB
and its initial complement of dissolved gas occupies at reservoir temperature and
pressure, i.e. it includes the volume of the gas which has evolved from the liquid
and is represented by:
Bg (R
sb - R
s)
i.e. Bt = B
o + B
g (R
sb - R
s) (1)
Rsb
= the solution gas to oil ratio at the bubble point
Institute of Petroleum Engineering, Heriot-Watt University 11
Oil
Oil
Gas
Hg
B0
Bt
B0b
Bg(Rsb-Rs)
Figure 8a Total formation volume factor or two phase volume factor
Its application comes from the Material Balance equation (Chapter 15) where it is
sometimes used to express the volume of oil and associated gas as a function of pres-
sure. It is important to note that Bt does not have volume significance in reservoir
terms since the assumption in Bt is that the system remains constant. As mentioned
earlier if the pressure drops below the bubble point in the reservoir then the gas
coming out of solution moves away from its associated oil because of its favourable
relative permeability characteristics.
Figure 8b gives a comparison of the total formation-volume factor with the oil for-
mation-volume factor. Clearly above Pb the two values are identical since no free
gas is released. Below Pb the difference between the values represents the volume
occupied by free gas.
BoBt
Pressure Pb
Figure 8b Total and oil formation volume factor
The value of BT can be estimated by combining estimates of B
O and calculation of
Bg and known solubility values for the pressures concerned.
Properties of Reservoir Liquids
12
5 BELOW THE BUBBLE POINT
Figure 9 depicts the behaviour below the bubble point when produced gas at the
surface comes from two sources, the solution gas associated with the oil entering the
wellbore plus free gas which has come out of solution in the reservoir and migrated to
the wellbore. The total producing gas to oil ratio is made up of the two components
solution gas Rs and the free gas which is the difference. The diagram illustrates the
volumes occupied by these two in the reservoir, the solution gas being part of Bo and
the free gas volume through Bg.
Free Gas
& Solution Gas
Stock Oil Tank
Oil Reservoir
rb (oil and dissolved gas) /stb
1 st b. oil
Bo
Pre
ssure
Temperature
R= Rs + (R - Rs)
+
(R - Rs) Bg
Gas Oil
Reservoir
rb (free gas) /stb
SurfacePi
P
Figure 9 Production of reservoir hydrocarbons below bubble point
6 OIL COMPRESSIBILITY
The volume changes of oil above the bubble point are very significant in the context
of recovery of undersaturated oil. The oil formation volume factor variations above
the bubble point reflect these changes but they are more fundamentally embodied
in the coefficient of compressibility of the oil, or oil compressibility.
The equation for oil compressibility is
cV
V
Po
T
1
in terms of formation volume factors this equation yields
Institute of Petroleum Engineering, Heriot-Watt University 13
cB
B
Po
o
o
T
1
Assuming that the compressibility does not change with pressure the above equation
can be integrated to yield ;
c P PV
Vo 2 1
2
1
ln
where P1 & P
2, and V
1 & V
2 represent the pressure and volume at conditions 1 & 2.
7 BLACK OIL CORRELATIONS
Over the years there have been many correlations generated based on the two com-
ponent based black oil model characterisation of oil. The correlations are based
on data measured on the oils of interest. These empirical correlations relate black
oil parameters, the variables of Bo and R
s to; reservoir temperature, and oil and gas
surface density. It is important to appreciate that these correlations are empirical and
are obtained by taking a group of data for a particular set of oils and finding a best fit
correlation. Using the correlation for fluids whose properties do not fall within those
for the correlation can result in significant errors. Danesh 2 has given an excellent
review of many of these correlations
A number of empirical correlations, based on largely US crude oils, and other loca-
tions across the world have been presented to estimate black oil parameters of gas
solubility and oil formation volume factor. The most commonly used is Standingʼs 3 correlation. Other correlations include, Lasater 4, and recently Glaso 6
Pb = f (R
s,
g, p
o, T)
where Pb = bubble point pressure at ToF
Rs = solution gas-oil ratio (cu ft/ bbl)
g= gravity of dissolved gas
o = density of stock-tank oil .(specific gravity)
Standingʼs correlation for the calculation of Pb, bubble point pressure is:
PR
T APIbs
g
. ( . . ( )) .
.
18 2 0 00091 0 0125 1 4
0 83
10
(2)
His correlation for the oil formation volume factor is;
B R To s
g
o
. . .
. .
0 9759 0 000120 1 25
0 5 1 2
(3)
Properties of Reservoir Liquids
14
Standing's correlations have been presented as nomographs enabling quick look
predictions to be made. Figures 10 & 11 give the nomogram forms of these correlations
for gas solubility and oil formation volume factor. Standingʼs correlation is based on
a set of 22 California crudes.
Other correlations have been presented by Lasater 4 based on 137 Canadian ,USA
and South American crudes, Vasquez and Beggs5 using 6000 data points, Glaso 6 us-
ing 45 North Sea crude samples, and Mahoun 7 who used 69 Middle Eastern crudes.
Danesh2 gives a very useful table showing the ranges covered by the respective black
oil correlations
Institute of Petroleum Engineering, Heriot-Watt University 15
Figure 10 Oil-formation volume factor as a function of gas solubility, temperature, gas
gravity and oil gravity (Standing)
20
30
40
50
60
70
8090100
150
200
300
400
500
600
700800
9001000
1500
2000
1.021.03
1.041.05
1.061.07
1.081.09
1.10
Formation volume of bubble-point liquid
Gas
-oil
ratio
, cu ft
per
bbl
bbl p
er b
bl o
f ta
nk
oil
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
1.10
1.20
1.30
1.40
1.50
0.5
0 0.6
0 0.7
0 0.8
0 0.9
0 1.0
0
Ga
s g
rav
ity
Air
=1
Tank oil gravity, ºAPI50 30 10
Temperature, ºF
100
140160
180200
220240
260
120
Properties of Reservoir Liquids
16
Fig
ure 11
Gas so
lub
ility as a fu
nctio
n o
f pressu
re. Tem
peratu
re, gas g
ravity
and
oil
grav
ity
600
500
400
300
200
20
30
40
50
60
7080
90100
150
200
300
400
500
600
700
700
800
900
1000
1500
2000
3000
4000
5000
6000
800900
1000
1500
2000
Tank oil gravity, ºAPI
Temperature, ºF
Gas gravity Air = 1
60
1.501.40
1.301.20
80
100
120
140
160
180
200
220240
260
1.101.000.900.80
1014
1618
2022
2426
2830
3234
3638
40
12
4244
4648
5052
5456
58
Bu
bb
le-p
oin
tPressure, psia
Gas-oil ratio, cu ft per bbl
60
(STA
ND
ING
)
0.700.600.50
Institute of Petroleum Engineering, Heriot-Watt University 17
Correlation Standing Lasater Vasquez-Beggs Glaso Marhoun
Ref 3 4 5 6 7
Bubble - point pressure (psia) 130-7000 45-5780 15-6055 165-7142 130-3573
Temperature, F 100-258 82-272 162-180 80-280 74-240
Bo 1.024-2.15 1.028-2.226 1.025-2.588 1.032-1.997
Gas - oil ratio (scf/stb) 20-1425 3-2905 0-2199 90-2637 26-1602
Oil Gravity, oAPI 16.5-63.8 17.9-51.1 15.3-59.5 22.3-48.1 19.4-44.6
Gas Gravity 0.59-0.95 0.574-1.22 0.511-1.651 0.65-1.276 0.752-1.367
Separator Pressure 265-465 15-605 60-565 415
Searator Temperature F 100 36-106 76-150 125
Table 1 Black oil correlation and their ranges at application 2
8 FLUID DENSITY
Liquids have a much greater density and viscosity than gases, and the density is affected
much less by changes in temperature and pressure. For petroleum engineers it is
important that they are able to estimate the density of a reservoir liquid at reservoir
conditions.
8.1 Specific Gravity of a Liquid
o
o
w (4)
The specific gravity of a liquid is the ratio of its density to that of water both at the
same T & P. It is sometimes given as 60˚/60˚, i.e. both liquid and water are measured
at 60˚ and 1 atmos.
The petroleum industry uses another term called ˚API gravity where
APIo
141 5131 5
..
(5)
where o is specific gravity at 60˚/60˚.
There are several methods of estimating the density of a petroleum liquid at reservoir
conditions. The methods used depend on the availability and nature of the data of
data. When there is compositional information on the reservoir fluid then the density
can be determined using the ideal solution principle. When the information we have
is that of the produced oil and gas then empirical methods can be used to calculate
the density of the reservoir fluid.
8.2 Density based on Ideal Solution PrinciplesMixtures of liquid hydrocarbons at atmospheric conditions behave as ideal solutions.
An ideal solution is a hypothetical liquid where no change in the character of the
liquids is caused by mixing and the properties of the mixture are strictly additive.
Properties of Reservoir Liquids
18
Petroleum liquid mixtures are such that ideal-solution principles can be applied for the
calculation of densities and this enables the volume of a mixture from the composi-
tion and the density of the individual components. The principle is illustrated using
the following exercise. Data for the specific components are given in the tables at
the end of the chapter
EXERCISE 1.
Calculate the density at 14.7psia and 60 ºF of the hydrocarbon liquid mixture with the composition given below:
Component Mol.
fract.
1b mol.
nC4 0.25
nC5 0.32
nC6 0.43
1.00
SOLUTION EXERCISE 1
Solution Component Mol. Mol. Weight Liquid Liquid
density
fract. weight 1b Density at volume
1b mol. 1b/1b at 60˚F and 14.7 cu ft
mol. psia
1b/cu ft
nC4 0.25 58.1 14.525 36.45 0.3985
nC5 0.32 72.2 23.104 39.36 0.5870
nC6 0.43 86.2 37.066 41.43 0.8947
____ _____ _____
1 74.695 1.8801
Liquids at their bubble point or saturation pressure contain large quantities of dis-
solved gas which at surface conditions are gases and therefore some consideration
for these must be given in the additive volume technique. This physical limitation
does not impair the mathematical use of a “pseudo liquid density “ for methane and
ethane since it is only a step in its application to determine a reservoir condition
density. This is achieved by obtaining apparent liquid densities for these gases and
determining a pseudoliquid density for the mixture at standard conditions which can
then be adjusted to reservoir conditions.
Standing & Katz 8 carried out experiments on mixtures containing methane plus other
compounds and ethane plus other compounds and from this were able to determine
a pseudo-liquid (fictitious) density for methane and ethane
Institute of Petroleum Engineering, Heriot-Watt University 19
Correlations have been obtained by experiment giving apparent liquid densities of
methane and ethane versus the pseudoliquid density (Figure 12).
0.1
0.2
0.3
0.4
0.5 0.6 0.7 0.8 0.9
0.3
0.4
0.5
0.6
0.40.3
Density of system, 60ºF B atm. pressure
Ap
pa
rre
nt
de
ns
ity
of
Me
tha
ne
, g
/cc
Ap
parr
en
t d
en
sit
y o
f o
f E
than
e, g
/cc
Ethane - N - Butane
Ethane - Heptane
Ethane - Crystal oil
Methane - Cyclo Hexane
Methane - Crude oil
Methane - Crystal oil
Methane - Propane
Methane - Hexane
Methane - Pentane
Methane - Heptane
Methane - Benzene
Figure 12 Variation of apparent density of methane and ethane with density of the system 8.
To use the correlations a trial and error technique is required whereby the density of
the system is assumed and the apparent liquid densities can be determined. These
liquid densities are then used to compute the density of the mixture by additive vol-
umes and the value checked against the initial assumption. The procedure continues
until the two values are the same.
When non hydrocarbons are present, the procedure is to add the mole fractions of
the nitrogen to methane, the mole fraction of carbon dioxide to ethane and the mole
fraction of hydrogen sulphide to propane.
Properties of Reservoir Liquids
20
EXERCISE 2:
Calculate the “surface pseudo liquid density” of the following reservoir
composition.
Component Mole percent
Methane 44.04
Ethane 4.32 Properties of
Propane 4.05 heptane +
Butane 2.84 API gravities = 34.2
Pentane 1.74 SG = 0.854
Hexane 2.9 Mol wt = 164
Heptane + 40.11
SOLUTION EXERCISE 2
Estimate 44.65 lb/cu ft. 0.716 gm/cc lb/cuft
From fig 12 Density 0.326 20.3424
C1
Density 0.47 29.328
C2
Component Mole Mol Weight Liq Liquid
fraction Weight Density Volume
lb/lb lb at 60 F &
mole 14.7 psia
lb/cu.ft cu ft.
z M zM o zM/ o
Methane 0.4404 16 7.0464 20.3424 0.34639
Ethane 0.0432 30.1 1.30032 29.328 0.04434
Propane 0.0405 44.1 1.78605 31.66 0.05641
Butane 0.0284 58.1 1.65004 35.78 0.04612
Pentane(n&i) 0.0174 72.2 1.25628 38.51 0.03262
Hexane(n&i) 0.029 86.2 2.4998 41.43 0.06034
Heptane+ 0.4011 164 65.7804 53.26 1.23508
Total 1 81.31929 1.8213
Density = 81.32 lb / 1.82 cu ft
= 44.65 lb/cu.ft
This trial and error method is very tedious so Standing and Katz devised a chart which
removes the trail and error required in the calculation. The densities have been con-
verted into the density of the heavier components, C3+
, and the weight percent of the
two light components, methane and ethane in the C1+
and C2+
mixtures. Figure 13.
Institute of Petroleum Engineering, Heriot-Watt University 21
70
60
50
40
30
10
20
30
40
50
60
70
De
nsity o
f syste
m in
clu
din
g m
eth
an
e a
nd
eth
an
e,
lb/c
u f
t
De
nsity o
f p
rop
an
e p
lus,
lb/c
u ft
Wt %
eth
ane in
eth
ane p
lus m
ate
rial
01020304050
Wt %
met
hane
in e
ntire
sys
tem
0
10
20
30
Figure 13 Pseudo-liquid density of systems containing methane and ethane 10.
We shall examine through examples various ways of calculating downhole reservoir
fluids densities dependant on the data available.
The three considered are:
1. The composition of the reservoir fluid is known.
2. The gas solubility , the gas composition and the surface oil gravity is known
3. The gas solubility, and gas and liquid gravities are known.
1. The composition of the reservoir fluid is known.
The procedure is illustrated using the following two exercises .
Properties of Reservoir Liquids
22
EXERCISE 3.
Calculate the surface density of the mixture in exercise 2 using the chart of figure 13
The pseudodensity is converted to reservoir conditions firstly by taking the effect
of pressure and secondly accounting for the effect of temperature. The variation of
density with respect to pressure and temperature has been investigated and it has
been demonstrated that thermal expansion is not affected by pressure. Standing &
Katz took National Petroleum Standards data and with supplementary data produced
correction factors for pressure and temperature to convert atmospheric density to
reservoir density.
The compressibility and thermal expansion effects have been expressed graphically
in Figures 14 and 15.
10
9
8
7
6
5
4
3
2
1
025 30 35 40 45 50 55 60 65
Density at 60ºF and 14.7 psia, lb/cu ft
De
nsity o
f p
ressu
re m
inu
s d
en
sity a
t 6
0ºF
1
4.7
psia
lb
/cu
ft
Pressure, psia
15,0
00
10,0
00 8
,000
5,000
6,000
4,000
3,000
2,000
1,000
Figure 14 Density correction for compressibility of liquids 8.
Institute of Petroleum Engineering, Heriot-Watt University 23
10
9
8
7
6
5
4
3
2
1
025 30 35 40 5045 55 60 65
Density at 60ºF and pressure P, lb/cu ft
De
nsity a
t 6
0ºF
min
us d
en
sity a
t te
mp
era
ture
, lb
/cu
ft
80
100
120
160
180
200
220
Temperature ºF
240
140
60
Figure 15 Density correction for thermal expansion of liquids 10.
EXERCISE 4.
Calculate the density of the reservoir liquid of exercise 3 at a reservoir temperature of 5,500 psia and 180 oF
Full compositional data may not always be available and the characterisation of the
produced fluids will vary from full compositional analysis to a description of the
fluids in terms of gas and oil gravity. The procedure just described is for the situa-
tion where the composition of the reservoir fluid is known. The procedures which
follow cover the situation where a less comprehensive analysis is available. These
methods make use of empirical correlations.
Properties of Reservoir Liquids
24
2. Reservoir Density when the Gas Solubility , the gas composition and the surface
oil gravity are known
By considering surface liquid as a single component and knowing the composition
of the collected gas the techniques previously discussed can be used to determine
reservoir liquid density. Again we will illustrate the procedure with an example
EXERCISE 5.
A reservoir at a pressure of 4,000 psia and a temperature of 200oF has a producing gas to oil ratio of 600 scf/STB. The oil produced has a gravity of 42 oAPI. Calculate the density of the reservoir liquid. The produced gas has the following composition
Component Mole Fraction Methane 0.71 Ethane 0.13 Propane 0.08 Butane 0.05 Pentane 0.02 Hextane 0.01
3. The Gas Solubility, and Gas and Liquid gravities are known.
Katz has produced a correlation (figure 16) to enable densities to be determined when
the only information on the gas is its solubility and its gravity. The figure gives ap-
parent liquid densities of gases against gravity for different API crudes
0.615
20
25
30
35
40
45
0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4
Gas Gravity
Ap
pa
ren
t L
iqu
id d
en
sity o
f D
isso
lve
d G
as a
t
60
F
an
d 1
4.7
psia
, lb
/cu
. ft.
20 API Crude
30
40
50
60
Figure 16 Apparent liquid densities of natural gases