5 - Behaviour of Gases

Behaviour of Gases F I V E
T1>Tc
Tc
Psat
T2<Tc
.008
Vsat (liq)
.006
Bg
rb/scf
Vsat (vap)
.004
.002
1000
2000
3000
PRESSURE (psig)
Reservoir Engineering
26/11/15
Petroleum Engineering
C O N T E N T S
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
26/11/15
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 &T 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 real gas and explain what Z is,
PV=ZnRT.

Define the pseudocritical pressure and psuedocritical temperature and be

able to use them to determine the Z 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.
Be aware of the development of EOSs to predict reservoir fluid properties.
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 is 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 Boyles Law
At constant temperature the pressure of a given weight of a gas is inversely proportional

to the volume of a gas.
i.e.
1
or PV = constant, T is constant (1)
P
P = pressure, V = volume, T = temperature.
1.2 Charles Law
At constant pressure, the volume of a given weight of gas varies directly with the
temperature:

i.e.
V T or
V
= constant, P is constant (2)
T
The pressure and temperature in both laws are in absolute units.
26/11/15
Institute of Petroleum Engineering, Heriot-Watt University
1.3 Avogadros Law
Avogadros 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 60F 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 Gas
By combining the above laws an equation of state relating pressure, temperature and
volume of a gas is obtained.

PV
= constant (3)
T
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;
cu ft psia
R in oilfield units = 10.732

lb mole R

Table 1 gives the values for different unit systems.
P
psia
atm
atm
atm
atm
mm Hg
in.Hg
cu ft
cu ft
cc
litre
cu ft
litre
cu ft
R
K
K
K
R
K
R
lb - mole
lb - mole
gm - mole
gm - mole
lb - mole
gm - mole
lb - mole
R
10.73
1.3145
82.06
0.08206
0.730
62.37
21.85
Table 1 Values of R for different unit systems.
For n moles the equation becomes:

PV = nRT
(4)
T= absolute temperature oK or oR where

K=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
PV
PV
is a constant so that 1 1 = 2 2
RT
T1
T2
EXERCISE 1.
A gas cylinder contains methane at 1000 psia and 70F. If the cylinder has a volume
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 Gas
Since density is defined as the weight per unit volume, the ideal gas law can be used
to calculate densities.
g = weight / volume =

where rg is the gas density

For 1 mole m = MW
V =
m
V
MW = Molecular weight
RT
P
g =
MW.P
(5)
RT
EXERCISE 2.
Calculate the density of the gas in the cylinder in exercise 1.
1.6 Standard Conditions
Oil 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 60F.
ie

26/11/15
Pres Vres
P V
= sc sc (6)
Tres
Tsc
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 Gases
Petroleum 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 Daltons
Law and Amagats Law.
1.7.1 Daltons Law of Partial Pressures
The 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 = PA + PB + PC + ..... (7)
where A, B and C are components.

therefore

P = nA
i.e.
RT
RT
RT
+ nB
+ nC
V
V
V
P =
RT
n j
V
Pj
n
= j = yj
P
n
where yj = mole fraction of jth component.

6
(8)
The pressure contribution of a component, its partial pressure, is the total pressure
times the mole fraction.
1.7.2 Amagats Law
Amagats 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 = VA + VB + VC (9)
V = nA
V =
i.e.
RT
RT
RT
+ nB
+ nC
P
P
P
RT
n j
P
Vj
n
= j = yj
V
n
(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.
1.8 Apparent Molecular Weight
A 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

26/11/15
If yj represents the mole fraction of the jth component:
AMW = y j MWj
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 Gas
The specific gravity of a gas, gg is the ratio of the density of the gas relative to that
of dry air at the same conditions.
g
(11)
air
g =
Assuming that the gases and air are ideal.
MgP
g =

RT = M g = M g
M air P
M air
29
RT

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.
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 Gases
The correction factor z 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:

P1V1
PV
= 2 2 (13)
z1T1
z 2 T2
z is an expression of the actual volume to what the ideal volume would be.
co
ns
ta
nt
Vactual
(14)
Videal
Compressibility factor, Z
1.0
at
ur
e
pe
r
i.e.
Te
m
0.5
PRESSURE, P
Figure 1 Typical plot of the compressibility factor as a function of pressure at constant

temperature.
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 States
The 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 (Pr) where:

26/11/15
Tr =
T
P
and Pr =
(15)
Tc
Pc
Where, Tc and Pc are the pure component critical temperature and pressure.
The compressibility factor z 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:

Tpc = y jTcj and Ppc = y j Pcj
(16)
where y is the mole fraction of component j and Tcj and Pcj 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 z is plotted as a function of pseudo reduced temperature, Tpr and
pseudo reduced pressure Ppr where
10
Compressibility Factors for Natural Gases as a

Function of Pseudoreduced Pressure and Temperature.
1.1
Pseudo Reduced Pressure, Pr
3.0
2.8
2.6
2.4
2.2
2.0
1.9
1.8
0.9
1.
4
1.
1.5
05
1.6
1.
1.1
1.0
1.05
1.2
0.95
1.7
0.8
1.3
1.1
1.
1.7
1.45
0.7
0.6
1.
3
1.2
1.6
1.8
1.15
0.4
2.0
1.7
1.9
3.0
1.3
2.6
1.2
1.05
0.25
2.8
1.1
3.0
2.2
2.0
1.8
1.7
1.6
0.9
7
1.9
1.2
1.1
1.4
1.3
1.1
Compressibility of
Natural Gases
(Jan. 1, 1941)
2.6 2.4
1.4
2.2
2.4
1.1
0.3
1.5
4
1.
1.5
1.25
0.5
1.6
1.4
1.35
1.3
1.
Compressibility Factor, z
Pseudo Reduced Temperature
1.0
1.0
1.0
1.05
0.9
9
10
11
12
13
Pseudo Reduced Pressure, Pr
14
15
Figure 2 Compressibility factors for natural gas1 (Standing & Katz, Trans AIME, 1942).
26/11/15
11
Tpr =
T
P
and Ppr =
Tpc
Ppc
(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
150F and a pressure of 3500psia.
2.3 Pseudocritical Properties of Natural Gases
Pseudocritical Pressure, psia
The 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.
Pseudocritical Properties of Natural Gases
700
Condens
650
Miscellaneou
s
ate Wel
l
Fluids
Gases
600
550
Pseudocritical Temperature, R
500
450
se
Ga
ids
Flu
ell
W
e
sat
de n
n
Co
400
350
300
an
ell
sc
Mi
s
ou
0.5
0.6
0.7
0.8
0.9
1.0
1.1
Gas Gravity (air = 1)
Figure 3 Pseudocritical properties of natural gases. 3
12
1.2
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 - e (18)
and

ppc =
p pcTpc
Tpc + yH 2 S 1 yH 2 S
(19)
T'pc and p'pc are used to calculate Tpr and Ppr. The value for e is obtained from
the Figure 4 from the Wichert and Aziz paper
80
15
70
PER CENT C02
60
50
20
40
25
30
30
20
30
25
10
20
15
10
10
20
30
40
34.5
50
60
70
80
PER CENT H2S
Figure 4 Adjustment factors for pseudocritiacl properties for non hydrocarbon

gases(Wichert & Aziz).
26/11/15
13
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
Gas
Components
1
2
3
4
5
6
Methane
Ethane
Propane
Hydrogen
sulphide
Carbon
Dioxide
Nitrigen
Total
Wgt
Mol
fraction weight
lb moles
Mole
fraction
Pc-psi
Tc R
Ppc
psia
Tpc
255.70
26.17
10.81
28.25
25
3
1.5
3
0.56
0.07
0.03
0.07
16.04
30.07
44.09
34.08
0.035
0.002
0.001
0.002
0.743
0.048
0.016
0.042
667.00
708.00
616.00
1306
344
550
666
673
495.8
33.7
10.0
54.8
10
0.22
44.01
0.005
0.108
1071
548
116.1 59.38
2.5
45
0.06
1.00
28.02
0.002
0.0466
0.043
1.000
493
227
21.0
731
9.66
390
From Wichert & Azis chart for compositions of H2S and CO2 e = 19
Tpc = Tpc - = 371o R

ppc =
Weight
p pcTpc
Ppc = 694.3
2.5 Standard Conditions for Real Reservoir Gases
As 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 60F (15.6C) 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.
14
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.

Bg =
volume occupied at reservoir temperature and pressure

volume occupied at STP
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 reservoir
barrels free gas/scf gas or rm3 reservoir cubic meters free gas/scm gas
.008
.006
Bg
rb/scf
.004
.002
1000
2000
3000
PRESSURE (psig)
Figure 5 Gas Formation Volume Factor, Bg.
For example Bg for a reservoir at condition 2 is;
26/11/15
15
V2
P Tz
= sc 2 2 (20)
Vsc P2 Tsc zsc
Bg =
sc 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
1
=
=E
volume in formation Bg
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.
VR
(21)
Vsc
Bg =
Subscripts R and sc are reservoir and standard conditions respectively.
VR =
T and P at reservoir conditions:
Vsc =
z = 1 for standard conditions
Bg = z
Since Tsc = 520Rm Psc = 14.7 psia for most cases
znRT
(22)
P
zsc nRTsc
(23)
Psc
T Psc cu. ft
. .
(24)
Tsc P SCF
Bg = 0.0283
Bg = 0.0283
or
zT cu. ft
P SCF
zT cu. ft
bbl
P SCF 5.615 cu ft

zT res bbl

B = 0.00504
g
16
P SCF
(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 150F (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 GASES

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

cg =
1 V
1 Vm
or =
V P
Vm P
(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 = nRT
dP
P2
1 nRT 1
cg =
=
V P 2 P (27)
For real gases:
26/11/15
V =
znRT
P
17
dz
P
z
dP
nRT
=
P T
P2
cg =
cg =
P nRT z
z
P
2
nRTz P P
1 1 z
.
P z P
(28)
z/P 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 = Ppc Ppr
z Ppr z
=
P P Ppr
Ppr
1
=
P Ppc
z 1 z
=
P Ppc Ppr
Combining this equation with eqn 28 above yields
cg =
1
1 z
Ppc Ppr zPpc Ppr

Tpr
c g Ppc =
1 1 z
Ppr z Ppr
Tpr
(29)
Units of cg = P-1, and cgPc is dimensionless

cgPpc is called pseudo reduced compressibility, cpr
18
Since the pseudo reduced compressibility is a function of z 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 Viscosity
Viscosity 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.
dynamic viscosity
density
kinematic viscosity =
Kinematic viscosity has units of cm2/100 sec and the term is called centistoke.
Viscosoty, micropoises
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.
1000
900
800
700
600
500
Viscosity of ethane
Pressure, psia
5000
400
3000
300
4000
2000
15000
200
750
100
90
80
70
50
1000
600
14.7
100
150
200
250
300
350
400
Temperature, deg F
Figure 6 Viscosity of ethane.
The viscosity of gases at low pressures can be obtained from correlations presented
by different workers.
26/11/15
19
0.024
0.022
He
m
liu
ro
Nit
Air
0.020
0.018
Ca
Viscosity, cp
0.016
d
Hy
en
rog
0.014
n
ge
nD
rbo
lfid
Su
iox
ide
e
han
Met
e
ylen
Eth
0.012
ane
Eth
e
pan
Pro
n
ta e
i-Bu
tane
n-Bu
0.010
0.008
e
ntan
n-Pe
ne
a
x
n-He
ane
t
p
e
n-H
ane
n-Oct
ane
n
o
N
nn
n-Deca
0.006
0.004
50
100
150
200
250
300
350
400
Temperature, oF
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 (Lee). Another way is by calculating the reduced
temperature and reduced pressure and use the chart developed by Carr6 which gives
a ratio of at reservoir conditions. This is given in Figure 9 in terms of pseudo
reduced conditions.
20
1.0
1.5
Gas Gravity (Air = 1)

2.0
2.5
0.015
Correction added to
Viscosity, c.p.
0.010
0.009
0.006
0.005
0.0010
G = 20
300
200
100
0
0
CO2
0.0015
G = 06
0.0005
5
10
15
Mole per cent N2
0
0
1.5
1.0
G = 20
G = 06
5
10
15
Mole per cent CO2
1.5
1.0
0.0005
0.004
10
400
H 2S
0.0015
Correction added to
Viscosity, c.p.
Viscosity, at 1 atm, m1, centipoise
0.011
0
0
3.5
0.0010
0.0005
0.012
0.007
1.5
1.0
G = 20
0.0010
0.013
0.008
N2
0.0015
0.014
3.0
Correction added to
Viscosity, c.p.
0.016
0.5
G = 06
5
10
15
Mole per cent H2S
20
30
40
50
60
70
Molecular Weight
80
90
100
Figure 8 Viscosity of gases at atmospheric pressure.6
6.0
5.0
Viscosity at operating temperature

and pressure, centipoises
A =
Viscosity at 14.7 psia (1atm) and

operating temperatures, centipoises
4.0
Viscosity, / A
3.5
3.0
20
15
2.5
2.0
6
3
1.5
Ps
eu
do
red
10
uc
ed
Pre
s
su
re,
P
2
1
1.0
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
Pseudoreduced Temperature, TR
Figure 9 Viscosity ratio vs pseudo reduced temperature and pseudo reduced pressure.
26/11/15
21
5.2 Viscosity of Mixtures
Another formula that is used for mixtures is:

mix
j y j M j
y j M j
(30)

j=1, n
where:
y j = mole fraction of jth component

M j = molecular weight of component
j = the viscosity of jth component

n = number of components
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 200F 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 150F and 3500 psia (ref ex 4, 7, & 8).
EQUATIONS OF STATE
6.1 Other Equations-of-State
As indicated at the start of section 2 the compressibility factor evolved out of the need
to use an equation derived from ideal gas behaviour and incorporating a correction
factor to suit real gas behaviour. One of the difficulties of the compressibility equation:

22
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 1873
The well known van der Waals equation was one of the earliest equations to represent
the behaviour of real gases. This most basic EOS, 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
A=
(32)
where
aP
bP
and B =
2
( RT )
RT
(33)
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, Tc is the critical isotherm and T2 gives the
isotherm below the critical temperature.
26/11/15
23
T1>Tc
Tc
Psat
T2<Tc
Vsat (liq)
Vsat (vap)
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:
2P
P
=0
=

V T = Tc V 2 T = T
c
That is the critical isotherm exhibits a horizontal inflection point at the critical point.
The application of these conditions to the van de Waals equation yields:

a=
RT
27 R2 Tc2
and b =
64 Pc
8 Pc
EXERCISE 17.
Calculate the critical constants for n- heptane.
24
(34)
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)
This 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.
PT Bo RT Ao Co / T 2 bRT a
+
+
+
V
V2
V3
a
C
+ 3 o 2 1 + 2 exp 2
6
V
V
V T V
P=

(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
Numerous 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 with the Redlich and
Kwong equation which involves only two empirical constants.

P =
RT
a( T )
V - b 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(T ) =
ac
TR

By applying the limiting condition at the critical points yields values of ac and b
related to critical constants. Such that ;
26/11/15
25
ac = 0.42748
R2 Tc2
RT
and b = 0.08664 c
Pc
Pc
(37)
6.5 Soave, Redlich Kwong equation
Soave modified the Redlick-Kwong (RK) equation and replaced the a/T0.5 term with
a temperature dependent term aT where aT = aca.
The Soave, Redlich-Kwong (SRK) equation is therefore:

P=
where
RT
ac
(V b) [V (V + b)]
(38)
a is a non dimensionless temperature dependent term which has a value of 1.0 at the
critical temperature.

a is obtained from
= 1 + m 1 Tr
)]
where m = 0.480 + 1.574 - 0.176 2

8
where is the Pitzer accentric factor .
6.6 Peng Robinson Equation of State
Peng 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.
RT
ac
V b [V (V + b) + b(V b)]
P=
ac = 0.457235
and
R2 Tc2
RT
and b = 0.0778 c
Pc
Pc
(39)
(40)
a is the same function as for the Soave equation except the a 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
26
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 Mixtures
When 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 j b j and a =
yi yj ai aj 1 kij (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.
EXERCISE 18.
A PVT cell contains 0.01 cu ft ( 300cc) of gas with a composition of : methane 0.67
mol.frac, ethane 0.235 and n-butane 0.05. The temperature is increased to 300C.
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
26/11/15
27
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
n
where n
m
M
= nRT
= m/M
= number of moles
= mass
= molecular weight
= PMV/RT
m=
lb
(3cuft )
lbmole
10.73 psia.cuft 530 o R
(
)
lbmole.o R
(1000 psia)16.04
Mass of methane, m = 8.46 lb
EXERCISE 2.
Calculate the density of the gas in the cylinder in exercise 1.
SOLUTION
g =
g =
28
MW.P
RT
(1000 psia)16.04
lb
lbmole
10.73 psia.cuft 530 0 R

(
)
lbmole.oR
Density of gas, g = 2.82
lb
cu. ft.
EXERCISE 3.
Assuming methane is at the conditions of exercise 1, calculate the volume the gas
would occupy at standard conditions.
SOLUTION
P1V1
PV
P V
= 2 2 = sc sc
T1
T2
Tsc
Vsc =
P1 Tsc V
Psc T1
Vsc =
1000 psia 520 o R

x3ft 3
o
14.7 psia 530 R
Vsc = 200.23 scf
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.
SOLUTION
Gas
Components
1 Methane
2 Ethane
3 Propane
Totals
A
Weight
25
3
1.5
29.05
B
Mol weight
16.04
30.07
44.09
C
lb moles
D
Mole fraction
1.559
0.100
0.034
0.921
0.059
0.020
1
EXERCISE 5.
What is the apparent molecular weight of the gas in exercise 4
26/11/15
29
SOLUTION
Gas
Components
1
2
3
A
Mol weight
mw
16.04
30.07
44.09
Methane
Ethane
Propane
B
Mol fraction
yi
0.921
0.059
0.020
1.000
C
A*B
14.77
1.77
0.89
17.43
Apparent Molecular weight= 17.43
EXERCISE 6.
What is the gas gravity of the gas in exercise 4 ?
SOLUTION
g =

Mg
M
= g
Mair 29
Mg = AMW = 17.43

Gas gravity = 0.6
EXERCISE 7.
Calculate the pseudo critical temperature and pseudocritical pressure of the mixture
in exercise 4 .

SOLUTION
Gas
Components
1
2
3
Methane
Ethane
Propane
Total
A
B
Mol weight Mole fraction
mw
16.04
30.07
44.09
yi
0.921
0.059
0.020
1.0
C
Pc-psi
667.00
708.00
616.00
D.
Tc R
344
550
666
Ppc
614.3
41.7
12.4
668.4
Tpc
316.81
32.42
13.39
362.6
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
o
F and a pressure of 3500psia.
30
SOLUTION
Ppr = P/Ppc, Tpr = T/Tpc

From exercise 7 Ppc = 668 psia, Tpc = 362.6R
P = 3500 psia, and T = 150F ie. 610R
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
Gas
Components
1
2
3
4
5
6
Methane
Ethane
Propane
Hydrogen
sulphide
Carbon
Dioxide
Nitrigen
Total
Wgt
Mol
fraction weight
lb moles
Mole
fraction
Pc-psi
Tc R
Ppc
psia
Tpc
255.70
26.17
10.81
28.25
25
3
1.5
3
0.56
0.07
0.03
0.07
16.04
30.07
44.09
34.08
0.035
0.002
0.001
0.002
0.743
0.048
0.016
0.042
667.00
708.00
616.00
1306
344
550
666
673
495.8
33.7
10.0
54.8
10
0.22
44.01
0.005
0.108
1071
548
116.1 59.38
2.5
45
0.06
1.00
28.02
0.002
0.0466
0.043
1.000
493
227
21.0
731
9.66
390
From Wichert & Azis chart for compositions of H2S and CO2 e = 19
Tpc = Tpc - = 371o R

ppc =
Weight
ppcTpc
Ppc = 694.3
EXERCISE 10.
Express the quantity of 1 lb mole of a gas as standard cubic feet.
SOLUTION

26/11/15
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.
31
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
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 200F and a pressure of
one atmosphere.
32
SOLUTION
Gas
Components
Mol Weight Mole fraction

yj
16.04
30.07
44.09
Methane
Ethane
Propane
mix =
0.921
0.059
0.020
1.000
Viscosity
from fig 7
j
0.013
0.0112
0.0098
Mj
yjMj
jyjMj
4.0050
5.4836
6.6400
SUM
3.6884
0.3233
0.1335
4.1451
0.0470
0.0036
0.0013
0.529
j y j M j
y j M j
mmix = 0.0529/4.1451
mmix =0.01275 cp
EXERCISE 15.
Use the gas gravity method to calculate the viscosity of the gas in exercise 4
SOLUTION
Gas
Components
Methane
Ethane
Propane
Mol Weight Mole fraction

mw
yj
16.040
30.070
44.090
0.000
0.921
0.059
0.020
1.000
14.7720
1.773
0.886
17.431
gg=AMW/Mair gg=AMW/29 Temperature = 150F

Mol weight air =
29.000
AMW of gas = 17.431
Gas Gravity = 0.601
mg =

0.01265 from Figure 8
EXERCISE 16.
Determine the viscosity of the gas in exercise 4 at 150oF and 3500 psia (ref ex 4, 7, &8)
26/11/15
33
SOLUTION
From exercise 7
Ppc = 668.4
Tpc = 362.6
3500
= 5.24
Pr = P P =
pc
668.4
610
Tr = T T =
= 1.68
pc
362.6
From Lee correlation

m / matmos = 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 Pc = 397 psia

Using equations above a = 115,872 cu ft 2/lb mole
and b = 3.2878 cu ft./lb mole
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 300C. 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
34
SOLUTION
Calculate the constants a and b for each component
ac = 0.42748
R2 Tc2
RT
and b = 0.08664 c
Pc
Pc
where m = 0.480 + 1.574
= 1 + m 1 Tr
- 0.176
)]
300
Vm
1.25
Components
TcR)
Methane
0.67
344
667
0.4759
8732
0.0104
0.4964
0.5757
5027 0.3212
ethane
0.235
550
708
0.7222
21028
0.0979
0.6324
0.7906
16625 0.1697
kijkijkijaijmethane ethane butane methane
aijethane
aijbutane
SUM
2256.5
1439.4
534.6
4230.5
534.6
344.5
131.9
1011.0
759.7
R
0.05
n-butane
methane
ethane
n-butane
pressure
Pc
766
551
0.02
0.01
0
bj
1.2924
1439.4
ac
52411
918.1
0.1995
344.5
0.7870
1.0065
a=a*
52751 0.0646
0.5556
2702.0
7943.4
8219
Now calculate the mixture values.
b = y j b j and a = yi y j ai a j 1 kij
j
where a ij = (1-k ij )(a i a j )0.5
Now use SRK to calculate pressure.

P=
RT
ac
(V b) [V (V + b)]
Vm = 1.25 cu ft / lb mole
b = 0.622 a c = 10201.9
P = 8617.6 psia
26/11/15
35
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 Zs 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
36

5 - Behaviour of Gases

Uploaded by

Copyright:

Available Formats

5 - Behaviour of Gases

Uploaded by

Document Information

Original Description:

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

5 - Behaviour of Gases

Uploaded by

Copyright:

Available Formats

What are the main laws relating to ideal gases?

What are the main laws relating to ideal gases?

What is the equation of state for real gases and what does 'Z' represent?

What is the equation of state for real gases and what does 'Z' represent?

Behaviour of Gases F I V E

Define the pseudocritical pressure and psuedocritical temperature and be

Express and calculate reservoir gas volumes in terms of standard cubic

Out of these assumptions come the following equations.

1.1 Boyles Law

At constant temperature the pressure of a given weight of a gas is inversely proportional

P = pressure, V = volume, T = temperature.

1.2 Charles Law

The pressure and temperature in both laws are in absolute units.

Institute of Petroleum Engineering, Heriot-Watt University

1.3 Avogadros Law

one lb mole of methane CH4 = 16 lb

1.4 The Equation of State for an Ideal Gas

R is the constant when the quantity of gas is equal to one mole.

R in oilfield units = 10.732

Table 1 Values of R for different unit systems.

For n moles the equation becomes:

T= absolute temperature oK or oR where

1.5 The Density of an Ideal Gas

1.6 Standard Conditions

Institute of Petroleum Engineering, Heriot-Watt University

res - reservoir conditions

1.7 Mixtures of Ideal Gases

1.7.1 Daltons Law of Partial Pressures

where A, B and C are components.

where yj = mole fraction of jth component.

1.7.2 Amagats Law

1.8 Apparent Molecular Weight

If yj represents the mole fraction of the jth component:

AMW for air = 28.97, a value of 29.0 is usually sufficiently accurate.

Institute of Petroleum Engineering, Heriot-Watt University

1.9 Specific Gravity of a Gas

Assuming that the gases and air are ideal.

Mg = AMW of mixture, Mair = AMW of air.

2 BEHAVIOUR OF REAL GASES

2.1 Compressibility Factor for Natural Gases

Figure 1 Typical plot of the compressibility factor as a function of pressure at constant

2.2 Law of Corresponding States

Institute of Petroleum Engineering, Heriot-Watt University

Tpc = y jTcj and Ppc = y j Pcj

Compressibility Factors for Natural Gases as a

Pseudo Reduced Pressure, Pr

Pseudo Reduced Temperature

Pseudo Reduced Pressure, Pr

Institute of Petroleum Engineering, Heriot-Watt University

2.3 Pseudocritical Properties of Natural Gases

Pseudocritical Pressure, psia

Gas Gravity (air = 1)

Figure 3 Pseudocritical properties of natural gases. 3

2.4 Impact of Nonhydrocarbon Components on z value.

PER CENT C02

PER CENT H2S

Figure 4 Adjustment factors for pseudocritiacl properties for non hydrocarbon

Institute of Petroleum Engineering, Heriot-Watt University

Tpc = Tpc - = 371o R