PE 459 LECTURE 2- natural gas basic concepts and properties

sarkmank1 337 views 108 slides May 09, 2024
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About This Presentation

This introduces to the concept of natural gas properties and applications


Slide Content

Natural gas engineering
Mr. K. Sarkodie

Properties of natural gas

Natural gas properties
•The overall physical properties of a natural gas
determine the behavior of the gas under
various processing conditions

Introduction
A gas is a homogeneous fluid
No definite volume
Completely fills the vessel in which it is
contained
Behaviour vital to petroleum engineers
Simple gas laws straightforward
Hydrocarbon gases at reservoir conditions are
more complicated.

Ideal Gases
Assumptions
Volume of molecules are insignificant with
respect to the total volume of the gas.
There are no attractive or repulsive forces
between molecules or molecules and
container walls.
No internal energy loss when molecules
collide

Ideal Gases
Boyle’s Law
1
V
P
α or
PV constant=
T is constant
P = pressure, V = volume, T = temperature

Ideal Gases
Charles’ Law
VTα
or
V
constant
T
=
P is constant
Pressure and temperature in both laws are in
absolute units

Absolute Units
Temperature
Kelvin K =
o
C + 273 Kelvin K =
o
C + 273 Kelvin K =
o
C + 273
Rankin
o
R =
o
F + 460

Avogadro’s Law
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.
2.73 x 10
26
molecules/lb.mole of ideal gas
1 lb.mole of any ideal gas at 60
o
F and 14.7
psia. occupies 379.4 cu.ft.
1 gm.mole at 0
o
C and atmos. pressure
occupies 22.4 litres

lb.mole
One lb.mol of methane CH
4 = 16 lb.
One kg.mole of methane CH
4 = 16 kg.

Ideal Gas Law
The Ideal Equation of State
Combining Boyle’s Law and Charles
Law gives an equation relating P,T & V PV
constant
T
=
Constant is termed R when quantity of gas is one mole
R is termed Universal Gas Constant

Universal Gas Constant
o
cu.ft.psia
R 10.73
l b.mol. R
=
o
cu.ft.psia
R 10.73
lb.mol. R
=
psfta

The Ideal Equation of State
For n moles equation becomes
A useful equation to compare conditions at two
conditions 1 & 2
PV = nRT
PV
n=
RT
therefore
11 22
12PV PV
TT
=

Density of an Ideal Gas
ρ
g is the gas density
g
m
weight/volume
V
ρ= =
For 1 mole m = MW MW= molecular weight
RT
V
P
=
g
MW.P
RT
ρ=

Standard Conditions
Oil and gas occur under a whole range of
temperatures and pressures
Convenient to express volumes at a reference
condition.
Common practise to relate volumes to surface conditions. 14.7 psia and 60
o
F

res res SC SC
res SC
PV PV
TT
=
res - reservoir conditions
SC - standard conditions
This equation assumes ideal behaviour. This
is NOT the case for real reservoir gases

Mixtures of Ideal Gases
Petroleum gases are mixtures of gases - Dalton’s Law
and Amagat’s Law
Dalton’s Law of Partial Pressures
Total pressure is the sum of the partial pressures
ABCD
P P P P P .........=++++
Therefore
ABC
RT RT RT
P n n n .....
VVV
=+++ i.e. j
RT
Pn
V
=

Therefore jj
j
Pn
y
Pn
= =
y
j =mole fraction of j
th
component

Amagat’s 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.
Law of additive volumes.
ABCV V V V ....=+++
ABC
RT RT RT
V n n n ...
PPP
=+++
i.e. j
RT
Vn
P
=

jj
jVn
y
Vn
= =
For ideal gas, volume fraction is equal to mole fraction

Apparent Molecular Weight
A mixture does not have a molecular weight.
It behaves as though it has a molecular weight.
Called Apparent Molecular Weight. AMW
jj
AMW yMW= ∑
MWj is the molecular weight of component j.
AMW for air = 28.97

Specific Gravity of a Gas
The specific gravity of a gas is the ratio of the density of
the gas relative to that of dry air at the same conditions.
g
g
air
ρ
γ=
ρ
Assuming that the gas and air are ideal
g
gg
g
air air
MP
MM
RT
MP M 29
RT
λ= = =
Mg = AMW of mixture, Mair = AMW of air

Behaviour of Real Gases
Equations so far for ideal gases
At high pressures and temperatures the volume of
molecules are no longer negligible
and attractive forces are significant.
Ideal gas law is therefore NOT applicable to light
hydrocarbons.
Necessary to use more refined equation.
Two general methods.
 Using a correction factor in equation PV=nRT
 By using another equation of state

Correction Factor for Natural Gases
A correction factor ‘z’ , a function of gas
composition, pressure and temperature is
used to modify ideal equation.
‘z’ is the compressibility factor PV znRT=
Equation known as the
compressibility equation of state.
‘Z’ is not the compressibility

Compressibility factor
To compare states the equation now takes the form
res res SC SC
res res SC SC
PV PV
zT zT
=
Z is an expression of the actual volume to
what the ideal volume would be. i.e.
To
Z = V
actual / V ideal

Compressibility factor

Law of Corresponding States
Law of corresponding states shows that the properties of
pure liquids and gases have the same value at the same
reduced temperature, T
r and reduced pressure, Pr.
r
c
T
T
T
=
and
r
c
P
P
P
=


T
c and Pc are the critical temperature and critical
pressure.
The compressibility factor follows this law.
Presented as a function of T
r, and Pr.

Law of Corresponding States as
Applied to Mixtures
The law of corresponding states does not apply to
hydrocarbon reservoir fluids.
The law has been modified to be used for mixtures
by defining parameters
Pseudo critical temperature, Tpc
and
Pseudo critical pressure, Ppc .
PC j cjT yT=∑
and
PC j cj
P yP=∑
Tcj and Pcj are the critical temperatures and
pressures of component j.

Pseudo critical temperature, Tpc and
Pseudo critical pressure, Ppc .
These pseudo critical temperatures and pressures
are not the same as the real critical temperature and
pressure.
By definition they must lie between the extreme
values of the pure components making up the
mixture.
Gas A B C D Ppc Tpc
Component Mol. WghtMol. Frac.Pc-psiTc-oR
Methane 16.04 0.921 667 344 614.3 316.8
Ethane 30.07 0.059 708 550 41.8 32.5
Propane 44.09 0.02 616 666 12.3 13.3
Total 1 668.4 362.6
Pseudo critical pressure = 668.4 psia
Pseudo critical temperature = 362.6 oR

Real Critical Pressures and Temperatures
These are not
average values
based on mole
fractions.
Averaged on
weight fraction
basis would give a
more real value.
Critical pressure much greater
than critical points of pure
components.
Particularly when methane is
involved.

Compressibility Factors for Natural
Gases
These are presented as a function of
pseudoreduced pressure, P
pr and
pseudoreduced temperature, T
pr.
PR
PC
T
T
T
=
and
PR
PC
P
P
P
=

Compressibility
Factors for
Natural Gases
(Standing & Katz)
From previous exercise
Ppc=668psia and Tpc =362
o
R
Z value for this mixture at
3500psia and 150
o
F
Ppr = 5.24 and Tpr = 1.68
Z=0.88

Pseudocritical Properties for Natural
Gases
Can be calculated
from basic
composition.
If data not available
can use correlations.
Do not use
composition to
calculate gravity and
hence Ppc & Tpc.

Impact of Nonhydrocarbons on ‘Z’ value
H
2S and CO
2 have significant impact on ‘z’.
Wichert & Aziz have developed equation to enable
correction.
'
pc pc
TT= −ε
and
( )
'
pc pc'
pc
pc H2S H2S
PT
P
T y 1y
=
+ −εε obtained from Wichert & Axix paper

Impact of Nonhydrocarbons on ‘Z’ value

Standard Conditions for Real Reservoir
Gases
Standard volumes are used to describe
quanitities of gas in the industry.
Standard cubic feet
Standard cubic metre.
Determined at standard temperature and
pressure.
60
o
F(15.6
o
C) & 14.7psia (1 atmos)
It is useful to consider a mass of gas in terms
of standard volumes.

Gas Formation Volume Factor
We need a conversion factor to convert
volumes in the reservoir to those at surface (
standard) conditions.
Termed Formation Volume Factors.
Gases, Gas Formation Volume Factor, Bg.
Is the ratio of the volume occupied at reservoir
conditions to the volume of the same mass
occupied at standard conditions.
g
volume occupied at reservoir temperature and pressure
B
volume occupied at STP
=

Gas Formation Volume Factor
Definition
Gas Formation Volume Factor is the volume
in barrels (cubic metres) that one standard
cubic foot (standard cubic metre) of gas
willoccupy as free gas in the reservoir at the
prevailing reservoir pressure and
temperature.

Gas Formation Volume Factor
Units:
Bg - rb free gas / SCF gas
Bg - rm
3
free gas / SCM gas

Gas Formation Volume Factor
Using equation of state PV znRT=
and
res res SC SC
res res SC SC
PV PV
zT zT
=
SC R RR
g
SC R SC SC
PTzV
B
V PTz
= =
Z at standard
conditions = 1.0
Reciprocal of Bg often used to reduce risk of misplacing
decimal point as B
g is less than 0.01
g
1 volume at surface
E
B volume in formation
= =
E is referred to as ‘Expansion Factor’

Gas Formation Volume Factor
R
g
SC
V
B
V
=
R
znRT
V
P
=
SC SC
SC
SC
z nRT
V
P
=
Z at standard conditions = 1.0
Therefore
SC
g
SCPT cu.ft
Bz
T P scf
=
Since Tsc=520
o
R and Psc= 14.7 psia for most cases
g
zTres.bbl
B 0.00504
P scf
=

Coefficient of Isothermal Compresibility of
Gases
Compressibility factor ,z, is not the compressibility.
Compressibility, cg, is the change in volume per unit
volume for a unit change in pressure.
m
g
m
1V 1V
c or
VP V P
∂∂  
=−= −
  
∂∂  
For an ideal gas PV nRT= or
2
dV nRT
dP P

=−


g 22
1 nRT P nRT 1
c = =
V P nRT P P

=−− − −



Viscosity of Gases
Viscosity is a measure of resistance to flow.
Units: centipoise - gm./100 sec.cm.
Termed: dynamic viscoisty.
Divide by density.
Termed kinematic viscosity
Units: centistoke -cm
2
/100sec

Viscosity of Gases
Gas viscosity reduces as pressure decreases
At low pressures, increase in temperature increases viscosity.
At high pressures, increase in temperature decreases viscosity.

Viscosity
of Gases
At low pressures
viscosity can be
obtained from
correlations.
Viscosity of pure components at 1 atmos.

Viscosity of Gases
At low pressures viscosity can be obtained from
correlations.
Viscosity of
gases (MW) at
atmospheric
pressure.

Viscosity of Gases
Carr presented a method to determine viscosity at higher
pressure and temperature.
 Uses pseudo reduced temperature and pseudo reduced
pressure.
Viscosity Ratio
µ/µ
atmos

Other equations of state, EOS
The ‘z’ factor is used to modify the ideal EOS
for real gas application.
PV=znRT
Rather than use this correction factor other
equations have been developed.
An irony is that many of these advanced
equations are used to generate ‘z’ for use in
the PV=znRT equation.

Van de Waal EOS, 1873
( )
2
a
P V b RT
V

+ −=


Two corrective terms used to overcome limiting
assumptions of ideal gas equation.
Internal pressure or cohesion term a/V
2
.
Co-volume term b. Represents volume
occupied by one mole at infinite pressure.
Can also be written as
32 RT a ab
Vb V V 0
P PP
  
−+ + −=
  
  
Termed cubic equations of state

Van de Waal EOS
When written to solve for ‘z’ becomes
()
32
Z Z 1 B ZA AB 0− ++−=
where
()
2
aP
A
RT
= and
bP
B
RT
=
Values for a and b are positive constants
for particular fluids.

Van de Waal EOS
Equation can be used to plot various P vs. V isotherms
T1>Tc is the single phase isotherm.
Tc is the critical isotherm.
At the critical point, for a pure
substance.

2
2
T Tc T Tc
PP
0
VV
= =
∂∂
= =
∂∂ 
This yields
22
c
cc
RT27 RT
a and b
64 P 8P
= =
T2<Tc is the two phase isotherm.

Benedict-Webb-Rubin EOS,BWR - 1940
Van de Waals equation not able to represent gas
properties over wide range of T&P.
BWR equation developed for light HC’s and found
application for thermodynamic properties of natural
gases
Constants which need to be determined by experiment
For mixtures mixing rules required.
o
oo 2
2 3 6 32 2 2
C
B RT A
RT bRt a a c
T
P 1 exp
V V V V VT V V
−−
−α γ γ
  
=+ + ++ + −
  
  
oooB , A , C , a, b, c, , and αγ

Redlich-Kwong EOS, 1949
Numerous equations with increasing number
of constants for specific pure components.
More recently a move to cubic EOS.
( )
1/2
RT a
P
Vb TVVb
= −
−+
The term a and b are functions of temperature
At the critical point
22
cc
cc
R T RT
a 0.42748 and b 0.08664
PP
= =

Soave,Redlich-Kwong EOS , SRK, 1972
Soave modified RK equation and replaced
a/T
0.5
term with a temperature dependant term
a
T.
a
T =a
cα ( )
caRT
P
V b VV b
α
= −
−+
α is a non dimensionless temperature dependent term.
Value of 1 at critical temperature
α is from ( )
2
r
1 m1 Tα= + −

where
2
m 0.480 1.574 0.176= + ω− ω
ω is the Pitzer
accentric factor from
tables

Peng Robinson EOS , PR, 1975
Peng and Robinson modified the attractive
term.
Predictions of liquid density are improved.
( )( )
caRT
P
Vb VV b bV b
α
= −
−++ −

22
c
c
c
RT
a 0.457235
P
=
and
c
cRT
b 0.0778
P
=
α is the same as for the SRK equation, except w function is
different.
2
m 0.37464 1.54226 0.26992= + ω− ω

Widely Used EOS
SRK and PR equations are widely used in the
industry.
Used in simulation software to predict
behaviour in reservoirs, wells and processing.
There are other EOS.
Reluctance to change because of investment
in associated parameters.

Gas Specific Gravity
•Gas specific gravity, gg, as commonly used in
the petroleum industry,
•is defined as the ratio of the molecular weight
of a particular natural gas to that of air

Specific Heat
•defined as the amount of heat required raising
the temperature of a unit mass of a substance
through unity.
•It is an intensive property of a substance.
It can be measured at constant pressure (Cp), or
at constant volume (Cv), resulting in two
distinct specific heat values.
In terms of basic thermodynamics quantities,

Joule-Thomson Effect
•The Joule-Thomson coefficient is defined as
the change in temperature upon expansion
which occurs without heat transfer or work
and is expressed with the formula,


In terms of reduced quantities the above
equation becomes,

Phase behavior of gases

DRY GAS RESERVOIRS
•The phase diagram of a dry gas reservoir is as shown
below:

WET GAS RESERVOIRS
•The phase diagram of a wet gas reservoir

GAS CONDENSATE RESERVOIRS

GAS RESERVE ESTIMATION

Reserves
•reserves are the amount of technically and
economically recoverable oil. Reserves may be
for a well, for a reservoir, for a field, for a
nation, or for the world. Different
classifications of reserves are related to their
degree of certainty.

Reserve estimation methods
•Analogy
•Volumetric
•Decline curve analysis
•Material balance

Volumetric

Decline curve analysis

ASSIGNMENT

GAS IN PLACE ESTIMATES FOR DRY
GAS RESERVOIRS
•GAS MATERIAL BALANCE
a) Volumetric depletion reservoirs
The term volumetric depletion, or simply depletion,
applied to the performance of a reservoir means
that as the pressure declines, due to production,
there is an insignificant amount of water influx
into the reservoir from the adjoining aquifer.
An expression for the hydrocarbon pore volume can
be obtained from equ.
HCPV = Vφ (1−Swc) = G/Ei

•where G is the initial gas in place expressed at
standard conditions.
• The material balance, also expressed at
standard conditions, for a given volume of
production Gp, and consequent drop in the
average reservoir pressure Δp = pi−p is then,

Gas Condensate Material Balance
•Case I - Reservoir Pressure above Dew-point
Pressure
where a2=5.615 ft3/bbl for
field units and a2=1 for pure
SI units

•In summary, the material balance for a
volumetric gas condensate reservoir above
the dew point is,

Case II - Reservoir Pressure Below Dew point
Pressure
•From the data obtained, a so called two-phase z-
factor (z2) is calculated assuming that the gas
condensate reservoir depletes according to the
material balance of a gas condensate reservoir
above the dew-point.
•The p/Z2 vs. Gpw for a volumetric gas condensate
reservoir is a straight line obtained from the gas
material balance:

GAS MBE
EXERCISE

•A dry gas reservoir has produced as follows:

Data
Reservoir Temperature T= 100°F
Gas Gravity SG= 0.68
1. Determine the original pressure and original gas
in place.
2. What will be the average reservoir pressure at
the completion of a contract calling for delivery
of 20 MM SCFD for 5 years (in addition to the
9,450 MM SCF Produced to 11-Jan-69?)

FIGURE 1

FIGURE 2

FIGURE 5

SOLUTION
•To construct the graphical material balance
plot we must first determine the P/Z values.
•Using Figures 1 and 2 below and the gas
gravity of: SG= 0.68
•The pseudo-critical parameters are found to
be:
Pseudo critical pressure (psia) Ppc= 667.5 psia
Pseudo critical temperature (oR) Tpc= 385.0 °R

FIGURE 1B

•From the straight line of Figure 1a
Slope = -0.0522
Intercept = 4,448.3
Equation = P/Z= -0.0522 Gp + 4448.34 (1a
Initial Pressure
From Figure 1a and b, at Gp = 0:
Pi/Zi= 4,448.3 psia
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