Advanced Thermodynamics
Course Administration
1
Advanced Thermodynamics
Unit Structure
The Virial Equation of States
Introduction
1
basic thermodynamic relations; phase
behaviour and phase diagrams
Cubic Equation of States
5
Advanced Equation of States
Process Thermodynamic
2
3
4
Applications of basic concepts to: Gas
compression; Refrigeration; Gas
liquefaction and Power generation cycles
Assignment: hydrogen liquefaction
Corresponding States and
Activity Coefficient Models
Phase Equilibrium,
and Flash Calculations
6
Models for Flow Assurance:
Solids and Hydrates
Thermodynamic Models:
7
Models for Transport Properties
8
Review and Mock Exam
Introduction: Statistical mechanics; the
partition function; the perfect molecular gas
and the intermolecular pair potential
2
Other information
Unit Delivery:
2hrs * 11 lectures + 2hrs
* 10 practical sessions
•
•
•
Unit Assessment:
 2 hour final exam: 60 %
 1 assignment (group): 20 %
 1 in-class test: 20 %
Advanced Thermodynamics
Tools:
Multiflash, Refprop
Excel Sheets and
Aspen Hysys
In-class test: Mixture of Multiple Choice and Calculation Qs
Assignment: process design of hydrogen liquefaction
Lecture recordings
• Lectures will be recorded
• Recording of practical sessions is not possible in the
computer lab
• 2013 Lectures available on LMS
Lecturer & Co-ordinator; Dr Saif Al Ghafri; office 2.08; saif.alghafri@uwa.edu.au
consultation by appointment: arrange via email
3
Text books
Advanced Thermodynamics
Recommended texts (short-hand author reference in bold)
1) Assael, Trusler & Tsolakis. "Thermophysical Properties of Fluids: An Introduction
to their Prediction"
2) Prausnitz, Lichtenthaler, de Azevedo. "Molecular Thermodynamics of Fluid-Phase
Equilibria"
4
Advanced Thermodynamics
Do Accurate Property Models Actually Matter?
Relevance of property models is controversially discussed in
scientific literature
Example: Booster compressor, LNG-like mixture
m1 = 10 kg/s
T1 = 280 K
p1 = 20 MPa
SRK
r1 = 181.7 kg/m3
EOS-LNG
r1 = 191.4 kg/m3 Dr1 = 5.1%
P12
P12 = 2.346 MW
P12 = 2.334 MW
h s = 0.88
p2 = 70 MPa
Methane
Ethane
Propane
Nitrogen
0.90
0.06
0.02
0.02
c2
u2
 DV = -5.1%
Blade
c1
r2
u1
r1
slides courtesy of professor Roland Span
DP12 = -0.5%
w
5
Advanced Thermodynamics
Do Accurate Property Models Actually Matter?
Relevance of property models is controversially discussed in
scientific literature
Example: Pre-cooling of a LNG-like mixture
mLNG = 10 kg/s
pLNG = 10 MPa
T1,LNG = 306 K
T2,LNG = 236 K
T2,Cool = 200 K
pCool = 1 MPa
Methane
Ethane
Propane
Nitrogen
0.90
0.06
0.02
0.02
Ethane
0.70
n-Butane
0.30
T1,Cool = 280 K
SRK
mCool = 5.11 kg/s
EOS-LNG
mCool = 5.28 kg/s Dm = 3.1%
The more detailed process simulations are,
the more relevant accurate property models become!
slides courtesy of professor Roland Span
6
Advanced Thermodynamics
The Laws of Thermodynamics
Some slides courtesy of professor Martin Trusler-ICL
7
Advanced Thermodynamics
What is Thermodynamics
An experimentally-based science that, without reference to
the microscopic nature of matter, establishes the
relationships among the variables describing systems at
equilibrium.
Laws of Thermodynamics are observed facts, not theories
based on models. The science of Thermodynamics is the
logical framework based on the application of these laws.
According to Einstein, Thermodynamics is:
“the only physical theory…that will never be overthrown…”
8
Advanced Thermodynamics
Compression Work: Closed System
• Gas contained within pistoncylinder assembly
pext
dx
pext +
dpext
• Massless and frictionless piston
• External pressure pext keeps
piston stationary: force is pextA
V
W -
V-dV

V2
V1
pext dV
• Increase pext by dpext and piston
moves down by dx.
• Volume changes by dV = -Adx.
• Work done is dW  - pext dV
9
Advanced Thermodynamics
Reversible Compression
m
pa
pext
• If the piston is frictionless and the
compression is carried out slowly
then:
• p = pext throughout process and
• Work is:
V2
W -
p
p
(a)
(b)

V1
p dV
• Process said to be reversible
• In (b) pext = mg/A + pa and have basis
for primary pressure measurement
10
Advanced Thermodynamics
First Law of Thermodynamics
• Energy may be transferred to a closed system from
the surroundings as heat Q and/or work W
• Conservation of total energy requires that the system
energy increases by the total amount of energy
transferred from the surroundings no matter how it is
transferred
• It follows that there exists a state function U such
that:
DU  Q  W

 First Law
dU  dQ  dW

11
Advanced Thermodynamics
Enthalpy
Constant
pressure p
Freely-sliding
piston
• For a process in an isobaric
system resulting in volume
change DV, work is: W = -pDV
• First law gives: DU = Q - pDV
• Define enthalpy H = U + pV
Isobaric
closed
system
Q
• DH = DU + D(pV), so:
• DH = Q for the isobaric process
• DU = Q for an isochoric process
If no non-pV work
12
Advanced Thermodynamics
Shaft Work: Steady-Open System
Shaft work Ws
Stationary Open system
1
2
p1
Heat Q

-
W  p1V1 - p2V2  Ws  -
⇒ Ws  - p1V1  p2V2
p2
Flow Device
V2
V1
V2
V1
Moving Closed system
p dV
p dV
Ws 

p2
p1
V dp
13
Advanced Thermodynamics
First Law for a Steady-Open System
Shaft work Ws
1
p1
Stationary Open system
2
Flow Device
Heat Q
p2
Moving Closed system
DU  Q  W
 Q  Ws  p1V1 - p2V2
 Q  Ws - D( pV )
hence
DH  Q  Ws
14
Advanced Thermodynamics
Reversible and Irreversible Processes
• Reversible Process: one that can be
made to retrace its path exactly leaving
both system and surroundings
indistinguishable (or at most infinitesimally
different).
• Irreversible Process: one that is not
reversible.
15
Advanced Thermodynamics
Example: An Irreversible Cycle
m
m
A
m
A
A
m
B
m
B
p1
V1
m
B
p2
V2
p1
V1
16
Advanced Thermodynamics
Work for Irreversible Compression
W -

V2
V1
pext dV
Compressions with Tfinal = Tinitial
(a)
pext,2
(c)
(b)
pext,2
pext,2
pext
pext
pext
pext,1
pext,1
V2
V1
V
pext,1
V2
V1
V
V2
V1
V
17
Advanced Thermodynamics
Work for Irreversible Compression
Conclude that:
•
If the piston-cylinder assembly is part of a
compressor, the mechanical power requirement
will be lowest if the process operates reversibly;
and
•
If the piston-cylinder assembly is part of an
engine, we will get more useful work from the
engine if we can operate it reversibly.
18
Advanced Thermodynamics
Reversible Compression
What are the criteria for a compression/expansion
to be reversible?
•
External pressure pext = p throughout the
process
•
No friction
•
Process slow compared with time to re-establish
full thermodynamic equilibrium
19
Advanced Thermodynamics
The Second Law
It is observed that all heat engines have efficiency less than
unity and that all refrigerators require work input. These
observations (and others) are consistent with the second law:
“It is impossible to construct a machine operating in a cyclic
manner whose sole effect is to absorb heat from its
surroundings and convert it into an equivalent amount of
work.” (Kelvin/Planck)
“It is impossible to construct a machine operating in a cyclic
manner which can convey heat from one reservoir at a lower
temperature to one at a higher temperature and produce no
effect on any other part of the surroundings.” (Clausius).
20
Advanced Thermodynamics
The Second Law
The second law can be summarised mathematically by
means of the fundamental equation:
dU = TdS – pdV
and the fundamental inequality:
DSad  0.
(adiabatic process)
Together these are equivalent to the statements of the
second law attributed to Kelvin, Planck and Clausius. Thus
they may be used to show that:
d Qrev
DS 
T

h
Thigh - Tlow
Thigh
21
Advanced Thermodynamics
Work of Compression/Expansion
General:
Dh  q  w s
Reversible:
w s  vdp
Isentropic:
w s  vdp and Ds  0


Isentropic perfect gas:
and hence
pv   C
ws   { p2v 2 - p1v1} /( - 1)
 cp (T2 - T1)
And various
other forms
Isothermal perfect gas: ws  RT ln(p2 / p1)
and
q  T (s2 - s1)  -RT ln(p2 / p1)
22
Advanced Thermodynamics
Non-Simple Thermodynamic Paths
•
Often we illustrate thermodynamic theory by consideration of
simple thermodynamic paths such as isothermal, isobaric,
isentropic etc.
•
Real processes follow more complicated paths and we need
to model these somehow.
•
Two approaches are commonly used:
 For reversible processes assume an approximate
analytical formula which fits the initial and final conditions
– this is called a “polytropic path”.
 For irreversible processes start with either adiabatic or
isothermal reversible ideals and apply “efficiency factors”.
23
Advanced Thermodynamics
Thermodynamic Efficiency
•
Irreversible processes cannot be represented by a path
because state variables are not known during the process.
•
In this case, start with an ideal isentropic or isothermal
process connecting p1, T1 and, say, p2.
•
Then define an efficiency factor which relates the actual
work to the work for the ideal reversible adiabatic or
isothermal process as follows.
Adiabatic compressor:
hs = (reversible work)/(actual work)
Adiabatic turbine:
hs = (actual work)/(reversible work)
Isothermal compressor:
hT = (reversible work)/(actual work)
24
Advanced Thermodynamics
Irreversible Adiabatic Paths
Expansion
Compression
h2
h2s
T
p2
p1
p1
h2
h1
p2
T h
2s
h1
S
S
25
Advanced Thermodynamics
Availability (Exergy)
•
•
•
•
In the preliminary design stage of a process, often want to know:
• What is the least work required to bring about a specified
outcome?
• What is the most work that can be extracted from a process
fluid?
Both questions amount to: what is the minimum W.
For some cyclic processes, comparison with Carnot’s cycle is
helpful: e.g. h for a power cycle
In other cases, the concept of availability or, for steady-open
systems, stream availability is more helpful.
26
Advanced Thermodynamics
Availability in a Steady-Open System
•
First law gives: H0 – H1 = Q + Ws
•
Second law gives: S0 – S1  Q/T0
•
Hence: Ws  (H0 – T0S0) - (H1 – T0S1)
•
Define stream availability B as
B = (H1 – T0S1) - (H0 – T0S0)
•
Hence Ws  -B
or ws  -b.
•
Ws = -B if the process is
reversible.
p0, T0
p1, T1
27
Advanced Thermodynamics
Example
What is the maximum work obtainable from N2 in a steady
flow process starting at 200 K and 50 bar, with ambient
conditions at 300 K and 1 bar?
Use thermodynamic tables for the properties of nitrogen:
h1 = 182.38 kJkg-1
h0 = 311.20 kJkg-1
s1 = 5.169 kJK-1kg-1
s0 = 6.846 kJK-1kg-1
b  (h1 - h0 ) - T0 (s1 - s0 )  374.28 kJ  kg-1
Hence gas can deliver up to 374.28 kJkg-1 work output.
28
Advanced Thermodynamics
Key Results
Reversible Work
(closed system)
Reversible Work
v2
w  -  pdv
v1
p2
(steady-open system)
ws   vdp
First Law
Du  q  w
p1
(closed system)
First Law
(steady-open system)
Second Law
Stream availability
Dh  q  w s
du = Tds – pdv
Dsad  0
b = (h1 – T0s1) - (h0 – T0s0)
29
Advanced Thermodynamics
The Fundamental Equation
CLOSED system, surrounded by reservoir at
(Tres, Pres). Combine 1st & 2nd Laws:
DU  Q  W
Q  Tres DS
DU  Tres DS - Pres DV
Restrict to
PDV work
For system undergoing infinitesimal change:
dU  TdS - PdV
For system undergoing reversible process:
dU  TdS - PdV
30
Advanced Thermodynamics
The Thermodynamic Potentials
Four functions that each contain all the
thermodynamic information about a system:
Internal Energy
Enthalpy
Helmholtz Energy
(symbol F sometimes used)
Gibbs Energy
U
H  U  PV
A  U - TS
G  H - TS
Also known as
“free energies” or
“work functions”
Each contains same information, but in different
forms – choose the one that’s convenient
31
Advanced Thermodynamics
Independent Variables
Consider the form of the Fundamental Equation
 U 
 U 
dU  
 dS  
 dV
dU  TdS - PdV
 S V
 V  S
This implies that: (1) S & V are the independent variables
for U and (2) T & P are the first partial derivatives of U
U  U ( S ,V )
 U 
T 

 S V
 U 
P  -

 V  S
We can consider U for a system at equilibrium as a surface and
its thermodynamic properties as the slopes, curvatures, etc. of
that surface; analogous results for H, A, G
33
Potentials & Equilibrium
Advanced Thermodynamics
Fundamental eqn, keeping 2nd law inequality: dU  TdS - PdV
For a process at constant S & V:
dU S ,V  0
At Equilibrium
dU S ,V  0
Similarly:
dH S , P  0
dAT ,V  0
dGT ,P  0
dH S , P  0
dAT ,V  0
dGT ,P  0
An equilibrium
state is a local
minimum of the
potential along
a specified path
These equations characterise what equilibrium is!
They establish requirements for phase equilibria & stability
35
Advanced Thermodynamics
Equilibrium Requirements
Consider 2 parts, & , of an isolated system
At equilibrium:
dU S ,V  0
dU  T
dS
( )
( )
dS
( )
 dS
( )
- P dV
( )
0
isolated system No heat
transfer

const. S
const. V

No work
done
( )
T
( )
dV
dS
( )
( )
( )
- P dV
 dV
( )
( )
0
0
Eliminating dS() & dV():
dU  (T
( )
-T
( )
)dS
( )
- ( P ( ) - P (  ) )dV ( )  0
T ( )  T (  )
Eqbm
requires: P ( )  P (  )
36
Advanced Thermodynamics
The Chemical Potential
From Calculus, the total differential becomes
N
 U
 U 
 U 
dU  
 dS  
 dV   
 S V ,ni
 V  S ,ni
i 1  ni
 U
 U 
U 

T 
 P  -
i  

 S V ,ni
 V  S ,ni
 ni


dni
 S ,V ,n j


 S ,V ,n j
Extensive properties
N
dU  TdS - PdV   i dni
i 1
Fundamental Eqn
for Open Systems
Intensive properties
38
Advanced Thermodynamics
Phase Equilibrium Requirements
1 component, 2 phase system:
Eqbm requires:
T
( )
T
( )
P
( )
P
( )
 ( )   (  )
N component, vapour-liquid system:
vapour
(open system)
N component
isolated system
(V )
i( L)  i(V )
…
liquid
(open system)
The extra N
equations needed to
solve the standard
flash problem
T T
P( L )  P(V )
1( L )  1(V )
( L)
…
mass transfer
Eqbm requires:
N( L)  N(V )
Eqns for an N component,  -phase system?
41
Advanced Thermodynamics
Phase Behaviour and Phase Diagrams
Some slides courtesy of professor Andrew Haslam and
professor Martin Trusler-ICL
42
Advanced Thermodynamics
What is a “Phase”?
A phase is a homogenous region in space where the
intensive properties are the same. i.e. uniform properties.
A heterogeneous system is made up of two or more phases
(homogenous systems).
Gibbs phase rule links the number of independent
properties (e.g. T, p, composition, …) required to completely
specify a mixture at phase equilibrium
F  C 2-
Number of independent
properties specified
(degrees of freedom)
Number of phases at
equilibrium
Number of components in
mixture (pure substance = 1)
43
Advanced Thermodynamics
p-T Projection of Phase Diagram
1000
1000
100
S+L
100
Solid-Liquid (Melting) Curve
Critical Point
10
Critical Point
L
S
1
p /MPa
p /MPa
10
Vapour-Liquid (Boiling) Curve
0.1
Triple Point
0.01
L
1
S
0.1
G
G
L+G
Three-Phase Line
0.01
S+G
Vapour-Solid (Sublimation) Curve
0.001
50
100
150
200
T /K
250
300
0.001
0.01
0.1
1
10
100
v /(dm3/mol)
Point to note: Methane used as example; Lines separate phases; p-T
projection of phase boundaries; semi-log scale1 MPa = 10 bar; Triplepoint tie line; Contracts on freezing; log-log scale for p-v diagram
44
Advanced Thermodynamics
p-T Projection with Isochores and Isotherms
185 K
3
3
L
G
1 mol/dm3
2
4
p /MPa
p /MPa
5
2 mol/dm
S
190.564 K
300 K
200 K
5 mol/dm3
4
3
6
15 10
150 K
3
G
L+G
S+L
5
20 mol/dm
25 mol/dm
3
6
L
2
1
1
0
0
0
75
100 125 150 175 200 225 250 275 300
T /K
5
10
15
20
25
30
3
r/(mol/dm )
• Points to note: Can read off p, r, T values; Methane again on linear
axes; Isochores nearly linear; Sublimation curve too low in pressure
to see on this scaleTie lines; Critical point
45
T-r Projection with Isobars
200
10
4.4992 MPa
10 MPa
3 MPa
180
Advanced Thermodynamics
Two-phase mixture, quality x :
v = (1-x )v l + xv g
Saturated liquid:
v = vl
100 MPa
G
Saturated vapour:
v = vg
p /MPa
T /K
160
1 MPa
1
140
L+G
L
L
S+L
120
0.01
0.1 MPa
S
0.1
0.01
100
0
5
10
L+G
15
20
3
r/(mol/dm )
25
30
35
G
0.1
0.1
0.5
0.9
1
10
3
v /(dm /mol)
Points to note: Can read off p, r, T values; Methane again on
linear axes; Two-phase L+G region; Tie lines; Critical point
46
Advanced Thermodynamics
T-s and p-h Charts
300
4
0.1 MPa
S
L
L+G
G
200 K
p /MPa
T /K
4.4992 MPa
200
190.564 K
300 K
250
1 MPa
185 K
5
10 MPa
120 K
100 MPa
150 K
6
3
2
S
L
L+G
G
150
1
100
-120
0
-100
-80
-60
-1
-40
-1
s /(J K mol )
-20
0
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
h /(kJ mol-1)
Points to note: Very useful for analysis of cycles; Isobars shown;
Typically also plot lines for constant h, v and quality; Methane
used here as example; Reference state: T = 298.15 K, p =
1.01325 bar;
47
T- S diagram for water
100
50
20
5
10
2
0.5
1
0.1
0.2
0.05
4000
200
700
500
Reference state:
s /(kJ·K-1·kg-1) = 0 and u /(kJ·kg-1) = 0
for saturated liquid at the triple point
0.02
T -s Diagram for Water
0.01
1000
800
Advanced Thermodynamics
0.01
0.02
0.05
0.
0.
0.
1
2
5
10
20
50
600
3800
100
 p /bar
 h /(kJ·kg-1)
 v /(m3·kg-1)
 Quality
3600
2400
500
T /C
2200
3400
2000
1800
400
1600
3200
1400
300
1200
3000
1000
200
800
2800
600
0.2
100
0.4
0.6
0.8
400
2600
200
0
0.0
2.0
4.0
6.0
8.0
10.0
12.0
s /(kJK kg )
-1
-1
48
Advanced Thermodynamics
1.66
1.62
1.58
1.54
1.50
1.46
1.42
1.38
1.34
1.30
1.26
1.22
1.18
1.14
1.10
1.06
1.02
0.98
0.94
0.86
0.90
0.82
0.74
100
0.78
P-h Diagram for R134a
120
p -h Diagram for R134a
0.002
1.70
1.74
1.78
110
Reference state:
h /(kJ·kg-1) = 200 and s /(kJ·K-1·kg-1) = 1.00
for saturated liquid at T = 0°C.
0.005
1.82
100
1.86
90
80
0.01
1.90
70
10
 T /°C
-1
-1
 s /(kJ·K ·kg )
3
-1
 v /(m ·kg )
 Quality
1.94
60
50
0.02
1.98
40
30
2.02
20
2.06 0.05
10
p /bar
2.10
0
0.1
2.14
-10
2.18 0.2
-20
-30
1
2.22
-40
2.26
-50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
2.30
0.9
0.5
1
0.1
100
200
300
150
130
140
120
110
100
90
70
80
60
40
50
30
20
0
400
10
-50
-40
-30
-20
-10
2.34
500
-1
h /(kJ·kg )
49
Advanced Thermodynamics
The p-T phase diagram of a pure substance
• For C = 1 and P = 1: F = 2
– At most 2 intensive variables need to
be specified in a pure-component
(unary) system to define fully its
thermodynamic properties and state
• For C = 1 and P = 2: F = 1
– The two-phase boundaries are fully
specified with one variable
• (recall Clausius-Clapeyron)
• For C = 1 and P = 3: F = 0
t
– At most three phases can be found at
coexistence in a pure-component
(unary) system
• (note, this is not the same as saying
that there can only be one triple point)
50
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
critical curve
(locus of V-L
critical points)
vapour-pressure
curve of pure C1
vapour-pressure
curve of pure C2
51
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
critical curve
(locus of V-L
critical points)
vapour-pressure
curve of pure C1
xC2 = xC2,1 (say);
(low xC2, high xC1)
vapour-pressure
curve of pure C2
52
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
critical curve
(locus of V-L
critical points)
vapour-pressure
curve of pure C1
xC2,2 > xC2,1
vapour-pressure
curve of pure C2
53
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
critical curve
(locus of V-L
critical points)
vapour-pressure
curve of pure C1
NB: the V-L critical point does not
(in general) lie at the maximum of a
constant-composition p-T curve.
vapour-pressure
curve of pure C2
54
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
vapour-pressure
curve of pure C1
xC2,3 > xC2,2
vapour-pressure
curve of pure C2
55
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
vapour-pressure
curve of pure C1
xC2,4 > xC2,3
vapour-pressure
curve of pure C2
56
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
vapour-pressure
curve of pure C1
xC2,5 > xC2,4
vapour-pressure
curve of pure C2
57
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
xC2,6 > xC2,5;
(high xC2, low xC1)
vapour-pressure
curve of pure C1
vapour-pressure
curve of pure C2
58
Advanced Thermodynamics
Constant-composition p-T isopleths
• Start with p-T projection
The location of the V-L critical point
on a constant-composition p-T
curve “rotates” in clockwise fashion
with increasing composition of the
heavier component.
59
Advanced Thermodynamics
Constant-composition p-T isopleths
• Increasing xC2, from left to right
vapour-pressure
curve of pure C1
vapour-pressure
curve of pure C2
60
Advanced Thermodynamics
Constant-composition p-T isopleths
• Phase envelopes are bounded
by the pure-component
vapour-pressure curve for
component A (here C1) on the
left, and that for B (here C2)
on the right.
• One component dominant
 relatively “narrow-boiling” systems
narrow boiling
narrow boiling
wide boiling
• Both components present in
comparable amounts  “wide-boiling” systems.
61
Advanced Thermodynamics
Constant-composition p-T isopleths
• Range of temperature of the
critical-point locus is bounded
by the critical temperatures of
the pure components.
– no binary mixture has a critical
temperature either below the lightest
component’s critical temperature or
above the heaviest component’s
critical temperature.
• NB: this is not true for critical
pressures.
– A mixture’s critical pressure can be higher than the critical pressures of both
pure components
• e.g., the concave shape for the critical locus.
62
Advanced Thermodynamics
Constant-composition p-T isopleths
dew
• NB: there are no tie lines on
constant-composition p-T
diagrams
– two points in coexistence share
the same p and the same T but
not the same composition
– constant-x diagrams are not the
most helpful for judging points
in coexistence
bubble
dew
bubble
two (superimposed)
points at coexistence
– intersection of bubble curve at one composition with dew curve at another
composition represents two points at coexistence
• perpendicular to the plane of the diagram
63
Advanced Thermodynamics
Binary mixture: critical locus
• In general, the more dissimilar
the two substances, the
farther the upward reach of the
critical locus.
– When the substances making up the
mixture are similar in molecular
complexity, the shape of the critical
locus flattens down.
from: J.W. Amyx, D.M. Bass, R.L. Whiting,
Petroleum Reservoir Engineering, McGraw Hill, 1960
64
Advanced Thermodynamics
Types of fluid phase behaviour in binary mixtures
• Classification of van Konynenburg and Scott
– Phil. Trans. Royal Soc. London A 298 (1980) 495–540
Type I
Type II
Type III
p
U
A
B
A
U
B
Type IV
U
L
B
Type V
U
A
A
Type VI
U
B
A
L
B
U
A
L
B
T
65
Advanced Thermodynamics
Types of fluid phase behaviour in binary mixtures
• Classification of van Konynenburg and Scott
– Phil. Trans. Royal Soc. London A 298 (1980) 495–540
Type I
Type II
Type III
p
U
A
B
A
U
V
Type IV
U
L
B
Type V
U
A
A
Type VI
U
B
A
L
B
U
A
L
B
T
66
Advanced Thermodynamics
Types of fluid phase behaviour in binary mixtures
• Classification of van Konynenburg and Scott
– Phil. Trans. Royal Soc. London A 298 (1980) 495–540
Type I
Type I
p
• Two substances of similar chemical type
U
- e.g., CH4 + N2; CO2 + O2; alkane + “similarsized” alkane
A U ◦ CH4 + up to n-C
A 5H12 is type I; CH4 + n-C6H14
B
B
is typeV V
◦ C2H6Type
+ up Vto n-C18H38 is typeType
I; C2VI
H6 + up to
Type IV
n-C19H40 is type V
U
U
A
A
II discussed so farType III
• What we’veType
mostly
U
L
B
A
L
B
U
A
L
B
T
67
Advanced Thermodynamics
Types of fluid phase behaviour in binary mixtures
• Classification of van Konynenburg and Scott
– Phil. Trans. Royal Soc. London A 298 (1980) 495–540
Type I
Type II
Type III
p
U
A
Type II
B
Type IV
A
U
V
Type V
A
B
Type VI
• L-L immiscibility
at T < Tc of the more-volatile
component
U
U
- e.g., CO2 + alkanes up to n-C12H26
• Type I can be thought of as “Type II, but where L-L immiscibility
is
U
A masked
L by the
A L
A
B solid phase”
B
B
U
L
◦ i.e., the L-L boundary lies at lower temperature than the
T
freezing temperature
68
Advanced Thermodynamics
Types of fluid phase behaviour in binary mixtures
• Classification of van Konynenburg and Scott
– Phil. Trans. Royal Soc. London A 298 (1980) 495–540
Type I
Type II
Type III
p
U
A
Type III
B
Type IV
A
U
V
Type V
A
B
Type VI
U
U is large enough, L-L region persists to
• If mutual immiscibility
of two substances
higher T, and Type II  Type III
- locus of L-L critical points merges with V-L critical curve
U
L
A n-C
L 13H28; methane
- Ae.g.,
from
+Asqualane; many
B
B
B alkane +
U CO2 + alkanes
L
polar-molecule mixtures; many hydrocarbon + water mixtures (including,
T
low-MW alkane + water)
69
Advanced Thermodynamics
Types of fluid phase Type
behaviour
in binary mixtures
V
•
• Instead of ending at the V-L critical point of the
more-volatile
component (as and
in TypeScott
I), the critical
Classification of van
Konynenburg
at the V-L critical point of the less– Phil. Trans. Royal Soc. Londoncurve
A 298beginning
(1980) 495–540
volatile component ends at a lower-critical end point
(LCEP) connected
Type I
Type II to a L-L-V three-phase
Type IIIline.
- e.g., some alkane + alkane mixtures
p
U
◦ CH4 + n-C6H14; CH4 + 2-methylpentane;
CH4
+ 3-methylpentane; CH4 + 2,3dimethylbutane; …
A
A U ◦ C H + up to n-C
A H
B
2 6 V
19 40
B
Type IV
Type V
U
A
U
L
Type VI
U
B
A
L
B
U
A
L
B
T
70
Advanced Thermodynamics
Constant-x p-T slices and retrograde behaviour
• Constant-temperature path: retrograde behaviour
– liquid evaporates to vapour upon an increase in pressure
• (or vapour condenses to liquid upon a decrease in pressure)
V  L+V  V  L
V  L+V  V  L
V  L  L+V  L
L
L
L+V
L
L+V
L+V
V
V
V
• The evaporation of liquid by an increase in pressure ((a) and (b) here) was termed
retrograde condensation by J. P. Kuenen in 1892
− retrograde: “the order of something reversed”
71
Advanced Thermodynamics
Constant-x p-T slices and retrograde behaviour
• Constant-pressure path: retrograde behaviour
– vapour condenses to liquid upon an increase in temperature
• (or liquid evaporates to vapour upon a decrease in temperature)
L  V L+V  V
L  L+V  L  V
L  L+V  L  V
L
L
L+V
L
L+V
L+V
V
V
V
A: cricondendbar (or maxcondendbar); B: cricondendtherm (or maxcondendtherm)
72
Advanced Thermodynamics
Phase Envelopes for Reservoir Fluids
Production
pathway
Reservoir
depletion
Phase envelope
completely determined by
(Tsep, Psep)
composition. Why?
Dry gas
Wet gas
GOR, [m3/m3]
no liquid
> 2500
C7+, [mol%]
< 0.7
0.7-4
Retrograde
Volatile oil
Black oil
600 to 2500
300 to 600
< 300
4-12.5
12.5 to 20
> 20
condensate
73
Advanced Thermodynamics
Issues for Engineers
•
•
•
•
•
•
Evaluate reserves & production strategy
Facilities Sizing
Gas-Oil Ratio – GOR & CGR
Condensate drop-out and Gas Recycling
Pipeline specifications: – HC dew point
Location of dew point curve VERY
sensitive to (small) amounts of heavy HCs
– Dew points key parameter for measurement
74