Working Guide to Vapor-liquid Phase Equilibria Calculations

136 Pages·2010·3.337 MB·English

by  Tarek Ahmed (Auth.)

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C H A P T E R
1
Vapor Pressure
A phase is defined as the part of a system that is uniform in physical
and chemical properties, homogeneous in composition, and separated
from other coexisting phases by definite boundary surfaces. The most
important phases occurring in petroleum production are the hydrocar-
bon liquid phase and the gas phase. Water is also commonly present
as an additional liquid phase. These can coexist in equilibrium when
the variables describing change in the entire system remain constant
over time and position. The chief variables that determine the state of
equilibrium are system temperature, system pressure, and composition.
Theconditionsunderwhichthesedifferentphasescanexistareamat-
ter of considerable practical importance in designing surface separation
facilities and developing compositional models. These types of calcula-
tions are based on the concept of equilibrium ratios.
A system thatcontainsonly onecomponentisconsideredthe simplest
typeofhydrocarbonsystem.Thewordcomponent referstothenumber of
molecularoratomicspeciespresentinthesubstance.Asingle-component
system is composed entirely of one kind of atom or molecule. We often
usethewordpuretodescribea single-componentsystem.Thequalitative
understandingoftherelationshipthatexistsbetweentemperatureT,pres-
sure p,and volume V of pure components can provide anexcellentbasis
for understanding the phase behavior of complex hydrocarbon mixtures.
Consider a closed evacuated container that has been partially filled
with a pure component in the liquid state. The molecules of the liquid
are in constant motion with different velocities. When one of these mole-
culesreachestheliquidsurface,itmaypossesssufficientkineticenergyto
overcometheattractiveforcesintheliquidandpassintothevaporspaces
above. Asthe number ofmolecules in the vapor phaseincreases, the rate
ofreturntotheliquidphasealsoincreases.Astateofequilibriumiseven-
tually reached when the number of molecules leaving and returning is
equal. The molecules in the vapor phase obviously exert a pressure on
1
Copyright#2006,ElsevierInc.Allrightsreserved.
2
1. VAPORPRESSURE
the wall of the container, and this pressure is defined as the vapor pres-
sure,p .Asthetemperatureoftheliquidincreases,theaveragemolecular
v
velocityincreases,withalargernumberofmoleculespossessingsufficient
energytoenterthevaporphase.Asaresult,thevaporpressureofapure
component in the liquid state increases with increasing temperature.
A method that is particularly convenient for expressing the vapor
pressure of pure substances as a function of temperature is shown in
Figure 1-1. The chart, known as the Cox chart, uses a logarithmic scale
600
500
Critical PointExtended Beyond Critical 400300
noitcarF F(cid:2)004 200 F
ainommA enaporP eneannoaNce-nD-n 140180160120 (cid:2)Temperature,
edifluS negordyH enaretctaOW-n 10080
enatpe 60
enahteM enelyhtEedixoiD nobraC enelyporP21 noerF enatuB-ienatuB-n 11 noerFenatnePe-niatneP-n enaxeH-n H-n 4020
enahtE
0
10,000.08000.06000.05000.04000.03000.0 2000.0 1000.0800.0600.0500.0400.0300.0 200.0 100.080.0Va60.0po50.0r 40.0pre30.0ssure20.0, psia10.08.06.05.04.03.0 2.0 1.00.80.60.50.40.3 0.2 0.1
FIGURE 1-1
Vapor pressures for hydrocarbon components. (Courtesy of the Gas
ProcessorsSuppliersAssociation,EngineeringDataBook,10thEd.,1987.)
3
1.VAPORPRESSURE
forthevaporpressureandanentirelyarbitraryscaleforthetemperature
(cid:1)
in F. Thevaporpressure curve for any particularcomponent, as shown
inFigure1-1,canbedefinedasthedividinglinebetweentheareawhere
vaporandliquidexist.Ifthesystempressureexistsatitsvaporpressure,
two phases can coexist in equilibrium. Systems represented by points
locatedbelowthatvaporpressurecurvearecomposedofonlythevapor
phase. Similarly, points above the curve represent systems that exist in
the liquid phase. These statements can be conveniently summarized by
the following expressions:
(cid:129) p < p ! system is entirely in the vapor phase.
v
(cid:129) p > p ! system is entirely in the liquid phase.
v
(cid:129) p ¼ p ! vapor and liquid coexist in equilibrium.
v
where p is the pressure exerted on the pure component. Note that these
expressionsarevalidonlyifthesystemtemperatureTisbelowthecriti-
cal temperature T of the substance.
c
The vapor pressure chart allows a quick determination of the p of a
v
pure component at a specific temperature. For computer and spread-
sheet applications, however, an equation is more convenient. Lee and
Kesler (1975) [29] proposed the following generalized vapor pressure
equation:
p ¼p expðAþoBÞ
v c
with
6:09648
A¼5:92714(cid:3) (cid:3)1:2886lnðT Þþ0:16934ðT Þ6
r r
T
r
15:6875
B¼15:2518(cid:3) (cid:3)13:4721lnðT Þþ0:4357ðT Þ6
r r
T
r
where p ¼ vapor pressure, psi
v
p ¼ critical pressure, psi
c
T ¼ reduced temperature (T / T )
r c
T ¼ system temperature, (cid:1)R
T ¼ critical temperature, (cid:1)R
c
o ¼ acentric factor
C H A P T E R
2
Equilibrium Ratios
In a multicomponentsystem, the equilibrium ratio K of a given com-
i
ponent is defined as the ratio of the mole fraction of the component in
the gas phase y to the mole fraction of the component in the liquid
i
phase x. Mathematically, the relationship is expressed as
i
y
K ¼ i (2-1)
i
x
i
where K ¼ equilibrium ratio of component i
i
y ¼ mole fraction of component i in the gas phase
i
x ¼ mole fraction of component i in the liquid phase
i
Atpressuresbelow100psia,Raoult’sandDalton’slawsforidealsolu-
tions provide a simplified means of predicting equilibrium ratios.
Raoult’s law states that the partial pressure p of a component in a mul-
i
ticomponent system is the product of its mole fraction in the liquid
phase x and the vapor pressure of the component p , or
i vi
p ¼xp (2-2)
i i vi
where p ¼ partial pressure of component i, psia
i
p ¼ vapor pressure of component i, psia
vi
x ¼ mole fraction of component i in the liquid phase
i
Dalton’s law states that the partial pressure of a component is the prod-
uct of its mole fraction in the gas phase y and the total pressure of the
i
system p, or
p ¼y p (2-3)
i i
where p ¼ total system pressure, psia.
Atequilibriumandinaccordancewiththeprecedinglaws,thepartial
pressure exerted by a component in the gas phase must be equal to the
5
Copyright#2006,ElsevierInc.Allrightsreserved.
6
2. EQUILIBRIUMRATIOS
partial pressure exerted by the same component in the liquid phase.
Therefore, equating the equations describing the two laws yields
xp ¼yp
i vi i
Rearrangingthisrelationshipandintroducingtheconceptoftheequilib-
rium ratio gives
y p
i ¼ vi ¼K (2-4)
i
x p
i
Equation2-4showsthat,foridealsolutionsandregardlessoftheoverall
composition of the hydrocarbon mixture, the equilibrium ratio is only
a function of the system pressure p and the temperature T since the
vapor pressure of a component is only a function of temperature (see
Figure 1-1).
It is appropriate at this stage to introduce and define the following
nomenclatures:
z ¼ mole fraction of component in the entire hydrocarbon mixture
i
n ¼ total number of moles of the hydrocarbon mixture, lb-mol
n ¼ total number of moles in the liquid phase
L
n ¼ total number of moles in the vapor (gas) phase
v
By definition,
n¼n þn (2-5)
L v
Equation 2-5 indicates that the total number of moles in the system is
equal to the total number of moles in the liquid phase plus the total
number of moles in the vapor phase.Amaterial balance on the ith com-
ponent results in
zn¼xn þyn (2-6)
i i L i v
where zn ¼ total number of moles of component i in the system
i
xn ¼ total number of moles of component i in the liquid phase
i L
yn ¼ total number of moles of component i in the vapor phase
i v
Also by the definition of mole fraction, we may write
X
z ¼1 (2-7)
i
i
X
x ¼1 (2-8)
i
i
X
y ¼1 (2-9)
i
i
It is convenient to perform all phase-equilibria calculations on the basis
of 1mol of the hydrocarbon mixture, i.e., n ¼ 1. That assumption
reduces Equations 2-5 and 2-6 to
7
2.EQUILIBRIUMRATIOS
n þn ¼1 (2-10)
L v
xn þyn ¼z (2-11)
i L i v i
Combining Equations 2-4 and 2-11 to eliminate y from Equation 2-11
i
gives
x n þðxKÞn ¼z
i L i i v i
Solving for x yields
i
z
x ¼ i (2-12)
i n þn K
L v i
Equation2-11canalsobesolvedfory bycombining itwithEquation
i
2-4 to eliminate x:
i
zK
y ¼ i i ¼xK (2-13)
i n þn K i i
L v i
Combining Equation 2-12 with 2-8 and Equation 2-13 with 2-19
results in
X X
z
x ¼ i ¼1 (2-14)
i n þn K
i i L v i
X X
zK
y ¼ i i ¼1 (2-15)
i n þn K
i i L v i
Since
X X
y (cid:2) x ¼0
i i
i i
therefore,
X X
zK z
i i (cid:2) i ¼0
n þn K n þn K
i L v i i L v i
or
XzðK (cid:2)1Þ
i i ¼0
n þn K
i L v i
Replacing n with (1 – n ) yields
L v
X zðK (cid:2)1Þ
fðn Þ¼ i i ¼0 (2-16)
v n ðK (cid:2)1Þþ1
i v i
Thissetofequationsprovidesthenecessaryphaserelationshipstoperform
volumetricandcompositionalcalculationsonahydrocarbonsystem.These
calculationsarereferredtoasflashcalculationsandarediscussedinChapter3.
C H A P T E R
3
Flash Calculations
Flash calculations are an integral part of all reservoir and process
engineering calculations. They are required whenever it is desirable to
know the amounts (in moles) of hydrocarbon liquid and gas coexisting
inareservoiroravesselatagivenpressureandtemperature.Thesecal-
culations are also performed to determine the composition of the exist-
ing hydrocarbon phases.
Given the overall composition of a hydrocarbon system at a specified
pressureandtemperature,flashcalculationsareperformedtodetermine
(cid:129) Moles of the gas phase n .
v
(cid:129) Moles of the liquid phase n .
L
(cid:129) Composition of the liquid phase x.
i
(cid:129) Composition of the gas phase y.
i
The computational steps for determining n , n , y, and x of a hydro-
L v i i
carbonmixturewithaknownoverallcompositionofz andcharacterized
i
by a set of equilibrium ratios K are summarized as follows:
i
Step 1. Calculation of n . Equation 2-16 can be solved for n using the
v v
Newton-Raphson iteration technique. In applying this iterative
technique,
(cid:129) Assume any arbitrary value of n between 0 and 1, e.g.,
v
n ¼ 0.5. A good assumed value may be calculated from the
v
following relationship, providing that the values of
the equilibrium ratios are accurate:
n ¼A=ðA(cid:2)BÞ
v
where
P
A ¼ ½zðK (cid:2)1Þ(cid:3)
Pi i i
B ¼ ½zðK (cid:2)1Þ=K(cid:3)
i i i i
9
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10
3. FLASHCALCULATIONS
(cid:129) Evaluatethefunctionf(n )asgivenbyEquation2-16usingthe
v
assumed value of n .
v
(cid:129) If the absolute value of the function f(n ) is smaller than a
v
presettolerance,e.g.,10(cid:2)15,thentheassumedvalueofn isthe
v
desired solution.
(cid:129) If the absolute value of f(n ) is greater than the preset
v
tolerance, then a new value of n is calculated from the
v
following expression:
ðn Þ ¼n (cid:2)fðn Þ=f0ðn Þ
v n v v v
with
( )
f0 ¼(cid:2)X ziðKi(cid:2)1Þ2
½n ðK (cid:2)1Þþ1(cid:3)2
i v i
where (n ) is the new value of n to be used for the next
v n v
iteration.
(cid:129) This procedure is repeated with the new values of n until
v
convergence is achieved.
Step 2. Calculation of n . Calculate the number of moles of the liquid
L
phase from Equation 2-10, to give
n ¼1(cid:2)n
L v
Step 3. Calculation of x. Calculate the composition of the liquid phase
i
by applying Equation 2-12:
z
x ¼ i
i n þn K
L v i
Step 4. Calculation of y. Determine the composition of the gas phase
i
from Equation 3-13:
zK
y ¼ i i ¼xK
i n þn K i i
L v i
Example 3-1
A hydrocarbon mixture with the following overall composition is
(cid:4)
flashed in a separator at 50 psia and 100 F.
Component z
i
C 0.20
3
i–C 0.10
4
n–C 0.10
4
11
3. FLASHCALCULATIONS
i–C 0.20
5
n–C 0.20
5
C 0.20
6
Assuming an ideal solution behavior, perform flash calculations.
Solution
Step 1. Determine the vapor pressure from the Cox chart (Figure 1-1)
and calculate the equilibrium ratios from Equation 2-4:
Component z p at100(cid:4)F K ¼p /50
i vi i vi
C 0.20 190 3.80
3
i–C 0.10 72.2 1.444
4
n–C 0.10 51.6 1.032
4
i–C 0.20 20.44 0.4088
5
n–C 0.20 15.57 0.3114
5
C 0.20 4.956 0.09912
6
Step 2. Solve Equation 2-16 for n using the Newton-Raphson method,
v
to give
Iteration n f(n )
v v
0 0.08196579 3.073E-02
1 0.1079687 8.894E-04
2 0.1086363 7.60E-07
3 0.1086368 1.49E-08
4 0.1086368 0.0
Step 3. Solve for n :
L
n ¼1(cid:2)n
L v
n ¼1(cid:2)0:1086368¼0:8913631
L
Step 4. Solve for x and y to yield
i i
Component z K x ¼z/(0.8914þ0.1086K) y ¼xK
i i i i i i i i
C 0.20 3.80 0.1534 0.5829
3
i–C 0.10 1.444 0.0954 0.1378
4
n–C 0.10 1.032 0.0997 0.1029
4
i–C 0.20 0.4088 0.2137 0.0874
5
n–C 0.20 0.3114 0.2162 0.0673
5
C 0.20 0.09912 0.2216 0.0220
6