Single-Stage Equilibrium Processes
• Degree of Freedom Analysis
• Binary Vapor-Liquid Equilibrium
• Azeotropes in binary systems
• Multicomponent Flash Calculations
• Ternary Liquid-Liquid Systems
SHR Chapter 4
Degrees of Freedom: Gibb’s Phase RuleExample: C=3 components in 𝒫=2 phases at equilibrium.
T, P
yi
xi
∴ we must specify 3 variables
Options:• Set T, P and one mole fraction.• Set 3 mole fractions and determine T, P.• Set T and 2 mole fractions • ...
Note: we may not be able to find a solution if we set an impossible choice of intensive variables.
Examples???
𝓕 = 𝒱 - ℰ = 3 degrees of freedom!Variables #
yi Cxi CT 1P 1
2C+2= 8
Equations #
∑yi = 1 1∑xi = 1 1
Ki = yi /xi CC+2= 5
F = C � P + 2
#degrees of freedom #species #phases
Gibbs’ Phase Rule:Variables #xi, yi, ... C𝒫
T 1P 1
C𝒫+2
Equations #∑xi = 1 𝒫
Ki = yi /xi C(𝒫-1)C(𝒫-1)+𝒫
In general:
SHR §4.1
Gibbs Phase Rule & Extensive VariablesExample: single-stage equilibrium with C components in 𝒫=2 phases
Why aren’t stream enthalpies counted as variables?
Variables #xi, yi, ... C𝒫
zi CL, V, ... 𝒫
F 1TF, PF 2T, P 2Q 1
C𝒫+C+𝒫+6
Equations #∑xi = 1 𝒫∑zi = 1 1
Ki = yi /xi C(𝒫-1)Fzi = Vyi + Lxi CFhF=VhV+LhL 1
C𝒫+𝒫+2
T, P
FziTFPF
Vyi
Lxi
Q
Could set: C-1 of the zi, TF, PF, T, P and then determine the remaining quantities from the equations.
F = V � E = C + 4
SHR §4.1.2
In general, we get C+𝓟+4 extra variables and C+2 extra equations when we
include extensive quantities.
Binary Vapor-Liquid Equilibrium
SHR §4.2
T, P
FzATFPF
VyA
LxA
Q
Tabulated Binary VLE DataF = C � P + 2 ⇒ 𝓕 = 2-2+2 = 2
Often data is obtained by fixing T or P and xA or yA.
xA(T, yA) at a given PxA(P, yA) at a given T
Water (A) - Glycerol (B) at 1 atm
T (C) yA xA αA,B
100 1 1 -104.6 0.9996 0.8846 333109.8 0.9991 0.7731 332128.8 0.998 0.4742 544148.2 0.9964 0.3077 627175.2 0.9898 0.1756 456207 0.9804 0.0945 481
244.5 9341 0.0491 275282.5 0.8308 0.025 191290 0 0 -
what do these points represent?
Para-xylene (A) - Meta-xylene (B) at 1 atm
T (C) yA xA αA,B
133.335 1 1 -138.491 0.8033 0.8 1.0041138.644 0.6049 0.6 1.0082138.795 0.4049 0.4 1.0123138.943 0.2032 0.2 1.016139.088 0 0 -
SHR §4.2
SHR Table 4.1c
SHR Table 4.1a
Saturated mixture
Binary VLE: Methanol/WaterMethanol (A) - Water (B) at 1 atm
T (C) yA xA αA,B
64.5 1 1 -66 0.958 0.9 2.53
69.3 0.87 0.7 2.8773.1 0.779 0.5 3.5278 0.665 0.3 4.63
84.4 0.517 0.15 6.0789.3 0.365 0.08 6.6193.5 0.23 0.04 7.17100 0 0 -
SHR Table 4.1bSHR Figure 4.2a
this “width” is directly related to the relative volatility, α.
This distance is related to the relative volatility, α
SHR Figure 4.2b
SHR Figure 4.5
SHR §4.2
↵ij ⌘Ki
Kj
T-x-y Diagrams
SHR Figure 4.3
SHR §4.2
n-hexane, n-octane at P=1 atm
Inverse lever-arm rule:
Given xhexane=0.3,• What is V/L at B-C?• What is V/L at D-F?• What is V/L at G?
• What state is G and B?• What state is H and A?• What state is E?
V/L = DE/EF
SHR Figure 4.4
FzA = V yA + LxA
F = V + L
Eliminate L and solve for yA...
“q-line”
species mole balance:
overall mole balance:
Given a molar vaporization % (V/F), we can graphically determine the product compositions.
Operating Line (q-line)
Example: for zhexane=0.6 and 60% vaporization, find the yi.
1. Locate zA on the 45∘ line (point “A”)2. From V/F, calculate the slope of the q-line.3. Follow the q-line to determine the
equilibrium composition (point “B”)
n-hexane, n-octane at P=1 atm
SHR §4.2
• goes through xA = yA = zA (45˚ line)• 0 ≤ V/F ≤ 1• Slope is bounded by -∞ (V=0) and 0 (V=F).• V=F gives us point “A” (always on 45˚ line).
yA =
V/F � 1V/F
�xA +
1
V/F
�zA
What does it represent?
Azeotropes in Binary Systems
SHR §4.3
What is an Azeotrope?
At an azeotrope, KA = KB = 1, α = 1.• No separation possible at this point!
• If you start on one side of the azeotrope, you can only recover at best a pure stream and the azeotrope - not two pure streams!
Typically occur in systems with close boiling point constituents of different chemical type (e.g. ethanol/water). Maximum or minimum boiling points.• zeoptropic systems have a monotonic relationship
between xA and T.
1 atm
1 atm
“Normal” (zeotropic)
system
Maximum-boiling
azeotropic system
Minimum-boiling
azeotropic system (most
common)
SHR Figure 4.7
SHR Figure 4.6
SHR Figure 4.3
Minimum-boiling AzeotropesFigure 4.6a:• What can we say about γA and γB?
• Positive deviation from Raoult’s law: Ki = Pi
s/P, (Ki = γiPi
s/P).
Figure 4.6b• What can we say about relative
volatility across the azeotrope point?
Note: we can change the pressure to shift the azeotrope a little bit.• Figure 4.6a (70˚C, 123 kPa)
shows composition of x=0.7. Figure 4.6b & 4.6c shows x=0.72 (at 66˚C & 101 kPa).
• ethanol-water azeotrope disappears at P < 9.3 kPa.
SHR Figure 4.6
isopropyl ether, isopropyl alcohol
T=70 ˚C P = 1 atm
1 atm
T = 100 ˚CP = 1 atm
Maximum-boiling Azeotrope
Figure 4.7a:• What can we say about γA and γB?
• Negative deviation from Raoult’s law: Ki = Pi
s/P, (Ki = γiPis/P).
Again, we see a shift in the azeotrope as we change the operating pressure.
• compare 4.7a with 4.7b & 4.7c.
SHR Figure 4.7
P = 1 atm
Predicting Azeotrope FormationKi = �i
P si
PModified Raoult’s law: For an azeotrope, Ki=1 �i =
P
P si
Modified Raoult’s law at an azeotrope
ln �1 =A12
[1 + (x1A12)/(x2A21)]2
ln �2 =A21
[1 + (x2A21)/(x1A12)]2
van-Laar: lnP
P s
1
=A12
[1 + (x1A12)/(x2A21)]2
lnP
P s
2
=A21
[1 + (x2A21)/(x1A12)]2
Solve these for x1 and x2.(Note: x2 = 1-x1.)
Example: does ethanol (1) and n-hexane (2) form an azeotrope?A12=2.409, A21=1.970
Solution: 1. Choose P.
2. Find Tbubble or Tdew and corresponding x, y.
3. See if Ki = 1.
Up next: flash calculations to get Tbubble or Tdew.
Multicomponent Flash Calculations
SHR §4.4
Flash Concepts
T, P
FziTFPF
Vyi
Lxi
Q
Two “modes”• partial vaporization (typically add
heat)• partial condensation (typically
remove heat)
“Isothermal flash”• add/remove heat such that T=TF.
• must determine the “duty” of the heat exchanger to maintain T.
“Adiabatic flash”• Q=0, T≠TF
• Temperature drops (vaporization) or rises (condensation).
• T is unknown, so this is more difficult than isothermal flash.
The Rachford-Rice EquationFzi = Lxi + V yi
Equations:Variables:xi Cyi Czi CF 1TF 1PF 1L 1V 1T 1P 1Q 1
3C+8
yi = Kixi
C species mole balances
C phase equilibrium
1
1
1 energy balance
1 total mole balance
2C+4
PCi=1 xi = 1
PCi=1 yi = 1
combine first two sets of equations
overall mole balance to eliminate L
Substitute into ∑xi and ∑yi and then subtract to get:
from second equation
Still 2C equations - just different form.
Rachford-Rice equation
xi =Fzi
L+ V Ki
=Fzi
F � V + V Ki
=zi
1 + (Ki � 1)VF
yi =Kizi
1 + (Ki � 1)VF
CX
i=1
0
BBB@K
i
zi
1 + (Ki
� 1)VF| {z }
yi
� zi
1 + (Ki
� 1)VF| {z }
xi
1
CCCA= 0
CX
i=1
zi (Ki � 1)
1 + (Ki � 1) = 0
⌘ V
F
T, P
FziTFPF
Vyi
Lxi
Q
F = V + L
hFF +Q = hV V + hLL
Ki(T, P, yj , xj)hV (T, P, yj) hL(T, P, xj)
hF (TF , PF , zj)Auxiliary equations:
• C+4 degrees of freedom• We typically know zi, TF and PF.• 2 “free” variables
SHR §4.4
Now what?• For ideal mixtures, Ki(T,p). If we know T, p
we can solve the Rachford-Rice equation for Ψ. After that, we can easily get xi, yi.
• For nonideal mixtures, Ki also depends on composition!➡ Either go back and solve a bigger system of
equations or use successive substitution.➡ Successive substitution is not highly robust,
but is usually good enough for these problems.
Non-Ideal Mixtures:1. Guess a value for Ki (e.g. from charts or Raoult’s law)2. Solve for Ψ.3. Calculate yi & xi 4. Calculate Ki using your favorite model.5. Solve for Ψ.6. If Ψ changed, return to step 3.7. Update yi & xi using Ψ calculated in step 6.
Succ
essi
ve
Subs
titut
ion
Once we know Ψ,
Ideal Mixtures:1. Determine Ki(T,p) (using Raoult’s
law, graphical techniques, etc.)2. Solve for Ψ.
3. Calculate yi & xi
CX
i=1
zi (Ki � 1)
1 + (Ki � 1) = 0
V = F
L = F � V = F (1� )
Q = hV V + hLL� hFF
Rachford-Rice equation
Typically, convergence is achieved when
�� (k+1) � (k)��
(k)< 10�4
If T = TF and P = PF then we only have one
equation to solve for Ψ.
xi =zi
1 + (Ki � 1)
yi =Kizi
1 + (Ki � 1)
Use Newtons’ method (or another nonlinear equation
solver) to solve for bubble/dew point temperature or pressure.
Bubble Point & Dew Point
• At the bubble point, V=0, L=F ⇒ Ψ=0.
CX
i=1
ziKi = 1
• At the dew point, V=F, L=0 ⇒ Ψ=1.
CX
i=1
ziKi
= 1
Recall Ki = Ki( xi, yi, T, P )
• Given P, find Tbubble
• Given T, find Pbubble
• Given P, find Tdew
• Given T, find PdewAt equilibrium, the vapor is at its dew point and the liquid is
at its bubble point.
CX
i=1
zi (Ki � 1)
1 + (Ki � 1) = 0
Rachford-Rice equation
SHR §4.4.2
Example: bubble point calculation
T, P
FziTFPF
Vyi
Lxi
Q
Component Feedkmol/h zi = xi
i-Butane 8.6 0.0319n-Butane 215.8 0.7992i-Pentane 28.1 0.1041n-Pentane 17.5 0.0648
270
Bubble point: V=0, L=F, Ψ=0.CX
i=1
ziKi = 1Solve to obtain T.
Empirical Correlation (dePriester Chart):
Compound a1 a2 a3 b1 b2 b3
Isobutane -1,166,846 0 7.72668 -0.92213 0 0
n-Butane -1,280,557 0 7.94986 -0.96455 0 0
Isopentane -1,481,583 0 7.58071 -0.93159 0 0
n-Pentane -1,524,891 0 7.33129 -0.89143 0 0
M. L. McWilliams, Chemical Engineering, 80(25), 1973 p. 138.
lnK =a1T 2
+a2T
+ a3 + b1 ln p+b2p2
+b3p
NOTE: T in °R and p in psia!
Need a model for Ki(T).
See SHR Example 4.2 for a graphical solution technique.
Example: equilibrium diagramUsing Raoult’s law, obtain a T-x-y diagram of the Propane-Benzene system.
Antoine Equation parameters
A B C
Propane 6.80398 803.81 246.99
Benzene 6.90565 1211.033 220.790 0.2 0.4 0.6 0.8 1220
240
260
280
300
320
340
360
Mole Fraction
T (K
)
liquidvapor Ki =
P si
PRaoult’s law:
Ki = �iP si
PModified
Raoult’s law:
Isothermal FlashGiven: T=TF, P=PF,
Find: xi, yi, V, L, Q.
CX
i=1
zi (Ki � 1)
1 + (Ki � 1) = 0
Rachford-Rice equation
SHR §4.4.1
Once we know Ψ,V = F
L = F � V = F (1� )
Q = hV V + hLL� hFF
1.Choose a model for Ki.
2.Given zi, solve Rachford-Rice to obtain Ψ.
3.Calculate xi, yi, V = ΨF, L = F(1-Ψ).4.Calculate hV, hL, hF at T and xi, yi
5.Calculate Q (heat exchanger duty).
T, P
FziTFPF
Vyi
Lxi
Q
hi = hi(T )
h =CX
i=1
hixi
(molar enthalpy)
xi =zi
1 + (Ki � 1)
yi = Kixi =Kizi
1 + (Ki � 1)
Adiabatic FlashGiven: Q=0, P=PF,
Find: xi, yi, T, V, L.
CX
i=1
zi (Ki � 1)
1 + (Ki � 1) = 0
Rachford-Rice equation:
SHR §4.4.3
Q = hV V + hLL� hFF
0 = F (hV + hL(1� )� hF )
Energy Balance:
F [hV + hL(1� )� hF ] = 0
2 equations, 2 unknowns
( Ψ, T )
1.Choose a model for Ki.
2.Given zi, solve Rachford-Rice & energy balance simultaneously to obtain Ψ and T.
3.Calculate xi, yi, V = ΨF, L = F(1-Ψ).
Energy balance:
T, P
FziTFPF
Vyi
Lxi
Q
Note: at each iteration, we must update values for hV and hL given the current value for T.
See SHR §4.4.3 for an alternative algorithm
Summary - Flash CalculationsIn
all
case
s, w
e ne
ed m
odel
s fo
r K i
( T, P
, xi,
y i ).
DESCRIPTION KNOWNS UNKNOWNS PROBLEM STATEMENT
Bubble point
pressure
T, z
i
P
Solve r(P) = 1 � ÂC
i=1
z
i
K
i
for P
Bubble point
temperature
P, z
i
T
Solve r(T) = 1 � ÂC
i=1
z
i
K
i
for T
Dew point
pressure
T, z
i
P
Solve r(P) = 1 � ÂC
i=1
z
i/K
i
for P
Dew point
temperature
P, z
i
T
Solve r(T) = 1 � ÂC
i=1
z
i/K
i
for T
Isothermal flash
T, P, z
i
Y, x
i
, y
i
Solve the Rachford-Rice equation
r(Y) =C
Âi=1
z
i
(Ki
� 1)1 + (K
i
� 1)Y
together with x
i
= z
i
1+(Ki
�1)Y and y
i
= K
i
x
i
.
Adiabatic flash
Q, P, z
i
Y, T, x
i
, y
i
Solve the coupled system of equations
r
1
(Y, T) =C
Âi=1
z
i
(Ki
� 1)1 + (K
i
� 1)Y,
r
2
(Y, T) = F [hV
Y + h
L
(1 � Y)� h
F
]
together with x
i
= z
i
1+(Ki
�1)Y and y
i
= K
i
x
i
.
1
Ternary Liquid-Liquid SystemsGraphical Methods
SHR §4.5
Introduction
“A” is insoluble in solvent “C” “A” is partly soluble in solvent “C”
“simple” mass balance. need to know phase equilibrium.
We have a carrier (A) which holds solute (B). We want to extract the solute into a solvent (C).
SHR §4.5
Case 1: Carrier (A) is insoluble in solvent (C)
Carrier (A) mole balance:x
(F )A F = x
(R)A R
Solute (B) mole balance:
S = x
(E)C E
Solvent (C) mole balance:
Use A & C balances to eliminate R & E in B balance:
Molar ratios of B to “other” component:
FA = x
(F )A F
FB = x
(F )B F
Note: we could do this whole thing on a mass basis as well.
in extract, only B & C
in raffinate, only B & A
X
(E)B ⌘ x
(E)B E
x
(E)C E
=x
(E)B
x
(E)C
X
(R)B ⌘ x
(R)B R
x
(R)A R
=x
(R)B
x
(R)A
x
(F )B F = x
(E)B
S
x
(E)C
+ x
(R)B
FA
x
(R)A
X
(F )B FA = X
(E)B S +X
(R)B FA
Distribution or partition coefficient
Extraction factor(large is good)
X(R)B
X(F )B
=
✓1 +
K 0DB
S
FA
◆�1
SHR §4.5
This is the familiar “K-value” (distribution
coefficient) for liquid-liquid.
x
(F )B F = x
(E)B E + x
(R)B R
KDB =x
(E)B
x
(R)B
K
0DB
=X
(E)B
X
(R)B
=x
(E)B /x
(E)C
x
(R)B /x
(R)A
= K
DB
1� x
(R)B
1� x
(E)B
SHR Figure 4.13a
1. Locate feed (M)
2. Follow tie-lines to miscibility boundary to determine product composition (E & R)
3. Inverse lever-arm rule to determine relative amount of E & R.
Case 2: Carrier (A) is soluble in solvent (C)SHR §4.5
Identify/describe:• How many degrees of
freedom? • Miscibility boundary• Two-phase region• Single-phase region• Tie line• Plait point• How is this diagram
made?
Note: these diagrams are for a specific T, P.
ExampleDetermine the extract and raffinate compositions when a 45 wt% glycol, 55 wt% water solution is contacted with twice its weight of pure furfural solvent at 25 ˚C and 101 kPa.
Assume F =100 g basis of feed• ∴ S = 200 g.
• Point “F” indicates F stream• Point “S” indicates S stream• Inverse lever arm rule to get point M: (SM)/
(MF) = 1/2 or (SM)/(SF) = 1/3• Follow tie line to get E & R.
• Extract: 4% A, 9% B, 88% C• Raffinate: 56% A, 34% B, 10% C
• Inverse lever arm rule on tie line to get ratio of E/M and R/M:• E = M (MR)/(ER) = 300 (5.5 cm / 7.6 cm)
= 220 g (measured from SHR figure 4.14)
• R = M (EM)/(ER) = M-E = 300-220 = 80 g.