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 #

y Cx CT 1P 1

2C+2= 8

Equations #

∑ 1∑ 1

K CC+2= 5

F = C � P + 2

#degrees of freedom #species #phases

Gibbs’ Phase Rule:Variables #x C𝒫

T 1P 1

C𝒫+2

Equations #∑ 𝒫

K 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 #x C𝒫

z CL, V, ... 𝒫

F 1T 2T, P 2Q 1

C𝒫+C+𝒫+6

Equations #∑ 𝒫∑ 1

K C(𝒫-1)Fz CFh 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) y x α100 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) y x α

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) y x α64.5 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:x Cy Cz CF 1T 1P 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 Pdew

At 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 Feed kmol/h z

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 a a a b b b

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.

Variations on the ThemeSHR Figure 4.13

“Right-triangle” diagram • 2 of the 3 components • same information as equilateral

triangle diagram • often simpler to deal with.

“Equilibrium solute” diagram • 2 of the 3 components • plots tie-line endpoint information.

Equilateral triangle diagram