Advanced Heat and Mass

CHEN90019 Semester 1, 2012
Lecture Materials for

ADVANCED HEAT & MASS
TRANSPORT PROCESSES
Part 1
1 of 3
CHEN90019 ADVANCED HEAT & MASS TRANSPORT PROCESSES
Coordinators: A/Prof. Ray Dagastine Prof. Geoffrey Stevens
Office: Rm 318, Chem. Eng. Building 1 Rm G03, Chem. Eng. Building 2
Email: rrd@unimelb.edu.au gstevens@unimelb.edu.au
Phone: 8344 4704 8344 6621
Lectures Times:
Monday 1:00 pm - 2:00 pm Redmond Barry-Medley Theatre
Wednesday 10:00 pm 11:00 pm Baldwin Spencer-Theatre
Friday 2:15 pm - 4:15 pm Redmond Barry-Lowe Theatre
Topics:
Mass Transport Processes
Multi-component Distillation
-short cut methods for sizing
column
-rigorous techniques for tray by
tray calculations
-Mass Transfer concepts
-Azetropic Distillation and
Residue curves
Axial dispersion
-Relevant to solvent extraction,
fixed and packed beds
Adsorption and ion exchange
-types of absorbents, fixed bed
absorber models,
-isothermal equilibrium and non-
equilibrium design and operation
Mass transfer with chemical reaction:
-homogeneous and
heterogeneous reactions, and
-application to equipment
performance and design
Solvent Extraction processes
Bioseparations
-Membranes
-Chromatography
Heat Transfer
Radiation
-Basic principles of radiation
-shape factors (viewfactors)
- radiation between grey surfaces
in the network approach
applications of networks for
various situations
Conduction
-Fourier's Law of heat
conduction
-multi-dimensional heat transfer
equations
-steady-state heat conduction
-steady-state conduction with
distributed heat source
- fins
-transient heat conduction and
the diffusion equation
-examples of simple solution of
transient heat conduction
-brief introduction to numerical
methods for heat conduction
problems
Application of simultaneous heat and
mass transfer to separation processes
Assessment:
A one hour (mid semester) written examination (20%)
Up to five problem set assignments (due throughout the semester) (20%)
One 3-hour (end of semester) written examination (60%). A pass of 40% out of the 60%
written examination must be attained in order to pass the subject.
1 x four hour laboratory class per semester assessment will be part of the problem sets
2 of 3
Generic Skills (you will use and develop)
- ability to apply knowledge of basic science and engineering fundamentals
- in-depth technical competence in chemical and biomolecular engineering
discipline
- ability to undertake problem identification, formulation and solution
- ability to utilise a systems approach to design and operational performance
- ability to learn, condense and take notes on technical materials in a lecture setting
Required Text:
Chemical Engineering Vol. 2
by Coulson & Richardson
Pergamon
Heat Transfer by Holman
McGraw-Hill
Recommended:
Mass Transfer Operations
Treybal, R.E.
McGraw Hill
Conceptual Design of Distillation
Systems
Doherty and Malone
Additional Text on Reserve in Library:
660.2842 SHER
Thomas K. Sherwood, Robert L.
Pigford, Charles R. Wilke.
Mass transfer
660.284248 HAND
Teh C. Lo, Malcolm H.I. Baird, Carl
Hanson.
Handbook of solvent extraction
660.284248 PRIN
Jan Rydberg, Claude Musikas, Gregory
R. Choppin.
Solvent extraction principles and
practice
660.2842 MCCA
McCabe, Warren L.
Unit operations of chemical engineering.
660.2842 GEAN
Geankoplis, Christie J.
Transport processes and separation
process principles (includes unit
operations)
660.2842 WANK
Wankat, Phillip C.
Equilibrium staged separations in
chemical engineering
Web Course Reference:
Black Board - Learning Management System
http://www.lms.unimelb.edu.au/
-Lecture materials index in by topic and date
-Assignments
-Announcements Page
-Discussion Forum for responses to questions
3 of 3
Staff-Student Expectations:
towards independent learning at Fourth Year
Even first year students are looking distinctly younger now
By now you should be an independent learner. Whilst some lecturers may at times
deliver knowledge, most will set up a framework which will encourage you to learn for
yourself. Lectures will offer less detail in terms of handouts, lecture materials and visual
presentations. You will often be required to do more extensive preparation before
attending lectures in order to benefit from them. You will certainly need to regularly
study after lectures, as they generally will only set the scene for your learning. You will
work on practice problems and tutorial outside of lectures, scheduling your own time to
work without tutorials sessions. Assignments will require you to critically analyse and
form your own opinionsvery different from reading a text and showing you know it!
In your group work you will be expected to be able to manage a team, motivate all to
contribute and show that you can make a team work efficiently and effectively.
Advanced Heat & Mass Transport Processes
CHEN90019 Semester 1, 2012
Tentative Schedule

WeekStarting Week Topics Lecturer
Feb th Nulticomponent
Bistillation Shoit Cut
Nethous
Bagastine
Naich Nulticomponent
Bistillation stagebystage
Bagastine
Bistillation iesiuue cuives
Bagastine
Bistillation iesiuue cuives
anu Aspen
Bagastine
Naich Solvent Extiaction Axial
Bispeision
Stevens
Apiil Nass Tiansfei with
Reaction
Bagastine
Apiil Sem Bieak
Apiil Nass Tiansfei with
Reaction BioSepaiations
Bagastine
Apiil Ausoiption Bagastine
Apiil Beat Tiansfei Rauiation Stevens
Nay Beat Tiansfei Rauiation Stevens
Nay Beat Tiansfei Conuuction Stevens
Nay Beat Tiansfei Conuuction Stevens
Nay SW0T vAC
}une to EXANS
Day Time Theater
Monday 1:00pm - 2:00pm Redmond Barry-Medley
Theatre
Wednesday 10:00am - 11:00am Baldwin Spencer-Theatre
Friday 2:15pm - 4:15pm Redmond Barry-Lowe
Theatre
No Lectures April 6
th
Good Friday
No Lectures April 25
th
ANZAC Day

18/02/09 MCD-part1
Section 1.0
MULTICOMPONENT DISTILLATION
Suggested References
:
Coulson & Richardson
Chemical Engineering
Volume II
Pergamon
Treybal, R.E.
Mass Transfer Operations
McGraw Hill
Smith, B.D.
Equilibrium Stage
Operations
McGraw Hill
660.2842 MCCA
McCabe, Warren L.
Unit operations of
chemical engineering.
660.2842 WANK
Wankat, Phillip C.
Equilibrium staged
separations in chemical
engineering
Doherty and Malone
Conceptual Design of
Distillation Systems
McGraw Hill
MD Content:
-VLE, Bubble Point and Dew Point
-short cut methods for sizing column
-rigorous techniques for tray by tray calculations
-Mass Transfer concepts
-Azetropic Distillation and Residue curves
18/02/09 MCD-part1
Multi-component Distillation is Important
Because:
- Binary Systems Rarely Encountered in Practice
- Extractive or Azeotropic Distillation is commonly three
component or more
Use of Entertainers
- Most common and preferable liquid-liquid separation
method whenever feasible
- Likely to be in the DESIGN PROJECT
KEY COMPONENTS [DEFINITIONS]
- Heavy Key - least volatile component present in
the distillate in appreciable amounts.
- Light Key - most volatile component present in
the residue/bottom in appreciable amounts.
- Light key - must be more volatile than the heavy
key.
- Non-Key A component the is not the light or
heavy key
- Distributed Component Component found in
tops and bottoms
- Non-Distributed Component In either tops OR
Bottoms
18/02/09 MCD-part1
Example:
Volatility : A > B > C > D > E
Feed Top Product Bottom
s
A A
B B B
C C C
D D D
E E
Light Key:__________
Heavy Key:___________
18/02/09 MCD-part1
Vapour-Liquid Equilibria (VLE) for
Multicomponent Distillation
Relative volatility used for equilibrium vapour-
liquid composition determinations defined
relative to a reference or key component
(usually the heavy key).
Relative Volatility Definition
o
ref
o
i
ref i
p
p
=
o
where
i p
k
i
o
i
ref i
component of pressure partial -
component to relative
component of y volatilit relative -
o
18/02/09 MCD-part1
Use Raoults Law for each component
i
i
i
ref
i
i ref
ref i
ref i
o
ref ref ref
o
i i i
x
y
K
K
K
x y
x y
p x p y
p x p y
=
=
= =
=
=
t coefficien
on distributi
and
o
Forms of VLE data for K
i
- Nomograph (see graph)
- Vapour Pressure Data
i
o
i
o
i i i
K
p
p
p x p y = = =
i
i
x
y
and
Using the Antoine Eq. for pure component saturated vapour
pressure:
C T
B
A P
sat
+
= log
For binary systems
( )
( )
x x
y y
y x
x y
1 2
1 2
1
1
1
1
+ =
+ = =
o
Remember
x y
p x p
i i
i i
o
_ _
_
= =
=
1
(Raoults Law)
18/02/09 MCD-part1 18/02/09 MCD-part1
P
Vap
for Benzene, Toluene and p-Xylene
0
100
200
300
400
500
600
700
800
25 35 45 55 65 75 85
Temp (C)
m
m
H
g
mmHg (B)
mmHg (T)
mmHg (X)
BP and DP??? BP and DP???
Bubble Point and Dew Point?
Have you looked at a thermo text
book
Bubble point -> What is boiling p g
point?
VLE of mixtures
Raults Law uses P
vap
(T) ( )
Non-ideal models van Laar,
Wilsonetc
K: one way to represent VLE
data
o: another way to represent
VLE data VLE data
When is alpha constant?
Vap Pres Vap Pres
Butane, Pentane, Octane
10
100
0.01
0.1
1
V
a
p

P
r
e
s

(
b
a
r
)
Butane (bar)
Pentane (bar)
Oct (bar)
0.001
0.01
60 70 80 90 100 110 120 130 140 150
Temp (C)
K
i
K
i
100
1
10
e
s

(
b
a
r
)
0.01
0.1 V
a
p

P
r
e
Butane (bar)
Pentane (bar)
Oct (bar)
0.001
60 70 80 90 100 110 120 130 140 150
Temp (C)
alpha
ij
alpha
ij
30
35
20
25
a
l
p
h
a
a (Butane)
a (Pentane)
a (Octane)
5
10
15
0
60 70 80 90 100 110 120 130 140 150
Temp (C)
Vap Pres Vap Pres
Phenol, o-cresol, m-cresol, Xylenols
Vapour Pressure
1
t
m
)
0.1
P
v
a
p

(
a
t
Phenol (ATM)
o-cresol (ATM)
l (ATM)
0.01
80 90 100 110 120 130 140 150 160 170 180
m-cresol (ATM)
Xylenols (ATM)
Temp (C)
alpha
ij
alpha
ij
Phenol, o-cresol, m-cresol, Xylenols
alpha
1.8
2
alpha - phenol
l h l
1
1.2
1.4
1.6
1.8
o
i
j
alpha - o-cresol
alpha - m-cresol
alpha - xylenols
0.2
0.4
0.6
0.8
o
0
80 90 100 110 120 130 140 150 160 170 180
Temp (C)
18/02/09 MCD-part1
Bubble Point & Dew Point
Bubble Point: Initial ______________ point of a
mixture.
Dew Point: Initial ______________ temperature
of a mixture.
Bubble Point
- Calculate vapour composition where liquid
compositions are known -
- Known Variables:
- Unknown Variables:
Bubble point can determined from distribution coefficient,
K
is
using the following equation:
0 . 1 = =
_ _ i i i
x K y
1. Assume Temperature
2. Get K
i
data (nomograph of P
vap
)
3. Calculate y
i
4. Check
0 . 1 =
_ i
y
5. If not, Choose new T
new
,
( )
( )
_
=
i i
old ref
new ref
x K
T K
T K
18/02/09 MCD-part1
Bubble point can determined from alphas
ref
i
ref i
ref
i
x
x
y
y
=o
(1)
sum over all components
ref
i iref
ref
x
x
y
_
=
o
1
(2)
Divide (1) and (2)
( )
_
=
i iref
i iref
i
x
x
y
o
o
(3)
Such that:
ref
i iref
K
x
1
= Eo
(4)
1. Assume temperature
2. Obtain ref i
o
from Equilibrium data
3. Calculate y
i
from (3)
4. Calculate
i
i o
i
x
py
p =
5. Check to see if the T from vapour pressure data is the
assumed T (same as checking 4)
18/02/09 MCD-part1
DEW POINT
Calculate liquid comp given vapour comp. if Ks are used
i
i
i
y
=
K
x
Or alphas
( )
( )
_
=
iref i
iref i
i
y
y
x
o
o

ref
iref
i
K
y
=
|
|
.
|
\
|
_
o
18/02/09 MCD-part1
Find The Bubble and Dew Point temperatures &
compositions for a mixture of 33 mole % n-hexane, 37%
n-heptane, & 30% n-octane at 1.2 ATM (1.22 kPa)
18/02/09 MCD-part1
Bubble Point: (see excel spread sheet)
Ref. comp =
Component C p
o o
i
@
ATM
p
K
o
i
i
2 . 1
= x
i
y
i
= K
i
x
i
Hexane
2.68 0.33
Heptane
1.21 0.37
Octane
0.554 0.30
E
New T =
Component C p
o o
i
@
ATM
p
K
o
i
i
2 . 1
= x
i
y
i
= K
i
x
i
y
i
Hexane
2.16 0.33
Heptane
0.93 0.37
Octane
0.41 0.30
E

E
18/02/09 MCD-part1
Dew Point:
Temp:
Component
ATM
p
K
o
i
i
2 . 1
=
Hexane
2.23 0.33 0.148
Heptane
1.01 0.37 0.366
Octane
0.458 0.30 0.655
E
1.169
New T =
Component C p
o o
i
@
ATM
p
K
o
i
i
2 . 1
=
Hexane
3.0 2.5 0.33 0.132 0.130
Heptane
1.38 1.15 0.37 0.3217 0.317
Octane
0.64 0.533 0.30 0.30 0.553
E
1.0162
E
1.00
18/02/09 MCD-part1
Set Up Table for Bubble Point using alpha:
18/02/09 MCD-part1
Flash Distillation
MultiComponent
For Each Component:
( )
i
i
i
B
F
D
x
f
f
f
x
y
=
1
(5)
In Flash D&B in Equilibrium:
|
|
.
|
\
|
+ = = 1
1
f
x
x
f
K
x
y
i
i
i
i
B
F
i
B
D
(6)
Solve for x
Bi
over all components N
c
( )
_ _
= =
+
= =
c
i
c
i
N
i i
F
N
i
B
K f
x
x
1 1
1 1
1
(7)
Find T and K
i
to satisfy (7), then use (6) to calculate
(y
Di
& x
Bi
)
Feed
x
Fi
x
Bi
liquid
(1-f)
vapour
(f)
y
Di
f = Fraction
Vaporized
Feed
x
Fi
x
Bi
liquid
(1-f)
vapour
(f)
y
Di
f = Fraction
Vaporized
18/02/09 MCD-part1
Example:
Using the previous Mixture at 1.2 ATM find the
temperature of the Flash for a flash distillation
where 60% of the feed is vaporized.
f = 0.6
x
1
= 0.19, x
2
=0.368 x
3
= 0.443
18/02/09 411-441 MulticomponetDistilaltion-part2.doc
MCD Part 2 Short Cut Methods
SHORT CUT METHODS FOR DISTILLATION
FENSKE TOTAL REFLUX EQUATION
o
ik
i k
k i
i
k
y x
y x
K
K
= =
or
y
y
x
x
i
k
ik
i
k
= o
for stage 1. (Reboiler) at total reflux mass balance
B V L
Bx Vy Lx
B i i
+ =
+ =
and
2 ,
at total reflux B = 0
( )
( )
= =
|
\
|
.
| =
|
\
|
.
|
|
\
|
.
| =
|
\
|
.
|
|
\
|
.
| =
|
\
|
.
|
|
\
|
.
| =
|
\
|
.
|
L V x y
x
x
x
x
x
x
y
y
x
x
x
x
ik
i
k
i
k
ik
i
k
i
k
i
k
i
k
and and
y
y
and
x
x
i
k
i
k
2 1
1
1
1
2
1
1
2
2 1
1
3
2 1
1
o
o
o o
o o
and so on
1 1
1
...
i i
c N N
k k
D
x x
x x
| | | |
= o o o o
| |
\ . \ .
if the os are replaced by an average o
1
1
1
for vapour overhead product
for liquid overhead product
N i i
r r
D
N i i
r r
D
y x
y x
x x
x x
+
| | | |
= o
| |
\ . \ .
| | | |
= o
| |
\ . \ .
N = No of stages in the column including the reboiler
r - reference component
average o obtained from
( ) o o o o
m T F B
=
1 3 /
T refers to top
F refers to feed
B refers to bottom
Fenske equations for total reflux
Assume - constant o
constant molar overflow
applied to simple 1 feed 2 product column

Example:
For a Total Condenser and partial reboiler with
the following compositions, calculate the
minimum number of theoretical stages.
Light Key:
Heavy Key:
Invariant Zones in MCD
- Invariant Zones (Pinch Points)
o -region where compositions from plate to plate are
equal.
o ie.
1 1
1 1
+
+
= =
= =
n n n
n n n
y y y
x x x
- In Binary Distillation one pinch point at feed
- In MCD two IV zones.
o In IV
Rect.
Heavy non-distributed compositions are zero
o In IV
strip
light non-distributed compositions are zero
- IV zones are used to Calculate Minimum Reflux Ratio
UNDERWOOD MINIMUM REFLUX EQUATIONS
Assumptions:
----
----
Mass Balance in Invariant Zone around top of column
D i n i n i
Dx Lx Vy
, , 1 ,
+ =
+
Note:
1 , , 1 ,
1 , , 1 ,
+
+
= =
= =
n i n i n i
n i n i n i
y y y
x x x
and 1 , 1 , + +
=
n i i n i
x K y
Using the Definition of K
i
D i n i
i
n i
Dx y
K
L
Vy
, 1 , 1 ,
+ =
+ +
Using:
r
i
i
K
K
= o
and
HK r
K K =
Solve for Vy
i,n+1
D i
HK i
n i
Dx
K V
L
Vy
, 1 ,
1 =
|
|
.
|
\
|
+
o
HK
i
D i i
n i
VK
L
Dx
Vy
=
+
o
o
,
1 ,
Sum Over All Components:
_
_
_
|
|
.
|
\
|
= =
+
C
i HK
i
C
i
D i i C
i
n i
VK
L
Dx
Vy V
o
o
,
1 ,
At minimum Reflux, V&L at min.
_
_
|
|
.
|
\
|
=
C
i HK
i
C
i
D i i
K V
L
Dx
V
min
min
,
min
o
o
Need to Know L
min
/V
min
and K
HK
What Underwood did: (simplification)
Let
HK
K V
L
min
min
= |
Sub in:
( )
_
=
C
i i
D i i
Dx
V
| o
o
,
min
Reflux is defined as:
R
L
D
L
V
R
R
D
V R
=
=
+
=
+ 1
1
1
a n d
Thus, In the Rectifying Section:
_
= +
C
i i
D i i
m
x
R
| o
o
,
1
(1)
In Stripping Section:
_
=
C
i i
B i i
m
x
R
| o
o
,
(2)
Combining (1) into (2)
q
x
C
i i
F i i
=
_
1
,
| o
o
(3)
q = fraction of feed that is liquid at the feed tray temperature
(3) is a polynomial in |
Position of Feed Plates
(Gilliland Criteria)
x
x
x
x
x
x
Lk
hk
Lk
hk
Lk
hk
|
\
|
.
| s
|
\
|
.
| s
|
\
|
.
|
+ on feed plate in feed on feed plate 1
General Rule for feed placement:
RIGOROUS MULTICOMPONENT
DISTILLATION
LEWIS MATHERSON METHOD
-column with total condenser
- partial reboiler
- single feed
- no side streams.
Computer vs hand calculations?
What we use from short cut methods:
- feed compositions specified
- overall mass balance methods
- specified recovery
- estimate of number of stages
NOMENCLATURE
Top of rectifying section Bottom of stripping section
v V y
i n n i n , ,
= v V y
i m m i m , ,
=
l L x
i n n i n , ,
=
l L x
i m m i m , ,
=
d Dx
i i D
=
,
b B x
i i B
=
,
v
i
= vapour component flow rate
l
i
= liquid component flow rate
PRELIMINARY ASSUMPTIONS
1. Number of stages KNOWN
2. Assume
(a) end product compositions
(b) temp profile
(c) liquid and vapour profile
Approach:
Calc. Stage by stage compositions from:
1. Top down in Rectifying Section
2. Bottom up in Stripping Section
-- COMPARE AT FEED
STAGE TO STAGE
CALCULATIONS
TOP DOWN
Total condenser
y x x
i N i D i N , , ,
= =
+1
For Composition in Tray N:
N i
y
, - Known
Use Dew Point Calculation to determine:
N i
x
, - Unknown
Also Sets Temperature form Dew Point Calc.

Assume Pressure
Dew Point ( Again)
( )
n
r i
r i
n
r i
y x
x y
|
|
.
|
\
|
=
,
o
or
n
r
n
r i
i
n i
K
y
x
|
|
.
|
\
|
|
|
.
|
\
|
=
1
,
,
o
Note
x
i
_
=1
so over x
i
K
y
r
i
i r i
c
n
=
|
\
|
.
|
=
_
o
1
(1)
Combine above
( )
( )
x
y
y
i n
i i r
n
i i r
n
i
c ,
/
/
=
=
_
o
o
1
(2)
Or in terms of vapour flow rates:
( )
( )
x
v
v
i n
i i r
i i r
n
i
c ,
/
/
=
=
_
o
o
1
(3)
SO
X
i,D
(in condenser) is specified
so is y
i,N
Dew Point gives x
i,N
Now for the next stage below the condenser: Need y
i,N-1
L
n
assumed l
i,n
from (3) (l
i,n
= L
n
x
i,n
)
Component balance over column
(Operating line in different form)
v l d
i N i N i , ,
= +
1 (4)
to give
v
i N , 1
remember:
( )
1 , 1 1 ,
=
N i N N i
y V v
Now Calculate x
i,N-1
and l
i,N-1
from Dew Point
Then Use (4) to get v
i,N-2
, i N i N i
d l v + =
1 , 2 ,
.and so on
From the TOP:
1 , , , +
= =
N i D i N i
x x y
From the Condenser
( )
( )
_
=
=
c
i
N r i i
r i i
N i
v
v
x
1
,
/
/
o
o
and (l
i,N
= L
N
x
i,N
) Dew Point
v l d
i N i N i , ,
= +
1 Operating Line
( )
1 , 1 1 ,
=
N i N N i
y V v
new vap frac. y
i,N-1
( )
( )
_
=

=
c
i
N r i i
r i i
N i
v
v
x
1
1
1 ,
/
/
o
o
and (l
i,N-1
= L
N
x
i,N-1
) Dew Point
i N i N i
d l v + =
1 , 2 ,
Operating Line
( )
2 , 2 2 ,
=
N i N N i
y V v
new vap frac. y
i,N-2
Dew Point
Operating Line
new vap frac.
BOTTOM UP
Same Procedure, but sue Bubble Point:
( )
( )
_
=
=
c
i
m i r i
m i r i
m i
l
l
y
1
,
o
o
(5)
and
i i i
b v l + =
1 , 2 , (reboiler = stage 1) (6)
i n i n i
b v l + =
+ , 1 , (general op line in stripping section)
TEMPERATURE AND Flow RATE PROFILES
Temperature Profiles from Dew and Bubble Points
Flow Rates require heat balance
Requires Enthalpy Balance !!!
Rectifying Section Heat Balance
c n n n n
q Dh h L H V + + =
+ + 0 1 1
since
D V L
n n
=
+1
( )
1
1 0
+
+
+
=
n n
c n
n
h H
q h h D
V
Stripping Section Heat balance
( )
1
1
+
+
+
=
m m
r B m
m
h H
q h h B
V
This assumes ideal vapour and liquid solutions, ie
n i
c
i
n i n
n i
c
i
in n
h x h
H y H
,
1
,
,
1
_
_
=
=
=
=
Convergence Criteria
Top and Bottom Compositions need to meet at feed.
- at roughly the feed compositions
( )
( ) ( )
i
B m
i
D m
i
D m B m
i
b x d x
x x
d
/ /
, 1 , 1
, 1 , 1
+ +
+ +
+
= A
Note
A A d b =
.
18/02/09 411-441 MulticomponetDistilaltion-part3.doc 18/02/09 411-441 MulticomponetDistilaltion-part3.doc
MULTI COMPONENT MASS TRANSFER
1. Equilibrium relation
n i
n
n in
in
l
L
V K
v
,
=
(1)
2. Component mass balance
0
, 1 , 1 , , ,
= +
+ n i n i n i n i n i
f v l v l
(2)
3. Enthalpy balance
0
,
1 1 1 1
=
+
+ +
n n F n
n n n n n n n n
q h F
H V h L H V h L
(3)
4. Summation equations
n n i
L l =
_ ,
(4)
n in
V v =
_
(5)
Matrix Solution
Eliminate v
i n ,
by combining equilibrium and component
balance equations, ie (1) and (2) gives
0
1 ,
1
1 1 ,
,
,
, 1 ,
=
+ +

+
n in n i
n
n n i
n i
n
n n i
n i n i
F Z l
L
V K
l
L
V K
l l
N N i N N i N
n n i n n i n n i n
i i i
i i
D l B l A
D l C l B l A
D l C l B l A
D l C l B
= +
= + +
= + +
= +
+
, 1 ,
1 , , 1 ,
2 3 , 2 2 , 2 1 , 2
1 2 , 1 1 , 1
where
N n
L
V K
A
n
n i,n
n
< < =

2
1
1 1
n
n n i
n
L
V K
B
,
1 + =
1 1 1 s s = N n C
n
n in n
F Z D =
or
(
(
(
(
(
(
(
(
(
=
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(

N
N
n
iN
iN
in
i
i
N N
N N N
n n n
D
D
D
D
D
L
L
L
L
L
B A
C B A
C B A
C B A
C B
1
2
1
1
2
1
1 1 1
2 2 2
1 1

18/02/09 SBS Example
Stage by Stage Example
Separate a mixture of n-butane, n-pentane and n-octane at in a single
feed column with a total condenser with a column pressure of 200
kPa. The feed is a saturated liquid at is boiling point with a
composition of C4 = 0.15, C5 = 0.25, and C8 = 0.60.
Recovery 99% of the C5 in the tops product and 98% of the C8 in the
bottoms.
The reflux ratio is 1. From earlier cal cautions using the Fenske and
Underwood equations you determined the number of ideal stages is 5
(N=5).
Find the outlet concentrations and the concentrations and temperature
for each stage.
Feed is 10,000 kg mol/hr
x
Fc4
= 0. 15
x
Fc5
= 0. 25
x
Fc8
= 0. 60
p = 200 kPa
L/D = 1
D
B
N = 5
N
feed
= 3
99% C
5
98% C
8
x
Fc4
= 0. 15
x
Fc5
= 0. 25
x
Fc8
= 0. 60
p = 200 kPa
L/D = 1
D
B
N = 5
N
feed
= 3
99% C
5
98% C
8
Thermodynamic Data
Need Ks or os
Vap Pres from Antoine Equation:
C T
B
A P
sat
+
= log
T is in Celsius, P
vap
is in bar
A B C
C
4
3.9327 935.77 238.79
C
5
3.9778 1064.84 232.014
C
8
4.051 1356.36 209.635
From Properties of Liquids and Gases, Prausnitz
Use Raults law and:
t
vap
i
i
p
p
K = =
i
i
x
y
where p
t
is the total pressure
See Excel Sheet for BP and DP Calculations!
Mass Balance
External:
F= D+B L/D = R =1
q=1 F= 10,000 kg mol/hr
From Fractional Recovery (kg mol/hr):
Feed Top Bot
C
4
1500
C
5
2500
C
8
6000
Tota
l
10000
D=
B=
Mole Fractions:
Feed Top Bot
C
4
0.15
C
5
0.25
C
8
0.60
Flow Rates in Column:
L Liquid Flow rate in Top of
Column
V- Vapour Flow Rate in Top of
Column
L
Liquid Flow rate in Bottom
of Column
V
- Vapour Flow Rate in Bottom
of Column
= = R
D
L

= = L D R*
=
+
=
1 R
R
V
L

= V
= + = qF L L
= =V V
Operating Lines:
Top of rectifying section Bottom of stripping section
V
D
x x
V
L
y
D i n i n i , , 1 ,
+ =

L
B
x y
L
V
x
B i n i n i , , 1 ,
+ =
+
C4:
C5:
C8:
TOP DOWN
Total condenser:
y x x
i N i D i N , , ,
= =
+1
For Composition in Tray N (=5):
N i
y
, - Known T=
Use Dew Point Calculation to determine: N i
x
, - Unknown
Vap Frac , N i
y
,
i
N i
N i
K
y
x
,
,
=
C
4
3.67
C
5
1.28
C
8
0.07
1. Calc 1 , N i
y
- From Operating Line T=
2. Use Dew Point Calculation to determine: 1 , N i
x
Vap Frac , 1 , N i
y
i
N i
N i
K
y
x
1 ,
1 ,
=
C
4
8.23
C
5
3.34
C
8
0.28
kPa
p
K
o
i
i
200
=
kPa
p
K
o
i
i
200
=
1. Calc 2 , N i
y
2. Use Dew Point Calculation to determine: 2 , N i
x
Vap Frac , 2 , N i
y
i
N i
N i
K
y
x
2 ,
2 ,
=
C
4
10.99
C
5
4.70
C
8
0.46
FEED is at 3 Stop!!!!
Bottom Up
Reboiler
1. Calc 4 , N i
y
- From Bubble Point T=
Liq Frac , 4 , N i
x
4 , 4 ,
=
N i i N i
x K y
C
4
-
C
5
7.86
C
8
0.97
kPa
p
K
o
i
i
200
=
kPa
p
K
o
i
i
200
=
1. Calc 3 , N i
x
2. Use Bubble Point Calculation to determine: 3 , N i
y
Liq Frac , 3 , N i
x
3 , 3 ,
=
N i i N i
x K y
C
4
-
C
5
7.24
C
8
0.86
1. Calc 2 , N i
x
2. Use Bubble Point Calculation to determine: 2 , N i
y
Liq Frac , 2 , N i
x
2 , 2 ,
=
N i i N i
x K y
C
4
-
C
5
5.39
C
8
0.56
kPa
p
K
o
i
i
200
=
kPa
p
K
o
i
i
200
=
Convergence Criteria
Top and Bottom Compositions need to meet at feed.
( )
( )
i
D m
i
D m B m
i
x
x x
d
, 1
, 1 , 1
+
+ +
= A
D Liq Frac
, 2 , N i
x
B Liq Frac ,
2 , N i
x
Ad
i
C
4
0.018 0 1
C
5
0.081 0.090 0.11
C
8
0.905 0.908 0.003
18/02/09 SBS Example2
Stage by Stage Example 2 Constant o

A feed consisting of 0.2 mf benzene, 0.3 mf toluene and the remainder o-xylene at its bubble
point at 1 atm is to be separated in a plate column. Calculate the number of theoretical plates
required to produce a bottoms of 0.99 mf o-xylene and a distillate of 0.005 o-xylene.

Neglect the pressure drop through the still
Assume constant relative volatility of
BX
= 6.4,
TX
= 2.7
Assume R = 5

Calculate also the minimum number of theoretical plates and the minimum reflux ratio.
Suggestions

Construct a mass balance as far as possible.
Decide on light and heavy keys and complete mass balance.
Calculate N
min
from Underwood-Fenske equation:
Hence estimate benzene in still (at total reflux) to justify choice of keys:
Calculating operating equations.
Use Lewis-Matheson plate to plate method, starting with condenser.
Estimate feed plate positions from Gillilands criterion.
When estimated feed plate reached (x
lK
/x
hK
= 0.6) check one plate either side to ensure x
R
/x
X

is a minimum further down column.
Still composition reached after 10 plates ( condenser + still) with feed on 6th plate down.
(B = 9 x 10
-6
, T = 0.00973, X = 0.9903 mf).

Sketch a graph of the composition change along column.

x
B,F
= 0. 2
x
T,F
= 0. 3
x
X,F
= 0. 5
p = 1 atm
L/D = R
N = ?
N
feed
= ???
x
B,D
=
x
T,D
=
x
X,D
= 0. 005
x
B,B
=
x
T,B
=
x
X,B
= 0. 99
x
B,F
= 0. 2
x
T,F
= 0. 3
x
X,F
= 0. 5
p = 1 atm
L/D = R
N = ?
N
feed
= ???
x
B,D
=
x
T,D
=
x
X,D
= 0. 005
x
B,B
=
x
T,B
=
x
X,B
= 0. 99
Mass Balance
F= D+B L/D = R =5 q=1 Let F= 1 kmol/hr
Need:
B D x x x x
D T B T B B D B
& , , , ,
, , , ,
Mole Fractions:
Feed Top Bot
Benzen
e
0.20
Toluene 0.30
Xylene 0.50
Fenske Eq
B
HK
LK
N
HK LK
D
HK
LK
x
x
x
x
|
|
.
|
\
|
=
|
|
.
|
\
|
min
o
43 . 9
7 . 2 ln
005 . 0
99 . 0
01 . 0
593 . 0
ln
min
=
(
|
.
|
\
|
|
.
|
\
|
= N
Checking Benzene Assumption
In Tops:
B HK
D HK N
D i F i
D i
x
x
Dx Fx
Bx
,
,
, ,
,
min
o =
99 . 0
005 . 0
4 . 6
4975 . 0 ) 2 . 0 )( 1 (
5025 . 0
43 . 9
,
,
=
D B
D B
x
x
402008 . 0
,
=
D B
x
In Bottoms:
B
HK
i
N
HK i
D
HK
i
x
x
x
x
|
|
.
|
\
|
=
|
|
.
|
\
|
min
o
4 . 6 ln
005 . 0
99 . 0 402008 . 0
ln
43 . 9
, (
(
|
.
|
\
|
|
|
.
|
\
|
=
B B
x
6
,
10 98894 . 1
= X x
B B
Gilliland Correlation
Flow Rates in Column:

L Liquid Flow rate in Top of
Column
V- Vapour Flow Rate in Top of
Column

L Liquid Flow rate in Bottom
of Column
V - Vapour Flow Rate in Bottom
of Column

5 = = R
D
L

4875 . 2
4975 . 0 * 5 *
=
= = L D R

6
5
1
=
+
=
R
R
V
L

( )
985 . 2
1
=
+
= L
R
R
V

4875 . 3 = + = qF L L

( ) 985 . 2 1 = + = F q V V

Position of Feed Plates (Gilliland Criteria)

1 plate feed on feed in plate feed on +
|
|
.
|
\
|
s
|
|
.
|
\
|
s
|
|
.
|
\
|
hk
Lk
hk
Lk
hk
Lk
x
x
x
x
x
x

match feed composition ratio -> minimum load on reboiler
TO use the feed criterion, do stage by stage from either the top or
the bottom, moving through the whole column.at feed plate
switch operating line.
TOP DOWN

Total condenser:

D i N i
x y
, ,
=

N i
x
,
B
0.402 0.2186
T
0.593 0.7640
X
0.005 0.0174

=
|
|
.
|
\
|
N
hk
Lk
x
x
( )
( )
_
=
iref i
iref i
i
y
y
x
o
o
Operating
Lines to
calculate
1 , N i
y
=
|
|
.
|
\
|
FEED
hk
Lk
x
x
For Composition in Tray N-1 (=10):
1 , N i
y

1 , N i
x
B
0.2490 0.1192
T
0.7353 0.8340
X
0.0153 0.0468

=
|
|
.
|
\
|
1 N
hk
Lk
x
x

N-2 (=9):
2 , N i
y
2 , N i
x
B
0.166 0.0722
T
0.793 0.817
X
0.0398 0.1107
N-3 (=8):
3 , N i
y
3 , N i
x
B
0.1272 0.0494
T
0.7797 0.7189
X
0.0930 0.2316

=
|
|
.
|
\
|
2 N
hk
Lk
x
x

=
|
|
.
|
\
|
3 N
hk
Lk
x
x

N-4 (=7):
4 , N i
y
4 , N i
x
B
0.1082 0.0360
T
0.6979 0.5509
X
0.1938 0.4131
N-5 (=6):
5 , N i
y
5 , N i
x
B
0.0970 0.0267
T
0.5579 0.3645
X
0.3450 0.6087

=
|
|
.
|
\
|
4 N
hk
Lk
x
x

=
|
|
.
|
\
|
5 N
hk
Lk
x
x

( )
( )
_
=
iref i
iref i
i
y
y
x
o
o
1 plate feed on feed in plate feed on +
|
|
.
|
\
|
s
|
|
.
|
\
|
s
|
|
.
|
\
|
hk
Lk
hk
Lk
hk
Lk
x
x
x
x
x
x
33 . 1 6 . 0 598 . 0 s s
N-5 (=6) is the feed plate!!!!
For N-6 (=5) Stage Switch to Bottom Operating Lines
6 , N i
y

6 , N i
x
B
0.0313 0.0047
T
0.4276 0.1521
X
0.8778 0.8432

=
|
|
.
|
\
|
6 N
hk
Lk
x
x

N-7 (=4):
2 , N i
y
2 , N i
x
B
0.00548 0.000968
T
0.1760 0.07371
X
0.8184 0.9253
N-8 (=3):
3 , N i
y
3 , N i
x
B
0.00113 0.0001
T
0.0844 0.0330
X
0.9143 0.9667

=
|
|
.
|
\
|
7 N
hk
Lk
x
x

=
|
|
.
|
\
|
7 N
hk
Lk
x
x

Operating
Lines to
calculate
6 , N i
y
( )
( )
_
=
iref i
iref i
i
y
y
x
o
o
N-9 (=2):
4 , N i
y
4 , N i
x
B
0.0002 0.0003
T
0.0369 0.0140
X
0.9628 0.9860
N-10 (=1):
5 , N i
y
5 , N i
x
B
0.0004 6.436x10
-6
T
0.0147 0.00549
X
0.9852 0.9945

=
|
|
.
|
\
|
9 N
hk
Lk
x
x

=
|
|
.
|
\
|
10 N
hk
Lk
x
x

Tray Efficiency
Cross Flow Sieve Tray (Treybal, Fig 6.15)
Bubble:
Sieve:
Valve:
Factors Effecting Efficiency
Flooding: Downcomer and tray fill up
Weeping: Liquid going through holes in plate
Factors Effecting Mass Transfer
- Vapour Flow Rate: E:
- Viscosity: E:
- Volatility: E:
MURPHREE PLATE EFFICIENCY
VAPOUR
E
y y
y y
y
n x
MV
n n
n n
n
n
=
1
1
*
*
composition of vapour in equilibrium
with the liquid leaving tray (ie )
LIQUID
E
x x
x x
x
y
ML
n n
n n
n
n
=
=
+
+
1
1
*
*
composition of liquid in equilibrium
with the vapour
MURPHREE POINT EFFICIENCIES
E
y y
y y
E
x x
x x
OG
n n
n n
OL
n n
n n
=
+
+
1
1
1
1
*
*
OVERALL EFFICIENCY
E
N
T
0
=
(no. of theoretical trays)
N (no. of actual trays)
A
AIChE technique for determining tray efficiency based on
1. vapour phase resistance to mass transfer
2. liquid phase resistance to mass transfer
3. hydrodynamics of the tray
4. entrainment
(Refer Perry; Chem. Eng. H/B)
18/02/09 411-441 MulticomponetDistilaltion-part3.doc 18/02/09 411-441 MulticomponetDistilaltion-part3.doc
Azeotropic Distillation
Reminder: What is an azeotrope:
For Pentane and Dicholormethane (Doherty and Malone, 2001)
at 750 mmHg
In a mixture an azeotrope is when the vapor and liquid
compositions are equal.
Which means:
Ways around an azeotrope
1. Is it pressure sensitive?
Does it change composition or go away completely with
changes in pressure?
2. If it is not pressure sensitive, use and entrainer
to break the azeotrope
Four Categories of Entrainers:
1. The entrainer does not induce liquid-phase separation in
multicompoent mixture
- Homogeneous azeotropic distillation
2. The entrainer does induce liquid-phase separation in
multicompoent mixture
- Heterogeneous azeotropic distillation
3. The entrainer reacts with one component in the binary
mixture
- Reactive Entrainer
4. Entrainers that ionically dissociate ( inorganic salts) and
move the composition of the binary azeotrope
Note: Azeotropes can exists for both binary or multi-
component mixture (ie. ternary azeotropes)
Approach: Develop a method to determine if
azeotropic separation is feasible in the Design stage
using Residue Curves
Simple Distillation Residues Curves
Simple batch distillation
Q Heat load
x and y liquid and vapor mole fractions
V vapor molar flow rate
H Instantaneous total number of moles in still
Overall mass balance:
V
dt
dH
=
(1)
Component mass balance:
( )
i
i
Vy
dt
Hx d
=
(2)
Expanding the derivative:
i i
i
Vy Vx
dt
dx
H =
(3)
Or
( )
i i
i
x y
H
V
dt
dx
=
(4)
A residue curve tracks changes in x
i
relative to
other components, but not time
Dont want to try to solve for time
Instead use a warped nonlinear time:
dt
H
V
d =
Applying to (4):
y x
x
=
d
d
(5)
Where x is a vector of (c-1) independent liquid phase
mole fractions, and y is for the vapor phase
c is the number of components
is a nonlinear transform of time defined over the interval 0

to
.
Time, t goes from 0 to t
max
, the point where the still boils away
(harder to deal with mathematically).
Solve (5) numerically using ODE solvers
-Need initial conditions for x at t=0
-Need VLE data to calculate y for each x vector
Example of solving (5) for a binary mixture
Ethanol and isopropanol
No azeotrope
Phase Line:
Trajectories on the phase line go from x =1 to x=0, show
changes in still composition with time.
(ethanol is boiling out of still over time)
Stable Node: The destination of the trajectories on a phase
line
Unstable Node: The origin of the trajectories on a phase line
AZEOTROPIC SYSTEM:
Pentane and Dicholormethane at 750 mmHg
Minimum boiling azeotrope at 52 mole % pentane
Boiling Points: Pentane (34.9 C) dichloromethane (38.9 C),
azeotrope (30.6)
Trajectories always move from low to high temperature
Methanol Ethanol Propanol
0 2 0 6 0 2
Propanol is
determined from 0.2 0.6 0.2
0.3 0.4 0.3
0.6 0.2 0.2
0.2 0.1 0.7
determined from
the other two
compositions (1-
X
meth
X
ethanol
). The
bottom axis is
methanol, the left methanol, the left
axis is ethanol and
the propanol is the
axis on the slope.
Azeotropic Distillation Part 2 Ternary Mixtures
Binary Phase line: straight line, but ternary systems are different:
- Each Vertex can be a NODE or a SADDLE POINT
- Nodes:
o Stable
When Arrows only go there
o Unstable
Where Arrows only leave there
- Saddle Point
o Some arrows going there,
some leaving
RESIDUE CURVE PROPERTIES:
Property 1
The residue curve through any given liquid composition point
is tangent to the vapor liquid equilibrium tie line through the same
point.
Property 2
Residue Curves do not cross each other, nor do they intersect
themselves
Property 3
The boiling temperature always increases along a residue
curve. (the only exception is at steady state where the boiling
temperature remains constant because the composition is
constant)
Property 4
Steady State solutions of the equations occur at all pure
components and azeotropes
Property 5
Steady State solutions are limited to one of the following types:
(1) stable node, (2) unstable node, (3) saddle point.

Property 6
Residue curves at nodes are tangent to a common direction. At
pure component nodes this common direction must be one of the
binary edges. The binary edge is determined for a Vertex, v (v=1, 2,
c) according to:
- Stable node if o
i,v
> 1 (i=1,2,c, i v)
- Unstable node if o
i,v
< 1 (i=1,2,c, i v)
- Otherwise it is a saddle
At a pure component node, the residue curves are tangent to the edge with the
relative volatility closest to unity, where the appropriate volatilities to check
are o
i,v
.
Methanol Vertex
o
n-propanol,methanol
= 0.25 o
ethanol,methanol
= 0.64
n-propanol Vertex
o
methanol,n-propanol
= 4.0 o
ethanol, n-propanol
= 2.1
Ethanol Vertex
o
methanol, ethanol
= 1.56 o
n-propanol,ethanol
= 0.48
Three Cases With one Minimum Boiling Binary Azeotrope:
(a)
(b)
(c)
Azeotrope
Property 7
Each distillation region must contain one unstable node, one
stable node and at least one saddle. Distillations regions are divided
by distillation boundaries.
Property 8
There must be a saddle on at lest one end of every distillation
boundary.
Two Min. Binary Boiling Azeotropes
How many distillation regions?

What is the product in each region?
Three Min. boiling binary Azeotropes and one Min. boiling Ternary
Azeotrope
How many distillation regions?

What is the product in each region?
Four Component Mixtures
Azeotropic Distillation Part 3
How RCM relates to packed and tray distillation:
RCM- apply to columns as well.
Sketching Residue Curves:
Property 2
Residue Curves do not cross each other, nor do they intersect
themselves
Property 3
The boiling temperature always increases along a residue
curve.
Property 4
Singular points occur at all pure components and azeotropes
Property 5
Singular Points are one of the following types: (1) stable node,
(2) unstable node, (3) saddle point.
Sketching Residue Curves:
Nomenclature:
Nodes:
N
1
Number of Pure Components
N
2
Number of Binary Azeo.
N
3
Number of Ternary Azeo.
Saddles
S
1
Number of Pure Components
S
2
Number of Binary Azeo.
S
3
Number of Ternary Azeo.
From Doherty and Malone
Examples:
Mass Balance and a Residues Curve:
Ideal
Non-Ideal
Separation Feasibility
Three Component Distillation:
The feed tops and bottoms are collinear on this type of graph
according to a lever rule:
F
F
F
F
z
z
z
z
, 1 D 1,
, 2 D 2,
, 1 B 1,
, 2 B 2,
z
z
z
z
(1)
z
i
mole fraction of either a vapour or liquid.
) 4 . 0 , 3 . 0 , 3 . 0 ( z
F
= , 049 . 0 , 95 . 0 , 05 . 0
, 2 , 1 , 1
= = =
D D B
z z z
From MB:
F B F D z
B
347 . 0 , 626 . 0 , 397 . 0
, 2
= = =
What a separation looks like:
47 . 2 , 35 . 6
3 , 2 3 , 1
= = o o ) 4 . 0 , 3 . 0 , 3 . 0 ( x
F
=
Spec:
Mole Frac in Tops: 0.95 pentane, 4.9% hexane, 0.1 % heptane
Mole Frac in Bot: 5% pentane
R = 2.5
n
s
= 5.0, n
R
= 4.3 n
T
= 8.3 Why?
Tie Line:
1 D
y x =
Pinch Points
Node Pinch Points Large Number of stages with no change in
composition
- At minimum reflux these the invariant zones (MCD, Lec 2)
- Position Sensitive to reflux and reboiler ratios
Saddle Pinch Points Asymptotically approach a particular point
- No analog in binary distillation
- Sensitive to composition specifications changes
Can be useful in determining minimum flows in MCD graphically
Node Pinch Points Another Use
Node Pinch Points move position with changes in condition.
Can be used to trace out the limiting distillate and bottoms
compositions in a separation.
Two Lever Rules:
Lever Rule for Stripping
Section Pinch Point
F 1, p 1,
F 2, p 2,
p 1, p 1,
p 2, p 2,
x x
x x
x y
x y
(2)
(2) Traces out the distillate limit
on the Right of the figure
Lever Rule for Rectifying
Section Pinch Point
p 1, F 1,
p 2, F 2,
p 1, p 1,
p 2, p 2,
x - x
x - x
x y
x y
=
(3)
(3) Traces out the bottoms limit
on the Left of the figure
A few more things:
Direct Split: Separates A from B and C (A/BC)
Indirect Split: Separate C from A and B (AB/C)
The line separating the Direct and Indirect lines is called the
Transition Line
- The transition split is AB/BC
- The dashed Tie line links the
F
y
vapor compositions that
would be in equilibrium with the liquid,
F
x .
Regions move with Feed!!!!
To Find Regions:
Eq (2) and (3) and:
( )
_
=
i iref
i iref
i
x
x
y
o
o
And
1 = =
_ _ i i
x y
Azeotropic Distillation
(No Distillation Boundaries)
Start from binary
mixture then add the
entrainer!
Residue Curve
Methyl Butyrate
Methanol
Toluene Azeo
F
o
Distillation Boundaries
i l (82 2
o
C)
Simple Distillation Boundaries (SDB)
i-propanol (82.2
o
C)
(80.4
o
C)
PDB Simple Distillation Boundaries (SDB)
are at infinite reflux
At finite reflux, the distillation region
overlap slightly.
Th b d i ll ti f
PDB
SDB
The boundary is a collection of
boundaries dependent on design
parameters.
This envelope is the region of overlap
between regions
Acetone (56.1
o
C) Water (100.0
o
C)
Chloroform(61.8
o
C)
between regions.
Region defined by SDB and a pitchfork
distillation boundary (PDB)
For curved boundaries, this envelope
i l h t d i
Chloroform(61.8 C)
(65.5
o
C)
PDB
is large enough to cross during
distillation
For straight boundaries this envelope
is too small and difficult if not
SDB
impossible to cross
Acetone (56.1
o
C) Benzene (80.1
o
C)
Some Comments about boundaries and splits
Straight boundaries are never crossed
Th t d b tt d t d f d t b i di l The tops and bottoms products and feed must be in a dingle
distillation region.
The entire separation series must also be in this region.
C d B d i b d i di till ti Curved Boundaries can be crossed in a distillation
Distillation Boundaries means some splits are infeasible,
some sharp splits that are feasible:
Unstable Node as distillate, bottoms determined by mass balance
Stable Node as bottoms, distillate determined by mass balance
One of more of the saddles as sidestreams (depending on feed ( p g
composition)
Entrainer is a third component used to break a binary
azeotrope p
Linear Distillation
Boundaries
Mass Balance for Case III
Start with binary azeo as Feed
AA
AB
F
1
F
2
AE
B E
Curved Distillation Boundaries
(exploiting the overlap region) (exploiting the overlap region)
Feed 1, F
1
, Compositions
1. Acetone 0.12
2. Chloroform 0.12
3. Benzene 0.76
F
1
Chloroform (61.8
o
C)
(65.5
o
C)
F
1 B
1
Acetone (56.1
o
C) Benzene (80.1
o
C)
1
D
1
B
1
Exploit the Overlap Region
B
1
lands in the overlap region
Curved Distillation Boundaries
(not exploiting the overlap region) (not exploiting the overlap region)
F
1
Feed 1, F
1
, Compositions
1. Acetone 0.4
2. Chloroform 0.1
3. Benzene 0.5
More Comments on Thermo:
For gases use an EOS, at least Peng-Robinson, but some more
sophisticated
For liquid phase: Activity Coefficient Model for moderately
non-ideal mixtures:
Two parameters for binary mixtures (fit to actual data):
- Van Laar, Margules, Wilson, UNIQUAC
Three parameters for binary mixtures
- NRTL (less desirable than two parameter models)
For mixtures of very different species (polar or associating compounds):
- Wilson, UNIQUAC and NRTL work better than van Laar
and Margules
For mixtures of very that form two liquid phases/partial solubility:
- UNIQUAC and NRTL work well, Wilson inapplicable
What does UNIQUAC stand for? (Universal quasichemical)
- Based on statistical mechanics of molecular structure
- Also has two parameters for binary mixtures fit to real data
UNIFAC (VLE or LLE) similar, but parameters not fit to data.
Same for ASOG (analytical solution of groups)
UNIFAC or ASOG or solubility parameter based data last resort
Why???????
Want to know more about these models?
READ Sandlers Thermodynamic book (its in the library)
THE ASPEN HELP FILE IS AMAZING!!!!!
Using Aspen for Distillation
(Fun with the Con Sep Block) (Fun with the Con Sep Block)
Rough directions for Residue Curve generation and
column design using Aspen are on LMS
Extractive distillation Extractive distillation
How can you tell which component comes out as distillate of the first
l ? column?
Product out top of first column Product out top of first column
Nonproduct out bottom of first column
Product has a higher relative volatility than nonproduct Product has a higher relative volatility than nonproduct
Extractive distillation
1
Extractive distillation
3 is the entrainer
Use the Scheibel Method:
Plot the isovolatility line o
1,2
= 1 on the composition triangle.
Thi li b i t th bi t d i t t th d
3
2
This line begins at the binary azeotrope and intersects one other edge
If it intersects the 1-3 edge, then component 1 has a higher volatility
than component 2 and component 1 is the distillate out of the first p p
(extractive column)
If it intersect the 1-2 edge, then component 2 has a higher volatility than
component 1 and component 2 is the distillate out the first column component 1 and component 2 is the distillate out the first column
The remaining component is separated from the entrainer in the second
(recovery) column.
Extractive distillation in Aspen Extractive distillation in Aspen
Binary Azeotrope between Benzene and Cyclohexane uses Aniline
h t i as a heavy entrainer
Plot the residue curve in Aspen Plot the residue curve in Aspen
Click the isovolatility line option
What goes out the top of the extractive column?
Stage by Stage Example 1 in Aspen Stage by Stage Example 1 in Aspen
Directions to make a residue curve in Aspen:
Go into Aspen Engineering Suite
Then Aspen Plus
Then Aspen Plus User Interface
It will prompt you on what to load. Initially load a template:
General with metric units
This will bring up a flow sheet.
You need to load Concept Design:
To do this go to Library, select references. In the dialog box, click on the Aspen Split
box. (if the aspen split box is not visible see directions below).
A new tab will appear at the bottom called conceptual design
Click on the column and create one on the flow sheet.
Then click on the stream button and put streams into and out of the column.
Double Click on the column to bring up the data browser. (things you need to enter for
the flow sheet to run have red icons next to them.)
First: Components
Enter three components. You may need to use the find function. Its similar to
Hysis.
Next: Properties
You need to select a fluid package. Go to: Specifications.
Process type: You can leave this on common, or you can specify depending on
the material. This affects what fluid packages are listed. ALL brings up all the
fluid packages available.
Base Method: Pick the fluid package you want to use. Remember the help file is
extensive and can be very helpful. It will also fill in the property method in the
column next to this one. Remember to choose wiselyjust like Indian Jones in
the search for the holy grail.
Go to: Parameters
You need to click on the parameters for the model you used so Aspen will use
them.
Then: Streams
You need to specify a few things in the feed stream to the column.
Click on Input:
Specify the pressure. Switch temperature to vapor fraction, then set the vapor
fraction (if q=1, then vapor fraction is 0).
Total Flow: At least put in 1, can specify more if you want.
Compositions: Feed compositions
Youve now specified enough to get the ternary map to work.
Goto: Tools, select Conceptual Design, then Ternary Maps
The default input options are fine, but you can always change things if you want.
The outputs give a lot of information. For the plot, just click on ternary plot.
Use the buttons on the right to put in curves. By clicking or by values.
You can print the plot and the other outputs.
To save all this, you can just save the simulation.
If Aspen Split is not listed in the library:
The ConSep block allows you to perform feasibility and design calculations for
distillation columns. Before you can use it, you must perform certain steps to configure
Aspen Plus to use it.
1 On the Library menu of Aspen Plus, select References.
2 If Aspen Split is listed, select the checkbox next to it. Otherwise, click Browse,
and navigate to the APrSystem installation folder, then the Gui\xeq subfolder, and select
Aspen Split.apm. Click OK.
Aspen Split should now be listed in the list of references, and the Conceptual Design tab
should appear following the other tabs in the Model Library.
Directions to make use conceptual design columns in Aspen:
This is not meant to be a complete step by step instruction sheet, but a guide for you to
figure out how to use this function in Aspen.
Go into Aspen Engineering Suite
Then Aspen Plus
Then Aspen Plus User Interface
It will prompt you on what to load. Initially load a template:
General with metric units
This will bring up a flow sheet.
You need to load Concept Design:
To do this go to Library, select references. In the dialog box, click on the Aspen Split
box.
A new tab will appear at the bottom called conceptual design
Click on the column and create one on the flow sheet.
Then click on the stream button and put streams into and out of the column.
Double Click on the column to bring up the data browser. (things you need to enter for
the flow sheet to run have red icons next to them.)
First: Components
Enter three components. You may need to use the find function. Its similar to
Hysis.
Next: Properties
You need to select a fluid package. Go to: Specifications.
Process type: You can leave this on common, or you can specify depending on
the material. This affects what fluid packages are listed. ALL brings up all the
fluid packages available.
Base Method: Pick the fluid package you want to use. Remember the help file is
extensive and can be very helpful. It will also fill in the property method in the
column next to this one. Remember to choose wiselyjust like Indian Jones in
the search for the holy grail.
Go to: Parameters
You need to click on the parameters for the model you used so Aspen will use
them.
Then: Streams
You need to specify a few things in the feed stream to the column.
Click on Input:
Specify the pressure. Switch temperature to vapor fraction, then set the vapor
fraction (if q=1, then vapor fraction is 0).
Total Flow: At least put in 1, can specify more if you want.
Compositions: Feed compositions
Youve now specified enough to get the ternary map to work.
Goto: Tools, select Conceptual Design, then Ternary Maps
The default input options are fine, but you can always change things if you want.
The outputs give a lot of information. For the plot, just click on ternary plot.
Use the buttons on the right to put in curves. By clicking or by values.
You can print the plot and the other outputs.
Next: Blocks: B1
Setup: Specify the pressure, components, model and phase.
To start leave Mode on mass balance.
Specifications: Enter specifications in either composition or percent recovery.
Once there are entered you are ready to run. Click the play button () on the main window
panel. This will launch the simulation in a separate dialog window.
In the new window, click on the ternary button to calculate the simulation. Once the
ternary diagram appears the calculation is complete. Click on the play button (triangle)
in the window to continue the simulation and return to the data browser. Or, click on the
menu button to explore the results.
Once you have returned to the data browser, witch the Mode to design. Now run the
simulation again.
In the new window you will need to enter the reflux ratio. Then click the ternary
plot/calculate button.
Use the tree menu button to browse through results and print output. When done, click
the triangle button to return to the data browser.

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CHEN90019 Semester 1, 2012

Lecture Materials for

Also Sets Temperature form Dew Point Calc.

is a nonlinear transform of time defined over the interval 0

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