Unit Operation and Separation Process Basic

Unit Operations: Introduction
Unit Operations: Introduction
What is chemical engineering? Chemical Engineering is a group of
industrial processes in which row materials are changed or separated
into useful products
Historical development: As the Industrial Revolution steamed along
certain basic chemicals quickly became necessary to sustain growth
- Example: Sulfuric acid was first among these "industrial chemicals".
http://www.pafko.com/
Chemistry:
- to create a new substance
- to study its properties
- to investigate all possible
pathways from one substance to
another
Chemical Engineering:
- to design the most optimal technology
for production of a specified substance
from row materials
- to develop and discover new
technological applications for materials

Process flowsheet: Example 1

Comparison of two processes
Comparison of two processes
Units:
Units:
- Heaters/heat exchangers
- Heaters/heat exchangers
- Pumps
- Pumps
- Distillation units
- Distillation units
- Reactors
- Reactors
- …
- …
Actions:
Actions:
- Heat exchange
- Heat exchange
- Material transport
- Material transport
- Separation
- Separation
- Mixing
- Mixing
- …
- …

Unit Operations:
Unit Operations:
- Unit Operations is a method of analysis and design of chemical
engineering processes in terms of individual tasks/operations
- It is a way of organizing chemical engineering knowledge into
groups of individual tasks/operations
- A unit operation: basic step in a chemical engineering process

Unit Operations: Classification
Unit Operations: Classification
Fluid flow processes
- fluid transport
- solids fluidization
- mixing
Heat transfer processes
- heating/cooling
- evaporation/condensation
Mass transfer processes
- absorption
- distillation
- extraction
- adsorption
- drying
Thermodynamic processes
- liquifaction
- refrigeration
Mechanical processes
- crushing
- sieving
- solid transportation

Section Lectures Tutorial
1. Introduction, revision,
binary distillation with non-
constant molar overflow
2. Separation of
multicomponent mixtures
3. Separations in packed
columns. Absorption
4. Adsorption processes
5. Humidification processes
6. Drying processes
7. Revision and problems
1.1 Introduction to Unit Operations;
Equilibrium stage operations (L1)
1.2 Thermodynamics of distillation (L2)
1.3 Binary distillations review (L3)
2.1 Multicomponent Distillation: Flash
distillation (L4)
2.2 Multicomponent Distillation: Short
Cut Methods (L5)
2.3 Multicomponent Distillation: Short
Cut Methods (L6)
3.1 Mass transport theories review (L7)
3.2 Mass transport theories review (L8)
3.3 Packed bed columns (L9)
3.4 Packed bed columns (L10)
4.1 Principles of adsorption (L11)
4.2 Principles of adsorption (L12)
5.1 Principles of humidification (L13)
5.2 Methods of humidification (L14)
6.1 Principles of drying (L15)
6.2 Methods of drying (L16)
7.1 Revision (L17)
7.2 Revision (L18)
http://www.see.ed.ac.uk/~lsarkiso/UnitOps/syllabus.htm

Total 18 lectures and at least 6 tutorials
Tutorials We will have a number of tutorials
focusing on specific examples of unit
operations.
Assessment Unit operations 1.5h exam. 2
questions
Text books 1) Warren L. McCabe, Julian C. Smith and
Peter Harriot, Unit Operations
of Chemical Engineering,
(Seventh Edition). McGrawHill, 2005.
2) Robert E. Treybal, Mass Transfer
Operations (McGraw-Hill Classic
Textbook Reissue Series) (Paperback)
3) J.D. Seader and Ernest J. Henley,
Separation Process Principles, John Wiley
& Sons, 1998.

Chemical separation processes:
Chemical separation processes:
required background
required background
B
B
V
V
D
D
- How do we know that at pressure P
and temperature T, vapour and liquid
phase are present in the system?
- What is the composition of the phases?
Chemical engineering thermodynamics
Chemical engineering thermodynamics
La,xa
Va,ya
Vb,yb Lb,xb
- How do we know the amount of mass
exchanged by two phases?
- What is the new composition of the
phases?
Mass transfer methods
Mass transfer methods

Chemical separation processes
- play a central role in chemical engineering
Use
Technology maturity
Distillation
Gas Absorption
Crystallization
Adsortion
Membranes
Chromatography

Objective:
take a mixture of components and
produce one or more products with
desired composition/purity
A+B
A B
B A
- A and B an be somehow different:
- boiling points
- size
- polarity
- etc.
)
(
/
)
(
1 A
n
A
n
SF F

nF
n1
n2 )
(
/
)
( 2
1 A
n
A
n
SR 
split factor
split ratio
)
(
/
)
( B
SR
A
SR
SP  separation power

Distillation process design
Step 1: Thermodynamics data and methods to predict equilibrium
phase compositions
- Separation utilizes the difference in volatility of
components

Thermodynamic considerations and phase
equilibria: Binary fluids
T
xA
Tb(B)
Tb(A)
V
L
y*
x*
T, P
V
L

T
xA
Tb(B)
Tb(A)
V
L
y*
x*
T, P
V
L
P=const

equilibria
equilibria
T
xA
Tb(B)
Tb(A)
V
L
y*
x*
P=const

equilibria
equilibria
T
xA
Tb(B)
Tb(A)
V
L
y*
x*
KA = yA/xA
KB = yB/xB
=(1-yA)/(1-xA)
Distribution of a component
among the two phases can
be characterized with a K-value
B
B
A
A
B
A
AB
x
y
x
y
K
K
/
/



Relative volatility

T
xA
Tb(B)
Tb(A)
V
L
T1
T2
T3
T4
x1 y1
x2 y2
x3 y3
x4 y4
yA
xA
T1
T2
T3
T4
Lets consider a binary mixture AB, where B is a heavy component (high boiling point) and
a is a light component (low boiling point). A T-x phase diagram of AB mixture, where x is a
mole fraction of component a might look like this at some constant pressure P. This phase diagram
can be also transformed in y-x diagram where composition of vapour phase in terms of mole
fraction of component A is plotted as function of the liquid phase composition.

T
xA
Tb(B)
Tb(A)
P1
P3
P2
P4 P1<P2<P3<P4

Thermodynamic data for mixtures
Thermodynamic data for mixtures
- graphs (T-y-x, P-y-x, y-x), tables (usually limited to binary mixtures)
- K-values, relative volatilities
DePriester charts
- Analytical methods (part of most chemical process design software)
ASPEN Tech
See a brief ‘Thermodynamics of multicomponent
phase equilibria file’
Promax
- Simplified models
Ideal gas/Ideal solution

equilibria: multicomponent mixtures
equilibria: multicomponent mixtures
For multicomponent mixtures simple graphical representations of
vapour-liquid equilibria data do not exist
Most often such data (including binary systems) is represented in terms of
K values defined as:
correlated empirically or theoretically in terms of temperature pressure
and composition
The ratio of two K-values, or relative volatility, indicates the relative ease
or difficulty of separating components i and j
Ki = yi/xi
j
j
i
i
j
i
ij
x
y
x
y
K
K
/
/




Light hydrocarbon mixtures: DePriester charts
Light hydrocarbon mixtures: DePriester charts
(1953)
(1953)

Thermodynamic data for mixtures:
Thermodynamic data for mixtures:
Simplified models
Simplified models
Raoult’s law (Ideal solution/ideal gas):
s
i
i
i P
x
p  pi is the partial pressure of component i
Dalton’s law (Ideal gas):
P
y
p i
i 
K-value for ideal gas/ideal solution system:
P
P
K s
i
i /

Relative volatility for ideal gas/ideal solution system:
s
j
s
i
j
i P
P
K
K /
/ 
Antoine equation:
i
i
i
s
i
C
T
B
A
T
P



)
(
ln
T, P
V
L

Thermodynamic calculations using K-values
Bubble
Bubble point
point
Dew point
Dew point
Two phase systems
Two phase systems
-Given P, T, V/(V+L), find
Given P, T, V/(V+L), find
x*, y*
x*, y*
- Given P, T, x*, y* , find
Given P, T, x*, y* , find
V/(V+L)
V/(V+L)
T
xA
Tb(B)
Tb(A)
V
L
y*
x*
P
Easy for 2 component system, if T-x-y diagram is available
(remember the lever rule?)
What about the multi-component system?

Bubble
Bubble point
point
T
V
L
- Model system: binary mixture A, B
- Consider the process in the figure: we start with
a mixture of composition 1 and temperature T1
and start increasing the temperature
- As we increase the temperature we are going to
reach a point where the first bubble forms
- The vapour in this bubble obeys:
- On the other hand:
- Thus as we increase the temperature we put new
K-values in the above equation until this condition is met
1

 B
A x
x
1

 B
A y
y
1
)
(
)
( 
 B
B
A
A x
T
K
x
T
K
1

Bubble point
Bubble point
Procedure:
a) Select T
b) Ki(T)
c)
d) if T is too high
e) Adjusting T
g) Final composition can be corrected using

i
i
i x
K
1


i
i
i x
K


i
i
i
i
i
i
x
K
x
K
y

Dew
Dew point
point
- Model system: binary mixture A, B
- Consider the process in the figure: we start with
a mixture of composition 1 and temperature T1
and start decreasing the temperature
- As we decrease the temperature we are going to
reach a point where the first drop of liquid forms
- The liquid in the droplet obeys:
- On the other hand:
- Thus as we decrease the temperature we put new
K-values the above equation until this condition is met
1

 B
A y
y
1

 B
A x
x
1
)
(
/
)
(
/ 
 T
K
y
T
K
y B
B
A
A
T
V
L
1

Procedure:
a) Select T
b) Ki(T)
c)
d) if T is too low
e) Adjusting T
g) Final composition can be corrected using

i i
i
K
y


i
i
i
i
i
i
K
y
K
y
x
/
/
1


i i
i
K
y
Dew
Dew point
point

Two phase system
Two phase system
F
i
z
1
1,T
P
Given the overall composition,
- How do you know that you a 2-phase system?
- How much vapour do you have per mole of the system?
- What is the composition of the vapour and liquid phases?
T
xA
Tb(B)
Tb(A)
V
L
y*
x*
1
1,T
P
1
1,T
P

B
i h
x
B ,
,
D
i h
y
D ,
,
F
F
i h
z
F ,
,
1
1,T
P
- The liquid mixture is partially vaporized
in a boiler (or vapour condensed in a cooler)
- How do you know that you a 2-phase system at a given T and P?
- How much vapour did you form per mole of feed?
- What is the composition of vapour and liquid phases?
T
xA
Tb(B)
Tb(A)
V
L
y*
x*
1
1,T
P
Isothermal flash separation

D
D
i h
y
D ,
,
F
F
i h
z
F ,
,
1
1,T
P
Objective: find D, B, and
their compositions
B
B
i h
x
B ,
,
F
D /



Isothermal multicomponent flash separation
D
D
i h
y
D ,
,
F
F
F
F
i T
P
h
z
F ,
,
,
,
1
1,T
P
their compositions
B
B
i h
x
B ,
,
 



i i
i
F
i
K
K
z
0
)
1
(
1
)
1
(

Rachford-Rice equation
- drums, condensers, reboilers etc

their compositions
Procedure:
1) Check the feasibility of the process: do you have two phases in coexistence
at given T, P?
a) all Ki > 1 – superheated vapour
b) all Ki < 1 – subcooled liquid
c) some Ki>1 and some Ki<1, then try
0

 0
)
1
(
)
1
(
1
)
1
(







 i
i
F
i
i i
i
F
i
K
z
K
K
z

subcooled liquid
overheated vapour
1

 0
)
1
1
(
)
1
(
1
)
1
(







 i i
F
i
i i
i
F
i
K
z
K
K
z


their compositions
Procedure:
 




i i
i
F
i
K
K
z
f 0
)
1
(
1
)
1
(
)
(


2) solve for F
D /


a) Guessing
)
(
f


their compositions
Procedure:
 




i i
i
F
i
K
K
z
f 0
)
1
(
1
)
1
(
)
(


2) solve for F
D /


b) Newton-Raphson
)
(
'
)
(
)
(
)
(
)
(
)
1
(
k
k
k
k
f
f



 


 
 



i i
i
F
i
K
K
z
f 2
2
)
1
(
1
)
1
(
)
(
'



Multicomponent flash separation (Adiabatic)
B
i h
x
B ,
,
D
i h
y
D ,
,
F
F
F
F
i T
P
h
z
F ,
,
,
,
1
1,T
P
- Liquid feed is heated under pressure and then adiabatically flashed through
the pressure reducing valve
F
F
T
T
P
P


1
1
F
i
i
i Fz
Dy
Bx 

F
D
B 

F
D
B Fh
Dh
Bh 
 0
)
1
( 


 F
D
B h
h
h 


their compositions
Procedure: 1) Guess T1
2) Isothermal flash procedure
3) Validate
0
)
1
( 


 F
D
B h
h
h 
 not
 




i i
i
F
i
K
K
z
f 0
)
1
(
1
)
1
(
)
(

 

their compositions
Procedure: 1) Guess
2) Isothermal flash procedure
find temperature of the flash drum
so that:
3) Validate
0
)
1
( 


 F
D
B h
h
h 

not
 




i i
i
F
i
i
K
K
z
K
f 0
)
1
(
1
)
1
(
)
(

F
D /


B
D
B
F
h
h
h
h




1
T

Distillation processes
Distillation is a process where a
feed mixture of two or more
components is separated into
products, of compositions different
from the feed. This process takes
advantage of the differences in
distribution of components between
the vapour and liquid phase.
F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
Overhead product D, xd
boiler
Vb, yb
Bottom product B, xb
R=La/D

The feed is material is introduced at
one or more points along the
column.
Liquid runs down the column from
tray to tray, where as vapour is
ascending along the column.
At each tray vapour and liquid
contact and mix with each other
F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
R=La/D

Liquid at the bottom of the column
is partially vaporized in a heated
reboiler.
The boil-up is send back to the
column.
The rest is withdrawn as bottoms,
or bottoms product
F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
R=La/D

Vapour at the top of the column is
cooled and condensed in the
overhead condenser.
Part of this liquid is returned back
to the column and the rest is
withdrawn as distillate or overhead
product
F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
R=La/D

At each stage of the column two
phases come in contact with each
other, mix, approach thermal and
composition equilibrium to the
extent which depends on the
efficiency of the contact stage F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
R=La/D
Lin,xin
Lout,xout
Vout,yout
Vin,yin

Definition of a stage in a process
Definition of a stage in a process
A single stage is a device or a subunit of the process,
where two (or more) phases of a different composition
come in contact with each other, exchange and leave
with new compositions
Lin,xin
Lout,xout
Vout,yout
Vin,yin
- Mass balance
• Overall
• Components
out
out
in
in V
L
V
L 


out
out
out
out
in
in
in
in y
V
x
L
y
V
x
L 


- Energy balance
Lin,hin
Lout,hout
Vout,hout
Vin,hin
Q
out
out
out
out
in
in
in
in h
V
h
L
Q
h
V
h
L 




Equilibrium stage processes
Equilibrium stage processes
Streams leaving the stage are in thermodynamic
equilibrium with each other
Lin,xin
Lout,xout
Vout,yout
Vin,yin
Streams coming to the stage: not in equilibrium
F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
R=La/D
The idea is then to consider a
hypothetical column, composed
of equilibrium stages
This idealistic design can be
converted to the actual design via
analysis of tray efficiency

The lighter component tends to
accumulate in the vapour phase
The heavier component tends to
accumulate in the liquid phase
F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
R=La/D

In general, the overall separation
process depends on:
- relative volatilities
relative volatilities
- number of contacting
number of contacting
stages
stages
- ratio of liquid and
ratio of liquid and
vapour flowrates
vapour flowrates
F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
R=La/D

If the feed is introduced at one point, it
divides the column into a rectifying
and stripping sections
But usually there are multiple feed
location and various side streams F, zf
Va, ya
La, xa=xd=y1=ya
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1
yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
R=La/D

Step 1: Thermodynamics data and methods to predict equilibrium
phase compositions
Step 2: Design of equilibrium stage separation
• Design problem type 1: To determine the number
of equilibrium stages required to accomplish the
desired separations
• Design problem type 2: Given a particular column
design, determine separation that can be accomplished
Step 3: Develop an actual design by applying the stage efficiency analysis
to equilibrium stage design

Design of equilibrium stage distillation: Binary
Design of equilibrium stage distillation: Binary
Mixtures Review
Mixtures Review
Va, ya
La, xa=xd
F, zf
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1 yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb

Operating lines
Operating lines
Va, ya
La, xa=xd=y1=ya
F, zf
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1 yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
Rectifying
section
R=La/D

Operating lines
Operating lines
D
n
n
n
n
n
D
n
n
n
n x
V
D
x
V
L
y
Dx
x
L
y
V
1
1
1
1
1




 




This equation is a straight line (V=const, L=const, L/V=const) if:
- Two components have similar and constant molar
enthalpies of vaporization (latent heats)
- Component sensible enthalpies changes and heats of mixing
are negligible (compared to latent heats)
- The column is well insulated (adiabatic)
- Pressure is uniform throughout the column

Operating lines
Operating lines
D
n
n x
V
D
x
V
L
y 

1
1
/
/
/
;







R
R
D
D
D
L
D
L
D
L
L
V
L
D
L
R Reflux ratio
Va, ya
La, xa
Ln xn Vn+1 yn+1
condenser
R=L/D
D
n
n x
R
x
R
R
y
1
1
1
1






Operating lines
Operating lines
D
n
n x
R
x
R
R
y
1
1
1
1





y
x
xD
slope=R/(R+1)
D
x
R 1
1


Operating lines
Operating lines
Va, ya
La, xa
F, zf
Lb, xN
Ln-1 xn-1
Vn yn
Ln xn
Vn+1 yn+1
Lm-1 xm-1
Vm ym
Lm xm
Vm+1 ym+1
condenser
boiler
Vb, yb
Stripping
section
R=La/D

B
m
m Bx
y
V
x
L 
 1
Operating lines
Operating lines
Lm xm
Vm+1 ym+1
boiler
B
m
m x
V
B
x
V
L
y 

1
y
x
xB
yB
slope= V
L

Operating line equation: Stage-by-stage
y
x
xa
ya
Va, ya
La, xa
Plate 1
Plate 2
Plate 3
L1, x1
x1

y
x
xa
ya
Va, ya
La, xa
Plate 1
Plate 2
Plate 3
Using the operating line equation we can
calculate y2 from x1. This step is depicted
by think green line in the graphs.
This process can be continued to
calculated the number of theoretical
stages. This method of graphical
construction of theoretical stages is called
McCabe Thiele method
L1, x1
x1
V2, y2

Feed stage considerations
L V
F
L V
F
bubble point
liquid feed
L V
F
dew point
vapour feed
L V
F
subcooled
liquid feed
F
L V
superheated
vapour feed
L V
F
partially vaporized
feed

F
V
V
F
L
L
q




 1
L V
F
L

L V
F
L V
F
bubble point
liquid feed
L V
F
dew point
vapour feed
L V
F
subcooled
liquid feed
F
L V
superheated
vapour feed
L V
F
partially vaporized
feed
q=1 q=0
q>1 q<0 0<q<1
F
L
L
q



L
L
F
V
V
F
D
B
q
z
x
q
q
y F




1
1
Feed line
F
L
L
q



q
z
x
q
q
y F




1
1
Feed line behavior
Feed line behavior
x=zf
F
F
F z
q
z
z
q
q
y 




1
1
y=zf
y
=
x
x=zf
q<0
q=0
0<q<1
q=1
q>1
y
x

Complete picture
Complete picture
D
n
n x
R
x
R
R
y
1
1
1
1





B
m
n x
V
B
x
V
L
y 

1
q
z
x
q
q
y F




1
1
y
x
zf
zf
xB xD
y1
yB
xN
D
x
R 1
1


Complete picture
Complete picture
D
n
n x
R
x
R
R
y
1
1
1
1





B
m
n x
V
B
x
V
L
y 

1
q
z
x
q
q
y F




1
1
y
x
zf
zf
xB xD
y1
yB
xN

Limiting cases
Limiting cases
y
x
xD
slope=R/(R+1)
R=L/D

Limiting cases
Limiting cases
y
x
slope=R/(R+1)
R=L/D

Total reflux
Total reflux
F=0
D=0
R=L/D=∞
L/V=1
B=0
Total reflux
D
n
n x
R
x
R
R
y
1
1
1
1





If R=L/D= ∞ then R/(R+1)=1; also L=V
n
n x
y 
1
y
x
zf
zf
xB xD
y1
yB
xN
Total reflux=Minimum number of stages
1
1





n
n
n
n
V
L
V
D
L

Minimum number of stages
Minimum number of stages
a) Graphical methods
R=L/D
F, z
D, xD
B, xB
y
x
xB xD
b) Short cut methods: Fenske Equation

Fenske Equation
Fenske Equation
B
B
AB
N
AB
AB
D
D
x
x
x
x




1
1
,
1
,
2
,
1 

 
B
B
N
D
D
x
x
x
x




1
)
(
1
1
min

1
ln
)]
1
(
/
)
1
(
ln[
min 



AB
D
B
B
D x
x
x
x
N

Fenske equation

Limiting cases: minimum reflux
y
x
xD
slope=R/(R+1)
R=L/D

y
x
slope=R/(R+1)
R=L/D
If we decrease the reflux ratio, then

y
x
zf
zf
xB xD
y1
yB
xN
If we decrease the reflux ratio, then
D
n
n x
R
x
R
R
y
1
1
1
1





B
m
n x
V
B
x
V
L
y 

1
q
z
x
q
q
y F




1
1

y
x
zf
zf
xB xD
y1
yB
xN
If we decrease the reflux ratio, then we
are arriving at a condition where both
the rectifying, stripping and feed line
intersect at the equilibrium line.
In order for this process to take place we
need an infinite number of plates
The minimum reflux
ratio condition

D
n
n x
R
x
R
R
y
1
1
1
1





y
x
zf
zf
xB xD
y1
yB
xN
At this point: xn=x* and yn+1=y*
x*
y*
*
*
*
1
1
*
1
*
min
min
min
min
x
y
y
x
R
x
R
x
R
R
y
D
D








Unit Operation and Separation Process Basic

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