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L/O/G/O
單元操作 (三)
Chapter 20
Equilibrium-Stage
Operations
化學工程系 李玉郎
Unit Operations (III)
Spring, 2014
 Instructors:
– Part 2: 李玉郎教授, Office number: 93612
Ext: 62693
 Teaching assistants:

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– Part 2: 凌煥松, Office number: 93952 Ext: 222
Part 2: (about 5.5 weeks, from March 26 to May 8)
B20
Equilibrium-stage operations
B21
Distillation
B22
Introduction to multi-component distillation
B23
Leaching and extraction
 Examination: May?19-23 or 26-30
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Equilibrium-Stage Operations
Mass transfer device
In(1)
Out(2)
contact
Mass transfer
device
mix
separate
Out(1)
In(2)
In(1) ,In(2) :not in equilibrium with each other
Out(1) ,Out(2) :usually not in equilibrium either
Equilibrium-Stage Operations
Simplification
assume the stage to be ideal
Out(1) and Out(2) are in equilibrium
correction factor, efficiency
Ideal stage 
 Real stage
Equilibrium-Stage Operations
Cascades(串級) : multistage systems
with two streams move counter currently.
Driving force for transfer:departure
from equilibrium
Vn
Vb
Ln=Lb
N
Ln-1
N-1
Vn-1
V3
Ln-2
L2
V1=Va
V2
2
L1
1
Stages
units
La
Equipment for Stage Contacts
Typical distillation equipment
decrease
Temp.
increase
FIGURE 20.1
(a) Reboiler with fractionating column: A, reboiler; B, heating
element; C, column; D,condenser.(b) Detail of sieve plate.
Equipment for Stage Contacts
Ln-1
Xn-1
Vn
yn
Ln
Xn
Vn+1
yn+1
FIGURE17.2
Sieve plate: A, tray or plate; B, perforations; C, downcomer to
plate below; D, downcomer from plate above.
Distillation
Reflux entering the top column  often at Tb (boiling point)
At each stage : T (vapor) = T (liquid) = Tb
Reflux increases the purity of the overhead product, but
increases the energy cost.
Energy cost is a larger part of the total cost of separation
by distillation.
Distillation
Mass transfer direction
V
(heat of vaporization)
low boiler
liquid
vapor
high boiler
(heat of condensation)
L
 heat of vaporization ≈ heat of condensation
 constant molar flow rate was always assumed
Distillation
rectifying section (精鎦段):
the vapor stream is enriched in low boilers.
feed
Stripping or enriching section(汽堤段):
The liquid is stripped of low boilers and
enriched in high boiling components.
Typical leaching equipment (溶提)
Soluble material is dissolved from its mixture with
an inert solid by means of a liquid solvent.
extracting solvent
FIGURE 20.2
Countercurrent leaching plant: A, launder; B, rake; C, slurry pump.
Principles of Stage Processes
L0
x0
VN+1
yN+1
N
total stage
n
stage number from top
V
molar flow rate of V phase
L
y
molar flow rate of L phase
x
mole fraction of A in L
mole fraction of A in V
Hv
enthalpy per mole of V phase
HL
enthalpy per mole of L phase
FIGURE 20.3
Material-balance diagram for plate column
Material balance
Consider the portion includes stage 1  n:
total:
La  Vn1  Ln  Va
(20.1)
component A:
La a  Vn1 yn1  Ln n  Va ya
(20.2)
Overall material balances of entire cascade:
total:
La  Vb  Lb  Va
(20.3)
component A:
La  a  Vb yb  Lb b  Va ya
(20.4)
Enthalpy balance
La  Vn1  Ln  Va
1n stages: La H L,a  Vn1HV ,n1  Ln H L,n  Va HV ,a
La  Vb  Lb  Va
entire cascade:La H L,a  Vb HV ,b  Lb H L,b  Va HV ,a
(20.5)
(20.6)
For systems containing only two components, it is possible to
solve many mass transfer problems graphically.
Based on: (1):material balance  operating line
(2):equilibrium relationship  equilibrium line
Operation-line diagram
from (20.2)
Ln
Va ya  La  a
yn 1 
n 
Vn1
Vn 1
(20.7)
Eq(20.7) is the operation-line equation, relates yn+1 and xn
A point on the operation line should be (xn ,yn+1)
If Ln ,Vn+1 are constants,
(20.7)
L
L
yn 1   n  ( ya   a )
V
V
a straight line,
slope: L V
intercept :
L
ya   a
V
Vn
yn
n
Ln
Vn+1
xn
yn+1
Overall the entire cascade
L
L
y N 1   N  ( ya   a )
V
V
y N 1  yb ,  N   b ; ya  y1 ,  a   0
La
a  0
Va
ya  y1
n=1
n=2
the operating line is the straight.
line connecting:
(xa , ya) and (xb , yb) ,
or
n=N
(x0 , y1) and (xN , yN+1)
Lb
Vb , yb  y N 1
 N  b
When the flow rates ( L,V ) are not constant,
operating line is not straight.
The material-balance calculation should be made to
establish a few intermediate points, and to locate the
operating line.
The equilibrium line
:a plot of equilibrium values of xe , ye ; (xn, yn)
The position of the operation line relative to the equilibrium
line determined the direction of mass transfer and how many
stages are required for a given separation.
FIGURE 20.4
Operating and equilibrium lines:(a) for rectification, (b) for gas absorption, (c)
for desorption.
For one component
transferred from L→V
yn+1 < yn
driving force for mass
transfer:(ye - yn+1 )
yn = ye
n
xn yn+1
( yn* = ye )
FIGURE 20.4
(b) for gas absorption
mass transferred from V→L
( gas absorption )
yn+1 > yn ( ≈ yn* )
driving force for mass transfer:
(yn+1 - ye )
yn = ye
xn
yn+1
Controlling the liquid rate L, to make the slope of OP line (L/V)
steeper than EQ line large driving force  relatively few stages.
FIGURE 20.4
(c) for desorption.
(c) desorption or stripping (氣提)
– the reverse of gas absorption
※direction of mass transfer
L→V
※To make the equilibrium curve
much steeper than OP line
 by changing of temperature or
pressure.
Ideal contact stages (perfect plates)
Ln-1
Xn-1
Vn
yn
The V phase leaving the stage is in
equilibrium with the L phase leaving the
same stage.
 n 
 yn
in equilibrium
Ln
Xn
Vn+1
yn+1
(  n , yn ) locates on the curve of
the equilibrium line (  e vs ye )
Ideal stages
stage efficiency
plate efficiency
actual stage
Determining the number of ideal stages
The McCabe-Thiele method
The example was shown for a typical gas absorber.
Operating line:connects two end points
a (  a , ya ) and b( b , yb )
The McCabe-Thiele method:
---can be used for gas absorption, distillation,
leaching, liquid extraction.
---the construction can be started at either end of
the column, and in general, the last step will not
exactly meet the terminal concentrations.
A fractional step may be assigned.
Determining the number of ideal stages
OP
EQ
Xa(X0) y1= ya
stage 1
(x2, y3)
(x1, y2)
x1
X2
y3
X3
y4
XN-2
yN-1
XN-1
yN
y2
stage 2
(x2, y2)
(x1, y1)
FIGURE20.5
Operating-line diagram for gas absorber.
XN= Xb yN-1= yb
stage N-1
stage N
Known
x0
y1 ( ya ) EQ
line

 1 OP
line

 y2 EQ
line

  2 OP
line

 y3 
point a
m
(X1, Y1)
one ideal stage
a,m,N
n (X1, Y2)
O
(X2, Y2)
one ideal stage
n,o,P
P (X2, Y3)
Example 20.1
By means of a plate column, acetone is absorbed
from its mixture with air into a nonvolatile
absorption oil. The entering gas contains 30 mole
percent acetone, and the entering oil is acetonefree. Of the acetone in the air 97 percent is to be
absorbed, and the concentrated liquor at the
bottom of the tower is to contain 10 mole percent
acetone, the equilibrium relationship is ye=1.9xe.
Plot the operating line and determine the number
of ideal stages.
oil
xa=0
ya= ?
yb= 0.3
xb=(0.1)
aceton
+air
Example 20.1
Solution
La=(261.9)
xa= 0
Set 100 mole entering gas as a basis
Vb=100
Acetone entering 100×0.3=30 (mole)
Air entering
100-03=70 (mole)
Acetone leaving 30(1-0.97)=0.9 (mole)
(97% absorbed)
Acetone
0.9
ya 

 0.0127
Air  Acetone 70.9
Acetone absorbed
xb=0.1
Lb=(291)
ya= 0.0127
Va=(70.9)
yb=0.3
Vb=(100)
30-0.9=29.1 (mole)
Acetone
29.1
Lb (contain 10% acetone) : xb 
, Lb 
 291
Lb
0.1
La=291-29.1=261.9 (mole)
L, V are not constant
Find an intermediate point of the operating line:
make acetone balance around the top part.
1. assume 10 mole acetone left , y=10/(10+70)=0.125
ya= 0.0127
xa=0
Ace=0.9
Air=70
Calculate x : Acetone lost by gas = gained by liquid
10-0.9=9.1
Ace=9.1
x=9.1/(9.1+261.9)=0.0336
oil=261.9
Ace=10
Air=70
op line point (0.0336, 0.125)
2. assume 20 mole acetone left , y=20/(20+70)=0.222
x=(20-0.9)/(19.1+261.9)=0.068
op line point (0.068, 0.222)
xb=(0.1)
yb= 0.3
Ace=30
Air=70
(0.0336, 0.125) , (0.068, 0.222)
slightly
curved
(x2,y3)
Construct the op line
EQ line:ye=1.9xe
ideal stage:
4 and a fraction
The fraction:l1/l2=0.27
x0,y1
a(0,0.0127)
b (0.1,0.3)
y2
(x2,y2)
if based on changes in y
fraction=0.33
Average=0.3
FIGURE20.6
Diagram for Example 20.1.
answer:4.3 ideal stage
Absorption factor method for calculating
the number of ideal stages
Applied when:
1. Operating and equilibrium lines are both straight.
2. Ideal stages and constant (L/V)
Equation of EQ line:
for stage n:
Eq(20.7)
yn 1
ye  m e  B
yn  m n  B
(20.8)
(20.9)
Ln
Va ya  La  a

χn 
Vn 1
Vn 1
Substitution xn (from 20.9) into (20.7)
L( y n  B )
L a
yn 1 
 ya 
mV
V
(20.10)
slope of OP line ( L / V )
Absorption factor A:
slope of EQ line (m)
yn 1 
L( y n  B )
L
 ya  a
mV
V
Eq (20.10):
Eq (20.8) :
ya*
L
xa
A
(20.11)
mV
yn 1  A( yn  B)  ya  Am a
 Ay n  A(m a  B)  ya (20.12)
ya*  m a  B
(20.13)
Concentration of V phase in
equilibrium with L phase (Xa)
Eq (20.12):
yn1  Ayn  Ay  ya
*
a
(20.14)
ya
yn1  Ayn  Aya*  ya
(20.14)
for stage 1, n=1, that y1=ya
y2  Aya  Aya*  ya  ya (1  A)  Aya*
for stage 2, using n=2,
y3  Ay 2  Ay  ya
*
a
FIGURE20.7
Derivation of absorption factor
equation.
 A[ ya (1  A)  Ay a* ]  Ay a*  ya
 ya (1  A  A2 )  ya* ( A  A2 )
These equations can be generalized for the nth stage, giving
yn1  ya (1  A  A2    An )  ya* ( A  A2    An ) (20.15)
yn1  ya (1  A  A2    An )  ya* ( A  A2    An )
For the entire cascade, n=N, the total number of stages, and
xa
ya
yn1  yN 1  y b
Then
yb  ya (1  A  A2    AN )  ya* ( A  A2    AN )
(20.16)
yn  yb* : in equilibriu m with L phase ( b )
xb
= xN
yb = yN+1
yb  ya (1  A  A2    AN )  ya* ( A  A2    AN )
The sum of such a series is
yN  y
*
b
a1 (1  r n )
Sn 
1 r
Equation (20.16) can then be written
N
1  A N 1
1

A
yb  ya
 ya* A
1 A
1 A
(20.17)
xb
=xN
yb
= yN+1
Equation (20.17) is a form of the Kremser equation.
A simpler form :
Equation (20.14) is, for stage N,
yn1  Ayn  Aya*  ya
yb  Ay N  Ay  ya
*
a
*
a
yb  Ay  Ay  ya
*
b
(20.18)
Figrue20.7 shows that yN=yb* , and Eq.(20.18) can be written
ya  yb  A( yb*  ya* )
yb  Ayb*  Aya*  ya
(20.19)
Collecting terms in Eq.(20.17) containing AN+1 gives
A
N 1
( ya  y )  A( yb  y )  y a  yb
*
a
*
a
(20.20)
Substituting ya-yb from Eq.(20.19) into Eq.(20.20) gives
A ( ya  y )  yb  y  y  y  y b  y
N
*
a
*
a
*
b
*
y

y
b
AN  b
ya  ya*
*
a
(20.21)
*
b
(20.21)
Taking logarithms of Eq.(20.21) and solving for N give
yb  yb*
A 
ya  ya*
ln[( yb  yb* ) /( ya  ya* )]
N
ln A
N
and from Eq. (20.19)
(20.22)
ya  yb  A( y  y )
*
b
yb  ya
OP  slope
 A (
)
*
*
yb  ya
EQ  slope
*
a
(20.23)
Equation (20.22) can be written
ln[( yb  yb* ) /( ya  ya* )]
N
ln[( yb  ya ) /( yb*  ya* )]
(20.24)
The various concentration differences in Eq. (20.24) are show
in Fig. 20.8
(20.23)
y y
A
b
*
b
y y
a
*
a
(20.24)
ln[( yb  yb* ) /( ya  ya* )]
N
ln[( yb  ya ) /( yb*  ya* )]
ln[( yb  yb* ) /( ya  ya* )]

ln A
FIGURE20.8
Concentration differences
in Eq.(20.24)
xa
xb=xN
When the operating line and the equilibrium line are parallel.
A=1
yb  ya yb  ya
N

*
ya  ya yb  yb*
(20.25)
If the operating line has a lower slope than the equilibrium
line, A is less than 1.0, but Eqs. (20.22) and (20.24) can still
be used by inverting both terms to give
ln[( ya  ya* ) /( yb  yb* )]
N
ln(1 / A)
(20.26)
ln[( ya  ya* ) /( yb  yb* )]
or N 
ln[( yb*  ya* ) /( yb  y a )]
(20.27)
L-phase form of Eq. (20.24)
Eq. (20.24):-----based on concentration of V phase (y )
-----commonly used in gas absorption
For stripping, equations in x are more common
ln[(  a   a* ) /(  b   b* )]
N
ln[(  a   b ) /(  a*   b* )]
(20.28)
ln[(  a   a* ) /(  b   b* )]

ln S
X*:equilibrium conc. corresponding to y
S :stripping factor
x1
xa*
x0=xa
1
mV
slope of EQ line , m
S

(
)
A
L
slope of OP line , L / V
ya= y1
ya
(xa, ya)
x1=xa*
yb
yb= yN+1
xN =xb
xb*
(xb, yb)
xb=xN
x1
FIGURE20.8
Concentration differences in Eq.(20.28) (for stripping)
xa=xo
In estimation the effect of change in operating conditions,
For absorption
For stripping
*
y

y
N
b
A  b
*
ya  ya
(20.21)
a  
S 
b  
(20.30)
N
*
a
*
b
Example 20.2
Ammonia is stripped from a dilute aqueous solution by
countercurrent contact with air in a column containing seven
sieve trays. The equilibrium relationship is ye=0.8xe, and when
the molar flow of air is 1.5 times that of the solution, 90
percent of the ammonia is removed. (a) How many ideal
stages does the column have, and what is the stage efficiency?
(b) What percentage removal would be obtained if the air rate
were increased to 2.0 times the solution rate?
Solution (a)
Stripping factor:
m
0.8
S

 1.2
L
1
V
1.5
?=xa*
base ←xa
ya= ?
(0.6xa)
Express all conc. interms of xa
xa : mole fraction of NH3 in entering sol.
90% NH3 is removed
b  0.1 a
b*  0 (since yb  0)
?=xb*
xb=(0.1xa)
yb= 0
pure
air
ya: calculate by mass balance
base ←xa
ya= ?
(0.6xa)
L (  a   b )  V ( y a  yb )
L
1
ya  (  a   b ) 
(0.9  a )  0.6  a
V
1.5
ya 0.6  a
*
a 

 0.75 a
0.8
0.8
ye=0.8xe
yb= 0
xb=(0.1xa)
Using Eq (20.28)
ln[(  a   ) (  b   )]
N
ln S
ln[(  a  0.75  a ) (0.1 a  0)]

 5.02
ln 1.2
*
a
*
b
5.02
stage efficiency 
 72%
7
(a)
Solution (b)
V/L increases to 2.0, if the stage efficiency does not change,
Nideal = 5.02
Using Eq (20.30)
(a)
S=m(V/L)=0.8×2=1.6
ln[(  a   ) ( b   )]  N  ln S  5.02 ln 1.6  2.36
*
a
*
b
 a   a*
 exp 2.36  10.59
b
or:
 a   a*
5.02=10.59
S 
* =1.6
b  b
N
(b)
Let b  (1  f )  a , f : fraction of NH3 removed
to get  a* , calculate ya first , by mass balance
xa
ya= ?
L
1
1
ya  (  a   b )  [  a  (1  f )  a]  f a
V
2
2
ya 0.5 f a
*
a 

 0.625 f a
m
0.8
Substitute into Eq.(b)
 a  0.625 f a
 10.59
(1  f )  a
 a   a*
 exp 2.36  10.59
b
f = 0.962
yb = 0
xb=(1-f)xa Xb*=0
96.2% is removed
FIGURE20.10
Diagram for example
※Higher V/L  decreases slope of OP line (L/V)
※For the stripping  increases the percentage of removal
The limiting flow rate (V or L) approaches:
The concentration driving force at the end point is zero.
For stripping operation
ya*
xa
L
V
EQ
ya
ya*
ya
y
ya
Slope = L/V
(xb,yb)
xb
yb
xb
x
xa
OP
ya increase
↑
V decrease
* As V decrease  ya increase
 driving force ( y  ya ) 
*
a
* at limiting V , driving force  0 , y  ya
*
a
theoretical plate N  
L(  a  b )  V ( ya  yb )
limiting case
L(  a  b )  Vlim ( y  yb )
*
a
Home work:
1.Derivation of Kremser Eq. in X form (eq. 20.28).
2.The problem given in the draft (after chapter 20)
3.The problems given in the text book.
problems 1,3,5 9 of Chapter 20.
L/O/G/O
Thank You!
化學工程系 李玉郎
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