Chapter 7
Equilibrium-Stage
Operations
1
Cascades 逐级接触设备
•
One class of mass-transfer devices consists of
assembliesof individual units, or
stages,interconnected so that the materials
being processed pass through each stage in
turn. The two streams move counter currently
through the assembly; in each stage they are
brought into contact, mixed, and then
separated. Such multistage systems are called
cascades.
2
xF
Ideal Stage理想级/Equilibrium
Stage
x1
平衡级/theoretical
Stage理论级
y1
xF
x1
t1
y1
x2
t1
y2
xF
x2
x F y 2x1
x1 t 2 y1
t2
y1 x3t
1
t1 y3
y3
x2
x2 t3
y 2 ( xyD 2)
t2
t2
•
y2 is in equilibrium
with x2.
x3
t3
( xD )
3
y1
t1
Ideal
Plate/Equilibrium
x
x2
x
Plate/Theoretical
Plate/Perfect
y2
y
plate
F
1
1
t2
t1
x3
x2
•
y2
y3
t2
t3
x3
( xD )
y3
xF
t3
x1
xF
( xD )
x1
y1
y1
t1
t1
x2
x2
y2 is in equilibrium
with x2.
y2
4
1. Equipment for stage contacts
•
•
1) Typical distillation equipment
Fig.20.1. Equipment for continuous
distillation.
5
Liq tu
1 id
x2
y3
F ro th
t3
t
2
( x Dx
)F
Rectifying
塔顶产品
section
3
xx
1
冷凝器
y3
Va p o r回流
t3
t1
Feed
( xD )
•
Equipment
x2
for y
2
continuous
t2
distillation.
x3
y3
上升蒸汽
塔顶产
Co n d en
塔顶产品
回流
冷凝器
进料
Overhead
reflux Overhead
上升蒸
product
product
进料 加热器
Vapor
Vapor
Feed plate
上升蒸汽
Liquid
Liquid
塔底产
Overhead
Overhead
加热器（再沸器
reboiler
product
reboiler
product
Overhead
Vapor
Bottom
塔底产品
Bottomprod
pr
Vapor
加热器（再沸器） product
Liquid
Liquid reboiler
塔底产品
Vapor
Stripping
/Enriching
section
reboilerBottom product
BottomLiquid
product
reboiler 6
7
•
•
2) Typical leaching equipment（自学）
Fig.20.2.
8
2. PRINCIPLES OF STAGE
PROCESSES
•
•
1) Terminology for stage-contact plants
Fig.20.3
9
Va 
V
V1
V
yyaa  yy11
yaa  y11
L
V
Laa1aVV1
V
VL
aa
1
Va  V
y1
ya xyxxyy1aaa1 Plate
y
1
a
1
ya  Ly1
La L
Lnnnn11
La
n 1
xa xxnn 1
xVa  xVnnn 11
Lnn 11 Plate n-1
Lna1 V
VL1 n 1
LVna1 VV
xyyna1xyynnyn11
xna1 xy1nn1
y
VVa aVV1 1
yya aVayy11 V1
y1a
LLa ya  V
a
xxa a La
LLn n11xa
xxn 1 Ln 1
n 1
VVn n xn 1
yyn n Vn
yn
Vbb  VNN 11
ybb  y NN 11
Fig.20.3
Materialbalance
diagram for
plate column
(Twocomponent
system)
Ln Vn n1 Plate n
V
L
Vnnn Vn 1
yyxxnnn y n 1
V

V
b
n
Vb  VN 1N 1
yn 1 Plate n+1
L
Lnn11
yb 
y N 1
y
y
b
N 1
xn 1
xn 1
Plate N L  L
Lb b LN N
Vn 1
Vn 1
x x
yn 1
b
N
10
•
•
•
•
1) Terminology for stage-contact plants
In this book,the stages are numbered in the
direction of flow of the L phase, and the last
stage is that discharging the L phase.
2)Material balances
Under steady flow, there is neither
accumulation nor depletion, the input and
the output are equal and
Total material
balance:
La  Vn1  Ln  Va
La xa  Vn1 yn1  Ln xn  Va ya
11
La  Vbalance
Va
n 1  Lfor
n component
Material
A:
La xa  Vn1 yn1  Ln xn  Va ya
Entire cascade:
Total material balance:
La  Vb  Lb  Va
La AVmaterial
 Lbalance
 Vya :  L x  V y
n
Ln1
x
V
component
a a
n 1 n 1
n n
a a
La xa  Vb yb  Lb xb  Va ya
12
•
3) Enthalpy balances
•
The general energy balance can be simplified by
neglecting mechanical potential energy and kinetic
energy . If in addition, the process is worklessand
adiabatic, a simple enthalpy balance applies.
•
For two-component system:
La H L,a  Vn1HV ,n1  Ln H L,n  Va HV ,a
Where HL and HV are the enthalpies per mole of L
phase and V phase, respectively.
overall cascade:
La H L,a  Vb HV ,b  Lb H L,b  Va HV ,a
13
•
•
4) Graphical methods for two-component
system to find stage numbers
The methods are based on material balances
and equilibrium relationships; some more
complex methods require enthalpy balances
as well.
14
•
•
5)Operating line diagram
La eq.(7.2-2):
Vn1  Ln  Va
From
La xa  Vn1 yn1  Ln xn  Va ya
•
Operating-line equation操作线方程:
Ln
Va ya  La xa
yn1 
xn 
(7.2-7)
Vn1
Vn1
•
When the flow rates are not constant in the
column, the operating line on a simple
arithmetic plot is not straight.
15
•
If Ln and Vn+1 are constant through the
column, the equation is that of a straight line
La slope斜率
Vn 1  Ln L/V
 Vaand intercept截距:
with
ya  ( L / V ) xa
•
Operating-line equation becomes:
L
L
yn 1  xn  ( ya  xa )
V
V
• Operating line:
When xn  xa , yn1  ya
When x  x  x , y  y
n
N
b
n 1
N 1  yb
16
xa
yb
Operating line
ya
y
b a
Equilibrium curve
xa
xb x a
xb
Operating-line diagram for gas absorber
y a xb
ya
yb y a
yb
（吸收）
17
•
•
x
The position1of the operating line relative to
the equilibrium
y1 line:
(1) For rectification（精馏）
in a distillation
t1
column, the operating line must lie below the
equilibriumx2line (Fig.7.2-4a, p.48), why?
yn
yn
xn 1
yn
x F yn is in equilibrium with xn
xn
yn 1
xn
yn 1
t2
x1 yn1  yn
yn 1  yn
x3
xny13
ytn31
y1
t1
xn
18
Va 
V
V1
V
yyaa  yy11
yaa  y11
L
V
Laa1aVV1
V
VL
aa
1
Va  V
y1
ya xyxxyy1aaa1 Plate
y
1
a
1
ya  Ly1
La L
Lnnnn11
La
n 1
xa xxnn 1
xVa  xVnnn 11
Lnn 11 Plate n-1
Lna1 V
VL1 n 1
LVna1 VV
xyyna1xyynnyn11
xna1 xy1nn1
y
VVa aVV1 1
yya aVayy11 V1
y1a
LLa ya  V
a
xxa a La
LLn n11xa
xxn 1 Ln 1
n 1
VVn n xn 1
yyn n Vn
yn
Vbb  VNN 11
ybb  y NN 11
For rectification:
y= mole fraction
of more volatile
component A
yb  ya
Ln Vn n1 Plate n
V
L
Vnnn Vn 1
yyxxnnn y n 1
V

V
b
n
Vb  VN 1N 1
yn 1 Plate n+1
L
Lnn11
yb 
y N 1
y
y
b
N 1
xn 1
xn 1
Plate N L  L
Lb b LN N
Vn 1
Vn 1
x x
yn 1
b
N
xb  xa
19
Equilibrium curve
ye  y n
yn 1
ye  y n
Operating line
yey yn
xn
ynx1
n 1
n
xn
yn
yn
yn is in equilibrium with xn
xn
xn
yn 
yn 1  yn
yn 1 
yn yn
xn
Driving
force:
Fig.7.2-4(a) for rectification
ye  yn 1
20
•
x
1
(2) Absorption:
When one component is to be
transferredyfrom
the V phase to L phase, as
1
in the absorption of soluble material from an
t
1 operating line must lie above
inert gas, the
the equilibrium
x2 line (Fig.7.2-4b), why?
yn
yn
xn 1
yn
x F yn is in equilibrium with xn
xn
yn 1
xn
yn 1
t2
x1 yn1  yn
yn 1  yn
x3
xny13
ytn31
y1
t1
xn
y=concentration of soluble
component in an inert gas
21
Va 
V
V1
V
yyaa  yy11 稀端Lean
yaa  y11 terminal
L
V
Laa1aVV1
V
VL
aa
1
Va  V
y1
ya xyxxyy1aaa1 Plate
y
1
a
1
ya  Ly1
La L
Lnnnn11
La
n 1
xa xxnn 1
xVa  xVnnn 11
Lnn 11 Plate n-1
Lna1 V
VL1 n 1
LVna1 VV
xyyna1xyynnyn11
xna1 xy1nn1
y
VVa aVV1 1
yya aVayy11 V1
y1a
LLa ya  V
a
xxa a La
LLn n11xa
xxn 1 Ln 1
n 1
VVn n xn 1
yyn n Vn
yn
Vbb  VNN 11
ybb  y NN 11
For absorption:
y=conc. of soluble
material in an
inert gas
yb  ya
xb  xa
Ln Vn n1 Plate n
V
L
Vnnn Vn 1
yyxxnnn y n 1
V

V
b
n
Vb  VN 1N 1
yn 1 Plate n+1
L
Lnn11
yb 
y N 1
y
y
b
N 1
xn 1
xn 1
Plate N L  L
Lb b LN N浓端Thick terminal
Vn 1
22
Vn 1
x x
yn 1
b
N
yn
yn
yn is in equilibrium with xn
xn
xn
yn 1
yn 1  yn
yn 1  yn
Operating line
yn1
yn1
y
n 1
curve
yn y Equilibrium
ye y
yn  ye n
e
xn x
xn
n
Fig.7.2-4(b) for gas absorption
Driving force:
yn+1 - yn
23
•
x
1
(3) Desorption/stripping:
the reverse of gas
absorption:yrecover
valuable solute from the
1
absorbing solution and regenerate the
t
1 operating line must lie below the
solvents. The
equilibriumx2line (Fig.7.24c), why?
yn
yn
yn
xn 1
yn
x F yn is in equilibrium withxn
xn
xn
yn 1
xn
t2
x1 xn1  xn yn1  yn yn1
yn 1  yn
x3
xny13
ytn31
y1
t1
xn
x=concentration of solute in
absorbing solution
24
Va 
V
V1
V
yyaa  yy11
yaa  y11
L
V
Laa1aVV1
V
VL
aa
1
Va  V
y1
ya xyxxyy1aaa1 Plate
y
1
a
1
ya  Ly1
La L
Lnnnn11
La
n 1
xa xxnn 1
xVa  xVnnn 11
Lnn 11 Plate n-1
Lna1 V
VL1 n 1
LVna1 VV
xyyna1xyynnyn11
xna1 xy1nn1
y
VVa aVV1 1
yya aVayy11 V1
y1a
LLa ya  V
a
xxa a La
LLn n11xa
xxn 1 Ln 1
n 1
VVn n xn 1
yyn n Vn
yn
Vbb  VNN 11
ybb  y NN 11
For desorption or
stripping:
y=conc. of soluble
material in an
inert gas
x=conc. of solute
in absorbing
solution
Ln Vn n1 Plate n
V
L
Vnnn Vn 1
yyxxnnn y n 1
V

V
b
n
Vb  VN 1N 1
yn 1 Plate n+1
L
Lnn11
yb 
y N 1
y
y
b
N 1
xn 1
xn 1
Plate N L  L
Lb b LN N
Vn 1
Vn 1
x x
yn 1
b
N
yb  ya
xb  xa
25
Equilibrium curve
ye  y n
yn 1
xn
ye Operating
yn
line
ye  y n
yn
yn 1
yn
yn
yn 1
yn is in equilibrium withxn
xn
xn
xn
xn
xn
yxnn 1  xn yn 1  yn yn 
yn 1  yn
xn
Driving
force:
Fig.7.2-4 (c) for stripping
ye  yn 1
26
xF
x1
•
6) Ideal contact stages
•
Ideal
Stage /Equilibrium Stage
t1
x
/theoretical
Stage
x2
x
Ideal
Plate /Theoretical
y 2 Plate/Equilibrium
y
t
Plate/
Perfect
plate
t2
y1
F
•
1
1
1
x2
x3
y2
y3
t2
•
t3
x3
( xD )
y3
xF
t3
x1
xF
( xD )
x1
y1
y1
t1
t1
x2
x
y2 is in equilibrium
with x2.
27
•
To use ideal stages in design, it is necessary
to apply a correction factor, called the
stage efficiency级效率 or plate efficiency板
效率, which relates the ideal stage to an
actual one. (See Chapter 9 and 12)
N
• Overall efficiency:  
N
Nactual
 
N actual
yn  yn1

  yn  yn1
M
• Plate (Murfree) efficiency: 
yMn yn1
yn  yn1
• （默弗里效率）
28
•
7) Determining the number of ideal
stages
•
The usual method of designing cascades:
Determining the number of ideal stages
Finding the stage efficiencies
Calculating the number of actual stages
29
•
•
A simple method of determining the number
of ideal stages when there are only two
components in each phase is a graphical
construction using the operating-line
diagram.
E.g.: Gas absorption:
30
Va  V1
VVa aVV1 1
Va  V1
yya aVayy11 V1 ya  y1
yaVVay1 V1
V
LLa a ya a y11 La
LaVyVyaEEVy1
y
V
xxa a La a a11 x1 a
x
LyO
1 y
aO
L
y
x
a
LLn n11 a Vaa 1E VyL12n 1
E
L
x
n L
1 a
x
L
L
2
y
a
xxn n11 n 1 aaO yx3n 1
O
x
L
n

1
nE
L
x
x
x
VVn n n 31n 1Laa 1 Vn
V
xnO
nL
1 y
x
L
V
n

1
yyn n n xan 1 n
y
V
n
V
xnnn11
ynxn 1L
Vyn Vxynnn1
Vbb  VNN 11y y
n Vnn
ybb  y NN 11 y
n
Plate N
For absorption:
y=conc. of soluble
material in an
inert gas.
From
yb  ya
( xa  xb )
V

V
b
Vb  VN 1N 1 •How many ideal
y yb 
y y N 1stages are needed?
b
N 1
Lb Lb LNLN
xb  xN
31
Utilize alternately the
yb
operating and equilibrium
yb
y3
lines
yb
y3
y2
yb
yb
Operating-line
y3 yb
y y3
2
y3
y3
y2
y2
y a  y1
Points (x1,y1),
(x2,y2), (x3,y3)
must lie on
equilibrium
curve.
y a  y1x3a
Every step, or
Equilibrium curve
triangle
2
y a  yx1a
x2
1
y2
yy2
y

y a  ya1
represents one
1
1
x1 2 x2
y

y
x
ideal stage.
a
1a
x
xa
a
3
xa x12 x2
x

x
x
3
b
x1
1
3
x1 x23 x3  xb
x
x2 Operating-line
2
Fig.20.5
diagram for gas absorber.
xx2 x3  xb
x

x3  b3
32
b
•
•
•
The same construction can be used for
determining the number of ideal stages
needed in any cascade, whether it is used for
gas absorption, distillation, leaching, or
liquid extraction.
The graphical step-by-step construction can
be started at either end of the column.
Fractional stage? (See example 7.4)
33
•
•
•
8) Absorption factor method吸收因数法 for
calculating the number of ideal stages
When the operating and equilibrium lines are
both straight:
Let the equation of the equilibrium line be
ye  m xe  B
•
y

m
x

B
n
n
Where, by definition, m and B are constant. If
ye  m xe  B
stage n is ideal,
yn  m xn  B
34
•
Substitution for xn into Eq.(20.7)[p.628] gives, for
ideal stages and constant L/V,
L
L
yn 1  xn  ( ya  xa )
V
V
L( yn  B )
L
yn 1 
 ( y a  xa )
mV
V
•
L
A
mV
yn 1  A( yn factor,
B)  yratio
• Where A=absorption
ofxathe slope of
a  Am
the operating
L/V
to
that
of
the
equilibrium
 Ayline

A
(
m
x

B
)

y
n
a
a
line m.
Define
35
L L
A A
• Therefore
mV mV
yn 1 ynA
(
y

B
)

y

Am
x

A
(
y

B
)

y

Am
xa
n
a
a
1
n
a
 Ayn AyAn(
mA
xa(m
B
xa) By)a  ya
•
Because

y

m
x

B
a
a
y  m xa  B


y

Ay

Ay

y
n

1
n
a
 y  Ay  Ay  y a

a
n 1
n
a
a
(
20
.
14
)
(20.14)



n

2
n
yn 1  ya (1  A  A2  

A
)

y
(
A

A



A
2
n a2 
2n
) n
yn 1  ya (1yn A1 
A  AA )
 Aya (1
 AA )A ya (
 yaA( A
nN
n  N , the
total
number of stages,and
n

N
yn 1  y N 1  yb
yn 1  y N 1  yb
yn 1  y N 1  yb
36

• Then
yb  ya (1  A  A2    A N )  ya ( A  A2    A N )
2
N

2
N
n

N
1  A  A    A )  ya ( A  A    A )
of geometric series is
y•n 1The
 y Nsum

y
1
b
1
a1 (1  r )
sn 
nn
n)
aa11 ((11r 
r
1
r )
1
ssnn 
n 
11 
s
There
n
 rr
a1 ssnnn =sum of first n terms of series
r aa111 =first term
rr =constant ratio of each term to
preceding term（公比）
 yb
(20.16)
n
37
•
Equation(20.16) can then be written
N 1
1 A
 1 A
yb  ya
 ya A
1 A
1 A
N
(20.17)
(Kremser equation克列姆塞尔方程)
Other Forms of Kremser equation[For absorption]:

a
A ( y a  y )  yb  y
N

b

b
(20.21)

a
ln[(yb  y ) /( ya  y )]
N
ln A


ln[(yb  yb ) /( ya  ya )]
N
ln[(yb  ya ) /( yb  ya )]
(20.22)
(20.24)
38
•
When A=1, (the operating line and the
equilibrium line are parallel):
yb  ya yb  ya
N

(20.25)


ya  ya yb  yb

a

b
( ya  y  yb  y )
39
yb
dy
•Question:
If the operating line and equilibrium line
N

oy


(
y

y
)
are straight
and parallel,
A=1,
ya
N ox  N oy  N
yb  y a
Where,
N oy 

N=NTP=Number
ya  ya of
theoretical plates
yb  y a
Why?
N oy 

yb
yb  yb
dydy
N oy N 

oy ( 
y

y
( y  y) )
ya
yb
yb
yb
y3
yb
y3
y2
yyb3
Operating line
yyyb32 yb
yyy3 y3y
y2
y

b
y a  y1
xa
xa
x1
Equilibrium
line x1
x2
2a
1
yyx2aa y 2y1
y
 y  y
xyxaa1  ay1 1
yb yb yaya
(18.18
.18))
 NoyNoy 
(18
 
y y y y
xx xa
a
1a2
x2
x3  xb40
x3 
yb
yb
xb
ya
dy
N oy   line is straight but steeper than
•When the operating
(y  y )
y a as in Fig.8.2-2b,
the equilibrium line,
 N oy  NTP
NTP=N=Number ofytheoretical
plates
y
N oy 
b
a

8.]
a
•Why? [Refer to chapter
ya  y

y a  yb
For case of N=NTP=1,y  y
a

b
 ya
yb  ya  yb  yb Nyoya 

 yb  ya  y L
Operating line
yb  yybb
yb
xb
yb  y a
 N oy  NTU 
 1  NTP
ya xybb
y L


y
y  yxba
a
b

a
aa
Equilibrium
line
yb  yb
yb  yy yyb b yb  ya  y41a
xa xb

•
9)L-phase form of Eq.(20.24):
ln[(xa  xa ) /( xb  xb )]
N


ln[(xa  xb ) /( xa  xb )]

a
•

b
ln[(xa  x ) /( xb  x )]

(20.28)
ln S
1 mV
S 
(20.29)
A
L
Where x*=equilibrium concentration
corresponding to y
• S=stripping factor
• Eq.(20.28) mainly for stripping.
42
•
•
The stripping factor is the ratio of the slope of
the equilibrium line to that of the operating
line.
It is not assumed that the linear extension of
the equilibrium line passes through the origin.
It is only necessary that the line be linear in
the range where the steps representing the
stages touch the line.
43
•
•
•
•
Summary:
In the design of a plant, N is calculated from
the proposed terminal concentrations and a
selected value of A or S.
For absorption, using Eq.(20.22) or (22.24) or
eq.(20.21);
For stripping, using eq.(20.28) or (20.30).
[Because equations in x are more common.]

a

b
xa  x
S 
xb  x
N
[EXAMPLE 20.2.]
(20.30)
44
•
*10)Equilibrium-Stage Calculations for
Multicomponent system(自学)
45