(7)Effect of pressure
•When total pressure P increases, or temperature T
decreases, the solubility of soluble components increases.
Increasing P and decreasing T are Favorable conditions
for absorption.吸收( 高压、低温有利)
•If the gas and liquid rates are kept constant so the
operating line is not changed, for given ZT=Hoy•Noy
=Constant,
P  m  Driving
N oy 
( yb  ya )
yL 
force  N A  y a  or
xb 
•For given V, yb, ya, P
x*b=yb/m liquid flow rates Lmin.
L min 
V ( yb  ya )
yb m   xa
1
(8)Temperature variations in packed towers
•When rich gas is fed to an absorption tower, the
temperature in the tower varies appreciably from
bottom to top.
•Heat of absorption of the solute temperature T;
•Evaporation of the solvent temperature T;
•Usually, overall effect: T. Tp*(equilibrium partial
pressure) Driving force. The heat of absorption of
the solute is disadvantageous to absorption Factors
influencing the temperature profile: 1)rates of solute
absorption; 2)evaporation or condensation of solvent;
3)heat transfer between the phases.
2
•When the gas inlet temperature is close to the exit
temperature of the liquid and the incoming gas is
saturated, there is little effect on solvent evaporation,
and the rise in liquid temperature is roughly
proportional to the amount of solute absorbed.
[Fig.18.15a, p.567]
x  T 
•When the gas enters the columns 10 to 20 0C below
the exit liquid temperature and the solvent is volatile,
evaporation will cool the liquid in the bottom part of
the column, and the temperature profile may have a
maximum, as shown in Figure 18.15b.
•*E.g.(18.4)
3
b
L
b

(9)Supplementary:
factor method
for
 y ~ y is straightStripping
line relationsh
ip
calculating
the
number
of
transfer
units
d (y ) y  y

b
a


•Assume
dy y  mx
y b  by a or

 y dy
 y  y y b  y dy
( mx  b )
 
 


x bb) )
yV ( yy  y b y) y L ( xmx 
y
yb
N oy

y  mx
a
(a )
a
V y


(
y

y
)
b
a
b
Operating-line
equation:
  y  y  mln ( y  y b )(18
 x. 19
b
 N oy 
b )
L



y


y
ya
V a y a b L a x aa
L
y
x   y ~ y is straight
(18 . 6 line
)
relationsh ip
V
V
That is:  d (  y )   y b   y a
dy
yb  ya
L
L
V
y
x  y a  y bx a  x  y bx a  ( y  y a )
(b )
dy
1 L( y b  y a ) d (  y ) 4
V
V
N oy  
 


dy
yb  ya
 V ( y y b y b )  L ( x yb x b )
dy
dy
N oy  
 
(a )

V

b )
y

y
y

(
mx

  y  yya  m  ( yy a  y b )  x b   b
L

( yb  ya )
yb
V


xy N~ oyxya is (straight
y  y a )ln line ( brelationsh
) (18 . 19 )ip
Ly b   y a
ya
d (y ) yb   ya


Substituting
x from Eq.(b) into Eq.(a) and
dy
yb  ya
rearrangement give
yb
N oy 

ya
dy
(1 
mV
L
)y [
mV
L
y a  ( mx a  b )]
mV ( y b  y a )   y b
Integration
 N oy   S , y aln mx a  b ,(18
. 19 )
Let
ya
Eq. gives
L yb  ya
(综合)
of above
5
N oy



1
yb  ya

ln  (1  S )
 S

1 S 
ya  ya

Slope of equilibrium line
S 
A
mV

脱吸因数
L
(L /V )
L
1
mV
N oy
m

Slope of operating line
S
 mV y b  y a 
 [ p . 115 , Figure 3  14 ]
 f 
,
 
ya  ya 
 L
[Similar to Eq.(3-84) in Chinese textbook]脱吸因数法
6
N oy
N oy



yb  ya

ln  (1  S )
 S

1 S 
ya  ya

1
 mV y b  y a 
 [ p . 115 , Figure 3  14 ]
 f 
,
 
ya  ya 
 L
S 
S 
N oy
N oy
yb  y
ya  y


a

a
S 
N oy

yb  ya
ya  y

a

y b  yya a  y 2 ; y b  y 1 ; y a  mx 2 ;

mq n ,V
ya  ya
N oy  N OG ; S  S  
q n ,L
7
Absorption efficiency A:
A 
yb  ya
yb
or
y a  y b (1   A )
When y   mx , x a  0  y a  0
N oy


yb

ln  (1  S )
 S
1 S 
ya

N oy


1

ln  (1  S )
 S
1 S 
1 A

1
1
8
•Two ways of increasing absorption efficiency A:
•1)Increasing the mass transfer driving force:
(a)Increasing the liquid-gas ratio液气比L/V driving
force; (b)Improving the equilibrium relationship: P
or T driving force; (c)xa driving force.
•2)Decreasing the resistance to mass transfer (1/Kya):
(a)V (Gas-film controls);L (liquid-film controls).
9
•Noy or Nox reflects the difficulty of absorption
process. The greater the value of Noy is, the more
difficulty is the absorption process.
•Hoy or Hox reflects the performance of the
absorption equipment.
10
•(10)Desorption or stripping(pp.81~85) 解吸
•1)Definition of Desorption/stripping: the reverse of gas
absorption: recover valuable solute from the absorbing
solution and regenerate the solvents.
•In stripping, the transfer is from the liquid to the gas
phase.
•2)Characteristics (comparing with absorption):
•Identical points: (a)The equipment used is the same as
in absorption(plate or packed tower); (b)Principles are
the same, both are of one way diffusion; (c)Methods
and equations for calculating ZT(height of a stripping
column) are the same as for absorption.
11
•Different points:
•(a)Purpose of absorption: Separating components
from gas mixtures; Purpose of stripping: Recover the
solute and regenerate the solvents.
•(b)Mass transfer direction: Gasliquid (in
absorption); Liquidgas (in stripping).
•(c)Mass transfer driving forces: y>y*,(y-y*)in
absorption; y<y*,(y*-y)in stripping.
•(d)Favorable conditions: P and T for absorption;
P and T for stripping.
•(e)Positions of equilibrium line and operating line: see
the following figures.
12
Operating line
y
y n 1
Equilibrium curve
y

ye
y n  nyy*
e
xn x
Equilibrium And Operating Lines for gas
absorption
Driving force:
y – y*
13
Equilibrium curve
y

y n 1
xn
y e Operating
yn
line
y ey  y n
y nx 1
n
x
yn
x n force:
Driving

For stripping
y  y
14
Va ya
V
V aa
y a La
•Calculations for stripping
yy a
Va x
La a
a
•Stripping media: Inert gas or superheated
V
L
steam(condensable).
L aa V a xy a aL
a
V
V
Lya Va V=?
•1)Molal flow rate of Inert gas or superheated
Vxxaaaa yaa steam
L
V
y a y a aL
L
L a x a ax
yLa
Vy
Va a ax
L
L
a
Higher conc. end LV
x a L ay
Va
xL
x a x ya a aL
x
L V V
xxa
y b
Inert gas or steam
x
L LVL a xV
a
y
Ly
Vb yb
L
x
V
V
a
x
VV
V bb x yy L b
b
V
•Figure. packed
L
Lower conc. end xyyxb xy V yb x
column for stripping.
L
b
b 15
b
x
V
V
Ly y
•Comparing absorption:
•By balancing the operating costs against the fixed
costs of the equipment, ( L / V ) opt  (1 . 1 ~ 1 . 5 )( L / V ) min
•That
L  (1 . 1 ~ 1 . 5 ) L min
is,
•For stripping, V  (1 . 1 ~ 1 . 5 )V min
•Vmin is determined by
calculating (V/L)min, i.e.,
(L/V)max:
Equilibrium curve
y a , max
yyaa, max
, max
Operating line
y a line
y a , maxfor Vmin
yy aOperating
a
y a , max  y b
L
xa
x

 
a

x a  xLb 
ya  yb
 V  max

xxbb
 
x a  xb
 V  max
xa
xb
Operating
line
ya
L
slope 
xa
V
16
y a , max  y b
L
L


 
V min
x a  xb
 V  max
•For case (a), y a , max  y a
Equilibrium curve
y a , max y
a , max
y , max
fory Va min
Operating
yfora , maxV line
y a line
a Operating
min
ya
x
y a line
x a Operating
Operating
line
a
ya
L
L
slope 
slope


x
V
V
y

ybab
L
x


x
a
x
xa 
a
b

x a  xxbb
 V  max
17
x
x
b
(a) b
(b)
y a , max
yyaa, max
, max
Operating line
y
y
linea , maxfor Vmin
a
yy aOperating
a

a
a
xxaa
 yb
xxbb
 xb
xa
xb
Equilibrium curve
for
•2)Calculating ZT(height of a stripping column)
[Equations are the same as for absorption]
Z T  H y N y  H x N x  H oy N oy  H ox N ox
Gas film: H y 
Liquid film: H x 
Overall gas: H oy 
Overall liquid: H ox 
V /S
k ya
L/S
kxa
V /S
K ya
L/S
K xa
yb
dy

Ny 
y  yi
ya
xb
Nx 
x
xa
dx

ya
dy
y y
xb
N ox 
x
xa
(18 . 22 )
x
i
yb
N oy 
(18 . 21 )

dx

x
(18 . 23 )
(18 . 24 )
18
ya
 N oy 
(y  y )
ya  yb
dy
yL
yb
N oy 
 (yy  y
N oy y a
b

a

a
b
ya  y
xa  x
 N ox 
y bx y a
L
N oy 

yybb  yy ab
N oy 

ya  ya
Or,
N oy
y L 

b

a
( y  yb )  ( y  ya )
ln
)
x L 

b

a
y  yb
y  ya

b

a
( xb  x )  ( xa  x )
ln
xb  x
xa  x

b

a



1
yb  ya
x

x
ln  (1  S )
 S
N oyb  a

ya  ya

y b  1y b S 
19
 y Ai  mxA Ai , yAi
 mxA , Ai
A
A


Ky
k y ( y A  y AimG
) k x ( x Ai  x A )
M
 H oy  H y 
H x (18.27)
y A , yyAi Lmx ,y Ai  y A
1y Ai  mx
M
Ai
A
 A
Ky
k y ( y A  y Ai ) k x ( x Ai  x A )
mV


H

H

H
oy
y
x

y A , yyAi Lmx ,y Ai  y A
1 y Ai  mx
Ai
A
 A
Ky
k y ( y A  y Ai ) k x ( x Ai  x A )
LM
 H ox  H x 
H y (18.29)
 y Ai  mxAi , y A mG
 mx
,
M A
 H ox  H x 
L
mV
(17.56)
(17.56)
(17.56)
Hy
[Example 18.5.]
20
•*Other absorption processes（自学）
•1)Multicomponent
（多组分）
absorption(pp.81)
•Definition: Absorption process in which more than one
solute is absorbed by a solvent from a gas mixture.
•Characteristics: Separate equilibrium and operating
lines are needed for each solute, but the slope of the
operating line, which is L/V, is the same for all the
solutes.
21
•Usually the packed height ZT and the liquid
flow rate L are determined according to the
absorption requirement of some key component
A, then the fractional removal of other
components are calculated from the given ZT
and L.
22
•2)Absorption from rich gases (pp.574~580)
（自学）
•Solute concentration is moderate or high in the gas.
•(a) V and L must be accounted for in the material
balance; (b)The correction factor for one-way
diffusion should be included; (c)Ky or Kx will not be
constant; (d)There may be an appreciable temperature
gradient in the column, which will change the
equilibrium line.
23
•3)Absorption with chemical reaction
(pp.588~589)（自学）
•Absorption followed by reaction in the liquid
phase is often used to get more complete removal
of a solute from a gas mixture.
•Characteristics: (a)Equilibrium partial pressure
of the solute p* and p*0; Driving force (pp*) . (b)Liquid film resistance to mass
transfer , that is kL (gas film controls).
(c)The solvent is hardly saturated by the solute.
Therefore, absorption with chemical reaction
can be used to substitute some difficult physical
absorption processes.
24
•*[Mass transfer correlations
•Empirical correlations for mass transfer
coefficients in absorption
•1)For NH3 absorption in water:
k g a  6 . 07  10
4
G
0 .9
W
0 . 39
( 2  78 )
•The absorption of NH3 in water is often cited as
an example of gas-film control, since the gas film
has about 80 to 90 percent of the total resistance.
25
•2)For CO2 absorption in water at 1 atm:
k L a  2 . 57 U
( 2  79 )
0 . 96
•Liquid film controls: KL kL.
•3)For SO2 absorption in water:
k g a  9 . 81  10
k L a  aW
0 . 82
4
G
0 .7
W
0 . 25
( 2  80 )
( 2  81 )
•With gases of intermediate solubility both
resistances are important.
26