Chapter 3
Convective Mass Transfer
3.1 Introduction
The mass transfer coefficient for the transport of species A between two locations within a
fluid may be defined from the following relations:
(Gases):
NA = kc(cA1 − cA2) = kG(pA1 − pA2) = ky(yA1 − yA2)
(Liquids):
NA = kc(cA1 − cA2) = kL(cA1 − cA2) = kx(xA1 − xA2)
In these equations, NA is the molar flux of species A and the mass transfer coefficient k has
different subscript and different units depending on the units of the driving force used in the
expression. Since many mass operations involve the transfer of material between two
contacting phases, different subscript for the mass transfer coefficient is also used to
distinguish between the phases.
The mass transfer coefficients might be obtained from the correlations given in chapter 8
where the Prandtl number (Pr = ν/α) is replaced by Schmidt number (Sc = ν/DAB) and the
Nusselt number (Nu = hL/k) is replaced by Sherwood number (Sh = kcL/DAB). Table 3.1-1
lists some correlations to determine the mass transfer coefficient for external forced
convection flow. The expressions for the flat plate are obtained from the solutions of the
boundary layer equations. The other formulas are experimental correlations.
--------- Table 3.1-1 Mass transfer coefficients for external forced convection flow. ------Correlations
Geometry
Conditions
1/2
1/3
Shx = 0.332Rex Sc
Flat plate
Laminar, local, Tf, 0.6 ≤ Sc ≤ 50
1/3
1/2
Flat plate
Laminar, average, Tf, 0.6 ≤ Sc ≤ 50
Shx = 0.664Rex Sc
Shx = 0.0296Rex4/5 Sc1/3
Flat plate
Turbulent, local, Tf, Rex ≤ 108
0.6 ≤ Sc ≤ 50
4/5
1/3
Flat
plate
Mixed, average, Tf, Rex,c = 5×105
Shx = (0.037Rex − 871) Sc
Rex ≤ 108, 0.6 ≤ Sc ≤ 50
Cylinder
Average, Tf, ReDSc > 0.2
ShD = 0.3 + [0.62 ReD1/2Sc1/3
Cross
flow
×[1 + (0.4/Sc)2/3]-1/4]
×[1 + (ReD/282,000)5/8]4/5
Sphere
Average, T∞, 3.5 < ReD < 7.6×104
ShD = 2 + (0.4ReD1/2 + 0.06ReD2/3)Sc0.4
0.71 < Sc < 380, 1.0 < (µ/µs) < 3.2
×(µ/µ )1/4
s
These correlations are valid for equimolar mass transfer or low mass transfer rate where the
mole fraction of species A is less than about 0.05. For higher mass transfer rate the
coefficients might be corrected by the log mean concentration difference. Instead of using kc
one should use kc/(1 − yA)lm where
3-1
(1 − yA)lm =
(1 − y A1 ) − (1 − y A2 )
 1 − y A1 
ln 

 1 − y A2 
Example 3.1-1 -----------------------------------------------------------------------------Air at 32oC is humidified by flowing over a 1.2-m-long container filled with water. The
interfacial temperature is 20oC. If the initial humidity of the air is 25% and its velocity is 0.15
m/s, calculate (a) the convective mass transfer coefficient, and (b) the amount of water
evaporated per unit width of the container.
(Ref. Fundamentals of Heat Transfer by Incropera and DeWitt, Wiley, 5th Edition, 2002)
Solution ---------------------------------------------------------------------------------------------The film temperature is Tf = (32 + 20)/2 = 26oC
Air at 26oC: ν = 1.51×10-5 m2/s, DAB = 2.77×10-5 m2/s.
Water vapor pressure: pAsat(20oC) = 0.02308 atm, pAsat(32oC) = 0.04696 atm.
Partial pressure of water vapor at the air-water interface is pAs = pAsat(20oC) = 0.02308 atm.
Mole fraction of water vapor at this location is yAs = 0.02308.
Partial pressure of water vapor in the ambient air is pA∞ = 0.25×pAsat(32oC) = 0.25×0.04696
atm = 0.01174 atm. Mole fraction of water vapor at this location is yA∞ = 0.01174.
Since both yAs and yA∞ are less than 0.05, we could use kc without the correction factor (1 −
yA)lm.
For parallel flow to a flat plate, laminar flow exists with Re < 300,000.
Re =
U∞L
ν
=
(0.15)(1.2)
= 1.192×104
−5
1.51 × 10
The average Sherwood number over the container may be obtained from the following
correlation:
1/ 3
1/2
1/3
ShL = 0.664Re Sc
4 1/2  1.51 
kc L
= 0.664(1.192×10 ) 
 = 59.22 =
DAB
 2.77 
The convective mass transfer coefficient is then
kc = 59.22
DAB
2.77 × 10−5
= 59.22
= 1.367×10-3 m/s
L
1.2
The molar flux of water is given by
NA = kc(cAs − cA∞)
3-2
The molar concentrations can be evaluated from the ideal gas law with gas constant R =
0.08205 atm·m3/(kmol·oK):
cAs =
p As
0.02308
=
= 9.6×10-4 kmol/m3.
RTs
(0.08205)(273 + 20)
cA∞ =
p A∞
0.01174
=
= 4.691×10-4 kmol/m3.
RT∞
(0.08205)(273 + 32)
NA = kc(cAs − cA∞) = 1.367×10-3(9.6×10-4 − 4.691×10-4) = 6.711×10-7 kmol/m2·s
The amount of water evaporated per m width of the container is
WA = (1)(1.2)(18) NA = 1.449×10-5 kg/s
Example 3.1-2 -----------------------------------------------------------------------------In a wetted-wall tower, an air-H2S mixture is flowing by a film of water which is flowing as
a thin film down a vertical plate. The H2S is being absorbed from the air to the water at a
total pressure of 1.50 atm abs and 30oC. The value of kc of 9.567×10-4 m/s has been predicted
for the gas-phase mass-transfer coefficient. At a given point the mole fraction of H2S in the
liquid at the liquid-gas interface is 2.0×10-5 and pA of H2S in the gas is 0.05 atm. The Henry’s
law equilibrium relation is pA(atm) = 609xA (mole fraction in liquid). Calculate the rate of
absorption of H2S.
(Ref: Transport Processes and Separation Process by C.J. Geankoplis, Prentice Hall, 4th Edition, 2003)
Solution ----------------------------------------------------------------------------------------------
Vapor
Liquid
Ai
yi
Vapor
Liquid
y
x
Ai
yi
xi
y
xi
x
Mass transfer from the liquid
to the gas phase
Mass transfer from the gas
to the liquid phase
The rate of absorption of H2S per unit area of the thin film is given by
NA =
ky
kc
kc c
(cA − cAi) =
(yA − yAi) =
(yA − yAi)
(1 − y A )lm
(1 − y A )lm
(1 − y A )lm
NA =
kc P
(yA − yAi)
(1 − y A )lm RT
3-3
The mole fraction of H2S in the gas phase is given by
yA =
0.05
pA
=
= 0.0333
p
1.5
The partial pressure of H2S in the gas phase at the interface is determined from Henry’s law
and the mole fraction of H2S in the liquid at the liquid-gas interface.
pAi = 609xAi = 609×2.0×10-5 = 1.218×10-2 atm
The mole fraction of H2S in the gas phase at the interface is then
yAi =
(1 − yA)lm =
0.01218
p Ai
=
= 0.00812
p
1.5
(1 − y A ) − (1 − y Ai ) (1 − y A ) + (1 − y Ai )
≈
= 0.979
 1 − yA 
2
ln 

 1 − y Ai 
NA =
kc P
kc
(yA − yAi) = NA =
(pA − pAi)
(1 − y A )lm RT
(1 − y A )lm RT
NA =
( 0.05 − .01218)
9.576 × 10 −4
= 1.486×10-3 kmol/m2·s
−3
(82.06
×
10
)(273
+
30)
0.979
3-4
Chapter 3
Example 3.1-3 -----------------------------------------------------------------------------Calculate the mass transfer from a sphere of naphthalene to air at 45oC and 1 atm abs flowing
at a velocity of 0.305 m/s. The diameter of the sphere is 25.4 mm. The diffusivity of
naphthalene in air at 45oC is 6.92×10-6 m2/s and the vapor pressure of solid naphthalene is
0.555 mmHg. The mass transfer coefficient may be obtained from the following correlation:
ShD = 2 + 0.552Re0.53Sc1/3
(Ref: Transport Processes and Separation Process by C.J. Geankoplis, Prentice Hall, 4th Edition, 2003)
Solution ---------------------------------------------------------------------------------------------Let A denote naphthalene and B denote air. Since the mole fraction of naphthalene is very
small, the physical properties of air at 45oC and 1 atm will be used for the gas mixture.
µ = µB = 1.93×10-5 Pa.s, ρ = ρB = 1.113 kg/m3
We now evaluate the dimensionless numbers:
Sc =
Re =
µ
ρ DAB
=
1.93 × 10−5
= 2.506
(1.113)(6.92 × 10−6 )
(1.113)(0.0254)(0.305)
ρVD
= 446.8
=
µ
1.93 × 10 −5
ShD = 2 + 0.552Re0.53Sc1/3 = 21 =
kc =
NA =
kc D
DAB
(21)(6.92 × 10 −6 )
= 5.72×10-3 m/s
0.0254
kc
kc
(cAi − cA) =
(pAi − pA)
(1 − y A )lm
(1 − y A )lm RT
(1 − yA)lm =
(1 − y A ) − (1 − y Ai )
≈1
 1 − yA 
ln 

 1 − y Ai 

5.72 × 10−3  0.555 × 1.013 × 105
− 0  = 1.60×10-7 kmol/m2·s
NA =

(8314)(318) 
760

The mass transfer rate from the sphere is then
WA = πD2NA = π(0.0254)2(1.60×10-7) = 3.24×10-10 kmol/s
3-5
Example 3.1-4 -----------------------------------------------------------------------------An experiment can be performed to determine the mass transfer coefficient by flowing pure
water through a tube constructed of solid benzoic acid. The saturated concentration of
benzoic acid is 2.0×10-2 g/cm3. The water velocity is 10 cm/s and the mass of the tube is
reduced by 0.62 g after 3 hr. Determine the mass transfer coefficient for the dissolution of
benzoic acid in water if the tube diameter is 1.0 cm and the tube length is 20 cm.
Solution ----------------------------------------------------------------------------------------------
V
V
x
Making a material balance of benzoic acid over the control volume πD2∆x/4 we have
D2
CVπ
4
x
D2
– CVπ
4
+ kc(C* – C)πD∆x = 0
(E-1)
x +∆x
In this expression, C is the bulk concentration of benzoic acid in water within the tube and
C* is the saturated benzoic acid concentration in water at the solid and liquid interface. kc is
the mass transfer coefficient for the dissolution of benzoic acid in water. Dividing equation
(E-1) by πD2∆x/4 gives
4
− C |x 
C |
=0
– V  x +∆x
 + kc(C* – C)
∆x
D


Taking the limit as ∆x → 0, we have
–V
dC
4
+ kc(C* – C)
=0
dx
D
Separating the variables and integrating to obtain
∫
CL
0
dC
4k c
=
C*− C
DV
∫
L
0
dx
 C * − CL  4k c L
– ln 
=> CL = C*
 =
DV
 C* 
The mass transfer rate to the water is then
W = Vπ
D2
D2
( CL – 0) = Vπ
CL
4
4
3-6

 4kc  
1 − exp  − DV L  



(E-2)
The mass change of the tube during the time t is given by
∆m = Wt = Vπ
D2

 4k  
tC* 1 − exp  − c L  
4
 DV  

For x << 1, exp(– x) ≈ 1 – x, therefore
∆m = Vπ
D2
4k L
tC* c = πDL tC*kc
4
DV
The mass transfer coefficient is evaluated
kc =
∆m
0.62
=
= 4.57×10-5 cm/s
π DLtC * π (1)(20)(3 × 3600)(0.02)
Now we need to check the condition that x =
4kc L
<< 1
DV
(4 × 4.57 × 10−5 )(20)
4kc L
=
= 3.65×10-4 << 1
(1)(10)
DV
Example 3.1-5 -----------------------------------------------------------------------------Liquid water at 25oC is to be aerated in a bubble column where finely air bubbles with
diameter dB of 0.5 mm are injected cocurrently with the liquid. The interfacial contact area,
a, between air and water can be calculated from the expression a = 6ε/dB, where ε is the
volume fraction of the injected air. The bubble column is 1.8 m high with a superficial liquid
velocity of 0.2 m/s. The oxygen concentration of the inlet water is 0.12×10-4 kmol/m3. The
saturated oxygen concentration is 2.67×10-4 kmol/m3. Determine the oxygen concentration of
the outlet water if the mass transfer coefficient for the transfer of oxygen from the liquid
interface to the bulk water is 5.8×10-6 m/s. The diffusivity of oxygen in water is 2.42×10-9
m2/s. The volume fraction of the injected air is 0.2.
Solution ---------------------------------------------------------------------------------------------Making a steady state material balance for dissolved oxygen in water we have
CVA|x – CVA|x+∆x + (1 – ε)A N A x – (1 – ε)A N A x +∆x + kc(C* – C)aA∆x = 0
In this expression, C is the bulk concentration of oxygen in water, C* is the saturated oxygen
concentration, A is the cross sectional area of the bubble column, V is the superficial liquid
velocity, NA is the molar flux of oxygen in the x direction, and kc is the mass transfer
coefficient. Dividing this equation by A∆x and taking the limit as ∆x → 0 we obtain
– (1 – ε)
dN A
dC
–V
+ kc(C* – C)a = 0
dx
dx
3-7
(E-1)
Substituting NA = – DAB
dC
into equation (E-1) we have
dx
(1 – ε)DAB
d 2C
dC
–V
+ kc(C* – C)a = 0
2
dx
dx
(E-2)
d 2C
, is usually much smaller than the
dx 2
other terms. If we neglect the axial diffusion, equation (E-2) becomes
The mass transfer contribution by axial diffusion, DAB
–V
dC
+ kc(C* – C)a = 0
dx
Separating the variables and integrating to obtain
∫
CL
C0
dC
ak
= c
C*− C
V
∫
L
0
dx
 C * − CL  ak c L
 ak
– ln 
=> CL = C* – (C*– C0) exp  − c
=
V
 V
 C * − C0 
We have

L

(6)(0.2)(5.8 × 10−6 )(1.8)
akc
6ε kc L
L=
=
= 0.1253
(0.5 × 10−3 )(0.2)
V
d BV
CL = 2.67×10-4 – (2.67 – 0.12)×10-4exp(– 0.1253) = 0.42×10-4 kmol/m3
3-8
3.2 Packed Column
Packed towers can be used for continuous countercurrent contacting of gas and liquid in
absorption and for vapor-liquid contacting in distillation. In a packed column used for gasliquid contact, the liquid flows downward over the surface of the packing and the gas flows
upward in the void space of the packing material. A low pressure drop and, hence, low
energy consumption is very important in the performance of packed towers. The packing
material provides a very large surface area for mass transfer, but it also results in a pressure
drop because of friction. The performance of packed towers depends upon the hydraulic
operating characteristics of wet and dry packing. In dry packing, there is only the flow of a
single fluid phase through a column of stationary solid particles. Such flow occurs in fixedbed catalytic reactor and sorption operations (including adsorption, ion exchange, ion
exclusion, etc.) In wet packing, two-phase flow is encountered. The phases will be a gas and
a liquid in distillation, absorption, or stripping. When the liquid flows over the packing it
occupies some of the void volume in the packing normally filled by the gas, therefore the
performance of wet packing is different from that of dry packing.
For dry packing, the pressure drop may be correlated by Ergun equation
 ∆P 


 h 
 Dp gc   ε 3 
1−ε
 = 150
+ 1.75

 
N Re
 ρ f v s   1 − ε 
(3.2-1)
where
∆P
h
Dp
= pressure drop through the packed bed
= bed height
= particle diameter
ρf = fluid density
vs = superficial velocity at a density averaged between inlet and outlet conditions
ε
= bed porosity
D vρ
NRe = average Reynolds number based upon superficial velocity p s f When the
µ
packing has a shape different from spherical, an effective particle diameter is defined
Dp =
6V p
Ap
=
6(1 − ε )
As
(3.2-2)
where
As
= interfacial area of packing per unit of packing volume, ft2/ft3 or m2/m3
The effective particle diameter Dp in Eq. (3.2-1) can be replaced by φsDp where Dp now
represents the particle size of a sphere having the same volume as the particle and φs the
shape factor. The bed porosity, ε, which is the fraction of total volume that is void is defined
as
ε
≡
volume voids
volume of entire bed
3-9
ε
≡
volume of
πR 2 h −
=
entire bed − volume of
volume of entire bed
particles
weight of all particles
particle density
πR 2 h
(3.2-3)
where R = inside radius of column, As and ε are characteristics of the packing. Experimental
values of ε can easily be determined from Eq. (3.2-3) but As for non-spherical particles is
usually more difficult to obtain. Values of As and ε can be found in various references6,7 for
the common commercial packing. As for spheres can be computed from the volume and
surface area of a sphere.
For wet packing, the pressure drop correlation is given by Leva8
(
 ∆P 
βL / ρ

 = α 10 L
h


)  Gρ
2
v

v



(3.2-4)
where ∆P is the pressure drop (psf), h is the packing height (ft), L is the liquid mass flow rate
per unit area (lb/hr-ft2), Gv is the gas mass flow rate per unit area (lb/hr-ft2), ρL is the liquid
density (lb/ft3), ρV is the gas density (lb/ft3), and α and β are packing parameters9.
The initial procedure for designing a packed column is similar to that for a plate column.
However we will need to follow different procedure in the calculation of the column
diameter and height.
Vapor
Liquid
Ai
yi
Vapor
Liquid
y
x
Ai
yi
xi
y
xi
x
Mass transfer from the liquid
to the gas phase
Mass transfer from the gas
to the liquid phase
Figure 3.2-1 Mass transfer across the interface.
6
Mc Cabe W. L. et al , Unit Operations of Chemical Engineering, McGraw-Hill, 1993, pg. 689
Perry, J. H., Chemical Engineers’ Handbook, McGraw-Hill, 1984, pg. 18-23
8
Leva M., Chem. Eng. Prog. Symp. Ser. 50(10): 51 (1954)
9
Wankat, P. C., Equilibrium Staged Separations, Elsevier, 1988, pg.420
7
3-10
Chapter 3
The mass transfer rate, mɺ A , of species A across the interfacial area for mass transfer Ai is
given by
mɺ A = (Area for mass transfer)(mass transfer coefficient)(driving force)
The driving force for mass transfer can be expressed in many different ways. It could be
based on the mole or mass fraction in the gas, or liquid phase, or both. The mass transfer rate
for mass transfer from the liquid to the gas phase can be written as
mɺ A = Aiky(yAi − yA) = Aikx(xA − xAi)
(3.2-5)
In this expression, ky and kx are the individual mass transfer coefficients based on the gas and
the liquid phase, respectively. The mole fractions yAi, yA, xA, and xAi are defined in Figure
3.2-1. The mass transfer rate for mass transfer from the gas to the liquid phase can be written
as
mɺ A = Aiky(yA − yAi) = Aikx(xAi − xA)
(3.2-6)
If we do not know the direction of mass transfer, we could use either equation (3.2-5) or
equation (3.2-6). If the mass transfer rate calculated to be positive, our assumption of the
mass transfer direction is correct. For example, if we use equation (3.2-5) and mɺ A is positive,
then species A is being transferred from the liquid to the gas phase.
Let a be the interfacial area per unit volume of packing (m2/m3). Multiplying both sides of
equation (3.2-5) by a, we obtain
mɺ A a = Aiaky(yAi − yA) ⇒
mɺ A
= aky(yAi − yA)
Ai / a
(3.2-7)
Ai/a = (Interfacial area)/(Interfacial area/volume of packing) = Vpack = Volume of packing.
Equation (3.2-7) can be written as
mɺ A
= aky(yAi − yA)
V pack
(3.2-8)
Since, the interfacial concentrations are difficult to measure, the mass transfer rate is usually
written in terms of the overall mass transfer coefficient that is based on the overall driving
force for mass transfer (yA − yA*) or (xA* − xA) as shown in Figure 3.2-2.
3-11
y
yA
m2
Equilibrium
data
yAi
m1
yA *
x
xA *
xAi
xA
Figure 3.2-2 Concentration driving forces in interphase mass transfer.
For mass transfer from the gas to the liquid phase (as shown in Figure 3.2-2)
mɺ A
= aKy(yA − yA*) = aKx(xA* − xA)
V pack
(3.2-9a)
For mass transfer from the liquid to the gas phase
mɺ A
= aKy(yA* − yA) = aKx(xA − xA*)
V pack
(3.2-9b)
1
1
is the total resistance to mass transfer based on the gas phase, and
the
aK y
aK x
total resistance to mass transfer based on the liquid phase. We have neglected the resistance
to mass transfer at the interface. In Figure 3.2-2, m1 is the average slope of the equilibrium
curve between two points (xA, yA*) and (xAi, yAi), m2 the average slope of the equilibrium
curve between two points (xAi, yAi) and (xA*, yA). The relation between the overall and the
individual mass transfer coefficients can be derived as follows:
The term
yA − yA*= (yA − yAi) + (yAi − yA*)
Since yA* = m xA and yAi = m xAi, we have
yA − yA*= (yA − yAi) + m(xAi − xA) ⇒
NA
N
N
= A +m A
Ky
ky
kx
1
1
m
=
+
Ky
ky
kx
(3.2-10a)
Similar derivation gives
3-12
1
1
1
=
+
Kx
mk y
kx
(3.2-10b)
yA,out
We now want to determine the height of a packed bed
required to change the concentration of the inlet gas from
yA,in to yA,out in a distillation column.
We will assume constant molar overflow so that the vapor
molar flow rate, V, and the liquid molar flow rate, L, are
constant over the height of the packed column. Let Ac be
the cross-sectional area of the column, the material balance
over the differential volume Acdz gives
xAL|z+dz + yAV|z = xAL|z + yAV|z+dz
V|z+dz L|z+dz
dz
V|z
L|z
z
yA,in
Figure 3.2-3 Material balance over Acdz
Rearranging the liquid and vapor flow rates gives
xAL|z+dz − xAL|z = yAV|z+dz − yAV|z
Dividing the equation by dz and letting the control volume Acdz approach zero, we have
d(LxA) = d(VyA)
For constant L and V
Ld(xA) = Vd(yA)
The molar flux of A across the interface is
NA = Ky(yA − yA*)
Multiplying the expression by aAcdz gives
NAaAcdz = Ky(yA − yA*)aAcdz
Since NAAiadz =
Mass transfer area
volume = rate of A transfer from volume Acdz, we
(area )(time) volume
have
Ky(yA − yA*)aAcdz = Vd(yA)
Solving for dz gives
dz =
V
dy A
K y a ( y A − y A *)
3-13
Integrating over the height of the packed bed we have
h=
∫
h
0
dz =
y
∫y
A , out
A , in
V
dy A
K y aAc ( y A − y A *)
(3.2-11)
The height of an overall gas transfer unit, HOG, is defined as
HOG =
V
K y aAc
If HOG is a constant, equation (3.2-11) becomes
h=
V
K y aAc
y
∫y
A , out
A , in
dy A
( y A − y A *)
(3.2-12)
For distillation column where species A is transferred from the liquid to the gas phase, h is
given by
h=
V
K y aAc
y
∫y
A , out
A , in
dy A
= HOGnOG
( yA * − yA)
(3.2-13)
In this expression, nOG is the number of overall gas transfer unit. If the driving force is based
on the driving force in the liquid phase
NAaAcdz = Kx(xA − xA*)aAcdz = Ld(xA)
(3.2-14)
The height of the packed bed is then given by
h=
L
K x aAc
∫
x
A , out
x
A , in
dx A
= HOLnOL
( x A − x A *)
(3.2-15)
HOL is the height of an overall liquid transfer unit and nOL the number of overall liquid
y
x
dx A
dy A
transfer unit. We now need to evaluate the integral ∫ A, out
or ∫ A, out
.
y
x
( yA * − yA)
( x A − x A *)
A , in
A , in
3-14
yA,out
yA
V
xA,in
xA
yA
Equilibrium curve
yA,out
yA*- yA
L
Operating line
yA,in
z
yA,in
xA,out
xA
xA
Figure 3.2-4 Material balance over the lower section of the tower.
Assuming L and V are constant and making an A balance over the lower section of the tower
as shown in Figure 3.2-4 we have
xA,outL + yAV = xAL + yA,inV
Solving for yA we obtain an equation called the operating line
yA =
L
L
xA + yA,in − xA,out
V
V
(3.2-16)
At any location in the packed bed the bulk concentration in the vapor and liquid phase are yA
and xA respectively. The point (xA, yA) is on the operating line shown in Figure 3.2-4. The
y
dy A
number of transfer unit nOG or the integral ∫ A, out
can be evaluated where (yA* −
y
( yA * − yA)
A , in
yA) is the difference between the concentration yA* that is in equilibrium with the liquid at xA
to the vapor concentration at the point (xA, yA). If the operating line and equilibrium curves
are straight, nOG can be evaluated analytically. This condition might occur in gas absorption
with dilute solution where the equilibrium curve is straight and L and G are constant. We will
drop the subscript A with the understanding that y and x are the mole fractions of the
diffusing species in the gas and liquid phase, respectively. For gas absorption, we use G for
the gas flow rate instead of V for the vapor flow used in distillation.
Making a material balance for the diffusing species over the top section of the column as
shown in Figure 3.2-5 we have
xinL + yG = xL + youtG
Solving for y we obtain a straight operating line since L/G = constant
y=
L
L
x + yout −
xin
G
G
(3.2-17)
3-15
yout
xin
y
Operating line
y
yout
x
G
L
yA - yA *
Equilibrium curve
yin
z
yin
xout
x
x
Figure 3.2-5 Material balance over the upper section of the tower.
For straight equilibrium line
y* = mx
The number of overall gas transfer unit can be written as
nOG =
y
∫y
out
in
dy A
=
c ( y * − y)
y
∫y
in
out
dy A
( y − y*)
Combining the operating line and the equilibrium line, we have
mG
G
mG 
G


yout − mxin
y − y* = y − mx = y − m  y − yout + xin  = 1 −
y+
L
L 
L
L


nOG =
y
∫y
in
out
dy A
mG
 mG 
yout − mxin
1 −
y +
L 
L

Performing the integration gives

  mG 
mG
yout − mxin 
1 −
 yin +

1
L 
L

nOG =
ln  
mG   mG 
mG
1−
1−
yout − mxin 
 yout +
 

L
L 
L
mG 
mG


yout − mxin 
1 −
 yin +

1
L 
L

nOG =
ln  
mG 
y
−
mx

out
in
1−


L
3-16
(3.2-18)
Since y*out = mxin, we have
mG
mG
mG 
mG 


yout − mxin = 1 −
yout − y*out
1 −
 yin +
 yin +
L 
L
L 
L


mG
mG
mG
mG
mG 
mG 


yout − mxin = 1 −
yout − y*out +
y*out −
y*out
1 −
 yin +
 yin +
L 
L
L 
L
L
L


mG
mG
mG 
mG 
mG 



yout − mxin = 1 −
(yout − y*out)
1 −
 yin +
 yin − 1 −
 y*out +
L 
L
L 
L 
L



Substituting the above expression into equation (3.2-18), the number of overall gas transfer
unit is finally
nOG =
 mG  yin − y *out mG 
1
+
ln 1 −

mG 
*
L
y
−
y
L 

out
out
1−
L
(3.2-19)
We can follow a similar procedure to obtain the number of overall liquid transfer unit
nOL =
∫
x
out
x
in
nOL =
dx A
( x − x*)

L  xin − x *out
L 
+
ln 1 −


L
 mG  xout − x *out mG 
1−
mG
1
(3.2-20)
In the above expression x*out = yin/m.
Example 3.2-110. ---------------------------------------------------------------------------------Acetone in air is being absorbed by water in a packed tower having a cross-sectional area of
0.186 m2 at 293 K and 1 atm. At these conditions, the equilibrium relation is given by ye =
mxe. The inlet air contains 2.6 mole % acetone and the outlet 0.5 %. The air flow is 13.65
kmol/hr and the pure water inlet flow is 43.56 kmol/hr. Film coefficients for the given flow
in the tower are kya = 3.8×10-2 kmol/s·m3 and kxa = 6.2×10-2 kmol/s·m3. Determine the tower
height.
Solution ----------------------------------------------------------------------------------------The tower height can be obtained from the following relation
h=
10
G
K y aAc
y
∫y
A , out
A , in
dy A
= HOGnOG
( yA * − yA)
C. J. Geankoplis, Transport Processes and Separation Process Principle, Prentice Hall, 2003, pg. 672
3-17
For straight operating and equilibrium lines, the number of overall gas transfer unit is given
by
nOG =
y
∫y
A , out
A , in
 mG  yin − y *out mG 
1
dy A
+
=
ln 1 −

L  yout − y *out
L 
( y A * − y A ) 1 − mG 
L
We have m = 1.186, G =13.65 kmol/hr, L = 43.56 kmol/hr, yin = 0.026, yout = 0.005, y*out =
mxin = 1.186(0) = 0.
mG
= 0.3569
L
nOG =
 mG  yin mG 
1
1
0.026


+
=
ln (1 − 0.3569 )
ln  1 −
+ 0.3569 


mG 
L  yout
L  1 − 0.3569 
0.005

1−
L
nOG = 2.0348
The overall mass transfer coefficient is determined from the film coefficients.
1
1
1
m
=
+
=> Kya =
m
1
K ya
k ya
kxa
+
k ya k xa
Kya =
1
1
1.186
+
0.038 .062
= 0.022 kmol/s·m3
Therefore
HOG =
G
(13.65 / 3600)
=
= 0.9264 m
K y aAc
(0.022)(0.186)
The tower height is then
h = HOGnOG = (0.9264)(2.0348) = 1.89 m
3-18
Chapter 3
Example 3.2-21. ---------------------------------------------------------------------------------In an experimental study of the absorption of ammonia by water in a wetted-wall column, the
value of KG was found to be 2.75×10-6 kmol/m2⋅s⋅kPa. At one point in the column, the
composition of the gas and liquid phases were 8.0 and 0.115 mole % NH3, respectively. The
temperature was 300 K and the total pressure was 1 atm. Eighty-five percent of the total
resistance to mass transfer was found to be in the gas phase. At 300 K, ammonia-water
solution follow Henry’s law up to 5 mole % ammonia in the liquid, with m = 1.64 when the
total pressure is 1 atm. Calculate the individual film coefficients and the interfacial
concentrations.
Solution ----------------------------------------------------------------------------------------NA = KG(pA − pA*) = KGP(yA − yA*) = Ky(yA − yA*)
Ky = KGP = (2.75×10-6)(101.3) = 2.786×10-4 kmol/m2⋅s
Since
m
1
1
=
+
for gas phase resistance that accounts for 85% of the total
kx
Ky
ky
resistance:
1/ k y
1/ K y
= 0.85 ⇒ ky = Ky/0.85 = 2.786×10-4/0.85 = 3.28×10-4 kmol/m2⋅s
m / kx
= 0.15 ⇒ kx = mKy/0.15 = 1.64×2.786×10-4/0.15 = 3.05×10-3 kmol/m2⋅s
1/ K y
yA* = mxA = 1.64×1.15×10-3 = 1.886×10-3
NA = Ky(yA − yA*) = 2.786×10-4(0.08 − 1.886×10-3) = 2.18×10-5 kmol/m2⋅s
2.18 × 10 −5
NA
NA = ky(yA − yAi) ⇒ yAi = yA −
= 0.080 −
= 0.01362
3.28 × 10 −4
ky
yAi = m xAi ⇒ xAi =
1
yAi
0.01362
=
= 8.305×10-3
m
1.64
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 169
3-19
Example 3.2-32. ---------------------------------------------------------------------------------A wetted-wall absorption tower is fed with water as the wall liquid and an ammonia-air
mixture as the central-core gas. At a particular point in the tower, the ammonia concentration
in the bulk gas is 0.6 mole fraction, that in the bulk liquid is 0.12 mole fraction. The
temperature is 300 K and the pressure is 1 atm. Ignoring the vaporization of water, calculate
the local ammonia mass-transfer flux. Data: kx = 3.5/(1 − xA)lm mol/m2⋅s, and ky = 2.0/(1 −
yA)lm mol/m2⋅s; equilibrium relation: yAi = 10.5xAi[0.156 + 0.622xAi(5.765xAi − 1)].
Solution ----------------------------------------------------------------------------------------The molar flux of ammonia (A) is given by
NA = ky
x Ai − xA
y A − y Ai
= kx
(1 − xA ) − (1 − x Ai )
(1 − y Ai ) − (1 − y A )
 1 − xA 
 1 − y Ai 
ln 
ln 


 1 − yA 
 1 − xAi 
 1 − y Ai 
 1 − xA
ky ln 
 = kx ln 
 1 − yA 
 1 − x Ai
 1 − xA 
yAi = 1 − (1 − yAi) 

 1 − x Ai 

 1 − xA 
1 − y Ai
=
⇒

1 − yA

 1 − x Ai 
kx / ky
kx / ky
1.75
 0.88 
= 1 − 0.4 

 1 − x Ai 
(E-1)
yAi and xAi are the solutions of Eq. (E-1) and the equilibrium relation:
yAi = 10.5xAi[0.156 + 0.622xAi(5.765xAi − 1)]
(E-2)
The following Matlab codes solve Eqs. (E-1) and (E-2) and plot out these equations. The
intersection of the two curves from these equations also provides the solution.
v=fminsearch('f3d2d3',[0.5 0.5]);
xi=v(1);yi=v(2);
fprintf('xAi = %8.3f, yAi = %8.3f\n',xi,yi)
x=0:0.01:0.28;
y1=1-0.4*(0.88./(1-x)).^1.75;
y2=10.5*x.*(0.156 + 0.622*x.*(5.765*x-1));
plot(x,y1,x,y2)
xlabel('x_A');ylabel('y_A')
grid on
function ff=f3d2d3(v)
xi=v(1);yi=v(2);
f1=yi-1+0.4*(0.88/(1-xi))^1.75;
f2=yi-10.5*xi*(0.156 + 0.622*xi*(5.765*xi-1));
2
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 171
3-20
ff=f1*f1+f2*f2;
>> e3d2d3
xAi = 0.231, yAi =
0.494
The interfacial concentrations are: xAi = 0.231 and yAi = 0.494. The molar flux of A is then
 1 − y Ai 
 1 − 0.494 
2
NA = ky ln 
 = 2 ln 
 = 4.7 mol/m ⋅s
1
−
y
1
−
0.600



A 
Example 3.2-43. ---------------------------------------------------------------------------------A mixture of methanol (substance 1, the more volatile) and water (substance 2) is being
distilled in a packed tower. At a point along the tower where the temperature is 360 K, the
methanol content of the bulk of the gas phase is 36 mol%; that of the bulk of the liquid phase
is 20 mol%. Assume equimolar counter diffusion with mass transfer coefficients ky = 0.0017
kmol/m2⋅s and ky = 0.0017 kmol/m2⋅s, estimate the local flux of methanol from the liquid to
the gas phase. Solve the problem by plotting equilibrium curve at 360 K. The activity
coefficients for this system are given by:
 ∆12
∆ 21 
V2
 −a 
−
ln γ1 = − ln(x1 +x2∆12) + x2 
exp  12 
 ; ∆12 =
V1
 RT 
 x1 + x2 ∆12 x2 + x1∆ 21 
3
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 174
3-21
 ∆12
∆ 21 
V1
 −a 
−
ln γ2 = − ln(x2 +x1∆21) + x1 
exp  21 
 ; ∆21 =
V2
 RT 
 x1 + x2 ∆12 x2 + x1∆ 21 
Recommended values of the parameter for this system are: V1 = 40.73 cm3/mol, V2 = 18.07
cm3/mol, a12 = 107.38 cal/mol, and a21 = 469.55 cal/mol. The vapor pressure for methanol
and water can be determined from the following equations:
ln P1vap = 16.5938 −
3644.3
T − 33
ln P2vap = 16.2620 −
3800.0
T − 47
Solution ----------------------------------------------------------------------------------------The equilibrium curve at 360 K is obtained from the following steps:
1)
2)
3)
4)
Choose a value of mole fraction in the liquid phase, x1, between 0.1 and 0.25;
Evaluate the activity coefficients;
Evaluate the vapor pressures
Evaluate the partial pressures:
p1 = γ1x1P1vap, and p2 = γ2x2P2vap
5) The mole fraction in the vapor phase, y1, is then p1/( p1 + p2)
The local flux is obtained from the following expressions:
N1 = kx(x − xi) = ky(yi − y)
yi = y +
0.0149
kx
(x − xi) = 0.36 +
(0.2 − xi)
0.0017
ky
(E-1)
The above equation is a straight line that can be plotted on the graph with the equilibrium
curve. The intersection of flux equation (E-1) and the equilibrium curve provides the
interfacial compositions xi and yi.
The Matlab codes, listed in Table E-1, plots the equilibrium curve and flux equation (E-1).
From the closed up view of the graph (Figure E-2), the interfacial compositions are
xi = 0.17884
and yi = 0.54545
The local flux is then
N1 = kx(x − xi) = 0.0149(0.2 − 0.17884) = 3.15×10-4 kmol/m2⋅s
3-22
--------- Table E-1 Matlab codes to plot equilibrium curve and flux equation ---------
R=1.987;T=360;v1=40.73;v2=18.07;a12=107.38;a21=469.55;
RT=R*T;
d12=v2*exp(-a12/(RT))/v1;d21=v1*exp(-a21/(RT))/v2;
x1=0.1:0.01:0.25;x2=1-x1;
con=d12./(x1+x2*d12)-d21./(x2+x1*d21);
gam1=exp(-log(x1+x2*d12)+x2.*con);
gam2=exp(-log(x2+x1*d21)-x1.*con);
pv1=exp(16.5938-3644.3/(T-33));
pv2=exp(16.2620-3800.0/(T-47));
p1=pv1*gam1.*x1;p2=pv2*gam2.*x2;
y1=p1./(p1+p2);
plot(x1,y1)
x=0.15;
y=0.36+(0.2-x)*.0149/.0017;
plot(x1,y1,[x 0.2],[y 0.36],'--')
xlabel('x_A');ylabel('y_A')
grid on
legend('Equilibrium','Flux equation')
---------------------------------------------------------------------------
Figure E-1 Graphical solution
3-23
Figure E-2 Graphical solution (closed-up view)
The interfacial compositions xi and yi can be obtained directly from the following Matlab
codes without graphing:
function ff=f3d2d4(v)
x1=v(1);y1=v(2);x2=1-x1;
R=1.987;T=360;v1=40.73;v2=18.07;a12=107.38;a21=469.55; RT=R*T;
d12=v2*exp(-a12/(RT))/v1;d21=v1*exp(-a21/(RT))/v2;
con=d12/(x1+x2*d12)-d21/(x2+x1*d21);
gam1=exp(-log(x1+x2*d12)+x2*con);
gam2=exp(-log(x2+x1*d21)-x1*con);
pv1=exp(16.5938-3644.3/(T-33));
pv2=exp(16.2620-3800.0/(T-47));
p1=pv1*gam1*x1;p2=pv2*gam2*x2;
f1=y1-p1/(p1+p2);
f2=y1-0.36-0.0149*(0.2-x1)/0.0017;
ff=f1*f1+f2*f2;
>> fminsearch('f3d2d4',[0.5 0.5])
ans =
0.1788
0.5455
3-24
Chapter 3
Example 3.2-5. ---------------------------------------------------------------------------------Sulfur dioxide produced by the combustion of sulfur in air is absorbed in water. Pure SO2 is
then recovered from the solution by steam stripping. Make a preliminary design for the
absorption column. The feed will be 5000 kg/hr of gas containing 8 mole percent SO2. The
gas will be cooled to 20oC. A 95 percent recovery of the SO2 is required2.
Solution ----------------------------------------------------------------------------------------Operation is at atmospheric pressure as the solubility of SO2 in water is high. The feed water
temperature will be taken as 20oC.
Table E-1. Equilibrium data for SO2 at 1 atm and 20oC.
0
.000564
.000842
.001403
.001965
.00279
0
.0112
.01855
.0342
.0513
.0775
x
y
At 95 percent recovery of SO2
xin = 0
.00420
.121
yout
yout = (0.05)(0.08) = 0.004
Slope of equilibrium line: y* = mx
G = 5000 kg/hr
0.0775 = m(0.00279) ⇒ m = 27.8
To decide the most economic water flow rate,
the stripper should be considered together with the
absorption design. For this example, the absorption
design will be considered alone.
xout
yin = 0.08
The number of gas transfer unit may be estimated from
NOG =
*


y − yout
1
ln (1 − mG / L) in
+ mG / L
*
y out − yout
1 − mG / L 

Where G =
5000
= 0.1055 lbmol/s
(0.454)(3600)( 29)
yin = 0.08, yout = 0.004, y*out = mxin = (27.8)(0) = 0
Evaluate as NOG a function of mG/L
mG/L
0.2
NOG
3.5
L,lbmol/s 14.66
0.3
3.8
9.78
0.4
4.2
7.33
0.5
4.7
5.87
0.6
5.4
4.89
3-25
0.7
6.3
4.19
0.8
7.8
3.67
0.9
10.6
3.26
0.99
17.4
2.96
The optimum will be between mG/L = 0.6 to 0.8. Below 0.6 there is only a small
decrease in the number of stages required with increasing liquid rate and above 0.8, the
number of stages increases rapidly with decreasing liquid rate (Figure E-1)
18
16
L (lbm ol/s), NOG
14
12
10
L
NOG
8
6
4
2
0.2
0.3
0.4
0 .5
0.6
0.7
0 .8
0.9
1
m G/L
Figure E-1. Optimum liquid flow rate for Sulfur Dioxide absorption.
Check the liquid outlet composition at mG/L = 0.6 and at mG/L = 0.8. Assuming dilute
solution, the material balance is
L(xout – xin) = G(yin – yout) ⇒ xout = G(yin – yout)/L + xin
xout = (0.08 – 0.004)
mG
27.8 L
At mG/L = 0.6, xout = 1.64×10-3
At mG/L = 0.8, xout = 2.19×10-3
Use mG/L = 0.8 as the higher concentration will favor the stripper design and operation
without significantly increasing the number of stages needed in the absorber. Therefore
NOG = 8
3-26
Estimate column diameter:: 2 methods.
First design method: chooses the pre
pressure
ssure drop per unit length of packing then use
Figure 13.41 to evaluate G'(lb/s
'(lb/s⋅ft2). Note: G in the ordinate of Figure E-2 is actually G'.
lb 
 lbmol 
G
 M .W .vapor

s 
lbmol 
Area = 
lb
G'
s ⋅ ft 2
Check the percent of flooding (65 to 90%).
Figure E-2 Generalized flooding and pressure drop correlation.
10
Second design method: Use the flooding curve of Figure E-2
2 or the Eq. (3.2-21)
(3.2
to
determine G'flood and operate at some percentage of G'flood.
G' = (.65 to .90) G'flood
Once G'' is known the column cross
cross-sectional area A can be determined.
10
Wankat, P. C., Equilibrium Staged Separations
Separations, Elsevier, 1988, pg. 420
3-27
 G ' 2 Fψµ 0.2 
log10 
= − 1.6678 − 1.085 log10(Flv) − 0.29655[log10(Flv)]2

 ρG ρ L gc 
1/ 2
L ρ 
L'  ρ G 
= m  G
where
Flv =


G'  ρL 
Gm  ρ L 
Gm are mass flow rates in this expression.
(3.2-11)
1/ 2
= the abscissa of Figure E--2. Note: Lm and
The first design method will be applied. The physical properties of the gas can be taken
as those for air as the concentration of SO2 is low.
G=
5000
= 0.1055 lbmol/s
(0.454)(3600)( 29)
At mG/L = 0.8, L = mG/0.8
0.8 = (27.8)(0.1055)/0.8 = 3.666 lbmol/s
Packing: Choose 1.5" Ceramic Intalox saddle (picture below) with packing parameters F =
32, α = 0.13, β = 0.15. Intalox saddle is one type of dumped packings.
 G ' 2 Fψµ 0.2 
F is the parameterr in the ordinate of Figure E
E-2 
 . α and β are the
 ρG ρ L gc 
parameters to determine the pressure drop ∆p in inches of water per foot of packing given by
 G' 2 

∆p = α(10 ) 
ρ
 G
βL'
Air (gas) density at 20oC: ρG =
(3.2-22)
(29)( 273)
= 0.0753 lb/ft3
(359)( 293)
Liquid density: ρL = 62.3 lb/ft3, liquid viscosity = 1 cp.
The column will be designed for a pressure drop of 0.5 in of water/ft of packing. Table
E-22 shows the recommended design values.
3-28
Table E-2. Recommended design value for ∆p (inches of water per foot of packing)
∆p (inches of water per foot of packing)
0.2 – 0.6
0.5 – 1.0
Application
Absorber and stripper
Distillation (atmospheric & moderate
pressure)
Vacuum columns
0.1 – 0.4
If very low bottom pressures are required, structured packings or special low pressure drop
dumped packings should be considered (Hyperfil, Multifil, or Dixon rings).
The column area may be estimated from the pressure drop ∆p using either Eq. (3.2-22)
or Figure E-2. The procedure for using Figure E-2 will be discussed
ψ=
Density of
Density of
water
=1
liquid
1/ 2
L ρ 
= m  G
Gm  ρ L 
L'  ρ G 
Flv =
 
G'  ρL 
1/ 2
1/ 2
(3.67)(18)  0.0753 
=


(0.1055)( 29)  62.3 
= 0.75
From Figure E-2 at Flv = 0.75 and ∆p = 0.5 in water/ft of packing
 G ' 2 Fψµ 0.2 
 ρ ρ g  = 0.018
 G L c 
 G ' 2f Fψµ 0.2 
At flooding using the flooding line or Eq. 3.2-21: 
 = 0.0294
 ρ G ρ L g c 
 G'

 G'
 f
2

G'
 = 0.018 = 0.612 ⇒
= 0.78 (O.K., between 65 and 90 %)

G
'
0
.
0294
f

 0.018 ρ G ρ L g c 
G' = 

Fµ 0.2


1/ 2
 (0.018)(0.0753)(62.3)(32.2) 
= 

(52)(1)


1/ 2
= 0.229 lb/s⋅ft2
Note: µ in the above expression is the liquid viscosity in centipoises.
Gas mass flow rate Gm = (0.1055)(29) = 3.06 lb/s = AcG'
3.06
 (4)(13.38) 
Ac =
= 13.38 ft2 ⇒ Dc = 

0.229
π


3-29
1/ 2
= 4.13 ft
Check packing size: Recommend size ranges are
Column diameter
< 1 ft
1 to 3 ft
> 3 ft
Use packing size
1 in.
1 to 1.5 in.
2 to 3 in.
In general, the largest size of packing that is suitable for the size of column should be used,
up to 2 in. Small sizes are appreciably more expensive than the larger sizes. Above 2 in., the
lower cost does not normally compensate for the lower mass transfer efficiency. Use of too
large a size in a small column can cause poor liquid distribution. Since 1.5 in. ceramic
Intalox saddle is used in this example, a larger size could be considered.
The height of packing may be determined from the following formula
hp = NOGHOG
The height of overall gas transfer unit, HOG, may be evaluated from the height of gas transfer
unit, HG, and the height of liquid transfer unit, HL
HOG = HG +
mG
HL
L
The correlation for HG is
HG =
Where
ψ ( D ' col ) b1 ( h p / 10)1/ 3 ( Scv )1/ 2
[(3600) L' ( µ L / µV ) −1.25 (σ L / σ V ) −0.8 ]b 2
b1 = 1.11 for saddles, b2 = 0.50 for saddles
D'col = lesser of column diameter in ft or 2
hp = height of packed bed in ft
Scv = Schmidt number for vapor = µv/ρvDv
L' = mass flux of liquid, lb/s⋅ft2
σ = surface tension of liquid (L) or water (W)
ψ = packing parameter given by Figure E-3.
3-30
0
20
40
60
80 90
Figure E-3. Packing parameter ψ (ft) as a function of percent flood11
Ceramic Berl saddles are used to make conservative estimate of packing parameter for
Intalox saddles since mass transfer efficiency of Intalox saddles is higher than that of the
equivalent size Berl saddles.
µL
ρ
σ
=1; L =1; L =1
µW
ρW
σW
102
103
104
105
Figure E-4. Packing parameter φ (ft) as a function of L'
Figure E-5. Vapor load coefficient
The correlation for HL is
HL = φCfL(hp/10)0.15(ScL)1/2
where
φ = packing parameter given in Figure E-412
CfL = vapor load coefficient given in Figure E-512
ScL = Schmidt number for liquid = µL/ρLDL
Estimate hp by assuming a value for so HOG that HG and HL can be evaluated. Let HOG =
2.2 ft, then
11
12
Wankat, P. C., Equilibrium Staged Separations
Separations, Elsevier, 1988, pg. 652
Wankat, P. C., Equilibrium Staged Separations
Separations, Elsevier, 1988, pg. 654
3-31
hp = NOG HOG = (8)(2.2) ≈ 18 ft
At 78% flooding and ∆p = 1.5 in, Figure E-3 gives a packing value ψ = 65 ft. From the
PROP program (T.K. Nguyen Website)
Diffusivity of SO2 in water at 20oC:
Diffusivity of SO2 in air at 20oC and 1 atm:
Viscosity of gas (air) at 20oC:
DL = 7.5×10-6 cm2/s = 8.07×10-9 ft2/s
Dv = 0.122 cm2/s = 1.31×10-4 ft2/s
µv = 1.82×10-5 kg/m⋅s = 1.22×10-5 lb/ft⋅s
Scv = µv/ρvDv = (1.22×10-5)/[(7.53×10-2)(1.31×10-4)] = 1.237
ScL = µL/ρLDL = (6.72×10-4)/[(62.3)(8.07×10-9)] = 1337
L' = (3.666)(18)/(13.38) = 4.93 lb/s⋅ft2 = 1.78×104 lb/hr⋅ft2
Flooding ratio = 0.78 ⇒ CfL ≈ 0.70 (Figure 19-8)
ψ ( D ' col ) b1 ( h p / 10)1/ 3 ( Scv )1/ 2
(65)(2)1.11 (18 / 10)1/ 3 (1.237)1 / 2
HG =
=
= 1.42 ft
[(3600) L' ( µ L / µV ) −1.25 (σ L / σ V ) −0.8 ]b 2
[(3600)( 4.93)]1 / 2
L' = 1.78×104 lb/hr⋅ft2 ⇒ φ ≈ 0.1 (Figure E-4)
HL = φCfL(hp/10)0.15(ScL)1/2 = (0.1)(0.7)(1.8)0.15(1337)1/2 = 2.80 ft
HOG = HG +
mG
HL = 1.42 + (0.8)(2.80) = 3.66 ft
L
hp = NOG HOG = (8)(3.66) = 29.3 ft
Repeat the calculation with hp = 30 ft
(65)(2)1.11 (30 / 10)1 / 3 (1.237)1/ 2
 3 
HG =
=

1/ 2
[(3600)(4.93)]
 1 .8 
 3 
HL = (0.1)(0.7)(3.0)0.15(1337)1/2 = 

 1.8 
HOG = HG +
1/ 3
(1.42) = 1.69 ft
0.15
(2.80) = 3.02 ft
mG
HL = 1.69 + (0.8)(3.02) = 4.1 ft
L
hp = NOG HOG = (8)(4.1 = 32.8 ft
3-32
Chapter 5
Absorption and Stripping
5.1 Introduction
In absorption (also called gas absorption, gas scrubbing, or gas washing), there is a transfer
of one or more species from the gas phase to a liquid solvent. The species transferred to the
liquid phase are referred to as solutes or absorbate. Absorption involves no change in the
chemical species present in the system. Absorption is used to separate gas mixtures, remove
impurities, or recover valuable chemicals. The operation of removing the absorbed solute
from the solvent is called stripping. Absorbers are normally used with strippers to permit
regeneration (or recovery) and recycling of the absorbent. Since stripping is not perfect,
absorbent recycled to the absorber contains species present in the vapor entering the
absorber. When water is used as the absorbent, it is normally separated from the solute by
distillation rather than stripping.
Exit gas
o
25 C, 90 kPa
Liquid absorbent
o
25 C, 101.3 kPa
kmol/h
Water 1943
Feed gas
o
25 C, 101.3 kPa
1
kmol/h
Argon
6.9
O2 144.291
N2 535.983
Water
22.0
Acetone
0.05
30
kmol/h
Argon
6.9
O2 144.3
N2 536
Water
5.0
Acetone 10.3
Exit liquid
o
22 C, 101.3 kPa
kmol/h
O2
0.009
N2
0.017
Water 1,926.0
Acetone 10.25
Figure 5.1-1 Typical absorption process.
A typical industrial operation for an absorption process is shown in Figure 5.1-11. The feed,
which contains air (21% O2, 78% N2, and 1% Ar), water vapor, and acetone vapor, is the gas
1
J. D. Seader and E. J. Henley, Separation Process Principles, , Wiley, 2006, pg. 194
5-1
leaving a dryer where solid cellulose acetate fibers, wet with water and acetone, are dried.
Acetone is removed by a 30-tray absorber using water as the absorbent. The percentage of
acetone removed from the air stream is
10.25
×100 = 99.5%
10.3
Although the major component absorbed by water is acetone, there are also small amounts of
oxygen and nitrogen absorbed by the water. Water is also stripped since more water appears
in the exit gas than in the feed gas. The temperature of the exit liquid decreases by 3oC to
supply the energy of vaporization needed to strip the water. This energy is greater than the
energy of condensation liberated from the absorption of acetone.
Three approaches have generally been employed to develop equations used to predict the
performance of absorbers and absorption equipment: mass transfer coefficients, graphical
solution, and absorption factor. The use of mass transfer coefficient is covered in Chapter
2.2. The graphical solution is simple to use for one or two components and provides explicit
graphical presentation of the interrelationships of the variables and parameters in an
absorption process. However the graphical technique becomes very tedious when several
solutes are present and must be considered. The absorption factor approach can be utilized
either for hand or computer calculation. Absorption and stripping are conducted mainly in
packed columns and plate columns (trayed tower) as shown in Figure 5.1-2.
Packed column2
Plate column3
Figure 5.1-2 Equipment for absorption and stripping.
2
3
www.mikropul.com/products/wscrubber/packed.htm (Aug. 25 2009)
http://www.cgscgs.com/ga_tt.htm (Aug. 25 2009)
5-2
5.2 Single-Component Absorption
Most absorption or stripping operations are carried out in counter current flow processes, in
which the gas flow is introduced in the bottom of the column and the liquid solvent is
introduced in the top of the column. The mathematical analysis for both the packed and
plated columns is very similar.
Vt
yA,t
Lt
xA, t
L
V
Lb
xA, b
Vb
yA, b
Figure 5.2-1 Countercurrent absorption process.
The overall material balance for a countercurrent absorption process is
Lb + Vt = Lt + Vb
where
(5.2-1)
V = vapor flow rate
L = liquid flow rate
t, b = top and bottom of tower, respectively
The component material balance for species A is
LbxA,b + Vt yA,t = LtxA,t + Vb yA,b
where
(5.2-2)
yA = mole fraction of A in the vapor phase
xA = mole fraction of A in the liquid phase
For some problems, the use of solute-free basis can simplify the expressions. The solute-free
concentrations are defined as:
XA =
YA =
xA
mole fraction of A in the liquid
=
1 − x A mole fraction of non-A components in the liquid
yA
mole fraction of A in the vapor
=
1 − y A mole fraction of non-A components in the vapor
(5.2-3a)
(5.2-3b)
If the carrier gas is completely insoluble in the solvent and the solvent is completely
nonvolatile, the carrier gas and solvent rates remain constant throughout the absorber. Using
5-3
L to denote the flow rate of the nonvolatile and V to denote the carrier gas flow rate, the
material balance for solute A becomes
L X A,b + V YA,t = L X A,t + V YA,b
or
YA,t =

LX A,b 
L
X A,t +  YA,b −

V 
V

(5.2-4)
(5.2-5)
The material balance for solute A can be applied to any part of the column. For example, the
material balance for the top part of the column is
YA,t =

LX A 
L
X A,t +  YA −

V
V 

(5.2-6)
In this equation, X A and YA are the mole ratios of A in the liquid and vapor phase,
respectively, at any location in the column including at the two terminals. Equation (5.2-6) is
L
called the operation line and is a straight line with slope
when plotted on X A - YA
V
coordinates.
The equilibrium relation is frequently given in terms of the Henry’s law constant which can
be expressed in many different ways:
PA = HCA = mxA = KxA
(5.2-7)
In this equation, PA is the partial pressure of species A over the solution and CA is the molar
concentration with units of mole/volume. The Henry’s law constant H and m have units of
pressure/molar concentration and pressure/mole fraction, respectively. K is the equilibrium
constant or vapor-liquid equilibrium ratio. Table 5.2-1 list Henry’s law constant m for
various gases in water.
Table 5.2-1 Henry’s Law constant for Gases in water4 (m×10-4 atm/mole fraction)
T(oC)
CO2
CO C2H6
C 2 H4
He
H2
H2 S
CH4
N2
O2
0
0.0728 3.52
1.26 0.552 12.9
5.79 0.0268 2.24
5.29
2.55
10
0.104
4.42
1.89 0.768 12.6
6.36 0.0367 2.97
6.68
3.27
20
0.142
5.36
2.63
1.02
12.5
6.83 0.0483 3.76
8.04
4.01
0.186
6.20
3.42
1.27
12.4
7.29 0.0609 4.49
30
9.24
4.75
40
0.233
6.96
4.23
12.1
7.51 0.0745 5.20
10.4
5.35
Example 5.2-1. 5---------------------------------------------------------------------------------A solute A is to be recovered from an inert carrier gas B by absorption into a solvent. The gas
entering into the absorber flows at a rate of 500 kmol/h with yA = 0.3 and leaving the
absorber with yA = 0.01. Solvent enters the absorber at the rate of of 1500 kmol/h with xA =
4
Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, pg.
988
5
Hines, A. L. and Maddox R. N., Mass Transfer: Fundamentals and Applications, Prentice Hall, 1985, pg. 255
5-4
0.001. The equilibrium relationship is yA = 2.8 xA. The carrier gas may be considered
insoluble in the solvent and the solvent may be considered nonvolatile. Construct the x-y
plots for the equilibrium and operating lines using both mole fraction and solute-free
coordinates.
Solution ----------------------------------------------------------------------------------------The flow rates of the solvent and carrier gas are given by
L = Lt(1 − xA,t) = 1500(1 − 0.001) = 1498.5 kmol/h
V = Vb(1 − yA,b) = 500(1 − 0.3) = 350 kmol/h
The concentration of A in the solvent stream leaving the absorber can be determined from the
following expressions:
xA,b =
Moles A in Lb
Moles A in Lb + L
Moles of A in Lb = Moles of A in Lt + Moles of A in Vb − Moles of A in Vt
Moles of A in Lb = 1500×0.001 + 500×0.3 − Moles of A in Vt
Moles A in Vt
Moles A in Vt
⇒ 0.01 =
Moles A in Vt + V
Moles A in Vt + 350
yA,t =
Moles of A in Vt = 350×0.01/(1 − 0.01) = 3.5354 kmol/h
Moles of A in Lb = 1.500 + 150 − 3.5354 = 147.965 kmol/h
xA,b =
Moles A in Lb
147.965
=
= 0.0898
Moles A in Lb + L 147.965 + 1498.5
For the solute free basis:
XA =
xA
yA
, YA =
1 − xA
1 − yA
X A ,t =
X A ,b =
x A, t
1 − x A ,t
x A ,b
1 − x A,b
=
0.0010
= 0.0010
1 − 0.0010
=
0.0898
= 0.0987
1 − 0.0898
5-5
YA,t =
y A ,t
1 − y A ,t
YA,b =
y A, b
1 − y A ,b
=
0.010
= 0.0101
1 − 0.010
=
0.30
= 0.4286
1 − 0.30
The equilibrium curves in both mole fraction and solute-free coordinates are calculated from
the following procedures:
1) Choose a value of xA between 0.001 and 0.10
xA
2) Evaluate the corresponding X A =
1 − xA
3) Evaluate yA = 2.8 xA
yA
4) Evaluate the corresponding YA =
1 − yA
The operating lines in both mole fraction and solute-free coordinates are calculated from the
following procedures:
1) Choose a value of xA between 0.001 and 0.0898
xA
2) Evaluate the corresponding X A =
1 − xA

LX A,b 
L
X A +  YA,b −

V
V 

YA
4) Evaluate the corresponding yA =
1 + YA
3) Evaluate YA =
The following Matlab codes plot the equilibrium and operating lines in both mole fraction
and solute-free coordinates.
------------------------------------------% Example 5.2-1
xe=linspace(0.001,0.1);
ye=2.8*xe;
Xe=xe./(1-xe);Ye=ye./(1-ye);
x=linspace(0.001,0.0898);
Lbar=1498.5;Vbar=350;
X=x./(1-x);Xb=.0898;Yb=0.4286;
LoV=Lbar/Vbar;
Y=LoV*X+Yb-LoV*Xb;
y=Y./(1+Y);
plot(xe,ye,x,y,'--')
legend('Equilibrium curve','Operating line',2)
xlabel('x');ylabel('y')
Title('Equilibrium and Operating lines on mole fraction coordinates')
figure(2)
plot(Xe,Ye,X,Y,'--')
legend('Equilibrium curve','Operating line',2)
5-6
xlabel('X');ylabel('Y')
Title('Equilibrium and Operating lines on solute free coordinates')
-------------------------------------------
The equilibrium relation in the mole fraction coordinates is a straight line while the operating
line in the solute-free coordinates is a straight line. Normally the equilibrium relation is not a
straight line in the mole fraction coordinates. Therefore it is advantage to use solute-free
coordinates because the operating line will always be straight.
5-7
(C)
(B)
(A)
Yb
Y
Yb
Yb
Yc
Yt
Yt
Yt
Xt
Xb
Xc
Xt
Xt
X
Xb
Figure 5.2-2 Limiting conditions for absorption process.
Xb
The driving force for mass transfer becomes zero whenever the operating line intersects or
touches the equilibrium curve. This limiting condition represents the minimum solvent rate to
recover a specified quantity of solute or the solvent rate required to remove the maximum
amount of solute. In Figure 5.2-2A, the intersection of the equilibrium and operating lines
occurs at the bottom of the absorber. This condition defines the minimum solvent rate to
recover a specified quantity of solute. This minimum solvent rate can be calculated from the
following expression:
L
Yb − Yt
  =
 V  min X b − X t
(5.2-8a)
In Figure 5.2-2B, the intersection of the equilibrium and operating lines occurs at the top of
the absorber. This condition represents the solvent rate required to remove the maximum
amount of solute. This solvent rate can be calculated from the following expression:
L
Yb − Yt
  =
 V  max X b − X t
(5.2-8b)
Equation (5.2-8b) is exactly the same as Eq. (5.2-8a) except in this case the bottom
compositions are fixed so that the maximum slope of the operating line occurs when the
operating line intersects the equilibrium curve at the top of the column.
Figure 5.2-2C shows the case when the operating line becomes tangent to the equilibrium
curve. The minimum liquid-to-vapor ratio for this case can be determined from
L
Yc − Yt
  =
 V  min X c − X t
(5.2-9)
In this equation, Yc and X c are the coordinates of the tangent point.
5-8
Chapter 5
Example 5.2-2. 6---------------------------------------------------------------------------------A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 20oC.
The entering gas contains 20 mol % SO2 and that leaving 2 mol % at a total pressure of 101.3
kPa. The inert gas flow rate is 150 kg air/h⋅m2, and the entering water flow rate is 6000 kg
water/h⋅m2. Assuming an overall tray efficiency of 25%, how many theoretical trays and
actual trays are needed? Assume that the tower operates at 20oC. Equilibrium data for SO2 –
water system at 20oC and 101.3 kPa are given:
x
y
x
y
0
0
.00420
.121
.0001403
.00158
.00698
.212
.000280
.00421
.01385
.443
.000422
.00763
.0206
.682
.000564 .000842 .001403 .001965 .00279
.01120 .01855 .0342
.0513
.0775
.0273
.917
Solution ----------------------------------------------------------------------------------------The vapor and liquid molar flow rates are calculated first
L X A,b + V YA,t = L X A,t + V YA,b
V = 150/29 = 5.18 kmol inert air/h⋅m2
L = 6000/18 = 333 kmol inert water/h⋅m2
We have yb = 0.20, yt = 0.02, and xt = 0. For the solute-free basis
XA =
xA
yA
, YA =
1 − xA
1 − yA
X A ,t =
YA,t =
YA,b =
x A, t
=
1 − x A ,t
y A ,t
1 − y A ,t
y A, b
1 − y A ,b
0
=0
1− 0
=
0.020
= 0.0204
1 − 0.020
=
0.20
= 0.250
1 − 0.20
X A,b can be determined from the component balance (A = SO2):
L X A,b + V YA,t = L X A,t + V YA,b
6
Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, p.
663
5-9
X A ,b = X A, t +
V
V
YA,b −
YA,t
L
L
5.18
5.18
×0.250 −
×0.0204 = 0.00357
333
333
The operating line and the equilibrium curve can be plotted using the following Matlab
codes:
X A ,b = 0 +
% Example 5.2-2
xe=[0 .0001403 .000280 .000422 .000564 .000842 .001403 .001965 .00279 .00420 .00698];
ye=[0 .00158 .00421 .00763 .01120 .01855 .0342 .0513 .0775 .121 .212];
Xe=xe./(1-xe);Ye=ye./(1-ye);
X=[0 .00357];Y=[.0204 .25];
plot(Xe,Ye,X,Y,'--')
legend('Equilibrium curve','Operating line',2)
xlabel('X');ylabel('Y')
Title('Equilibrium and Operating lines on solute free coordinates')
grid on
Figure E-1 Theoretical number of trays.
The number of theoretical trays is determined by simply stepping off the number of trays as
shown in Figure E-1. This gives 2.4 theoretical trays. The actual number of trays is 2.4/.25 =
10 trays.
5-10
Example 5.2-3. 6---------------------------------------------------------------------------------A waste airstream from a chemical process flows at the rate of 1.0 m3/s at 300 K and 1 atm,
containing 7.4% by volume of benzene vapors. It is desired to recover 85% of the benzene in
the gas by a three-step process. First, the gas is scrubbed using a non-volatile wash oil to
absorb the benzene vapors. Then, the wash oil leaving the absorber is stripped of the benzene
by contact with steam at 1 atm and 373 K. The mixture of benzene vapor and steam leaving
the stripper will then be condensed. Because of the low solubility of benzene in water, two
distinct liquid phases will form and the benzene layer will be recovered by decantation. The
aqueous layer will be purified and returned to the process as boiler feedwater. The oil leaving
the stripper will be cooled to 300 K and returned to the absorber. Figure E5.2-3 is a
schematic diagram of the process.
Gas out
Wash oil
Absorber
Benzene product
Gas in
Cooler
Separator
Condenser
Stripper
To water
treatment plant
Steam
Figure E5.2-3 Schematic diagram of the benzene-recovery process.
The wash oil entering the absorber will contain 0.0476 mole fraction of benzene; the pure oil
has an average molecular weight of 198. An oil circulation rate of twice the minimum will be
used. In the stripper, a steam rate of 1.5 times the minimum will be used.
Compute the oil-circulation rate and the steam rate required for the operation. Wash oilbenzene solutions are ideal. The vapor pressure of benzene at 300 K is 0.136 atm, and is 1.77
atm at 373 K.
Solution ----------------------------------------------------------------------------------------For calculations in this example, subscript “a” will be used to indicate absorber and subscript
“s” will be used to indicate stripper.
Molar rate of the gas entering the absorber is
Va,bot =
6
101, 300 × 1.0
= 40.61 mol/s
8.314 × 300
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 183
5-11
Molar rate of the carrier gas is given by
Va = Va,bot(1 − ya,bot) = 40.61×(1 − 0.074) = 37.61 mol/s
Liquid in
Gas out
ya,top
xa,top = 0.0476
Absorber
85% recovery
Liquid out
Gas in
ya,bot = 0.074
xa,bot
Converting the entering-gas mole fraction to mole ratio:
Ya,bot =
ya,bot
1 − ya,bot
=
0.074
= 0.0799
1 − 0.074
Converting the entering-liquid mole fraction to mole ratio:
Xa,top =
xa,top
1 − ya,top
=
0.0476
= 0.0500
1 − 0.0476
Since the absorber will recover 85% of the benzene in the entering gas, the concentration of
the gas leaving will be
Ya,top = 0.15×0.0799 = 0.0120
The equilibrium data for the conditions prevailing in the absorber can be generated in the
solute free basis from the following equation:
ya = 0.136xa [ From yaP = (Pbenzene)vap xa ⇒ ya(1 atm) = (0.136 atm) xa]
Ya
Xa
= 0.136
1 + Ya
1+ X a
The following Matlab codes generate and plot the equilibrium data for the conditions in the
absorber:
X=0:0.02:1;
a=0.136*X./(1+X);
Y=a./(1-a);
plot(X,Y)
xlabel('X_a');ylabel('Y_a'); grid on
line([0.05 0.3], [0.012 0.0799])
line([0.05 0.88], [0.012 0.08])
5-12
The following table shows some of the equilibrium values generated from the Matlab codes.
Xa
Ya
0
0
0.1000 0.0125
0.2000 0.0232
0.3000 0.0324
0.4000 0.0404
0.5000 0.0475
0.6000 0.0537
0.7000 0.0593
0.8000 0.0643
0.9000 0.0689
1.0000 0.0730
Figure E-1 XY diagram for the absorber
Figure E-1 shows the equilibrium curve and the operating line for the absorber. Starting with
any operating line above the equilibrium curve, such as DE, rotate it toward the equilibrium
curve using D as a pivot point until the operating line touches the equilibrium curve for the
first time. In this case the operating line DM touches the equilibrium curve at point P, a
location between the two end points of the operating line. The operating line DM corresponds
to the minimum solvent (oil) rate. From the diagram Xa,bot(max) = 0.88. Then
Ya,bot − Ya,top
La (min)
0.0799 − 0.012
=
= 0.0818
=
X a,bot (max) − X a,top
0.88 − 0.050
Va
The minimum solvent rate is then:
La (min) = 0.818×37.61 = 3.08 mol oil/s
For an actual oil flow rate which is twice the minimum La = 6.16 mol/s. The actual
concentration of the liquid phase leaving the absorber is
Xa,bot = Xa,top +
Va
37.61
(Ya,bot − Ya,top) = 0.050 +
(0.0799 − 0.012) = 0.4646
La
6.16
We now consider the conditions in the stripper. Figure E5.2-3 shows that the wash oil cycles
continuously from the absorber to the stripper, and through the cooler back to the absorber.
Therefore Ls = La = 6.16 mol/s. The concentration of the liquid entering the stripper is the
same as that of the liquid leaving the absorber (Xs,top = Xa,bot = 0.4646 mol of benzene/mol
oil), and the concentration of the liquid leaving the stripper is the same as that of the liquid
5-13
entering the absorber (Xs,bot = Xa,top = 0.05 mol of benzene/mol oil). The gaseous phase
entering the stripper is pure steam, therefore Ys,bot = 0.
Liquid in
Xs,top = 0.4646
Gas out
Ys,top
Stripper
Liquid out
Gas in
Ys,bot = 0
Xs,bot = 0.05
To determine the minimum amount of steam needed, we need to generate the equilibrium
distribution curve for the stripper at 373 K from the following equation:
ys = 1.77xs [ From ysP = (Pbenzene)vap xs ⇒ ys(1 atm) = (1.77 atm) xa]
Ys
Xs
= 1.77
1 + Ys
1+ Xs
The following table shows some of the equilibrium values generated from equilibrium
relation.
Xs
Ys
0
0
0.0400 0.0730
0.0600 0.1113
0.0800 0.1509
0.1000 0.1918
0.1400 0.2777
0.1600 0.3230
0.1800 0.3699
0.2000 0.4184
0.2400 0.5211
0.2600 0.5754
0.3000 0.6905
0.3400 0.8152
0.3600 0.8816
0.4000 1.0231
0.4400 1.1779
0.4600 1.2608
0.5000 1.4390
Figure E-2 XY diagram for the stripper
Figure E-2 shows the equilibrium curve and the operating line for the stripper. Starting with
any operating line below the equilibrium curve, such as DE, rotate it toward the equilibrium
curve using D as a pivot point until the operating line touches the equilibrium curve for the
5-14
first time. In this case the operating line DM touches the equilibrium curve at point P, a
location between the two end points of the operating line. The operating line DM corresponds
to the minimum steam rate. From the diagram Ys,top(max) = 1.13. Then
Ys,top (max) − Ys,bot
Ls
1.13 − 0
=
= 2.726
=
Vs (min)
X s,top − X s,bot
0.4646 − 0.050
The minimum steam rate is then:
Vs (min) =
6.16
= 2.26 mol steam/s
2.726
For an actual steam flow rate which is 1.5 times the minimum Vs = 1.5×2.26 mol/s = 3.39
mol/s. The actual concentration of the gas stream leaving the stripper is
Ys,top = Ys,bot +
Ls
6.16
(Xs,top − Xs,bot) = 0 +
(0.4646 − 0.05) = 0.753
Vs
3.39
5-15