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Chapter 6: Rate-based Absorption 1
Chapter 6
Rate-Based Absorption
Chapter 6: Rate-based Absorption 2
Absorption Equipment
Liquid out
Liquid in
Vapor in
Vapor out
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N–1N
Trayed Tower
Liquid in
Vapor in
Vapor out
Liquid out
Spray Tower
Liquid in
Vapor in
Vapor out
Liquid out
Bubble Column
Liquid inVapor out
Vapor in
Liquid out
Centrifugal Contactor
Liquid in
Vapor in
Vapor out
Liquid out
Packed Column
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Chapter 6: Rate-based Absorption 3
Absorption Equipment
Ammonia Absorption Unit
Chapter 6: Rate-based Absorption 4
Packing in Packed Beds
The material of construction can be metal, plastic, or ceramic
The choice of materials depend on the corrosiveness of the system and the cost of the material as well as its efficiency
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Chapter 6: Rate-based Absorption 5
Packed Beds
Liquid inlet
Liquid outlet Gas inlet
Gas outlet
Chapter 6: Rate-based Absorption 6
Gas Absorption
Definition:Transfer of a gaseous component (absorbate) from the gas phase to a liquid (absorbent) phase through a gas-liquid interface.
What are the key parameters that affect the effectiveness?How can we improve absorption efficiency?
Mass transfer rate:gas phase controlled absorptionliquid phase controlled absorption
Does it matter if it’s gas phase or liquid phase controlled?
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Chapter 6: Rate-based Absorption 7
Liquidm
ole
frac
tion
x
Gasy
Liquid
mol
e fr
actio
n
x
Gas
yixiy
Mass transfer of A
BA
In absorption process material has to diffuse from one phase (gas) to another (liquid). The rate of diffusion in both phases impacts the overall rate of mass transfer
Two-Film Theory (Gas-Liquid)
Chapter 6: Rate-based Absorption 8
Two-Film Theory (Gas-Liquid)
• The two-film theory is based on two basic assumptions:
– The rate of mass transfer is controlled by the rates of diffusion through the phases on each side of the interface
– No resistance is offered to the transfer of the diffusing component across the interface
• If PAG and CAL are the partial pressure and concentration of A in the gas and liquid bulk phases (respectively), and PAi, CAi the corresponding values at the interface, we can say:
• Before equilibrium is established“A” will diffuse:
1. through the bulk of one phase2. through the interface3. through the bulk of the other phase
If A is diffusing from, e.g., the gas to liquid bulk phase, CAL< CAi and PAi< PAG
– PAG > PAi: driving force from bulk gas to interface– CAi> CAL: driving force from interface to bulk liquid
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Chapter 6: Rate-based Absorption 9
Two-Film Theory (Gas-Liquid)
• At Equilibrium:– If equilibrium exists between the 2 phases, then mass transfer does not take place.
Condition for equilibrium: e.g., PAi = H CAi (Henry’s law). Note that this condition is true only when there is no resistance at the interface (i.e., two-film theory)
• The interfacial partial pressure, PAi, can be lower, equal, or greater than CAi. The relation is dictated by the value of Henry’s constant (H).
Concentration gradients between two contacting phases
Chapter 6: Rate-based Absorption 10
Liquid
mol
e fr
actio
n
x
Gas
yix
Mass transfer of A
Two film theory:
resistance to the overall mass transfer is viewed as a combinedresistance of liquid and gas films at the interface
iy
Two-Film Theory (Gas-Liquid)
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Chapter 6: Rate-based Absorption 11
MT Between Two Fluid Phases (Two-Film Theory)
• This concept has found extensive application in steady-state, gas–liquid, and liquid–liquidseparation processes.
• It is an extension of the film theory to two fluid films in series, with each film presents a resistance to mass transfer,
• Concentrations in the two fluids at the interface are assumed to be in phase equilibrium. So, no additional interfacial resistance to mass transfer.
Gas–Liquid Case
(a) film theory (b) more realistic gradients.
Chapter 6: Rate-based Absorption 12
Two-Film Theory (Gas-Liquid)
• Consider the steady-state mass transfer of A from a gas phase, across an interface, into a liquid phase as shown:
• Two possible MT conditions at the interface
– Equimolar counter-diffusion (EMCD)
– Diffusion of A through stagnant B (UMD)
• For the gas phase, under dilute or (EMCD) conditions:
• For the liquid phase, we use molar concentrations:
• An equilibrium equation applies at the interface :
Problem!:
– Equilibrium equations are written in terms of interfacial compositions
– The interfacial compositions are very difficult to measure
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Chapter 6: Rate-based Absorption 13
Two-Film Theory (Gas-Liquid)
Recall from the previous slide:
Substitute
Elimination of gives:
Through defining:
a fictitious liquid-phase concentration “the concentration that would be in equilibrium with the partial pressure in the bulk gas;
an overall mass-transfer coefficient based on liquid phase, KL.
where
Chapter 6: Rate-based Absorption 14
Two-Film Theory (Gas-Liquid)
Remark: Notice that there are many different forms of Henry’s law such as:
Therefore, always check the units of the Henry’s constant
Similarly, we can define an overall mass-transfer coefficient, KG, based on the gas phase.
where
Another common way for V (G)–L mass transfer is through using mole fraction-driving forces,
In this case, phase equilibrium at the interface can be expressed in terms of the K-value for vapor–liquid equilibrium “V-L equilibrium ratio, see Chapter 2, slide 20”
By eliminating yAi and xAi
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Chapter 6: Rate-based Absorption 15
Two-Film Theory (Gas-Liquid)
Through defining fictitious concentration quantities and overall mass-transfer coefficients for mole-fraction driving forces:
Where Kx and Ky are overall mass-transfer coefficients based on mole-fraction driving forces
and
When using correlations to estimate mass-transfer coefficients for the use in the above equations, it is always important to check the units
Liquid phase:
Ideal-gas phase:
Chapter 6: Rate-based Absorption 16
Two-Film Theory (Gas-Liquid)
Table: Relationships among Mass-Transfer Coefficients
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Chapter 6: Rate-based Absorption 17
Two-Film Theory (Large Driving Forces for Mass Transfer)
Large driving forces do exist for mass transfer when one or both phases are not dilute with respect to the diffusing solute,
If mole-fraction driving forces are used, we can write
But we know that:
Therefore:
Which could be re-arranged as:
Therefore:
…..(1)
…..(2)
By substitution of equation (2) in (1), we could obtain: Similarly:
Thus, the phase equilibria ratios such as HA, KA, and KDA may not be constant across the two phases.
Chapter 6: Rate-based Absorption 18
Two-Film Theory (Large Driving Forces: Graphical Representation)
Atypical curved equilibrium line is shown in the figure below
Because the line is curved, the V–L equilibrium ratio, KA = yA/xA, is not constant across the two phases
KA, the slope of the curve, decreases with increasing concentration of A.
The two slopes of the equilibrium line can be represented by:
But from the previous slide, we obtained:
By substituting the slopes and
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Chapter 6: Rate-based Absorption 19
Liquid
mol
e fr
actio
n
x
Gas
yix
Mass transfer of A
Two film theory:
resistance to the overall mass transfer is viewed as a combinedresistance of liquid and gas films at the interface
iy
Two-Film Theory (Gas-Liquid)
Chapter 6: Rate-based Absorption 20
Mass transfer rate (per unit area)
Liquid
mol
e fr
actio
n
x
Gas
yixiy
iA yN k y y⎡ ⎤= −⎣ ⎦
iA xN k x x⎡ ⎤= −⎣ ⎦
x(mole fraction of A in L)
(mole fraction of A in V)
y equilibrium line
iy
ix
y
x
Two-Film Theory (Large Driving Forces: Graphical Representation)
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Chapter 6: Rate-based Absorption 21
Mass transfer rate (per unit area)
Liquidm
ole
frac
tion
x
Gasyix
iy
iA yN k y y⎡ ⎤= −⎣ ⎦
iA xN k x x⎡ ⎤= −⎣ ⎦
x(mole fraction of A in L)
(mole fraction of A in V)
y equilibrium line
iy
ix
y
x
Two-Film Theory (Large Driving Forces: Graphical Representation)
Chapter 6: Rate-based Absorption 22
Two-Film Theory (Large Driving Forces: Graphical Representation)
Liquid
mol
e fr
actio
n
x
A
C
Gas mixture C is in equilibrium with the liquid system A:
xTHy )(* = (in Henry’s law regime)
*y
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Chapter 6: Rate-based Absorption 23
Two-Film Theory (Large Driving Forces: Graphical Representation)
Liquidm
ole
frac
tion
x
ix
Gas
yBA
C
*y
Mass transfer rate (per unit area)
*A xN K x x⎡ ⎤= −⎣ ⎦
*A yN K y y⎡ ⎤= −⎣ ⎦
x(mole fraction of A in L)
y equilibrium line
iy
ix
y
x
*y*x
Chapter 6: Rate-based Absorption 24
Two-Film Theory (Large Driving Forces: Graphical Representation)
Liquid
mol
e fr
actio
n
x
ix
Gasy
B
A
C
*y
Mass transfer rate (per unit area)
*A xN K x x⎡ ⎤= −⎣ ⎦
*A yN K y y⎡ ⎤= −⎣ ⎦
x(mole fraction of A in L)
y equilibrium line
iy
ix
y
x
*y
*x
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Chapter 6: Rate-based Absorption 25
Two-Film Theory (Large Driving Forces: Graphical Representation)
x(mole fraction of A in L)
y equilibrium line
iy
ix
y
x
*y*x
yi
x
i
y kxxkyy
K1
)(1 *
+−
−=
1 1x
y x y
mK k k
= +
Resistance of gas film
Resistance of liquid film
Overall gas resistance m
Chapter 6: Rate-based Absorption 26
Two-Film Theory (Large Driving Forces: Graphical Representation)
x(mole fraction of A in L)
y equilibrium line
iy
ix
y
x
*y*x
1 1 1
x x y yK k m k= +
Resistance of gas film
Resistance of liquid film
Overall liquid resistance
1 1 * i
ix x y
x xK k k ( y y )
−= +
−
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Chapter 6: Rate-based Absorption 27
Two-Film Theory (Large Driving Forces: Graphical Representation)
1 1x
y x y
mK k k
= +
Resistance of gas film
Resistance of liquid film
Overall gas resistance
- when coefficients ky and kx are of the same order of magnitude and m is much greater than 1 the liquid phaseresistance is controlling
- in the opposite situation when solubility is very high, the gas film resistance is controlling
Chapter 6: Rate-based Absorption 28
Two-Film Theory (Ratio of Resistances)
• The ratio of resistance in an individual phase to the total resistance may be expressed in the following ways:
• A relation between the overall and the individual phase coefficients can be obtained when the equilibrium relation is linear (Henry’s law, partition coefficient):
PAi = m CAi• Then we have:
• The previous two equations stipulate that the relative magnitudes of the individual phase resistances depend on the solubility of the gas, as indicated by the magnitude of the proportionality constant, m
y
y
1 / kGas phase resis tanceTotal resis tance 1 / K
= x
x
1 / kLiquid phase resis tan ceTotal resis tan ce 1 / K
=
y y x
1 1 mK k k
= +x y x
1 1 1K m k k
= +
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Chapter 6: Rate-based Absorption 29
Two-Film Theory (Ratio of Resistances)
Gas-phase controlled mass transfer
For systems involving a soluble gas, such as ammonia in water, m is very small. From the previous equations we can then deduce that the overall resistance is equal to the gas-phase resistance, i.e., (1 / Ky = 1 / ky).
Liquid-phase controlled mass transfer
Systems involving a gas of low solubility, such as CO2 in water, have such a large value of m then, the previous equation stipulates that the gas-phase resistance may be neglected. This means that (1 / Kx =1 / kx), or in other words, we have liquid-phase controlled mass transfer.
Chapter 6: Rate-based Absorption 30
Tray operations• Surface area for mass transfer:
Bubble/liquid interface
• Equilibrium:Vapour and liquid phases leaving a stage are assumed to be in equilibrium; non-equilibrium is accounted for with stage efficiencies
• Operating points are given by set of ( )1, +nn yx
Packed operations• Surface area for mass transfer:
Surface area for packing
• Equilibrium:Vapour and liquid are not at equilibrium; non-equilibrium provides the force for mass transfer
Mass Transfer Operations in Packed Columns
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Chapter 6: Rate-based Absorption 31
Advantages of packed columns
• Lower pressure drop for the gas phase, e.g. for vacuum distillation
• Lower capital cost if the diameter (function of vapour flow rate) is less than 0.6 m.
• Can be made of corrosion resistant material, e.g. ceramics.
Mass Transfer Operations in Packed Columns
Advantages of plate columns over packed columns:
• More economical at higher vapour flow rates (i.e. diameter).
• More suitable for large numbers of theoretical stages (because of redistribution issue).
• Better for large fluctuations of temperature (leading to packing attrition).
• More suitable for highly exothermic/endothermic operations (easier to fit heat transfer surface).
• Better for highly fouling conditions (if the column size allows for man-way access for cleaning).
Chapter 6: Rate-based Absorption 32
Absorption of Dilute Gases in Packed Columns
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Chapter 6: Rate-based Absorption 33
Absorption of Dilute Gases in Packed Columns
• We have seen that the flux of a component A trough an interface inside a continuous contacting tower (distillation, absorption, stripping column) can be expressed as:
• In both expressions, the flux is expressed in moles of A transferred per unit time per unit area and per unit driving force (here, mole fraction). In order to use these equations in mass transfer operations, the contact area between the two phases must be known. This, however, is technically impossible for most operations. For this reason the factor “a”must be introduced to represent the interfacial surface area per unit volume of the mass transfer equipment (the mass-transfer area, A, per unit volume, V, of tower)
• Since “a” is not easy to measure, we normally lump ka: individual capacity coefficient, Ka: overall capacity coefficient
Chapter 6: Rate-based Absorption 34
Inter-phase Mass Transfer in a Packed-Column
Differential mass transfer rate of species i at a given position in the liquid on a certain level.
i idn V dy= −
( )( )( )*i y cross i idn K a A dl y y= −
idy 0 , absorber<
( )( )( )*i y cross i iV dy K a A dl y y− = −
Material balance on species i:
Combine material balance and mass transfer equations:
Consider an absorption column; For diluted gases the change in flow rates is neglected
Consider a mass transfer process in a section of the column dl (cross-section of the column is Across)
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Chapter 6: Rate-based Absorption 35
Material Balance and MT in a Packed Column
( )( )
1,icol
N 1,i
yly i
cross *0 y i i
K a dyA dlV y y
+
= −−∫ ∫
• Separate variables and integrate
( )( )( )*i y cross i iV dy K a A dl y y− = −
• Assume Kya ≈ constant• Absorption of dilute solutions (V ≈ constant)
( )( )
j ,i
j 1,i
yy cross col i
*y i i
K a A l dyV y y
+
= −−∫
( ) ( )1,i
N 1,i
yi
col *yy cross i i
dyVlK a A y y
+
⎡ ⎤= ⎢ ⎥
−⎢ ⎥⎣ ⎦∫
yN+1 xN
y1 x0
lcol
Chapter 6: Rate-based Absorption 36
Material Balance and MT in a Packed Column
( )1,i
N 1,i
yi
*y i i
dyy y
+−∫
yN+1 xN
y1 x0
lcol
cross
y
V AK a
Change in the concentration divided
by driving force. This property is
called NOG the Overall Number of Transfer
Units (NTU)
This property has units of length, is constant for constant L/V and is called
HOG the Overall height of Transfer Units(HTU)
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Chapter 6: Rate-based Absorption 37
NOG and HOG
NOG is generally evaluated by numerical integration (Trapezoidal Rule or Simpson’s Rule)
( )1,i
N 1,i
yi i
OG **i iy i i
dy Overall change in y across the stageNAverage driving force (y - y )y y
+
= − =−∫
( ) ( )1,i
N 1,i
yi
col *yy cross i i
dyVlK a A y y
+
⎡ ⎤= ⎢ ⎥
−⎢ ⎥⎣ ⎦∫
col OG OGl H N= ⋅
( ) [ ] ( ) [ ]OG 3 2y cross
V mole / secH mK a A mole / m sec m
= = =⋅
Chapter 6: Rate-based Absorption 38
Analytical Solutions
• For dilute solutions with linear operating and equilibrium curves :
OGy cross
VHK a A
=( )1 N 1
OG *AG A LM
y yNy y
+−=
−
• For dilute solutions with linear operating and equilibrium curves we can express NOG in terms of
– The Absorption Factor: A– The inlet and outlet stream mole fraction– The slope of the equilibrium curve: m
LAmV
=• Where :
• Absorption :N 1 0
OG1 0
y m xA A 1 1N LnA 1 y m x A A
+⎡ ⎤− −⎛ ⎞= +⎢ ⎥⎜ ⎟− − ⎝ ⎠⎣ ⎦
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Chapter 6: Rate-based Absorption 39
Slightly Curved Equilibrium / Operating Lines
• If the equilibrium curve and/or the operating curve are not quite linear, you can obtain a good approximation of the number of equilibrium stages using the following method
– Evaluate the Absorption Factor under the conditions that exist at the top and at the bottom of the column
toptop
top top
LA
m V= bottom
bottombottom bottom
LAm V
=
top bottomA A A=
– Use the geometric average value of A :
– Use this value of A in the analytical solution equations on the previous slide.
Chapter 6: Rate-based Absorption 40
Other Types of Transfer Units
• Basis: Overall Gas MT Coefficients , Kya : col OG OGl H N=
• Basis: Gas Film MT Coefficient, kya : col G Gl H N=
• Basis: Overall LiquidMT Coefficients , Kxa : col OL OLl H N=
• Basis: Liquid FilmMT Coefficient, kxa : col L Ll H N=
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Chapter 6: Rate-based Absorption 41
Which K’s and Transfer Units to Use ?
Choose HOG NOG if the resistance to MT is greater in the gas phase than in the liquid phase
MT resistance in the gas phase is greater when…
The solute is highly soluble in the liquid phase (Example: NH3 being absorbed into water)
The solute reacts chemically when it dissolves in the liquid phase (Example: CO2 dissolving into aqueous NaOH)
MT is often greater in the gas phase than in the liquid phase.
Therefore, HOG & NOG are most common
Film MT coefficients are difficult to measure
You are more likely to know the value of one of the overall MT coefficients
You are more likely to use HOG & HOL than HG or HL.
Choose HOL NOL if the resistance to MT is greater in the liquid phase than in the gas phase
MT resistance in the liquid phase is greater when the solute is not very soluble in the liquid phase Example: O2 being absorbed into water
Chapter 6: Rate-based Absorption 42
HOG NOG vs. Nstages
To understand the meaning of thesedefinitions consider a specific case when both the equilibrium and operating lines arestraight and parallel
The driving force is then constant throughout the process and can be moved outside the integral, leading to:
b aOG
y yNy y*−
=−
x(mole fraction of A in L)
y equilibrium line
bx
by
axay
operating line
HETP = Height equivalent of a theoretical plate or equilibrium stage
OG OG
stages
H NPacked HeightHETPNo.of equivalent equilibrium stages N
⋅= =
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Chapter 6: Rate-based Absorption 43
To understand the meaning of thesedefinitions consider a specific case when both the equilibrium and operating lines arestraight and parallel
The driving force is then constant throughout the process and can be moved outside the integral, leading to:
*yyyyN ab
Oy −−
=
x(mole fraction of A in L)
y equilibrium line
bx
by
axay
operating line
Similar to the number of stages in the tray process
HOG NOG vs. Nstages
Chapter 6: Rate-based Absorption 44
HOG NOG vs. Nstages
• Absorption with dilute solutions and linear operating and equilibrium curves :
OG stagesA 1N N Ln
A 1 A⎛ ⎞ ⎡ ⎤= ⎜ ⎟ ⎢ ⎥−⎝ ⎠ ⎣ ⎦
OGA 1HETP H Ln
A 1 A⎛ ⎞ ⎡ ⎤= ⎜ ⎟ ⎢ ⎥−⎝ ⎠ ⎣ ⎦
NOG = Nstages
NOG > Nstages
NOG < Nstages
• HETP = Height equivalent of a theoretical plate or equilibrium stage
• If the operating and equilibrium lines are also parallel, then…
– A = 1
– NOG = Nstages
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Chapter 6: Rate-based Absorption 45
Homework
First: Practice through solving Examples (6.9, 6.10, and 6.11)
Homework
Problem 6.24 and Problem 6.25
Due Date: 3rd April, 2006