Chapter 6
Rate-Based Absorption
Chapter 6: Rate-based Absorption
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Absorption Equipment Trayed Tower
Packed Column
Vapor out
Liquid in
Liquid in
Vapor out
Spray Tower Vapor out
Liquid in
1 2
N–1 N Liquid out
Vapor in
Liquid out
Vapor in
Bubble Column Liquid in
Vapor out
Liquid out
Vapor in
Centrifugal Contactor Vapor out
Liquid in
Vapor in Liquid out Liquid out Chapter 6: Rate-based Absorption
Vapor in 2
1
Absorption Equipment
Ammonia Absorption Unit Chapter 6: Rate-based Absorption
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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 Chapter 6: Rate-based Absorption
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2
Packed Beds Gas outlet Liquid inlet
Liquid outlet
Gas inlet
Chapter 6: Rate-based Absorption
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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: ¾ gas phase controlled absorption ¾ liquid phase controlled absorption
Does it matter if it’s gas phase or liquid phase controlled? Chapter 6: Rate-based Absorption
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3
Two-Film Theory (Gas-Liquid) 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
mole fraction
A
Liquid
B
Gas
y x
mole fraction
Liquid
Gas
x
y
i
y
x
i
Mass transfer of A
Chapter 6: Rate-based Absorption
7
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 phase 2. through the interface 3. 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 Chapter 6: Rate-based Absorption
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4
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
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Two-Film Theory (Gas-Liquid)
Two film theory: resistance to the overall mass transfer is viewed as a combined resistance of liquid and gas films at the interface
mole fraction
Liquid
Gas
x x
y
i
yi
Mass transfer of A
Chapter 6: Rate-based Absorption
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5
MT Between Two Fluid Phases (Two-Film Theory) • This concept has found extensive application in steady-state, gas–liquid, and liquid–liquid separation 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
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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:
Problem!: – Equilibrium equations are written in terms of interfacial compositions
•
An equilibrium equation applies at the interface :
Chapter 6: Rate-based Absorption
– The interfacial compositions are very difficult to measure
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6
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
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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 Chapter 6: Rate-based Absorption
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7
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
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Two-Film Theory (Gas-Liquid) Table: Relationships among Mass-Transfer Coefficients
Chapter 6: Rate-based Absorption
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8
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, Thus, the phase equilibria ratios such as HA, KA, and KDA may not be constant across the two phases. If mole-fraction driving forces are used, we can write Therefore: Which could be rearranged as:
…..(1)
But we know that: …..(2)
Therefore: By substitution of equation (2) in (1), we could obtain:
Similarly:
Chapter 6: Rate-based Absorption
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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 Chapter 6: Rate-based Absorption
and 18
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Two-Film Theory (Gas-Liquid)
Two film theory: resistance to the overall mass transfer is viewed as a combined resistance of liquid and gas films at the interface
mole fraction
Liquid
Gas
x
y
i
y
x
i
Mass transfer of A
Chapter 6: Rate-based Absorption
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Two-Film Theory (Large Driving Forces: Graphical Representation)
mole fraction
Liquid
Gas
x
y
i
y
x
i
(mole fraction of A in V)
Mass transfer rate (per unit area)
N A = k y ⎡⎣ y − y i ⎤⎦ N A = k x ⎡⎣ x − x ⎤⎦ i
y
equilibrium line
y
yi
x Chapter 6: Rate-based Absorption
xi (mole fraction of A in L)
x 20
10
Two-Film Theory (Large Driving Forces: Graphical Representation)
mole fraction
Liquid
Gas
x
y
i
y
x
i
(mole fraction of A in V)
Mass transfer rate (per unit area)
y
N A = k y ⎡⎣ y − y i ⎤⎦
y
N A = k x ⎡⎣ x − x ⎤⎦
yi
i
equilibrium line
x Chapter 6: Rate-based Absorption
xi
x
(mole fraction of A in L)
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Two-Film Theory (Large Driving Forces: Graphical Representation)
mole fraction
A
Liquid
x
y* C
Gas mixture C is in equilibrium with the liquid system A:
y * = H (T ) x
Chapter 6: Rate-based Absorption
(in Henry’s law regime)
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11
Two-Film Theory (Large Driving Forces: Graphical Representation)
mole fraction
A
Liquid
Gas
x
B
y
i
x
y*
y
equilibrium line
C
Mass transfer rate (per unit area)
y
N A = K y ⎡⎣ y − y* ⎤⎦
yi
N A = K x ⎡⎣ x* − x ⎤⎦
y* x
xi
x*
x
(mole fraction of A in L)
Chapter 6: Rate-based Absorption
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Two-Film Theory (Large Driving Forces: Graphical Representation)
mole fraction
A
Liquid
xi x
Gas
B
y y*
y
equilibrium line
C
Mass transfer rate (per unit area)
N A = K y ⎡⎣ y − y ⎤⎦ N A = K x ⎡⎣ x* − x ⎤⎦
y
*
yi y* x
Chapter 6: Rate-based Absorption
xi
x*
(mole fraction of A in L)
x 24
12
Two-Film Theory (Large Driving Forces: Graphical Representation)
y i − y* 1 1 = + i K y kx (x − x ) k y
m 1 1 = x+ K y kx k y Resistance of gas film Resistance of liquid film Overall gas resistance
y
equilibrium line
y yi m
y* x
xi
x*
x
(mole fraction of A in L)
Chapter 6: Rate-based Absorption
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Two-Film Theory (Large Driving Forces: Graphical Representation)
1 1 x* − x i = + K x kx k y ( y − yi )
1 1 1 = + K x k x my k y Resistance of gas film Resistance of liquid film Overall liquid resistance
y
equilibrium line
y yi y* x
Chapter 6: Rate-based Absorption
xi
x*
(mole fraction of A in L)
x 26
13
Two-Film Theory (Large Driving Forces: Graphical Representation)
m 1 1 = x+ K y kx k y 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 phase resistance is controlling - in the opposite situation when solubility is very high, the gas film resistance is controlling Chapter 6: Rate-based Absorption
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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:
Gas phase resis tan ce 1 / k y = Total resis tan ce 1/ Ky
Liquid phase resis tan ce 1 / k x = Total resis tan ce 1 / Kx
•
A relation between the overall and the individual phase coefficients can be obtained when the equilibrium relation is linear (Henry’s law, partition coefficient):
•
Then we have:
PAi = m CAi
1 1 m = + K y k y kx •
1 1 1 = + K x m k y kx
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
Chapter 6: Rate-based Absorption
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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
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Mass Transfer Operations in Packed Columns
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 ( xn , y n +1)
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
Chapter 6: Rate-based Absorption
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Mass Transfer Operations in Packed Columns 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.
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
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Absorption of Dilute Gases in Packed Columns
Chapter 6: Rate-based Absorption
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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
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Inter-phase Mass Transfer in a Packed-Column 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) Material balance on species i:
dni = − V dyi dyi < 0 , absorber Differential mass transfer rate of species i at a given position in the liquid on a certain level.
dni = ( K y a ) ( Across dl ) ( yi − y*i ) Combine material balance and mass transfer equations:
− V dyi = ( K y a ) ( Across dl ) ( yi − yi* ) Chapter 6: Rate-based Absorption
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Material Balance and MT in a Packed Column − V dyi = ( K y a ) ( Across dl ) ( yi − yi* ) •
( K a ) dl = −
lcol
Across
∫
y1,i
y
V
0
• •
y1
Separate variables and integrate
x0
dyi
∫ (y − y )
y N +1 ,i
* i
i
lcol
Assume Kya ≈ constant Absorption of dilute solutions (V ≈ constant)
( K a) A y
l
y j ,i
cross col
V
=−
y j +1,i
⎡ ⎤ V ⎥ lcol = ⎢ ⎢⎣ ( K y a ) Across ⎥⎦
dyi
∫ (y − y ) i
* i
y1,i
dyi ∫ * y ( y N +1 ,i i − yi )
yN+1
xN
Chapter 6: Rate-based Absorption
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Material Balance and MT in a Packed Column
y1 y1,i
dyi ∫ * y N +1,i ( yi − yi )
x0
Change in the concentration divided by driving force. This property is called NOG the Overall Number of Transfer Units (NTU) lcol
V Across K ya
Chapter 6: Rate-based Absorption
This property has units of length, is constant for constant L/V and is called HOG the Overall height of Transfer Units (HTU)
yN+1
xN
36
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NOG and HOG ⎡ ⎤ V ⎥ lcol = ⎢ ⎢⎣ ( K y a ) Across ⎥⎦
y1,i
dyi
∫ (y − y )
y N +1,i
* i
i
lcol = H OG ⋅ N OG H OG =
V K a ( y ) Across
y1,i
N OG = −
dyi
∫ (y − y ) =
y N +1,i
i
* i
[ =]
mole / sec mole ( / m3 ⋅ sec ) m2
[ =]
m
Overall change in yi across the stage Average driving force (yi - yi* )
NOG is generally evaluated by numerical integration (Trapezoidal Rule or Simpson’s Rule) Chapter 6: Rate-based Absorption
37
Analytical Solutions •
For dilute solutions with linear operating and equilibrium curves :
N OG =
y1 − yN +1 y ( AG − y*A )
H OG =
LM
V K y a Across
•
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
•
Absorption :
•
Where :
Chapter 6: Rate-based Absorption
N OG =
⎡ y − m x0 ⎛ A − 1 ⎞ 1 ⎤ A Ln ⎢ N +1 ⎜ ⎟+ ⎥ A−1 ⎣ y1 − m x0 ⎝ A ⎠ A ⎦ A=
L mV 38
19
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
Atop =
Ltop
Abottom =
mtop Vtop
Lbottom mbottom Vbottom
– Use the geometric average value of A :
A = Atop Abottom – Use this value of A in the analytical solution equations on the previous slide.
Chapter 6: Rate-based Absorption
39
Other Types of Transfer Units
•
Basis: Overall Gas MT Coefficients , Kya :
lcol = H OG N OG
•
Basis: Overall Liquid MT Coefficients , Kxa :
lcol = H OL N OL
•
Basis: Gas Film MT Coefficient, kya :
lcol = H G N G
•
Basis: Liquid Film MT Coefficient, kxa :
lcol = H L N L
Chapter 6: Rate-based Absorption
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20
Which K’s and Transfer Units to Use ? ¾
¾
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 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… 9
The solute is highly soluble in the liquid phase (Example: NH3 being absorbed into water)
9
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
9 ¾
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
41
HOG NOG vs. Nstages HETP =
H ⋅N Packed Height = OG OG No.of equivalent equilibrium stages N stages
HETP = Height equivalent of a theoretical plate or equilibrium stage To understand the meaning of these definitions consider a specific case when both the equilibrium and operating lines are straight and parallel The driving force is then constant throughout the process and can be moved outside the integral, leading to:
N OG =
yb − ya y − y*
y
equilibrium line operating line
yb
ya xa
xb
x
(mole fraction of A in L) Chapter 6: Rate-based Absorption
42
21
HOG NOG vs. Nstages
To understand the meaning of these definitions consider a specific case when both the equilibrium and operating lines are straight and parallel The driving force is then constant throughout the process and can be moved outside the integral, leading to:
N Oy =
yb − y a y − y*
Similar to the number of stages in the tray process
y
equilibrium line operating line
yb
ya xa
xb
x
(mole fraction of A in L) Chapter 6: Rate-based Absorption
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HOG NOG vs. Nstages •
Absorption with dilute solutions and linear operating and equilibrium curves :
⎛ A ⎞ ⎡1⎤ N OG = N stages ⎜ ⎟ Ln ⎢ ⎥ ⎝ A−1⎠ ⎣ A⎦ ⎛ A ⎞ ⎡1⎤ HETP = H OG ⎜ ⎟ Ln ⎢ ⎥ ⎝ A−1⎠ ⎣ A⎦ • •
HETP = Height equivalent of a theoretical plate or equilibrium stage If the operating and equilibrium lines are also parallel, then… – A=1 – NOG = Nstages
Chapter 6: Rate-based Absorption
NOG = Nstages
NOG > Nstages
NOG < Nstages 44
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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
Chapter 6: Rate-based Absorption
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