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Chapter 6: Rate-based Absorption 1

Chapter 6

Rate-Based Absorption


Absorption Equipment

Liquid out

Liquid in

Vapor in

Vapor out

12

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

2


Absorption Equipment

Ammonia Absorption Unit


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

3


Packed Beds

Liquid inlet

Liquid outlet Gas inlet

Gas outlet


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?

4


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)



• 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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• 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


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


6


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.



• 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

7



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



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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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:



Table: Relationships among Mass-Transfer Coefficients

9


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.


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

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



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


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Liquidm

ole

frac

tion

x

Gasyix

iy

iA yN k y y⎡ ⎤= −⎣ ⎦

iA xN k x x⎡ ⎤= −⎣ ⎦


(mole fraction of A in V)

y equilibrium line

iy

ix

y

x




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

12



Liquidm

ole

frac

tion

x

ix

Gas

yBA

C

*y


*A xN K x x⎡ ⎤= −⎣ ⎦

*A yN K y y⎡ ⎤= −⎣ ⎦


y equilibrium line

iy

ix

y

x

*y*x



Liquid

mol

e fr

actio

n

x

ix

Gasy

B

A

C

*y


*A xN K x x⎡ ⎤= −⎣ ⎦

*A yN K y y⎡ ⎤= −⎣ ⎦


y equilibrium line

iy

ix

y

x

*y

*x

13




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




y equilibrium line

iy

ix

y

x

*y*x

1 1 1

x x y yK k m k= +



Overall liquid resistance

1 1 * i

ix x y

x xK k k ( y y )

−= +

−

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1 1x

y x y

mK k k

= +



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


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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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.


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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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).


Absorption of Dilute Gases in Packed Columns

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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


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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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


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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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

= = =⋅


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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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.


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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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


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*−

=−


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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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 −−

=


y equilibrium line

bx

by

axay

operating line

Similar to the number of stages in the tray process

HOG NOG vs. Nstages


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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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

Date post:	06-Mar-2018
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