Absorption Notes

7/23/2019 Absorption Notes

1/79

ABSORPTION

eg. absorbing NH3from air using liquid water,

Acetone from air using liquid water

liquid phase is immiscible in the gas phase

solute is removed from a liquid by contacting it

with a gas

mass transfer process

separating a solute (A) or several solutes from a

gas phase by contacting the gas with a liquid

phase

STRIPPING/DESORPTION


2/79

GAS-LIQUID EQUILIBRIUM

yA= HxAor

pA= partial pressure of component A (atm)

Henrys laws:pA= HxA

where

H= Henrys law constant (atm/mol fraction) in Appendix A.3-18

H = Henrys law constant (mol fraction gas/mol fraction liquid) = H/P

xA= mole fraction of component A in liquid

yA= mole fraction of component A in gas = pA/P

P= total pressure (atm)


3/79

SINGLE-STAGE EQUILIBRIUM CONTACT

Liquid & gas phases are brought into contact and separated

Long enough for equilibrium

Gas phase solute A & inert gas B

Liquid phase solute A & inert liquid/solvent C

xA1xA0

yA1 yA2

Total material balance: L0+ V2= L1+ V1

Balance on A:

L0xA0+ V2yA2= L1xA1+ V1yA1

Balance on A can also be written as

L + V = L + V

where L = moles inert C

V = moles inert B


4/79

Example 10.31

A gas mixture at 1.0 atm pressure abscontaining air and CO2 is contacted in a single

stage mixer continuously with pure water at 293

K. The two exit gas and liquid streams reach

equilibrium. The inlet gas flow rate is 100kgmol/h, with a mole fraction of CO2 of

yA2=0.20. The liquid flow rate entering is 300 kg

mol water/h. Calculate the amounts and

compositions of the two outlet phases. Assumethat water does not vaporize to the gas phase.


5/79

Example 10.3-1

Gas phase= CO2+ air

Inert C = pure water


L + V = L + V

where

L = moles water =L0(1-xA0) = 300 (1-0) = 300 kmol/h

V = moles air = V2(1-yA2) = 100 (1-0.2) = 80 kmol/h

L1

xA1

L0= 300 kmol/h

xA0= 0

V1

yA1

V2= 100 kmol/h

yA2=0.21 atm

293K

Balance on A:

L0xA0+ V2yA2= L1xA1+ V1yA1

300 + 80 = 300 + 80


6/79

Example 10.3-1

300 + 80 = 300 + 80 (1)

At 293K, Henrys law constant from App. A.3-18 = 0.142 x 104atm/mol frac.

yA1= HxA1= (H/P)xA1= 0.142 x 104xA1 (2)

Substitue yA1= 0.142 x 104xA1into eq. (1):

Solving for xA1:

300 + 80 = 300 + 80

Substituting xA1= 1.41 x 10-4into eq. (2):

Outlet flow rates:

L =L1(1-xA1) = L1(1-1.41 x 10-4) = 300 kmol/h

V = V1(1-yA1) = V1(1-0.2) = 80 kmol/h

xA1= 1.41 x 10-4

yA1= 0.142 x 104(1.41 x 10-4) = 0.2

L1= 300 kmol/h

V1= 100 kmol/h


7/79

COUNTERCURRENT MULTIPLE-CONTACT STAGES

Total number of ideal stages = N

Total material balance:

L0+ VN+1= LN+ V1

Balance on A:

L0xA0+ VN+1yAN+1= LNxAN+ V1yA1


operating line

L + V = L + V


8/79

COUNTERCURRENT MULTIPLE-CONTACT STAGES

Graphical determination of N:

Dilute Concentrated


9/79

Example 10.3-2

It is desired to absorbed 90% of the acetone in

a gas containing 1.0 mol% acetone in air in a

countercurrent stage tower. The total inlet gas

flow to the tower is 30.0kgmol/h, and the total

inlet pure water flow to be used to absorb theacetone is 90 kgmol H2O/h. The process is to

operate isothermally at 300 K and a total

pressure of 101.3 kPa. The equilibrium relation

for the acetone(A) in the gas-liquid isyA=2.53xA. Determine the number of theoretical

stages required for this separation.


10/79

Example 10.3-2

A = acetone, B = air, C = water 90% acetone absorb

Acetone in VN+1= 0.01(30) = 0.3 kmol/h

Acetone in LN= 0.9(0.3) = 0.27 kmol/h

Balance of Acetone in V1= 0.03 kmol/h

V1= 29.7 + 0.03 = 29.73 kmol/h

Entering air = 30 0.3 = 29.7 kmol/h

300K, 101.3 kPa

L0 = 90 kmol/h

VN+1= 30 kmol/h

xA0=0

yAN+1= 0.01

Entering water= 90 kmol/h

LN= 90 + 0.27 = 90.27 kmol/h

yA1= 0.03/29.73 = 0.00101

xAN= 0.27/90.27 = 0.003


11/79

Example 10.3-2

L0 = 90 kmol/h

VN+1= 30 kmol/h

xA0=0

yAN+1= 0.01

V1= 29.73 kmol/h

xAN= 0.003

yA1

= 0.00101

LN= 90.27 kmol/h

Given: equlibrium relation for acetone-water : yA= 2.53 xA

N theoretical stages = 5.2


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ANALYTICAL/ KREMSER EQUATION

Absorption:

where

m = slope of equilibrium line

A = absorption factor = Aav.= !(A1AN)A1=L0/(mV1)

AN= LN/(mVN+1)

When A = 1

Kremser equations valid only when operating & equilibrium lines are straight

L0

VN+1

xA0

yAN+1

V1

xAN

yA1

LN


13/79


Stripping:

where


A = absorption factor = Aav.= !(A1AN)A1=L0/(mV1)

AN= LN/(mVN+1)

When A = 1

L0

VN+1

xA0

yAN+1

V1

xAN

yA1

LN


14/79

Example 10.3-3

Repeat Example 10.3-2 but use the

Kremser analytical equations for

countercurrent stage processes.


15/79

Example 10.3-3

L0 = 90 kmol/h

VN+1= 30 kmol/h

xA0=0

yAN+1= 0.01

V1= 29.73 kmol/h

xAN= 0.003

yA1= 0.00101

LN= 90.27 kmol/h

Given: equlibrium relation for acetone-water : yA= 2.53 xA

A1=L0/(mV1) = 90/(2.53x29.73) = 1.2

AN= LN/(mVN+1) = 90.27/(2.53x30) = 1.19

A = !(A1AN)= !(1.2 x 1.19) = 1.95

N = 5.04


16/79

Types of Plate Towers

1.Sieve Tray

2.Valve Tray

3.Bubbe-cap Tray

Design of Plate Absorption Towers

1.Operating Line

2.Graphical Determination of The Number

of Tray


17/79

Design of Tray/Plate

Absorption Towers

1.Operating Line Derivation

-same as the countercurrent multiple-

stage

2.Graphical Determination of the Number

of Trays


18/79

Example 10.6-2

A tray tower is to be designed to absorb SO2

from an air stream by using pure water at 293K(68oF). The entering gas contains 20 mol%

SO2and that leaving 2 mol% at a total pressure

of 101.3 kPa. The inert air flow rate is 150 kg

air/h.m2, and the entering water flow rate is6000 kg water/h.m2.Assuming an overall tray

efficiency of 25%, how many theoretical trays

and actual trays are needed? Assume that thetower operates at 293 K (20oC).


19/79

PACKED TOWER


20/79

PRESSURE DROP & FLOODING IN PACKED TOWERS

Flooding point liquid no longer flow downward

- blown out with the gas

Loading point gas starts to hinder liquid downflow

- local accumulations of liquid start to appear in packing

Actual operation gas velocity below flooding

Optimum economic gas velocity = half or more of the flooding velocity


21/79

PRESSURE DROP IN RANDOM PACKINGS

Prediction of pressure drop in random packings:

where:

G= superficial gas velocity (ft/s) = GG/G

G= gas density (Ibm/ft3)

L

= liquid density (Ibm

/ft3)

Fp= packing factor (ft-1) in Table 10.6-1

v= kinematic viscosity (centstokes) = L/(L/62.4)

L= liquid viscosity (cp)

GL= liquid mass velocity (Ibm/s.ft2)

GG= gas mass velocity (Ibm/s.ft2)

Accuracy = 11%


22/79

PRESSURE DROP IN STRUCTURED PACKING


23/79

FLOODING PRESSURE DROP IN PACKED & STRUCTURED

PACKINGS

Prediction of limiting pressure drop at flooding:

#Pflood= 0.115FP0.7

where:

#Pflood= pressure drop at flooding (in. H2O/ft height of packing)

Fp= packing factor (ft-1) in Table 10.6-1

Conversion: 1 in H2O/ft height packing= 83.33 mm H2O/m height of packing

FPfrom 9 - 60

Accuracy = 10-15 %

FP60 or higher, #Pflood= 2.0 in. H2O/ft (166.7 mm H2O/m)


24/79

PRESSURE DROP & TOWER DIAMETER IN PACKED &

STRUCTURED PACKINGS

1. From the type of packing used, get FPfrom Table 10.6-1

2. Determine #Pfloodfrom #Pflood= 0.115FP0.7or #Pflood=2 in H2O/ft packing

height when FP"60

3. Calculate flow parameter using the gas and liquid flows in the bottom of

the tower. Using Fig. 10.6-5 or 10.6-6, read off the capacity parameter

4. Calculate $Gfrom the capacity parameter which is equal to GG/Gwhere

GG= maximum value of gas mass velocity at flooding

5. Using a given % of the floodingGG, obtain new GLand GGbased on the

given liquid-to-gas ratio GL/GG

6. Calculate the cross-sectional area of the tower (%D2

/4) from the given gasflow rate and hence, the diameter of the tower

7. Calculate the total flow rates of the outlet and inlet water assuming all the

solute is absorbed

Procedure:


25/79

Example 10.6-1

Ammonia is being absorbed in a tower using

pure water at 25oC and 1.0 atm abs. The feed

rate is 1440 lbm/h and contains 3.0 mol%

ammonia in air. The process design specifies a

liquid to gas mass ratio GL/GG=2/1 and the useof 1-in.metal Pall Rings.

Calculate the pressure drop in the packing and

gas mass velocity at flooding. Using 50% of the

flooding velocity, calculate the pressure drop,gasand liquidflows, and the tower diameter.


26/79

Example 10.6-1

1.From Table 10.6-1, for 1-in Pall rings,

Fp=56 ft-1

2.GL/GG= 2/1=2.0

3.Pressure drop: use equation 10.6-1.

Dpflood=0.115Fp0.7=0.115(56)0.7=1.925 in

4.Calculate flow parameter:

Air density: from Appendix A.3-3.Fromcapacity parameter, GGcan be calculated.


27/79

Example 10.6-1

6. Use 50% of flooding velocity, 0.5GG.

GL= 2GG . This new GGand GLand 50%

of capacity parameter can be used to

obtain the pressure drop.

Tower diameter is determined from GG.


28/79

PACKED TOWERS FOR ABSORPTION

Balance on A :

Operating line

L + V = L + V

L + V = L + V

Absorption

Stripping


29/79

Absorption

Stripping

LIMITING & OPTIMUM L/V RATIOS

Balance on A:

Entering liquid flow L2or L open to choice

At point P, liquid flow L = Lmin& x1= x1max

Lmin + V =Lmin + V

L + V = L + V

Equilibrium line curved concavely downward, operating line becomes

tangent to the equilibrium line


30/79

ANALYTICAL EQUATION FOR THEORETICAL TRAYS

Stripping:

where

m = slope of equilibrium line (m2is used for absorption, m1is

used for stripping)

A = absorption factor = Aav.= !(A1A2)A1=L1/(m1V1)

A2= L2/(m2V2)

Absorption:


31/79


Absorption:

where


A = absorption factor = Aav.= !(A1AN)A1=L0/(mV1) & AN= LN/(mVN+1)

Kremser equations valid only when operating & equilibrium lines are straight

L0

VN+1

xA0

yAN+1

V1

xAN

yA1

LN

Stripping:


32/79

Example 10.6-3

A tray is absorbing ethyl alcohol from an inertgas stream using pure water at 303 K and

101.3 kPa.The inlet gas stream flow rate is

100.0 kgmol/h and it contains 2.2 mol% alcohol.

It is desired to recover 90% of the alcohol.Theequilibrium relationship is y=0.68x for this dilute

stream. Using 1.5 times the minimum liquid flow

rate, determine the number of tray needed. Do

this graphically and also using the analyticalequations.


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y1A

x0A

yN+1A

xNAmax

Example 10.6-3

A = ethyl alcohol , B = inert gas, C = pure water

90% recovery of alcohol

Equilibrium line: y = 0.68x

No. of trays needed = ?

V = VN+1(1-yAN+1) = 100 (1-0.022) = 97.8 kmol/h

Alcohol in VN+1= 0.022(100) = 2.2 kmol/h

Alcohol in LN= 0.9(2.2) = 1.98 kmol/h

Balance of alcohol in V1= 0.22 kmol/h

L0 kmol/h

VN+1= 100 kmol/h

xA0=0

yAN+1= 0.022

V1kmol/h

xAN

yA1

LNkmol/h

T= 303K, P = 101.3 kPa, L = 1.5 Lmin

V1= V + alcohol in V1= 97.8 + 0.22 = 98.02 kmol/h

y1A= 0.22/98.02 = 0.002244xNAmax= 0.03235


34/79

Example 10.6-3

Lmin= 59.24 kmol/h

xA0= 0, yAN+1= 0.022, yA1= 0.002244, xNAmax= 0.03235

L + V = L + V

Lmin + 97.8 = Lmin + 97.8

Lmin + V = Lmin + V

L =1.5Lmin= 1.5(59.24) = 88.86 kmol/h = L0

88.86 + 97.8 = 88.86 + 97.8

xNA= 0.0218

V= 97.8 kmol/h

LN= 88.86 +1.98 = 90.84 kmol/h

Alcohol in LN= 0.9(2.2) = 1.98 kmol/h


35/79

xNA= 0.0218x0A

y1A

yN+1A

Example 10.6-3xNA= 0.0218

Number of theoretical trays = 4

A = absorption factor = Aav.= !(A1AN)A1=L0/(mV1) & AN= LN/(mVN+1)

Equilibrium line: y = 0.68x

V1= 98.02 kmol/h, VN+1= 100 kmol/h,

L0= 88.86 kmol/h, LN= 90.84 kmol/h

xA0= 0, yAN+1= 0.022, yA1= 0.002244, xNA= 0.0218

A1=L0/(mV1) = 88.86/[(0.68)(98.02)] = 1.333

AN= LN/(mVN+1) = 90.84/[(0.68)(100)] = 1.336

A = !(A1AN) = ![(1.333)(1.336)] = 1.335


36/79

MASS TRANSFER BETWEEN PHASES

NH3from air to water

2 phases (immiscible in each other) in direct contact

concentration gradient exist in each phase

equilibrium at interface

resistance at interface = negligible

Mass transfer of solute A from one fluid phase by convection &

then through a second phase by convection


37/79

MASS TRANSFER USING FILM MASS-TRANSFER

COEFFICIENTS

A diffusing from the gas to liquid & B from liquid to gas

ky= gas-phase mass-transfer coefficient (kmol/s.m2.mol frac)

Equimolar counterdiffusion

Determination of interface compositons

where

kx= liquid-phase mass-transfer coefficient (kmol/s.m2.mol frac)

NA= ky(yAG yAi) = kx(xAi xAL) or

Line PM =slope = -kx/ky


38/79


COEFFICIENTS

Dilute solutions, (1-yA)iM& (1-xA)iM#1

Diffusion of A through stagnant B in the gas phase &

then through a non-diffusing liquid

Determination of interface compositons(by trial-and-error method):

Rearranging,


39/79


COEFFICIENTS

1sttrial: assume (1-yA)iM& (1-xA)iM=1, determine slope

Determination of interface compositons(by trial-and-error method):

Slope =

Draw line PM , get values of yAi& xAi

2ndtrial: determine slope & get new values of yAi& xAi

Repeat until interface compositions do not change


40/79

Example 10.4-1

At a certain point in the tower, yAG= 0.38 & xAL= 0.1

Equilibrium data:

ky= 1.465 x 10-3kmol A/s.m2.mol frac.

kx = 1.967 x 10-3kmol A/s.m2.mol frac.

Interface concentrations yAi ,xAi & NA= ?

Solute A absorbed from a gas mixture A-B in a wetted wall tower at

298K and 1.013 x 105Pa

xA yA xA yA

0

0.05

0.10

0.15

0

0.022

0.052

0.087

0.2

0.25

0.3

0.35

0.131

0.187

0.265

0.385

Solute A diffuses through stagnant B in the gas phase &through a non-diffusing liquid


41/79

Example 10.4-1yAG= 0.38 & xAL= 0.1 ky= 1.465 x 10

-3kmol A/s.m2.mol frac.

kx = 1.967 x 10-3

kmol A/s.m2

.mol frac.1sttrial: assume (1-yA)iM& (1-xA)iM=1, determine slope

Slope = -

yAi=0.183

xAi=0.247

On yAvs xA draw equilibrium line & line PM1with slope = -1.342


42/79

2ndtrial:

New yAi= 0.197 & xAi= 0.257

Example 10.4-1 1sttrial:yAi=0.183 & xAi = 0.247

Slope = -

Repeat the above calculation

using the latest yAi& xAi

(1-yA)iM= 0.709

(1-xA)iM= 0.82

Slope = -1.16

Previous slope #new slope

yAi = 0.197 & xAi= 0.257


43/79

Example 10.4-1

yAG= 0.38, xAL= 0.1

yAi = 0.197 & xAi= 0.257

(1-yA)iM= 0.709

(1-xA)iM= 0.82




44/79

OVERALL MASS-TRANSFER

COEFFICIENTS

NA= Ky(yAG y*A) NA= Kx(x*A xAL)

where

Ky= overall gas mass-transfer coefficient (kmol/s.m2.mol frac.)

Overall mass-transfer coefficients Ky& Kx

Kx= overall liquid mass-transfer coefficient (kmol/s.m2

.mol frac.)

y*Ain equilibrium with xAL

x*Ain equilibrium with yAG


45/79

EQUIMOLAR COUNTERDIFFUSION AND/OR

DIFFUSION IN DILUTE SOLUTIONS

where

m = slope of the equilibrium line between points E & M

when m= very small

gas solute A very soluble in liquid phase

major resistance in gas phase/gas phase controlling


46/79

EQUIMOLAR COUNTERDIFFUSION AND/OR

DIFFUSION IN DILUTE SOLUTIONS

where

m = slope of the equilibrium line between points E & M

when m= very large

gas solute A very insoluble in liquid phase

major resistance in liquid phase/liquid phase controlling


47/79

DISSUSION OF A THROUGH STAGNANT OR

NONDIFFUSING B

where

Similarly,

where


48/79

Example 10.4-2 (similar to example 10.4-1)

At a certain point in the tower, yAG= 0.38 & xAL= 0.1



Ky, NA,% resistance in the gas & % resistance in the liquid films= ?

Solute A diffuses through stagnant B in the gas phase &

through a non-diffusing liquid

y*A= 0.052, yAi= 0.197, xAi = 0.257


49/79

Example 10.4-2

yAG= 0.38 & xAL= 0.1



From example 10.4-1,

y*A= 0.052, yAi= 0.197, xAi = 0.257


50/79

Example 10.4-2

yAG= 0.38 & xAL= 0.1



y*A= 0.052, yAi= 0.197, xAi = 0.257

Solving,

Ky= 8.90 x 10-4kmol/s.m2. mol frac.

% resistance in gas film = (484/868.8) x 100% = 55.7%

% resistance in liquid film = (100 55.7)% = 44.3%


51/79

Example 10.4-2

yAG= 0.38 & xAL= 0.1


kx= 1.967 x 10-3kmol A/s.m2.mol frac.

y*A= 0.052, yAi= 0.197, xAi = 0.257

Ky= 8.90 x 10-4kmol/s.m2. mol frac.

Same flux value as calculated in example 10.4-1


52/79

Design Method for Packed Towers using Mass-Transfer

Coefficients

It is very difficult to measure the interfacial

area bet. Liquid and gas phase, kxand ky

In packed tower, the mass transfer

coefficients that were measuredexperimentally, were already taken into

account the interfacial area, a.

a = m2per m3(volume of packed section)

It is called volumetric film and overall mass-

transfer coefficients: kya, kxa, Kxa, Kya


53/79

For absorption A through stagnant B:

L[(x)/(1-x)]+V[y1/(1-y1)]=L[(x1)/(1-x1)]

+V[(y)/(1-y)]

For differential height dz:

d(Vy) = d(Lx)=kgmol A transferred/s

V=kgmol total gas/s, L=kgmol total liquid/s


54/79

Equation (10.4-8)


55/79

dA = aSdz --------(10.6-9)


56/79

SIMPLIFIED DESIGN METHODS FOR ABSORPTION

OF DILUTE GAS MIXTURES IN PACKED TOWERS

Equilibrium & Operating lines = straight

Height of packed tower, z :

V = Vave= (V1+V2)/2 L =Lave= (L1+L2)/2


57/79



Operating lines = straight

Height of packed tower, z :

$ $

$ $

Dilute : (1-y)iM, (1-x)iM, (1-y)*M& (1-x)*M#1


58/79

Procedure:

1. Plot operating line & equilibrium line

3. Plot y vs 1/(y-yi) or x vs 1/(xi-x)

4. Calculate area under plot (for equilibrium line = curve)

5. Calculate height of tower, z



2. By trial-and-error, determine yi, xior y*, x*: 1sttrial,

Using values of yi, xior y*, x*, calculate new slope:

Compare latest values of yi, xior y*, x* with former values

Slope = -

Slope = -


59/79

Example 10.6-4

A = Acetone

B = air

C = water

kya = 3.78 x 10-2kmol A/s.m3.mol frac.

kxa= 6.16 x 10-3kmol A/s.m3.mol frac.

Calculate height of tower, z, using

a) kya

b) kxa

c) Kya

T = 293KP = 101.3 kPa

S = 0.186m2

L =45.36 kmol/h

x2=0

V = 13.65 kmol/h

y1A

= 0.026

V2

y2A= 0.005

L1


60/79

Example 10.6-4



Material balance on A:

T = 293KP = 101.3 kPa

S = 0.186m2

L =45.36 kmol/h

x2=0

V = 13.65 kmol/h

y1A

= 0.026

V2

y2A= 0.005

L1

L + V = L + V

45.36 + 13.65 = 45.36 + 13.65

x1= 0.00648


61/79

Example 10.6-4



T = 293KP = 101.3 kPa

S = 0.186m2

L =45.36 kmol/h

x2=0

V = 13.65 kmol/h

y1A

= 0.026

V2

y2A= 0.005

L1

x1= 0.00648

1. Plot operating line & equilibrium line


62/79

Example 10.6-4 kya = 3.78 x 10-2kmol A/s.m3.mol frac.


V = 13.65 kmol/h, L = 45.36 kmol/h , S = 0.186 m2

y1= 0.026 , x1= 0.00648, y2= 0.005, x2= 0

Slope = -

2. By trial-and-error, determine yi, xior y*, x*: 1sttrial, Slope = -

For point 1: (y1= 0.026 , x1= 0.00648)


63/79

Example 10.6-4

From the plot, yi1= 0.0154 , xi1= 0.013, y*1= 0.0077

For 2sttrial, Slope = -

For point 1:


64/79

Example 10.6-4

Slope = -

For 2sttrial, Slope = -

Since the latest slope and the former slope is approximate close, the values

yi1= 0.0154 , xi1= 0.013, y*1= 0.0077 are accurate enough.


65/79

Example 10.6-4

For the slope at point 2 (x2= 0,y2= 0.005), 1sttrial:

#-

Slope #-

Since the slope at point 2 and point 1 changes little in the tower, the valueof the slope -1.62 from the 1sttrial is acceptable. Plotting the slope at point

2 gives yi2= 0.002, xi2= 0.0018 and y*2= 0


66/79

Example 10.6-4

where

Since both the operating and equilibrium lines are straight,the height of the tower is determined using



V = 13.65 kmol/h, L = 45.36 kmol/h , S = 0.186 m2

y1= 0.026 , x1= 0.00648, y2= 0.005, x2= 0

yi2= 0.002, xi2= 0.0018 and y*2= 0

yi1= 0.0154 , xi1= 0.013, y*1= 0.0077


67/79

Example 10.6-4

V1= V/(1-y1) and V2= V/(1-y2)

V1= 13.65/(1-0.026) = 14.014 kmol/h = 3.893 x 10-3kmol/s

V2= 13.65/(1-0.005) = 13.719 kmol/h = 3.811 x 10-3 kmol/s



V = 13.65 kmol/h, L = 45.36 kmol/h , S = 0.186 m2

y1= 0.026 , x1= 0.00648, y2= 0.005, x2= 0

yi2= 0.002, xi2= 0.0018 and y*2= 0

yi1= 0.0154 , xi1= 0.013, y*1= 0.0077


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Example 10.6-4

V1= 3.893 x 10-3kmol/s V2= 3.811 x 10

-3 kmol/s



V = 13.65 kmol/h, L = 45.36 kmol/h , S = 0.186 m2

y1= 0.026 , x1= 0.00648, y2= 0.005, x2= 0


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HEIGHT & NUMBER OF TRANSFER (HTU & NTU)

Equilibrium & Operating lines = straight

Operating lines = straight

$ $

$ $


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When major resistance to mass transfer is in gas phase, NOG

or NG

should be used (absorption)


When major resistance to mass transfer is in liquid phase,

NOLor NLshould be used (stripping)

A = absorption factor = Aav.= !(A1A2)

Where A1=L1/(mV1) A2= L2/(mV2)

Equilibrium & Operating lines = straight & not parallel

Operating line = straight

Operating line = straight


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HOG= HG+ HL

HOL= HL+ HG

HOG (HOL) is related to HG(HL) by

where


HOG= height of transfer unit based on overall gas phase

HOL= height of transfer unit based on overall liquid phase

L,V = molar flowrate of liquid & gas, respectively (kmol/s.m2)


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Height of a theoretical tray or stage, HETP (m) is related to HOG

(m)

by

Height of tower, z

where A = absorption factor = Aav.= !(A1A2)

Where A1=L1/(mV1) A2= L2/(mV2)

Example 10 6 5


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Example 10.6-5

Repeat Example 10.6-4 using transfer units

and height of a transfer unit as follows:

(a) Use HG and NG to calculate tower height

(b) Use HOG and NOG to calculate tower

height (c) Use Eq (10.6-52) to calculate NOG and

tower height

(d) Using the analytical equations, calculate

HETP from Eq (10.6-55), number oftheoretical steps N (Eq 10.6-7), and tower

height.


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ESTIMATION OF MASS-TRANSFER COEFFICIENTS

FOR PACKED TOWERS (dilute mixtures)

where

= viscosity of liquid (kg/m.s)

fP= relative mass transfer coefficient (Table 10.6-1)

Gx, Gy= liquid and gas mass flowrate per m2tower cross-section (kg/s.m2)

HG= height of transfer unit based on gas film =

HL= height of transfer unit based on liquid film =


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Example 10.8-1

Predict HG, HLand HOLfor absorption of CO2from air by

water in a dilute solution in a packed tower with 1%-in metal

Pall rings at 303K and 101.32 kPa pressure. The flow rates areGx= 4.069 kg/s.m

2and Gy= 0.5424 kg/s.m2.

fP= 1.34 fromTable 10.6-10

At 303K and 101.32kPa, from A.3-3, air= 1.666 kg/m3 and = 1.866 x 10-5kg/m.s

NSc = Schmidt number =

From Table 6.2-1, for Air-CO2system, at 276K DAB= 0.142 x 10-4m2/s

Correcting for 303K, DAB303K= DAB276K

DAB303K= 0.142 x 10-4


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Example 10.8-1


At 303K and 101.32kPa, from A.3-3, air= 1.666 kg/m3

andair= 1.866 x 10

-5kg/m.s


DAB303K=0.167 x 10-4m2/s

Gx= 4.069 kg/s.m2and Gy= 0.5424 kg/s.m

2.


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Example 10.8-1

At 303K and 101.32kPa, from A.2-4, water= 0.8007 x 10-3kg/m.s

water= 995.68 kg/m3


2.

From Table 6.3-1, for water-CO2system, at 298K DAB= 2.0 x 10-9m2/s

Correcting for 303K, DAB303K= DAB298K

waterat 298K from A.2-4 = 0.8937 x 10-3kg/s.m



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Example 10.8-1

At 303K and 101.32kPa, from A.2-4, water= 0.8007 x 10-3kg/m.s


2.

NSc =354.3



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Example 10.8-1


2.

HG= 0.2426 m & HL= 0.2306 m

HOL= HL+ HG

where


From A.3-18, for CO2at 1 atm, pA= 0.186 x 104xA,

yA= (pA/P)xA= (0.186 x 104/1)xA

L,V = molar flowrate of liquid & gas, respectively (kmol/s.m2)

L = Gx/Mwater= 4.069/18 = 0.2261 kmol/s.m2

V= Gy

/Mair

= 0.5424/29 = 0.01872 kmol/s.m2

HOL= HL+ HG

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Absorption Notes