1CHE3165

Separation Processes

S1 2015

GAS ABSORPTION IN PACKED

TOWERS

2Leaning Outcomes

By the end of the lecture you should be able to:

1. Define rate based method for packed columns

2. Define Height Equivalent to a Theoretical Plate (HETP) and explain

how it and the number of equilibrium stages differ with the height of a

transfer unit (HTU) and number of transfer units (NTU)

3. Explain the differences between loading point and flooding point in a

packed column

4. Estimate the packed height

3Contents

1 Introduction

2 Mole balance

3 Absorption of very dilute mixtures

4 Transfer unit

5 Absorption of concentrated mixtures

6 Absorption of semi-dilute mixtures

7 Controlling resistance

41. Introduction

Absorption:

Solute A absorbed from the gas phase into the liquid phase OR a

process involves molecular/mass transfer of solute A through a

stagnant, non-diffusing gas B into a stagnant liquid C

Gas-Liquid System: Solute transfer from Gas Liquid

Adsorption:

Components of a liquid or gas stream adsorbed on the surface or in the

pores of a solid adsorbent.

Gas/Liquid Solid System: Solute transfer from Gas/LiquidSolid

5 Stripping:

Reverse of absorption

Liquid Gas System: Solute transfer from Liquid Gas

6Tray and packed bed columns

7yo, V, Vo

yi, V, Vi

xi, L, Li

xo, L, Lo

L = Liquid stream total flow rate

V = Gas stream total flow rate

L = Inert (carrier) liquid flow rateV = Inert gas flow rateyA = Mass/mole fraction of A in gas

stream

xA = Mass/mole fraction of A in liquid

stream

8Example 1 (Geankoplis: Example 10.3-2)

It is desired to absorb 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 30.0 kg mol/h and the total inlet pure water flow to be used

to absorb the acetone is 90kg mol H2O/h.

yo, V, Vo

yi, V, Vi

xi, L, Li

xo, L, Lo

Identify all the unknowns: xi, L, Li, yi, V, Vi, yo, Vo, xo, Lo

Vi = 30 kgmol/hyi = 0.01V = 30 (1-0.01) = 29.7 kgmol/hL = 90 kgmol/hxi = 0 Li = 90 kgmol/hAc in V0 = 30 (0.01) (1-0.9) = 0.03 kgmol/hAc in L0 = 30 (0.01) (0.9) = 0.27 kgmol/hV0 = 29.7 + 0.03 kgmol/h = 29.73 kgmol/hL0 = 90 + 0.27 kgmol/h = 90.27 kgmol/hy0 = 0.03/29.73 = 0.00101x0 = 0.27/90.27 = 0.003

Solution:

9Example 2 (Geankoplis: Example 10.3-1)

A gas mixture at 1.0 atm pressure abs containing 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 100 kgmol/h, with a mole fraction of CO2 of yA2 = 0.20. The liquid flow rate entering is 300

kgmol water/h. Calculate the amounts and compositions of the two outlet phases. Assume that water

does not vaporize to the gas phase.

V1?yA1?

V=100yA2=0.2V =?

L=300xA2=0

L1?XA1?

Solution:

10

Gas

V1, y1

V2, y2Gas

L2, x2Liquid

Liquid

L1, x1

z1=0

z2=Z

(1)

(2)

V and L = total molar flow rates

Packing

A total interfacial area (m2)

a interfacial area per unit volume of packed tower (m2/m3)

S cross sectional area of tower (m)

Z Bed height (m)

Function of packing is to generate largest

possible interfacial area for the smallest

possible gas pressure drop.

A = a x vol. of packed tower

= a x (S x Z)

11

Packed Tower Operation

Flooding velocity:

Upper limit to the rate of gas

flow. Liquid can no longer flow

Optimum economic: V 0.5 flooding velocity = f(equipment

cost, P, processing variables)

Loading point:

Gas flow rate where liquid down

flow starts to be hindered by

gas

12

Packing Materials

13

Packing Materials

14

Packing Materials

http://finepacstructures.tradeindia.com

15

Packing Materials

Random Structured

Raschig rings

and saddles

Through flow

Relative cost Low Moderate High

Pressure drop Moderate Low Very low

Efficiency Moderate High Very high

Vapor capacity Fairly high High High

Example 3 (Geankoplis: Example 10.6-2)

A tray tower is to be designed to absorb SO2 from an air stream by using pure

water at 293K. The entering gas contains 20 mol% SO2 and 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 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 293 K. Lxo

Vyn+1

y1

xn ??

Solution

17

2. Mole Balance

Mole balance for solute over differential volume of tower:

- We can choose either gas or liquid phase.

VY = V(Y+dY) + d(in at z) (out at z+dz) (transfer to liquid)

VdY = - d

18

Write in terms of flux and area, and Y in terms of y:

nd- dY V '

N.dA- y-1

ydV '

dz NaS- y)-(1

dyV

2

'

dz NaS- y)-(1

dyV

Volume

19

Separating variables

Integrating from (1) to (2)

This is the most general equation relating total packed height (z) to gas-

phase variables.

y)-(1

dy

NaS

Vdz

Zy)-(1

dy

NaS

Vdz

z2

z1

2

1

y

y

20

If we balance the liquid phase alone, an analogous equation is obtained:

To evaluate the integrals, we need V(y), N(y) or L(x), N(x)

x2

x1x)-(1

dx

NaS

LZ

21

For gas phase analysis:

Solute flux

Choose overall Ky as typical case

AiAG

*AAG

iy

*y

P - Pk N

or P-PK N

or yykN

or yyKN

**LM

'y*

BLM

'y

yyy-1

K yy

y

KN

Across gas film

22

varies up the tower

f(y)

*y-1

y-1ln

*y-1-y-1y-1 *LM

*LMy-1

23

yyKN *y

2

1

y

yy)-(1

dy

NaS

Vz

*

*LM

'y

yyy-1

KN

y2

y1

*

*LM

'y )y(y

dy

y)-(1

y-1

aSK

Vz

y2

y1

*2'y

*LM'

)y(yy)-aS(1K

dyy-1

S

Vz

y1V

V '

a may vary with V (i.e. with y)

Ky is a function of flowrate V

V solvent-only flow and S tower cross section area are the only true

constant

To further progress, need to make

simplifying assumptions!!!

24

Assumptions:

1. Very dilute solutions (x, y less than 0.1 or 10%)

2. Linear equilibrium (i.e. Henry)

Start with

V varies only slightly from (1) to (2)

Use

y2

y1

*

*LM

'y )y(y

dy

y)-(1

y-1

aSK

Vz

y1V

V '

V = f(y)

2

VVV 21AVE

Kya will be nearly constant

because V ~ constant

3. Absorption of very dilute mixtures

25

(1 y) ~ 1

(1 y)*LM ~ 1

1

y)-(1

y-1 *LM

y2

y1

*

*LM

'y )y(y

dy

y)-(1

y-1

aSK

Vz

1

y2

y1

*'y

AVE

)y(y

dy-

aSK

Vz Approximate

26

This integral can be evaluated analytically if Henrys law holds

Mass balance (operating line):

straight line if V & L are constant

from (1) to (2)

y

y*)(y

y-y

)y-(y)y-(y

)y y*,y(y & )y y*,y(y put

dy y*)-d(y

y y*)-(ySay

line stright is f(y) y*)(y

12

*1

*22

*22

*1

1

1

y*)-d(ydy

27

(2)-(1) LM

12

'y

AVE

*11

*22

*11

*22

12

'y

AVE

*11

*22

'y

AVE

y2

y1

'y

AVE

y*)-(y

)y(-1)(y

aSK

Vz

yy

y-y

)y-(y)y-(y

1))(y(y

aSK

Vz

yy

y-y

aSK

Vz

y*)(y

y*)-d(y-

aSK

Vz

ln

ln1 12

*1

*22

y-y

)y-(y)y-(y 1

These terms make up a log mean

28

Similar expressions hold for the various forms of N

(2)-(1) LMi

12

'y

AVE

)y-(y

)y(-1)(y

aSk

Vz

(2)-(1) LMi

12

'x

AVE

x)-(x

)x(-1)(x

aSk

Lz

(2)-(1) LM*

12

'x

AVE

x)-(x

)x(-1)(x

aSK

Lz

We usually choose the phase which is controlling the rate

(i.e. the phase with the greatest mass transfer resistance)

29

Re-arrange

M'x21

M'y21

Mi'x21

Mi'y21

-x)*az(xKx-xS

L

y*)az(yKy-yS

V

x)az(xkx-xS

L

)yaz(yky-yS

V

30

How to find mole fraction at interface and equilibrium ???

y1,x1

y2,x2

y

x

y*1

y*2xi1

yi1

xi2

yi2

Dilute solution

31

Example 4 (Geankoplis: Example 10.6-4)

Acetone is being absorbed by water in a packed tower having a cross-sectional

area of 0.186 m2 at 293K and 101.32 kPa. The inlet air contains 2.6 mol%

acetone and outlet 0.5%. The gas flow is 13.65 kgmol inert air/h. Pure water inlet

flow is 45.36 kg mol water/h. Film coefficients for the given flows in the tower at

kya = 3.78 x 10-2 kgmol/s.m3.mol frac and kxa = 6.16 x 10

-2 kgmol/s.m3.mol frac.

(a) Calculate the tower height using kya.(b) Repeat using kxa.(c) Calculate Kya and the tower height.

Solution

M'x21

M'y21

Mi'x21

Mi'y21

-x)*az(xKx-xS

L

y*)az(yKy-yS

V

x)az(xkx-xS

L

)yaz(yky-yS

V

32

4. Transfer Units

y2

y1 i'y

AVE

)y(y

dy-

aSk

Vz

(m) Dimensionless

(m)

HEIGHT of a

gas film

transfer unit

Number of a

gas film

transfer unit

Z = Hy Ny

Z = Hx Nx

If OVERALL coefficients are

used to describe the flux:

Z = Hoy Noy

Z = Hox Nox

Depending on concentration

units and coefficients used

Z = HG NG = HL NL

Z = HOG NOG = HOL NOL

HG Hy; NG Ny; HOG Hoy; NOG Noy

Heights of transfer units have been

correlated with operating variables

(i.e. V, L, packing type and size)

33

Sx)-a(1K

L

aSK

LH

Sy)-a(1K

V

aSK

VH

Sx)-a(1k

L

aSk

LH

Sy)-a(1k

V

aSk

VH

*Mx'x

OL

*My'y

OG

iMx'x

L

iMy'y

G

M

1OL

M

1OG

Mi

1L

Mi

1G

-y)*(x

x-xN

y*)-(y

y-yN

x)-(x

x-xN

)y-(y

y-yN

2

2

2

2

34

35

Example 5 (Geankoplis: Example 10.6-5)

Repeat Example 5 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

Solution

36

0.5

LL

0.5

G

ScS

LH

ScS

L

S

VH

Correlation form:

Constants , , , , for different packings and sizes(Geankoplis 10.8b)

37

Slope ~ 0.43 on log-log graph

eg. correlation for inch Raschig

rings with NH3/H2O

yS

VH

38

Absorption Column Size

Height Mass transfer (Hy Ny)

Diameter Hydraulics Two phase flow over packing; gas

pressure drop - Design (Flooding

and loading)

Difficult to locate precisely

(deviation from standard line)

Loading curve nearly vertical

Operate close to loading, ~1/2

flooding

39

40

Analytical equation to calculate theoretical number of trays

For transfer of the solute from phase V to phase L (absorption)

For transfer of the solute from phase L to phase V (stripping)

Aln

1/A1/A)(1mx-y

mx-y ln

N22

21

(1/A) ln

AA)(1/my-x

/my-x ln

N11

12

2122221111 AA A and V/mL A,V/mLA

41

Analytical equation to calculate packed bed height

Absorption

Stripping

Operating and/or equilibrium lines are slightly curved

A/my-x

/my-xA)(1 ln

A-1N

1/Amx-y

mx-y1/A)(1 ln

1/A-1N

11

12OL

22

21OG

1

1

42

Analytical equation to calculate packed bed height

When the operating and equilibrium lines are straight

z = HOG x NOGz = N x HETP

See Example 10.6-5

1/A-1lnA

NOGN

The height of a theoretical tray or stage, HETP, in m is related to HOG by

/AA-1ln(1/A)

HHETP OG

43

Example 6:

Experimental data have been obtained for air containing 1.6% of SO2being scrubbed by pure water in a packed bed column of 1.5m2

cross-sectional area and 3.5 m in packed height. Entering gas and

liquid flow rates are 0.062 and 2.2 kmol/s, respectively. If the outlet

mole fraction of SO2 in the gas is 0.004. m = 40. Calculate NOG.

Solutions

44

5. Absorption of concentrated mixtures

y2

y1 i

iLM

'y

)y(y

dy

y)-(1

y-1

aSk

Vz

1

y2

y1

*'y

AVE

)y(y

dy-

aSK

Vz

Approximate

Dilute mixtures

y2

y1 i

iLM

'y

)y(y

dy

y)-(1

y-1

aSk

Vz

Concentrated mixtures

The only constant

Each term (or bracketed terms) needs to

be calculated for a range of values of y

between y1 and y2. Then, evaluate the

integral numerically (or graphically) in

order to find yi between y1 and y2.

45

We need the whole operating curve from (1) to (2). This is obtained from a

series of mass balances between (1) and a range of points up to (2).

46

Sherwood suggests that (1-y)*LM and (1-y*) can often replace the true interface

values if the liquid resistance is small. The * values are much easier to get

since ky need not be known.

47

48

Example 6 (Geankoplis: Example 10.7-1)

A tower packed with 25.4 mm ceramic rings is to be designed to absorb SO2from air by using pure water at 293K and 1.013 x 105Pa abs pressure. The

entering gas contains 20 mol% SO2 and that leaving 2 mol%. The inert air

flow is 6.53 x 10-4 kg mol air/s and the inert water flow is 4.20 x 10-2 kg mol

water/s. The tower cross-sectional area is 0.0929 m2. For dilute SO2, the

film mass-transfer coefficients at 293 K are, for 25.4 mm rings:

kya = 0.0594Gy0.7Gx

0.25 kxa = 0.152Gx0.82

Gx and Gy are kg total liquid/gas per sec per m2 tower cross section.

Calculate the tower height.

Solution

49

6. Absorption of semi-dilute mixtures

Dilute enough to use VAVE or LAVE and constant Kya or Kxa

y2

y1

*LM

'y

y*)(y

dy

y)-(1

y-1

aSK

Vz

y2

y1

*LM

'y

AVE

y*)(y

(-dy)

y)-(1

y-1

aSK

Vz

Always check this for ~ 1.0. If yes, need only evaluate

Which is relatively simple (numerical or graphical)

y2

y1*yy

dy

50

7. Controlling resistance

i) Liquid film controlling (eg. gas almost insoluble O2/H2O)

51

7. Controlling resistance

ii) Gas film controlling (eg. gas very soluble or react with liquid NH3/H2O)