Packed absorption and stripping columns

Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik

Packed columns are continuous contacting devices that do not have the physically distinguishable stages found in trayed columns.

1. HETP - approach

stagesmequilibriuequivalentofnumberheightpackedHETP

HETPnH

In practice, packed columns are often analyzed on the basis of equivalent equilibrium stages using a Height Equivalent to a Theoretical Plate (HETP):

Knowing the value of the HETP and the theoretical number of stages n of a trayed column, we can easily calculate the height H of the column :

The HETP concept, unfortunately, has no theoretical basis. HETP values can only be calculated using experimental data from laboratory or commercial-size columns.

For packed columns, it is preferable to determine packed height from a more theoretically based method using mass transfer coefficients.

2. Absorption: Mass transfer approach (HTU, NTU)y2< y spec

Furthermore, we introduce the coordinate z, which describes the height of the column.

The absorption problem is usually presented as follows. There is a polluted gas stream coming out from a process. The pollutant must be recovered in order to clean the gas.

At the bottom and the top of the column, the compositions of the entering and leaving streams are:

)y,x( 11 )y,x( 22

L, x1G, y1

T, p

Process

G, y2 L, x2

z = 0

z = H

xy

The green, upper envelope is needed for the operating line of the absorption column.

D'abord, nous avons besoin d'un bilan de matière autour du vert, enveloppe supérieure de la colonne. C'est la ligne d'opération, passant par le point (x2, y2):

22 yxxGLy

),(fGLf

GL

min

21

)(1

Alors nous avons besoin de l'état d'équilibre :

xm*y )(2

Nous pouvons maintenant tracer l'équilibre et la ligne d'opération dans le diagramme. De la ligne d'opération avec la plus petite pente (Lmin/G), nous pouvons obtenir (L/G) avec la formule connue :

GyLxGyLx 22

y* = m xy1

GLmin

GL

x1 x

y2

x2

y

As a third equation, we need a mass transfer rate equation. We take a small slice of the column. The material balance over the “gas side” of this slice gives:

transfermassgasgas OUTOUTIN

S is the cross-sectional area of the tower. Please note that N, G and L are defined as fluxes and not as molar flow rates [mol/s]:

scmmolG

StionseccolumnflowratemolarG 2

Determination of the packed height of a column most commonly involves the overall gas-phase coefficient Ky because the liquid usually has a strong affinity for the solute. Its driving force is the mole fraction difference (y-y*):

zSaN)zz(yGS)z(yGS

N

G

G

L

L

zz

z

3

2

cmcma

columntheofvolumesurfacetransfermassa

smol

scmmolN*yyKN y 2

Dividing the mass transfer rate equation by S and z, we get: z

)z(y)zz(yGaN

Because we want a differential height of the slice, we let z 0.

Introducing the definition of N:

Separating variables and integration gives:

dzdyGaN

*yyaKdzdyG y )(3

*yydy

aKGdzH

H y

y y

0

2

1

Taking constant terms out of the integral and changing the integration limits:

H y

yy *yydy

aKGdzH

0

1

2

The right-hand side can be written as the product of the two terms HOG and NOG:

HOG NOG

OGOGNHH

aKGH

yOG

1

2

y

yOG *yy

dyN

The term HOG is called the overall Height of a Transfer Unit (HTU) based on the gas phase. Experimental data show that the HTU varies less with G than with Kya. The smaller the HTU, the more efficient is the contacting.

The term NOG is called the overall Number of Transfer Units (NTU) based on the gas phase. It represents the overall change in solute mole fraction divided by the average mole fraction driving force. The larger the NTU, the greater is the extent of contacting required.

Now we would like to solve the integral of NOG. Therefore we replace y* by equation (2):

1

2

y

yOG xmy

dyN

Introducing the result into the equation for NOG:

1

2 221

y

y*OG Ayyy)A(

AdyN

Solving (1) for x, knowing that A=L/(Gm): mAyx

mAyx 2

2

Integration of NOG gives: 1

2

2211

y

y

*

OG AAyyyAln

AAN

*

*

OG yyAAyyyAln

AAN

22

22111

Splitting the inner part of the logarithm into two parts:

We already know the fraction of absorption :

*

*

OG yyyy

AA

Aln

AAN

22

21111

Introducing and doing some transformations, we finally get for NOG:

*yyyy

amountabsorbedmaxamountabsorbed

21

21

11

1Aln

AANOG

The NTU and the HTU should not be confused with the HETP and the number of theoretical equilibrium stages n, which can be calculated with the Kremser Equation:

111 Aln

Alnn

When the operating and equilibrium lines are not only straight but also parallel, NTU = n and HTU = HETP. Otherwise, the NTU is greater than or less than n.

The height of the column can be calculated in two ways: HETPnNHH OGOG

3. Comparison between HTU / NTU and HETP

op. line

x

y

eq. line

op. line

x

y

eq. line

nNTU

op. line

x

y

eq. line

nNTU nNTU

When the operating and equilibrium lines are straight but not parallel (NTU n), we need a formula to transform them. We can write:

1

AAlnAHHETP OG

nNHHETP OG

OG

Replacing NOG and n by the formulas found earlier, we get for HETP:

Doing the same calculation for NOG, we find: 1

AAlnAnNOG

Finally we want to calculate the volumetric overall mass transfer coefficient Kya. We know that: aK

GNHH

yOGOG

Solving for Kya, we find: HNGaK OG

y

4. Stripping: Mass transfer approach (HTU, NTU)

L, x1G, y1

T, p

ProcessG, y2

L, x2

Now we want to focus on a stripping problem, which is usually presented as follows. There is a polluted liquid stream coming out from a process. The pollutant must be recovered in order to clean the liquid.

z = 0

z = H

xyFirst, we need a material balance around the green, upper envelope of the column. It is the operating line, going through the point (x1,y1):

11 yxxGLy )(1

Then we need the equilibrium condition:

my*x )(2

GyLxLxGy 11

y* = m xy2

),.(fGL

fGL

max

2211

maxGL

GL

x2

We can now draw the equilibrium and operating line into the diagram. From the operating line with the largest slope (L/G)max, we can get (L/G) with the known formula:

x

y1

x1

y

N

G

G

L

L

zz

z

As a third equation, we need a mass transfer rate equation. We take a small slice of the column. The material balance over the “liquid side” of this slice gives:

transfermassliqliq OUTOUTIN

zSaN)z(xLS)zz(xLS

smol

The flux N involves the overall liquid-phase coefficient Kx and the driving force (x-x*):

*xxKN x

*xxaKdzdxL x )(3

Dividing the mass transfer rate equation by S and z, we get:

We let z 0 and introduce the definition of N:

Separating variables and integration gives:

H x

xx *xxdx

aKLdzH

0

2

1

HOL NOL

The term HOL is called the overall Height of a Transfer Unit (HTU) based on the liquid phase.

aNz

)z(x)zz(xL

The term NOL is called the overall Number of Transfer Units (NTU) based on the liquid phase.

aKLHx

OL

2

1

x

xOL *xx

dxN

We already know the fraction of stripping σ:

12

12

xxxx

strippableamountmaxstrippedamount

Furthermore, we know the stripping factor S: LGmS

The solution of the integral of NOL can be found if one proceeds exactly as in the case of absorption:

11

12111 xx

xxS

SS

lnS

SNOL

Finally, after some transformations, we find:

11

1Sln

SSNOL