# abstrip-packed columns (1)

Post on 10-Apr-2015

284 views

Category:

## Documents

Embed Size (px)

TRANSCRIPT

Packed absorption and stripping columns

Prof. Dr. Marco Mazzotti - Institut fr Verfahrenstechnik

1. HETP - approachPacked columns are continuous contacting devices that do not have the physically distinguishable stages found in trayed columns.

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

HETP !

packed height number of equivalent equilibriu m stages

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 :

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

2. Absorption: Mass transfer approach (HTU, NTU)y2< y spec For packed columns, it is preferable to determine packed height from a more theoretically based method using mass transfer coefficients. G, y2 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. L, x2

z=H

At the bottom and the top of the column, the compositions of the entering and leaving streams are:( x1 , y1 ) ( x2 , y 2 )

T, p

y Furthermore, we introduce the coordinate z, which describes the height of the column. G, y1

x

z=0

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

Process

L, x1

First, we need a material balance around the green, upper envelope of the column. It is the operating line, going through the point (x2,y2):Lx Gy 2 ! Lx 2 GyL y ! x x 2 y 2 G

y

L G

y1(1) Lmin G

y* = m x

Then we need the equilibrium condition:y* ! m x ( 2)

y2 x2 x1 x

We can now draw the equilibrium and operating line into the diagram. From the operating line with the smallest slope (Lmin/G), we can get (L/G) with the known formula:L L p f (1 ,2 ) !f G G min

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:IN ! OUT OUT mass tr

G N

Lz (z

as

as

S G y ( z ) ! S G y ( z (z ) N a S (z

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]:G! molar flowrate column sec tion S

mass transfer surface a! volume of the column

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*):N ! K y y y *

nsfer

mol s

z

G

L

?G A!

mol 2 cm s

cm 2 ?aA ! 3 cm

?N A! mol 2

cm s

Dividing the mass transfer rate equation by S and (z, we get: Because we want a differential height of the slice, we let (z p 0.

Na! G

y ( z (z ) y ( z ) (z dy dz

Na! G

Introducing the definition of N:

G

dy ! K y ay y * dz

( 3)

Separating variables and integration gives:

H

H ! dz ! 0

y2

G dy y1 K y a y y *

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

G H ! dz ! Kya 0HOG

H

y1

y2

y y * NOG

dy

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

H ! H OG NOG

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.

HOG !

G Kya

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.

y1

NOG !

y2

y y *

dy

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

NOG !

dy y2 y m x

y1

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

x!

y y x2 2 Am Am

Introducing the result into the equation for NOG:

NOG !

Ady ( A 1)y y 2 Ay 2* y2

y1

Integration of NOG gives:

NOG

A A 1y y 2 Ay *2 ! ln A 1 A

y1

y2

NOG !

* A A 1y 1 y 2 Ay 2 ln * A 1 A y2 y2

Splitting the inner part of the logarithm into two parts:

NOG !

* 1 A 1 y1 y 2 A ln * A 1 A A y2 y2

We already know the fraction of absorption E:

E!

absorbed amount y y2 ! 1 * max absorbed amount y1 y 2

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

NOG !

A 1 E A ln A 1 1E

3. Comparison between HTU / NTU and HETPThe height of the column can be calculated in two ways:H ! HOG NOG ! n HETP

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:

n!

1 1E A ln ln A 1 E

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.

y

op. line

y

op. line

y

op. line

eq. line

eq. line

eq. line

x

x

xNT n

NT

!n

NT

"n

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

HETP ! HOG

NOG n

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

HETP ! HOG

A ln A A 1

Doing the same calculation for NOG, we find:

NOG ! n

A ln A A 1

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

HOG !

H G ! NOG K y a

Solving for Kya, we find:

Kya !

G NOG H

4. Stripping: Mass transfer approach (HTU, NTU)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. G, y2 Process z=H

T, pFirst, we need a material balance around the green, upper envelope of the column. It is the operating line, going through the point (x1,y1):Gy1 Lx ! Lx1 GyL x x1 y 1 G

y

x

z=0(1)

y!

G, y1

L, x1

Then we need the equilibrium condition:x* ! y m ( 2)

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:L 1 L p f ( 1 .2 , 2 ) ! G f G max

yL G

y2 y* = m x

L G max

y1 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:IN li ! OUTli OUTmass transfer

x1

x2

x

G mol s

Lz (z

S L x ( z (z ) ! S L x ( z ) N a S (z

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

Nz

G

L

Dividing the mass transfer rate equation by S and (z, we get: We let (z p 0 and introduce the definition of N:

L

x( z (z ) x( z ) !N a (z L dx ! K x ax x * dz( 3)

Separating variables and integration gives:

L H ! dz ! K xa 0

H

x2

x1

x x*NOL

dx

HOL

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

HOL !

L K xa

x2

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

NOL !

x1

x x *

dx

We already know the fraction of stripping :

W !

amount stripped x x1 ! 2 max amount strippable x 2 x1

Furthermore, we know the stripping factor S:

S!

mG L

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

NOL

1 S 1 x 2 x1 S ! ln S 1 S S x1 x1

Finally, after some transformations, we find:

NOL !

S 1W S ln S 1 1 W