Transcript No Slide Title

Packed absorption and stripping columns
Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik
1. HETP - approach
Packed 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 equilibrium 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 HETP
The 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.
L, x2
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.
z=H
At the bottom and the top of the column, the
compositions of the entering and leaving streams are:
( x1 , y1 )
T, p
( x2 , y 2 )
y
x
Furthermore, we introduce the coordinate z, which
describes the height of the column.
The green, upper envelope is needed for the
operating line of the absorption column.
z=0
Process
G, y1
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  Gy2  Lx2  Gy
y
L
x  x 2   y 2
G
L
G
y
y1
y* = m x
Lmin
G
(1)
Then we need the equilibrium condition:
y*  m x
( 2)
y2
x2
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
f
G
L
 f (1, 2 )
 
 G min
x1
x
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:
L
G
z  z
N
INgas  OUTgas  OUTmass transfer
 mol 
 s 
S G y ( z )  S G y ( z  z )  N a S z
z
G
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
G  
mol 
2 
 cm s 
a
mass transfer surface
volume of the column
a   cm3 
2
 cm 
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 * 
N    mol2

 cm s 
L
Dividing the mass transfer rate equation by S
and z, we get:
NaG
y ( z  z )  y ( z )
z
Because we want a differential height of the
slice, we let z  0.
NaG
dy
dz
Introducing the definition of N:
G
Separating variables and integration gives:
H   dz   
dy
 K y ay  y * 
dz
H
0
Taking constant terms out of the integral and
changing the integration limits:
y2
G
dy
K y a y  y * 
y1
H
G
H   dz 
Kya
0
HOG
The right-hand side can be written as the
product of the two terms HOG and NOG:
H  HOGNOG
( 3)
y1
dy
 y  y * 
y2
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.
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):
Solving (1) for x, knowing that A=L/(Gm):
Introducing the result into the equation for NOG:
HOG 
NOG 
G
Kya
y1
dy
 y  y * 
y2
NOG 
y1
dy
 y  m x 
y2
x
y
y
 x2  2
Am
Am
NOG 
y1
Ady
 ( A  1 )y  y
y2
2
 Ay 2*
Integration of NOG gives:
A  1y  y 2  Ay 2*
A

ln
A 1
A
NOG
NOG
y1
y2

A
A  1y1  y 2  Ay 2*

ln
A 1
A y 2  y 2*

Splitting the inner part of the logarithm into two parts:
NOG 
We already know the fraction of absorption :

Introducing  and doing some transformations, we
finally get for NOG:
NOG 

 1 A  1 y1  y 2* 
A

ln 
A 1  A
A y 2  y 2* 
absorbed amount
y  y2
 1
max absorbed amount y1  y 2*
A
1 A 
ln

A 1  1 
3. Comparison between HTU / NTU and HETP
The height of the column can be calculated in two ways:
H  HOGNOG  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
1 A 
ln

ln A  1   
y
op. line
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
y
op. line
op. line
eq. line
eq. line
eq. line
x
x
x
NTU  n
NTU  n
NTU  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
Finally we want to calculate the volumetric overall mass
transfer coefficient Kya. We know that:
HOG 
Solving for Kya, we find:
A ln A
A 1
H
G

NOG K y a
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, p
First, we need a material balance around the
green, upper envelope of the column. It is the
operating line, going through the point (x1,y1):
y
x
Gy1  Lx  Lx1  Gy
y
L
x  x1   y1
G
z=0
(1)
Then we need the equilibrium condition:
x* 
y
m
( 2)
G, y1
L, x1
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:
y
L
G
y2
y* = m x
L 1 L 
  
 f (1.2 , 2 )
G f  G max
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:
INliq  OUTliq  OUTmass transfer
S L x( z  z )  S L x( z )  N a S z
 mol 
 s 
x
L
G
The flux N involves the overall liquid-phase
coefficient Kx and the driving force (x-x*):
N  K x x  x * 
x2
x1
z  z
N
z
G
L
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:
L
x( z  z )  x( z )
N a
z
L
dx
 K x ax  x * 
dz
H
L
H   dz 
K xa
0
x2
The term NOL is called the overall Number of Transfer
Units (NTU) based on the liquid phase.
HOL 
x1
NOL 
L
K xa
x2
dx
 x  x * 
x1
dx
 x  x*
HOL
The term HOL is called the overall Height of a Transfer
Unit (HTU) based on the liquid phase.
( 3)
NOL
We already know the fraction of stripping σ:

amount stripped
x  x1
 2
max amount strippable x2  x1
Furthermore, we know the stripping factor S:
S
mG
L
 1 S  1 x 2  x1 
S


ln  
S  1  S
S x1  x1 
The solution of the integral of NOL can be found if
one proceeds exactly as in the case of absorption:
NOL
Finally, after some transformations, we find:
NOL 
S
1  S 
ln 

S 1  1  