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
stagesmequilibriuequivalentofnumber
heightpackedHETP
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)
L, x1G, y1
T, p
Process
G, y2
y2< y spec
L, x2
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
z = 0
z = H
xy
The green, upper envelope is needed for the operating line of the absorption column.
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):
22 yxxG
Ly
y* = m xy1
),(fG
Lf
G
L
min
21
)(1
Then we need the equilibrium condition:
xm*y )( 2
G
Lmin
G
L
x1
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:
x
y2
x2
y
GyLxGyLx 22
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]:
scm
molG
Stionseccolumn
flowratemolarG
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
cm
cma
columntheofvolume
surfacetransfermassa
s
mol
scm
molN*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:
dz
dyGaN
*yyaKdz
dyG y )(3
*yy
dy
aK
GdzH
H y
y y
0
2
1
Taking constant terms out of the integral and changing the integration limits:
H y
yy *yy
dy
aK
GdzH
0
1
2
The right-hand side can be written as the product of the two terms HOG and NOG:
HOG NOG
OGOGNHH
aK
GH
yOG
1
2
y
y
OG *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
y
OG 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): mA
yx
mA
yx 2
2
Integration of NOG gives: 1
2
221
1
y
y
*
OG A
AyyyAln
A
AN
*
*
OG yyA
AyyyAln
A
AN
22
2211
1
Splitting the inner part of the logarithm into two parts:
We already know the fraction of absorption :
*
*
OG yy
yy
A
A
Aln
A
AN
22
2111
1
Introducing and doing some transformations, we finally get for NOG:
*yy
yy
amountabsorbedmax
amountabsorbed
21
21
1
1
1
Aln
A
ANOG
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:
1
11 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
A
AlnAHHETP OG
n
NHHETP OG
OG
Replacing NOG and n by the formulas found earlier, we get for HETP:
Doing the same calculation for NOG, we find: 1
A
AlnAnNOG
Finally we want to calculate the volumetric overall mass transfer coefficient Kya. We know that: aK
G
N
HH
yOGOG
Solving for Kya, we find:H
NGaK 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 yxxG
Ly )(1
Then we need the equilibrium condition:
m
y*x )( 2
GyLxLxGy 11
y* = m xy2
),.(fG
L
fG
L
max
2211
maxG
L
G
L
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
s
mol
The flux N involves the overall liquid-phase coefficient Kx and the driving force (x-x*):
*xxKN x
*xxaKdz
dxL 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 *xx
dx
aK
LdzH
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.
aK
LH
xOL
2
1
x
x
OL *xx
dxN
We already know the fraction of stripping σ:
12
12
xx
xx
strippableamountmax
strippedamount
Furthermore, we know the stripping factor S:L
GmS
The solution of the integral of NOL can be found if one proceeds exactly as in the case of absorption:
11
1211
1 xx
xx
S
S
Sln
S
SNOL
Finally, after some transformations, we find:
1
1
1
Sln
S
SNOL