Absorption & Stripping Introduction
Graphical Methods
Packed Towers
SHR Chapter 6
Introduction
Trayed tower
Goals:
minimize mass transfer resistance to
achieve equilibrium on each tray
minimize bubble carry-over to tray below
minimize liquid entrainment to tray above
minimize weeping of liquid through
holes in tray
SHR 6.1
Types of Traysperforated valve cap bubble cap
Regimes in a Trayed Tower
Spray
gas phase is continuous
(low liquid depths, high gas flow rates)
Froth
gas passes through liquid as jets or a series of bubbles
Emulsion
!
Bubble
low gas flow rates - swarms of bubbles
Cellular Foam
(think blowing bubbles
in chocolate milk)
SHR 6.1.1
Packed Towers
Stru
ctur
ed P
acki
ng
Uns
truc
ture
d Pa
ckin
gSHR 6.1.2
Other Configurations
Spray tower
very low pressure drop
use for absorption only when
solute is highly soluble in the liquid (e.g. SO2 in flue gas)
Bubble column
absorption
high pressure drop
use when solute is poorly
soluble in liquid
use when slow chemical
reactions occur that require long residence time
Centrifugal contactor
short residence time
compact
SHR 6.1.3
Heuristics & Design Considerations
Trayed towers
reliable design
low liquid velocities
liquid phase (the continuous phase) is typically mass-transfer limiting
Unstructured Packed towers
corrosive environments
small towers (
Analysis Approach
Trayed towers:
analyze each tray as an equilibrium problem
what assumption here???
write coupled equations for mass & energy balances between trays
Other towers (packed, etc.):
Height Equivalent of a Theoretical Plate (HETP)
effective height acts as one tray
vendors of packing report this value
Graphical Methods
SHR 6.3
Some Terminology
Stripper
Solute enters in liquid.
Stripping agent enters
bottom of column.
L
Molar flow rate of solute-free liquid
V
Molar flow rate of solute-free gas
x
mole fraction of solute in liquid
y
mole fraction of solute in gas
X
mole ratio of solute to solute-free liquid
Y
mole ratio of solute to solute-free gas
Assume that only solute is transferred from one phase to another (no vaporization of liquid or condensation of gas carriers).
What does XL and YV represent?
X =x
1 xY =
y
1 y
Absorber
Solute enters in gas.
Liquid absorbent enters from top of column.
Note different tray ordering convention...
Stage "i"
Xi
Xi-1 Yi
Yi+1
For the absorber:
Ki =yi
xi=
Yi/(1+Yi)
Xi/(1+Xi)
Streams leaving the tray are assumed to be in equilibrium
Mole Balances & Operating LinesSolute balance around arbitrary # of trays in the top section of the absorber:
X0L0 + Yn+1V
0 = XnL0 + Y1V
0
solute flow rate in solute flow rate out
Yn+1 = (Xn X0)L0
V 0+ Y1
L0
V 0= slope of operating line
(absorber)
(stripper)
What happens as ?L0/V 0 ! 1
Absorber
Solute enters in gas.
Liquid absorbent enters from top of column.
We know these.
What happens as gets small?L0/V 0
Yn = (Xn+1 X1)L0
V 0+ Y0
SHR 6.3.2
For an absorber, we typically know YN+1, X0 and V. Therefore, we get to choose L to achieve desired Y1.
design variable
Absorber: Minimum Flow RateYn+1 = (Xn X0)
L0
V 0+ Y1
Absorber
Solute enters in gas.
Liquid absorbent enters from top of column.
L0 =V 0 (YN+1 Y1)
XN X0
Over the whole tower (n=N):
Lmin corresponds to equilibrium with XN and YN+1.
KN =yN+1
xN=
YN+1/(1+YN+1)
XN/(1+XN )
For dilute solutes (Y y and X x):
If X0 0 then:
L
0min = V
0
yN+1 y1yN+1
/KN x0
Corresponding equation for a
stripper:
L0min =V 0 (YN+1 Y1)
YN+1/[YN+1(KN1)+KN ] X0
As V , Lmin.
SHR 6.3.3
Why are these in equilibrium?
L0min = V0KN (fraction absorbed)
V 0min =L0
KN (fraction stripped)
This is the best we can achieve given the inlet constraints.
Examples
Given: feed stream composition & flow rates, recovery (Y1)
Find: XN.
Given: feed stream composition & flow rates,
solvent loading (XN)
Find: Y1.
Given: feed stream composition, gas flow rate, solvent loading (XN), and recovery (Y1),
Find: solvent flow rate.
Yn+1 = (Xn X0)L0
V 0+ Y1
X
Y
XNX0
YN+1
Y1 Y =
KX
L'
V'
XY
XNX0
YN+1
Y1Y =
KX
L'
V'X
Y
XNX0
YN+1
Y1
Y = KX
L'
V'
Number of Equilibrium StagesSHR 6.3.4
1. Locate the point for the solvent feed (X0) and desired Y1 on the graph.
2. Determine operating line from V and L.
3. March off to determine the stages
(assuming each stage is in equilibrium)
Stage "i"
Xi
Xi-1 Yi
Yi+1
For the absorber:
Streams leaving the tray are assumed to be in equilibrium
Streams passing one another are on the operating line.
Yn+1 = (Xn X0)L0
V 0+ Y1
equilibrium relates Yi and Xi.
Operating line:
What happens to the number of stages as L/V
approaches or its minimum?
Algebraic ApproachYn+1 = (Xn X0)
L0
V 0+ Y1
Operating line:
Given: X0, Y1, L/V,
1. K1 = Y1/X1 solve for X1.
2. Find Y2 from the operating line.
3. K2 = Y2/X2 solve for X2.
4. Find Y3 from the operating line.
Must have a model for Ki.
If Ki is not a function of composition:
1. Calculate Ki at given T and P.
2. Follow steps outlined above.
If Ki is a function of composition:
1. Guess Xi. (note that Yi is known from
operating line).
2. Calculate Ki.
3. Update guess for Xi and return to step 2 if
not converged.
SHR 6.4 presents an alternative to this formulation.
Number of Stages for StrippersYn = (Xn+1 X1)
L0
V 0+ Y0
Stage EfficiencyComplex function of:
tray design/geometry
fluid dynamics on trays
Typically less than 50% efficient (10%-50%)
trays are not at equilibrium!
more viscous liquids typically lead
to lower efficiencies (inhibit mass transfer)
logEo
= 1.597 0.199 logKM
L
L
L
0.0896log
KM
L
L
L
2
Empirical correlation for stage efficiency
Data over a wide range of column diameters, pressures, temperatures and liquid viscosities.
Eo
NtN
a
# theoretical (equilibrium) stages
# actual stages
Other (less empirical) methods exist - see SHR 6.5.4.
Packed ColumnsSHR 6.7
SHR 6.7
Analysis Options
Option 1: graphical techniques
HETP is known
HETP = (height) / (number of theoretical equilibrium stages)
Use methods previously discussed to get number of trays/stages
solve for height given number of stages
HETP is typically found empirically & supplied by packing vendors.
!
Option 2: rate-based techniques
Use mass transfer coefficients (and a few hefty assumptions)
See SHR 6.7 for more details.
lT = HETP Nt
Operating Lines
x
in
L
in
+ yV` = xL` + youtVout
y = x
L
V
+ y
out
xin
L
V
Solute mole balance:
For dilute solutions, V and L are approximately constant:
Packed absorber operating line
y = x
L
V
+ y
in
xout
L
V
Solute mole balance:
For dilute solutions, V and L are approximately constant:
xL` + yinVin = xoutLout + yV`
Packed stripper operating line
Here, x and y are
bulk compositions.
Finite-Rate Mass Transfer (Back to Two-Film Theory)
Often we dont know the surface area for mass transfer from all of the packing.
r = mass transfer rate per unit volume,
a = surface area per unit volume of packing
mol/(m3s)
r = Ja = ky
a(y yI)= k
x
a(xI x)
y = yI k
x
a
k
y
a
(x xI)
relative resistance of mass transfer between
the two phases
Gas Interface Liquid
liquid film composition
gas film composition yI or pI
xI or cI
bulk liquid composition
bulk gas compositiony or p
x or c
x*
y*
r = Ky
a (y y) = Kx
a (x x)Overall mass transfer coefficient approach:
1
Ky
a=
1
ky
a+
K
kx
a1
Kx
a=
1
kx
a+
1
Kky
a
J = ky(y yI)
y = yI k
x
a
k
y
a
(x xI)AB line:
Making Connections...
y = x
L
V
+ y
out
xin
L
V
Applicable for small x, y.
Gas Interface Liquid
liquid film composition
gas film composition yI or pI
xI or cI
bulk liquid composition
