absorption 2nd1415
DESCRIPTION
Absorption 2nd1415TRANSCRIPT
-
ABSORPTION
-
ABSORPTION AND STRIPPING
Absorption (or scrubbing) is the removal of a component (the solute or absorbate) from a gas stream via uptake by a nonvolatile liquid (the solvent or absorbent).
Desorption (or stripping) is the removal of a component from a liquid stream via vaporization and uptake by an insoluble gas stream.
Thus, absorption and stripping are opposite unit operations, and are often used together as a cycle.
Both absorption and stripping can be operated as equilibrium stage processes using trayed columns or, more commonly, using packed columns.
-
Absorber/Stripper Cycle
-
Absorption Systems Physical
Examples:CO2 and water Acetylene and acetic acidCO and water NH3 and acetoneH2S and water Ethane and carbon disulfideNH3 and water N2 and methyl acetateNO2 and water NO and ethanol
Physical absorption relies on the solubility of a particular gas in a liquid.
This solubility is often quite low; consequently, a relatively large amount of liquid solvent is needed to obtain the required separation.
This liquid solvent containing the solute is typically regenerated by heating or stripping to drive the solute back out.
Because of the low solubility and large solvent amounts required in physical absorption, chemical absorption is also used
-
Absorption Systems Chemical
Chemical absorption relies on reaction of a particular gas with a reagent in a liquid.
Examples:CO2 / H2S and aqueous ethanolaminesCO2 / H2S and aqueous hydroxidesCO and aqueous Cu ammonium saltSO2 and aqueous dimethyl anilineHCN and aqueous NaOHHCl / HF and aqueos NaOH
This absorption can often be quite high; consequently, a smaller amount of liquid solvent/reagent is needed to obtain the required separation.
However, the reagent may be relatively expensive, and it is often desirable to regenerate when possible.
-
Absorption and Stripping the Problem
The principal difference in handling adsorption and stripping, compared todistillation, how the equilibria (equilibrium curve) and mass balances (operating lines) are represented.
In absorption, consider removal of the solute from the gas stream and uptake by the solvent liquid stream; thus, the total liquid and gas stream amounts or flow rates can change.
If we use mole fractions of the solute and assume that the gas and liquid stream amounts or flow rates remain constant, significant error can result if the solute concentration in the inlet gas stream is greater than about 1%.
If we can set up our equilibrium curve and operating line to account for this change in the overall gas and liquid flow rates, we can use the McCabe-Thiele method to solve absorption and stripping problems.
-
Absorption and Stripping Assumptions
We assume that: The carrier gas is insoluble (or it has a very low solubility), e.g, N2 or Arin water. The solvent is nonvolatile (or it has a low vapor pressure), e.g., water
in air at low temperatures. The system is isothermal. e.g., the effects of heat of solution or
reaction are low or there is a cooling or heating system in the column. The system is isobaric. While the total gas and liquid streams can change in absorption, the
flow rate of the carrier gas, which we assume to be insoluble in the solvent, does not change.
define the equilibrium curve and operating line in terms of mole ratios with respect to the carrier gas and solvent, instead of mole fractions
Similarly, the flow rate of the solvent, which we assume to be nonvolatile, does not change.
The concentration of the solute is low, say
-
MATERIAL BALANCE IN AN ABSORPTION TRAY TOWER
V1, y1 Lo, x0
1
2
Vn+1, yn+1n
Ln, xn
n + 1 N 1
N
VN+1, yN+1 LN, xN
-
DESIGN OF PLATE ABSORPTION TOWERS
OPERATING-LINE DERIVATION
L(xo) V (yn+1) L(xn) V (y1)
(1 xo ) (1 yn+1) (1 xn ) (1 y1)+ +=
Where: x is the mole fraction A in the liquidy is mole fraction of A in the gasL is the total moles liquidV the total moles gas
yn+1= xn (L/V) + y1 xo(L/V)
-
DESIGN OF PLATE ABSORPTION TOWERS
Graphical determination of the number of trays
If x an y are very dilute, the denominators 1 x and 1 y will be close to 1.0 and the OL will be approximately straight with a slope L/V
The number of theoretical trays is determined by stepping off the number of trays.
-
McCabe-Thiele Plot Absorber
-
EXAMPLE 1
A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 293K. The entering gas contains 25%mol SO2 and that the leaving 3 mol% at a total pressure of 101.3kPa. The inert air flow rate is 100 Kg air/h-m, and the entering water flow rate is 4000 kg water/h-m. Assuming an overall tray efficiency of 30%, how many theoretical trays and actual trays are needed? Assume that the tower operates at 293K.
*Equilibrium data at A.3-19
2
2
-
00.05
0.1
0.15
0.2
0.25
0.3
0 0.002 0.004 0.006 0.008 0.01
-
DESIGN OF PACKED TOWERS FOR ABSORPTION
V2, y2 L2, x2
V, y
dz
L, x
z
V1, y1 L1, x1
-
DESIGN OF PACKED TOWERS FOR ABSORPTION
OPERATING-LINE DERIVATION
L(x) V (y1) L(x1) V (y)
(1 x ) (1 y1) (1 x1 ) (1 y)+ +=
Where: x1 is the mole fraction A in the liquidy1 is mole fraction of A in the gasL is Kgmol inert liquid/s or kgmol inert liquid/s-mV is Kgmol inert liquid/s or kgmol inert liquid/s-m
2
2
-
LOCATION OF OPERATING LINES
y1
y2
x2 x1
Tower bottom
Operating line
Equilibrium line
Top
Mole fraction, x
Mo
le f
ract
ion
,y
Absorption of A from V to L stream
y2
y1
x2x1
Tower bottom
Operating lineEquilibrium line
Top
Mole fraction, x
Stripping of A from V to L stream
-
DESIGN OF PACKED TOWERS FOR ABSORPTION
OPERATING LINE DERIVATION
The equation of the line can also be written in terms of partial pressure p1 of A. Where
y1/(1 y1) = p1/(P p1)
If x and y are very dilute, (1 x) and (1 y) can be taken as 1.0 and the equation of OL becomes
Lx + Vy1 Lx1 + Vy
-
LIMITING AND OPTIMUM L/V RATIOSABSORPTION
y1
y2
x2 x1
Operating line for actualliquid flow
Equilibrium line
Operating line for minimum liquid flow
Mole fraction, x
Mo
le f
ract
ion
,y
x1 max
-
LIMITING AND OPTIMUM L/V RATIOSSTRIPPING
y2
y1
x2x1
Operating line for minimumgas flows
Operating lineFor actual gas flow
Equilibrium line
-
Lecture 21
Minimum Absorbent Rate Lmin
-
ANALYTICAL EQUATIONS FOR THEORETICAL NUMBER OF STEPS OR TRAYS
y1 mx2y2 mx2
(1 1/A) + 1/A
ln AN = ABSORPTION
ln
x2 y1/mx1 y1/m
(1 A) + 1/A
ln (1/A)N = STRIPPING
ln
Where : A = L/mV
-
ANALYTICAL EQUATIONS FOR THEORETICAL NUMBER OF STEPS OR TRAYS
For absorption, at the bottom concentrated end tray, the slope m1 or tangent at the point x1 on the equilibrium line is used. Then A1 = L1/m1V1
At the dilute top tray, the slope m2 of the equilibrium line at point y2 is used. Then A2 = L2/m2V2
A = (A1A2)1/2
-
ANALYTICAL EQUATIONS FOR THEORETICAL NUMBER OF STEPS OR TRAYS
For stripping, at the top concentrated stage, the slope m2 or tangent at point y2 on the equilibrium line is used. Then A2 = L2/m2V2
At the bottom or dilute end, the slope m1 at point x1 on the equilibrium line is used. Then A1 = L1/m1V1
A = (A1A2)1/2
-
Analytical Minimum Absorbent Flowrate
L = V(yN+1 y1)xN x0
absorbent flowrate
KN = yN+1 /(1 + yN+1)xN /(1 + xN)
Minimum absorbent rate
Lmin =V(yN+1 y1)
yN+1 x0
KN
Minimum absorbent flowrate for Dilute solutions
Lmin = V KN (fraction of solute absorbed)
Minimum absorbent flow rate if xo = 0
-
Example 2
It is desired to absorb 90% of the acetone in a gas containing 1 mol% acetone in air in a tray tower. The total inlet gas flow to the tower is 30.0 kmol/h. The process is operating isothermally at 300K and a total pressure of 101.3 kPa. The gas-liquid equilibrium relation for this dilute stream is y = mx = 2.53x. Using 1.5 times the minimum liquid flowrate, determine the number of trays needed a) graphically b) analytically.