TRAPHE3- Gas Absorption

MASS TRANSFER MASS TRANSFER MASS TRANSFER MASS TRANSFER

Gas AbsorptionGas AbsorptionGas AbsorptionGas Absorption

Absorption and StrippingAbsorption and StrippingAbsorption and StrippingAbsorption 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).

• Stripping (or desorption) 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 CycleAbsorber/Stripper CycleAbsorber/Stripper CycleAbsorber/Stripper Cycle

Absorption operationsAbsorption operations

Absorption Systems Absorption Systems Absorption Systems Absorption Systems –––– PhysicalPhysicalPhysicalPhysical

• Examples:

CO2 and water Acetylene and acetic acid

CO and water NH3 and acetone

H2S and water Ethane and carbon disulfide

NH3 and water N2 and methyl acetate

NO2 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 Absorption Systems Absorption Systems Absorption Systems –––– ChemicalChemicalChemicalChemical• Chemical absorption relies on reaction of a particular gas with a

reagent in a liquid.

• Examples:

– CO2 / H2S and aqueous ethanolamines

– CO2 / H2S and aqueous hydroxides

– CO and aqueous Cu ammonium salt

– SO2 and aqueous dimethyl aniline

– HCN and aqueous NaOH

– HCl / 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 AssumptionsAbsorption and Stripping AssumptionsAbsorption and Stripping AssumptionsAbsorption and Stripping Assumptions

Assumptions:

– The carrier gas(inert gas) is insoluble (or it has a very low solubility), e.g, N2 or Ar in 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.

– The concentration of the solute is low, say <10% –this is the limit for the use of Henry’s Law,

Equilibrium Data for AmmoniaEquilibrium Data for AmmoniaEquilibrium Data for AmmoniaEquilibrium Data for Ammonia----Water SystemWater SystemWater SystemWater System

Equilibrium Data for SOEquilibrium Data for SOEquilibrium Data for SOEquilibrium Data for SO2222----HHHH2222O SystemO SystemO SystemO System

Material Balance in an Absorption Tray TowerMaterial Balance in an Absorption Tray TowerMaterial Balance in an Absorption Tray TowerMaterial Balance in an Absorption Tray Tower

Material Balance Diagram for Plate ColumnMaterial Balance Diagram for Plate ColumnMaterial Balance Diagram for Plate ColumnMaterial Balance Diagram for Plate Column

Absorption: Design considerationsAbsorption: Design considerations

y

x

(mole fraction of A in L)

(mole fraction of A in V)

xa xb

ya

yb

equilibrium line

x*b

1+−

neyy

The driving force for mass transfer

xn

Operating and equilibrium lines: (a) for gas absorption (b) foOperating and equilibrium lines: (a) for gas absorption (b) foOperating and equilibrium lines: (a) for gas absorption (b) foOperating and equilibrium lines: (a) for gas absorption (b) for r r r

desorptiondesorptiondesorptiondesorption

No. of Theoretical Stages in Plate AbsorbersNo. of Theoretical Stages in Plate AbsorbersNo. of Theoretical Stages in Plate AbsorbersNo. of Theoretical Stages in Plate Absorbers

Graphical Solution – Mc Cabe-Thiele Method

(a) Using mole fractions (x, y)

To determine the theoretical number of stages, plot

the equilibrium diagram (EC) and the operating line

(OL). From the point (x0,y1) or (xa, ya) draw triangular

steps using alternately the OL and EC until the point

(xN, yN+1) or (xb, yb) is reached or overrun. The

theoretical no. of stages is equivalent to the no. of

triangles formed.

(b) Using mole ratios ( X, Y)

McCabeMcCabeMcCabeMcCabe----Thiele Plot AbsorberThiele Plot AbsorberThiele Plot AbsorberThiele Plot Absorber

McCabeMcCabeMcCabeMcCabe----Thiele Plot Thiele Plot Thiele Plot Thiele Plot –––– StripperStripperStripperStripper

Absorption Factor MethodAbsorption Factor MethodAbsorption Factor MethodAbsorption Factor Method

Concentration DifferencesConcentration DifferencesConcentration DifferencesConcentration Differences


Analytical Solution – If OL and EC are both linear use

Absorption Factor or Kremser Equation

(a) based on the gas phase (V phase)

*a

*b

ab

*aa

*bb

*aa

*bb

yy

yyln

yy

yyln

Aln

yy

yyln

N

−−−−

−−−−

−−−−

−−−−

====

−−−−

−−−−

====

where : N = no. of theoretical stages or plates

*a

*b

ab

yy

yy

mV

LA

−−−−

−−−−

========A = absorption factor

y* = equilibrium concentration corresponding to xy* = equilibrium concentration corresponding to xy* = equilibrium concentration corresponding to xy* = equilibrium concentration corresponding to x


Analytical Solution – If OL and EC are both linear use

Absorption Factor or Kremser Equation

(a) based on the liquid phase (L phase)

*b

*a

ba

*bb

*aa

*bb

*aa

xx

xxln

xx

xxln

Sln

xx

xxln

N

−−−−

−−−−

−−−−

−−−−

====−−−−

−−−−

====

where : N = no. of theoretical stages or plates

*b

*a

ba

xx

xx

L

mV

A

LS

−−−−

−−−−============

S= stripping factor

x* = equilibrium concentration corresponding to yx* = equilibrium concentration corresponding to yx* = equilibrium concentration corresponding to yx* = equilibrium concentration corresponding to y

Minimum Absorbent Rate Minimum Absorbent Rate Minimum Absorbent Rate Minimum Absorbent Rate –––– LLLLminminminmin

Limiting Liquid to Gas RatioLimiting Liquid to Gas RatioLimiting Liquid to Gas RatioLimiting Liquid to Gas Ratio

Minimum Liquid to Gas Ratio for AbsorptionMinimum Liquid to Gas Ratio for AbsorptionMinimum Liquid to Gas Ratio for AbsorptionMinimum Liquid to Gas Ratio for Absorption


• For a given gas flow, a reduction in liquid flow decreases the

slope of the operating line

• As the upper end of the operating line (OL) is shifted in the

direction of the equilibrium line the concentration of the strong

liquor, xb or xn increases.

• The maximum possible liquor concentration and the minimum

liquid rate (Lmin or L’min) are obtained when the operating line

just touches the equilibrium diagram

• At Lmin or L’min an infinite no. of plates or an infinitely deep

packed section is necessary. At this condition the concentration

difference for mass transfer becomes zero at the bottom of the

tower


• In any actual tower the liquid rate must be greater than this

maximum to achieve the specified change in gas composition

Problem # 1 Problem # 1 Problem # 1 Problem # 1 –––– Plate AbsorberPlate AbsorberPlate AbsorberPlate Absorber

A tray tower is absorbing ethyl alcohol from an inert gas streamA tray tower is absorbing ethyl alcohol from an inert gas streamA tray tower is absorbing ethyl alcohol from an inert gas streamA tray tower is absorbing ethyl alcohol from an inert gas stream

using pure water at 303 K and 101.3 using pure water at 303 K and 101.3 using pure water at 303 K and 101.3 using pure water at 303 K and 101.3 kPakPakPakPa. The inert gas stream . The inert gas stream . The inert gas stream . The inert gas stream

flow rate is 100.0 kg/h and it contains 2.2 mol% alcohol. It is flow rate is 100.0 kg/h and it contains 2.2 mol% alcohol. It is flow rate is 100.0 kg/h and it contains 2.2 mol% alcohol. It is flow rate is 100.0 kg/h and it contains 2.2 mol% alcohol. It is

desired to recover 90% of the alcohol. The equilibrium desired to recover 90% of the alcohol. The equilibrium desired to recover 90% of the alcohol. The equilibrium desired to recover 90% of the alcohol. The equilibrium

relationship is y = relationship is y = relationship is y = relationship is y = mxmxmxmx = 0.68x for the dilute stream. Calculate = 0.68x for the dilute stream. Calculate = 0.68x for the dilute stream. Calculate = 0.68x for the dilute stream. Calculate

the number of trays graphically and analytically for an the number of trays graphically and analytically for an the number of trays graphically and analytically for an the number of trays graphically and analytically for an operaingoperaingoperaingoperaing

flow rate of 1.3 times the minimum.flow rate of 1.3 times the minimum.flow rate of 1.3 times the minimum.flow rate of 1.3 times the minimum.

Problem # 2 Problem # 2 Problem # 2 Problem # 2 –––– Plate AbsorberPlate AbsorberPlate AbsorberPlate Absorber

A mixture of 5% butane and 95% air is fed to a sieve-plate absorber containing eight ideal plates. The absorbing liquid is a heavy, nonvolatile oil

having a molecular weight of 250 and a specific gravity of 0.90. The absorption takes place at 1 atm, and 15oC. The butane is to be recovered to the extent of 95%. The vapor pressure of butane at 15oC is 1.92, and liquid butane has a density of 580 kg/m3

at 15oC. (a) Calculate the cubic meters of fresh absorbing oil per cubic meter of butane recovered. (b) Repeat, on the assumption that the total pressure is 3 atm and all other factors remain constant. Assume that Raoult’s Law and Dalton’s Law apply

Problem # 3 Problem # 3 Problem # 3 Problem # 3 ---- Plate AbsorberPlate AbsorberPlate AbsorberPlate Absorber

A relatively nonvolatile hydrocarbon oil contains 4.0 mol% propane and is being stripped by direct superheated steam in a stripping tray tower to reduce

the propane content to 0.2%. The temperature is held constant at 422 K by internal heating in the tower at 2.026 x 105 Pa pressure. A total of 11.42 kgmole direct steam is used for 300 kgmol of total entering liquid. The vapor-liquid equilibria can be represented by y = 25x, where y is mole fraction propane in the steam and x is the mole fraction propane in the oil. Steam can be considered an inert gas and will not condense. Plot the operating and equilibrium and determine the number of theoretical trays needed.

Notes on AbsorbersNotes on AbsorbersNotes on AbsorbersNotes on Absorbers• Note that the operating line for an absorber is above the equilibrium curve.

For a given solute concentration in the liquid, the solute concentration in the gas is always greater than the equilibrium value, which provides the driving force for the separation.

• The solute is transferred from the gas to the liquid in absorption. In distillation, we plotted the more volatile component, which was transferred from the liquid to the gas. In distillation, if we had plotted the less volatile component, which was transferred from the gas to the liquid, the OL would also lie above the equilibrium curve.

• Also note that the OL is linear. This results because of the form of the operating line where L’/V’ is a constant. L’ and V’ are based upon the nonvolatile solvent and insoluble carrier gas, respectively, which do not change.

• If we had used mole fractions and total gas and liquid rates, the OL would be curved because the total gas and liquid rates would change since we are removing the solute from the gas and absorbing it into the liquid.

• One could use mole fractions and the total gas and liquid streams in our calculations only if the solute is in low concentrations, say < 1%, in most systems. Don’t confuse this requirement with that for the use of Henry’s Law, which requires low solute concentrations, < 10%, to be valid.

Final Notes on StrippersFinal Notes on StrippersFinal Notes on StrippersFinal Notes on Strippers

• We use the same assumptions and mole ratio methods that we used for absorbers for the design of strippers.

• The OL will be the same as that used for absorbers.

• The difference, compared to an absorber, is that the equilibriumcurve will be below the operating line.

• This is analogous to the stripping section of a distillation column.

• Just as we stepped up from the bottom of a distillation column’s stripping section, we step up from the bottom of the stripper.

• Thus, one uses the same McCabe-Thiele algorithm method that we used for stepping up in distillation.

• In the algorithm, the equilibrium relationship is expressed as YEq = YEq(XEq), and the operating line is expressed in the form YOL = XOL(YOL).

• Otherwise, the design approach for strippers is the same as that for absorbers.

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TRAPHE3- Gas Absorption