1. Gas absorption and stripping
Prepared by
Bikash Kumar Mahato
Assistant Professor
IOE, Pulchowk Campus
2. Absorption
• Gas absorption is dissolving one or more components of the gas into a
liquid system and to provide a solution.
• Examples:
• Gas from by-product coke ovens is washed with water to remove ammonia
and again with an oil to remove benzene and toluene vapor.
• Objectionable hydrogen sulfide gases is removed from natural hydrocarbon
gases by washing with various alkaline solution.
• Sometimes the absorber gas undergoes a chemical reaction in the liquid phase.
• Removal of H2S from natural gas using alkanol amine solutions.
3. Conti.
• Absorption occurs via:
• Chemical absorption- A reactive process.
Chemical reaction between CO2 and amine solution.
• Physical absorption- A non-reactive process.
• Oxygen present in air dissolves in water..
• Absorption require mass transfer of a substance from gas phase to the liquid
phase.
• When mass transfer occurs in the opposite direction (i.e. from the liquid to
the gas) is called desorption or stripping.
• Very often the loaded solvent is regenerated for further use via desorption
process.
4. Equilibrium curve and Equilibrium Solubility
• The rate at which a gaseous constituent of a mixture will dissolve in an absorbent
liquid depends on
• The departure from equilibrium which exist.
• So, the departure from the equilibrium states will define the rate of diffusion or
rate of absorption process or the rate at which the gas will dissolve.
• Therefore, it is necessary to consider the equilibrium characteristics of the gas
liquid systems.
For two component system:
• If a quantity of single gas and a relatively nonvolatile liquid are brought to equilibrium, then:
• The resulting dissolved gas in the liquid is said to be the solubility of gas at the prevalling temperature
and pressure.
• At a fixed temp. the solubility concentration will increase with pressure in the manner as
shown in fig.
5. Conti.
• Different gases and liquids yield
separate solubility curves.
• If the equilibrium pressure for a
given liquid concentration is
high,(curve B) the gas is
relatively insoluble in liquid.
• While if it is low as for curve C,
the solubility is high.
• But all these are relative matters
only.
6. Conti.
• It is possible to produced any ultimate gas concentration in the liquid if
sufficient pressure is applied.
• The liquefied form of the gas is completely soluble in the liquid. So, if we
increase the pressure in the gas phase and if we make it liquid and it is
completely miscible with the liquid solvent. So, we can get mole fractions
to 1.
• The solubility of any gas is influenced by the temperature in a manner van’t
Hoffs law of mobile equilibrium.
• If the temperature of a system at equilibrium is raised, that changes will
occur which will absorb heat.
• Usually the solubility of the gas decreases with increasing temperature.
7. Selection of Solvent for Absorption
• The principal purpose of the absorption operation is to produce a
specific solution.
• E.g: In the production of HCl acid, the solvent is specified by the nature of the
product.
• But, if the principle purpose is to remove some constituents from the
gas, some choice frequently possible. Water is of course, the cheapest
and most plentiful solvent but the following properties are important
consideration.
8. Conti.
• Solubility:
• the solvent in which the solubility of the solute is very high is the prefer
solvent but is should not dissolve the carrier gas.
• It should be selective to the particular solute.
• a chemical reactions of the solvent with the solute will frequently result in high
solubility, if the solvent is required to be reused the reaction must be
reversible.
• E.g: H2S can be removed from mixture using alkanolamine solution and easily stripped at
high temperature.
• But, caustic soda also removed H2S excellently but not release it in a stripping operation.
9. Conti.
• Volatility: A good solvent should have a low volatility or vapor pressure so
that the solvent loss with the carrier gas due to vaporization remain small.
• Cost: A solvent should not be very expensive so that the vaporization and
other losses are not significant in terms of economy.
• Corrosiveness: Use of a non-corrosive or less corrosive solvent reduces the
material cost of the equipment as well as its maintenance cost.
• Viscosity: The low viscosity is preferred so that it flows easily. Low
pressure drop on pumping improve flooding and good heat transfer
characteristics.
• Hazardous and toxicity: A solvent should not be hazard as toxic and it
should be preferably nonflammable.
10. Selection of Stripping Medium
• For desorption or stripping of a solute from a loaded solvent a carrier gas
which is the stripping medium has to be brought in contact with the
solution.
• Steam is a common stripping medium. Use of steam is very convenient
because the stripped solute can be easily recovered by condensing the steam
leaving the stripping tower.
• If the solute is not soluble in water it can be separated from the condensate
by decantation.
• E.g. in case of stripping of a hydrocarbon from an absorption oil.
• On the other hand, if the solute is soluble in water, it can be recovered from
the condensate by distillation.
• Sometimes nitrogen gas is also used as a carrier for stripping.
12. Conti.
• A suitable solvent has to be selected for the absorption process and the equilibrium data at the
chosen rating conditions that is temperature and pressure conditions have to be obtained from the
literature or from any other source.
• Now, how to determine the required solvent flow rate for a given absorption problem?
• The overall material balance for above fig:
GS (Y1-Y2) = LS(X1-X2) (1)
so, the operating line eq. is
(LS/GS)=(Y1-Y2)/(X1-X2) (2)
• There still remain two unknown quantities i.e. LS and X1.
• If one of these quantities (say LS ) is fixed the other (X1.) can be calculated from the material
balance eq. given earlier (and vice versa).
• Now, the operating line can then be drawn by locating and joining the terminal points (X1, Y1) and
(X2, Y2 ) on X-Y plane.
• The slope of the operating line that is LS/GS. Two types of equilibrium lines and a set of operating
lines of different slopes are shown in fig. a and b.
14. Conti.
• If LS is reduced (i.e. a lower liquid rate is used), the slop decreases and the operating line PQ have
one fixed terminal at the point (X2, Y2), comes closer to the equilibrium line (see fig. a).
• As a result the driving force will decrease and a large height of packing, or a large number of
plates(if tray tower is used), has to be provided to achieve the desired degree separation.
• But we cannot keep on reducing the liquid flow rate indefinitely. As the liquid rate is gradually
reduced the orientation of the operating line will change eventually a situation will arise when the
operating line meets the equilibrium line at a certain point.
• The driving force at this point is zero. The corresponding liquid rate is the theoretical minimum
liquid rate.
• On further reduction of the liquid rate beyond the minimum the operating line will cross the
equilibrium line which means that absorption of the gas occurs in one section of the column and
desorption in the other (see PQ’’’ in fig. b).
• This is physically impossible . So, a column cannot operate, even theoretically, at a liquid rate less
than the minimum as defined above.
• So, the slope of the operating line at minimum liquid rate is (Ls)min /Gs
15. Conti.
• Since, Gs (the gas flow rate of the carrier gas) is known (Ls)min can be calculated from the slope of
the PQ’’.
• The corresponding concentration of the exit liquid is Xmax, that is the maximum liquid
concentration we can achieve.
• The driving force is 0 at point M(i.e. the point of tangency for the particular shape of the
equilibrium line) which is called the pinch point.
• What result do you expect if you want to find out the number of ideal plates required for the
desired separation using the minimum liquid rate?
• In order to address this question we redraw the equilibrium line and operating line PQ’’ for the
minimum liquid in figure c.
• To find out the number of ideal stages construct the steps between the equilibrium line and the
operating line starting from the from either end.
• Let us starts at point P. As we approach Q’’ which is the pinch point the driving force tends to 0 and
it becomes impossible to reach the point Q’’ by drawing a finite number of steps.
• So, theoretically it will need an infinite number of plates or an infinite height of packing if packed
tower is used to achieve the desired separation using the minimum liquid rate.
17. Problem
• In a petrochemical plant, a gas containing 4% cyclo-hexane and 96%
inerts has to be treated with a non-volatile absorption oil in a packed
tower. It is required to remove 98% of the cyclo-hexane of the feed
gas. The feed solvent is free from cyclo-hexane. If the feed gas rate is
80 kmol per hour, calculate the minimum solvent rate. The equilibrium
relation is given as
𝑌 =
0.2𝑋
1+0.8𝑋
