This is unit operation of chemical engineering for separation processes. Separation of one or more gas components from a gas mixture by dilution in a liquid.
Of course, the more gas components one desires to separate from the mixture, the more complex the problem becomes. Also, only soluble components in the liquid will separate.
The inverse process in which a gas diluted in a liquid is transferred to a gas stream is called desorption or stripping. There is, in fact, an equipment called a strip, in the form of a column, in which desorption occurs.
Important assumptions
Most situations can be easily analyzed when some key assumptions simplify the physics. For example, during the process, the temperature remains constant. This could be a reasonable assumption since no external heating is provided to the equipment.
However, either the gas or the liquid can be hot at the inlet, so that its temperature could not be the same at the outlet. You must be careful with this assumption. When isothermal conditions are not met, different mathematical tools must be considered.
![]() |
| Fig. 01 A schematic of an absorption equipment. $L$ stands for the liquid phase and $G$ for the gas mixture. The $x$ indicates the fraction of the gas component to be absorbed. |
There are several ways of carrying out a gas-liquid operation of this kind:
- in bubble columns,
- in agitated vessels,
- in tray towers
- in packed beds.
Mechanism of absorption
One crucial factor to consider is the solubility of the gas component in the liquid. Once the gas components' molecules are dissolved in the liquid, they also tend to return to the gas mixture. This means that the vapor pressure is key.
![]() |
| Fig. 02 A simplified sketch of gas molecules trapped in a liquid phase. Notice that not all soluble molecules are passed into the liquid. |
In Fig. 02, only one component is soluble in the liquid while the others remain in the mixture. The transfer of mass ceases once equilibrium has been reached. In the gas mixture, the pressure exerted by the molecules is a fraction of the vapor pressure. This is,
$p^*=p\,x$ Eq. (01)
where $p^*$ is the partial pressure exerted by the molecules of a given component, $p$ is the vapor pressure, and $x$ is the fraction. Every component exerts a different partial pressure. The summation of all partial pressures is the total pressure (not the vapor pressure). Equation (01) is valid for ideal solutions at a fixed temperature (the isothermal case).
Of course, there are no ideal solutions, but Eq. (01) is helpful in cases where no experimental data are available.
Solubility of the gas components
Solubility refers to the capacity of transfer of gas molecules into the liquid phase, so it depends on:
- the nature of both the gas and the liquid phase,
- the temperature,
- the vapor pressure of the gas component and
- the already existing concentration of dilute gas molecules in the liquid.
For the liquid working as a solvent, some features need to be watched closely,
- its volatility,
- its corrosion,
- its viscosity,
- and cost.
One common solvent is water due to its universal properties, but in chemical processes, other options may need to be considered, such as alcohols.
The solubility of a substance is usually presented as data tables or as plots of the partial pressure versus concentration. For example, in the table below, the solubility of NH$_3$ is given, showing that the partial pressure changes with temperature and concentration.
| Weight NH$_3$ per 100 weights H$_2$O |
Partial pressure of NH$_3$, mm Hg | |||||||
|---|---|---|---|---|---|---|---|---|
| 0 °C | 10 C | 20 °C | 25 °C | 30 °C | 40 °C | 50 °C | 60 °C | |
| 100 | 947 | |||||||
| 90 | 785 | |||||||
| 80 | 636 | 987 | 1450 | 3300 | ||||
| 70 | 500 | 780 | 1170 | 2760 | ||||
| 60 | 380 | 600 | 945 | 2130 | ||||
| 50 | 275 | 439 | 686 | 1520 | ||||
| 40 | 190 | 301 | 470 | 719 | 1065 | |||
| 30 | 119 | 190 | 298 | 454 | 692 | |||
| 25 | 89.5 | 144 | 227 | 352 | 534 | 825 | ||
| 20 | 64 | 103.5 | 166 | 260 | 395 | 596 | 834 | |
| 15 | 42.7 | 70.1 | 114 | 179 | 273 | 405 | 583 | |
| 10 | 25.1 | 41.8 | 69.6 | 110 | 167 | 247 | 361 | |
| 7.5 | 17.7 | 29.9 | 50.0 | 79.7 | 120 | 179 | 261 | |
| 5 | 11.2 | 19.1 | 31.7 | 51 | 76.5 | 115 | 165 | |
| 4 | 16.1 | 24.9 | 40.1 | 60.8 | 91.1 | 129.2 | ||
| 3 | 11.3 | 18.2 | 23.5 | 29.6 | 45 | 67.1 | 94.3 | |
| 2.5 | 15.0 | 19.4 | 24.4 | (37.6)* | (55.7) | 77.0 | ||
| 2 | 12.0 | 15.3 | 19.3 | (30.0) | (44.5) | 61.0 | ||
| 1.6 | 12.0 | 15.3 | (24.1) | (35.5) | 48.7 | |||
| 1.2 | 9.1 | 11.5 | (18.3) | (26.7) | 36.3 | |||
| 1.0 | 7.4 | (15.4) | (22.2) | 30.2 | ||||
| 0.5 | 3.4 | |||||||
Different gases will yield different partial pressures, so different solubility data tables are required. The book Perry's Chemical Engineering Handbook is a good source of this kind of data.
In the equilibrium reached when a gas component dissolves in a liquid phase, different partial pressures shall be exerted.
When the equilibrium partial pressure of the gas component in the liquid is significant at low gas concentrations, the solubility is low. In the other case, high solubility would imply a high concentration of the gas component in the liquid phase, along with low equilibrium partial pressures.
![]() |
| Fig. 03 General representation of the idea of the different pressures exerted at both sides of the interface gas-liquid as the driving force of dilution of a gas component into a liquid phase. |
As you may have already imagined, the driving force at the heart of the gas absorption is a pressure gradient. Of course, solubility is key, but without a gradient pressure, no dilution of the gas component into the liquid phase can occur.
One crucial fact not apparent from the solubility data table is that, as gas-phase pressure increases, the fraction of the gas component in the mixture decreases. This is not just a curious fact from the data but something useful in practice.
This is the end of the post. I hope you find it useful.
Other stuff of interest
- LE01 - AC and DC voltage measurement and continuity test
- LE 02 - Start and stop push button installation 24V DC
- LE 03 - Turn on/off an 24V DC pilot light with a push button
- LE 04 - Latch contact with encapsulated relay for turning on/off an AC bulb light
- LE 05 - Emergency stop button installation
- About PID controllers
- Ways to control a process
- About pilot lights
- Solving the Colebrook equation
- Example #01: single stage chemical evaporator
- Example #02: single stage process plant evaporator
- Example #03: single stage chemical evaporator
- Example #04: triple effect chemical evaporator
==========
Ildebrando.



No comments:
Post a Comment