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, more complex becomes the problem. Also, only soluble components in the liquid will separate.
The inverse situation in which a gas dilute in a liquid passes to a gas stream is called desorption or stripping. There is in fact, an equipment called strip, in the from of a column, in which desorption occurs.
Important assumptions
Most situations can be easily analyzed when some key assumptions simplify the physics. For example, that during the process, the temperature remains constant. This could be a reasonable assumption since no external heating is provided into 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 asumption. When no isothermal conditions are met, different mathematical tools need to be considered.
 |
| Fig. 01 An schematics of an absorption equipment. $L$ stands for the liquid phase and $G$ for the gas mixture. The $x$ is used to indicate 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 important factor to consider is the soutbility of the gas component into the liquid. Once the molecules of the gas components are disolved into the liquid there is also a tendency of these molecules for returning to the gas mixture. This means that the vapor pressure is key.
 |
| Fig. 02 A simplified sketch of gas molecules trapped into 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 component, $p$ the vapor pressure and $x$ the fraction. Every component exerts a different partial pressure. The summation of all partial pressures is thetotal 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 useful for cases in which no experimental data is available.
Solubility of the gas components
Solubility refers to the capacity of transfer of gas molecules into the liquid phase, so that 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 solvent, some features need to be watch closely,
- its volatility,
- its corrosion,
- its viscosity,
- and cost.
One common solvent is water due to its universal properties but in chemical processes there can be a situation in which other options 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 soluility for NH$_3$ is presented in which can be seen 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 cause different partial pressures so that different data tables for solubility are required. The book Perrys Chemical Engineering Handbook is a good source of this kind of data.
In the equilibrium reached for the solution of a gas component into a liquid phase different partial pressures shall be exerted too.
Whe the equilibrium partial pressures are large for small concentrations of the gas component in the liquid, it is said that there is low solublity. For the other case, high solubility would imply high concentration of the gas component in the liquid phase along with low partial pressures at equilibrium.
 |
| Fig. 03 General representation of the idea of the different pressures exerted at both side 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 gradient pressure. Of course, the solubility is key but without gradient pressure, no dilution of gas component into the liquid phse may occur.
One important fact not seen from the table of solubility data is that as the pressure exerted in the gas increases, the fraction of the gas component in the mixture dicreases. 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