Absorption process design

1. Absorption process design Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik

2. y1< y spec G, y1 L, x0 T, p G, yn+1 L, xn 1. Variables An absorption problem is usually presented as follows. There is a polluted gas stream coming out from a process. The pollutant must be recovered in order to clean the gas or because it's a valuable compound. In general, the conditions of the gas stream are known, the compositions can be measured, and the gas flow rate can be set. The temperature and the pressure at which the process takes place must be chosen. The thermodynamics of the process at different conditions must be studied in order to decide the most convenient. The solvent can be used pure or can come from a recycle, and thus contain some pollutant. The initial composition of the solvent (xo) must be obtained by analytical methods, or can be set by mixing pure and recycled solvent. Process

3. Once temperature, pressure and initial composition of the solvent are set we can consider them as data. This is a list of known variables: Data and set values Specifications Unknowns G L y1 x0 Temperature, T Gas final composition, y1 Solvent flow rate, L Pressure, p Number of stages, n Gas flow rate, G Solvent final composition, xn Gas initial composition, yn+1 Solvent initial composition, x0 T p Equilibrium data, y=m x Y = m x n yn+1 xn G L

4. y yn+1 y = m x x0 x 2. Diagram We have already seen in the counter-current cascade of ideal stages that a graphical representation can help to visualize and solve the problem. Considering the initial compositions of the gas and solvent flows and the specification made for the gas outlet composition, we can draw in the diagram: y1

5. yn+1 L1 /G L2 /G y = f(x) L3 /G y1 x x0 Now, there are infinite number of possible segments passing by the point (xo, y1) and ending in a point with y coordinate yn+1. We need to choose one of those segments as operating line. The operating line is the equation of all those lines: The slope of the operating line depends on the solvent flow-rate (L/G). When one of the lines is chosen as operating line, then the slope is set and thus, the required solvent flow-rate can be calculated.

6. 3. Choosing an operating line There is a special operating line among the infinite number we can choose. This is the one that touches the equilibrium line at the point (yn+1/m, yn+1) yn+1 The mass exchange is maximum because the equilibrium is reached, but the required number of stages is infinite. y = m x Lmin /G y1 x x0 The slope corresponding to this line is the smallest possible slope, because the process cannot go beyond he equilibrium line. The minimum slope can also be written in terms of the fractional absorption for a linear equilibrium: In the practice, the slope for the operating line is taken as 1 to 2 times the minimum slope:

7. y y = m x x 4. Graphical construction Once the operating line is set, we can proceed with the construction seen in the counter-current stage configuration. This gives us the number of stages, n. yn+1 y4 L/G y3 y2 y1 xo x1 x2 x3 x4 We already know all the initial unknowns: L, n and xn.

8. 5. Kremser equation The next two concepts are very important. You can review them in 4.3 Ideal equilibrium stage 1. Absorption factor 2. Fraction of absorption As we saw before, in the case of a counter-current configuration, the fraction of absorption can also be written in terms of the absorption factor, A. This gives us the Kremser equation for the case in which A is not equal to 1: Therefore, the number of stages required, n, can be calculated using the two forms: the fraction of absorption definition and the Kremser equation. The number of stages obtained by this method should be equal to the one obtained by the graphical construction.

9. Have a look to the design Flowsheet