Absorption design with nonlinear equilibrium Prof Dr Marco

Absorption design with nonlinear equilibrium Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik

1. Introduction to nonlinear equilibrium Up to now, we always treated the case of linear equilibrium. In the linear case, the equilibrium can be described with the simple equation y = mx. However, we don’t always have such an easy relation between the gas and the liquid phase. The equilibrium relation can only be described by the general formula: y y y=mx y = f(x) x x

When y = mx does not apply, then Kremser`s equation is useless. However, the graphical stage-by-stage construction (and the corresponding calculation) can still be applied. Considering the initial compositions of the gas yn+1 and of the solvent x 0 and the specification made for the gas outlet composition y 1, we can draw in the diagram: y yn+1 y = f(x) y 1 x 0 x

The operating line is not influenced by a nonlinear equilibrium. Therefore, the same operating line equation is still valid. The slope of the operating line depends on the solvent flow-rate L. 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. y yn+1 L 1 /G L /G 2 L 3 /G y = f(x) y 1 x 0 x

2. 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 (xn*, yn+1). Therefore, the following equation is valid at that point: y yn+1 Lmin /G The slope corresponding to this line is the smallest possible slope, because the process cannot go beyond the equilibrium line. The slope of this operating line can be calculated: y = f(x) y 1 x 0 In practice, the slope for the operating line is taken as 1 to 2 times the minimum slope: x

3. Graphical construction Once the operating line is set, we can proceed with the construction seen in the linear case. The usual stage-by-stage construction is adopted, switching between equilibrium line and operating line. This gives us the number of stages, n. y yn+1 y 3 3 L /G y = f(x) y 2 y 1 2 1 x 0 x 1 x 2 x 3 x In the beginning, we had the following unknowns: L, n and xn. The liquid flowrate L can be calculated with the slope of the operating line L/G, knowing the gas flow rate G. The number of stages n and the mole fraction xn can be read out of the graph. In our example the stage number is at least n = 3.