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

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: 1. Introduction to nonlinear equilibrium y = m x y x y = f(x) y 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 y n+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 = f(x) y x y n+1 x0x0 y1y1

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. x y = f(x) y1y1 x0x0 y n+1 L 1 /G L 2 /G L 3 /G y

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 (x n *, y n+1 ). Therefore, the following equation is valid at that point: The slope of this operating line can be calculated: The slope corresponding to this line is the smallest possible slope, because the process cannot go beyond the equilibrium line. 2. Choosing an operating line x y = f(x) y1y1 x0x0 y n+1 L min /G In practice, the slope for the operating line is taken as 1 to 2 times the minimum slope: y

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. 3. Graphical construction y = f(x) y x y n+1 x0x0 y1y1 x1x1 y2y2 y3y3 x2x2 x3x3 In the beginning, we had the following unknowns: L, n and x n. 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 x n can be read out of the graph. In our example the stage number is at least n = 3. L /G 1 2 3