Absorption design with real stages Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik

Up to now, we always treated a case where the equilibrium is reached in every stage. However, this assumption is often not correct. Therefore we need to use an efficiency to describe the non-ideal behaviour. The Murphree efficiency for the gas phase can be defined for every stage i: 1. Murphree Efficiency The Murphree efficiency compares the actually absorbed amount in stage i (y i+1 – y i ) to the maximum possible amount absorbed in stage i if the equilibrium was reached (y i+1 – y i * ). In the equation, the M in E i MG stands for Murphree, the G stands for gas phase. We assume, that the Efficiency is constant during the whole column and doesn’t depend on the stage i. Therefore we get:

For the operating line we need the mass balance for the column from stage 1 to stage i: The equation for the equilibrium is our third equation to solve the problem. Please note, that the equation is for y i * and not for y i. Eq. line y x First we draw the equilibrium line and the operating line into the diagram: Op. line Now we need to draw a third line using the Murphree efficiency. This line is drawn between equilibrium and operating line at a fractional vertical distance from the equilibrium line equal to the Murphree gas efficiency. A B C Thus the value of E is is the ratio of the length of lines, AB/AC.

y = m x y x y n+1 xoxo y1y1 x1x1 y2y2 y3y3 y4y4 x2x2 x3x3 x4x4 L /G 2. The stage-by-stage construction Op. line Since the stage composition line represents the real effluent compositions from the trays, it is used instead of the equilibrium line to complete the tray construction. Using this line describing the stage compositions of the column, we can apply the usual stage-by- stage construction, switching from operating line to the stage composition line. Eq. line

3. Mathematical solution Now we want to calculate the number of stages instead of a graphical solution. That’s why we take the three equations seen before: If we combine the second and the third equation, we get: Combining the last two equations: Multiplying the first equation with E/A and knowing that A = L / (mG), we get:

Solving for y i+1 : We can introduce two new variables A’ and B’: We therefore get the following equation: If we set the Efficiency E = 1, we get for A’ and B’:

As a next step, we would like to calculate the number of real stages. This calculation needs some mathematical effort. At first, we take the result from the last slide: However, we need an equation for i instead of i+1: This equation is a First Order Difference Equation. For a general equation the solution has the following form: If we look into our equation we can identify the coefficients as: And using those coefficients in the general solution: Furthermore we need equation (3a) for i=1:

Combining the last two equations, we get: The fractional absorption α can now be calculated. After some mathematical transformations, we finally get: The number of real stages n’ is then: The Kremser equation for the ideal case where equilibrium is reached on every stage: Dividing the second equation by the first one, we get the overall efficiency E 0 : It is important to note that the overall efficiency E 0 and the Murphree Efficiency E are not the same.