Absorption of a Concentrated Vapor

Inserting the values of m, Ky, G, and L into Eq. 10.3-12 gives the desired result. 

In this section, we want to extend the preceding analysis to the case of a concentrated vapor. As before, we plan to accomplish this absorption using a packed tower. As before, we must decide on an appropriate tower packing and on liquid and gas fluxes that will avoid flooding. As before, we depend on a variety of mole balances, though now for concentrated solutions. 

Before we develop these new mass balances, we can benefit by looking at our analysis for a dilute vapor in a somewhat different way. This analysis depended on three key equations. The first key equation, the operating line, came from a mole balance for solute in both gas and liquid. The second key equation, the equilibrium line, gave the concentration which would exist if the gas were in equilibrium with the liquid. The third key equation involved the rate of mass transfer and was a mass balance written on only one phase, which in our case was the gas. 

We could represent the first and second key equations graphically, as shown in Fig. 10.4-1. The operating line in this figure plots y vs. x and the equilibrium line plots y vs. X. Thus we can read off values of the driving force (y - y ) and integrate the rate equation. Of course, in the dilute case, all the equations are linear so integration is easily analytical. 

For concentrated absorption, the analysis depends on the same three key equations. However, because a lot of solute is transferred, the gas flux G gets smaller as the gas flows up the column, and the liquid flux L gets larger as it flows down the column. As a result. 

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