11.2. SINGLE-STAGE EQUILIBRIUM CONTACT FOR VAPOR–LIQUID SYSTEM

If a vapor–liquid system is being considered, where the stream V2 is a vapor and L0 is a liquid, and the two streams are brought into contact in a single equilibrium stage which is quite similar to Fig. 10.3-1, the boiling point or the x-y equilibrium diagram must be used, because an equilibrium relation similar to Henry's law is not available. Since we are considering only two components, A and B, only Eqs. (10.3-1) and (10.3-2) are used for the material balances. If sensible heat effects are small and the latent heats of both compounds are the same, then when 1 mol of A condenses, 1 mol of B must vaporize. Hence, the total moles of vapor V2 entering will equal V1 leaving. Also, moles L0 = L1. This case is called one of constant molal overflow. An example is the benzene–toluene system.

EXAMPLE 11.2-1. Equilibrium Contact of Vapor–Liquid Mixture

A vapor at the dew point and 101.32 kPa containing a mole fraction of 0.40 benzene (A) and 0.60 toluene (B) and 100 kg mol total is brought into contact with 110 kg mol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molal overflow. Calculate the amounts and compositions of the exit streams.

Solution: The process flow diagram is the same as in Fig. 10.3-1. The given values are V2 = 100 kg mol, yA2 = 0.40, L0 = 110 kg mol, and xA0 = 0.30. For constant molal overflow, V2 = V1 and L0 = L1. Substituting into Eq. (10.3-2) to make a material balance on component A,

Equation 10.3-2


Equation 11.2-1


To solve Eq. (11.2-1), the equilibrium relation between yA1 and xA1 in Fig. 11.1-1 must be used. This is by trial and error, since an analytical expression is not available that relates yA and xA.

First, we assume that xA1 = 0.20 and substitute into Eq. (11.2-1) to solve for yA1:


Solving, yA1 = 0.51. The equilibrium relations for benzene–toluene are plotted in Fig. 11.2-1. It is evident that yA1 = 0.51 and xA1 = 0.20 do not fall on the curve. This point is plotted on the graph. Next, assuming that xA1 = 0.40 and solving, yA1 = 0.29. This point is also plotted in Fig. 11.2-1. Assuming that xA1 = 0.30, yA1 = 0.40. A straight line is drawn between these three points which represents Eq. (11.2-1). At the intersection of this line with the equilibrium curve, yA1 = 0.455 and xA1 = 0.25, which agree with Eq. (11.2-1).

Figure 11.2-1. Solution to Example 11.2-1.



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