Single-Stage Extractor

If point M in the previous example lies in the 1-phase region of the phase diagram, then we are done because we know the composition xc of the single phase leaving the mixer. If the point M lies in the two-phase region and the two phases are allowed to reach equilibrium and separate in the settler, we will have (not one, but) two product streams. 

Let s re-label the two input and two output streams as in the figure at right. In a typical extraction problem, stream L is an aqueous solution containing a solute (component A) we are trying to extract by contact with an organic solvent (component C). So let s assume that the second feed stream (labelled V2) is pure solvent. Let s try to determine the composition of the two product streams. 

Solution Here is a receipe for a graphical solution using a rectangular diagram (see figure at right). 

Step 3 Find the tie line which passes through M 

Step 5 The flowrates Lj and Fj can be determined using the reverse of the procedure for determining the mixing point M in Example 1 

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The optimum recoverability of solute by a single-stage extractor is determined based on the K value and the E/R ratio. To increase the recoverability further, several extractors can be connected crosscurrently or countercurrently. 

TABLE 2. Single-Stage Extractor Mass Balance. 

If we are satisfied with the amount of solute we have been able to extract from the solution in the single-stage extractor, then we are done. If we desire to extract more of the solute (i.e. we wish to achieve a smaller value of x j) then multiple stages might be used. Let s specify an arbitrarily small value for xj, the concentration of solute remaining after N stages. 

A single-stage extractor is used to extract component E from a feed stream. The feed and solvent stream compositions and flow rates are as follows ... 

A solvent is used to recover component E from a hquid feed containing components R and E by extraction in a single-stage extractor. The compositions and flow rates of the feed and solvent are given below ... 

A pure solvent 5 is used to remove component E from a binary liquid solution containing 30% mole E and 70% mole R. In a single-stage extractor how many kmol solvent per 100 kmol feed are required to bring the concentration of E in the raffinate to 3% mole What are the resulting rates and compositions of the raffinate and extract Use liquid-liquid equilibrium data from Figure 11.2. 

It is required to recover 93% of component E from a binary mixture of components E and R by treating with a solvent in a single-stage extractor. What is the required solvent rate per 100 kmol of feed, and what are the product rates and compositions Assume the liquid-liquid equilibrium... 

As the volumetric mass-transfer coefficient of liquid-liquid systems is very large, the module can be used as a single-stage extractor but requires a separator. The modules can be configured to form a countercurrent extractor using suitable phase separators. 

Phase path Single-stage extractor data ... 

Figure 5.19 Early designs for single-stage extractors ...

A feed (2.78 kg/sec) of 30 percent acetone and 70 percent methyl isobutyl ketone is mixed with pure water (1.39 kg/sec). Find the compositions and flow rates, leaving a single-stage extractor (see Figure 12-6). 

Fig. A9. Response ofxA to a step change in Fb of 25% for single-stage extractor.

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