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Chapter 6: Problem 1

Two-Stage Countercurrent Extraction In a two-stage countercurrent extraction of a pharmaceutical product, what is the relationship between the feed concentration and the final raffinate concentration in terms of the extraction factor \(E\) if both stages are at equilibrium? For a partition coefficient \(K\) of \(5.0\) and a solvent-to-feed ratio \((S / F)\) of \(0.5\), what will be the ratio of the raffinate concentration to the feed concentration?

Short Answer

Expert verified
The ratio of raffinate concentration to the feed concentration is \(1/(1 + 5.0*0.5)^2\).

Step by step solution

01

Understanding countercurrent extraction

Countercurrent extraction is a separation process where a solution or a solute is passed through a solvent in such a way that the solute is concentrated in one of them. In simple terms, the solute moves from a phase where it is more concentrated to another where it is less concentrated.
02

Deriving the relationship between feed and raffinate concentrations

For a single stage extraction, the raffinate concentration \(C_R\) can be related to the feed concentration \(C_F\) as \(C_R = C_F / (1 + ES)\) where E is the extraction factor, S is the solvent-to-feed ratio. When another extraction stage is applied, the formula becomes \[ C_{R2} = C_{R1} / (1+ES) \], which in turn gives \[ C_{R2} = C_{F} / (1 + ES)^2 \] . The exponent denotes that two stages were used.
03

Calculating the ratio of raffinate concentration to the feed concentration

Applying given values of E=5.0 and S=0.5 in the above relation, we have \(C_{R2} / C_{F} = 1 / (1 + 5.0*0.5)^{2}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Feed Concentration
When discussing countercurrent extraction in chemical engineering, feed concentration, denoted as \( C_F \), is a critical starting point. It represents the initial amount of solute present in the phase that is being extracted from. In the context of a two-stage countercurrent extraction system, it is paramount to understand how this initial concentration influences the process's effectiveness.

As the solute interacts with the solvent, it transfers according to its solubility and affinity towards the solvent, impacting the final concentrate and raffinate concentrations significantly. Understanding the relationship between feed and raffinate concentrations allows us to optimize the process to achieve the desired level of extraction.
Raffinate Concentration
In contrast to feed concentration, raffinate concentration, represented as \( C_R \), refers to the concentration of solute left in the feed phase after the extraction process has taken place. It is the residual concentration that did not transfer to the solvent phase.

In the exercise provided, the final raffinate concentration after two stages is especially important as it reflects the efficiency of the extraction process. By understanding the relationship between the feed concentration and the raffinate concentration, we can determine how much of the solute was removed, which is crucial in industries where purity levels are strictly regulated.
Extraction Factor
The extraction factor, denoted by \( E \), is a numerical expression of the extraction efficiency. It indicates the relative ability of the solvent to extract the solute from the feed. In a mathematical context, it's a part of the formula that relates feed concentration to raffinate concentration after each extraction stage.

Higher extraction factors signify a more efficient extraction process, assuming the other variables remain constant. Thus, this factor is pivotal when designing and analyzing multi-stage extraction systems. In practical terms, by manipulating this factor, you can directly control how much solute is removed per stage, which is essential for achieving specific concentration targets.
Partition Coefficient
The partition coefficient, represented by \( K \), is a ratio of concentrations of a compound in the two phases of an extraction system at equilibrium. Specifically, it is the concentration of solute in the solvent phase to that in the feed phase. A high partition coefficient indicates a preference of the solute for the solvent phase over the feed phase.

In the context of the given exercise, with a partition coefficient of \( K = 5.0 \), the solute is significantly more concentrated in the solvent than in the feed, favoring extraction. The partition coefficient helps predict the distribution of the solute between phases and is vital for calculating the necessary solvent volume for efficient extraction.
Solvent-to-Feed Ratio
The solvent-to-feed ratio, often expressed as \( S/F \), is a measure of the proportion of solvent used relative to the feed material. Adjusting this ratio is a common method of controlling the operation of the extraction process.

A higher solvent-to-feed ratio can lead to a greater transfer of solute into the solvent, thus increasing the extraction efficiency. Conversely, a lower ratio might not be as effective but could be more economical or required for process constraints. In the given problem, a solvent-to-feed ratio of \( 0.5 \) allows for a concise prediction of how it affects the overall extraction efficiency, especially when combined with the other variables to determine the final concentration ratios.

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Most popular questions from this chapter

Required Solvent-to-Feed Ratio in Countercurrent Extraction In a countercurrent, equilibrium staged extractor with four equilibrium stages, determine the necessary ratio of extract to feed \((S / F)\) to purify bioproduct A to \(90 \%\) purity from contaminant B. The feed contains two components, \(60 \% \mathrm{~A}\) and \(40 \% \mathrm{~B}\), and the partition coefficients are \(K_{\mathrm{A}}=13.0\) and \(K_{\mathrm{B}}=1.0\).

Purification of an Antibiotic by Countercurrent Extraction A countercurrent extraction unit with three equilibrium stages is used for the separation of a desired antibiotic (partition coefficient \(=6.0\) ) from a major contaminant (partition coefficient \(=1.0\) ) in an aqueous feed stream. The feed (or raffinate) and extract phase flow rates are equal. What fraction of each is discarded from the raffinate? Assuming that the feed contains only these two components at an antibiotic-to- contaminant ratio of \(3: 1\), what is the purity of the antibiotic in the exit extract from stage \(3 ?\) What do you conclude about the effectiveness of this extraction as a means of purification of the antibiotic?

Scale-up of Pilot Plant Tests of a Reciprocating-Plate Extraction Column The pilot plant data in Table P6.9 are for the extraction of an antibiotic from whole \begin{tabular}{lccc} & \multicolumn{2}{c}{ Flow rates \((\mathbf{m l} / \mathbf{m i n})\)} & \\ \cline { 2 - 4 } Run number & Broth & Chloroform & Antibiotic concentration in raffinate (mg/liter) \\ \hline 1 & 45 & 135 & 2 \\ 2 & \(67.5\) & 135 & 3 \\ 3 & 125 & 135 & 30 \\ 4 & 80 & 120 & 5 \\ 5 & 100 & 150 & 7 \\ 6 & 120 & 180 & 9 \\ 7 & 150 & 225 & 25 \end{tabular} fermentation broth using the solvent chloroform in a reciprocating-plate extraction column. The concentration of the antibiotic was \(1.4 \mathrm{~g} / \mathrm{liter}\) in the broth. The column had a diameter of \(2.54 \mathrm{~cm}\), and the height of the plates was \(3.05 \mathrm{~m}\). The partition coefficient \(K\) for the antibiotic is known to be \(2.68\). For each pilot run, determine the diameter and height of the plates for the plant column that would be required for processing 50,000 liters of broth in \(12 \mathrm{~h}\) to give an exit raffinate concentration of antibiotic of \(10 \mathrm{mg} / \mathrm{liter}\), assuming a concentration of antibiotic in the feed broth of \(1.0 \mathrm{~g} / \mathrm{liter}\) (a spreadsheet is convenient for these calculations). Without doing a complete economic analysis, in your judgment which scaled-up pilot run appears to be optimum? (Data from A. E. Karr, W. Gebert, and M. Wang, Can. J. Chem. Eng., vol. 58, p. \(249,1980 .)\)

Purification Factor for Extraction of an Enzyme You are preparing an industrial enzyme, alcohol dehydrogenase, from yeast using two-phase aqueous extraction. The crude, clarified extract contains only proteins and has a specific activity of 200 units/g protein. After a single stage of affinity extraction using a Cibacron dye-PEG complex, the extract contains 400,000 units of enzyme activity and \(20 \mathrm{~g}\) of protein. What is the purification factor for this step?

Graphical Equilibrium Stage Calculations for Extraction of a Peptide The equilibrium partitioning of a peptide between an aqueous feed phase and an organic solvent extract phase has been found to be nonlinear and can be represented by the following equation: $$ y=1.47 \ln x+3.96 $$ where \(y\) and \(x\) are concentrations of the peptide in the extract and aqueous feed (or raffinate) phases, respectively, in grams per liter. It is desired to extract \(95 \%\) of the peptide from a feed stream having a peptide concentration of \(4.0 \mathrm{~g} / \mathrm{liter}\). For a feed stream at a flow rate of \(5.0 \mathrm{liters} / \mathrm{min}\) and an extract stream at a flow rate of \(3.3 \mathrm{liters} / \mathrm{min}\), graphically estimate how many equilibrium stages will be required for countercurrent flow of the phases. What is the concentration of the peptide in the exit extract stream? As the concentration of the peptide in the raffinate decreases, does the partitioning of the peptide into the extract become more or less favorable?
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