Year: 2016-17 Subject: Process Calculation (2130504)

Topic: EXTRACTION

Name of the Students:Patil Mayur 160283105009Rohit Chetan 160283105010Sindhav Jaydrath 160283105011Vasava Yogesh 160283105012

Gujarat Technological University L.D. College of Engineering

• Normally, the term extraction is used for liquid-liquid separation. • When a mixture of liquid is not easily separable by distillation,

Extraction is employed. • In this process, a solvent is added to the liquid-liquid mixture. As

a result, two immiscible layers are formed, both of which contain varying amounts of different compounds. • These isolated layers are removed as extract phase and Raffinate

phase using the density difference. • But, distillation has to follow extraction for the solvent recovery

of the solvent for re-use.

Extraction

• For example, Furfural is common solvent in the extraction operations in the petroleum industries.• Another liquid phase according to the solubility. Extraction

becomes a very useful when a suitable extraction solvent is chosen. • Extraction can be used to separate a substance selectively from a

mixture Liquid-liquid extraction is based on the transfer of a solute substance from one liquid phase into, or to remove unwanted impurities from a solution.

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• It is a useful method to separate components (compounds) of a mixture

Liquid-Liquid Extraction

• Suppose that you have a mixture of sugar in vegetable oil (it tastes sweet!) and you want to separate the sugar from the oil.• You observe that the sugar particles are too tiny to filter and you

suspect that the sugar is partially dissolved in the vegetable oil.

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• How about shaking the mixture with water • Will it separate the sugar from the oil? Sugar is much more

soluble in water than in vegetable oil, and, as you know, water is immiscible (=not soluble) with oil.

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• Did you see the result? The water phase is the bottom layer and the oil phase is the top layer, because water is denser than oil.• *You have not shaken the mixture yet, so sugar is still in the oil

phase.

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• By shaking the layers (phases) well, you increase the contact area between the two phases. The sugar will move to the phase in which it is most soluble: the water layer• Now the water phase tastes sweet,because the sugar is moved to

the water phase upon shaking

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• the original oil-sugar mixture was the solution to be extracted; and sugar was the compound extracted from one phase to another. Separating the two layers accomplishes the separation of the sugar from the vegetable oil

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• When a compound is shaken in a separatory funnel with two immiscible solvents, the compound will distribute itself between the two solvents.• Normally one solvent is water and the other solvent is a water-

immiscible organic solvent.

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Most organic compounds are more soluble in organic solvents, while some organic compounds are more soluble in water.

• The efficiency of a liquid liquid extraction can be enhanced by adding one or more extractants to the solvent phase. • The extractant interacts with component I increasing the

capacity of the solvent for i.• To recover the solute from the extract phase the extractant-

solute complex has to be degraded.

Extractants

• A mixture containing 47.5 % acetic acid and 52.5 % water (by mass) is being separated by the extraction in a counter multistage unit. The operating temperature is 297 K and the solvent used is iso-propyle ether. On the solvent free basis is to be found to be 82% by mass of the acetic acid. The Raffinate is to be found to contain 14% by the mass of acetic acid on a solvent free basis. Calculate the percentage of acetic acid of the original feed which remains unextracted.

Example

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• Solution: Basis – 100 kg of Feed mixture

• Let E and R be the masses in kg of extract phase and Raffinate phase respectively.• Overall balance,

Feed F = E + R E + R = 100 (i)

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• Balance of acetic acidXF . F = XE . E + XR . R

• So, 0.82 E + 0.14 R = 0.475 x 100 = 47.5 (ii)• Where, XF , XR & XE is mass fraction of acetic acid in feed,

Raffinate and extract resp.• Solving equation (i) and (ii),

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• E= 49.2 kgR=50.8kg

• Acetic acid leftover in raffinate = 50.8 x 0.14 = 7.11 kg

• Acetic acid which is remain unextracted = (7.11/47.5) x 100 = 15%

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Thanks…