Problem_4_7(1) - Problem 4-7 Problem Statement: Extractor...

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Unformatted text preview: Problem 4-7 Problem Statement: Extractor Basis: 400 (g/min) Acetic Acid Fraction 0.12 Water Fraction 0.89 Notice how unit labels are in a different cell Liquid extraction is an operation used to separate the components of a liquid mixture contains to components: a solute (A) and a liquid solvent (B). The mixture is contacte that has 2 key properties: A dissolves in it and B is immiscible or nearly immiscible wi oil, and A a species that dissolves in both water and oil). Some of the A transfers from the C-rich phase (the extract) separate from each other in a settling tank. If the raffina A will be transferred from it. This process can be repeated until essentially all of the A Shown below is a flowchart of a process in which acetic acid (A) is extracted from a m into 1-hexanol (C ), a liquid immiscible with water. m C (g/min C 6 H 13 OH) (g CH 3 COOH/g) (g H 2 O/g) For example: the basis has "400" ty label of "(g/min)" typed into (g/min) Extract Acetic Acid Fraction 0.1 1-Hexanol Fraction 0.9 (g/min) Raffinate Acetic Acid Fraction 0.01 Water Fraction 1 l than the numbers are! of 2 or more species. In the simplest case, the mixture ed in an agitated vessel with a second liquid solvent (C) ith it. (For example, B may be water, C a hydrocarbon m B to C, and then the B-rich phase (the raffinate) and ate is then contacted with fresh C in another stage, more A has been extracted from the B. mixture of acetic acid and water (B) m E (g CH 3 COOH/g) (g C 6 H 13 OH/g) m R (g CH 3 COOH/g) (g H 2 O/g) typed into one cell, and the o the cell next to it. Extractor Basis: 400 (g/min) Acetic Acid Fraction 0.12 Water Fraction 0.89 (a) What is the maximum number of independent material balances that can be written for this process? m C (g/min C 6 H 13 OH) (g CH 3 COOH/g) (g H 2 O/g) (g/min) Extract Acetic Acid Fraction 0.1 1-Hexanol Fraction 0.9 (g/min) Raffinate Acetic Acid Fraction 0.01 Water Fraction 1 m E (g CH 3 COOH/g) (g C 6 H 13 OH/g) m R (g CH 3 COOH/g) (g H 2 O/g) 3 You can either do three material balances by: 1) a balance on each component (ace 2) an overall mass balance, and 2 com Extractor Basis: 400 (g/min) Acetic Acid Fraction 0.12 Water Fraction 0.89 (a) What is the maximum number of independent material balances that can be written for this process? m C (g/min C 6 H 13 OH) (g CH 3 COOH/g) (g H 2 O/g) etic acid, 1-hexanol, water) mponent balances (g/min) Extract Acetic Acid Fraction 0.1 1-Hexanol Fraction 0.9 (g/min) Raffinate Acetic Acid Fraction 0.01 Water Fraction 1 m E (g CH 3 COOH/g) (g C 6 H 13 OH/g) m R (g CH 3 COOH/g) (g H 2 O/g) Extractor Basis: 400 g/min Acetic Acid Fraction 0.12 Water Fraction 0.89 (b) Calculate m C , m E , and m R , using the given mixture feed rate as a basi writing balances in an order such that you never have an equation that in more than one unknown variable....
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Problem_4_7(1) - Problem 4-7 Problem Statement: Extractor...

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