AMERICAN UNIVERSITY OF RAS AL KHAIMAH 3) For the extraction process below, aceti

Question

AMERICAN UNIVERSITY OF RAS AL KHAIMAH 3) For the extraction process below, acetic Acid (A) with a boiling point of CHEN 201: PRINCIPLES OF CHEMICAL ENGINEERING 118.1°C is extracted from a mixture of acetic acid and water (B) using 1- hexanol (C) as a solvent with a boiling point of 157°C. The extract is then sent to a distillation column where the solvent 1-hexanol is separated from acetic acid. Calculate the flow rate and composition of all streams Extract M g/min 0.096 g CH,COOH/E 0.904 g CgHiOH/g Solvent Distillate Mo R/min 0.991 g CH,COOH/ 0.009 g C,H OH/ M (sCH OH/min) Extrac tion Distilla tion Solution 400 g/min 0.115 & CH,COOH/s 0.885 g H,O/g Raffinate M, g/min 0.005 & CH,COOM/E Bottom Me g/min 0.006 & CH,COOH/g 0.994 gCH1OH/g 0995 g H,0/g

Explanation / Answer

Mass fractions of all the streams are already provided.

By mass balance of H2O:

H2O in feed solution = H2O in the raffinate

0.885(400) = 0.995*MR

.: MR = 355.7789 g/min

Consider Extraction unit:

mass balance of acetic acid:

0.115*400 = 0.096*ME + 0.005*MR

.: 0.115*400 = 0.096*ME + 0.005*355.7789

.: ME = 460.6365 g/min

Now, by mass balance around extraction unit,

M + 400 = ME + MR

.: M = 460.6365 + 355.7789 - 400

.: M = 416.415 g/min

Considering the entire system,

mass balance of acetic acid:

0.115*400 = 0.991*MD + 0.006*MB ................(1)

by mass balance around distillation column:

MB + MD = ME = 460.6365 g/min...................(2)

Solving (1) and (2), we get

MD = 43.8946 g/min

MB = 416.7419 g/min

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