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Procesos de etapas por extraccion liq-liq, Apuntes de Procesos Químicos

Universidad DunamisProcesos Químicos

Resolución de algunos ejercicios de tarea

Tipo: Apuntes

2018/2019

Subido el 24/08/2022

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2019 Spring ChE338 - Homework #6 (Assigned 03/28)
Save your solution as a PDF file (maximum size 10 MB) and submit on Moodle by Midnight 4/5
(Friday)
Problem 1.
One thousand kg/h of a 45 wt% acetone-in-water solution is to be extracted at 25°C in a continuous,
countercurrent system with pure 1,1,2-trichloroethane to obtain a Raffinate containing 10 wt%
acetone. Using the given equilibrium data.
The minimum flow rate of solvent is 233 kg/h. Determine the below with an equilateral-triangle
diagram:
1.1. The number of stages required for a solvent rate equal to 1.5 times minimum;
1 pt: Draw a ternary phase diagram with given tie-line info
1 pt: Draw the first (F-E1=P) and the last (RN-S=P) operating lines and find P (operating point)
1 pt: Find a tie-line that end E1 by interpolation and resultant R1
1 pt: Repeated staging until R merged to RN
1 pt: Determine how many stages
pf3
pf4
pf5
pf8
pf9
pfa
pfd

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2019 Spring ChE338 - Homework #6 (Assigned 03/28)

Save your solution as a PDF file (maximum size 10 MB) and submit on Moodle by Midnight 4/ 5

(Friday)

Problem 1.

One thousand kg/h of a 45 wt% acetone-in-water solution is to be extracted at 25°C in a continuous,

countercurrent system with pure 1,1,2-trichloroethane to obtain a Raffinate containing 10 wt%

acetone. Using the given equilibrium data.

The minimum flow rate of solvent is 233 kg/h. Determine the below with an equilateral-triangle

diagram:

1.1. The number of stages required for a solvent rate equal to 1.5 times minimum;

1 pt: Draw a ternary phase diagram with given tie-line info

1 pt: Draw the first (F-E 1 =P) and the last (RN-S=P) operating lines and find P (operating point)

1 pt: Find a tie-line that end E 1 by interpolation and resultant R 1

1 pt: Repeated staging until R merged to RN

1 pt: Determine how many stages

About 5 stages

1.2. The flow rate and composition of each stream leaving each stage

1 pt: Determine solute (A), carrier (C), and solvent (S) compositions in each equilibrium stage

Extract Raffinate

Stage Solute (A) Carrier (C) Solvent (S) Solute (A) Carrier (C) Solvent (S) 1 0.5 0.04 0.46 0.4 0.57 0. 2 0.43 0.03 0.54 0.34 0.64 0. 3 0.35 0.02 0.63 0.29 0.7 0. 4 0.28 0.02 0.72 0.2 0.79 0. 5 0.18 0.01 0.81 0.1 0.89 0.

1 pt: Determine solvent flow: S = Smin x 1.1 S=349.5 kg

1 pt: Determine the mixing point

1 pt: Determine E 1 and R 5 the same way in problem 1.2.

Total mass balance F + S = M = E 1 + R 5 1349.5 = E 1 + R 5

Solute mass balance F(XA)F + S(XA)S = M(XA)M = E 1 (XA)E + R 5 (XA)R

1000(0.45) + 350(0) = E 1 (0.5) + R 5 (0.1)

E 1 = 787.62 kg, R 5 = 561.88 kg

1 pt: Mass balance each stage

** Depending on this reading, the results can be significantly different. Possibly get unreasonable

values. If the approach is correct, full grade will be given.

Check attached solutions

Problem 2.

Isopropyl ether (E) is used to separate acetic acid (A) from water (W). The liquid–liquid equilibrium data at

25 °C and 1 atm are below:

One hundred kilograms of a 30 wt% A–W solution is contacted with 120 kg of ether (E).

2 .1. What are the compositions of the resulting Extract and Raffinate?

1 pt: Determine mixing point composition

1 pt: Ternary phase diagram with given tie-line info and find the mixing point.

1 pt: Find a tie-line that passes the mixing point (interpolation)

1 pt: Read Extraction and Raffinate points

2 .2. What are the weights of the resulting Extract and Raffinate?

1 pt: Total mass balance F + S = M = E + R

220 = E + R

Solute mass balance F(XA)F + S(XA)S = M(XA)M = E(XA)E + R(XA)R

Problem 3.

Why is liquid–liquid extraction preferred over distillation for the separation of a mixture of formic acid

and water?

At 1 atm, the boiling point of formic acid is 100.6°C, almost identical to the normal boiling point of

water. A maximum-boiling azeotrope is formed at 107.1 C with 43.4 mol% water. Thus, it is

impossible to separate formic acid from water at 1 atm by ordinary distillation. Liquid-liquid-extraction

overcomes azeotrope and requires less energy to separate formic acid from water better than

distillation.

1 pt: Identify close boiling points or azeotrope

1 pt: Discuss overcome azeotrope and less energy requirement

Problem 4.

Discuss similarities and differences between distillation and liquid-liquid-extraction processes.

1 pt: Discussion about thermodynamic phase equilibrium

1 pt: Discussion about mass transport

1 pt: Unique application: eg. Energy intensive, hydrocarbon separation (Distillation), Require

additional separation, biological material separation (Extraction)

1 pt: Operating parameters and limiting conditions (eg. Reflux and solvent flow)

Refer Table (attached)

100(0.3) + 120(0) = E(0.09) + R(0.21)

1 pt: Solve linear algebra

E = 133.2 kg, R = 86.6 kg

2 .3. What would the concentration of acid in the (ether-rich) extract be if all ether were removed?

1 pt: Extraction phase when all solvent is removed.

9 / (9+3) x 100 = 75 wt%

ChE338 Separation: Distillation vs. Extraction

Distillation Extraction General strategies Phase creation (require ESA: Energy Separating Agent) Phase addition (require MSA: Material Separating Agent) Representative examples Table 7. Thermodynamically reversible components e.g. hydrocarbons, crude oils Table 8. Temperature irreversible components e.g. biological molecules Thermodynamic phase equilibrium Vapor Liquid Equilibrium (Ideal & less ideal: Equation Of State (EOS)) (Non idea: Experimental measurement, Graphs) Each point in V L E indicates an equilibrium point Ternary Phase Diagram: Solute Solvent Carrier (Experimental measurement) (Additional tie lines can be interpolated) Single phase vs Binary phase regions Tie line indicates an equilibrium Impractical condition to separate Azeotrope Plait point alpha phase: Extract, beta phase: Raffinate Quantitative info about how a target component distributed between two phases Relative volatility Relative selectivity

Single stage process Flash drum Operating line: Mass balance (Feed, Vapor, Liquid), Feed composition (zF) & condition (q line slope) Comparison process: Separating power Operating line: Mixing point, Tie line passing a mixing point Comparison process: Extraction factor Graphical integration equilibrium and operation (Single stage) Operation (change thermodynamic state) Degree of freedom : Determine how many parameters and how much adjusted to move a thermodynamic state to a desired state Degree of freedom : Determine how many parameters and how much adjusted to move a thermodynamic state to a desired state Cascade (Multi staging) General process General process For both distillation and extraction Overcome single stage separation that is mainly restricted by thermodynamic equilibrium. (Chemistry and analytical separation) Achieve higher purity products Continuous process Scale up process (Unique aspect of chemical engineering separation) Critical consideration in separation process design Energy balance: Phase creation requires significant amount of energy. Energy efficiency is critical consideration. Solvent: Solvent is critical consideration to determine thermodynamic equilibrium, economic aspect (cheap), safety aspect (stable), and down stream separation (required for extraction) Mass transport Specific density: Vapor (light move up), Liquid (heavy move down) Bubble cap to facilitate mass transport Specific density: different density between carrier and solvent Agitation / Mixing: Increase interface between two liquid phases