CHEE 4704 - Problem Set 2a - Solutions

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CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 1 of 10 Problem Set 2a Solutions Liquid-Liquid Extraction Problem 1: Extraction with Immiscible Solvents A water solution of 100 kg/h containing 1.0 wt% nicotine in water is extracted with a kerosene stream containing 0.05 wt% nicotine in a countercurrent stage tower. The exit water is to contain only 0.1 wt% of the original nicotine, that is, 90% is removed. Equilibrium data is provided in the table below. a) Determine the minimum solvent kerosene flow rate to perform the desired extraction. ( 103.17 kg/h ) b) Using 1.5 times the minimum solvent flow rate, determine graphically the number of theoretical stages required. ( 6.01 stages ) c) Determine analytically the number of theoretical stages required. ( 6.8 stages ) Solution a) An appropriate block flow diagram for the process: The nicotine recovery can be used to calculate the concentration in the exiting water stream: 𝑅? 𝑁 = (1 − 0.9)? 0 𝑅 → ? 𝑁 = 0.001 The slope of the equilibrium curve can be found by fitting a linear trendline to the equilibrium data. Note that the equilibrium curve is not a perfectly straight line; however, we can approximate the slope using a linear trendline.

CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 2 of 10 The minimum solvent flow rate is, 𝐸 ?𝑖? = 𝑅 ?? 0 − ? 𝑁+1 ? 0 − ? 𝑁 = 100 kg/h [(0.9224)(0.01) − 0.0005] (0.001 − 0.001) = 103.17 kg/h Alternatively, the minimum solvent flow rate can be found by varying 𝐸 ?𝑖? until the operating line pinches the equilibrium line. Using this method 𝐸 ?𝑖? = 99 kg/h . Either method is acceptable. b) The actual solvent flow rate is, 𝐸 ′ = 1.5𝐸 min ′ = 1.5( 103.17 kg/h ) = 154.75 kg/h The concentration in the exiting kerosene stream can be determined using a nicotine mass balance on the overall system: 𝑅? 0 + 𝐸? 𝑁+1 = 𝑅? 𝑁 + 𝐸? 1 ? 1 = 𝑅? 0 + 𝐸? 𝑁+1 − 𝑅? 𝑁 𝐸 = (100 ?𝑔 ℎ ) (0.01) + ( 103.17 kg h ) (0.0005) − 𝑅? 𝑁 𝐸 ? 1 = (100 kg/h)(0.01) + (154.75 kg/h) (0.0005) − (100 kg/h)(0.001) (154.75 kg/h) = 0.005816 The operating line can be plotted between the following points: (? 𝑁 , ? 𝑁+1 ) → (0.001, 0.0005) (? 0 , ? 1 ) → (0.01, 0.005816)

CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 3 of 10 The number of stages can now be stepped off. The stages can be stepped off using the following procedure: 1) Start at (? 𝑁 , ? 𝑁+1 ) → (0.001, 0.0005) . 2) Set ? = ? ??? and calculate ? = [polynomial fit to equilibrium curve] . 3) Set ? = ? ??? and calculate ? from the operating line: ? = (𝑦−𝑏) 𝑠???? . 4) Repeat 2 to 3 until ? ≥ ? 1 . The McCabe-Thiele plot shows that the required number of equilibrium stages is approximately 6.01. c) The equilibrium line can be divided into two separate lines with a linear slope:

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CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 4 of 10 The slope at the bottom of the equilibrium line: ? 𝑁 = 0.79626 The slope at the top of the equilibrium line: ? 1 = 0.9178 The flow rate of raffinate at the bottom of the operating line (or equilibrium line): 𝑅 𝑁 = 𝑅(1 − ? 0 ) 1 − ? 𝑁 = (100 kg/h)(1 − 0.01) 1 − 0.001 = 99.1 kg/h The flow rate of the solvent at the bottom of the operating line (or equilibrium line): 𝐸 = 154.75 kg/h The flow rate of raffinate at the top of the operating line (or equilibrium line) is just the feed raffinate flow rate. 𝑅 = 100 kg/h The flow rate of the solvent at the top of the operating line (or equilibrium line): 𝐸 1 = 𝐸(1 − ? 𝑁+1 ) 1 − ? 1 = (154.75 kg/h)(1 − 0) 1 − 0.0045 = 155.58 kg/h The extraction factor at the bottom of the equilibrium line: 𝐴 𝑁 = 𝑅 𝑁 ? 𝑁 𝐸 = 99.1 kg/h (0.79626)(154.75 kg/h) = 0.8042 The extraction factor at the top of the equilibrium line: 𝐴 1 = 𝑅 ? 1 𝐸 1 = 100 kg/h (0.9178)(155.58 kg/h) = 0.7003 The extraction factor:

CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 5 of 10 𝐴 = (𝐴 𝑁 𝐴 1 ) 0.5 = [(0.8042)(0.7003)] 0.5 = 0.7505 The number of theoretical stages: 𝑁 = ?? [ ? 0 − (? 𝑁+1 /?) ? 𝑁 − (? 𝑁+1 /?) (1 − 𝐴) + 𝐴] ??(1/𝐴) 𝑁 = ln [ 0.01 − (0.0005/0.79626) 0.001 − (0.0005/0.79626) (1 − 0.7505) + 0.7505] ln(1/0.7505) = 6.80 The Kremser method shows that the required number of equilibrium stages is approximately 6.80. The result matches closely to the McCabe-Thiele approach. Problem 2: Ternary Equilibrium Data A mixture weighing 200 kg and containing 50 kg of isopropyl ether, 20 kg of acetic acid, and 130 kg of water is equilibrated in a mixer-settler and the phases are separated. Determine the amounts and compositions of the raffinate and extract layers using the phase data below. Solution

CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 6 of 10 Problem 3: Ternary Equilibrium Data Use the provided acetone (A), water (C), and trichloroethane (S) ternary equilibrium data provided below to draw a rectangular diagram with tie lines. Phase Envelope Data Acetone (wt fr.) Water (wt fr.) Trichloroethane (wt f r.) Extract 0.60 0.13 0.27 0.50 0.04 0.46 0.40 0.03 0.57 0.30 0.02 0.68 0.20 0.015 0.785 0.10 0.01 0.89 Raffinate 0.55 0.35 0.10 0.50 0.43 0.07 0.40 0.57 0.03 0.30 0.68 0.02 0.20 0.79 0.01 0.10 0.895 0.005 Tie-line Data Raffinate Acetone wt fr. Extract Acetone wt fr. 0.44 0.56 0.29 0.40 0.12 0.18

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CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 7 of 10 Solution Problem 4: Extraction with Partially Miscible Solvents An aqueous feed solution of 1000 kg/h of acetic acid-water solution contains 30 wt% acetic acid and is to be extracted in a countercurrent multistage process with pure isopropyl ether to reduce the acid concentration to 2.0 wt% acid in the final raffinate. a) Calculate the minimum solvent flow rate that can be used. (1900 kg/h) b) Using 1.5 times the minimum solvent flow rate, determine the number of theoretical stages required. (6.72 stages) Solution a) An appropriate block flow diagram for the process:

CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 8 of 10 Since this is a concentrated system, we will work in terms of solvent flow rates and mass ratios instead of mass fractions and total flow rates. This way the operating line will still be linear. The water flow rate is, 𝑅 ′ = 𝑅(1 − ? 0 ) = (1000 kg h ⁄ )(1 − 0.3) = 700 kg W h ⁄ The given mass fractions can be converted into mass ratios, ? 0 = ? 0 1 − ? 0 = 0.30 1 − 0.30 = 0.4286 ? 𝑁+1 = ? 𝑁+1 1 − ? 𝑁+1 = 0 1 − 0 = 0 ? 𝑁 = ? 𝑁 1 − ? 𝑁 = 0.02 1 − 0.02 = 0.02041 ? 1 can be calculated using, ? 1 = (𝑅 ′ ? 0 + 𝑅 ′ ? 𝑁+1 − 𝑅 ′ ? 𝑁 ) 𝐸 ′ = [(700 kg/h)(0.4286) − (105kg/h)(0.02041)] 1900 kg/h = 0.1504 The minimum solvent flow rate can be found by varying the solvent flow rate until the operating line pinches the equilibrium line.

CHEE 4704: Separation Processes II Fall 2023 Problem Set 2a Solutions: Liquid-Liquid Extraction Page 9 of 10 From the McCabe-Thiele diagram, the minimum solvent flow rate is, 𝐸 ?𝑖? ′ = 1900 kg/h b) The actual solvent flow rate is, 𝐸 ′ = 1.5𝐸 min ′ = 1.5( 1900 kg/h ) = 2850 kg/h ? 1 can be calculated using, ? 1 = (𝑅 ′ ? 0 + 𝑅 ′ ? 𝑁+1 − 𝑅 ′ ? 𝑁 ) 𝐸 ′ = [(700 kg/h)(0.4286) − (700 kg/h)(0.02041)] 2850 kg/h = 0.1003 The operating line can be plotted using the following points: (? 𝑁 , ? 𝑁+1 ) → (0.02041, 0) (? 0 , ? 1 ) → (0.4286, 0.1003) The number of stages can now be stepped off. The stages can be stepped off using the following procedure: 1) Start at (? 𝑁 , ? 𝑁+1 ) → (0.02041, 0) . 2) Set ? = ? ??? and calculate ? from polynomial of the equilibrium line. 3) Set ? = ? ??? and calculate ? from the operating line: ? = (𝑦−𝑏) ? . 4) Repeat 2 to 3 until ? ≥ ? 1 . The McCabe-Thiele plot shows that the required number of equilibrium stages is approximately 6.72.

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