CHEE 4704 - Problem Set 2a - Solutions

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Chemistry

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Jan 9, 2024

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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 ) (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:
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:
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