Liquid-Liquid Extraction. Experiment 350 ECH4404L: Unit Operations Lab II Instructor : Dr. Loren B. Schreiber Senior Lab Engineer : Mr. Richard Crisler Teaching Assistant : Mike Kirkpatrick. Team 8. Team Leader : Calvin Alleyne Safety Officer : Tehra Bouldin Team Members :
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
ECH4404L: Unit Operations Lab II
Instructor: Dr. Loren B. Schreiber
Senior Lab Engineer: Mr. Richard Crisler
Teaching Assistant: Mike Kirkpatrick
Team Leader: Calvin Alleyne
Safety Officer: Tehra Bouldin
The countercurrent extraction of ethanol from heptane to water occurs mostly because of the solubility differences of ethanol in these two solvents. Ethanol is more miscible in water than in heptane.
The reciprocator plates housed in the Karr extractor column have an impact on the extent of extraction as well. Physically the plates disperse the ethanol-heptane bubbles as they rise through the water. The reduction in the volume of the individual bubbles allows for more surface area of the bubbles to be in contact with the solvent.
Due to the increase of surface area more mass transfer can occur, as the ethanol leaves a non-aqueous phase liquid (heptane) and dissolves into the water.
Ternary system data provided by Dr. Schreiber in the laboratory was highly helpful, allowing for a plot of the equilibrium curve. Also, "tie-line" data was provided for the components of the ternary system.
Using a procedure outlined in Transport Process and Unit Operations (Geankopolis) an equilibrium plot was created with Microsoft Excel using the "tie line" data. The "D point" was then extrapolated from the "tie line" data. From this plot, the theoretical number of equilibrium stages of the extractor column was calculated.
Let L0 = initial mass flow rate of feed;
xethanol = mass fraction of ethanol in feed;
xheptane = mass fraction of heptane in feed;
Vn = the mass flow rate of the solvent (water) entering the column
Vn+1 = the mass flow rate of the raffinate;
yheptane = mass fraction of heptane in raffinate;
yethanol = mass fraction of ethanol in raffinate
Ln = the mass flow rate of the extract leaving the column;
x'ethanol = mass fraction of ethanol in the extract;
x'heptane = mass fraction of heptane in extract;
x'water = mass fraction of water in the extract
§Special note: 1- x'ethanol - x'heptane = x'water
L0 + Vn = Vn+1 + Ln
Species Balance on Ethanol: xethanol L0 = yethanol Vn+1 + x'ethanol Ln
Species Balance on Heptane: xheptane L0 = yheptane Vn+1 + x'heptane Ln
Mass % Conversion of Ethanol:
100% + ((yethanol Vn+1 + x'ethanol Ln) - (xethanol L0) / (xethanol L0)) = mass % conversion of ethanol
(Area %) = (% Ethanol)*0.723 -0.0721