Chapter 2 - Lle Edited

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<ul><li><p>CHAPTER 2: LIQUID LIQUID EXTRACTION</p></li><li><p>CHAPTER / CONTENTDefinition &amp; ApplicationLLE for Partially Miscible SolventLLE for Immiscible SolventLiquid liquid extraction equipment</p></li><li><p> LLE= Separation of constituents (solutes) of a liquid solution by contact with another insoluble liquid. </p><p> Solutes are separated based on their different solubilities in different liquid. </p><p> LLE= Separation process of the components of a liquid mixture by treatment with a solvent in which one or more desired components is soluble.</p><p> There are two requirements for liquid liquid extraction to be feasible:</p><p> component (s) to be removed from the feed must preferentially distribute in the solvent.</p><p> the feed and solvent phases must be substantially immiscibleDefinition &amp; Application</p></li><li><p> The simplest LLE involves only a ternary (i.e 3 component system)</p><p> Important terms you need to know:</p><p> Feed-The solution which is to be extracted (denoted by component A)</p><p> Solvent-The liquid with which the feed is contacted(denoted by component C)</p><p> Diluent-Carrier liquid (denoted by component B) Extract-The solvent rich product of the operation Raffinate-The residual liquid from which solutes has been removed.Definition &amp; Application</p></li><li><p> In some operations, the solutes are the desired product, hence the extract stream is the desirable stream. In other applications, the solutes my be contaminants that need to be removed, and in this instance the raffinate is the desirable product stream. </p><p> Extraction processes are well suited to the petroleum industry because of the need to separate heat sensitive liquid feeds according to chemical type (e.g aromatic, naphthenic) rather than by molecular weight or vapor pressure.</p><p> Application:</p><p> Major applications exist in the biochemical or pharmaceutical industry, where emphasis is on the separation of antibiotics and protein recovery.</p><p> In the inorganic chemical industry, they are used to recover high boiling components such as phosphoric acid, boric acid and sodium hydroxide from aqueous solution.Definition &amp; Application</p></li><li><p> Examples:</p><p> Extraction of nitrobenzene after reaction of HNO3 with toluene in H2SO4</p><p> Extraction of methylacrylate from organic solution with perchlorethylene</p><p> Extraction of benzylalcohol from a salt solution with toluene.</p><p> Removing of H2S from LPG with MDEA</p><p> Extraction of caprolactam from ammonium sulfate solution with benzene</p><p> Extraction of acrylic acid from wastewater with butanol</p><p> Removing residual alkalis from dichlorohydrazobenzene with waterDefinition &amp; Application</p></li><li><p> Examples:</p><p> Extraction of methanol from LPG with water</p><p> Extraction of chloroacetic acid from methylchloroacetate with water.Definition &amp; Application The difference between LLE and distillation process in the separation of liquid mixtures:</p><p> LLE depends on solubilities between the liquid components and produces new solution which in turn has to be separated again, whereas;</p><p> Distillation depends on the differences in relative volatilities / vapor pressures of substances. Furthermore, it requires heat addition.</p></li><li><p>Definition &amp; Application Advantages of LLE over distillation process:</p><p> Where distillation requires excessive amount of heat</p><p> Presence of azeotropes or low relative volatilities are involved ( value near unity and distillation cannot be used)</p><p> Removal of a component present in small concentrations, e.g hormones in animal oil.</p><p> Recovery of a high boiling point component present in small quantities in waste stream, e.g acetic acid from cellulose acetate.</p><p> Recovery of heat sensitive materials (e.g food) where low to moderate processing temperatures are needed. Thermal decomposition might occur. </p><p> Solvent recovery is easy and energy savings can be realized. </p></li><li><p>SINGLE STAGE CALCULATIONSMULTISTAGE COUNTER CURRENT SYSTEMLLE for Partially Miscible Solvent</p></li><li><p> Solvent and the solution are in contact with each other only once and thus the raffinate and extract are in equilibrium only once.</p><p> The solution normally binary solution containing solute (A) dissolved in a diluent or carrier (B). The extracting solvent can be either pure solvent C or may content little A. Raffinate (R) is the exiting phase rich in carrier (B) while extract is exiting phase rich in solvent (C).</p><p> When liquid solution mixed with solvent (C), an intermediate phase M momentarily forms as the light liquid moves through the heavy liquid in the form of bubbles. These bubbles provide a large surface area for contact between the solution and the solvent that speed up mass transfer process. </p><p> The raffinate and extract are in equilibrium with each other. Single stage calculations</p></li><li><p>Single stage calculationsLiquid-Liquid Extractiony* (A)ys (A)Intermediate, MRaffinate phase, Rx* (A) Extract phase, ExM (A)FMass of feed solutionSMass of extracting solventEMass of extract phaseRMass of raffinate phaseMMass of intermediatexFMass fraction of A in FySMass fraction of A in SxMMass fraction of A in Mx*Equilibrium mass fractiony*Equilibrium mass fraction of A in Eof A in R</p><p>Note: Intermediate shown just for purpose of demonstration. Dont have to draw it when answering the questionFeed Solution, FxF (A) Extracting Solvent, S</p></li><li><p> In most single extraction, we are interested to determine the equilibrium composition and masses of raffinate and extract phases by using ternary phase diagram and simple material balances.</p><p> Using material balance, Single stage calculations Calculate the mass of intermediate M using total material balance: Eq. (1) Determine mass fraction of solute A in intermediate M using material balance for solute A : Eq. (2) Use both Eq. 1 and 2 to find xM value </p></li><li><p> On a right angle triangular diagram or equilateral triangular diagram for A-B-C system:Single stage calculations Locate point F (xF) and S (yS)</p><p> Draw a straight line from F to S</p><p> Using the calculated value of xM, locate point M (xM) on the FS line. Note that point M must be on FS line.</p><p> Draw a new tie line that pass through point M. This new tie line must take shape of the nearest given tie lines.</p><p> From the new tie line, you can locate point E and R and hence you can determine the composition of raffinate, R and extract, E that are in equilibrium.</p></li><li><p> Once you have determine composition of R and E, you can determine the masses of E and R using the material balance as follows:Single stage calculations Using the total material balance:Eq. (3) Using the material balance for solute A:Eq. (4) Solve those Eq 3 and 4 to determine the masses of E and R </p></li><li><p> Example 1Single stage calculations100 kg of a solution containing 0.4 mass fraction of ethylene glycol (EG) in water is to be extracted with equal mass of furfural 250C and 101 kPa. Using the ternary phase equilibrium diagram method, determine the followings: the composition of raffinate and extract phases </p><p> the mass of extract and raffinate</p><p> the percent glycol extracted Use the following equilibrium tie line to construct the ternary phase diagram</p><p>Furfural rich layerWater rich layer% EG% water% furfural% EG% water% furfural0.05.095.00.092.08.08.54.587.02.089.68.414.54.581.05.586.08.521.06.073.07.084.48.629.07.064.08.083.38.742.08.549.514.077.28.850.014.036.031.060.09.051.033.016.051.033.016.0</p></li><li><p> Solution 1Single stage calculations Calculate the mass of intermediate M using total material balance F = 100 kgS = 100 kgxF=0.4yS=0kg Determine mass fraction of solute A in intermediate M using material balance for solute A: Locate point F &amp; S, draw line FS. Locate point xM on FS line. Draw new tie line that pass through point xM. From that tie line, locate point E and R hence you can determine the composition of R (x*) and E (y*) which is in equilibrium. From the graph, y* = 0.26, x* = 0.075 (Solution for point 1) </p></li><li><p>Single stage calculations Right angle methodFSMER</p></li><li><p>Single stage calculations Equilateral methodSMERF</p></li><li><p> Solution 1 (cont)Single stage calculations Using the total material balance Using the material balance for solute A: Eq. (i)Insert eq (i) into eq above Solution for point 2% of EG extracted = (Mass of EG in extract / Mass of EG in feed) x 100%% of EG extracted =Solution for point 3</p></li><li><p>Tutorial12.5-212.5-4</p></li><li><p>Tutorial1. 12.5-2 (Textbook-page 832)A single stage extraction is performed in which 400kg of a solution containing 35 wt% acetic acid in water is contacted with 400 kg of pure isopropyl ether. By using equilateral diagram, determine the composition of raffinate and extract phases the mass of extract and raffinate the percent acetic acid extracted </p></li><li><p>Tutorial2. 12.5-4 (Textbook-page 832)A mixture weighing 1000kg contains 23.5 wt% acetone and 75.5 wt% water and is to be extracted by 500 kg methyl isobutyl ketone in a single stage extraction. By using equilateral diagram, determine the composition of raffinate and extract phases the mass of extract and raffinate the percent acetone extracted </p></li><li><p> Solvent and solution which flow opposite (countercurrent) to each other, come into contact more than once and mix on stages inside the reactor. </p><p> Normally numbering of the stages begin at the top down to the bottom. Thus the top most stage is named as stage 1, stage directly below stage 1 is stage 2 and so on. Multi stage counter current system</p></li><li><p> The analysis of multistage extraction can be performed using right angle or equilateral triangular diagram to determine the number of ideal stages required for a specified separation.</p><p> Using material balance, Multi stage counter current system Calculate the mass of intermediate M using total material balance: Eq. (1) Determine mass fraction of solute A in intermediate M using material balance for solute A : Eq. (2) Use both Eq. 1 and 2 to find xM value </p></li><li><p> On a right angle triangular diagram or equilateral triangular diagram for A-B-C system: Locate point F (xF) and S (yS)</p><p> Draw a straight line from F to S</p><p> Using the calculated value of xM, locate point M (xM) on the FS line. Note that point M must be on FS line.</p><p> Locate point E1 (Point M must be on E1RN line).Multi stage counter current system Operating Points and Lines. Locate the Operating Point by finding the intersection of operating lines for the left most and right most stage. Draw a line through E1 and F. Draw a line through S and RN. Locate the intersection P. This point is the operating point P. </p></li><li><p>Multi stage counter current systemPlait PointCarrierSoluteFeedRNM E1SOperating PointP</p></li><li><p>Multi stage counter current system Operating Lines and Tie Lines: Stepping Off Stages: Locate point R1 from the tie line intersecting E1</p><p> Draw a line from the operating point P through R1 to the extract side of the equilibrium curve. The intersection locates E2 </p><p> Locate point R2 from a tie line.</p><p> Repeat Steps 2 and 3 until RN is obtained </p><p> Summary: E1 R1 : Tie line, R1 E2 : Operating line. Stop until E value is slightly below RN value </p></li><li><p>Multi stage counter current systemPlait PointSolvent CCarrierSoluteE1R1FeedRNE2E3E4E5E6M Operating PointP</p></li><li><p> Minimum solvent amount / minimum solvent flow rate Minimum solvent flow rate is the lowest rate / amount at which solvent could be theoretically used for a specified extraction.</p><p> Occurs when operating line touches the equilibrium curve at which the separation requires infinite number of ideal stages. </p><p> Point M is dependent upon the solvent flow rate / amount. The larger the rate / amount, the closer is point M to point S on the FS line. Multi stage counter current system</p></li><li><p>Multi stage counter current system On a right angle triangular diagram or equilateral triangular diagram for A-B-C system: Locate point F (xF) and S (yS)</p><p> Draw a best tie line that originate from F. The intersection of this line with extract half dome is point Emin (minimum extract flow rate / amount).</p><p> Draw a straight line from Emin to point R. The intersection of this line with FS gives point Mmin. From point Mmin you can read the value of xmin. </p><p> Use the value of xmin and material balance to calculate the Smin . % Overall efficiency of multi stage extraction column:% Overall efficiency = (number of ideal stage / number of real stage) x 100% </p></li><li><p> Example 25300 kg/h of a solution containing 30% by weight of ethylene glycol (EG) in water is to be reduced to 4.5% (solvent free) by a continuous extraction in a countercurrent column using recycled furfural that contains 1.5% EG as the extracting solvent: Determine the minimum solvent flow rate for the extraction above If the solvent enters at 1.25 times the minimum solvent rate, how many ideal stages are required?</p><p> Determine the number of real stages if the overall efficiency of the column is 60% Multi stage counter current system</p></li><li><p>Multi stage counter current system From the graph above, XMmin = 0.25 </p><p> From material balance: Smin=1127.66 kg/h Solution for point 1 S = 1.25 x Smin=1.25 x 1127.66 kg/hS = 1409.58 kg/h</p><p> Calculate the mass of intermediate M using total material balance : Determine mass fraction of solute A in intermediate M using material balance for solute A: </p><p> From material balance: </p></li><li><p>Multi stage counter current systemSolution for point 2 FSE1RSFMPE2E3E4E5From figure above, no of ideal stages = 5 Number of real stages = 5 / 0.60 = 8.33 = 9 stages. Solution for point 3</p></li><li><p>Multi stage counter current systemSolution for point 2 FSE1RSFMPE2E3E4E5From figure above, no of ideal stages = 5 Number of real stages = 5 / 0.60 = 8.33 = 9 stages. Solution for point 3</p></li><li><p>Exercise12.7-3 (Textbook, pages 833)An aqueous feed solution of 1000 kg/hr containing 23.5 wt% acetone and 76.5 wt% water is being extracted in a countercurrent multistage extraction system using pure methylisobutyl ketone solvent at 298-299 K. The outlet water raffinate will contain 2.5 wt% acetone. By using equilateral method, Determine the minimum solvent that can be used (Answer: Xmin = 0.18, Smin = 305.56 kg/hr , Mmin = 1305.56 kg/hr)Using a solvent flow rate of 1.5 times the minimum, determine the number of theoretical stages. (Answer: S= 458.34 kg/hr, Xm=0.16 , 5 stages)</p></li><li><p>LLE for Immiscible Solvent Sometimes extraction use a solvent C that is only slightly soluble in B or the solvent C used is in range where the solubility in B is so low that for all practice purpose, it can be assumed to be completely insoluble / immiscible in B and vice versa. </p><p> Bancroft weight fractions or mass ratio, x and y are defined as follows:x (in raffinate phase) = mass of solute A / mass of diluent B y (in extract phase) = mass of solute A / mass of solvent C </p></li><li><p>SINGLE STAGE CALCULATIONSMULTISTAGE COUNTER CURRENT SYSTEMLLE for Immiscible Solvent</p></li><li><p> Solvent C is used in such a range that it is considered insoluble in B.</p><p> Material balance of the solute (A) are:Feed solution M kg A in feedS kg solvent C in Extracty kg A/kg solvent CSolventN kg A in feedB kg diluent B in Raffinatex kg A/kg diluent BEq. (3)Single stage calculations</p></li><li><p> Example 3Single stage calculations An aqueous solution of acetic acid is to be extracted in a single extractor with isopropyl ether. The solution contains 24.6 kg...</p></li></ul>