{"id":35,"date":"2019-12-13T21:59:43","date_gmt":"2019-12-13T21:59:43","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/chapter\/liquid-liquid-extraction-2\/"},"modified":"2026-03-16T01:12:53","modified_gmt":"2026-03-16T01:12:53","slug":"liquid-liquid-extraction-2","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/chapter\/liquid-liquid-extraction-2\/","title":{"raw":"Liquid-liquid Extraction","rendered":"Liquid-liquid Extraction"},"content":{"raw":"<h2>Staged Liquid-Liquid Extraction and Hunter Nash Method<\/h2>\n$E_n$\u00a0= extract leaving stage $n$. This could refer to the mass of the stream or the composition of the stream.\n\n$F$ = solvent entering extractor stage 1. This could refer to the mass of the stream or the composition of the stream.\n\n$n$ = generic stage number\n\n$N$ = Final stage. This is where the fresh solvent S enters the system and the final raffinate $R_N$\u00a0leaves the system.\n\n$M$ = Composition of the mixture representing the overall system. Points ($F$ and $S$) and ($E_1$ and $R_N$) must be connected by a straight line that passes through point $M$. $M$ will be located within the ternary phase diagram.\n\n$P$ = Operating point. $P$ is determined by the intersection of the straight line connecting points ($F$, $E_1$) and the straight line connecting points ($S$, $R_N$). Every pair of passing streams must be connected by a straight line that passes through point $P$. $P$ is expected to be located outside of the ternary phase diagram.\n\n$R_n$ = raffinate leaving stage $n$. This could refer to the mass of the stream or the composition of the stream.\n\n$S$ = solvent entering extractor stage $N$. This could refer to the mass of the stream or the composition of the stream.\n\n$S\/F$ = mass ratio of solvent to feed\n\n$(x_i)_n$ = Mass fraction of species $i$ in the raffinate leaving stage $n$\n\n$(y_i)_n$\u00a0= Mass fraction of species $i$ in the extract leaving stage $n$\n\n[caption id=\"attachment_600\" align=\"aligncenter\" width=\"1024\"]<img class=\"wp-image-600 size-large\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Staged-LLE-1024x346-1.jpg\" alt=\"Schematic for multistage liquid-liquid extraction process.\" width=\"1024\" height=\"346\"> Process schematic for multistage liquid-liquid extraction.[\/caption]\n\n&nbsp;\n\nDetermining number of stages $N$ when (1) feed rate; (2) feed composition; (3) incoming solvent rate; (4) incoming solvent composition; and (5) outgoing raffinate composition have been specified\/selected.\n<ol>\n \t<li>Locate points $F$ and $S$ on the ternary phase diagram. Connect with a straight line.<\/li>\n \t<li>Do a material balance to find the composition of one species in the overall mixture. Use this composition to locate point $M$ along the straight line connection points $F$ and $S$. Note the position of point $M$.<\/li>\n \t<li>Locate point $R_N$ on the ternary phase diagram. It will be on the equilibrium curve. Draw a straight line from $R_N$ to $M$ and extend to find the location of $E_1$\u00a0on the equilibrium curve.<\/li>\n \t<li>On a fresh copy of the graph, with plenty of blank space on each side of the diagram, note the location of points $F$, $S$, and $R_N$ (specified\/selected) and $E_1$\u00a0(determined in step 3).<\/li>\n \t<li>Draw a straight line between $F$ and $E_1$. Extend to both sides of the diagram. Draw a second straight line between $S$ and $R_N$. Note the intersection of these two lines and label as \u201c$P$\u201d.<\/li>\n \t<li>Determine the number of equilibrium stages required to achieve the desired separation with the selected solvent mass.<\/li>\n<\/ol>\n- Stream $R_N$ is in equilibrium with stream $E_N$. Follow the tie-lines from point $R_N$ to $E_N$.\n\n- Stream $E_N$ passes stream $R_{N-1}$. Connect point $E_N$ to operating point $P$ with a straight line, mark the location of $R_{N-1}$.\n\n- Stream $R_{N-1}$ is in equilibrium with stream $E_{N-1}$. Follow the tie-lines from stream $R_{N-1}$ to $E_{N-1}$.\n\n- Stream $E_{N-1}$ passes stream $R_{N-2}$. Connect $E_{N-1}$ to operating point $P$ with a straight line, mark the location of $R_{N-2}$.\n\n- Continue in this manner until the extract composition has reached or passed $E_{1}$. Count the number of equilibrium stages.\n\n&nbsp;\n\nWatch this two-part series of videos from <a href=\"http:\/\/www.learncheme.com\/\">LearnChemE<\/a> that shows how to use the Hunter Nash method to find the number of equilibrium stages required for a liquid-liquid extraction process.\n<ul>\n \t<li><a href=\"https:\/\/youtu.be\/-yW0jIcH0_E\">Hunter Nash Method 1: Mixing and Operating Points<\/a> (9:30)<\/li>\n \t<li><a href=\"https:\/\/youtu.be\/JZuavrt8ksQ\">Hunter Nash Method 2: Number of Stages<\/a> (6:30)<\/li>\n<\/ul>\n&nbsp;\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">Example<\/header>\n<div class=\"textbox__content\">\n\n1000 kg\/hr of a feed containing 30 wt% acetone, 70 wt% water. The solvent is pure MIBK. We intend that the raffinate contain no more than 5.0 wt% acetone. \u00a0How many stages will be required for each proposed solvent to feed ratio in the table below?\n<table class=\"alignleft\" style=\"border-collapse: collapse; width: 100%; height: 59px;\" tabindex=\"0\" border=\"0\">\n<tbody>\n<tr style=\"height: 15px;\">\n<td style=\"width: 147.390625px; height: 15px;\"><strong>$\\bold{S\/F}$<\/strong><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><strong>$\\bold{S}$ (kg\/hr)<\/strong><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><strong>$\\bold{(x_A)_M}$<\/strong><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><strong>target $\\bold{(y_A)_1}$<\/strong><\/td>\n<td style=\"width: 147.390625px; height: 15px;\"><strong>$\\bold{N}$<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 147.390625px; height: 14px;\">1.0<\/td>\n<td style=\"width: 147.890625px; height: 14px;\"><\/td>\n<td style=\"width: 147.890625px; height: 14px;\"><\/td>\n<td style=\"width: 147.890625px; height: 14px;\"><\/td>\n<td style=\"width: 147.390625px; height: 14px;\"><\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 147.390625px; height: 15px;\">2.0<\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.390625px; height: 15px;\"><\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 147.390625px; height: 15px;\">0.2<\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.390625px; height: 15px;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<img class=\"alignnone wp-image-355\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic2-300x273-1.jpg\" alt=\"\" width=\"305\" height=\"278\"> <img class=\"alignnone size-medium wp-image-356\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic3-300x280-1.jpg\" alt=\"\" width=\"300\" height=\"280\"> <img class=\"alignnone size-medium wp-image-357\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic4-300x275-1.jpg\" alt=\"\" width=\"300\" height=\"275\"> <img class=\"alignnone wp-image-358\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic5-300x263-1.jpg\" alt=\"\" width=\"313\" height=\"274\"><img class=\"alignnone wp-image-359 size-large\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic6-1024x329-1.jpg\" alt=\"\" width=\"1024\" height=\"329\">\n\n<\/div>\n<\/div>\n<h2>Hunter Nash Method for Finding Smin, Tank Sizing and Power Consumption for Mixer-Settler Units<\/h2>\n<h3><strong>Staged LLE: Hunter-Nash Method for Finding the Minimum Solvent to Feed Ratio<\/strong><\/h3>\n$E_n$ = extract leaving stage $n$. This could refer to the mass of the stream or the composition of the stream.\n\n$F$ = solvent entering extractor stage 1. This could refer to the mass of the stream or the composition of the stream.\n\n$n$ = generic stage number\n\n$N$ = Final stage. This is where the fresh solvent $S$ enters the system and the final raffinate $R_N$\u00a0leaves the system.\n\n$M$ = Composition of the overall mixture. Points ($F$ and $S$) and ($E_1$ and $R_N$) are connected by a straight line passing through $M$.\n\n$P$ = Operating point. Every pair of passing streams must be connected by a straight line that passes through $P$.\n\n$R_n$ = raffinate leaving stage $n$. This could refer to the mass of the stream or the composition of the stream.\n\n$S$ = solvent entering extractor stage $N$. This could refer to the mass of the stream or the composition of the stream.\n\n$S\/F$ = mass ratio of solvent to feed\n\n$S_{\\rm min}\/F$ = Minimum feasible mass ratio to achieve the desired separation, assuming the use of an infinite number of stages.\n\n$(x_i)_n$ = Mass fraction of species $i$ in the raffinate leaving stage $n$\n\n$(y_i)_n$\u00a0= Mass fraction of species $i$ in the extract leaving stage $n$\n\n&nbsp;\n\n$P_{\\rm min}$ = Point associated with the minimum feasible $S\/F$ for this feed, solvent and (raffinate or extract) composition. $P_{\\rm min}$ is the intersection of the line connecting points ($R_N$, $S$) and the line that is an extension of the upper-most equilibrium tie-line.\n\n&nbsp;\n\nDetermining minimum feasible solvent mass ratio ($S_{\\rm min}\/F$) when (1) feed composition; (2) incoming solvent composition; and (3) outgoing raffinate composition have been specified\/selected.\n<ol>\n \t<li>Locate points $S$ and $R_N$ on the phase diagram. Connect with a straight line.<\/li>\n \t<li>Extend the upper-most tie-line in a line that connects with the line connecting points ($S$ and $R_N$). Label the intersection $P_{\\rm min}$.<\/li>\n \t<li>Find point $F$ on the diagram. Draw a line from $P_{\\rm min}$ to F and extend to the other side of the equilibrium curve. Label $E_1$@$S_{\\rm min}$.<\/li>\n \t<li>On a fresh copy of the phase diagram, label points $F$, $S$, $R_N$ and $E_1$@$S_{\\rm min}$. Draw one line connecting points $S$ and $F$ and another line connecting points $E_1$@$S_{\\rm min}$<\/li>\n \t<li>\u00a0and $R_N$. The intersection of these two lines is mixing point $M$. Note the composition of species $i$ at this location.<\/li>\n \t<li>Calculate<\/li>\n<\/ol>\n\\begin{displaymath}\n<p style=\"padding-left: 40px;\">\\tag{5.1}\n\\frac{S_{\\rm min}}{F}=\\frac{(x_i)_F-(x_i)_M}{(x_i)_M-(x_i)_S}\n\\end{displaymath}<\/p>\n\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">Example<\/header>\n<div class=\"textbox__content\">\n\nWe have a 1000 kg\/hr feed that contains 30 wt% acetone and 70 wt% water. We want our raffinate to contain no more than 5.0 wt% acetone. What is the minimum mass of pure MIBK required?\n\n<\/div>\n<\/div>\n<h2><strong>Liquid-Liquid Extraction: Sizing Mixer-settler Units<\/strong><\/h2>\n$\\Phi_C$\u00a0= volume fraction occupied by the continuous phase\n\n$\\Phi_D$ = volume fraction occupied by the dispersed phase\n\n$\\mu_C$ = viscosity of the continuous phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)\n\n$\\mu_D$ = viscosity of the dispersed phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)\n\n$\\mu_M$ = viscosity of the mixture (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)\n\n$\\rho_C$ = density of the continuous phase (mass volume<sup>-1<\/sup>)\n\n$\\rho_D$ = density of the dispersed phase (mass volume<sup>-1<\/sup>)\n\n$\\rho_M$ = average density of the mixture (mass volume<sup>-1<\/sup>)\n\n&nbsp;\n\n$D_i$\u00a0= impeller diameter (length)\n\n$D_T$\u00a0= vessel diameter (length)\n\n$H$ = total height of mixer unit (length)\n\n$N$ = rate of impeller rotation (time<sup>-1<\/sup>)\n\n$N_{\\rm Po}$ = impeller power number, read from Fig 8-36 or Perry\u2019s 15-54 (below) based on value of $N_{Re}$\u00a0(unitless)\n\n$(N_{\\rm Re})_C$\u00a0= Reynold\u2019s number in the continuous phase = inertial force\/viscous force (unitless)\n\n$P$ = agitator power (energy time<sup>-1<\/sup>)\n\n$Q_C$\u00a0= volumetric flowrate, continuous phase (volume time<sup>-1<\/sup>)\n\n$Q_D$\u00a0= volumetric flowrate, dispersed phase (volume time<sup>-1<\/sup>)\n\n$V$ = vessel volume (volume)\n\n&nbsp;\n\nTank and impeller sizing\n\n\\begin{displaymath}\n<p style=\"padding-left: 40px;\">\\tag{5.2}<\/p>\n<p style=\"padding-left: 40px;\">{\\rm residence\\; time} = \\frac{V}{Q_C+Q_D}<\/p>\n\\end{displaymath}\n\nGeometry of a cylinder\n\n\\begin{displaymath}\n<p style=\"padding-left: 40px;\">\\tag{5.3}<\/p>\n<p style=\"padding-left: 40px;\">V = \\frac{{\\pi}D_T^2H}{4}<\/p>\n\\end{displaymath}\n\nGeneral guidelines\n\n\\begin{displaymath}\n<p style=\"padding-left: 40px;\">\\tag{5.4}<\/p>\n<p style=\"padding-left: 40px;\">\\frac{H}{D_T}=1<\/p>\n\\end{displaymath}\n\n\\begin{displaymath}\n<p style=\"padding-left: 40px;\">\\tag{5.5}<\/p>\n<p style=\"padding-left: 40px;\">\\frac{D_i}{D_T}=\\frac{1}{3}<\/p>\n\\end{displaymath}\n\nImpeller power consumption:\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{5.6}<\/p>\n<p style=\"padding-left: 40px;\">P=N_{Po}N^3D_i^5{\\rho}_m<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{5.7}<\/p>\n<p style=\"padding-left: 40px;\">N_{Re}=\\frac{D_i^2N{\\rho}_M}{{\\mu}_M}<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{5.8}<\/p>\n<p style=\"padding-left: 40px;\">{\\rho}_M={\\rho}_C{\\Phi}_C+{\\rho}_D{\\Phi}_D<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{5.9}<\/p>\n<p style=\"padding-left: 40px;\">{\\mu}_M=\\frac{{\\mu}_C}{{\\Phi}_C}\\left[1+\\frac{1.5{\\mu}_D{\\Phi}_D}{{\\mu}_C+{\\mu}_D}\\right]<\/p>\n\\end{equation}\n\n<img class=\"alignnone size-medium wp-image-364\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.6-Pic1-300x244-1.jpg\" alt=\"\" width=\"300\" height=\"244\"> <img class=\"alignnone size-medium wp-image-365\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.6-Pic2-300x147-1.jpg\" alt=\"\" width=\"300\" height=\"147\">\n<h2>Modeling Mass Transfer in Mixer-Settler Units<\/h2>\n$\\Delta\\rho$ = density difference (absolute value) between the continuous and dispersed phases (mass volume<sup>-1<\/sup>)\n\n$\\phi_C$\u00a0= volume fraction occupied by the continuous phase\n\n$\\phi_D$ = volume fraction occupied by the dispersed phase\n\n$\\mu_C$\u00a0= viscosity of the continuous phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)\n\n$\\mu_D$ = viscosity of the dispersed phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)\n\n$\\mu_M$ = viscosity of the mixture (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)\n\n$\\rho_C$ = density of the continuous phase (mass volume<sup>-1<\/sup>)\n\n$\\rho_D$ = density of the dispersed phase (mass volume<sup>-1<\/sup>)\n\n$\\rho_M$ = average density of the mixture (mass volume<sup>-1<\/sup>)\n\n$\\sigma$ = interfacial tension between the continuous and dispersed phases\n(mass time<sup>-2<\/sup>)\n\n$a$ = interfacial area between the two phases per unit volume (area volume<sup>-1<\/sup>)\n\n$c_{D,\\rm in}$, $c_{D,\\rm out}$\u00a0= concentration of solute in the incoming or outgoing dispersed streams (mass volume<sup>-1<\/sup>)\n\n$c^*_D$\u00a0= concentration of solute in the dispersed phase if in equilibrium with the outgoing continuous phase (mass volume<sup>-1<\/sup>)\n\n$D_C$\u00a0= diffusivity of the solute in the continuous phase (area time<sup>-1<\/sup>)\n\n$D_D$ = diffusivity of the solute in the dispersed phase (area time<sup>-1<\/sup>)\n\n$D_i$\u00a0= impeller diameter (length)\n\n$D_T$\u00a0= vessel diameter (length)\n\n$d_{vs}$\u00a0= Sauter mean droplet diameter; actual drop size expected to range from $0.3d_{vs}-3.0d_{vs}$\u00a0(length)\n\n$E_{MD}$\u00a0= Murphree dispersed-phase efficiency for extraction\n\n$g$ = gravitational constant (length time<sup>-2<\/sup>)\n\n$H$ = total height of mixer unit (length)\n\n$k_c$\u00a0= mass transfer coefficient of the solute in the continuous phase (length time<sup>-1<\/sup>)\n\n$k_D$\u00a0= mass transfer coefficient of the solute in the dispersed phase (length time<sup>-1<\/sup>)\n\n$K_{OD}$\u00a0= overall mass transfer coefficient, given on the basis of the dispersed phase (length time<sup>-1<\/sup>)\n\n$m$ = distribution coefficient of the solute, $\\Delta c_C\/\\Delta c_D$\u00a0(unitless)\n\n$N$ = rate of impeller rotation (time<sup>-1<\/sup>)\n\n$(N_{\\rm Eo})_C$\u00a0= Eotvos number = gravitational force\/surface tension force (unitless)\n\n$(N_{\\rm Fr})_C$\u00a0= Froude number in the continuous phase = inertial force\/gravitational force (unitless)\n\n$N_{\\rm min}$\u00a0= minimum impeller rotation rate required for complete dispersion of one liquid into another\n\n$(N_{\\rm Re})_C$\u00a0= Reynold\u2019s number in the continuous phase = inertial force\/viscous force (unitless)\n\n$(N_{\\rm Sh})_C$\u00a0= Sherwood number in the continuous phase = mass transfer rate\/diffusion rate (unitless)\n\n$(N_{\\rm Sc})_C$\u00a0= Schmidt number in the continuous phase = momentum\/mass diffusivity (unitless)\n\n$(N_{\\rm We})_C$\u00a0= Weber number = inertial force\/surface tension (unitless)\n\n$Q_D$\u00a0= volumetric flowrate of the dispersed phase (volume time<sup>-1<\/sup>)\n\n$V$ = vessel volume (volume)\n\n&nbsp;\n\nCalculating $N_{\\rm min}$\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.1}<\/p>\n<p style=\"padding-left: 40px;\">\\frac{N_{\\rm min}^2{\\rho}_MD_i}{g{\\Delta}{\\rho}}=1.03\\left(\\frac{D_T}{D_i}\\right)^{2.76}({\\phi}_D)^{0.106}\\left(\\frac{{\\mu}_M^2{\\sigma}}{D_i^5{\\rho}_Mg^2({\\Delta}{\\rho})^2}\\right)^{0.084}<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.2}<\/p>\n<p style=\"padding-left: 40px;\">{\\rho}_M={\\rho}_C{\\phi}_C+{\\rho}_D{\\phi}_D<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.3}<\/p>\n<p style=\"padding-left: 40px;\">{\\mu}_M=\\frac{{\\mu}_C}{{\\phi}_C}\\left(1+\\frac{1.5{\\mu}_D{\\phi}_D}{{\\mu}_C+{\\mu}_D}\\right)<\/p>\n\\end{equation}\n\n&nbsp;\n\nEstimating Murphree efficiency for a proposed design\n<p style=\"padding-left: 40px;\">Sauter mean diameter<\/p>\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.4}<\/p>\n<p style=\"padding-left: 40px;\">{\\rm if}\\;\\; N_{\\rm We} &lt; 10,000,\\; d_{vs}=0.052D_i(N_{\\rm We})^{-0.6}\\exp({4{\\phi}_D})<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.5}<\/p>\n<p style=\"padding-left: 40px;\">{\\rm if}\\;\\; N_{\\rm We}\u00a0&gt;10,000,\\; d_{vs}=0.39D_i(N_{\\rm We})^{-0.6}<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.6}<\/p>\n<p style=\"padding-left: 40px;\">N_{\\rm We}=\\frac{D_i^3N^2{\\rho}_C}{\\sigma}<\/p>\n\\end{equation}\n<p style=\"padding-left: 40px;\">mass transfer coefficient of the solute in each phase<\/p>\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.7}<\/p>\n<p style=\"padding-left: 40px;\">k_D=\\frac{6.6D_D}{d_{vs}}<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.8}<\/p>\n<p style=\"padding-left: 40px;\">k_C=\\frac{(N_{\\rm Sh})_CD_c}{d_{vs}}<\/p>\n\\end{equation}\n\n\\begin{align}\n<p style=\"padding-left: 40px;\">\\tag{6.9}<\/p>\n<p style=\"padding-left: 40px;\">&amp; (N_{\\rm Sh})_C = \u00a01.237\\times 10^{-5}(N_{\\rm Sc})_C^{1\/3}(N_{\\rm Re})_C^{2\/3}(\\phi_D)^{-1\/2}\\\\\n&amp; (N_{\\rm Fr})_C^{5\/12}\\left(\\frac{D_i}{d_{vs}}\\right)^2\n\\left(\\frac{d_{vs}}{D_T}\\right)^{1\/2}(N_{\\rm Eo})_C^{5\/4}<\/p>\n\\end{align}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.10}<\/p>\n<p style=\"padding-left: 40px;\">(N_{\\rm Sc})_C=\\frac{{\\mu}_C}{{\\rho}_CD_C}<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.11}<\/p>\n<p style=\"padding-left: 40px;\">(N_{\\rm Re})_C=\\frac{D_i^2N{\\rho}_C}{{\\mu}_C}<\/p>\n\\end{equation}\n\n\\begin{displaymath}\n<p style=\"padding-left: 40px;\">\\tag{6.12}<\/p>\n<p style=\"padding-left: 40px;\">(N_{\\rm Fr})_C=\\frac{D_iN^2}{g}<\/p>\n\\end{displaymath}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.13}<\/p>\n<p style=\"padding-left: 40px;\">(N_{\\rm Eo})_C=\\frac{{\\rho}_Dd_{vs}^2g}{{\\sigma}}<\/p>\n\\end{equation}\n<p style=\"padding-left: 40px;\">Overall mass transfer coefficient for the solute<\/p>\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.14}<\/p>\n<p style=\"padding-left: 40px;\">\\frac{1}{K_{OD}}=\\frac{1}{k_D}+\\frac{1}{mk_C}<\/p>\n\\end{equation}\n\nMurphree efficiency\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.15}<\/p>\n<p style=\"padding-left: 40px;\">E_{MD}=\\frac{K_{OD}aV}{Q_D}\\left(1+{\\frac{K_{OD}aV}{Q_D}}\\right)^{-1}<\/p>\n\\end{equation}\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.16}<\/p>\n<p style=\"padding-left: 40px;\">a=\\frac{6\\phi_D}{d_{vs}}<\/p>\n\\end{equation}\n\nExperimental assessment of efficiency\n\n\\begin{equation}\n<p style=\"padding-left: 40px;\">\\tag{6.17}<\/p>\n<p style=\"padding-left: 40px;\">E_{MD}=\\frac{c_{D,\\rm in}-c_{D,\\rm out}}{c_{D,\\rm in}-c^*_D}<\/p>\n\\end{equation}\n\n&nbsp;\n\n&nbsp;\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">Example<\/header>\n<div class=\"textbox__content\">\n\n1000 kg\/hr of 30 wt% acetone and 70 wt% water is to be extracted with 1000 kg\/hr of pure MIBK. Assume that the extract is the continuous phase, a residence time of 5 minutes in the mixing vessel, standard sizing of the mixing vessel and impeller. Find the power consumption and Murphree efficiency if the system operates at $N_{\\rm min}$, controlled at the level of 1 rev\/s. Ignore the contribution of the solute and the co-solvent to the physical properties of each phase.\n<ul>\n \t<li>MIBK\n<ul>\n \t<li>density = 802 kg m<sup>-3<\/sup><\/li>\n \t<li>viscosity = 0.58 cP<\/li>\n \t<li>diffusivity with acetone at 25\u00b0C = 2.90x10<sup>-9<\/sup> m<sup>2<\/sup> s<sup>-1<\/sup><\/li>\n<\/ul>\n<\/li>\n \t<li>Water\n<ul>\n \t<li>density = 1000 kg m<sup>-3 <\/sup><\/li>\n \t<li>viscosity = 0.895 cP<\/li>\n \t<li>diffusivity with acetone at 25\u00b0C = 1.16x10<sup>-9<\/sup> m<sup>2<\/sup> s<sup>-1<\/sup><\/li>\n<\/ul>\n<\/li>\n \t<li>The interfacial tension of water and MIBK at 25\u00b0C = 0.0157 kg s<sup>-2<\/sup>. Use the ternary phase diagram to find $m$.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h2>Liquid-Liquid Extraction Columns<\/h2>\n$\\Delta \\rho$ = density difference (absolute value) between the continuous and dispersed phases (mass volume<sup>-1)<\/sup>\n\n$\\mu_C$\u00a0= viscosity of the continuous phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)\n\n$\\rho_C$ = density of the continuous phase (mass volume<sup>-1<\/sup>)\n\n$\\rho_D$ = density of the dispersed phase (mass volume<sup>-1<\/sup>)\n\n$\\sigma$ = interfacial tension between the continuous and dispersed phases\n(mass time<sup>-2<\/sup>)\n\n&nbsp;\n\n$D_T$\u00a0= column diameter (length)\n\n$H$ = total height of column (length)\n\n${\\rm HETS}$ = height of equilibrium transfer stage (length)\n\n$m^*_C$\u00a0= mass flowrate of the entering continuous phase (mass time<sup>-1<\/sup>)\n\n$m^*_D$ = mass flowrate of the entering dispersed phase (mass time<sup>-1<\/sup>)\n\n$N$ = required number of equilibrium stages\n\n$u_0$\u00a0= characteristic rise velocity of a droplet of the dispersed phase (length time<sup>-1<\/sup>)\n\n$U_i$\u00a0= superficial velocity of phase $i$ (C = continuous, downward; D = dispersed, upward)\u00a0 (length time<sup>-1<\/sup>)\n\n$V^*_i$\u00a0= volumetric flowrate of phase $i$ (volume time<sup>-1<\/sup>)\n\n&nbsp;\n\n\\begin{displaymath}\n\n\\tag{7.1}\n\nU_i=\\frac{4V^*_i}{{\\pi}D_T^2}\n\n\\end{displaymath}\n\ndefinition of superficial velocity\n<p style=\"padding-left: 40px;\">\u00a0\\begin{displaymath}<\/p>\n\\tag{7.2}\n\n\\frac{U_D}{U_C}=\\frac{m^*_D}{m^*_C}\\left(\\frac{{\\rho}_C}{\\rho_D}\\right)\n\n\\end{displaymath}\n<p style=\"padding-left: 40px;\">\\begin{displaymath}<\/p>\n\\tag{7.3}\n\n(U_D+U_C)_{\\rm actual}=0.50(U_D+U_C)_f\n\n\\end{displaymath}\n\nfor operation at 50% of flooding\n\n\\begin{displaymath}\n\n\\tag{7.4}\n\nu_0=\\frac{0.01{\\sigma}{\\Delta}{\\rho}}{{\\mu}_C{\\rho}_C}\n\n\\end{displaymath}\n\nfor rotating-disk columns, $D_T$\u00a0= 8 to 42 inches, with one aqueous phase\n\n&nbsp;\n\n\\begin{displaymath}\n\n\\tag{7.5}\n\nD_T=\\left(\\frac{4m^*_D}{{\\rho}_DU_D{\\pi}}\\right)^{0.5}=\\left(\\frac{4m^*_C}{{\\rho}_CU_C{\\pi}}\\right)^{0.5}\n\n\\end{displaymath}\n\n\\begin{displaymath}\n\n\\tag{7.6}\n\nH = {\\rm HETS}*N\n\n\\end{displaymath}\n\n&nbsp;\n\n&nbsp;\n\n&nbsp;\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">Example<\/header>\n<div class=\"textbox__content\">\n\n1000 kg\/hr of 30 wt% acetone and 70 wt% water is to be extracted with 1000 kg\/hr of pure MIBK in a 2-stage column process. Assume that the extract is the dispersed phase. Ignoring the contribution of the solute and the co-solvent to the physical properties of each phase, find the required column diameter and height.\n<ul>\n \t<li>MIBK\n<ul>\n \t<li>density = 802 kg m<sup>-3 <\/sup><\/li>\n \t<li>viscosity = 0.58 cP<\/li>\n<\/ul>\n<\/li>\n \t<li>Water\n<ul>\n \t<li>density = 1000 kg m<sup>-3 <\/sup><\/li>\n \t<li>viscosity = 0.895 cP<\/li>\n<\/ul>\n<\/li>\n \t<li>The interfacial tension of water and MIBK at 25\u00b0C = 0.0157 kg s<sup>-2<\/sup>.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n&nbsp;","rendered":"<h2>Staged Liquid-Liquid Extraction and Hunter Nash Method<\/h2>\n<p>$E_n$\u00a0= extract leaving stage $n$. This could refer to the mass of the stream or the composition of the stream.<\/p>\n<p>$F$ = solvent entering extractor stage 1. This could refer to the mass of the stream or the composition of the stream.<\/p>\n<p>$n$ = generic stage number<\/p>\n<p>$N$ = Final stage. This is where the fresh solvent S enters the system and the final raffinate $R_N$\u00a0leaves the system.<\/p>\n<p>$M$ = Composition of the mixture representing the overall system. Points ($F$ and $S$) and ($E_1$ and $R_N$) must be connected by a straight line that passes through point $M$. $M$ will be located within the ternary phase diagram.<\/p>\n<p>$P$ = Operating point. $P$ is determined by the intersection of the straight line connecting points ($F$, $E_1$) and the straight line connecting points ($S$, $R_N$). Every pair of passing streams must be connected by a straight line that passes through point $P$. $P$ is expected to be located outside of the ternary phase diagram.<\/p>\n<p>$R_n$ = raffinate leaving stage $n$. This could refer to the mass of the stream or the composition of the stream.<\/p>\n<p>$S$ = solvent entering extractor stage $N$. This could refer to the mass of the stream or the composition of the stream.<\/p>\n<p>$S\/F$ = mass ratio of solvent to feed<\/p>\n<p>$(x_i)_n$ = Mass fraction of species $i$ in the raffinate leaving stage $n$<\/p>\n<p>$(y_i)_n$\u00a0= Mass fraction of species $i$ in the extract leaving stage $n$<\/p>\n<figure id=\"attachment_600\" aria-describedby=\"caption-attachment-600\" style=\"width: 1024px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-600 size-large\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Staged-LLE-1024x346-1.jpg\" alt=\"Schematic for multistage liquid-liquid extraction process.\" width=\"1024\" height=\"346\" \/><figcaption id=\"caption-attachment-600\" class=\"wp-caption-text\">Process schematic for multistage liquid-liquid extraction.<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p>Determining number of stages $N$ when (1) feed rate; (2) feed composition; (3) incoming solvent rate; (4) incoming solvent composition; and (5) outgoing raffinate composition have been specified\/selected.<\/p>\n<ol>\n<li>Locate points $F$ and $S$ on the ternary phase diagram. Connect with a straight line.<\/li>\n<li>Do a material balance to find the composition of one species in the overall mixture. Use this composition to locate point $M$ along the straight line connection points $F$ and $S$. Note the position of point $M$.<\/li>\n<li>Locate point $R_N$ on the ternary phase diagram. It will be on the equilibrium curve. Draw a straight line from $R_N$ to $M$ and extend to find the location of $E_1$\u00a0on the equilibrium curve.<\/li>\n<li>On a fresh copy of the graph, with plenty of blank space on each side of the diagram, note the location of points $F$, $S$, and $R_N$ (specified\/selected) and $E_1$\u00a0(determined in step 3).<\/li>\n<li>Draw a straight line between $F$ and $E_1$. Extend to both sides of the diagram. Draw a second straight line between $S$ and $R_N$. Note the intersection of these two lines and label as \u201c$P$\u201d.<\/li>\n<li>Determine the number of equilibrium stages required to achieve the desired separation with the selected solvent mass.<\/li>\n<\/ol>\n<p>&#8211; Stream $R_N$ is in equilibrium with stream $E_N$. Follow the tie-lines from point $R_N$ to $E_N$.<\/p>\n<p>&#8211; Stream $E_N$ passes stream $R_{N-1}$. Connect point $E_N$ to operating point $P$ with a straight line, mark the location of $R_{N-1}$.<\/p>\n<p>&#8211; Stream $R_{N-1}$ is in equilibrium with stream $E_{N-1}$. Follow the tie-lines from stream $R_{N-1}$ to $E_{N-1}$.<\/p>\n<p>&#8211; Stream $E_{N-1}$ passes stream $R_{N-2}$. Connect $E_{N-1}$ to operating point $P$ with a straight line, mark the location of $R_{N-2}$.<\/p>\n<p>&#8211; Continue in this manner until the extract composition has reached or passed $E_{1}$. Count the number of equilibrium stages.<\/p>\n<p>&nbsp;<\/p>\n<p>Watch this two-part series of videos from <a href=\"http:\/\/www.learncheme.com\/\">LearnChemE<\/a> that shows how to use the Hunter Nash method to find the number of equilibrium stages required for a liquid-liquid extraction process.<\/p>\n<ul>\n<li><a href=\"https:\/\/youtu.be\/-yW0jIcH0_E\">Hunter Nash Method 1: Mixing and Operating Points<\/a> (9:30)<\/li>\n<li><a href=\"https:\/\/youtu.be\/JZuavrt8ksQ\">Hunter Nash Method 2: Number of Stages<\/a> (6:30)<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">Example<\/header>\n<div class=\"textbox__content\">\n<p>1000 kg\/hr of a feed containing 30 wt% acetone, 70 wt% water. The solvent is pure MIBK. We intend that the raffinate contain no more than 5.0 wt% acetone. \u00a0How many stages will be required for each proposed solvent to feed ratio in the table below?<\/p>\n<table class=\"alignleft\" style=\"border-collapse: collapse; width: 100%; height: 59px;\" tabindex=\"0\">\n<tbody>\n<tr style=\"height: 15px;\">\n<td style=\"width: 147.390625px; height: 15px;\"><strong>$\\bold{S\/F}$<\/strong><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><strong>$\\bold{S}$ (kg\/hr)<\/strong><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><strong>$\\bold{(x_A)_M}$<\/strong><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><strong>target $\\bold{(y_A)_1}$<\/strong><\/td>\n<td style=\"width: 147.390625px; height: 15px;\"><strong>$\\bold{N}$<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 147.390625px; height: 14px;\">1.0<\/td>\n<td style=\"width: 147.890625px; height: 14px;\"><\/td>\n<td style=\"width: 147.890625px; height: 14px;\"><\/td>\n<td style=\"width: 147.890625px; height: 14px;\"><\/td>\n<td style=\"width: 147.390625px; height: 14px;\"><\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 147.390625px; height: 15px;\">2.0<\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.390625px; height: 15px;\"><\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 147.390625px; height: 15px;\">0.2<\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.890625px; height: 15px;\"><\/td>\n<td style=\"width: 147.390625px; height: 15px;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img decoding=\"async\" class=\"alignnone wp-image-355\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic2-300x273-1.jpg\" alt=\"\" width=\"305\" height=\"278\" \/> <img decoding=\"async\" class=\"alignnone size-medium wp-image-356\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic3-300x280-1.jpg\" alt=\"\" width=\"300\" height=\"280\" \/> <img decoding=\"async\" class=\"alignnone size-medium wp-image-357\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic4-300x275-1.jpg\" alt=\"\" width=\"300\" height=\"275\" \/> <img decoding=\"async\" class=\"alignnone wp-image-358\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic5-300x263-1.jpg\" alt=\"\" width=\"313\" height=\"274\" \/><img decoding=\"async\" class=\"alignnone wp-image-359 size-large\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.5-Pic6-1024x329-1.jpg\" alt=\"\" width=\"1024\" height=\"329\" \/><\/p>\n<\/div>\n<\/div>\n<h2>Hunter Nash Method for Finding Smin, Tank Sizing and Power Consumption for Mixer-Settler Units<\/h2>\n<h3><strong>Staged LLE: Hunter-Nash Method for Finding the Minimum Solvent to Feed Ratio<\/strong><\/h3>\n<p>$E_n$ = extract leaving stage $n$. This could refer to the mass of the stream or the composition of the stream.<\/p>\n<p>$F$ = solvent entering extractor stage 1. This could refer to the mass of the stream or the composition of the stream.<\/p>\n<p>$n$ = generic stage number<\/p>\n<p>$N$ = Final stage. This is where the fresh solvent $S$ enters the system and the final raffinate $R_N$\u00a0leaves the system.<\/p>\n<p>$M$ = Composition of the overall mixture. Points ($F$ and $S$) and ($E_1$ and $R_N$) are connected by a straight line passing through $M$.<\/p>\n<p>$P$ = Operating point. Every pair of passing streams must be connected by a straight line that passes through $P$.<\/p>\n<p>$R_n$ = raffinate leaving stage $n$. This could refer to the mass of the stream or the composition of the stream.<\/p>\n<p>$S$ = solvent entering extractor stage $N$. This could refer to the mass of the stream or the composition of the stream.<\/p>\n<p>$S\/F$ = mass ratio of solvent to feed<\/p>\n<p>$S_{\\rm min}\/F$ = Minimum feasible mass ratio to achieve the desired separation, assuming the use of an infinite number of stages.<\/p>\n<p>$(x_i)_n$ = Mass fraction of species $i$ in the raffinate leaving stage $n$<\/p>\n<p>$(y_i)_n$\u00a0= Mass fraction of species $i$ in the extract leaving stage $n$<\/p>\n<p>&nbsp;<\/p>\n<p>$P_{\\rm min}$ = Point associated with the minimum feasible $S\/F$ for this feed, solvent and (raffinate or extract) composition. $P_{\\rm min}$ is the intersection of the line connecting points ($R_N$, $S$) and the line that is an extension of the upper-most equilibrium tie-line.<\/p>\n<p>&nbsp;<\/p>\n<p>Determining minimum feasible solvent mass ratio ($S_{\\rm min}\/F$) when (1) feed composition; (2) incoming solvent composition; and (3) outgoing raffinate composition have been specified\/selected.<\/p>\n<ol>\n<li>Locate points $S$ and $R_N$ on the phase diagram. Connect with a straight line.<\/li>\n<li>Extend the upper-most tie-line in a line that connects with the line connecting points ($S$ and $R_N$). Label the intersection $P_{\\rm min}$.<\/li>\n<li>Find point $F$ on the diagram. Draw a line from $P_{\\rm min}$ to F and extend to the other side of the equilibrium curve. Label $E_1$@$S_{\\rm min}$.<\/li>\n<li>On a fresh copy of the phase diagram, label points $F$, $S$, $R_N$ and $E_1$@$S_{\\rm min}$. Draw one line connecting points $S$ and $F$ and another line connecting points $E_1$@$S_{\\rm min}$<\/li>\n<li>\u00a0and $R_N$. The intersection of these two lines is mixing point $M$. Note the composition of species $i$ at this location.<\/li>\n<li>Calculate<\/li>\n<\/ol>\n<p>\\begin{displaymath}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.1}<br \/>\n\\frac{S_{\\rm min}}{F}=\\frac{(x_i)_F-(x_i)_M}{(x_i)_M-(x_i)_S}<br \/>\n\\end{displaymath}<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">Example<\/header>\n<div class=\"textbox__content\">\n<p>We have a 1000 kg\/hr feed that contains 30 wt% acetone and 70 wt% water. We want our raffinate to contain no more than 5.0 wt% acetone. What is the minimum mass of pure MIBK required?<\/p>\n<\/div>\n<\/div>\n<h2><strong>Liquid-Liquid Extraction: Sizing Mixer-settler Units<\/strong><\/h2>\n<p>$\\Phi_C$\u00a0= volume fraction occupied by the continuous phase<\/p>\n<p>$\\Phi_D$ = volume fraction occupied by the dispersed phase<\/p>\n<p>$\\mu_C$ = viscosity of the continuous phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)<\/p>\n<p>$\\mu_D$ = viscosity of the dispersed phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)<\/p>\n<p>$\\mu_M$ = viscosity of the mixture (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)<\/p>\n<p>$\\rho_C$ = density of the continuous phase (mass volume<sup>-1<\/sup>)<\/p>\n<p>$\\rho_D$ = density of the dispersed phase (mass volume<sup>-1<\/sup>)<\/p>\n<p>$\\rho_M$ = average density of the mixture (mass volume<sup>-1<\/sup>)<\/p>\n<p>&nbsp;<\/p>\n<p>$D_i$\u00a0= impeller diameter (length)<\/p>\n<p>$D_T$\u00a0= vessel diameter (length)<\/p>\n<p>$H$ = total height of mixer unit (length)<\/p>\n<p>$N$ = rate of impeller rotation (time<sup>-1<\/sup>)<\/p>\n<p>$N_{\\rm Po}$ = impeller power number, read from Fig 8-36 or Perry\u2019s 15-54 (below) based on value of $N_{Re}$\u00a0(unitless)<\/p>\n<p>$(N_{\\rm Re})_C$\u00a0= Reynold\u2019s number in the continuous phase = inertial force\/viscous force (unitless)<\/p>\n<p>$P$ = agitator power (energy time<sup>-1<\/sup>)<\/p>\n<p>$Q_C$\u00a0= volumetric flowrate, continuous phase (volume time<sup>-1<\/sup>)<\/p>\n<p>$Q_D$\u00a0= volumetric flowrate, dispersed phase (volume time<sup>-1<\/sup>)<\/p>\n<p>$V$ = vessel volume (volume)<\/p>\n<p>&nbsp;<\/p>\n<p>Tank and impeller sizing<\/p>\n<p>\\begin{displaymath}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.2}<\/p>\n<p style=\"padding-left: 40px;\">{\\rm residence\\; time} = \\frac{V}{Q_C+Q_D}<\/p>\n<p>\\end{displaymath}<\/p>\n<p>Geometry of a cylinder<\/p>\n<p>\\begin{displaymath}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.3}<\/p>\n<p style=\"padding-left: 40px;\">V = \\frac{{\\pi}D_T^2H}{4}<\/p>\n<p>\\end{displaymath}<\/p>\n<p>General guidelines<\/p>\n<p>\\begin{displaymath}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.4}<\/p>\n<p style=\"padding-left: 40px;\">\\frac{H}{D_T}=1<\/p>\n<p>\\end{displaymath}<\/p>\n<p>\\begin{displaymath}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.5}<\/p>\n<p style=\"padding-left: 40px;\">\\frac{D_i}{D_T}=\\frac{1}{3}<\/p>\n<p>\\end{displaymath}<\/p>\n<p>Impeller power consumption:<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.6}<\/p>\n<p style=\"padding-left: 40px;\">P=N_{Po}N^3D_i^5{\\rho}_m<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.7}<\/p>\n<p style=\"padding-left: 40px;\">N_{Re}=\\frac{D_i^2N{\\rho}_M}{{\\mu}_M}<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.8}<\/p>\n<p style=\"padding-left: 40px;\">{\\rho}_M={\\rho}_C{\\Phi}_C+{\\rho}_D{\\Phi}_D<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{5.9}<\/p>\n<p style=\"padding-left: 40px;\">{\\mu}_M=\\frac{{\\mu}_C}{{\\Phi}_C}\\left[1+\\frac{1.5{\\mu}_D{\\Phi}_D}{{\\mu}_C+{\\mu}_D}\\right]<\/p>\n<p>\\end{equation}<\/p>\n<p><img decoding=\"async\" class=\"alignnone size-medium wp-image-364\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.6-Pic1-300x244-1.jpg\" alt=\"\" width=\"300\" height=\"244\" \/> <img decoding=\"async\" class=\"alignnone size-medium wp-image-365\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/2\/2019\/12\/Lecture-1.6-Pic2-300x147-1.jpg\" alt=\"\" width=\"300\" height=\"147\" \/><\/p>\n<h2>Modeling Mass Transfer in Mixer-Settler Units<\/h2>\n<p>$\\Delta\\rho$ = density difference (absolute value) between the continuous and dispersed phases (mass volume<sup>-1<\/sup>)<\/p>\n<p>$\\phi_C$\u00a0= volume fraction occupied by the continuous phase<\/p>\n<p>$\\phi_D$ = volume fraction occupied by the dispersed phase<\/p>\n<p>$\\mu_C$\u00a0= viscosity of the continuous phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)<\/p>\n<p>$\\mu_D$ = viscosity of the dispersed phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)<\/p>\n<p>$\\mu_M$ = viscosity of the mixture (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)<\/p>\n<p>$\\rho_C$ = density of the continuous phase (mass volume<sup>-1<\/sup>)<\/p>\n<p>$\\rho_D$ = density of the dispersed phase (mass volume<sup>-1<\/sup>)<\/p>\n<p>$\\rho_M$ = average density of the mixture (mass volume<sup>-1<\/sup>)<\/p>\n<p>$\\sigma$ = interfacial tension between the continuous and dispersed phases<br \/>\n(mass time<sup>-2<\/sup>)<\/p>\n<p>$a$ = interfacial area between the two phases per unit volume (area volume<sup>-1<\/sup>)<\/p>\n<p>$c_{D,\\rm in}$, $c_{D,\\rm out}$\u00a0= concentration of solute in the incoming or outgoing dispersed streams (mass volume<sup>-1<\/sup>)<\/p>\n<p>$c^*_D$\u00a0= concentration of solute in the dispersed phase if in equilibrium with the outgoing continuous phase (mass volume<sup>-1<\/sup>)<\/p>\n<p>$D_C$\u00a0= diffusivity of the solute in the continuous phase (area time<sup>-1<\/sup>)<\/p>\n<p>$D_D$ = diffusivity of the solute in the dispersed phase (area time<sup>-1<\/sup>)<\/p>\n<p>$D_i$\u00a0= impeller diameter (length)<\/p>\n<p>$D_T$\u00a0= vessel diameter (length)<\/p>\n<p>$d_{vs}$\u00a0= Sauter mean droplet diameter; actual drop size expected to range from $0.3d_{vs}-3.0d_{vs}$\u00a0(length)<\/p>\n<p>$E_{MD}$\u00a0= Murphree dispersed-phase efficiency for extraction<\/p>\n<p>$g$ = gravitational constant (length time<sup>-2<\/sup>)<\/p>\n<p>$H$ = total height of mixer unit (length)<\/p>\n<p>$k_c$\u00a0= mass transfer coefficient of the solute in the continuous phase (length time<sup>-1<\/sup>)<\/p>\n<p>$k_D$\u00a0= mass transfer coefficient of the solute in the dispersed phase (length time<sup>-1<\/sup>)<\/p>\n<p>$K_{OD}$\u00a0= overall mass transfer coefficient, given on the basis of the dispersed phase (length time<sup>-1<\/sup>)<\/p>\n<p>$m$ = distribution coefficient of the solute, $\\Delta c_C\/\\Delta c_D$\u00a0(unitless)<\/p>\n<p>$N$ = rate of impeller rotation (time<sup>-1<\/sup>)<\/p>\n<p>$(N_{\\rm Eo})_C$\u00a0= Eotvos number = gravitational force\/surface tension force (unitless)<\/p>\n<p>$(N_{\\rm Fr})_C$\u00a0= Froude number in the continuous phase = inertial force\/gravitational force (unitless)<\/p>\n<p>$N_{\\rm min}$\u00a0= minimum impeller rotation rate required for complete dispersion of one liquid into another<\/p>\n<p>$(N_{\\rm Re})_C$\u00a0= Reynold\u2019s number in the continuous phase = inertial force\/viscous force (unitless)<\/p>\n<p>$(N_{\\rm Sh})_C$\u00a0= Sherwood number in the continuous phase = mass transfer rate\/diffusion rate (unitless)<\/p>\n<p>$(N_{\\rm Sc})_C$\u00a0= Schmidt number in the continuous phase = momentum\/mass diffusivity (unitless)<\/p>\n<p>$(N_{\\rm We})_C$\u00a0= Weber number = inertial force\/surface tension (unitless)<\/p>\n<p>$Q_D$\u00a0= volumetric flowrate of the dispersed phase (volume time<sup>-1<\/sup>)<\/p>\n<p>$V$ = vessel volume (volume)<\/p>\n<p>&nbsp;<\/p>\n<p>Calculating $N_{\\rm min}$<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.1}<\/p>\n<p style=\"padding-left: 40px;\">\\frac{N_{\\rm min}^2{\\rho}_MD_i}{g{\\Delta}{\\rho}}=1.03\\left(\\frac{D_T}{D_i}\\right)^{2.76}({\\phi}_D)^{0.106}\\left(\\frac{{\\mu}_M^2{\\sigma}}{D_i^5{\\rho}_Mg^2({\\Delta}{\\rho})^2}\\right)^{0.084}<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.2}<\/p>\n<p style=\"padding-left: 40px;\">{\\rho}_M={\\rho}_C{\\phi}_C+{\\rho}_D{\\phi}_D<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.3}<\/p>\n<p style=\"padding-left: 40px;\">{\\mu}_M=\\frac{{\\mu}_C}{{\\phi}_C}\\left(1+\\frac{1.5{\\mu}_D{\\phi}_D}{{\\mu}_C+{\\mu}_D}\\right)<\/p>\n<p>\\end{equation}<\/p>\n<p>&nbsp;<\/p>\n<p>Estimating Murphree efficiency for a proposed design<\/p>\n<p style=\"padding-left: 40px;\">Sauter mean diameter<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.4}<\/p>\n<p style=\"padding-left: 40px;\">{\\rm if}\\;\\; N_{\\rm We} &lt; 10,000,\\; d_{vs}=0.052D_i(N_{\\rm We})^{-0.6}\\exp({4{\\phi}_D})<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.5}<\/p>\n<p style=\"padding-left: 40px;\">{\\rm if}\\;\\; N_{\\rm We}\u00a0&gt;10,000,\\; d_{vs}=0.39D_i(N_{\\rm We})^{-0.6}<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.6}<\/p>\n<p style=\"padding-left: 40px;\">N_{\\rm We}=\\frac{D_i^3N^2{\\rho}_C}{\\sigma}<\/p>\n<p>\\end{equation}<\/p>\n<p style=\"padding-left: 40px;\">mass transfer coefficient of the solute in each phase<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.7}<\/p>\n<p style=\"padding-left: 40px;\">k_D=\\frac{6.6D_D}{d_{vs}}<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.8}<\/p>\n<p style=\"padding-left: 40px;\">k_C=\\frac{(N_{\\rm Sh})_CD_c}{d_{vs}}<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{align}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.9}<\/p>\n<p style=\"padding-left: 40px;\">&amp; (N_{\\rm Sh})_C = \u00a01.237\\times 10^{-5}(N_{\\rm Sc})_C^{1\/3}(N_{\\rm Re})_C^{2\/3}(\\phi_D)^{-1\/2}\\\\<br \/>\n&amp; (N_{\\rm Fr})_C^{5\/12}\\left(\\frac{D_i}{d_{vs}}\\right)^2<br \/>\n\\left(\\frac{d_{vs}}{D_T}\\right)^{1\/2}(N_{\\rm Eo})_C^{5\/4}<\/p>\n<p>\\end{align}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.10}<\/p>\n<p style=\"padding-left: 40px;\">(N_{\\rm Sc})_C=\\frac{{\\mu}_C}{{\\rho}_CD_C}<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.11}<\/p>\n<p style=\"padding-left: 40px;\">(N_{\\rm Re})_C=\\frac{D_i^2N{\\rho}_C}{{\\mu}_C}<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{displaymath}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.12}<\/p>\n<p style=\"padding-left: 40px;\">(N_{\\rm Fr})_C=\\frac{D_iN^2}{g}<\/p>\n<p>\\end{displaymath}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.13}<\/p>\n<p style=\"padding-left: 40px;\">(N_{\\rm Eo})_C=\\frac{{\\rho}_Dd_{vs}^2g}{{\\sigma}}<\/p>\n<p>\\end{equation}<\/p>\n<p style=\"padding-left: 40px;\">Overall mass transfer coefficient for the solute<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.14}<\/p>\n<p style=\"padding-left: 40px;\">\\frac{1}{K_{OD}}=\\frac{1}{k_D}+\\frac{1}{mk_C}<\/p>\n<p>\\end{equation}<\/p>\n<p>Murphree efficiency<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.15}<\/p>\n<p style=\"padding-left: 40px;\">E_{MD}=\\frac{K_{OD}aV}{Q_D}\\left(1+{\\frac{K_{OD}aV}{Q_D}}\\right)^{-1}<\/p>\n<p>\\end{equation}<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.16}<\/p>\n<p style=\"padding-left: 40px;\">a=\\frac{6\\phi_D}{d_{vs}}<\/p>\n<p>\\end{equation}<\/p>\n<p>Experimental assessment of efficiency<\/p>\n<p>\\begin{equation}<\/p>\n<p style=\"padding-left: 40px;\">\\tag{6.17}<\/p>\n<p style=\"padding-left: 40px;\">E_{MD}=\\frac{c_{D,\\rm in}-c_{D,\\rm out}}{c_{D,\\rm in}-c^*_D}<\/p>\n<p>\\end{equation}<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">Example<\/header>\n<div class=\"textbox__content\">\n<p>1000 kg\/hr of 30 wt% acetone and 70 wt% water is to be extracted with 1000 kg\/hr of pure MIBK. Assume that the extract is the continuous phase, a residence time of 5 minutes in the mixing vessel, standard sizing of the mixing vessel and impeller. Find the power consumption and Murphree efficiency if the system operates at $N_{\\rm min}$, controlled at the level of 1 rev\/s. Ignore the contribution of the solute and the co-solvent to the physical properties of each phase.<\/p>\n<ul>\n<li>MIBK\n<ul>\n<li>density = 802 kg m<sup>-3<\/sup><\/li>\n<li>viscosity = 0.58 cP<\/li>\n<li>diffusivity with acetone at 25\u00b0C = 2.90&#215;10<sup>-9<\/sup> m<sup>2<\/sup> s<sup>-1<\/sup><\/li>\n<\/ul>\n<\/li>\n<li>Water\n<ul>\n<li>density = 1000 kg m<sup>-3 <\/sup><\/li>\n<li>viscosity = 0.895 cP<\/li>\n<li>diffusivity with acetone at 25\u00b0C = 1.16&#215;10<sup>-9<\/sup> m<sup>2<\/sup> s<sup>-1<\/sup><\/li>\n<\/ul>\n<\/li>\n<li>The interfacial tension of water and MIBK at 25\u00b0C = 0.0157 kg s<sup>-2<\/sup>. Use the ternary phase diagram to find $m$.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h2>Liquid-Liquid Extraction Columns<\/h2>\n<p>$\\Delta \\rho$ = density difference (absolute value) between the continuous and dispersed phases (mass volume<sup>-1)<\/sup><\/p>\n<p>$\\mu_C$\u00a0= viscosity of the continuous phase (mass time<sup>-1<\/sup> length<sup>-1<\/sup>)<\/p>\n<p>$\\rho_C$ = density of the continuous phase (mass volume<sup>-1<\/sup>)<\/p>\n<p>$\\rho_D$ = density of the dispersed phase (mass volume<sup>-1<\/sup>)<\/p>\n<p>$\\sigma$ = interfacial tension between the continuous and dispersed phases<br \/>\n(mass time<sup>-2<\/sup>)<\/p>\n<p>&nbsp;<\/p>\n<p>$D_T$\u00a0= column diameter (length)<\/p>\n<p>$H$ = total height of column (length)<\/p>\n<p>${\\rm HETS}$ = height of equilibrium transfer stage (length)<\/p>\n<p>$m^*_C$\u00a0= mass flowrate of the entering continuous phase (mass time<sup>-1<\/sup>)<\/p>\n<p>$m^*_D$ = mass flowrate of the entering dispersed phase (mass time<sup>-1<\/sup>)<\/p>\n<p>$N$ = required number of equilibrium stages<\/p>\n<p>$u_0$\u00a0= characteristic rise velocity of a droplet of the dispersed phase (length time<sup>-1<\/sup>)<\/p>\n<p>$U_i$\u00a0= superficial velocity of phase $i$ (C = continuous, downward; D = dispersed, upward)\u00a0 (length time<sup>-1<\/sup>)<\/p>\n<p>$V^*_i$\u00a0= volumetric flowrate of phase $i$ (volume time<sup>-1<\/sup>)<\/p>\n<p>&nbsp;<\/p>\n<p>\\begin{displaymath}<\/p>\n<p>\\tag{7.1}<\/p>\n<p>U_i=\\frac{4V^*_i}{{\\pi}D_T^2}<\/p>\n<p>\\end{displaymath}<\/p>\n<p>definition of superficial velocity<\/p>\n<p style=\"padding-left: 40px;\">\u00a0\\begin{displaymath}<\/p>\n<p>\\tag{7.2}<\/p>\n<p>\\frac{U_D}{U_C}=\\frac{m^*_D}{m^*_C}\\left(\\frac{{\\rho}_C}{\\rho_D}\\right)<\/p>\n<p>\\end{displaymath}<\/p>\n<p style=\"padding-left: 40px;\">\\begin{displaymath}<\/p>\n<p>\\tag{7.3}<\/p>\n<p>(U_D+U_C)_{\\rm actual}=0.50(U_D+U_C)_f<\/p>\n<p>\\end{displaymath}<\/p>\n<p>for operation at 50% of flooding<\/p>\n<p>\\begin{displaymath}<\/p>\n<p>\\tag{7.4}<\/p>\n<p>u_0=\\frac{0.01{\\sigma}{\\Delta}{\\rho}}{{\\mu}_C{\\rho}_C}<\/p>\n<p>\\end{displaymath}<\/p>\n<p>for rotating-disk columns, $D_T$\u00a0= 8 to 42 inches, with one aqueous phase<\/p>\n<p>&nbsp;<\/p>\n<p>\\begin{displaymath}<\/p>\n<p>\\tag{7.5}<\/p>\n<p>D_T=\\left(\\frac{4m^*_D}{{\\rho}_DU_D{\\pi}}\\right)^{0.5}=\\left(\\frac{4m^*_C}{{\\rho}_CU_C{\\pi}}\\right)^{0.5}<\/p>\n<p>\\end{displaymath}<\/p>\n<p>\\begin{displaymath}<\/p>\n<p>\\tag{7.6}<\/p>\n<p>H = {\\rm HETS}*N<\/p>\n<p>\\end{displaymath}<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">Example<\/header>\n<div class=\"textbox__content\">\n<p>1000 kg\/hr of 30 wt% acetone and 70 wt% water is to be extracted with 1000 kg\/hr of pure MIBK in a 2-stage column process. Assume that the extract is the dispersed phase. Ignoring the contribution of the solute and the co-solvent to the physical properties of each phase, find the required column diameter and height.<\/p>\n<ul>\n<li>MIBK\n<ul>\n<li>density = 802 kg m<sup>-3 <\/sup><\/li>\n<li>viscosity = 0.58 cP<\/li>\n<\/ul>\n<\/li>\n<li>Water\n<ul>\n<li>density = 1000 kg m<sup>-3 <\/sup><\/li>\n<li>viscosity = 0.895 cP<\/li>\n<\/ul>\n<\/li>\n<li>The interfacial tension of water and MIBK at 25\u00b0C = 0.0157 kg s<sup>-2<\/sup>.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n","protected":false},"author":1,"menu_order":4,"template":"","meta":{"pb_show_title":"","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-35","chapter","type-chapter","status-publish","hentry"],"part":20,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/pressbooks\/v2\/chapters\/35","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":1,"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/pressbooks\/v2\/chapters\/35\/revisions"}],"predecessor-version":[{"id":36,"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/pressbooks\/v2\/chapters\/35\/revisions\/36"}],"part":[{"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/pressbooks\/v2\/parts\/20"}],"metadata":[{"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/pressbooks\/v2\/chapters\/35\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/wp\/v2\/media?parent=35"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/pressbooks\/v2\/chapter-type?post=35"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/wp\/v2\/contributor?post=35"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/wp-json\/wp\/v2\/license?post=35"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}