{"id":42,"date":"2019-06-27T18:22:31","date_gmt":"2019-06-27T18:22:31","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/chapter\/absorption-and-stripping\/"},"modified":"2026-03-16T01:12:55","modified_gmt":"2026-03-16T01:12:55","slug":"absorption-and-stripping","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/chemicalengineeringseparations\/chapter\/absorption-and-stripping\/","title":{"raw":"Absorption and Stripping","rendered":"Absorption and Stripping"},"content":{"raw":"

Overview of Absorption and Stripping Processes<\/h2>\n$\\phi_A$\u00a0= fraction not absorbed (unitless)\n\n$\\phi_S$ = fraction not stripped (unitless)\n\n$A$ = absorption factor (unitless)\n\n$K$ = equilibrium constant, (mole fraction in gas)\/(mole fraction in liquid)\n\n$L$ = total liquid flow rate (mol time-1<\/sup>)\n\n$V$ = total gas flow rate (mol time-1<\/sup>)\n\n$S$ = stripping factor (unitless)\n\n$x$ (lowercase) = mol fraction of solute in liquid phase (mol solute\/mol total liquid)\n\n$X$ (uppercase) = mol ratio of solute in liquid phase (mol solute\/mol solvent)\n\n$y$ (lowercase) = mol fraction of solute in gas phase (mol solute\/mol total vapor)\n\n$Y$ (uppercase) = mol ratio of solute in gas phase (mol solute\/mol gaseous carrier)\n\n\\begin{equation}\n

\\tag{8.1}<\/p>\n

A=\\frac{(L\/V)}{K}<\/p>\n\\end{equation}\n\n\\begin{equation}\n

\\tag{8.2}<\/p>\n

S=\\frac{K}{(L\/V)}=\\frac{1}{A}<\/p>\n\\end{equation}\n

<\/p>\n \n\n \n

\n

Example<\/p>\n\n<\/header>\n

\n\nK for acetone in an air\/water system is 2.0. We intend to absorb 90% of the acetone entering in the gaseous phase into the liquid phase. How many theoretical stages would be required for (L\/V) = 1, (L\/V) = 2, and (L\/V) = 10?\n\n<\/div>\n<\/div>\nFinding mole ratio from mole fraction:\n\n\\begin{displaymath}\n\n\\tag{8.3}\n\nY=\\frac{y}{1-y}\n\n\\end{displaymath}\n\n\\begin{displaymath}\n\n\\tag{8.4}\n\nX=\\frac{x}{1-x}\n\n\\end{displaymath}\n\nFinding mole fraction from mole ratio:\n\n\\begin{displaymath}\n\n\\tag{8.5}\n\ny=\\frac{Y}{1+Y}\n\n\\end{displaymath}\n\n\\begin{displaymath}\n\n\\tag{8.6}\n\nx=\\frac{X}{1+X}\n\n\\end{displaymath}\n

Graphical Method to Find Equilibrium Stages for Absorption Columns<\/h2>\n$K$ = equilibrium constant, (mole fraction in gas)\/(mole fraction in liquid)\n\n$L$ = liquid flow rate (mol time-1<\/sup>)\n\n$L'$ = liquid flow rate on a solute-free basis (mol time-1<\/sup>)\n\n$N$ = total number of stages. For an absorption column, $N$ is the bottom stage. For a stripping column, $N$ is the top stage.\n\n$n$ = generic stage number\n\n$V$ = gas flow rate (mol total time-1<\/sup>)\n\n$V\u2019$ = gas flow rate on a solute-free basis (mol gaseous carrier time-1<\/sup>)\n\n$x$ = mole fraction of solute in liquid phase (mol solute\/mol total liquid)\n\n$X$ = mole ratio of solute in liquid phase (mol solute\/mol liquid absorbent)\n\n$X_0$\u00a0= mole ratio of solute\/absorbent in liquid entering the top of the absorption column\n\n$X_N$\u00a0= mole ratio of solute\/absorbent in liquid leaving the bottom of the absorption column\n\n$y$ = mole fraction of solute in gas phase (mol solute\/mol total gas)\n\n$Y$ = mole ratio of solute in gas phase (mol solute\/mol gaseous carrier)\n\n$Y_1$\u00a0= mole ratio of solute\/gaseous carrier in gas leaving the top of the absorption column\n\n$Y_{N+1}$\u00a0= mole ratio of solute\/gaseous carrier in gas entering the bottom of the absorption column\n\n \n\nOperating line for absorption column:\n\n\\begin{equation}\n

\\tag{9.1}<\/p>\n

Y_{n+1}=X_n\\left(\\frac{L'}{V'}\\right)+Y_1-X_0\\left(\\frac{L'}{V'}\\right)<\/p>\n\\end{equation}\n\nWatch a video from LearnChemE<\/a> that demonstrates how to determine the number of equilibrium stages required for an absorption column when given inlet and outlet specifications: Absorption of a Dilute Species (10:43): https:\/\/youtu.be\/BoPKngZZwVI<\/a>\n

\n

Example<\/p>\n\n<\/header>\n

\n\nThe following equilibrium data is available for water, ammonia, and air at 20\u00b0C.\n\n\"\"\n\nWe intend to use fresh water to absorb the ammonia from a stream of air containing 28.5 mol% ammonia, with both streams at 20\u00b0C. We intend to absorb 80% of the incoming ammonia. How many equilibrium stages are required for each of the following design cases?\n\n(a) $L\u2019\/V\u2019$ = 1.0\n\n\"\"\u00a0\"Schematic\n\n(b) $L\u2019\/V\u2019$ =\u00a0 1.6\n\n\"\"\u00a0 \"Schematic\n\n(c) $L\u2019\/V\u2019$= 2.0\n\n\"\" \"Schematic\n\n<\/div>\n<\/div>\n

Determining the Minimum Ratio for Absorbent to Gas Flow Rates and a Mathematical Approach for Finding the Number of Equilibrium Stages<\/h2>\n$K$ = equilibrium constant, (mole fraction solute in gas)\/(mole fraction solute in liquid)\n\n$L\u2019$ = liquid flow rate on a solute-free basis (mol liquid absorbent time-1<\/sup>)\n\n$n$ = generic equilibrium stage\n\n$N$ = bottom equilibrium stage for an absorption column\n\n$V\u2019$ = gas flow rate on a solute-free basis (mol gaseous carrier time-1<\/sup>)\n\n$(L\u2019\/V\u2019)_{\\rm min}$ = relative molar flowrates of solute-free absorbent and solute-free carrier gas at which an infinite number of equilibrium stages is required in order to achieve the desired separation\n\n$x$\u00a0= mole fraction of solute in liquid phase (mol solute\/mol total liquid)\n\n$X$ = mole ratio of solute in liquid phase (mol solute\/mol liquid absorbent)\n\n$X_0$\u00a0= mole ratio of solute\/absorbent in liquid entering the top of the absorption column\n\n$X_N$\u00a0= mole ratio of solute\/absorbent in liquid leaving the bottom of the absorption column\n\n$y$ = mole fraction of solute in gas phase (mol solute\/mol total gas)\n\n$Y$ = mole ratio of solute in gas phase (mol solute\/mol gaseous carrier)\n\n$Y_1$\u00a0= mole ratio of solute\/gaseous carrier in gas leaving the top of the absorption column\n\n$Y_{N+1}$\u00a0= mole ratio of solute\/gaseous carrier in gas entering the bottom of the absorption column\n\n \n\nEquilibrium:\n\n\\begin{equation}\n

\\tag{10.1}<\/p>\n

K_n=\\frac{y_n}{x_n}=\\frac{Y_n\/(1+Y_n)}{X_n\/(1+X_n)}<\/p>\n\\end{equation}\n\nOperating line:\n\n\\begin{equation}\n

\\tag{10.2}<\/p>\n

Y_{n+1}=X_n\\left(\\frac{L'}{V'}\\right)+Y_1-X_0\\left(\\frac{L'}{V'}\\right)<\/p>\n\\end{equation}\n\nwhen $X$ ~ $x$ and $Y$ ~ $y$,\n\n\\begin{equation}\n

\\tag{10.3}<\/p>\n

\\left(\\frac{L'}{V'}\\right)_{\\rm min}=\\frac{y_{N+1}-y_1}{\\frac{y_{N+1}}{K_N}-x_0}<\/p>\n\\end{equation}\n\nwhen $X$ ~ $x$, $Y$ ~ $y$ and $x$0<\/sub> = 0\n\n\\begin{equation}\n

\\tag{10.4}<\/p>\n

\\left(\\frac{L'}{V'}\\right)_{\\rm min}=K_N\\left(1-\\frac{y_1}{y_{N+1}}\\right)<\/p>\n\\end{equation}\n

\n

Example<\/p>\n\n<\/header>\n

\n\n180 kmol\/hr of off-gas from a fermentation unit contains 98 mol% CO2<\/sub> and 2 mol% ethanol. We would like to recover 97% of this ethanol and have proposed the use of a staged absorption column with water as the absorbent. The incoming water contains no ethanol and is at 30\u00b0C. The incoming gaseous stream is at 30\u00b0C, 110 kPa. Assume there is no absorption of CO2<\/sub> into the water. The activity coefficient at infinite dilution for ethanol in water at 30\u00b0C is 6.0 and Psat<\/sup> is 10.5 kPa. Thus, K = \u03b3i<\/sub>Pi<\/sub>sat<\/sup>\/P = 0.57.\n

(a) What is $(L\u2019\/V\u2019)_{\\rm min}$?<\/p>\n

(b) How many equilibrium stages are required if we operate at 1.5*$(L\u2019\/V\u2019)_{\\rm min}$?<\/p>\n\"Schematic\n\n<\/div>\n<\/div>\n

Stage Efficiency and Column Height<\/h2>\n$\\lambda$ = inverse absorption factor (unitless)\n\n$\\mu_L$\u00a0= liquid-phase viscosity (mass time-1<\/sup> length-1<\/sup>)\n\n$\\rho_L$ = liquid-phase density (mass volume-1<\/sup>)\n\n \n\n$a$ = vapor-liquid interfacial area per volume of combined gas\/liquid on the tray (area volume-1<\/sup>)=(length-1<\/sup>)\n\n$A_b$\u00a0= portion of the column cross-sectional area dedicated to mixing of the gas\/liquid (area)\n\n$D_T$\u00a0= column diameter (length)\n\n$E_{MV}$\u00a0= Murphree efficiency, calculated on the basis of the vapor phase composition\n\n$E_O$\u00a0= overall stage efficiency\n\n$K$ = equilibrium constant (gas-phase composition\/liquid-phase composition)\n\n$K_G$\u00a0= overall gas mass-transfer coefficient, partial pressure driving force (mol time-1<\/sup> area-1<\/sup> pressure-1<\/sup>)\n\n$L$ = molar flow rate of liquid phase (mol time-1<\/sup>)\n\n$M_L$\u00a0= molecular weight of the liquid phase\n\n$N_a$\u00a0= actual number of stages required to achieve the desired separation\n\n$N_{OG}$\u00a0= number of overall gas phase mass transfer units\n\n$N_t$\u00a0= number of theoretical stages required to achieve the desired separation\n\n$P$ = pressure (force area-1<\/sup>)\n\n$T$ = temperature (temperature)\n\n$V$ = molar flow rate of gaseous phase (mol time-1<\/sup>)\n\n$x_{i,n}$\u00a0= the actual mole fraction of species i in the liquid leaving stage n and entering stage n+1\n\n$y_i$\u00a0= the mole fraction of species i in a gas at some fixed point in the column\n\n$y_{i,n+1}$\u00a0= the actual mole fraction of species i in the gas leaving stage n+1 and entering stage n\n\n$y_{i,n}$\u00a0= the actual mole fraction of species i in the gas leaving stage n and entering stage n-1\n\n$y^*_{i,n}$ = the mole fraction of species i in the gas leaving stage n if it had reached equilibrium with the liquid leaving that stage, where the liquid phase composition is $x_{i,n}$\n\n$Z_f$\u00a0= height of combined gas and liquid holdup on the tray (length)\n\n \n\n\\begin{equation}\n

\\tag{11.1}<\/p>\n

E_O=\\frac{N_t}{N_a}<\/p>\n\\end{equation}\n\n\\begin{equation}\n

\\tag{11.2}<\/p>\n

E_O=19.2-57.8\\log_{10}{\\mu_L} \\;\\;\\;\\; 0.2 < \\mu_L < 1.6\\; {\\rm cP}, \\mu_L {\\rm \\;must\\; be\\; in\\; cP}<\/p>\n\\end{equation}\n\n0.2 < $\\mu_L$ < 1.6 cP, $\\mu_L$\u00a0must be in cP\n\n\\begin{equation}\n

\\tag{11.3}<\/p>\n

\\log_{10}{E_O}=1.597-0.199\\log_{10}{\\left(\\frac{KM_L{\\mu}_L}{{\\rho}_L}\\right)}-0.0896\\left[\\log_{10}{\\left(\\frac{KM_L{\\mu}_L}{{\\rho}_L}}\\right)\\right]^{2}<\/p>\n\\end{equation}\n\nEq 11.3 is applicable for this range of conditions: $\\mu_L$ must be in cP, $\\rho_L$\u00a0must be in lb ft-3<\/sup>; restricted to $D_T$ = 2 in \u2013 9 ft, average $P$ = 14.7 \u2013 485 psi, average $T$ = 60 \u2013 138\u00b0F, $E_O$\u00a0= 0.65% - 69%\n\n \n

\n

Example<\/p>\n\n<\/header>\n

\n\nFor our CO2<\/sub>\/ethanol\/water system, we found that for 97% removal of the ethanol using water as the absorbent and operating at 30\u00b0C, 110 kPa, with $K$ = 0.57 and $(L\u2019\/V\u2019)_{\\rm actual}$ = $(L\u2019\/V\u2019)_{\\rm min}$\u00a0= 0.828, 7 theoretical stages required. How many actual stages are needed, according to equations 11.2 and 11.3?\n