Mass Transfer Operations β Robert TreybalSample Problems 1. If 100 kg of a solution of acetic acid (C) and water (A) containing 30% acid is to be extracted three times with isopropyl ether (B) at 20Β°C, using 40kg of solvent in each stage, determine the quantities and compositions of the various streams. How much solvent would be required if the same final raffinate concentration were to be obtained with one stage? Solution. The equilibrium data at 20Β°C are listed below [Trans. AIChE, 36, 628 (1940), with permission]. The horizontal rows give the concentrations in equilibrium solutions. The system is of the type shown in Fig 10.9, except that the tie lines slope downward toward the B apex. The rectangular coordinates of Fig. 10.9a will be used, but only for acid concentrations up to x=0.30. These are plotted in Fig. 10.15. Stage 1 F=100 kg, xF=0.30, yS=0, S1=B1=40 kg OMB: M1=100+40=140 kg x bal: 100(0.30)+40(0)=140xM1 => xM1=0.214 Point M1 is located on line FB. With the help of a distribution curve, the tie line passing through M1 is located as shown, and x1=0.258, y1=0.117 wt fraction acetic acid. 140(0.214β0.258) πΈ1 = = 43.6 ππ 0.117β0.258 R1=140-43.6=96.4 kg Stage 2 S2=B2=40 kg M2=R1+B2=96.4+40=136.4 kg Point M2 is located on line R1B and the tie line R2E2 through M2.x2=0.227, y2=0.095 become π2 (π₯π2 βπ₯2 ) 136.4(0.1822β0.227) πΈ2 = = = 46.3 ππ π¦2 βπ₯2 0.095β0.227 R2=M2-E2=136.4-46.3=90.1 kg Stage 3. In a similar manner, B3=40, M3=130.1, xM3=0.1572, x3=0.20, y3=0.078, E3=45.7 and R3=84.4. The acid content of the final raffinate is 0.20(84.4)=16.88 kg. The composited extract is E1+E2+E3=43.6+46.3+45.7=135.6 kg and its acid content = E1y1+E2y2+E3y3=13.12 kg. If an extraction to give the same final raffinate concentration, x=0.20, were to be done in one stage, the point M would be at the intersection of tie line R3E3 and line BF or at xM=0.12. The solvent required would then be, by, S1=100(0.30-0.12)/(0.12-0)= 150 kg, instead of the total of 120 required in the three-stage extraction. 2. Nicotine (C) in a water (A) solution containing 1% nicotine is to be extracted with kerosene (B) at 20Β°C. Water and kerosene are essentially insoluble. (a) Determine the percentage extraction of nicotine if 100 kg of feed solution is extracted once with 150 kg solvent. (b) Repeat for three theoretical extractions using 50 kg solvent each. Solution. Equilibrium data are provided by Claffey et al., Ind. Eng. Chem., 42, 166 (1950), and expressed as kg nicotine/kg liquid, they are as follows: (a) xF= 0.01 wt fraction nicotine, xβF= 0.01/(1-0.01)= 0.0101 kg nicotine/ kg water. F= 100 kg. A= 100(1-0.01)= 99 kg water, A/B= 99/150 = 0.66. Refer to the figure which shows the equilibrium data and the point F representing the composition of the feed. From F, line FD is drawn of slope - 0.66, intersecting the equilibrium curve at D where xβ1=0.00425 and yβ1=0.00380 kg nicotine/kg liquid. The nicotine removed from the water is therefore 99(0.0101-0.00425)= 0.580 kg, or 58% of that in the feed. (b) For each stage, A/B=99/50= 1.98. The construction is started at F, with operating lines of slope -1.98. The final raffinate composition is xβ3=0.0034 and the nicotine extracted is 99(0.0101-0.0034)= 0.663 kg or 66.3% of that in the feed. 3. If 8000 kg/h of an acetic acid (C)- water (A) solution, containing 30% acid is to be counter currently extracted with isopropyl ether (B) to reduce the acid concentration to 2% in the solvent-free raffinate product, determine (a) the minimum amount of solvent which can be used and (b) the number of theoretical stages if 20 000 kg/h of solvent is used. Solution. The equilibrium data are plotted on triangular coordinates in the figure. The tie lines have been omitted for clarity. (a) F= 8000 kg/h; xF=0.30 wt fraction acetic acid, corresponding to point F on the figure RβN, as shown. In this case the tie line J, which when extended passes through F, provides the conditions for minimum solvent, and this intersects line RNB on the right of the figure nearer B than any other lower tie line. Tie line J provides the minimum E1 as shown at y1=0.143. Line E1mRN, intersects line FB at Mm, for which xM=0.114 with ys=0 and S=B: πΉπ₯πΉ 8000(0.30) π΅π = π₯π βπΉ = 0.114 β 8000 = 13 040 kg/h, min solvent rate (b) For B= 20 000 kg solvent/h with yS=0 and S=B πΉπ₯πΉ 8000(0.30) π₯π = = = 0.0857 πΉ + π΅ 8000 + 20 000 and point M is located as shown on line FB. Line π ππΉ π extended provides E1 at y1=0.10. Line FE1 is extended to intersect line RNB at ΞR. Random lines such as OKL are drawn to provide yS+1 at K and xS at Ls as follows: These are plotted on the figure as the operating curve, along with the tie-line data as the equilibrium curve. There are required 7.6 theoretical stages. The weight of extract can be Obtained by an acid balance, π(π₯π β π₯ππΉ ) 28 000(0.01. determine (a) the minimum kerosene rate and (b) the number of theoretical stages required if 1150 kg of kerosene is used per hour. The equilibrium data of number 2 are plotted in the figure below (a) F=1000 kg/h. ys=0 0.02 π ππΉ = π β πΈ1 = 28 000 β 23 000 = 5000 ππ/β 4.02) πΈ1 = = = 23 000 ππ/β π¦1 β π₯ππΉ 0.0101 1 β 0. If 1000 kg/h of a nicotine (C)-water (A) solution containing 1% nicotine is to be counter currently extracted with kerosene at 20Β°C to reduce the nicotine content to 0.1%.01 ππ πππππ‘πππ π₯β²πΉ = = 0. Solution.01 ππ π€ππ‘ππ . xF=0.10 β 0.01)= 990 kg water/h. A=1000(1-0.0857 β 0. 0 50.110)]]0.02. The average is π₯βππΉ 0.798.xβF) and 8.5=1.021= 990/1.0 50.860 π¦β²1 π¦β²1 = = 0. and πβ²π΅ 0.099 and the figure indicates 8.30 0.2.001001 = 1.000kg/h of solvent is used.01.0101 Illustration 10.9 Determine the number of transfer units NTU for the extraction of illustration 10.230 0.192 0.05 0. xβ=0. y2=0.00782 kg nicotine/kg kerosene The operating line is drawn through (yβ1. Alternatively.25 20. The operating diagram is already plotted in Fig.928 π΄ 990 At the concentrated end.0 π₯βπ₯β .001 π₯β²ππΉ = = 0.30 17.928(1. at the dilute end of the system. mβ=0.0101β0. values of x and x* are taken from the operating line and equilibrium curve at various values of y.0093β0 = 0. Since yK=0.030 0 1 14.114 0.001001 yβ1=0.021= 969 kg kerosene/h (b) B=1150 kg/h.75 27.0101 β 0.3 if 20.798(1150) = = 0.001 001) and for infinite stages passes through K on the equilibrium curve at xβF. A/B=990/1150= 0.30.154 0.15 0. = = 0.02 x* 0.4 theoretical stages.0093.10. 10. π₯β²πΉ 0.10 0. x2=0. Solution: Define x and y in terms of weight fractions acetic acid. x1=xF=0. π΄ 0.953. From this plot.001 ππ π€ππ‘ππ The operating line starts at point L (yβ=0.860 π¦β²πΉ βπ₯β²ππΉ 0.8 40. 0.021 π΅π Bm=A/1.001001 [0.35 0.30 0. mβ=dyβ*/dxβ=0.953(1150)/990=1.001 ππ πππππ‘πππ π₯ππΉ = 0.3 theoretical stages are determine graphically.001 001 1 β . and mβB/A=0. as follows: x 0.075 0. y1=0.110. yβ2=0.10 Determine the number of transfer units NTU for the extraction of illustration 10.3β1)+1 πππ = 8.0782. find the number of stages required.2-trichloroethane at 25β.30. 8.(10.116) of the equivalent.8 times the minimum (solvent rate)/(feed rate).40 + ln 1β0.1. 100 kg/h of a 40:60 acetone water solution is to be reduced to 10 percent acetone by extraction with pure 1.4 if 1150 kg/h of kerosene are used.6 Solution: .20 π₯β²2 β π¦2 β²β πβ² = 0. Eq.4) From fig.4. Data are given in Table 23. the mutual solubility ofwater and 18 isopropyl ether is very small.8 Mc Cabe 23. yβ1=0. 1 The area under a curve of x as abscissa against π₯βπ₯β as ordinate (not shown) between x= 0.110): 1 1β0.20.30(0. 8.3β1)+1 = π. (a) Find the minimum solvent rate.0101. In these solutions. xβ1=xβF=0. ππ 2 Illustration 10.40. In a continuous countercurrent train of mixer-settlers.001001 = 0. Fig. so that r can be taken as 60 = 0. xβ2=0.02 1 0. Solution: Use weight-ratio concentration as in Illustration 10.02(0.(10.0909 π₯β²1 β π¦β²2 0. The calculation can be done through Eq.001001.30 + 2 ln 0. (b) At 1.01 (Illustration 10.5. NTU = 8.02 is determined to be 8. (c) For conditions of part (b) find the mass flowrats of all streams.0101 β πβ² The average mE/R = mβB/A = 1.30 and x = 0. Vb = 26.46 (b) Vb = 1.53Va = xaLa = 40 Water: ywaVa + xwbLb = 60 When xb = 0. (a) Basis: 100 kg feed solution Let x.y refer to the wt.46. fraction of acetone: xw. ye Establish the ends of the operating line from overall material balances.63 As before.53 Hence 0. Plot the equilibrium line xe.8 times the minimum = 1.46 = 47.6. xT. from a water balance: YwaVa + xwbLb = 60 Overall: Lb = 100 .1 Lb + yaVa = xaLa =40 At minimum solvent rate.61 Lb = 62.63 β Va Xwb is unchanged at 0.1Lb + 0. yaβ is found from the equilibrium curve at xa = 0.47. yw to water.85 The minimum solvent rate is 26. xTb = 0.8 x 26. from the first part of Table 23. Balances: Overall: Lb + Va = Vb +100 Acetone: 0.40 to be 0. Va = 63.0061 From these.10.8939 .yT to trichloroethane. 404 The lower end is at xb = 0.0089 From an acetone balance: yV= (0.63.01 = 0. yw = 0.74L + (0.8939 β ywa) Va = 147.028 x 83.0. (A) and (C): L = 78.25 Estimate xT to be 0. From the diagram 2. The coordinates of the intermediate point are x= 0. Use 3 stages.11) + 0.404 x 83.63ywa)/(0.Lb By trial.404 From the equilibrium data for ya = 0.80 From Eq. yw = 0. Hence the upper end of the operating line is at xa = 0.10. V=61. Set x=0.8 stages are needed.0089. yb = 0 Intermediate point. ywa = 0.028 Lb=64.25L β 40 y = (0.25L β 6.89 (A) Estimate: y = 0. y = 0.028 (as estimated).52.42)/V (B) Water balance: ywV = 0. xw = 1-.74 Overall balance from feed end: V = L + Va β La = L + 83.11) β 60 (C) From Eqs.25.01. . Va = 83.22.1Lb)/Va = 0. y= 0.404. (B).11 ya = (40 β 0.11 β 100 =L β 16.214. From the equilibrium curve.69.Hence Lb = (60 β 147.25. ya=0. as estimated.214 The operating line is almost straight.40. estimate ywa to be 0. 3 H2O. A tie line through this point on Fig.0 0.2 acetone. 0.232E + 0.232=0. 0.4β0. 0.8 shows the extract to be 0.5 MIK.6 A mixture containing 40 weight percent acetone and 60 weight percent water is contacted with an equal amount of MIK.4 = 0.52 23.023 MIK. (c) The flow rates. (a) What fraction of the acetone can be extracted in a single-stage process? (b) What fraction of the acetone could be extracted if the fresh solvent were divided into two parts and two successive extractions used? Solution: (a) Adding an equal amount of MIK to the feed gives a mixture with 0.4 = 0. and 0.043 H2O.36 .63 Raffinate: 64.725 MIK. The raffinate composition is 0.23E + 0. are Feed: 100 Extract: 83.132R E+R= 1 + 1 = 2.232 acetone. 23. Per unit mass of feed.132 acetone.264 E=0. in kg/h.11 Solvent: 47.132(2-E) 0. and 0. an acetone balance gives 0.845 H2O.132= 1. and 0. 32(0.21(1.755 By a material balance.325E + 0.21 Acetone extracted = 0.105 Total extracted: 0.105 = .24 Adding 0. 1.4 H2O.24 + 0.755(0.523 Acetone extracted: 0.345/ 0.761 parts of raffinate gives a mixture with the following compostions 0.21(0.261-E) E = 0.739.4 = 0.789 0.060 0. and 0.615 0.127 1.261 0.761) Acetone = = 0. 0. E + R = 1.5 parts MIK 50 0. The phase compositions are Extract Raffinate MIK 0.075 acetone.325) = 0.4 β 0.5 0.761) H2O = = 0.267 acetone. π = 0.455 1.35) MIK = = 0.16 = 0.325 β 0.4 = 0.523 x 0.333 MIK.20E + 0.232) Fraction acetone extracted = = 0.210 Water 0.261 0.418 1.315 πΈ= = 0.5+0.325 0.261 This separates to give an extract with 0.075(1.863 .20 = 0.035 Acetone 0.5-E) 0.345 Fraction acetone extracted: 0.4 (b) If onlye half the MIK is added in the first step.739 (0.76(0. the mixture is 0.761 0. A acetone balance give 0.20 acetone and a raffinate with 0. 67 5.25kg/s (17800 lb/h) with chlorobenzene to reduce the pyridine concentration to 2% in the final raffinate.08 99.55 65.24 88.0 3.82 18.0 6. (b) If 2.05 61.25 25.05 0. what are the number of theoretical stages and the saturated weights of extract and raffinate Solution: Pyridine Chlorobenzene Water Pyridine Chlorobenzene Water 0 99.72 28.10 74.58 2.90 37. Solved Problems 10.60 53.05 0 0.95 44. is to be continuously and counter-currently extracted at the rate of 2.40 53.5.92 31.50 0.85 62.58 73.87 36.15 2.1: (a) Determine the minimum solvent rate required. Using the coordinate systems plotted in (b) and (c) of prob.3 kg/s (18250 lb/h) is used.60 69. A pyridine-water solution.90 1.28 1.38 80.62 18. 10.71 24.28 0.1 1.20 8.05 35.02 0.18 50. 50% pyridine.87 40.90 0.95 4.05 88.90 .15 11.95 0.16 94.95 79.92 11. 3 = 4. E.247 Mark point M on FS at xm=0. Mark the point S at the B-apex as the solvent is pure.5) + 2.022 ππππ ππ 2.25(0.02 Solvent flowrate: π = 2.25 + 2. y v/s weight fraction of chlorobenzene Mark point F on y-axis at xf=0.55π₯π π₯π = 0.2 49 37.55ππ/π πΉπ₯π + ππ¦π = ππ₯π 2.6 π’πππ‘π So locate M on FS such that line MF=56. Make a line using point F and S.022 ππ We have: πΉπ = ππΉ + ππ = 112 π’πππ‘π (ππ.3(0) = 4.8 13.8 13.2 Basis: 2.4 π’πππ‘π ππΉ = 112 β 55.6 units and line MS =55.022ππ + 1ππ = 2. ππππ π’πππ) 1.247(read from x-axis) .3 ππ/π Refer to Fig.25 kg/s of pyridine-water solution feed π₯π = 0.2.5 Weight fraction at final raffinate: π₯π = 0.25 ππΉ = 1. 49 37.5 units OR: π = πΉ + π = 2.4 = 56.022ππ = 112 ππ = 55. Now locate M on FS such that: π ππππ ππΉ = πΉ ππππ ππ ππππ ππΉ 2.3 = = 1.14 (a) and (b) Plot x v/s y and x.50. . Locate Rn(raffinate from nth stage) on the bimodal solubility curve corresponding to 2% pyridine(i. . Join FE1 and project these lines to meet at Ξg.02) Join RnM and project it to meet the equilibrium curve at E1.e. at xn=0.Both the above mentioned procedures give the same location for point M on FS. 02) and then count the theoretical stages required.Read y1 corresponding to E1 and from the x vs.8 Water-dioxane solutions form a minimum-boiling azeotrope at atmospheric pressure and cannot be separated by ordinary distillation methods. get the value of x2 corresponding to y2. get the value of x1 (in equilibrium with y1) Mark R1 on the bimodal solubility curve at location corresponding to x1 value. Chem. Benzene forms no azeotrope with dioxane and can be used as an extraction solvent.e. water and benzene are substantially insoluble. 282 (1944)] is as follows: Wt% dioxane in water 5. The line R2Ξg meets the curve at E3. xn=0. which is a tie line for stage-2. E 2.. 66. The benzene is dioxane-free.0 At these concentrations. number of theoretical stages required = n = 3 10. The working is continued in this way till we cover Rn(i.9 25. y plot. y plot. and 1000 kg of a 25% dioxane-75% water solution is to be extracted with benzene to remove 95% of dioxane. In our case. join R2Ξg. Am. At 25Β°C. Join Re1E1 which is a required tie line for first stage. .14(a). the tie line R3E3 is such that R3 is exactly xn=0.1 18. the equilibrium distribution of dioxane between water and benzene [J. Soc. Join R2E2.5 32. Mark R2 on the bimodal curve at location corresponding to x2 value.02 From Fig. From the x vs.2 Wt% dioxane in benzene 5.2 22. 25 Γ 1000 = 250ππ Water in feed solution = 0. A-water. how much fsolvent would be required? Solution: (a) single-stage operation: Basis: 1000kg of solution containing 25% dioxane. (a) Calculate the solvent requirement for a single batch operation (b) If the extraction were done with equal amounts of solvent in five crosscurrent stages. C-dioxane Dioxane in feed solution = 0.75 Γ 1000 = 750ππ . B-benzene. πππ solvent required per liter feed Cross-current operation: Number of stages=3.0155 mol/l Amount of solvent (benzene) used = B liters Picric acid in extract =0. So read the value of CS1 from the plot corresponding to CA1=0. We have to construct/draw three operating lines parallel to each other (as the slope is the same) starting from point F(CF.08 = 0.0) that represents the feed solution and draw the operating line through it which will cut equilibrium curve at P(CA1.1 = 0.2 Γ 0. The operating line for the first stage passes through the point F(CF.1 mol/liter 80% of the picric acid is removed. the operating lines will have the same slope (-A/B).CS1 and through CA1 on x- axis. From the plot: CS1=0. Picric acid in final raffinate=20% of its original value = 0.02mol/liter.02 πππ Concentration of picric acid in final raffinate = = 0. draw the operating line parallel to the first one and so on and construct exactly three stages.e.02πππ 0. Feed solution = 1 liter Concentration of picric acid in feed solution =Cf=0.02πππ/πππ‘πππΆπ3 = 0.1.CS3) So locate point F(CF. Cs1 and CA1 are the equilibrium values of picric acid concentration as the effluent streams leaving theoretical stage are in equilibrium.0155 = π.02 1 πππ‘ππ πππ πΆπ΄3 = 0. the final raffinate concentration. F(0.Cs1 is the equilibrium value of picric acid concentration in extract.08 mol πππππ ππ ππππππ ππππ Concentration of picric acid in extract = CS1= ππππ’ππ ππ π πππ£πππ‘ πππππ ππ ππππππ ππππ Volume of solvent B: π΅π£πππ’ππ = πΆπ1 0.CS0) and the operating line of the third stage passes through the point Q(CA3.CS0) i.0155 (πππ’πππππππ’π π£πππ’π) πππ‘ππ For stage 1: π΄ πΆπ1 β πΆπ 0 β = = π ππππ ππππππ‘πππ ππππ π΅ πΆπ΄1 β πΆπΉ π΄ πΆπ1 βπΆπ0 For stage 3: βπ΅ = πΆπ΄1 βπΆπΉ As the equal amount of solvent is used in each stage.CS0) and ending/covering exactly CA3. . For this.3 = π. we have to adopt a trial and error procedure. measure the slope of any one of these operating lines. ππ³ .3L Total benzene requirement: πππ¨πππ₯ = 3 Γ 1.1 β 0. From the graph: slope of operating line FP 0.773 π΅ π΄ π΅= .3πΏ πππ πππ π π‘πππ 0.773 0.773 1 π΅= = 1. π΄ = 1πΏ 0.773 Benzene required per stage= 1.0255 β 0 π=β = 0.067 π΄ = 0. Once we construct exactly three stages. the results are not a straight line and do not follow the Langmuir Equation (12.117 0. The equilibrium data at room temperature are shown in Table 12.1-1.1-1 Equilibrium Data for Example 12.0011 0.122 0.094 0. TABLE 12.322 0.1-3). (kg phenol / m3 solution) q.059 0. Determine the isotherm that fits the data.045 Solution: Plotting the data as 1/q versus 1/c.0061 0.039 0.1-1 Adsorption Isotherm for Phenol in Wastewater Batch tests were performed in the laboratory using solutions of phenol in water and particles of granular activated carbon (R5). qo c q= K+c . (kg phenol / kg carbon) 0. Principles of Transport Processes and Separation Processes β John Geankoplis Sample Problems 12.1-1 (R5) c.150 0. 229 12.199c 0. follow the Freundlich isotherm Eq.21 kg phenol/m3 of solution (0. and what percent of phenol of phenol is extracted? Solution: cF = 0.2-1 Batch Adsorption on Activated Carbon (Geankoplis Example) A wastewater solution having a volume of 1.21 g/L).A plot of log q versus log c in Fig. A total of 1.0 m3 M = 1.1-1.1-2 gives a straight line and. (12.199.229 and the constant K is 0. 12. to give q=0. which is then mixed thoroughly to reach equilibrium.0 m3 contains 0.0) = q(1. Using the isotherm from Example 12. hence.1-2). q=Kc n The slope n is 0.21(1.40) + 0.40 kg carbon qF = is assumed zero 0(1.40 kg of fresh granular activated carbon is added to the solution.0) . what are the final equilibrium values.21 kg phenol/m3 S = 1.40) + c(1. q = 0.396 5. Also determine the saturation loading capacity of the carbon.155 5 0.062 kg phenol / m3.062 % extracted = (100) = (100) = 70.210β0.030 4. (a.658 6 0. Data in Table 12.5 0.3-1.903 6.8 0.106 kg phenol / kg carbon and c = 0. At the intersection.00115 g/cm3 entered the bed at a flow rate of 754 cm3 per s.5 0.002 4 0.5 0. the fraction of total capacity used up to the break point. The break-point concentration is set at c/co = .933 6. Do as follows. h c/co 0 0 3 0 3.975 6.5 ππΉ 0. and the length of the unused bed.2 g of carbon.210 12.3-1 Scale-up of Laboratory Absorption Column (Geankoplis Example) A waste stream of alcohol vapor in air from a process was absorbed by activated carbon particles in a packed bed having a diameter of 4cm and length of 14cm containing 79.) Determine the break-point time.5 0.2 0. Breakthrough Concentration for Example 12. The percent of phenol extracted is ππΉβπ 0.3-1 Time.3-1 give the concentrations of the breakthrough curve.01. The inlet gas stream having a concentration co of 600 ppm and a density of .993 . Table 12. 707(14) = 9.95 h approximately A1 = 3. ππ π‘ 0 ππ The time equivalent to the usable capacity of the bed up to the break-point time is. The length of the used bed is HB = (tu /tt)HT = 0.16 = 0. what is the new total length of the column required? Solution (a): c/co = 0.00 h.65 h td = 6.707.01 tb = 3.51 = π.(b.51 h graphically integrating The time equivalent to total or stoichiometric capacity of the bed is β π π‘π‘ = β« (1 β ) ππ‘ = π΄1 + A2 = 3.65 + 1. π ππ¦ π‘π‘ . π‘π= 3.65 h graphically integrating A2 = 1.65/5.9 cm. the fraction of total capacity used up to the break point is tu /tt = 3. ππ π‘ 0 ππ Hence.707)(14) = π.65 π π‘π’ = β« (1 β ) ππ‘ = π΄1 = π. π‘π’ π»π’ππ = (1 β ) π»π = (1 β 0.) If the break-point time required for a new column is 6. 1 = 20. the new HB is obtained simply from the ratio of the break-point times multiplied by the old HB.0 HB = 3.3 + 4.4 cm We determine the saturation capacity of the carbon.799 12.1220 g alcohol/ g carbon The fraction of the new bed used up to the break point is now 16.67 g alcohol Saturation capacity = 9.0015 g/cm3) = 3122 g air/h 600 π ππππβππ πππ Total alcohol adsorbed = ( ) (312 π )(5.9) = 16. 6.3/20.16 β) 106 π πππ β = 9.2 g carbon = 0.5-1 Material Balance for Equilibrium Layers (Geankoplis Example) .67 g alcohol/79. Air flow rate = (754 cm3/s)(3600 s)(0.65(9.Solution (b): For a new tb of 6.0 h.3 cm π»π = HB + π»π’ππ = 16.4 = 0. The raffinate (water) layer composition is xC = 0. .94. The composition of the extract (ether) layer is yA = 0. (B) is equilibrated and the equilibrium phases separated. xA = 0.10.30. 10 kg of acetic acid (A).02 mass fraction.86. and xB = 0. and xB = 0. and 60 kg water. and yB = 0.60.An original mixture weighing 100 kg and containing 30 kg of isopropyl ether (C). What are the compositions of the two equilibrium phases? Solution: The composition of the original mixture is xC = 0.04. yC = 0.02. xA = 0.12. 7-1 Material Balance for Countercurrent Stage Process. Amounts of Phases in Solvent Extraction (Geankoplis Example) From the compositions obtained in Example 12.04) + L(0. The desired acetic acid concentration in the aqueous phase is 4%. determine the amounts of V and L. xBO=0. The original mixture contained 100 kg and xAM = 0.12.5-1.3.0 V = 25. xAO=0. Use equilibrium data from Appendix A.70.0. Pure solvent isopropyl either at the reate of Vn+1= 600kg/h is being used to extract an aqueous solution of Lo=200kg/h containing 30 wt% acetic acid (A) by countercurrent multistage extraction. yAN+1=1. xCO=0.5-2. yAN+1=0. . Lo=200.10 Solution: V + L = M = 100 V(0.30.12) = 100(0.0 12. Calculate the compositions and amounts of the ether extract V1 and the aqueous raffinate LN.10) L = 75.04. Solution: The given values are Vn+1= 600. LN=136kg/h and V1=664kg/h.017 is obtained. substituting into Eqs.7-3) and (12. since LN is on the phase boundary.7-2 Number of Stages in Countercurrent Extraction. We locate V1 by drawing a line from LN through M land extending it until it intersects the phase boundary. For the mixture point M.0) π₯πΆπ = = = 0. the point M is plotted in Fig 12. yCN+1=1. Calculate the number of stages required. Solution: The known values are VN+1=450. πΏπ π₯πΆπ + ππ+1 π¦πΆπ+1 200(0)+600(1. and xAN=0. By substituting into Eqs.7-3.30)+600(0) π₯π΄π = = = 0.In figure 12. yAN+1=0. Also. Pure isopropyl ether of 450 kg/h is being used to extract an aqueous solution of 150 kg/h with 30 wt % acetic acid (A) by countercurrent multistage extraction.7-3.04. xBO=0. Vn+1 and Lo are plotted.10.30.70.08 and yC1=0. xCO=0. xAO=0.7-2) and solving. (12. (12.0.75 πΏπ + ππ+1 200+600 πΏπ π₯π΄π + ππ+1 π¦π΄π+1 200(0.7-1) and (12. This gives yA1=0.7-4).90. 12.075 πΏπ + ππ+1 200+600 Using these coordinates. For LN a value of xCN=0. LO=150. it can be plotted at xAN=0. The exit acid concentration in the aqueous phase is 10 wt %. . (12. about 2.5 theoretical stages are needed. 12. which locates V1.7-5.7-3) and (12. Alternatively. Hence. 12.7-10) to locate point β. A tie line through V2 gives L2. 12. A final tie line gives L3.5-3.75 and xAM=0. An inlet water solution of 100 kg/h containing 0. (12. Starting at LO we draw line LOβ. LO. and LN are plotted in Fig. substituting into Eqs. For the mixture point M. Then a tie line through V1 locates L1 in equilibrium with V1. The points VN+1. It is desired to reduce the concentration of the exit water . (The tie-line data are obtained from an enlarged plot such as the bottom of Fig.0005 wt fraction nicotine in a countercurrent stage tower. This construction is not shown.072 and yC1=0. the coordinates of β can be calculated from Eq.7-4).075. The lines LO V1 and LN VN+1 are drawn and the intersection is the operating point β as shown. The water and kerosene are essentially immiscible in each other. the point M is plotted and V1 is located at the intersection of line LN M with the phase boundary to give yA1=0. A line L2β gives V3.7-3 Extraction of Nicotine with Immiscible Liquids.010 wt fraction nicotine (A) in water is stripped with a kerosene stream of 200 kg/h containing 0. xCM=0.895. which has gone beyond the desired LN.) Line L1β is next drawn locating V2. 12. giving N=3. xO =0.00497.0005.7-6. with x the weight fraction of nicotine in the water solution and y in the kerosene. The equilibrium line is also shown. the line is straight. Determine the theoretical number of stages needed.0 kg water/hr Vβ = V(1 β y) = VN+1 (1 β yN+1) = 200(1 β 0.xO) = 100(1 β 0. xN =0.010.010) = 99. VN+1 =200 kg/h. Solution: The given values are LO =100 kg/h.0005) = 199. Since the solutions are quite dilute. These end points on the operating line are plotted in Fig. The equilibrium data are as follows (C5).0010.7-12) and solving. The number of stages are stepped off. y1 =0.9 kg kerosene/hr Making an overall balance on A using Eq (12.to 0.0010 wt fraction nicotine. yN+1 =0. Solved Problems .8 theoretical stages. The inert streams are Lβ = L(1 β x) = LO (1 . 255 xB1= 0.03 xC1= 0.12. Calculate the amounts and compositions of the extract and raffinate layers.50 Plot point M on graph.35 xB0= 0. 4th Ed) A single-stage extraction is performed in which 400 kg of a solution containing 35 wt % acetic acid in water is contacted with 400 kg of pure isopropyl ether. Tie line through M gives L1 and V1. What percent of the acetic acid is removed? Given: V1 V2 = 400 kg isopropyl ether L0= 400 kg acetic acid solution L1 xA0= 0.35) + (400)(0) = 800xAM xAM = 0.715 V1 (Extract Layer): yA1= 0.5-2 (Geankoplis.11 .65 OMB: L0 + V2 = V1 + L1 V1 + L1 = 400 + 400 V1 + L1 = 800 = M Acetic Acid Balance: L0xA0 + V2yA2 = MxAM (400)(0.175 Isopropyl Ether Balance: L0yC0 + V2yC2 = MyCM (400)(0) + (400)(1) = 800yCM yCM= 0. L1 (Raffinate Layer): xA1= 0. 62 kg 48.yB1= 0. V1= 358 kg L1= 442 kg Amount of acetic acid in feed = 400(0.5-2: .255) + V1(0.175) V1 + L1 = 800 Solving simultaneously.11) = 48.62 % πππππ£ππππ = π₯ 100 = 34.11) = (800)(0.86 yC1= 0.7% 140 Graph for problem 12.35) = 140 kg Amount of acetic acid in extract = 442(0.03 Calculate L1 and V1: L1xA1 + V1yA1 = MxAM L1(0. . 765 OMB: L0 + V2 = V1 + L1 Let: V1 + L1 = M 1000 + 500 = M M = 1500 kg Acetone Balance: L0xA0 + V2yA2 = MxAM 1000(0.748 .120 xC1= 0.235 xB0= 0. Determine the amounts and composition of the extract and raffinate phases.5-4 (Geankoplis. and V2 on graph.1567 Methyl-isobutyl Ketone Balance: L0yC0 + V2yC2 = MyCM 1000(0) + 500(1) = 1500yCM yCM = 0. 4th Ed) A mixture weighing 1000 kg contains 23.205 yC1= 0.5 wt % acetone and 76.12. Given: V1 V2 = 500 kg methyl-isobutyl ketone L0= 1000 kg acetone solution L1 xA0= 0. L1 (Raffinate Layer): xA1= 0.022 V1 (Extract Layer): yA1= 0.235) + 500(0) = 1500xAM xAM = 0. tie line through M gives L1 and V1 compositions.333 Plot points M. L0.5 wt % water and is to be extracted by 500 kg methyl-isobutyl ketone in a single stage extraction. L0.03 xA0= 0.0625 πΏ0 + ππ+1 200 + 600 Plot points M.017 V1 (Extract Layer): yA1= 0. Calculate the compositions and amounts of the exit extract and raffinate streams. 4th Ed) An aqueous feed of 200 kg/h containing 25 wt % acetic acid is being extracted by pure isopropyl ether at the rate of 600 kg/h in a countercurrent multistage system.120) + (1500.25 xB0= 0.030 xCN= 0.0 wt % acetic acid. Given: VN+1 = 600 kg/h isopropyl ether L0= 200 kg/h acetic acid solution xAN= 0.L1) (0.75 πΏ0 + ππ+1 200 + 600 πΏ0 π₯π΄0 + ππ+1 π¦π΄π+1 (200)(0. LN on phase B. The exit acid concentration in the aqueous phase is to contain 3.7-2 (Geankoplis.070 .Calculate L1 and V1: L1xA1 + V1yA1 = MxAM L1(0. Locate V1 by drawing line MLN to intersect extract phase boundary at V1. and VN+1 on graph. Ln (Raffinate Layer): xAN= 0.75 πΏ0 π₯πΆ0 + ππ+1 π¦πΆπ+1 (200)(0) + (600)(1.1567) V1= 852 kg L1= 648 kg 12.0) π₯πΆπ = = = 0.25) + (600)(0) π₯π΄π = = = 0.205) = (1500)(0. 75) V1= 660.0 wt % acid in the final raffinate. 4th Ed) An aqueous feed solution of 1000 kg/h of acetic acid-water solution contains 30. (a) Calculate the minimum solvent flow rate that can be used.017) + (800.02 xA0= 0.6 kg/h 12. Given: VN+1 = Pure isopropyl ether L0= 1000 kg/h acetic acid solution xAN= 0.905 Calculate L1 and V1: OMB: L0 + VN+1 = V1 + LN Let: V1 + LN = M 200 + 600 = M M = 800 kg Isopropyl Ether Balance: L0xC0 + VN+1yCN+1 = MxCM LNxCN + V1yC1 = MxCM LN(0.yC1= 0. Minimum solvent Tie line through L0 gives V1min.7-4 (Geankoplis.4 kg/h LN= 139. Line LNVN+1 and L0V1min (tie line) gives Ξ pt.LN)(0.905) = (800)(0.70 a. Draw line LNV1min and L0VN+1 and intersection gives Mmin.0 wt % acetic acid and is to be extracted in a countercurrent multistage process with pure isopropyl ether to reduce the acid concentration to 2. determine the number of theoretical stages required. (b) If 2500 kg/h of ether solvent is used. . min.30 xB0= 0. Number of Stages VN+1 = 2500 kg/h isopropyl ether L0= 1000 kg/h acetic acid solution xAN= 0.114 = = πΏ0 + ππ+1 1000 + ππ+1 VN+1 = VN+1min = 1630 kg/h b.714 πΏ0 + ππ+1 1000 + 2500 πΏ0 π₯π΄0 + ππ+1 π¦π΄π+1 (1000)(0.30) + (2500)(0) π₯π΄π = = = 0.010 (off graph) xC1 = 0.30 xB0= 0.25) + ππ+1 (0) π₯π΄ππππ = 0.015 yA1 = 0.02 xCN = 0. Algebraic calculation of Ξ operating point: xAN = 0.70 πΏ0 π₯πΆ0 + ππ+1 π¦πΆπ+1 (1000)(0) + (2500)(1. 114 πΏ0 π₯π΄0 + ππ+1 π¦π΄π+1 (1000)(0.02 xA0= 0.0) π₯πΆπ = = = 0.0857 πΏ0 + ππ+1 1000 + 2500 This point checks graphical determination of M on lines LNV1 and L0VN+1.From graph: xA= 0. Draw line LNVN+1 and L0V1 to obtain Ξ operating point graphically.863 OMB: L0 + VN+1 = V1 + LN Let: V1 + LN = M 1000 + 2500 = M M = 3500 kg . Then. Nmin = 7. determined by lines from L to Ξ operating point. Plot stages up to V4.5 stages .32 πΏ0 + π1 1000 + 2874 πΏ0 π₯π΄0 + π1 π¦π΄1 (1000)(0. V2.0857 = = πΏπ + π1 3500 V1= 2874 kg/h LN= 626 kg/h πΏ0 π₯πΆ0 + π1 π¦πΆ1 (1000)(0) + (2874)(0. Use Ξ op. plot expanded scale as shown.30) + (2874)(0.863) π₯βπΆ = = = +1. L2V3.0067 πΏ0 + π1 1000 + 2874 This checks the graphical Ξ op. point.010) π₯π΄π = 0.02) + (3500 β πΏπ )(0. point to determine the number of stages.10) π₯π΄π = = = β0. πΏπ π₯π΄π + π1 π¦π΄1 (πΏπ )(0. etc. The operating points on expanded scales as L1. 7-4: .Graph for problem 12. 172 x0 1+E Recovery is 1 .8 L 100 x1 1 1 Thus = = 5. Smith. Unit Operations of Chemical Engineering β McCabe.06? Solution: (a) By a material balance.172 = 0.828.2 Penicillin F is recovered from a dilute aqueous fermentation broth by extraction with amyl acetate using 6 volumes of solvent per 100 volumes of the aqueous phase. At pH = 3.0.8 = 0.2 the distribution coefficient KD is 80.8% (b) With the same value of E. since y0 = 0. or 82. x2 1 = x1 1+E . Harriott Sample Problems Example 23. L(x0 β x1) = Vy1 y1 = KD x1 VKD x1 ( + 1) = x0 L The extraction factor is VKD 6 x 80 E= = = 4. (a) What fraction of the penicillin would be recovered in a single ideal stage? (b) What would be the recovery with two-stage extraction using fresh solvent in both stages?(c) How many ideal stages would be needed to give the same recovery as in part (b) if a counterflow cascade were used with V/L = 0. 2 KD 80 xb β = 0 100β20.2 ln[ ] 3 N = = 2. x2 1 = = 0.8 Example 23.09 ln 4. The feed consists of 40 percent acetone and 60 percent water.3 A countercurrent extraction plant is used to extract acetone (A) from its mixture with water by means of methyl isobutyl ketone (MIK) at a temperature 25β°C.0 and ya = 97(100)/6 = 1617 ya 1.617 xa β = = = 20. the stripping factor S.0% (c) With KD and V/L constant. the number of ideal stages can be calculated from the stripping form of Kremser equation using E in place of its equivalent.9703 or 97.0297 x0 (1 + E)2 Recovery is 1 β 0. Pure solvent equal in mass to the feed is used as the extracting liquid. ln[(xa βxa β )/(xb βxb β ) N = ln E Let xa = x0 = 100 Then xb = 3.0297 = 0. How many ideal stages are required to extract 99 percent of the acetone fed? What is the extract composition after removal of the solvent? Solution: . 64.6/140 = 0.006.99 x 40 = 39.a β 39.44.4 + m β n Since n and m are small and tend to cancel in the summation for Va and La. which is the upper operating line in Fig 23. The terminal points for the operating line are determined by material balances with allowances for the amounts of water in the extract phase and MIK in the raffinate phase.10. The total flows are At the top.4A + (60 β n)H2 O + mMIK = 60. Basis: F = 100 mass units per hour Let n = mass flow rate of H2O in extract M = mass flow rate of MIK in raffinate For 99% recovery of A.6A + nH2 O + (100 β m)MIK = 139.7 to prepare a plot of the equilibrium relationship yA versus xA.4/60 = 0. The value of XA. which would make yA. and the raffinate has 0.6 + n β m At the bottom.Use the data in Fig 23.283. the total extract flow Va is about 140. La = F = 100 = 40A +60 H2O Va = 39. the extract has 0. Vb =100MIK Lb = 0. . These estimates are adjusted after calculating values of n and m.b is about 0. From Fig.7 for xA = 0.85 = 117.1 = 145. yH2O = 0.4 + 1.6 β 100 = 72.0074 54.6 39.951) (139.4 + 117.02 0.4 + 117. 23.98 Revised n = (0.0074.4 Plot points (0.From Fig 23.02.6(0. assume that the amount of MIK in the extract is 100.02 m= (0.4 xA.12) β 0 . 0.1 0. yH2O = 0.1 Va = 139.6 β 1.272 145.6 yA.049 0.2 From Fig 23.6) = 7.283. ππ + πΏ = πΏπ + π πΏ β 54.03. For an intermediate point on the operating line. Since the raffinate phase has only 2 to 3 percent of MIK.02 β (0.6 + 7.1) = 7.6 Lb = 60. 0.049 0..4 + 60 β n) 1 β 0.b = = 0.4.049/0.007.272) to establish the ends of the operating line.7 for yA = 0.049 n= (39. 0) and (0.0 A balance on A over the same secion gives xA: πΏπ₯π΄ + ππ π¦π = πΏπ π₯π + ππ¦π΄ πΏπ₯π΄ β 0.4 + 52.1 β 1.951)( 139.049 If m is very small n β (0.8) = 1.1 = 54.4 0.6 By an overall balance from the solvent inlet (bottom) to the intermediate point. the same as the solvent feed: 100 β ππ¦ππΌπΎ 100 π β 0. xMIK = 0.1 β 7.a = = 0.12 and calculate V and L.6 +100 β m) 1β0. pick yA = 0.85.7. and yMIK = 0. 03(Fig 23. KDV E= L .10.201 72 This value is probably accurate enough.9(0.7).4 stages Solved Problems 23.5 π₯π΄ β = 0.4 + 118.1 Revised V = = 118.15.12) Revised xA = = 0. 12.02yb Neglect changes in L and V Operating line slope = 0.3 0.9 β 100 = 73. For xA = 0.45V Ya = 0. 14. N = 3. which gives a slightly curved operating line. From Fig 23.45 = L/V ya* = 0 For extraction. yA = 0. L and xA can be determined.03) β 1.4+ 118.200 73.3 Plot xA = 0.9 0.85 Revised L = 54. If the water flow rate is set at 0. but corrected values of V. how many ideal stages would be needed for 98 percent recovery of the antibiotic in a countercurrent cascade? Solution: KD = 0.20. A balance on MIK from the solvent inlet to the intermediate point gives ππ + πΏπ₯MIK = Lb xMIK. π₯ππΌπΎ β 0. where Kp = 0.15 = ye / xe xa = 0 LΞx = VΞy L = 0.201.b + VyMIK VyMIK = 100 + 72(0.7 An antibiotic that has been extracted from a fermentation broth using amyl acetate at low pH is to be extracted back into clean water at pH = 6.1 101.45 times the solvent rate. 82 percent oil and 6.18)yb Solving for N.20 stages ππ 3.5)].45 = = 3.5 7.015 = 240 m2/m3 .327yb ln[ 0. Here solute is extracted from the organic phase into the water. Lxb = V(0. by contact with hexane.6) / 0.15 = xe / ye and appropriate factor is LK'D 0. Assume that a large excess of hexane was used and that the external mass transfer resistance was negligible. yb β 0. Time. but this is for transfer into the organic phase. 11. The particles originally contained 43.18π¦π 0. Dp = 0.35 3.0 23.8 Oil was extracted from small particles of rapeseed. the oil content of the dried meal was reported for different extraction times. min 75 90 105 120 Oil in dry meal.15(2.98)π¦π = 2.[(Eq.0 V 0. After drying.02yb ] N= = 3.97 4.43 percent moisture. 23.015 m av = 6(1-Ξ΅) / Dp = (6 x 0.45 yb* = 0.15 This factor corresponds to absorption factor A in absorption. averaging 0.98)yb 1 π₯π = (0. so KβD = 1 / KD = 1 / 0. Determine the effective diffusion coefficient for oil in the hexane- soaked particles using the equation for transient diffusion in spheres.58mm in size.88 kg/100 kg inert material Answer: Assume the packing 40% voids.15xb = 0. The results are given in the table below. s Ο = 36. calculate the terminal velocity of the drops.c =0.30 (close to flooding)? For different ratios of flow rates. For spheres.018 = 995 kg/m3 ΞΌc = 0.041.2 240 8.35= 0.c g 0.1 x 10-3 = 0. at what flow rate of the dispersed phase would the fraction holdup be 0.0615 Vs.0 (continuous wetting).3 0. If the average drop size is 2.01 x 10-4 Pa.15 136.64 x 16.3 x 0.0 mm.54 ( ) [ ( ) 3] x [ ] 859.01 x 10-4 136. Ξ± = 1.32 x 9.805 1 0.208.3 8. is either the sum of the flow rates or the sum of their square roots nearly a constant at a given holdup? (Note the correlation in Figure below) .01 x 10-4 =1.0361 0.0361 N/m Οd = 53.801 cP = 8.3 kg/ m3 Assume R =2.0 x 0.41 1 136.15 x 16. The density of the hydrocarbon phase is 53 lb/ft3.5 ( 240 ) = 0. Then 2 av 0.35Vs.5 Vs.00225 m/s 23.0615/27.018 = 859. If the slip velocity is assumed to be independent of the holdup and if the dispersed-phase volumetric flow rate is taken to be twice that of the continuous phase.Οc = 62.208 x 1.5 ) ( ) =27.2 kg/m3 ΞΟ = 136. use C1 = 0.9 (From Unit Operation of Chemical Engineering by McCabe and Smith) A spray-tower extractor operates at 30Β°C with hydrocarbon drops dispersed in water.c (1+R0. 2)2 (0. Ο = 0. ΞΌ = 0.2(39. Guess Rep = 100.996 g / cm3.996) Rep = =994 0.2 cm. which gives CD = 1.996-0. so ΞΌt must be calculated using Eq. Οp = 53 / 62.6. 7.0081) 0.0 . try Stokeβs Law 980(0.97)(0. (7.Fig.7 Flooding velocities in packed extraction towers Solution: Dp = 0.37) and Fig. 23.4 = 0.849) ΞΌt = =39.801 cP To get ΞΌt.0081 This is well beyond the range of Stokeβs Law.97 cm/s 18(0.849 g / cm3 For water at 30Β°C. 21 ππ/π 3(1.74 x =796 ft/h s 30.9 0. 4π(ππβπ)π·π Eq.00801 Guess CD = 0.85 (OK) cm 3600 ΞΌt =6. CD = 0.996) 0. πΆπ· = 0.85 ΞΌt = 6. (7.2(6.996) π ππ = = 154.2) 1/2 ππ‘ = ( ) = 6.5 .21)(0.147)(0.0)(0.74 cm / s Re = 167.37) ππ‘ = β 3πΆπ· π 4(980)(0. c and h = 0.5 15% higher than for Vs.c = 197 + 0.c =11.30 796 Vs.9 Second Part For Vs.Let h = volume fraction holdup of drops Vd .d(3.d = Vs.1=27.d = 0.c =14.33+0.9=23.c =334 and βVs.c h = 0.d + βVs.d + Vs.c and h = 0.c / (1-h) Vs.d / h Vc = Vs.c = 387 31% higher than for Vs.d Vs.4+16.30 796 Vs.c but βVs.33+2.d = 2Vs.30 Vd = Vs.d +βVs.c 3.0+9.86 Vs.c =258 Vs.d +βVs.d = 2Vs.33+1.d + Vs.d = 197 ft / h Vs.d = =129 3.8 For Vs.5Vs.c .c =25.7) Vs.d + Vs.43 Vs.3 2(0. Vc = actual velocities of drops or continuous phase (both positive) Then ΞΌt = Vd + Vc = 796 ft/h Given: First Part Vs.d + =796 0.714) = 796 Vs.d = 2Vs.5(197) = 296 βVs.d = =167=Vs. .The sum of the square roots is more nearly constant than the sum of the velocities.