Solutions Manual

Exercise 9.1 Subject: Selection of type condenser and operating pressure for distillation of propionic and n-butyric acids. Given: Distillate of 95 mol% propionic acid. Bottoms of 98 mol% n-butyric acid. Assumptions: Ideal solutions so that Raoult's K-values apply. Find: type condenser and operating pressure Analysis: From Perry's Handbook or other source, the normal boiling points are: Propionic acid 141.1oC n-Butyric acid 163.5oC Using Fig. 7.16, an operating pressure of 30 psia or less, with a total condenser, might be used. Could also consider using subcooled reflux with a top pressure of just above atmospheric pressure. Exercise 9.2 Subject: Selection of type condenser and operating pressure for distillation of a mixture of light paraffin hydrocarbons in two columns Given: Feed and product compositions as follows: kmol/h: Component Feed Product 1 Product 2 Product 3 C1 160 160 0 0 C2 370 365 5 0 C3 240 5 230 5 nC4 25 0 1 24 nC5 5 0 0 5 Total: 800 530 236 34 Find: Type condensers and operating pressures for: (a) Direct sequence. (b) Indirect sequence. Analysis: Use Chemcad or other simulator to make the bubble-point and dew-point calculations, using the SRK equation of state. The results should be close to hand calculations using K-values from Figs. 2.8 and 2.9. Use Fig. 7.16 to determine pressure and type condenser. Column 1 of direct sequence: Distillate is Product 1. Can not condense at 120oF because this temperature is above the critical temperatures of methane and ethane. At 415 psia, the dew point temperature is 16oF. Therefore, use a partial condenser with a refrigerant boiling at a temperature of about 6oF. One choice of refrigerant is propane. Bottoms temperature would not be in the decomposition range. Column 2 of direct sequence: Distillate is Product 2, which is mainly propane. Bubble-point pressure at 120oF is 256 psia, which is greater than 215 psia. Dew-point pressure at 120oF is 248 psia. Use partial condenser with about 250 psia pressure at the top. Use cooling water in the condenser. Bottoms conditions present no problems. Column 1 of indirect sequence: Distillate is a mixture of Products 1 and 2. At 120oF, pressure is very high, even for a partial condenser. Dew-point temperature at 415 psia is 83oF. Raise pressure a little and use cooling water in the condenser. Bottoms conditions present no problems. Column 2 of indirect sequence: Conditions at the top are the same as Column 1 of the direct sequence. Therefore, use a partial condenser with a refrigerant for removing heat at 16oF. Exercise 9.3 Subject: Selection of type condenser and operating pressure for the distillation of a mixture of methane, ethane, benzene, and toluene by the direct sequence in two columns. Given: Feed and product conditions as follows from Fig. 9.20: kmol/h: Component Feed Product 1 Product 2 Product 3 Methane 20 20.000 0.000 0.0 Ethane 5 4.995 0.005 0.0 Benzene 500 5.000 485.000 10.0 Toluene 100 0.000 0.500 99.5 Total: 625 29.995 485.505 109.5 Find: Type condenser and operating pressure in each column. Analysis: Use Chemcad or other simulator to make the bubble-point and dew-point calculations, with the SRK equation of state. Use Fig. 7.16 to determine pressure and type condenser. Column 1: Distillate is Product 1. Bubble point on Product 1 at 120oF gives an impossible very high pressure because the critical temperatures of methane and ethane are less than 120oF. A dew point on Product 1 at 120oF gives 31.4 psia because of the presence of benzene. Therefore, can use a partial condenser with cooling water. No problems occur at the bottom. Column 2: Distillate is Product 2. Bubble point on Product at 120oF gives 5.1 psia. No need to operate at a vacuum. Raise condenser pressure to 30 psia or just above atmospheric pressure and use a total condenser, possibly with subcooled reflux. No problems at the column bottom. Exercise 9.4 Subject: Number of equilibrium stages for a deethanizer. Given: Deethanizer with feed and average relative volatilities as follows from Fig. 9.21: Component Feed, kmol/h Avg. Relative volatility C1 160 8.22 C2 370 2.42 C3 240 1.00 nC4 25 0.378 nC5 5 0.150 Distillate is to contain 2 kmol/h of C3. Bottoms is to contain 2 kmol/h of C2. Therefore, LK is C2 and HK is C3. Assumptions: Relative volatilities are relative to C3. Find: Number of equilibrium stages for 2.5 times minimum number of stages. Analysis: To compute the minimum number of stages, use component flow rate form of the Fenske equation given by (9-12): N d d b b min C C C C C C 2 3 3 2 2 = = - - =log log log log . .,a 3 370 2 2 240 2 2 2 42 113 N = 2.5Nmin = 2.5(11.3) = 28.3 Exercise 9.5 Subject: Determination of the minimum number of stages by the Fenske equation for each of three sections in a distillation column with a vapor sidestream Given: The following feed and product compositions for a distillation column: kmol/h: Component Feed Distillate Vapor sidestream Bottoms Benzene, B 260 257.0 3.0 0.0 Toluene, T 80 0.1 79.4 0.5 Biphenyl, BP 5 0.0 0.2 4.8 Pressure, kPa 165 130 180 200 Assumptions: Raoult's law for K-values. Find: Using the Fenske equation, the minimum number of stages for sections between: (a) Distillate and feed. (b) Feed and vapor sidestream. (c) Sidestream and bottoms. Analysis: (a) For this section, take the LK as benzene and the HK as toluene. At the top, have almost pure benzene at 130 kPa = 19 psia. Therefore, from Fig. 2.4, the distillate temperature = boiling point of benzene at 19 psia = 195oF. At the feed point, the pressure = 165 psia = 23.9 psia. Assume a saturated liquid feed. The bubble-point temperature is estimated to be 215o F. Take an average temperature = 205oF, at which the vapor pressures for benzene and toluene, respectively, are 23 psia and 9 psia respectively. From Eq. (2-44), the corresponding K-values at the average section pressure of 21.5 psia are 1.07 and 0.42. The relative volatility, from Eq. (2- 21) = 1.07/0.42 = 2.55. From the Fenske equation (9-12), applied between the distillate and feed compositions: B T T B B,T min 257 80log log 7.10.1 260log log 2.55 a = = = d f dN f stages (b) For the section between the feed and the vapor sidestream, again take the LK as benzene and the HK as toluene. The pressure at the vapor sidestream = 180 kPa = 26.1 psia. A dew-point on the vapor sidestream gives a temperature of 265oF. Assume an average temperature for the section of 240oF, at which the vapor pressures for benzene and toluene, respectively, are 37 psia and 16 psia respectively. From Eq. (2-44), the corresponding K-values at the average section pressure of 25 psia are 1.48 and 0.64. The relative volatility, from Eq. (2-21) = 1.48/0.64 = 2.31. Exercise 9.5 (continued) Analysis: (b) (continued) From the Fenske equation (9-12), applied between the feed and vapor sidestream compositions: B T T B B in ,T m 260 79.4log log 80 3 log log 2. 1 .33 5 = = =a f vs N f vs stages (c) In the section between the vapor sidestream and bottoms, take toluene as the LK and biphenyl as the HK. At the bottom of the column, the pressure = 200 kPa = 29 psia. Biphenyl boils at 255oC = 491oF at 14.7 psia, which is much higher than toluene. From Perry's Handbook, the following vapor pressure data are obtained for biphenyl: T, oF 243 274 307 165.2 329 400 445 Vapor pressure, psia 0.19 0.39 0.77 1.16 1.93 3.87 7.74 A bubble point on the bottoms pressure gives 465oF. At this temperature, the relative volatility, aT,BP = vapor pressure of T/vapor pressure of BP = 220/10.6 = 20.8. At the temperature of the vapor sidestream, 265oF, aT,BP = vapor pressure of T/vapor pressure of BP = 27.9/0.37 = 75.4. Use a geometric mean for this wide range. Thus, aT,BP = [20.8(75.4)]1/2 = 40. From the Fenske equation (9-12), applied between the vapor sidestream and bottoms compositions: T BP BP T B,T min 79.4 4. 2.3 8log log 0.2 0.5 log log 40 = ==a vs b v bN s stages Exercise 9.6 Subject: Comparison of minimum number of stages by Fenske equation and McCabe- Thiele method for a non-ideal system of acetone (A) and water (W). Given: A feed of 25 mol% A and 75 mol% W. Distillation at 130 kPa (975 torr) to obtain a distillate of 95 mol% A and a bottoms of 2 mol% A. Infinite-dilution liquid-phase activity coefficients of 8.12 for acetone and 4.13 for water. Find: Minimum stages by the Fenske equation and by the McCabe-Thiele method. Analysis: Because the pressure is close to 1 atm, estimate vapor-liquid equilibria from the modified Raoult's law, Eq. (4) in Table 2.3. Obtain vapor pressures from Antoine equations and liquid-phase activity coefficients over the entire composition range from fitting the infinite- dilution coefficients to the van Laar equation, Eq. (3) in Table 2.9, using Eqs. (2-75). Thus, K yx PP K yx PP A x A x A A x A x A A A s s A A A A A W W W W W A AW A WA W AW W WA W AW A WA AW A WA W (1) and (2) 1+ (3) and 1+ (4) = = = = = = = = = = = = ¥ ¥ g g g g g g exp exp ln ln(8. ) . ln ln( . ) . 2 2 12 2 094 4 13 1418 Substituting these van Laar coefficients into Eqs. (3) and (4), g gA A A W A A 1+1.477 (5) and 1+ 0.6772 (6)= - = - exp . exp .2 094 1 1418 12 2x x x x From the 7th edition of Perry's Chemical Engineers' Handbook, page 13-21, log . . . log . . .P T P T P i T s s i s A W o (7) and (8) where vapor pressure of component in torr and C = - + = - + = = 7 11714 1210 595229 664 8 07131 1730 630233 426 Minimum stages by the Fenske equation: First determine the relative volatilities at the top and bottom and take the geometric average for use in the Fenske equation. Exercise 9.6 (continued) Analysis: Fenske equation (continued) Bubble point for the distillate composition. From Eqs. (4-12), (1), and (2): x K x PPD i i D i i s i i i = = g 1 (9) In the distillate, xA = 0.95 and xW = 0.05. From Eqs. (3) and (4), g gA W 1+ and 1+ = = = =exp . . . . . . exp . . . . . .2 094 0 95 2 094 0 05 1418 1003 1418 0 05 1418 0 95 2 094 3 752 2 Eq. (9) becomes: 0 95 1003975 0 05 375975 1. . . .P P s s A A + = where vapor pressure is in torr. Using Eqs. (7) and (8) with a spreadsheet, a trial and error calculation gives a distillate temperature of 64oC. Then from Eqs. (1) and (2), aA,W = KA/KW = 1.48 Bubble point for the bottoms composition. In the bottoms, xA = 0.02 and xW = 0.98. From Eqs. (3) and (4), g gA W 1+ and 1+ = = = =exp . . . . . . exp . . . . . .2 094 0 02 2 094 0 98 1418 7 19 1418 0 98 1418 0 02 2 094 10012 2 Eq. (9) becomes: 0 02 719975 0 98 1001975 1. . . .P P s s A A + = where vapor pressure is in torr. Using Eqs. (7) and (8) with a spreadsheet, a trial and error calculation gives a distillate temperature of 95oC. Then from Eqs. (1) and (2), aA,W = KA/KW = 27.7 The geometric mean relative volatility = [(1.48)(27.7)]1/2 = 6.40 From the Fenske equation (9-11), WA A W A,W min 0.95 0.98log log 0.02 0.05 log log 6.40 3.7 = = a = BD B D xx x N x Minimum stages by the McCabe-Thiele method: To obtain a y-x equilibrium curve at 975 torr in terms of acetone mole fractions, we can run bubble-point temperature calculations, as above, for a set of points in the range of liquid compositions between the distillate and bottoms compositions. For each point, the K-value and vapor mole fraction for acetone are computed from Eq. (1). The results from a spreadsheet are as follows: Exercise 9.6 (continued) Analysis: McCabe-Thiele method (continued) T, oC yA xA ------------------------------------- 63.6 1.000 1.000 64.0 0.965 0.950 64.5 0.936 0.900 65.0 0.911 0.850 65.5 0.889 0.800 66.0 0.870 0.750 66.5 0.855 0.700 67.0 0.842 0.650 67.5 0.833 0.600 67.9 0.824 0.550 68.3 0.817 0.500 68.6 0.811 0.450 68.9 0.806 0.400 69.2 0.801 0.350 69.6 0.796 0.300 70.2 0.789 0.250 71.2 0.777 0.200 73.1 0.754 0.150 76.8 0.705 0.100 84.9 0.578 0.050 95.1 0.361 0.020 107.1 0.000 0.000 The McCabe-Thiele plot for these data are shown on the next page. The equilibrium stages for total reflux are stepped off between the equilibrium curve and the 45oline, which represents the operating line, from the acetone xD = 0.95 down to the acetone xB = 0.02. From the plot, there are 5 minimum stages, which is 35% higher than the 3.7 minimum stages computed from the Fenske equation. The Fenske equation is not reliable for highly nonideal binary systems. Exercise 9.6 (continued) Analysis: McCabe-Thiele method (continued) Exercise 9.7 Subject: Minimum equilibrium stages and distribution of nonkey components using the Fenske equation for distillation of a paraffin hydrocarbon mixture. Given: Column pressure of 700 kPa. Feed composition and split for two key components in Fig. 9.23. K-values from Figs. 2.8 and 2.9. Find: Minimum number of equilibrium stages. Distribution of nonkey components at total reflux. Analysis: To apply the Fenske equation, the geometric mean relative volatility between the distillate and bottoms is needed. From the following crude estimate of the split of all nonkey components, compute the bubble-point temperatures of the distillate and bottoms and use the K- values at those temperatures to get the relative volatilities. Use the bubble-point equation, Eq. (4-12). The following results are obtained for a distillate bubble point of 68oF (20oC) and a bottoms bubble point of 257oF (125oC): Component Feed, kmol/h Distillate, kmol/h K-value, distillate KxD Bottoms, kmol/h K-value, bottoms KxB Propane 2500 2500 1.250 0.8895 0 5.70 Isobutane 400 399 0.490 0.0557 1 3.25 0.0078 n-Butane 600 594 0.340 0.0575 6 2.65 0.0382 Isopentane 100 15 0.138 0.0006 85 1.45 0.2955 n-Pentane 200 5 0.108 0.0002 195 1.23 0.5751 n-Hexane 40 0 0.038 40 0.65 0.0623 n-Heptane 50 0 0.013 50 0.33 0.0222 n-Octane 40 0 0.005 40 0.19 0.0177 Total 3930 3513 1.0035 417 1.0188 For the two key components, nC4 and iC5, the relative volatilities at the top and bottom are, respectively (0.34/0.138) = 2.46 and (2.65/1.45) = 1.83. The geometric mean anC4,iC5 = [(2.46)(1.83)]1/2 = 2.12. From the Fenske equation, (9-12), 54 5 4 4 5 iCnC iC nC nC ,iC min 594 85log log 6 15 log log 2.12 8.4 = = = aN bd d b To compute the distribution of nonkey components at total reflux, use for the lighter than light key components, LLK, Eq. (9-15), and use Eq. (9-16) for the heavier than heavy key, HHK, with iC5 as the reference component, r, and the above value of Nmin. Exercise 9.7 (continued) Analysis: (continued) Thus, for the LLK, i, b fd b f i i N i= + = + 1 1 1585 5 5 8 4iC iC i,iC i,iC 5 5 mina a . (1) For the HHK, i, d f db d b f i i i N i N i i i = = iC iC ,iC iC iC ,iC ,iC ,iC 5 5 5 min 5 5 5 min 5 51+ 1+ a a a a 15 85 15 85 8 4 8 5 . . (2) Using the above K-values to compute geometric mean values of ai, iC5 , followed by use of Eqs. (1) and (2) with the material balance, fi = di + bi , the following results are obtained: Component ai, iC5 at top ai, iC5 at bottom ai, iC5 average di , kmol/h bi , kmol/h Propane 9.06 3.93 5.97 2499.996 4.2x10-3 Isobutane 3.55 2.24 2.82 399.63 0.37 n-Butane 594 6 Isopentane 1.00 1.00 15 85 n-Pentane 0.783 0.848 0.815 6.12 193.88 n-Hexane 0.272 0.448 0.349 0.001 39.999 n-Heptane 0.094 0.228 0.146 8.2x10-7 50 n-Octane 0.038 0.128 0.070 1.92x10-10 40 Exercise 9.8 Subject: Type condenser and operating pressure. Minimum equilibrium stages and distribution of nonkey components using the Fenske equation for distillation of a light hydrocarbon mixture. Given: Feed composition and split of two key components in Fig. 9.24. K-values from Figs. 2.8 and 2.9. Find: Type condenser. Operating pressure. Minimum equilibrium stages. Distribution of nonkey components at total reflux. Analysis: To determine the type condenser and operating pressure, follow the algorithm in Fig. 7.16. The estimated distillate and bottoms compositions from the given data are: Component Feed, lbmol/h Distillate, lbmol/h Bottoms, lbmol/h Methane 1000 1000 0 Ethylene 2500 2500 0 Ethane 2000 1999 1 Propylene 200 5 195 Propane 100 0 100 n-Butane 50 0 50 Using Fig. 2.8, it is quickly shown that at 120oF, both the bubble-point and the dew-point pressures for the estimated distillate composition are greater than 1000 psia because the K-value for ethane is greater than 1.0 at 120oF and 1000 psia. Therefore, select the operating pressure as 415 psia and use a partial condenser. Run a dew point on the distillate at 415 psia. Use Eq. (4-13) in the following form: dK D K K K Ki ii = = + + + = 1000 2500 1999 5 5504 C C C C1 2= 2 3= (1) By trial and error, using Fig. 2.9 for K-values, the temperature that satisfies Eq. (1) is 0oF, which gives: dK Di ii = = + + + = 100050 2500105 19990 69 5018 5505. . . . The relative volatility of the two key components is aC2, C3= = 0.69/0.18 = 3.83 Exercise 9.8 (continued) Analysis: (continued) Now run a bubble-point on the estimated bottoms composition at 415 psia. Use Eq. (4-12) in the following form: b K B K K K Ki i i = = + + + = C C C nC 2 3= 3 4 195 100 50 346 (2) By trial and error, using Fig. 2.8 for K-values, the temperature that satisfies Eq. (1) is 155oF, which gives: b K Bi i i = = + + + = 2 7 195 115 100 101 50 0 39 347. ( . ) ( . ) ( . ) The relative volatility of the two key components is aC2, C3= =2.7/1.15 = 2.35 The geometric mean relative volatility = [(3.83)(2.35)]1/2 = 3.0 From the Fenske equation, (9-12), = 32 = 2 3 = 2 3 CC CC C , in C m 1999 195log log 5 1 log 10. log 3.0 3 = =a =N bd d b To compute the distribution of nonkey components at total reflux, use for the lighter than light key components, LLK, Eq. (9-15), and use Eq. (9-16) for the heavier than heavy key, HHK, with C3= as the reference component, r, and the above value of Nmin. Thus, for the LLK, i, b fd b f i i N i= + = + 1 1 5195 10 3C C i,C i,C 3 = 3 = 3 = min 3 =a a . (3) For the HHK, i, d f d b d b f i i i N i N i i i = = C C ,C C C ,C ,C ,C 3 = 3 = 3 = min 3 = 3 = 3 = min 3 = 3 =1+ 1+ a a a a 5 195 5 195 10 3 10 3 . . (4) Using the above K-values to compute geometric mean values of ai, C3= , followed by use of Eqs. (3) and (4) with the material balance, fi = di + bi , the following results are obtained: Component ai, C3= at top ai, C3= at bottom ai, C3= average di , lbmol/h bi , lbmol/h Methane 27.8 8.70 15.55 1000 2.1x10-8 Ethylene 5.83 3.22 4.33 2499.973 0.027 Ethane 3.83 2.35 3.00 1999 1 Propylene 1.00 1.00 1.00 5 195 Propane 0.78 0.88 0.83 0.375 99.625 n-Butane 0.18 0.34 0.25 8.1x10-7 50 Exercise 9.9 Subject: Recovery of a key component as a function of distillate flow rate for the distillation of a paraffin hydrocarbon mixture when the minimum number of equilibrium stages is fixed. Given: Feed composition and flow rate of 1000 lbmol/h. Column pressure of 250 psia. Minimum equilibrium stages = 15. K-values from Figs. 2.8 and 2.9. Find: Percent recovery of propane in the distillate as a function of distillate flow rate. Analysis: Assume that the average relative volatility between propane and n-butane is not sensitive to the separation. If a perfect separation were made between propane and n-butane at 250 psia, the distillate temperature would be approximately 110oF and the bottoms temperature would be approximately 270oF. Assume an average temperature of 190oF. From Fig. 2.8, the K- values and relative volatilities at 250 psia and 190oF, with the light key as propane and the heavy key as n-butane are: Component Feed rate, lbmol/h K-value a referred to nC4 Ethane 30 4.3 5.97 Propane 200 1.74 2.42 n-Butane 370 0.72 1.00 n-Pentane 350 0.30 0.417 n-Hexane 50 0.133 0.185 Total: 1000 Based on these relative volatility values, assume that little of the nonkey components will distribute. Therefore, D d d d d d d d= + + + + = + +C C nC nC nC C nC 2 3 4 5 3 46 30 (1) The Fenske equation, (9-12) becomes: N d d b b d d d d min C nC nC C C nC avg C nC nC C 3 4 4 3 3 4 3 4 4 3 log 200 - = = = - 15 370 2 42 log log log .,a Rearranging, 370 200 2 42 571800 20015- = - = - dd dd ddnC nC C C C C 4 4 3 3 3 3 . (2) Assume a value of d for C3. Calculate d for nC4 from Eq. (2). Then compute D from Eq. (1). Then compute recovery of C3 = d for C3/200 and make plot. Exercise 9.9 (continued) Analysis: (continued) d of C3, lbmol/h % recovery of C3 d of nC4, lbmol/h D, lbmol/h 198 99 0.064 228.1 190 95 0.012 220.0 180 90 0.006 210.0 150 75 0.002 180.0 100 50 0.00065 130.0 We see that with 15 minimum stages, the distribution of nC4 to the distillate is almost negligible. Exercise 9.10 Subject: Minimum reflux ratio by the Underwood equation for the separation of a binary mixture as a function of feed vaporization. Given: Binary feed of 30 mol% propane and 70 mol% propylene. Distillate to contain 99 mol% propylene, and bottoms to contain 98 mol% propane. Column pressure of 300 psia. K- values in Figs. 2.8 and 2.9. Find: Minimum reflux ratio by the Underwood equation for: (a) Bubble-point liquid feed. (b) Feed of 50 mol% vapor. (c) Dew-point vapor feed. Analysis: Because we have a binary mixture, this is a Class I separation, where all components distribute. Therefore, the Underwood equation that applies is (9-20): LD x x x x x x D D min C , C C C C , C C C C C C C C C 3 = 3 = 3 = 3 3 3 3 = 3 3 = 3 = 3 3 3 = 3 = - - = - - ¥ ¥ ¥ ¥, , , , , , , , . .a a a a1 0 99 0 01 1 (1) (a) For a bubble-point liquid feed, the liquid-phase mole fractions in the pinch zone are those of the total feed. Thus, xC 3 = ,¥ = 0.70 and xC 3 ,¥ = 0.30. The relative volatility is that at the pinch. A bubble-point temperature on the feed composition at 300 psia gives 126oF, with a = 1.035/0.92 = 1.125. Substitution in Eq. (1) gives Lmin/D = 11.0 (b) For 50 mol% vaporization, a flash gives about the same temperature of 126oF, and therefore the same a, with liquid-phase mole fractions of 0.688 for propylene and 0.312 for propane. Substitution into Eq. (1) gives Lmin/D =11.2. (c) For a dew-point feed, the temperature still is 126oF, with liquid-phase mole fractions of 0.68 for propylene and 0.32 for propane. Substitution into Eq. (1) gives Lmin/D =11.4 Because the relative volatility is close to 1, the % vaporization of the feed has only a small effect on the minimum reflux ratio. Exercise 9.11 Subject: Minimum reflux rate and distribution of nonkey components for the distillation of a paraffin hydrocarbon mixture. Given: Column pressure of 700 kPa. Feed composition and split for two key components in Fig. 9.23. Feed is a bubble-point liquid. K-values from Figs. 2.8 and 2.9. Find: Minimum external reflux rate. Distribution of nonkey components at minimum reflux ratio. Analysis: Because of the wide range of volatility of the components in the feed, and the relative sharpness of the separation between the LK, n-butane, and the HK, isopentane, assume a Class 2 separation for estimating the minimum reflux ratio by the method of Underwood. Compute the minimum reflux using relative volatilities at the feed temperature. Because the feed is at the bubble point, use Eq. (4-12). For the pressure of 700 kPa (102 psia), a trial-and-error calculation for the bubble-point temperature gives the following results for 80oF, where nC8 is by extrapolation: Component z, feed mole fraction K at 102 psia and 800F Kz a referred to iso-pentane Propane 0.6361 1.37 0.8715 8.56 iso-Butane 0.1018 0.56 0.0570 3.50 n-Butane 0.1527 0.39 0.0596 2.44 iso-Pentane 0.0254 0.16 0.0041 1.00 n-Pentane 0.0509 0.12 0.0061 0.75 n-Hexane 0.0102 0.038 0.0004 0.24 n-Heptane 0.0127 0.013 0.0002 0.081 n-Octane 0.0102 0.004 0.0000 0.025 Total: 1.0000 0.9989 The Underwood equations that apply are Eqs. (9-28) and (9-29). For a bubble-point feed, 1 - q = 0 and Eq. (9-28) becomes: a a q q q q q q q q q i r i F i ri z q, , , . ( . ) . . ( . ) . . ( . ) . . ( . ) . . ( . ) . . ( . ) . . ( . ) . . ( . ) . - = - = = - + - + - + - + - + - + - + - 1 0 8 56 0 63618 56 35 0101835 2 44 015272 44 10 0 025410 0 75 0 0509 0 75 0 24 0 0102 0 24 0 081 0 0127 0 081 0 025 0 0102 0 025 (1) Eq. (1) has 8 roots for q. However, only 3 roots are probably necessary. These are those between values of a for: (1) iso-butane and n-butane, (2) n-butane and iso-pentane, and (3) iso- Exercise 9.11 (continued) Analysis: (continued) pentane and n-pentane, because iso-butane and n-pentane boil closely to n-butane and iso- pentane, respectively. Using a spreadsheet, the 3 roots of q are: (1) 2.7159 between 3.50 and 2.44, (2) 1.02575 between 2.44 and 1.00, and (3) 0.78274 between 1.00 and 0.75. Eq. (9-29) is applied in the following form for each of the three values of q, in terms of the three unknowns, diC4 , dnC5 , and Lmin. The form used is obtained by multiplying Eq. (9-29) by the total distillate rate, D, which converts distillate mole fractions to component distillate rates, and the reflux ratio, R to L. D L d d d d d d D L d + = - + - + - + - + - = + - - - = - - + = - + - + - min iC nC iC nC iC nC min iC .50 .44 .00 .75 .50 .44 4 5 4 5 4 5 4 8 56 2500 856 2 7159 3 350 2 7159 2 2 44 2 7159 1 15 100 2 7159 0 0 75 2 7159 36618 4 4637 52532 8 7418 0 3815 4 4637 160014 0 3815 8 56 2500 856 102575 3 350 102575 2 2 44 1 . ( ) . . ( ) . . (594) . . ( ) . . ( ) . . . . ( ) . . . ( ) . ( ) . . ( ) . ( ) . . ( ) . . (594) . .02575 1 15 100 102575 0 0 75 102575 2840 36 141457 1024 83 582 52 2 71986 141457 3282 67 2 71986 856 2500 856 0 78274 3 350 0 78274 2 2 44 0 78274 1 15 100 0 78274 0 0 75 0 78274 275161 128806 + - + - = + + - - = + - + = - + - + - + - + - = + .00 .75 .50 .44 .00 .75 nC iC nC iC nC min iC nC iC 5 4 5 4 5 4 5 4 ( ) . . ( ) . . . . ( ) . . . ( ) . ( ) . . ( ) . ( ) . . ( ) . . (594) . . ( ) . . ( ) . . . . ( d d d d d D L d d d ) . . . ( ) . ( ) . . ( )+ + - = + -874 55 69 04 22 9078 128806 369520 22 9078d d dnC iC nC 5 4 5 Because the total distillate rate, D, is also an unknown, the following equation is also needed: D = dC3 + diC4 + dnC4 + diC5 + dnC5 = 2500 + diC4 +594 + 15 + dnC5 = 3109 + diC4 + dnC5 Solving these 4 linear equations, the following results are obtained: D = 4712.8 kmol/h, Lmin = 795.4 kmol/h, diC4 = 1593.4 kmol/h, dnC5 = 10.45 kmol/h Because the flow rate of iC4 in the distillate is greater than its feed rate, iC4 does not distribute. Recalculate omitting the first of the four equations that uses q = 2.7159. Replace diC4 = fiC4 = 400 kmol/h. The equations to be solved now are: D L d D L d D d + + = + + = - = min nC min nC nC 5 5 5 2 71986 3848 5 22 9078 4210 4 3509 0 . . . . . Exercise 9.11 (continued) Analysis: (continued) Solving, D = 3526.9 kmol/h, Lmin = 272.8 kmol/h, and dnC5 = 17.9 kmol/h Because the distillate rate of nC5 is positive and less than its feed rate, it does distribute. If we had assumed that only the two key components distributed, then it is only necessary to solve the following two equations using only q = 1.02575, the one between the two key components, with dnC5 = 0.0 : D L D + = = min 3848 5 3509 0 . . Then, Lmin = 3848.5 - 3509.0 = 339.5 kmol/h = internal reflux rate, which is high and incorrect. Assume that the minimum external reflux rate = minimum internal reflux rate = 272.8 kmol/h The product compositions at minimum reflux rate are: Component Feed, kmol/h Distillate, kmol/h Bottoms, kmol/h Propane 2500 2500 0 iso-Butane 400 400 0 n-Butane 600 594 6 iso-Pentane 100 15 85 n-Pentane 200 17.9 182.1 n-Hexane 40 0 40 n-Heptane 50 0 50 n-Octane 40 0 40 Total: 3930 3526.9 403.1 : Exercise 9.12 Subject: Minimum reflux ratio and minimum number of equilibrium stages as a function of product purity for the distillation of a binary mixture. Given: Equimolar bubble-point feed of isobutane and n-butane at column pressure of 100 psia. Purity of iC4 in the distillate equal to purity of nC4 in the bottoms. Purity range from 90 mol% to 99.99 mol% Find: Minimum reflux ratio. Minimum number of equilibrium stages. Discuss significance of results. Analysis: From the vapor pressure plot of Fig. 2.4, for 100 psia, the boiling points are 120oF for iC4 and 146oF for nC4. Take an average temperature of 133oF and assume a relative volatility equal to the vapor pressure ratio. Thus, again using Fig. 2.4, aiC4, nC4 = 115/85 = 1.35. For Nmin , use the Fenske equation, (9-11), which simplifies to: N x x x x D B D D min iC iC iC iC 4 4 4 4= = - log log . log . 2 2 2 2 135 1 01303 (1) For Rmin , use the Class 1 form of the Underwood equation, (9-20), which simplifies to: R x xD D min iC iC4 4= - -, ,/ . . / . . 05 135 1 0 5 0 35 (2) Using a spreadsheet, the following values are computed: % Purity Nmin Rmin 90 14.6 4.37 92 16.3 4.64 94 18.3 4.91 95 19.6 5.04 96 21.2 5.18 98 25.9 5.45 99 30.6 5.58 99.5 35.3 5.65 99.9 46.0 5.70 99.95 50.7 5.71 99.99 61.4 5.71 Exercise 9.12 (continued) Analysis: (continued) A plot of the results is shown below. It is clearly evident from the table above or the plot below that the minimum reflux increases little and less and less as the purity approaches 100%, while the minimum number of equilibrium stages increases greatly and more and more as the purity approaches 100%, Exercise 9.13 Subject: Reflux ratio by the FUG method for the distillation of a propylene-propane mixture. Given: Binary feed of 360 kmol/h of propylene and 240 kmol/h of propane at the bubble point at column pressure. Distillate to contain 347.5 kmol/h of propylene and 3.5 kmol/h of propane. Average relative volatility = 1.11. N/Nmin = 2. Total condenser and partial reboiler. Assumptions: External reflux ratio = internal reflux ratio at the upper pinch. Find: Operating reflux ratio by the FUG method. Analysis: By material balance, the bottoms product contains 12.5 kmol/h of propylene and 236.5 kmol/h of propane. Using the Fenske equation (9-12): N d d b b min C C C C C C avg 3 = 3 3 3 = 3 = 3 log = = =log log .. .. log . . ,a 347 5 35 236 5 12 5 111 72 2 N = 2 Nmin = 2(72.2) = 144.4 For minimum reflux, use the Class 1 Underwood equation (9-20), which applies for a binary mixture, where the pinch composition is that of the liquid feed. The distillate mole fractions are 347.5/351 = 0.99 for propylene and 0.01 for propane. In the feed, the mole fractions are 360/600 = 0.60 for propylene and 0.40 for propane. Thus, R LD x x x x D F D F min min C , C C C C , C C C 3 = 3 = 3 = 3 3 3 3 = 3 = = - - = - - = , , , , . . . . . . . a a 1 0 99 0 60 111 0 01 0 40 111 1 14 75 Use either the Gilliland graphs of Figs. 9.10 or 9.11, or the Gilliland correlation equation, (9-34). The latter is more accurate, but requires iteration when solving for X. Apply the equation. Y N NN= - + = -+ =min1 144 4 72 2144 4 1 0 497. .. . Using a spreadsheet to solve Eq. (9-34), which is nonlinear in X, we obtain X = 0.159 min 0.159 Solving, +1 17. 3 7 -= = =R RX R R and R/Rmin = 17.73/14.75 = 1.202 Exercise 9.14 Subject: Number of equilibrium stages by the FUG method for the distillation of a binary mixture of dichlorobenzene (DCB) isomers. Given: Feed of 62 mol% p-DCB and 38 mol% o-DCB at approximately 1 atm. Distillate to contain 98 mol% p-DCB, and bottoms to contain 96 mol% o-DCB. Total condenser and partial reboiler. Average relative volatility = 1.154 with the para isomer as the LK. Feed condition is q = 0.9 (10 mol% vaporized). Want R/Rmin = 1.15. Assumptions: External reflux ratio = internal reflux ratio at the upper pinch. Find: Number of equilibrium stages by the FUG method. Analysis: Using the Fenske equation (9-11): N x x x x D B B D min p-DCB p-DCB o-DCB o-DCB p-DCB,o-DCB log = = =log log log ( . )( . ) .. ( . ) . a 0 98 0 04 0 96 0 02 1154 49 3 For minimum reflux, use the Class 1 Underwood equation (9-20), which applies for a binary mixture, where the pinch composition is that of the liquid portion of the feed. From the feed composition, determine the equilibrium liquid composition by a flash calculation for 10 mol% vaporization. Do this by combining Eqs. (7-26) for the q-line with (4-8) for equilibrium at a constant relative volatility. Thus, applying these equations to the LK, p-DCB, aaxx x x qq x zq x xF1 1 11541 0154 1 1 9 0 6201 6 2 9+ - = + = - - - = - + = -( ) . . .. . Solving this nonlinear equation for a positive root of x between 0 and 1 gives x = xp-DCB = 0.617 Then, R LD x x x x D F D F min min p-DCB, p-DCB p DCB,o DCB o-DCB, o DCB p DCB,o DCB = = - - = - - = - - - - - , , . . . . . . . a a 1 0 98 0 617 1154 0 02 0 387 1154 1 9 93 R = 1.15 Rmin =1.15(9.93) = 11.42 In the Gilliland equation, (9-34), X = (R - Rmin)/(R + 1) = (11.42-9.93)/(11.42+1) = 0.120. Using Eq. (9.34), Y = 0.534 = (N - Nmin)/(N + 1). Solving, N = 107 stages. Exercise 9.15 Subject: Shortcomings of the Gilliland correlation Given: Column with a small ratio of rectifying to stripping stages. Find: Explanation for possible inapplicability of the Gilliland correlation. Analysis: When a column has a small ratio of rectifying to stripping stages, stripping of a light key in the stripping section may dominate over absorption of the heavy key in the rectifying section, in determining stage requirements. The Gilliland correlation is based on reflux considerations in the rectifying section. Alternatively, a correlation could have been developed based on boilup considerations in the stripping section, using a modified Underwood equation for minimum boilup ratio. Such a correlation, since it would be empirical, would seem to be favored over the Gilliland correlation, which emphasizes rectification, rather than stripping. Exercise 9.16 Subject: Use of the FUG method for the distillation of a normal paraffin hydrocarbon mixture. Given: Bubble-point liquid feed at 300 psia, with composition in mole fractions: Component C2 C3 nC4 nC5 nC6 nC7 Mole fraction 0.08 0.15 0.20 0.27 0.20 0.10 Find: (a) For a sharp separation between nC4 and nC5, determine column pressure and type condenser for condenser outlet temperature of 120oF. (b) At total reflux, determine the separation for 8 equilibrium stages, with 0.01 mole fraction of nC4 in the bottoms. (c) Minimum reflux ratio for separation in Part (b). (d) Number of equilibrium stages for R/Rmin = 1.5 using Gilliland correlation. Analysis: (a) For a sharp split, the distillate composition is as follows: Component xd K at 120oF, 250 psia Kxd C2 0.186 2.8 0.520 C3 0.349 1.13 0.394 nC4 0.465 0.34 0.158 Total: 1.000 1.072 In this table, a distillate bubble-point pressure of 250 psia is assumed and the bubble-point equation (4-12) is applied, using K-values in Fig. 2.8. The summation is 1.072. In order for the sum to be 1.0, an even higher pressure is needed, since the K-value is almost inversely proportional to pressure. But, in Fig. 7.16, we are already > 215 psia, therefore use a partial condenser. To determine the pressure, run a dew point using Eq. (4-13). Component yD K at 120oF, 140 psia yD/K K at 120oF, 120 psia yD/K C2 0.186 4.7 0.040 5.4 0.030 C3 0.349 1.58 0.221 1.80 0.194 nC4 0.465 0.53 0.877 0.60 0.775 Total: 1.000 1.138 0.999 On the second trial in the above table, the dew-point pressure is found to be 120 psia. Set the column pressure at this value with a partial condenser. Exercise 9.16 (continued) Analysis: (continued) (b) Run a bubble point on an assumed bottoms composition, using Eq. (4-12): Component xB K at 320oF, 120 psia xB K K at 290oF, 120 psia xB K nC4 0.010 3.3 0.033 2.75 0.028 nC5 0.469 1.7 0.797 1.38 0.647 nC6 0.347 0.92 0.319 0.72 0.250 nC7 0.174 0.48 0.084 0.36 0.063 Total: 1.000 1.233 0.988 A third trial gives a bottoms temperature of 295oF The K-values and relative volatilities referred to the nC5, the heavy key, at distillate and bottoms conditions are as follows for a pressure of 120 psia: Component K at 120oF, 120 psia aI, nC5 at 120oF, 120 psia K at 295oF, 120 psia aI, nC5 at 295oF, 120 psia Geometric average a C2 5.40 25.7 12.0 8.28 14.6 C3 1.80 8.6 5.8 4.0 5.9 nC4 0.60 2.86 2.8 1.93 2.35 nC5 0.21 1.00 1.45 1.00 1.00 nC6 0.074 0.352 0.74 0.51 0.424 nC7 0.028 0.133 0.38 0.262 0.187 From the Fenske Eq. (9-12), N d d b b d d b b min nC nC nC nC nC nC avg nC nC nC nC 4 5 5 4 4 5 4 5 5 4 log= = = 8 2 35 log log log .,a (1) Take as a basis, 1 mol of feed. Assume that the distribution of the nonkeys is negligible. That is, only the LK and the HK are found in both the distillate and bottoms. Assume the total bottoms, B, is 0.57 mol. Then bnC4 = 0.01(0.57) = 0.0057. Therefore, dnC4 = 0.20 - 0.0057 = 0.1943 mol. Solving Eq. (1), bnC5/dnC5 = 27.3. Since bnC5 + dnC5 = fnC5 = 0.27, bnC5 = 0.2605 mol and dnC5 = 0.0095. Therefore, B = 0.0057 + 0.2605 + 0.20 + 0.10 = 0.5662. Repeating the calculations with this value of B, the result is bnC4 = 0.00566, dnC4 = 0.19434, bnC5 = 0.2604, and dnC5 = 0.0096. Using Eq. (1) for each of the nonkey components shows that they do not distribute to any extent. Therefore the distillate and bottoms compositions at total reflux are: Component f, mol d, mol xD b, mol xB C2 0.08 0.0800 0.184 0.0000 0.000 C3 0.15 0.1500 0.346 0.0000 0.000 nC4 0.20 0.1943 0.448 0.0057 0.010 nC5 0.27 0.0096 0.022 0.2604 0.460 nC6 0.20 0.0000 0.000 0.2000 0.353 nC7 0.10 0.0000 0.000 0.1000 0.177 Total 1.00 0.4339 1.000 0.5661 1.000 Exercise 9.16 (continued) Analysis: (c) Because the split of the two key components is quite sharp, assume a Class 2 separation for the purpose of minimum reflux. Thus, only components methane to n-pentane will distribute to the distillate. To determine the phase condition of the feed, perform an adiabatic flash with bubble-point conditions at 300 psia upstream of the feed valve to the column, and 120 psia downstream of the valve, using Chemcad. The result is a V/F ratio for the feed of 0.334. From Eq. (7-19), q = 1 - (V/F) = 1 - 0.334 = 0.666 or 1 - q = 0.334. From Underwood Eq. (9-28), a a q q q q q q q i i F ii z q, , , . . ( . ). . ( . ). . ( . ). . ( . ). . ( . ). . ( . ).nC nC 5 5 - = - = = - + - + - + - + - + - 1 0 334 14 6 0 08 14 6 5 9 015 015 2 35 0 20 2 35 100 0 27 100 0 424 0 20 0 424 0187 010 0187 Solving this equation for the root between 2.35 and 1.00 gives q = 1.5962 From Underwood Eq. ((9-29), a a q i i D ii x R, , , . ( . ) . . . ( . ) . . . ( . ) . . . ( . ) . . . nC nC min 5 5 - = - + - + - + - = = + 14 6 0184 14 6 15962 59 0 346 59 15962 2 35 0 448 2 35 15962 100 0 022 100 15962 2 04 1 Therefore, the internal minimum reflux ratio = 2.04 - 1.00 = 1.04 Assume the external minimum reflux ratio = 1.04 (d) For the number of theoretical stages, use the Gilliland equation, (9-34), where for this exercise, R = 1.5R = 1.5(1.04) = 1.56. X = (R - Rmin)/(R + 1) = (1.56 - 1.04)/(1.56 + 1) = 0.203. Using Eq. (9.34), Y = 0.458 = (N - Nmin)/(N + 1). Solving, N = 15.6 equilibrium stages. Exercise 9.17 Subject: Use of the FUG method for the distillation of a light paraffin hydrocarbon mixture. Given: Feed mixture of: Component C3 iC4 nC4 iC5 nC5 lbmol/h 5 15 25 20 35 Column pressure = 120 psia. Liquid distillate contains 92.5% of the nC4 in the feed. Bottoms contains 82 mol% of iC5 in the feed. Use Chemcad with SRK instead of Figs. 2.8 and 2.9 for K- values. Find: (a) Minimum number of equilibrium stages. (b) Distribution of nonkey components. (c) Minimum reflux ratio for a bubble-point liquid feed. (d) Number of equilibrium stages for R = 1.2 Rmin. (e) Feed-stage location. Analysis: (a) To obtain the average relative volatilities, run bubble points on the following assumed distillate and bottoms compositions, using the given key-component recoveries: lbmol/h: Component Feed Distillate Bottoms C3 5 5.000 0.000 iC4 15 15.000 0.000 nC4 25 23.125 1.875 iC5 20 3.600 16.400 nC5 35 0.000 35.000 Total: 100 46.725 53.275 The results with K-values and relative volatilities are as follows, using the SRK equation of state for K-values and the bubble-point equation, (4-12): Component K, distillate, 140oF ai,iC5 , 140oF K, bottoms, 230oF ai,iC5 , 230oF ai,iC5 , geometric mean C3 2.100 5.24 3.880 3.63 4.36 iC4 1.056 2.63 2.260 2.12 2.36 nC4 0.818 2.04 1.863 1.74 1.88 iC5 0.401 1.00 1.068 1.00 1.00 nC5 0.329 0.82 0.922 0.86 0.84 Exercise 9.17 (continued) Analysis: (continued) (a) Using the Fenske equation (9-12), with nC4 as the LK and iC5 as the HK, ( ) ( ) ( ) ( ) ( )nC4 iC5iC5 nC4 nC4,iC5 av min g 23.125 16.4loglog 3.6 1.875 log1.88log 6.38 = = a = d N b d b (b) To compute the distribution of nonkey components at total reflux, use for the lighter than light key components, LLK, Eq. (9-15), and use Eq. (9-16) for the heavier than heavy key, HHK, with iC5 as the reference component, r, and the above value of Nmin. Thus, for the LLK, i, b fd b f i i N i= + = + 1 1 3 616 4 6 38iC iC i,iC i,iC 5 5 5 min 5a a.. . (1) For the HHK, i, d f db d b f i i i N i N i i i = = iC iC ,iC iC iC ,iC ,iC ,iC 5 5 5 min 5 5 5 min 5 51+ 1+ a a a a 3 6 16 4 3 6 16 4 6 38 6 38 . . . . . . (2) Using the above geometric mean values of ai, iC5 , followed by use of Eqs. (1) and (2) with the material balance, fi = di + bi , the following results are obtained: lbmol/h: Component Feed Distillate Bottoms C3 5 4.998 0.002 iC4 15 14.720 0.280 nC4 25 23.125 1.875 iC5 20 3.600 16.400 nC5 35 2.356 32.644 Total: 100 48.799 51.201 (c) Because the split of the LK and HK components is not sharp and because iC4 boils close to nC4, and iC5 boils close to nC5, it is possible that both iC4 and nC5 will distribute at minimum reflux. It is questionable that C3 will distribute. Nevertheless, first assume a Class 1 Underwood minimum reflux and use the following rearrangement of Eq. (9-21) for a feed at the bubble point: L F df df min nC nC nC iC iC iC nC iC 4 4 5 5 5 5 = - - = - - = a a 4 4 1 100 2312525 188 3620 188 1 66 7 , , . . . . . l Exercise 9.17 (continued) Analyze: (c) (continued) Now, use the Class 2 Underwood equation, Eq. (9-28),which for a bubble-point feed is: a a q q q q q q i i F ii z q, , , . ( . ) . . ( . ) . . ( . ) . . ( . ) . . ( . ) . iC iC 5 5 - = - = = - + - + - + - + - 1 0 4 36 0 05 4 36 2 36 015 2 36 188 0 25 188 100 0 20 100 0 84 0 35 084 (3) There are four roots to Eq. (3), which from a spreadsheet calculation are: q = 0.9248, 1.317, 2.173, 3.996 Assume that the C3 does not distribute. Then, the 3.996 root is not needed and dC3 = 5 lbmol/h. Eq. (9-29) is applied in the following form for each of the three remaining values of q, in terms of the three unknowns, diC4 , dnC5 , and Lmin. The form used is obtained by multiplying Eq. (9-29) by the total distillate rate, D, which converts distillate mole fractions to component distillate rates, and the reflux ratio, R to L. D L d d d d d d D L d + = - + - + - + - + - = + + + - = + - + = - + - + min iC nC iC nC iC nC min iC .36 1.88 .00 .84 .36 1.88 4 5 4 5 4 5 4 4 36 4 36 0 9248 2 2 36 0 9248 23125 188 0 9248 1 36 100 0 9248 0 0 84 0 9248 6 346 1644 45514 47 872 9 9057 1644 93386 9 9057 4 36 4 36 1317 2 2 36 1317 23125 188 . (5) . . ( ) . . ( . ) . . ( . ) . . ( ) . . . . ( ) . . . ( ) . ( ) . . ( ) . (5) . . ( ) . . ( . ) . - + - + - = + + - - = + - + = - + - + - + - + - = + 1317 1 3 6 100 1317 0 084 1317 7 164 2 263 77 22 1136 1761 2 263 7302 1761 4 36 4 36 2 173 2 2 36 2173 23125 188 2173 1 3 6 100 2 173 0 0 84 2173 9 968 12 62 . ( . ) . . ( ) . . . . ( ) . . . ( ) . ( ) . . ( ) . (5) . . ( ) . . ( . ) . . ( . ) . . ( ) . . . . .00 .84 .36 1.88 .00 .84 nC iC nC iC nC min iC nC 5 4 5 4 5 4 5 d d d d d D L d d ( ) . . . ( ) . ( ) . . ( )d d d diC nC iC nC 4 5 4 5 - - - = - -148 38 3 07 0 63 12 62 14148 0 63 Because the total distillate rate, D, is also an unknown, the following equation is also needed: D = dC3 + diC4 + dnC4 + diC5 + dnC5 = 5 + diC4 +23.125 + 3.6 + dnC5 = 31.725 + diC4 + dnC5 Solving these 4 linear equations, the following results are obtained: D = 53.27 lbmol/h, Lmin = 64.74 lbmol/h, diC4 = 20.61 lbmol/h, dnC5 = 0.93 lbmol/h Note that the distillate rate for iC4 is impossible because it exceeds the feed rate of 15 lbmol/h. Therefore, iC4 does not distribute, and the calculations must be repeated, eliminating diC4 as an unknown by replacing it with a value of 15 lbmol/h. Also, the q root of 2.173 is not used. The following three equations result: Exercise 9.17 (continued) Analyze: (c) (continued) D L d D L d D d + = - + = - = + min nC min nC nC 5 5 5 106 96 1761 118 05 9 9057 46 725 . . . . . Solving these three equations: D = 48.085 lbmol/h, Lmin = 56.49 lbmol/h, and dnC5 = 1.36 lbmol/h This is the correct value of internal minimum reflux rate. Assume the external minimum reflux ratio = internal minimum reflux ratio = 56.49/48.085 = 1.175. Note that this value is about 15% less than that obtained by assuming a Class 1 separation. (d) The operating reflux ratio = 1.2(1.175) = 1.41. In the Gilliland equation, (9-34), X = (R - Rmin)/(R + 1) = (1.41 - 1.175)/1.41+1) = 0.0975. Using Eq. (9.34), Y = 0.556 = (N - Nmin)/(N + 1). Solving, N = 15.6 stages. (e) Apply the Kirkbride equation (9-36): From above, using the Fenske distribution, D = 48.799 lbmol/h and B = 51.201 lbmol/h NN zz xx BDR S F F B D = = =iC nC nC iC 5 4 4 5 , , , , . . . . . . . . . 2 0 206 2 0 2060 20 0 25 0 0366 0 0738 51201 48 799 0 723 Therefore, of 15.6 equilibrium stages, (0.723/1.723)(15.6) = 6.5 stages are in the rectifying section. Thus, the feed stage is equilibrium stage 6 or 7 from the top. Exercise 9.18 Subject: Use of the FUG method for the distillation of a chlorination effluent. Given: Bubble-point liquid feed with the following composition and average K-values: Component Mole fraction Aver. K-value C2H4 , A 0.05 5.10 HCl , B 0.05 3.80 C2H6 , C 0.10 3.40 C2H5Cl , D 0.80 0.15 Partial condenser and partial reboiler. Column pressure = 240 psia. (xD/xB) = 0.01 for C2H5Cl and 75 for C2H6. Find: Minimum equilibrium stages Component distribution Minimum reflux ratio Number of equilibrium stages for R = 1.5 Rmin Feed stage location Analysis: For minimum equilibrium stages, use the Fenske equation (9-11): ( ) ( ) 2 6 2 5 C H C H Cl C in D m , / 75log log/ 0.01 log log(3.4 / 0.1 2.8) 65 = = = a D B D Bx xN x x For component distribution, use the following rearrangements of the Fenske equation: x x x x x x x x D B D B N D B D B N = = = = = =C H C H Cl C H ,C H Cl HCl C H Cl HCl,C H Cl 2 4 2 5 2 4 2 5 min 2 5 2 5 min a a 0 01 51015 240 0 01 38015 103 2 86 2 86 . .. . .. . . Because the separation between the two key components is quite sharp, assume that the two nonkeys do not distribute at minimum reflux. The relative volatilities with respect to the heavy key are: 34 for A, 25.3 for B, and 22.7 for C. Now, use the Class 2 Underwood equation, Eq. (9- 28),which for a bubble-point feed is: Exercise 9.18 (continued) Analysis: (continued) aa q q q q qi D i F i Di z q, , , ( . ) . ( . ) . . ( . ) . . ( . ) .- = - = = - + - + - + - 1 0 34 0 05 34 253 0 05 25 3 22 7 010 22 7 100 080 100 (1) There are three roots to Eq. (1). The one of interest is the one between 22.7 and 1.0, which from a spreadsheet calculation is q = 4.318. The composition of the distillate by material balances is: Component distillate, mole fraction C2H4 , A 0.25405 HCl , B 0.25405 C2H6 , C 0.48195 C2H5Cl , D 0.00995 Apply Underwood Eq. (9-29): a a q i D i D i Di x R R , , , ( . ) . . ( . ) . . . ( . ) . . . ( . ) . . . . . . . - = - + - + - + - = + = + + - = = + 34 0 2540534 4 318 25 3 0 25405253 4 318 22 7 0 4819522 7 4 318 100 0 00995100 4 318 1 0 291 0 306 0595 0 003 1189 1 min min Therefore, Rmin = 0.189. Assume this is equal to the external minimum reflux ratio. Operating reflux ratio = 1.5(0.189) = 0.284. In the Gilliland equation, (9-34), X = (R - Rmin)/(R + 1) = (0.284 - 0.189)/0.284 + 1) = 0.074. Using Eq. (9.34), Y = 0.581 = (N - Nmin)/(N + 1). Solving, N = 8.2 stages. For feed stage location, apply the Kirkbride equation (9-36): From above, using the Fenske distribution with a material balance, D/B = 0.245. Also, mole fractions in the bottoms product are: xC = 0.0064 and xD = 0.9936 NN zz xx BDR S F F B D = = =D C C D , , , , . . . . . . . . 2 0 206 2 0 206080 010 0 0064 0 00995 1 0 245 171 Therefore, of 8.2 equilibrium stages, (1.71/2.71)(8.2) = 5.2 stages are in the rectifying section. With a partial condenser acting an equilibrium stage, the feed is about stage 4 from the top stage in the column. Exercise 9.19 Subject: Use of the FUG method for the distillation of a ficticious ternary mixture. Given: 100 kmol/h of a bubble-point feed of: Component Feed rate, kmol/h ai,C A 40 5 B 20 3 C 40 1 Find: (a) Product distribution for a distillate rate of 60 kmol/h and 5 minimum theoretical stages. (b) Minimum reflux and boilup ratios for separation of part (a). (c) Number of equilibrium stages and feed-stage location for R = 1.2Rmin. Analysis: (a) With a distillate rate of 60 kmol/h, the LK is B and the HK is C. From the rearranged form of the Fenske equation, given by Eq. (9-14), using component C as the reference, d b d b d b d b d b d b A A C C C C B B C C C C = = = = 5 3125 3 243 5 5 The following 4 material balances also apply: dA + bA = 40 dB + bB = 20 dC + bC = 40 dA + dB + dC = 60 Solving these 6 equations, two of which are nonlinear, in 6 unknowns gives the following results: Component zF xD xB A 0.4 0.662 0.007 B 0.2 0.307 0.039 C 0.4 0.031 0.954 (b) Assuming that component A does not distribute, the Underwood Eq. (9-28) for Class 2 applies: aa q q q qi D i F i Di z q, , , ( . ) ( . ) ( . ) - = - = = - + - + - 1 0 5 0 40 5 3 0 20 3 1 0 40 1 (1) Exercise 9.19 (continued) Analysis: (continued) There are two roots to Eq. (1). The one of interest is the one between 3 and 1.0, which from a spreadsheet calculation is q = 1.425. If component A does not distribute, the composition of the distillate by material balances is as follows, if the total distillate rate of 60 kmol/h is maintained: kmol/h: Component Feed Distillate Bottoms xD xB A 40 40 0.0 0.6667 0.0000 B 20 18.3 1.7 0.3050 0.0425 C 40 1.7 38.3 0.0283 0.9575 Total: 100 60 40 1.0000 1.0000 Apply Underwood Eq. (9-29): a a q i D i D i Di x R R , , , ( . ) . ( . ) . ( . ) . . . . . - = - + - + - = + = + - = = + 5 0 66675 1425 3 0 30503 1425 1 0 02831 1425 1 0 9324 05810 0 0666 1447 1 min min Therefore, Rmin = 1.447 - 1.000 = 0.447. Assume this is also the external reflux ratio. Assume constant molar overflow. Then, the reflux rate in the rectifying section = 0.447(60) = 26.8 kmol/h. The reflux rate in the stripping section = 100 + 26.8 = 126.8 kmol/h. The boilup rate in the stripping section = 126.8 - 40 = 86.8 kmol/h. The boilup ratio = 86.8/40 = 2.17. (c) Operating reflux ratio = 1.2(0.0.447) = 0.536. In the Gilliland equation, (9-34), X = (R - Rmin)/(R + 1) = (0.536 - 0.447)/0.536 + 1) = 0.058. Using Eq. (9.34), Y = 0.60 = (N - Nmin)/(N + 1). Solving, N = 14 stages. For feed stage location, apply the Kirkbride equation (9-36): From above, using the Fenske distribution with a material balance, D/B = 1.5. NN zz xx BDR S F F B D = = =C B B C , , , , . . . . . . . . 2 0 206 2 0 2060 40 0 20 0 039 0 031 1 15 1166 Therefore, of 14 equilibrium stages, (1.166/2.166)(14) = 7.5 stages are in the rectifying section. Assuming a total condenser, the feed stage is stage 7 or 8 from the top. Exercise 9.20 Subject: Comparison of the Kirkbride, Fenske, and McCabe-Thiele methods for determining feed-stage location. Given: A feed mixture of 25 mol% acetone (A) and 75 mol% water (W). Distillate to contain 95 mol% A, and bottoms to contain 2 mol% A. Infinite dilution activity coefficients from Exercise 9.6. Find: Ratio of rectifying to stripping stages by: (a) Fenske equation. (b) Kirkbride equation. (c) McCabe-Thiele method. Analysis: (a) Fenske equation: From Exercise 9.6: Component zF xD xB ai,W, D ai,W, B Acetone 0.25 0.95 0.02 1.48 27.7 Water 0.75 0.05 0.98 1.00 1.0 The relative volatility at feed conditions is also needed. Assume a bubble-point liquid feed. Also from vapor-liquid, y-x table of Exercise 9.6, ai,W = [(0.789/0.250)]/[(0.211/0.750)] = 11.2 From Eq. (9-35), ( )( )( )( ) 1/ 2 1/ 2 log 0.95 / 0.25 0.75/ 0.05 log[(27.7)(11.2)] 1.756(1.246) 1.213(0.6097)log 0.25/ 0.02 0.98 / 0.75 log[(1.48)(11.2) 2. 96] = = = R S N N (b) Kirkbride equation: By material balances on A and W, B/D = 0.753/0.247 = 3.05 0.206 0.2062 2 W, A, A, W, 0.75 0.02 3.05 0. 1.0825 0.05 = = = R S F B F D z x N z x D N B (c) McCabe-Thiele method: From the McCabe-Thiele graph of Exercise 9.6, shown on the next page, where the total number of minimum stages = 5. 4.2 0.8 5.25== R S N N The McCabe-Thiele method is the most accurate. The Fenske equation is better than the Kirkbride equation because the Kirkbride equation does not take into account the drastic change in relative volatility from the distillate to the bottoms. Exercise 9.20 (continued) Analysis: McCabe-Thiele method (continued) Exercise 9.21 Subject: Derivation of Eq. (9-37) for the effective stripping factor Given: Equations (5-46) and (5-47) Analysis: By analogy to Eqs. (5-46) and (5-47), the fraction of species in the entering feed liquid that is not stripped is given for just 2 stages by: fS N N e e e e N N e e e N e N N S S S S S S S S S S S S S S S S S S S S = + + = + + + + = + + + - + = = - + + + = + + - 1 1 1 1 1 1 1 0 1 1 4 1 2 0 25 1 05 1 2 2 1 2 1 1 1 Therefore, Solving for , . . : Which is Eq. (9-37). Exercise 9.22 Subject: Effect of number of stages and pressure on multicomponent absorption Given: Feed gas of 2,000 lbmol/h of light paraffin hydrocarbons at 60oF, and 500 lbmol/h of nC10 absorbent at 90oF. K-values in Fig. 2.8. Find: Separation at following conditions: (a) 6 stages and 75 psia. (b) 3 stages and 150 psia. (c) 6 stages and 150 psia Analysis: Apply Kremser's method, with absorption factors, A = L/KV, computed for L = entering absorbent rate = 500 lbmol/h, V = entering gas rate = 2,000 lbmol/h, and K-values from Fig. 2.8 at the average of the two entering temperatures, (90 + 60)/2 = 75oF and each of the two pressures. The feed gas composition and the resulting K-values and values of A are as follows: Component Feed, lbmol/h K at 75 psia A at 75 psia K at 150 psia A at 150 psia C1 1660 29 0.00862 15 0.0167 C2 168 5.8 0.0431 3.1 0.0806 C3 96 1.68 0.149 0.93 0.269 nC4 52 0.48 0.521 0.27 0.926 nC5 24 0.145 1.724 0.08 3.13 Total: 2000 The fraction of a gas component not absorbed, fA, is given by the Kremser Eq. (5-48) and by material balance the amount absorbed follows: f fA A and = -- = -+AA l fN N1 1 11 The results for all three cases are as follows: Case (a): Case (b): Case (c): Component fA lN, lbmol/h fA lN, lbmol/h fA lN, lbmol/h C1 0.991 14.9 0.983 28.2 0.983 28.2 C2 0.957 7.2 0.919 13.6 0.919 13.6 C3 0.851 14.3 0.731 25.8 0.731 25.8 nC4 0.484 26.8 0.280 37.4 0.178 42.7 nC5 0.0164 23.6 0.0224 23.5 0.00072 23.98 Total: 86.8 128.5 134.28 The results show that doubling the pressure is much more effective than doubling the number of equilibrium stages. Even Case (c) results in little additional absorption over Case (b) when the stages are doubled for the same pressure. Exercise 9.23 Subject: Effect of the number of stages on the absorption of a light normal paraffin hydrocarbon gas by the Kremser method. Given: 1000 kmol/h of gas at 70oF of composition below, and 500 kmol/h of nC10 absorbent at 90oF. Absorber operating at 4 atm. K-values in Fig.2.8. Find: Percent absorption for each component for (a) 4 stages, (b) 10 stages, and (c) 30 stages. Analysis: : Apply Kremser's method, with absorption factors, A = L/KV, computed for L = entering absorbent rate = 500 kmol/h, V = entering gas rate = 1,000 kmol/h, and K-values from Fig. 2.8 at the average of the two entering temperatures, (90 + 70)/2 = 80oF and a pressure of 4 atm = 59 psia. The feed gas composition and the resulting K-values and values of A are as follows: Component f, kmol/h K-value A = L/KV C1 250 35 0.0142 C2 150 7.7 0.065 C3 250 2.2 0.227 nC4 200 0.65 0.769 nC5 150 0.20 2.50 Total: 1000 The fraction of a gas component not absorbed, fA, is given by the Kremser Eq. (5-48), and by material balance the amount absorbed follows: f fA A and Percent absorbed = 100 1-= --+AAN 111 The results for all three cases are as follows: N = 4 stages: N = 10 stages: N = 30 stages: Component fA % absorbed fA % absorbed fA % absorbed C1 0.986 1.4 0.986 1.4 0.986 1.4 C2 0.935 6.5 0.935 6.5 0.935 6.5 C3 0.773 22.7 0.773 22.7 0.773 22.7 nC4 0.316 68.4 0.245 75.5 0.231 76.9 nC5 0.0155 98.4 0.0001 99.99 0.0000 100.0 The above results show that as N increases, the percent absorption of the heavier components increases, but the absorption of the lighter components does not increase. However, beyond 10 stages, little change occurs. Exercise 9.24 Subject: Flow rate of stripping steam by the Kremser method Given: Feed of composition below enters a flash drum operating at 150oF and 2 atm. Equilibrium liquid is sent to a 5-equilibrium-stage stripper operating at 2 atm to give 0.5 kmol/h of nC5 in stripper bottoms. Find: Flow of steam needed in the stripper. Analysis: Use Chemcad with the SRK equation of state to flash the feed to determine the liquid feed to the stripper. The result is: Component Feed, kmol/h Distillate, kmol/h Bottoms, kmol/h C1 13.7 13.6 0.1 C2 101.3 98.2 3.1 C3 146.9 134.9 12.0 nC4 23.9 19.1 4.8 nC5 5.7 3.4 2.3 nC12 196.7 0.4 196.3 Total: 488.2 269.6 218.6 The stripping steam enters the stripper at 2 atm and 300oF. For the stripper, the average temperature of the two feeds = (150 + 300) = 225oF. From Fig. 2.8, the K-value of nC5 at 2 atm and 225oF = 3.0. The average stripping factor for n-pentane can be taken as: S = KV/L = 3V/218.6 = 0.0137 V The fraction of nC5 not stripped = fS = 0.5/2.3 = 0.217 Use Kremser Eq. (5-50) to compute the value of S needed for N = 5 theoretical stages. By trial and error, the result from: fS NSS= --+ 111 = 0.217 is S = 0.89. Therefore, Flow rate of stripping steam = V = S/0.0137 = 0.89/0.0137 = 65 kmol/h Exercise 9.25 Subject: Stripping of a hydrocarbon liquid with superheated steam. Given: 1000 kmol/h of feed liquid at 250oF with composition below. 100 kmol/h of superheated steam at 300oF and 50 psia. Stripper operates at 50 psia and has 3 equilibrium stages. Assumptions: Negligible stripping of nC10 and negligible condensation of steam Find: Flow rates and compositions of the exiting streams by the group (Kremser) method. Analysis: Apply Kremser's method, with stripping factors, S = KV/L , computed for L = entering feed = 1000 kmol/h, V = entering steam rate = 100 kmol/h, and K-values from Fig. 2.8 at the average of the two entering temperatures, (250 + 300)/2 = 275oF and a pressure of 50 psia. The feed gas composition and the resulting K-values and values of S are as follows: Component f, kmol/h K-value S = KV/L C1 0.3 67 6.7 C2 2.2 26 2.6 C3 18.2 12.2 1.22 nC4 44.7 6.0 0.60 nC5 85.9 2.85 0.285 nC10 848.7 Total: 1000.0 The fraction of a gas component not stripped, fS, is given by the Kremser Eq. (5-50), and by material balance the amounts stripped and not stripped follow: f fS S 3 and and = -- = = -+SS l f v f lN 1 11 1 1 The results are as follows: Component fS l1, kmol/h v3, kmol/h C1 0.0028 0.0 0.3 C2 0.036 0.1 2.1 C3 0.181 3.3 14.9 nC4 0.46 20.6 24.1 nC5 0.72 61.8 24.1 Total: 85.8 65.5 The exiting gas is 65.5 + 100 = 165.5 kmol/h. The exiting liquid is 85.8 + 848.7 = 934.5 kmol/h. Exercise 9.26 Subject: Liquid-liquid extraction of a hydrocarbon mixture by diethylene glycol (DEG) by the group (Kremser) method. Given: 100 kmol/h of an equimolar mixture of benzene (B), toluene (T), n-hexane (C6), and n-heptane (C7). Extraction at 150oC with 300 kmol/h of DEG. Extractor has 5 equilibrium stages. Distribution coefficients given below. Find: Flow rates and compositions of extract and raffinate. Analysis: Treat the problem like a stripper. Apply Kremser's method, with extraction factors, Eq. (9-39), E = KDV/L , computed for L = entering feed = 100 kmol/h, V = entering DEG solvent rate = 300 kmol/h, and given values of KD = mole fraction in extract/mole fraction in raffinate. The extract is like the vapor in stripping and the raffinate is like the liquid in stripping. The feed composition and the K-values and values of S are as follows: Component f, kmol/h KD E = KDV/L Benzene 25 0.33 0.99 Toluene 25 0.29 0.87 n-Hexane 25 0.050 0.15 n-Heptane 25 0.043 0.129 Total: 100 The fraction of a feed component not extracted, fE , is given by the Kremser Eq. (9-44), and by material balance the amounts stripped and not stripped follow: f fE E 3 and and = -- = = -+EE l f v f lN 1 11 1 1 The results are as follows: Component fE l1, kmol/h v3, kmol/h Benzene 0.177 4.3 20.7 Toluene 0.230 5.8 19.2 n-Hexane 0.850 21.2 3.8 n-Heptane 0.871 21.8 3.2 Total: 53.1 46.9 Now compute the transfer of DEG to the raffinate using Eqs. (9-40) and (9-43), using KD = 30 U = L/KDV = 100/[30(300)] = 0.0111. fU = 0.9889. Therefore 0.9889(300) = 296.7 kmol/h of DEG leaves in the extract and 300 - 296.7 = 3.3 kmol/h in the raffinate. The exiting extract flow rate = 46.9 + 296.7 = 343.6 kmol/h The exiting raffinate flow rate = 53.1 + 3.3 = 56.4 kmol/h. Exercise 9.27 Subject: Reboiled stripping of a normal paraffin hydrocarbon feed liquid by the group method. Given: Feed liquid at 39oF and 300 psia and of composition given below is adiabatically flashed to 150 psia before entering a stripper containing 7 equilibrium stages and a partial reboiler. The bottoms flow rate from the stripper = B = 99.3 lbmol/h. Find: Compositions of the vapor and liquid products from stripper. Analysis: First, adiabatically flash the feed given below, using Chemcad with the SRK equation of state. The results of the flash are as follows, where the calculated flash temperature = 26.3oF: Lbmol/h: Component Feed Vapor Liquid C1 59.5 37.67 21.83 C2 73.6 14.83 58.77 C3 153.2 8.90 144.30 nC4 173.5 2.58 170.92 nC5 58.2 0.23 57.97 nC6 33.6 0.04 33.56 Total: 551.6 64.25 487.35 The liquid from the adiabatic flash is sent to the reboiled stripper where it is separated into an overhead vapor and a liquid bottoms. The overhead vapor is mixed with the vapor from the adiabatic flash to give the final vapor product. To solve the reboiled stripper by the group method, apply Edmister's method, rather than Kremser's method, because we do not know the composition and flow rate of the vapor leaving the reboiler and entering the column. For a reboiled stripper, Eq. (5-64) applies: lb S S K VBF B AX SX B B B= + =f f 1 (1) where (2) For fAX and fEX, use average absorption and stripping factors computed for L = entering flash liquid = 487.35 kmol/h, V = vapor leaving the reboiler, which can be taken as L - B = 487.35 - 99.30 = 388.05 kmol/h, and K-values from Fig. 2.8 at an average temperature of say 200oF and a pressure of 150 psia. The feed gas composition and the resulting K-values, values of S = KV/L and A = 1/S, and corresponding values of fAX and fEX , computed from Eqs. (5-48) and (5-50), f fAX N SX NAA SS= -- = --+ +1 1 1 11 1 and are as follows: Exercise 9.27 (continued) Analysis: (continued) Component lF , lbmol/h K-value S = KV/L A = 1/S fSX fAX C1 21.83 22 17.5 0.0571 0.000 0.943 C2 58.77 7.1 5.65 0.177 0.000 0.823 C3 144.30 2.9 2.31 0.433 0.004 0.569 nC4 170.92 1.23 0.979 1.021 0.152 0.134 nC5 57.97 0.52 0.414 2.415 0.587 0.003 nC5 33.56 0.23 0.183 5.46 0.817 0.000 Total: 487.35 To compute values of SB from Eq. (2), the assumed value of VB is critical. We seek the value of VB that will give values of SB , which when substituted into Eq. (1) will give values of bi that will sum to the specified value of B = 99.3 lbmol/h. Assume the temperature at the bottom is 250oF and obtain from Fig. 2.8 values of K at 250oF and 150 psia. To begin, assume a value of VB = 500 lbmol/h, calculate B = bi , and then adjust the values of bi so they sum to the specified B. The results are as follows: lbmol/h: Component KB SB lF/b b Adj. b C1 23 116 -- 0.00 0.00 C2 8.8 44.3 --- 0.00 0.00 C3 4.0 20.1 3100 0.05 0.07 nC4 1.85 9.31 14.8 11.55 25.92 nC5 0.88 4.43 1.73 33.51 45.87 nC5 0.45 2.27 1.22 27.44 27.44 Total: 72.55 99.30 The final product distribution is as follows, where it is compared to the result obtained by using Chemcad with the Tower model to also calculate the reboiled stripper: Edmister method: Rigorous Chemcad: Component Feed, lbmol/h Vapor product, lbmol/h Stripped liquid, lbmol/h Vapor product, lbmol/h Stripped liquid, lbmol/h C1 59.5 59.5 0.0 59.5 0.0 C2 73.6 73.6 0.0 73.6 0.0 C3 153.2 153.1 0.1 153.1 0.1 nC4 173.5 147.6 25.9 150.8 22.7 nC5 58.2 12.3 45.9 12.7 45.5 nC5 33.6 6.2 27.4 2.6 31.0 Total: 551.6 452.3 99.3 452.3 99.3 The Edmister method is uncertain because average temperature and vapor rate are difficult to estimate. Exercise 9.28 Subject: Comparison of FUG method with Edmister group method for distillation. Given: Bubble-point liquid feed in kmol/h of 100 ethylbenzene, 100 paraxylene, 200 metaxylene, and 100 orthoxylene. Column pressures of 25 psia at the top and 35 psia at the bottom. Distillate to contain 1 kmol/h of orthoxylene and bottoms to contain 2 kmol/h metaxylene. Column to be equipped with a total condenser and partial reboiler. Assumption: Applicability of Raoult's law for estimating K-values. Find: (a) Number of stages and reflux ratio by the FUG method for R/Rmin = 1.1 Feed-stage location by the Kirkbride equation. (b) Distribution of components between distillate and bottoms by Edmister method. Analysis: Need K-values at distillate, feed, and bottoms conditions. Because metaxylene and paraxylene have very close boiling points, assume the d/b ratio of paraxylene is the same as that for metaxylene. Therefore, have 1 kmol/h of paraxylene in the bottoms. Use Chemcad to obtain K-values by running bubble points on the feed, and estimated distillate and bottoms compositions. The results are: K-values: Component d, kmol/h b, kmol/h 25 psia, 159.3oC 30 psia, 168.4oC 35 psia, 180.5oC Ethylbenzene, EB 100 0 1.051 1.080 1.206 Paraxylene, PX 99 1 0.986 1.012 1.131 Metaxylene, MX 198 2 0.982 1.012 1.136 Orthoxylene, OX 1 99 0.855 0.884 0.996 Total: 398 101 distillat e feed bottoms (a) Using the Fenske equation (9-12), with MX as the LK and OX as the HK, with a geometric mean relative volatility = [(0.982/0.855)(1.136/0.996)]1/2 = 1.144 ( ) ( ) ( ) ( ) ( )OXMXOX MX MX,OX min avg 198 99loglog 1 2 log1.144log 68. 3 = = a = bd d bN To compute the distribution of nonkey components at total reflux, use for the lighter than light key components, LLK, Eq. (9-15), with OX as the reference component, r, and the above value of Nmin. Exercise 9.28 (continued) Analysis: (a) (continued) Thus, for the LLK, i, b fd b f i i N i= + = + 1 1 199 68 3OX OX i,OX i,OX mina a . (1) Using geometric mean values of ai, OX, with Eq. (1) and the material balance, fi = di + bi , the following results are obtained: kmol/h: Component Feed Distillate Bottoms EB 100 99.99 0.01 PX 100 99.00 1.00 MX 200 198.00 2.00 OX 100 1.00 99.00 Total: 500 397.99 102.01 Because it is highly likely that PX, and possibly EB, will distribute, use the Class 2 Underwood equation, Eq. (9-28),which for a bubble-point feed, using relative volatilities at the feed conditions is: aa q q q q qi i F ii z q, , , . ( . ) . . ( . ) . . ( . ) . . ( . ) . OX OX - = - = = - + - + - + - 1 0 12217 0 2012217 11448 0 2011448 11448 0 4011448 100 0 20100 (2) There are two roots to Eq. (2), which from a spreadsheet calculation are: q = 1.0280 and 1.2027 Eq. (9-29) is applied in the following form for each of the two values of q, in terms of the two unknowns, dEB and Lmin. The form used is obtained by multiplying Eq. (9-29) by the total distillate rate, D, which converts distillate mole fractions to component distillate rates, and the reflux ratio, R to L. The distillate rate of PX is not an unknown because it has the same relative volatility as MX. Thus, it must have the same d/b ratio as MX, making its distillate rate as 99 kmol/h. D L d d D L d d + = - + - + - = + + = - + - + - = - min EB EB min EB EB 1.1448 1.00 1.1448 1.00 12217 12217 10280 297 11448 10280 1 100 10280 6 307 2875 3 12217 12217 12027 297 11448 12027 1 100 12027 64 30 5877 2 . ( ) . . ( ) . . ( ) . . . ( ) . . ( ) . . ( ) . . ( ) . . . ( ) . Exercise 9.28 (continued) Analysis: (a) (continued) Because the total distillate rate, D, is also an unknown, the following equation is also needed: D = dEB + 99 + 198 + 1 = dEB + 298 Solving these 3 linear equations, the following results are obtained: D =448.9 kmol/h, Lmin = 3378 kmol/h, and dEB = 150.9 kmol/h. Thus, the assumption that EB distributes is incorrect because only 100 kmol/h of EB is in the feed. Therefore discard the root of q = 1.2027. Also, D now equals 100 + 99 + 198 + 1 = 398 kmol/h. From above for the q root of 1.0280, D + Lmin = 6.307 dEB + 2875.3. Therefore, Lmin = 6.307(100) + 2875.3 - 398 = 3108 kmol/h or Rmin = Lmin/D = 3108/398 = 7.81 The operating reflux ratio = 1.1(7.81) = 8.59. In the Gilliland equation, (9-34), X = (R - Rmin)/(R + 1) = (8.59 - 7.81)/(8.59 + 1) = 0.0813. Using Eq. (9.34), Y = 0.573 = (N - Nmin)/(N + 1). Solving, for Nmin = 68.3, N = 161.3 equilibrium stages. The Kirkbride Eq. (9-36) is used to determine the feed-stage location. NN zz xx BDR S F F X B D = = =OX MX M OX , , , , . . . . . . . . . 2 0 206 2 0 2060 20 0 40 0 0196 0 00251 102 01 397 99 153 Therefore, the feed stage is located at (1.53/2.53) 161.3 = stage 97.5 from the top. (b) The Edmister group method is applied by using Eq. (5-66) to estimate the b/d ratio for each component: b d A A SF C SE AE B AX SX = + + f f f f 1 1 (3) For a total condenser, AC = the reflux ratio = 8.59 for all components. To compute values of f for each component, assume the following values of flow rates and stages for the enriching (rectifying) section and exhausting (stripping) section, based on constant molar overflow and noting that the feed and reboiler stages are not included in these two sections. Variable Enriching Section Exhaustin g Section Number of stages 97 63 Vapor rate, kmol/h 3817 3817 Liquid rate, kmol/h 3419 3919 Exercise 9.28 (continued) Analysis: (b) (continued) Because in the above table the K-values do not vary much, use arithmetic mean values for each section. These values and those for the feed stage and partial reboiler are as follows along with the corresponding values of the absorption and stripping factors, as defined by equations (5-38), (5-51), below (5-64), and (5-65), where VB/B = 3817/102 = 37.4 Variable EB PX MX OX KE 1.066 0.999 0.997 0.870 AE 0.840 0.897 0.898 1.030 SE = 1/AE 1.190 1.115 1.114 0.971 KF 1.080 1.012 1.012 0.884 AF 0.951 1.015 1.015 1.161 KX 1.143 1.072 1.074 0.940 AX 0.898 0.958 0.956 1.092 SX = 1/AX 1.114 1.044 1.046 0.918 KB 1.206 1.131 1.136 0.996 SB 45.1 42.3 42.5 37.3 Values of fA and fS are obtained from Eqs. (5-48) and (5-50), respectively: Variable EB PX MX OX fAE 0.160 0.103 0.102 0.00175 fSE 0.0000 0.0000 0.0000 0.0307 fAX 0.102 0.0449 0.0466 0.0003 fSX 0.0001 0.0030 0.00274 0.0823 Now apply Eq. (3) above to obtain the component distribution: Variable EB PX MX OX f, kmol/h 100 100 200 100 b/d 0.0001 0.0102 0.0091 68.2 d, kmol/h 99.99 98.99 198.20 0.68 b, kmol/h 0.01 1.01 1.80 99.32 Comparing these Edmister method results with the FUG method, we see that the Edmister method predicts a somewhat better split of the two key components than does the FUG method for this problem. J. Seader Microsoft Word - Chapter 9 Exercises.doc