Solutions Manual

Exercise 15.1 Subject: Estimation of adsorbent characteristics. Given: Porous particles of activated alumina. BET area = 310 m2/g = Sg. Particle porosity = 0.48 = ep. Particle density = 1.30 g/cm3 = rp. Assumptions: Straight pores of circular cross-section with uniform diameter. Find: (a) Vp = specific pore volume in cm3/g. (b) rs = true solid density, g/cm3. (c) dp = average pore diameter, angstoms. Analysis: (a) From Eq. (15-3), Vp = ep/rp = 0.48/1.30 = 0.369 cm3/g (b) From a rearrangement of Eq. (15-5), rs = rp/(1 - ep) = 1.30/(1 - 0.48) = 2.50 g/cm3 (c) From a rearrangement of Eq. (15-2), dp = 4 ep/Sgrg = 4(0.48)/[310 x 104(1.30)] = 4.76 x 10-7 cm or 47.6 angstroms. Exercise 15.2 Subject: Estimation of surface area of two forms of molecular sieves. Given: Form A with Vp = 0.18 cm3/g and average dp = 5 angstroms. Form B with Vp = 0.38 cm3/g and average dp = 2.0 microns. Assumptions: Straight pores of circular cross-section with uniform diameter. Find: Surface area, Sg, of each form. Analysis: By substitution of Eq. (15-3) into (15-2), Sg = 4 Vg/dp (1) Form A: From Eq. (1), Sg = 4(0.18)/(5 x 10-8) = 14.4 x 106 cm2/g or 1,440 m2/g Form B: From Eq. (1), Sg = 4(0.38)/(2 x 10-4) = 7,600 cm2/g or 0.760 m2/g This is a very large difference in surface area. Exercise 15.3 Subject: Consistency of the characteristics of a small-pore silica gel. Given: Small-pore silica gel. Pore diameter = 24 angstroms = dp. Particle porosity = 0.47 = ep. Particle density = 1.09 g/cm3 = rp. Specific surface area = 800 m2/g = Sg. Assumption: Straight pores of circular cross-section with uniform diameter. Find: (a) Reasonableness of the above values. (b) Fraction of a monolayer adsorbed if the adsorption capacity for water vapor at 25oC and 6 mmHg partial pressure is 18% by weight. Analysis: (a) From Eq. (15-2), based on the above assumption, Sg = 4 ep/dprg = 4(0.47)/[(1.09 x 106)(24 x 10-10)] = 719 m2/g This is reasonable compared to the given value of 800 m2/g. The true density of silica gel, assuming it is SiO2, depends on its crystalline form. From a handbook, the density is from 2.20 to 2.65 g/cm3, with 3 of 4 forms from 2.20 to 2.26 g/cm3. Therefore, using a value of rs = 2.20 g/cm3 with Eq. (15-5),ep = 1 - rp/ rs = 1 - 1.09/2.20 = 0.505. This is reasonable compared to the given value of 0.47. (b) From the problem statement, it is not clear whether the 18 wt% refers to a dry basis or a wet basis, so consider both possibilities. Dry basis: Adsorb 0.18 grams water per 1.0 gram of water-free silica gel. Therefore, for 1.0 gram of dry silica gel, the surface area for adsorption = 800 m2. From Eq. (15-8) and Example 15.1, the projected surface area per molecule = a = 10.51 x 10-16 cm2/molecule. The number of water molecules adsorbed = 0.18(6.023 x 1023)/18.02 = 6.02 x 1021. The number of molecules that could be adsorbed to cover 800 m2 = 800(104)/10.51 x 10-16 = 7.61 x 1021. Therefore, the fraction of the area covered = 6.02/7.61 = 0.79 Wet basis: Adsorb 0.18 grams water per 0.82 gram of water-free silica gel (1.00 gram total). Therefore, for 0.82 gram of dry silica gel, the surface area for adsorption = (0.82)800 = 656 m2. From Eq. (15-8) and Example 15.1, the projected surface area per molecule = a = 10.51 x 10-16 cm2/molecule. The number of water molecules adsorbed = 0.18(6.023 x 1023)/18.02 = 6.02 x 1021. The number of molecules that could be adsorbed to cover 656 m2 = 656(104)/10.51 x 10-16 = 6.89 x 1021. Therefore, the fraction of the area covered = 6.02/6.89 = 0.87 Exercise 15.4 Subject: Estimation of the specific surface area from BET data. Given: Data for adsorption equilibrium of nitrogen on silica gel (SG) at -195.8oC: P, N2 partial pressure, torr u, Volume of N2 adsorbed, cm3 (0oC, 1 atm) per gram SG 6.0 6.1 24.8 12.7 14.3 17.0 230.3 19.7 285.1 21.5 320.3 23.0 430 27.7 505 33.5 Find: Specific surface area in m2/g of SG. Compare to Table 15.2. Analysis: From Eq. (15-6), PP P c c c PP m mu u u0 0 1 1 - = + - ( ) (1) where: P = total pressure = partial pressure of N2 in the above table in mmHg P0 = vapor pressure of N2 at -195.8oC = 760 mmHg u = volume of gas adsorbed at STP u m = volume of monomolecular layer of gas adsorbed at STP c = constant Eq. (1) is of the form of a straight line, y = mx + b, where, y = PP Pu 0 - , x = PP 0 , m = ( )c cm -1 u (2) and b = 1 umc (3) If a least-squares fit of the data is made, using a spreadsheet or POLYMATH, m = 0.08557 and b = -0.0016905 Combining Eqs. (2) and (3) to eliminate c, u m = 1/(m + b) = 1/(0.08557 - 0.0016905) = 11.92 cm3/g c = -49.62 From Eq. (15-7), S NVg m A= = · = ·a a au ( . )( . ), .1192 6 023 1022 400 3 205 10 23 20 From Eq. (15-8), a r= = · = · -1091 1091 286 023 10 0808 1626 10 2 3 23 2 3 15. . . ( . ) . / /M N A L cm 2 Therefore, Sg = 3.205 x 1020 (1.626 x 10-15) = 5.21 x 105 cm2/g or 52.1 m2/g This is much smaller than the values given for silica gel in Table 15.2. Exercise 15.5 Subject: Maximum ion-exchange capacity of a resin. Given: Ion-exchange resin made from 8 wt% divinylbenzene and 92 wt% styrene. Find: Maximum ion-exchange capacity in meq/g resin. Analysis: Compute the moles of each aromatic component per 100 grams of resin: MW grams gmol Styrene 104.14 92 0.8834 Divinylbenzene 130.18 8 0.0615 Total 100 0.9449 Therefore, need 0.9449 mol H2SO4 or 0.9449(81.07) = 76.6 g. Total resin weight after sulfonation = 100 + 76.6 = 176.6 g. Maximum ion-exchange capacity = 0.9449(1000)/176.6 = 5.35 meq/g of resin. Exercise 15.6 Subject: Fitting adsorption data to linear, Freundlich, and Langmuir isotherms and computing the heat of adsorption. Given: Silica gel adsorbent with: surface area = Sg = 832 m2/g, pore volume = Vp = 0.43 cm3/g, particle density = rp = 1.13 g/cm3, and average pore diameter = dp = 22 angstroms. Equilibrium adsorption data for pure benzene vapor at 4 different temperatures as follows: q, moles adsorbed /g gel x 105: p, partial pressure, atm 70oC 90oC 110oC 130oC 0.0005 14.0 6.7 2.6 1.13 0.0010 22.0 11.2 4.5 2.0 0.0020 34.0 18.0 7.8 3.9 0.0050 68.0 33.0 17.0 8.6 0.0100 88.0 51.0 27.0 16.0 0.0200 78.0 42.0 26.0 Find: (a) For each temperature, best fits of the data to (1) linear, (2) Freundlich, and (3) Langmuir isotherms. Which give a reasonable fit? (b) Do data represent less than a monolayer? (c) Heat of adsorption, with comparison to heat of vaporization of benzene. Analysis: The regression program of POLYMATH can be used to do nonlinear curve fits or a spreadsheet program can be used to do least squares curve fits on the linearized isotherms. For the former, the following results are obtained for fitting Eqs. (15-16), (15-19), and (15-24). Somewhat different results would be obtained with the linearized equations (15-20) and (15-25). (1) q = kp Linear (2) q = kp1/n Freundlich (3) q = Kqmp/(1 + Kp) Langmuir Temperature, oC Linear: k Freundlich: k n Langmuir: K qm 70 0.1011 0.0132915 1.71882 188.18 0.0013584 90 0.0431033 0.00944163 1.57209 67.55 0.0013337 110 0.0229401 0.00653397 1.43190 50.0327 0.00083383 130 0.0138306 0.00595422 1.25461 27.4455 0.00073471 The linear equation is a poor fit for all 4 temperatures. The Freundlich equation a reasonably good fit for all 4 temperatures. The Langmuir equation gives the best fit for all 4 temperatures. Exercise 15.6 (continued) Analysis: (continued) (b) The surface area covered by one adsorbed molecule is given by Eq. (15-8), where for benzene, M = 78.11 and the liquid density, from Fig. 2.3, ranges from 0.82 cm3/g at 90oC to 0.76 cm3/g at 130oC. At 90oC, a r= = · = · -1091 1091 78116 023 10 082 319 10 2 3 23 2 3 15. . . . ( . ) . / /M N A L cm 2 From the data, the maximum experimental adsorption is 0.00088 moles benzene/g silica gel. This is a coverage of 0.00088(6.023 x 1023)(3.19 x 10-15) = 1.69 x 106 cm2/g or 169 m2/g. This is far less than the given Sg = 832 m2/g. At higher temperatures, the experimental values for adsorption are even less, with just a slight increase in a. Therefore, the measured values of adsorption are equivalent to far less than a monolayer. (c) The heat of adsorption is related to adsorption isotherm data by a rearrangement of Eq. (15- 17), which is a form of the Clausius-Clapeyron equation. For a constant amount adsorbed, - =DH RT d pdTads 2 ln (1) where, R = 1.987 cal/mol-K Take an arbitrary value of 0.00025 mols of benzene adsorbed per gram of silica gel. For each temperature, calculate the partial pressure of benzene from the above fits of the Freundlich equation (the Langmuir equation could also be used). Evaluate d ln p/dT numerically and apply Eq. (1). The results are as follows. Temperature, oC Temperature, T, K Benzene pressure, p, atm ln p d ln p/dT Tavg, K - DHads, cal/mol 70 343 0.00108 -6.831 0.05615 353 13,900 90 363 0.00332 -5.708 0.05180 373 14,300 110 383 0.00935 -4.672 0.0347 393 10,650 130 403 0.01873 -3.978 The average heat of adsorption = - 13,000 cal/mol. From the ChemCad simulation program, the heat of condensation for benzene varies from -7,500 cal/mol at 70oC to -6,620 cal/mol at 130oC. Thus, the heat of adsorption is almost twice as great as the heat of condensation. Exercise 15.7 Subject: Use of adsorption equilibria data to select the best adsorbent. Given: Langmuir adsorption equilibrium constants, K and qm in Eq. (15-24), for 3 zeolite molecular sieves and activated carbon for propylene (C3) and propane (C3=) at 25oC: Adsorbent Sorbate qm K ZMS 4A C3 0.226 9.770 C3= 2.092 95.096 ZMS 5A C3 1.919 100.223 C3= 2.436 147.260 ZMS 13X C3 2.130 55.412 C3= 2.680 100.000 Activated carbon C3 4.239 58.458 C3= 4.889 34.915 Find: (a) Most strongly adsorbed component for each adsorbent. (b) Adsorbent with greatest adsorption capacity. (c) Adsorbent with greatest selectivity. (d) Best adsorbent for the separation of propylene from propane. Analysis: (a) The Langmuir equation is: q = Kqmp/(1 + Kp) At high pressure, q approaches qm. Thus, qm is a measure of capacity. Because of the high values of K, the value of qm is reached at about 1 bar. Thus, we can make the comparisons on the basis of qm. Therefore, propylene is the most strongly adsorbed for all 4 adsorbents. (b) The activated carbon has the highest values of qm and, therefore, has the highest capacity. (c) The selectivity is measured by the ratio of qm for propylene to that of propane. With a value of 2.092/0.226 = 9.26, zeolite ZMS 4A is by far the best. (d) Because of its very high selectivity of ZMS 4A, where the second best is only 1.27, and the high capacity of ZMS 4A, although only about 40% of that of activated carbon, ZMS 4A is the best adsorbent for the separation of propylene and propane. Exercise 15.8 Subject: Fitting zeolite adsorption data to linear, Freundlich, and Langmuir isotherms. Given: NaX zeolite adsorbent. Adsorption equilibrium data at 547 K for 1,2,3,5- tetramethylbenzene (TMB) as follows: q, wt% 7.0 9.1 10.3 10.8 11.1 11.5 p, torr 0.012 0.027 0.043 0.070 0.094 0.147 Find: Best fit of the data to linear, Freundlich, and Langmuir isotherms, with q in mol/g and p in atm. Which isotherm fits best? Analysis: First convert the given adsorption data to the desired units: q in mols TMB/gram NaX = (wt% q/MW of TMB)/(100 - wt% q) (1) MW of TMB = 134.1 Substitution of the data into Eq. (1) plus converting pressure from torr to atm gives: q, mol/g 0.000562 0.000747 0.000856 0.000903 0.000931 0.000969 p, atm 0.0000158 0.0000355 0.0000566 0.0000921 0.0001237 0.0001934 The regression program of POLYMATH can be used to do nonlinear curve fits or a spreadsheet program can be used to do least squares curve fits on the linearized isotherms. For the former, the following results are obtained for fitting Eqs. (15-16), (15-19), and (15-24). Somewhat different results would be obtained with the linearized equations (15-20) and (15-25). Linear: q = kp = 7.12557 p Freundlich: q = kp1/n = 0.00532 p1/5.15605 Langmuir: q = Kqmp/(1 + Kp) = (75410)(0.001036) p/(1 + 75410 p) The fits are shown on the next page, where it is clear that: Linear model is a very poor fit. Freundlich model is a fairly good fit. Langmuir model is a very good fit and is the best fit. Exercise 15.8 (continued) Analysis: (continued) 0 0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0014 0.0016 0 0.00005 0.0001 0.00015 0.0002 TMB Pressure, atm T M B L o a d i n g , m o l / g Exercise 15.9 Subject: Fitting pure component and mixture adsorption data for gaseous propane (C3) and propylene (C3=) on silica gel (SG) at 25oC to Freundlich and Langmuir isotherms. Given: Pure component adsorption equilibrium data for C3 and C3= on SG in the exercise statement. Adsorption equilibrium data for mixtures of C3 and C3= on SG in the exercise statement. Surface area of SG = 832 m2/g, pore volume = 0.43 cm3/g, particle density = 1.13 g/cm3, and average pore diameter = 22 angstroms. Find: (a) Fit of the data to Freundlich and Langmuir isotherms. Which isotherm is best? Which component is most strongly adsorbed? (b) Prediction of mixture adsorption from the extended Langmuir isotherm based on the pure component fits. (c) Fit of the mixture data to the extended Langmuir isotherm. Comparison to Part (b). (d) Fit of the mixture data to the extended Langmuir-Freundlich isotherm. Comparison to Part (c). (e) Relative selectivity from the mixture data. Does it vary widely? Analysis: (a) The regression program of POLYMATH can be used to do nonlinear curve fits or a spreadsheet program, such as Excel, can be used to do least squares curve fits on the linearized isotherms. For the latter, the following results are obtained for fitting Eqs. (15-20) and (15-25). Somewhat different results would be obtained with the nonlinear Eqs. (15-19) and (15-24). Component Freundlich: k n Langmuir: K qm Propane 0.01062 1.311 0.001853 2.5213 Propylene 0.06242 1.8306 0.003916 2.8201 For C3, the Freundlich equation, with R2 = 0.995, is a slightly better fit than the Langmuir equation with R2 = 0.972. For C3=, the Langmuir equation, with R2 = 0.986, is about the same goodness of fit as the Freundlich equation with R2 = 0.984. Considering both components, the Freundlich isotherm appears to be the best. However, as shown in the comparisons of the fits to the data (shown as markers) on the next page, the data, at the higher pressures, seems to be more in agreement with the asymptotic nature of the Langmuir isotherm. Probably, the Langmuir isotherm is best. For a given pressure, the amount adsorbed is greater for C3=, so it is more strongly adsorbed. Exercise 15.9 (continued) Analysis: (a) (continued) Propane adsorption 0.0 0.5 1.0 1.5 2.0 2.5 0 200 400 600 800 1000 p, Pressure, torr q , A d s o r b a t e , m m o l / g Propylene adsorption 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0 100 200 300 400 500 600 700 800 p, Pressure, torr q , A d s o r b a t e , m m o l / g Exercise 15.9 (continued) Analysis: (continued) (b) Using the above pure component data fits, Eq. (15-32), the extended Langmuir isotherm becomes: For C3: ( )C3 C3 C3 C33 C3 C3 C3= C3= C3 C3= 2.5213(0.001853) 1+ 1+0.001853 0.003916 m C q K p pq K p K p p p= =+ + (1) For C3=: ( )C3= C3= C3= C3=3 C3 C3 C3= C3= C3 C3= 2.8201(0.003916) 1+ 1+0.001853 0.003916 m C q K p pq K p K p p p= = =+ + (2) Using a spreadsheet, Eqs. (1) and (2) are used to predict the millimoles of components adsorbed from the given mixture data at 25oC, where the partial pressures are computed from the given total pressure and vapor mole fractions by Dalton's law, pi = Pyi , and component adsorbate loadings are compared to the data using, qi = qtotal xi. The results are as follows, where for measured partial pressures, the measured loadings are compared to those predicted by the extended Langmuir isotherm using constants fitted from pure component adsorption data. Plots are shown on the following page, where the points are the experimental data. Experimental data: Predicted: Total pressure, torr p of C3, torr p of C3=, torr q of C3, mmole/g q of C3=, mmole/g q of C3, mmole/g q of C3=, mmole/g 769.20 188.07 581.13 0.237 1.960 0.242 1.771 760.90 227.51 533.39 0.519 1.494 0.303 1.678 767.80 310.19 457.61 0.607 1.445 0.430 1.501 761.00 403.33 357.67 0.575 1.466 0.599 1.255 753.60 401.89 351.71 0.717 1.246 0.601 1.244 766.30 410.43 355.87 0.614 1.353 0.608 1.246 754.00 462.96 291.04 0.709 1.265 0.722 1.072 753.60 468.74 284.86 1.027 0.824 0.734 1.054 754.00 471.40 282.60 1.192 0.509 0.739 1.047 760.00 568.48 191.52 1.219 0.467 0.947 0.754 760.00 681.26 78.74 0.504 1.489 1.238 0.338 760.00 699.96 60.04 0.572 0.854 1.291 0.262 From the above table and the plots on the following page, it is seen that the agreement is mostly poor. Exercise 15.9 (continued) Analysis: (b) (continued) Propane adsorption from mixture 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0 100 200 300 400 500 600 700 800 p, Partial pressure, torr q , L o a d i n g , m m o l / g Propylene adsorption from mixture 0.0 0.5 1.0 1.5 2.0 2.5 0 200 400 600 800 p, Partial pressure, torr q , L o a d i n g , m m o l / g Exercise 15.9 (continued) Analysis: (continued) (c) To fit the mixture data to the extended Langmuir equation, to determine the best values of the four coefficients (qC3)m, (qC3=)m, KC3, and KC3=, Eqs. (1) and (2) above must be used simultaneously. Polymath does not permit more than one equation to be fitted at a time. Alternatively, the two equations can be added together to give one equation in the total loading, qt = qC3 + qC3=. However, this does not insure a good fit to the data for the individual loadings. Nevertheless, this technique was applied with Polymath. It was found that the minimum sum of squares of the deviations of the fit from the experimental data was not very sensitive to the values of the four coefficients. One reasonable set that was close to the set determined from the pure component data was: (qC3)m = 2.5213, (qC3=)m = 2.8201, KC3 = 0.002116, and KC3= = 0.004752 Unfortunately, this set showed little improvement in the fit of the mixture data as compared to the fit of the set determined in Part (b) from the pure component data. The plots are not much different from those on the previous page. Another procedure that can be applied is to use the program MathCAD with the Minerr function. The MathCAD program and result is shown on the next page. Minerr searches for the values of the four constants that provide the best fit to the data. The objective function is the sum of the squares of the deviations of the data from the predictions (SSE) by the Eqs. (1) and (2). SSE C3 C3 C3 C3 C3 C3 C3= C3= C3= C3= C3= C3= C3 C3 C3= C3= = - + + + - + + q q K p K p K p q q K p K p K pi m i i i i m i i ii 1 12 2 (3) where the four constants are varied to find the smallest value of SSE. Initial quesses for the four constants must be provided, which, as shown on the next page for the MathCAD program were: (qC3)m = 2.5, (qC3=)m = 2.8, KC3 = 0.002, and KC3= = 0.002 The values computed are seen to be: (qC3)m = 2.153, (qC3=)m = 4.042, KC3 = 0.001642, and KC3= = 0.001997, with SSE = 3.501, which does not indicate a very good fit. Unfortunately, different initial guesses give other results, none of which gives much, if any improvement to the fit. (d) The procedure outlined in Part (c), although not very satisfactory, was applied to the extended Langmuir-Freundlich isotherm, Eq. (15-33), which for the total loading is: q q k Py k Py k Py q k Py k Py k Pyt n n n n n n= + + + C3 C3 C3 C3 C3 C3= C3= C3= C3= C3= C3 C3 C3= C3= C3 C3 C3= C3= C3 C3=1+ 1+ 0 1 1 1 0 1 1 1 / / / / / / (4) When used with the nonlinear regression program of Polymath, the results obtained were not satisfactory because some coefficients were negative for the best fit. When MathCAD was used with the Minerr function according to the program and result is shown on a following page, the following values were obtained for the six constants: (qC3)m = 1.706, (qC3=)m = 4.507, KC3 = 0.0270, KC3= = 0.03773, nC3 = 1.59, and nC3= = 1.90, with SSE = 2.583, which is a better fit than for the extended Langmuir equation. However, other initial guesses for the six constants gave other results, because of the the great sensitivity. Exercise 15.9 (continued) Analysis: (c) (continued) MathCAD Program p1 188.07 227.51 310.19 403.33 401.89 410.43 462.96 468.74 471.40 568.48 681.26 699.96 := p2 581.13 533.39 457.61 357.67 351.71 355.87 291.04 284.86 282.60 191.52 78.74 60.04 := q1 0.237 0.519 0.607 0.575 0.717 0.614 0.709 1.027 1.192 1.219 0.504 0.572 := q2 1.960 1.494 1.445 1.466 1.246 1.353 1.265 0.824 0.509 0.467 1.489 0.854 := SSE Q1 Q2, K1, K2,( ) 1 12 i q1i Q1 K1 p1i 1 K1 p1i + K2 p2i +( ) - 2 q2i Q2 K2 p2i 1 K1 p1i + K2 p2i + - 2 + = := Q1 := 2.5 Q2 := 2.8 K1 := 0.002 K2 := 0.004 Given SSE Q1 Q2, K1, K2,( ) 0 1 1 1 1 1 1 Q1 Q2 K1 K2 Minerr Q1 Q2, K1, K2,( ):= Q1 = 2.153 Q2 = 4.042 K1 = 0.001642 K2 = 0.001997 ( )SSE Q1, Q2, K1, K2 3.501= Exercise 15.9 (continued) Analysis: (d) (continued) MathCAD Program p1 188.07 227.51 310.19 403.33 401.89 410.43 462.96 468.74 471.40 568.48 681.26 699.96 := p2 581.13 533.39 457.61 357.67 351.71 355.87 291.04 284.86 282.60 191.52 78.74 60.04 := q1 0.237 0.519 0.607 0.575 0.717 0.614 0.709 1.027 1.192 1.219 0.504 0.572 := q2 1.960 1.494 1.445 1.466 1.246 1.353 1.265 0.824 0.509 0.467 1.489 0.854 := SSEQ1 Q2, K1, K2, N1, N2,( ) 1 12 i q1i Q1K1 p1i( )N1 1 K1 p1i( )N1 + K2 p2i( )N2 + - 2 q2i Q2K2 p2i( )N2 1 K1 p1i( )N1 + K2 p2i( )N2 + - 2 + = := Q1 1:= Q2 4:= K1 0.001:= K2 0.001:= N1 0.5:= N2 0.5:= Given SSE Q1 Q2, K1, K2, N1, N2,( ) 0 1 1 1 1 1 1 1 1 1 1 Q1 Q2 K1 K2 N1 N2 Minerr Q1 Q2, K1, K2, N1, N2,( ):= Q1 = 1.706 Q2 = 4.507 K1 = 0.027 K2 = 0.037734 N1 = 0.628 N2 = 0.527 ( )SSE Q1, Q2, K1, K2, N1, N2 = 2.583 Exercise 15.9 (continued) Analysis: (continued) (e) The relative selectivity is given by aC3,C3= = yC3(1 - xC3)/[xC3(1 - yC3)] (4) Using the mixture data with Eq. (4), the following results are obtained: Pressure, torr y, C3 x, C3 a C3 to C3= 769.2 0.2445 0.1078 2.678 760.9 0.2990 0.2576 1.229 767.8 0.4040 0.2956 1.615 761.0 0.5300 0.2816 2.877 753.6 0.5333 0.3655 1.984 766.3 0.5356 0.3120 2.543 754.0 0.6140 0.3591 2.839 753.6 0.6220 0.5550 1.319 754.0 0.6252 0.7007 0.713 760.0 0.7480 0.7230 1.137 760.0 0.8964 0.2530 25.547 760.0 0.9210 0.4010 17.415 For the mixture, propane adsorption is favored over propylene adsorption and a varies widely. Exercise 15.10 Subject: Use of the extended Langmuir-Freundlich isotherm to fit mixture data for the system acetone (1), propionitrile (2) on activated carbon at 25oC. Given: Mixture adsorption data in terms of solution concentrations in mol/L and loadings on the adsorbent in mmol/g. Find: Coefficients in the extended Langmuir-Freundlich isotherm Analysis: From Eq. (6) in Example 15.6, the extended Langmuir-Freundlich equations are: q q k ck c k c q q k ck c k c n n n n n n 1 0 1 1 1 1 1 1 1 2 2 1 2 0 2 2 2 1 1 1 1 2 2 1 1 1 2 1 1 2 1 1 = + + = + + / / / / / / (1) (2) Fit mixture adsorption equilibrium data for c1, c2, q1, and q2 to obtain the coefficients (q0)1, (q0)2, k1, k2, n1, and n2. Rather than fitting both Eqs. (1) and (2) simultaneously, the following procedure was used. First, fit Eq. (1) with the nonlinear regression program of Polymath. Then fit Eq. (2). The following results are obtained, with both equations giving good fits: Coefficient Eq. (1) Eq. (2) Refit of Eq. (2) Refit of Eq. (1) (q0)1 3.456 1.855 (q0)2 37.4 2.492 k1 14.92 13.65 14.92 14.92 k2 14.95 2.14 20.48 20.48 n1 1.295 3.906 1.295 1.295 n2 2.469 1.290 1.136 1.136 Next, Eq. (2) is refitted, holding k1 and n1 at the values above that were obtained from fitting Eq. (1). The refit gives the above values. Eq. (1) is refitted to determine only (q0)1. The results are also given above. The final fits of Eqs. (1) and (2) are as follows with sum of squares of deviations of 0.0144 and 0.0163, respectively. A simultaneous fit would give better values. 1/1.295 1 1 1/1.295 1/1.136 1 2 1/1.136 2 2 1/1.295 1/1.136 1 2 1.855(14.92) 1 14.92 20.48 2.492(20.48) 1 14.92 20. 4 8 cq c c cq c c = + + = + + The fits of the data are shown in the bar charts on the following page. Exercise 15.10 (continued) Analysis: (continued) Exercise 15.11 Subject: Analysis of liquid adsorption data for a mixture of cyclohexane (1) and ethyl alcohol (2) on activated carbon at 30oC. Given: Data table for loading of (1) as a function of its mole fraction. Assumption: No adsorption of (2). Find: (a) Plot of loading as a function of mole fraction. Explain shape of curve. (b) Fit of Freundlich equation over the low mole fraction region. Analysis: (a) Using Polymath, the data for cyclohexane liquid adsorption are shown below for the entire composition region, with fitting to a cubic equation. The curve is of the type of Fig. 15.12c. The explanation for the shape of the curve is obtained from Fig. 15.13d. At low concentrations of (1), it begins to adsorb, with increasing adsorption for increasing mole fraction of (1), much like a Freundlich isotherm. The solvent, (2), contrary to the assumption of no adsorption is highly adsorbed at low concentrations of (1), but its adsorption gradually diminishes as the concentration of (2) decreases. (b) For the cyclohexane mole fraction region from 0 to 0.250, the data are fitted with the Regress program of Polymath to a modification of the Freundlich Eq. (15-35), using mole fraction instead of concentration, with a plot on the following page. 1 3.8130.94511 q x= Exercise 15.11 (continued) Analysis: (b) (continued) Exercise 15.12 Subject: Fitting adsorption equilibria data for small concentrations of toluene in water with activated carbon, and small concentrations of water in toluene with activated alumina, to Langmuir and Feundlich isotherms. Given: Table of adsorption equilibrium data for toluene in water with activated carbon at 25oC. Table of adsorption equilibrium data for water in toluene with activated alumina at 25oC. Assumptions: Negligible adsorption of water on activated carbon. Negligible adsorption of toluene on activated alumna. Find: Best fitting isotherms from Langmuir, Freundlich, and linear. Analysis: Use the nonlinear regression program of Polymath. However, for toluene in water, the data point at c = 2 for q = 70.2 appears to be suspect, so discard it. The following results are obtained using Eq. (15-35) for the Freundlich isotherm: q = k c1/n and (15-36) for the Langmuir isotherm: q = qmKc/(1 + Kc) Component Freundlich: k n Langmuir: K qm Toluene in water 74.0218 2.86424 1.30434 161.863 Water in toluene 0.199624 1.42732 0.00214855 28.0673 The plots of the fits are shown below and on the next page. For toluene in water, the Freundlich isotherm gives the best fit. For water in toluene, the Freundlich isotherm gives a very good fit. The linear isotherm gives a poor fit for both cases. Exercise 15.12 (continued) Analysis: (continued) Exercise 15.13 Subject: Ion exchange of sulfate ions with chloride ions, and regeneration of the resin. Given: 60 L of water with sulfate ion and chloride ion concentrations of 0.018 eq/L and 0.002 eq/L, respectively. Remove sulfate ion by chloride ion exchange on 1 L of a strong-base resin. Regeneration of the resin with 30 L of 10 wt% aq. NaCl. Resin ion-exchange capacity of 1.2 eq/L. From Table 15.6, molar selectivities are: 1.0 for Cl- and 0.15 for SO4=. Find: Derivation of Eq. (15-44). (a) Ion-exchange reaction. (b) Value of chemical equilibrium constant for KSO4=,Cl-. (c) Equilibrium concentrations of SO4= and Cl- ions in solution and in resin in eq/L for the ion-exchange step. (d) Concentration of Cl- in eq/L for the regenerating solution. (e) Equilibrium concentrations of SO4= and Cl- ions in solution and in resin in eq/L for the regeneration step. (f) Whether the steps are sufficiently selective. Analysis: =4A=SO and BLet =Cl- Eq. (15-44) is K CQ y x x y n n nA,B A A A A = - - -1 1 1 (1) This a case of unequal charges, so use Eq. (15-37) for the reaction: (a) ( ) 2 ( ) 4 (s) 4 2(s)SO 2Cl R SO R 2Cll l - -+ « + Thus, n = 2 and by the law of mass action: K q cq cA,B SO R Cl ClR 2 SO 4 4 2 = - - 2 (2) The counterion valences are: zA = 2 and zB = 1 From Eq. (15-39), cA = CxA/2 and cB = CxB/1 (3) From Eq. (15-40), qAR = QyA/2 and qBR = QyB/1 (4) Substitution of Eqs. (2) and (3) into (1), gives: K Q y C x Q y C x C Q y x x y K CQ y x x y n n nA,B A B 2 B A A A A A A,B A A A A = = -- = = -- -2 2 1 1 1 1 2 2 2 2 2 1 since, n =2. This is Eq. (15-44). Exercise 15.13 (continued) Analysis: (continued) (b) From Eq. (15-45), using values of Ki from Table 15.6, KA,B = KA/KB = 0.15/1.0 = 0.15 (5) (c) For the initial ion-exchange step, Q = 1.2 eq/L and C = cA + cB = 0.018 + 0.002 = 0.02 eq/L Let: a = eq of SO4= exchanged. The equivalent fractions of the ions for 1.0 L of resin, 60 L of aq. solution, and n = 2 are: x a a y a a SO SO 4 = 4 = (6) (7) = - = - = = 0 018 60 0 02 0 9 0833 1 12 12 . . . . / . . Combining Eqs (1), (5), (6), and (7), K CQ y x x y a a a a n n nA,B SO Cl SO Cl 4 = - 4 = - = = = - + - - - -1 2 1 2 2015 0 02 12 12 1 0 9 0833 0 9 0 833 1 12 . . . . . . . . . (8) Solving Eq. (8) with a nonlinear solver, a = 0.755 eq. From Eqs. (6) and (7), x y SO SO 4 = 4 = = 0.629 = - = = 0 9 0833 0 755 0 271 0 755 12 . . ( . ) . . . From Eq. (15-39), = 4 = 4 - - = 4 - SO Cl SO SO Cl Cl 0.00271 mol/L or 0.00542 eq/0.02(0.271)2 0.02(1 0.271 L 0.01458 mol/L o) r 0.01458 e L1 q/ c C Cc x z x z = = = = -= = From Eq. (15-40), = 4 - = 4 = 4 - - SO Cl SO SO Cl Cl 0.377 mol/L or 0.754 eq/L1.2(0.629)2 1.2(1 0.62 0.445 mol/L or 0.449) 1 5 eq/L q q Qy z Qy z = = -= = = = Exercise 15.13 (continued) Analysis: (continued) (d) The regenerating solution is 30 L of 10 wt% aq. NaCl. From Perry's Handbook, assuming a temperature of 25oC, the density is 1.0689 g/cm3 or 1068.9 g/L. The MW of NaCl = 58.45. Therefore, the concentration of chloride ion = cCl- = 0.10(1068.9)/58.45 = 1.829 eq/L. (e) From Part (c), 1 L of resin includes: 0.445 eq of Cl- and 0.754 eq From Part (d), 30 L of regenerating solution includes: 30(1.829) = 54.870 eq. Cl- Thus, we have a total of 30 L of solution and 1 L of resin containing a total of: 0.754 eq. of SO4= and 54.870 + 0.445 = 55.315 eq. of Cl- Let: a = eq of SO4= on the resin at equilibrium. The equivalent fractions of the ions for 1.0 L of resin, 30 L of aq. solution, and n = 2 are: x a a y a a a SO SO 4 = 4 = (9) = 0.833 (10) = - = - = = 0574 54 870 0 01046 0 01822 1 12 12 . . . . / . . Combining Eqs (1), (5), (9), and (10), with C = 1.829 eq/L and Q = 1.2 eq/L, with n = 2, K CQ y x x y a a a a n n nA,B SO Cl SO Cl 4 = - 4 = - = = = - + - - - -1 2 1 2 2015 1829 12 0833 1 0 01046 0 01822 0 01046 0 01822 1 0 833 . . . . . . . . . (11) Solving Eq. (11) with a nonlinear solver, a = 0.00123 eq. From Eqs. (9) and (10), x y SO SO 4 = 4 = = 0.001025 = - = = 0 01046 0 01822 0 00123 0 01044 0 833 0 00123 . . ( . ) . . ( . ) From Eq. (15-39), = 4 = 4 4 - - = - SO SO C C l O l C S l 0.00955 mol/L or 0.0191 eq1.829(0.01044)2 1.829(1 0 /L 1.810 mol/L or 1.810.0 1044)1 eq/L Cx z Cx z c c = = = = -= = Exercise 15.13 (continued) Analysis: (e) (continued) From Eq. (15-40), = 4 - = 4 = 4 - - SO Cl SO SO Cl Cl 0.000615 mol/L or 0.00123 eq1.2(0.001025)2 1.2(1 0.0 /L 1.199 mol/L or 1.199 eq01025 /)1 L q q Qy z Qy z = = -= = = = (f) For the ion-exchange step, (0.755/1.080) x 100% = 70% of the sulfate ion is transferred from the solution to the resin. This is reasonably high considering the low and unfavorable chemical equilibrium constant of 0.15. The regeneration step is very effective, with (0.754 - 0.00123)/0.754 x 100% = 99.84% of the sulfate ion removed from the resin. Exercise 15.14 Subject: Equivalent fractions at equilibrium for the ion exchange of silver ion with sodium ion. Given: Ion-exchange resin of Dowex 50 cross-linked with 8% divinylbenzene. Exchange of silver ion in methanol with sodium ion in the resin. Wet capacity of the resin = 2.5 eq/L. Resin is initially saturated with sodium ion. Experimental data for the molar selectivity coefficient as a function of equivalent fraction of sodium ion in the resin: xNa+ 0.1 0.3 0.5 0.7 0.9 Kag+, Na+ 11.2 11.9 12.3 14.1 17.0 Find: Equivalent fractions if 50 L of 0.05 M Ag+ in methanol is treated with 1 L of wet resin. Analysis: 1 L of resin contains 2.5 eq of Na+. 50 L of solution contains 0.05(5) = 2.5 eq Ag+ Because both ions have a charge of one, Eq. (15-38) applies at equilibrium, with n = 1. Let x = eq of Ag in resin at equilibrium. Since we begin with 2.5 eq of each ion, then, also, x = eq of Na+ in the solution at equilibrium. Therefore, K q cq c x x x x x x x= = - - = - + Ag Na Na Ag + + 1 50 2 5 1 2 5 50 6 25 5 2 2. . . (1) To solve Eq. (1), the need the molar selectivity coefficient, which depends on xNa+ in the table above. Assume a value of 12. Eq. (1), which is a quadratic equation in x becomes: 11 x2 - 60 x + 75 = 0. Solving, x = 1.94 eq of Na+ in solution. Therefore, the equivalent fraction of Na+ in solution = 1.94/2.5 = 0.776. So we need to assume a higher value of K. Using the above table of data, assume a value of 15.2 for the molar selectivity coefficient. Then Eq. (1) becomes: 14.2 x2 - 76 x + 95 = 0. Solving, x = 2.0 and the equivalent fraction of Na+ in solution = 2/2.5 = 0.80. From the above table, K = 15.6. Then Eq. (1) becomes: 14.6 x2 - 78 x + 97.5 = 0. Solving x = 1.995. The value of K = 15.6 is close enough. Therefore, the equivalent fractions at equilibrium are: In the resin, yNa = 0.202 and yAg = 0.798 In the methanol solution, xNa+ = 0.798 and xAg+ = 0.202 Exercise 15.15 Subject: Recovery of glycerol from an aqueous solution of NaCl by ion exclusion. Given: Feed solution of 1,000 kg containing 6 wt% NaCl, 35 wt% glycerol, and 47 wt% water. Dowex 50 ion-exchange resin in the sodium form, prewetted to contain 40 wt% water. The equilibrium distribution coefficient for glycerol is given by: Kd = mass fraction glycerol in solution inside the resinmass fraction glycerol in solution outside the resin Values of Kd depend on the mass fraction of glycerol in the solution outside the resin and on the wt% NaCl in the solution outside the resin, as given in a table in the problem statement. Assumptions: The remainder of the feed solution (12 wt%) is inert material. During ion exclusion, no additional water or NaCl will transfer to the solution inside the resin. Find: The kg of resin on a dry basis needed to transfer 75% of the glycerol to the solution inside the resin. Analysis: Let R = kg of dry resin needed. Then the water in the prewetted resin = (40/60) R. The original solution contains 0.35(1000) = 350 kg of glycerol. For a transfer of 75%, the equilibrium solution inside the resin will contain 0.75(350) = 262.5 kg of glycerol, leaving 87.5 kg of glycerol in the equilibrium solution outside the resin. Therefore, we can write Kd as: K R R R K K d d d = + = + = - = - 262 5 40 60 262 5 87 5 100 3000 0 6667 262 5 3000 262 5 0 6667 4500 39375 . ( / ) . . . . . . .Solving, (1) At equilbrium, the mass fraction of glycerine in the solution outside the resin is: 87.5/(1000-262.5) = 0.1186 The wt% NaCl in the solution outside the resin is: 0.06(1000)/(1000-262.5) x 100% = 8.13 wt% Interpolate the given table for Kd to obtain: (0.759)(0.467) + (0.914)(0.533) = 0.842 = Kd Therefore, from Eq. (1), R = 4500 - 393.75(0.842) = 4,170 kg dry resin. Exercise 15.16 Subject: Estimation of external gas-to-particle mass-transfer and heat-transfer coefficients in a fixed-bed adsorption column. Given: Fixed bed of 2-foot diameter packed with 4 x 6 mesh silica gel with an external void fraction of 0.5. Gas is benzene vapor in air, which flows through the bed at a benzene-free flow rate of 25 lb/min. Find: External gas-to-particle mass-transfer and heat-transfer coefficients at a location in the bed where the pressure is 1 atm, temperature is 70oF, and benzene bulk mole fraction = 0.005. Analysis: From Perry's Handbook, 4-mesh and 6-mesh screens have openings of 0.476 cm and 0.336 cm, respectively, with an average of (0.476 + 0.336)/2 = 0.406 cm. Aasssuming the silica gel is crushed particles, the sphericity = y = 0.65. Therefore, the effective particle diameter = 0.65(0.406) = 0.264 cm = 0.00264 m = Dp. Provided that the dimensionless groups are in the specified ranges of the correlations, Eqs. (15-65) and (15-66) can be used to estimate the coefficients: k DD D G D h DD D G Ck c i p p i i p p P = + = + 2 11 2 11 0 6 1 3 0 6 1 3 . . . / . / m m r m m (1) (2) Because the gas is dilute in benzene, use the properties of air in the coefficient correlations. At 70oF, m = 183 micropoise = 1.83 x 10-5 kg/m-s k = 0.0256 J/m-s-K CP = 1.09 kJ/kg-K = 1090 J/kg-K From Perry's Handbook, at 32oF, the diffusivity of benzene in air at 1 atm = 0.077 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power. Therefore, Di = 0.077[(70 + 460)/(32 + 460)]1.75 = 0.0877 cm2/s = 0.0877 x 10-4 m2/s. From the ideal gas law, r = = =PMRT 1013 298 314 18 120. ( ). (530 / . ) . kg / m3 N Ck N D P i Pr Sc = = · = = = · · = - - - m m r 1090 183 10 0 0256 0 779 183 10 120 0 0877 10 174 5 5 4 ( . ) . . . . ( . ) . Exercise 15.16 (continued) Analysis (continued) In the Reynolds number, G = mass flow rate/bed cross-sectional area bed cross-sectional area = Ab = pD2/4 = 3.14(2)2/4 = 3.14 ft2 or 0.292 m2 Air flow rate = 25 lb/min = 25/60 = 0.417 lb/s or 0.417/29 = 0.0144 lbmol/s Benzene flow rate = (0.005/0.995)0.0144 = 0.000072 lbmol/s or 0.000072(78.11) = 0.0056 lb/s Total gas mass flow = 0.417 + 0.0056 = 0.423 lb/s or 0.192 kg/s Gas mass velocity = G = 0.192/0.292 = 0.658 kg/m2-s NRe = 0.00264(0.658)/1.83 x 10-5 = 94.9 All dimensionless groups and the particle diameter are in the specified ranges of applicability of the correlations of Eqs (1) and (2). Therefore, using those two equations, ( ) ( ) ( ) ( ) 4 0.6 1/3 0.6 1/3 2 0.0741 m/s0.0877 10 2 1.1 94.9 1.740.00264 0.0256 2 1.1 94.9 0.779 0.00 170.1 J/m - K264 s- ck h -· = + = + = = Exercise 15.17 Estimation of external gas-to-particle mass-transfer and heat-transfer coefficients in a fixed-bed adsorption column. Given: Fixed bed of 12.06-cm inside diameter packed with 3.3-mm-diameter Alcoa F-200 activated alumina beads with an external void fraction of 0.442. Gas is air containing water vapor, which flows through the bed at a flow rate of 1.327 kg/min. Find: External gas-to-particle mass-transfer and heat-transfer coefficients at a location in the bed where the pressure is 653.3 kPa, temperature is 21oC , and the dew-point temperature = 11.2oC. Analysis: Dp = 3.3 mm = 0.0033 m. Provided that the dimensionless groups are in the specified ranges of the correlations, Eqs. (15-65) and (15-66) can be used to estimate the coefficients: k DD D G D h DD D G Ck c i p p i i p p P = + = + 2 11 2 11 0 6 1 3 0 6 1 3 . . . / . / m m r m m (1) (2) At the dew-point temperature, the vapor pressure of water = 0.192 psi or 1.32 kPa. Therefore, at 653.3 kPa, the mole fraction of water in the air is only 1.32/653.3 = 0.002. Because the gas is so dilute in water vapor, use the properties of air in the coefficient correlations. At 21oC = 70oF, m = 183 micropoise = 1.83 x 10-5 kg/m-s k = 0.0256 J/m-s-K CP = 1.09 kJ/kg-K = 1090 J/kg-K From Perry's Handbook, at 32oF, the diffusivity of water vapor in air at 1 atm = 0.22 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power and inversely proportional to the pressure. Therefore, Di = 0.22(101.3/653.3)[(70 + 460)/(32 + 460)]1.75 = 0.0388 cm2/s = 0.0388x10-4 m2/s. At the high pressure of 653.3 kPa, the compressibility factor of air = 0.998. From the generalized gas law, r = = =PMZRT 653 3 290 998 314 18 7 75. ( ). (8. )(530 / . ) . kg / m3 N Ck N D P i Pr Sc = = · = = = · · = - - - m m r 1090 183 10 0 0256 0 779 183 10 7 75 0 0388 10 0 609 5 5 4 ( . ) . . . . ( . ) . Exercise 15.17 (continued) Analysis (continued) In the Reynolds number, G = mass flow rate/bed cross-sectional area Bed cross-sectional area = Ab = pD2/4 = 3.14(0.1206)2/4 = 0.0114 m2 Gas flow rate = 1.327 kg/min = 1.327/60 = 0.0221 kg/s Gas mass velocity = G = 0.0221/0.0114 = 1.94 kg/m2-s NRe = 0.0033(1.94)/1.83 x 10-5 = 350 All dimensionless groups and the particle diameter are in the specified ranges of applicability of the correlations of Eqs (1) and (2). Therefore, using those two equations, ( ) ( ) ( ) ( ) 4 0.6 1/3 0.6 1/3 2 0.0392 m/0.0388 10 2 1.1 350 0.6090.0033 0.0256 2 1.1 350 0.779 0. s 280 J/m -s-K0033 ck h -· = + = + = = Exercise 15.18 Subject: Effective diffusivity of acetone in the pores of activated carbon. Given: Activated carbon with rp = 0.85 g/cm3, ep = 0.48, dp = 25 Angstroms = 25 x 10-8 cm, and tortuosity = t = 2.75. Diffusion of acetone vapor in nitrogen, where from Example 15.7, temperature = 297 K, pressure = 136 kPa, and molecular (bulk) diffusivity = Di = 0.085 x 10-4 m2/s or 0.085 cm2/s. Assumptions: Negligible surface diffusion. Find: Effective diffusivity of acetone vapor through nitrogen in the pores, accounting for bulk (molecular) diffusion and Knudsen diffusion. Analysis: From a modification of Eq. (15-75), omitting surface diffusion, D D D D D e p i K K K = + = + = + e t 1 1 1 0 48 2 75 1 1 0 085 1 0175 1176 1 . . . . . (1) From Eq. (14-21), using 58 for the molecular weight of acetone, DK = 4850 dp(T/Mi)1/2 = 4850(25 x 10-8)(297/58)1/2 = 0.00274 cm2/s Substituting this value into Eq. (1) gives: 4 20.175 0.175 1 11.76 36411.76 0.002 4.66 10 cm / 74 seD -== ·= + + It is noted that Knudsen diffusion is very important and largely controls diffusion. Exercise 15.19 Subject: Effective diffusivity of benzene vapor in the pores of silica gel. Given: Silica gel with rp = 1.15 g/cm3 = 71.8 lb/ft3 , ep = 0.48, dp = 30 Angstroms = 30 x 10-8 cm, and tortuosity = t = 3.2. Diffusion of benzene vapor in air in the pores, where from Exercise 15.16, temperature = 70oF and pressure = 1 atm. Differential heat of adsorption = - 11,000 cal/mol. Adsorption equilibrium constant from Eq. (1) of Example 15.10 = K = 5,120 (lb/ft3 particles)/(lb/ft3 gas). Find: Effective diffusivity of benzene vapor through air in the pores, accounting for bulk (molecular) diffusion, Knudsen diffusion, and surface diffusion. Analysis: From Eq. (15-75), for all three mechanisms of internal diffusion, D D D D K D D D Ke p i K i s p p i K i s p= + + = + + e t r e r1 1 1 0 48 3 2 1 1 1 0 48 . . . (1) From Perry's Handbook, at 32oF, the diffusivity of benzene in air at 1 atm = 0.077 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power. Therefore, Di = 0.077[(70 + 460)/(32 + 460)]1.75 = 0.0877 cm2/s. From Eq. (14-21), using 78 for the molecular weight of benzene and T = 70oF = 294 K , DK = 4850 dp(T/Mi)1/2 = 4850(30 x 10-8)(294/78)1/2 = 0.00282 cm2/s From Eq. (15-76), using m =1 because silica gel is an insulating adsorbent, D HmRTi s = - - = - = · -0 016 0 45 0 016 0 45 11 0001 1987 294 3 34 10 6. exp . . exp . ,( )( . )( ) .D ads 2 cm / s In Eq. (1), the units of K are (g adsorbed/g adsorbent)/(g solute/cm3 of gas). Therefore, converting the units of the given K, K = (5,120/rp)[62.4 (lb/ft3)/(g/cm3)] = 5,120(62.4)/71.8 = 4,450 (g adsorbed/g adsorbent)/(g solute/cm3 of gas). Substituting the above values of Di , DK, (Di)s, and K into Eq. (1) gives: ( )6 2 0.48 1 (1.15)(4,450)3.12 10 1 13.2 0.48 0.0877 0.00282 1 0.15 0.0333 0.15(0. 0.0054 cm /00273 0.0333) 11.40 3 s55 eD - = + · + = + = + + = Note that the internal diffusion process is controlled by surface diffusion. Exercise 15.20 Subject: Effective diffusivity of water vapor in the pores of activated alumina. Given: Activated alumina with rp = 1.38 g/cm3, ep = 0.52, dp = 60 Angstroms = 60 x 10-8 cm, and tortuosity = t = 2.3. Diffusion of water vapor in air in the pores, where from Exercise 15.17, temperature = 21oC and pressure = 653.3 kPa. Find: Effective diffusivity of water vapor through air in the pores, accounting for bulk (molecular) diffusion and Knudsen diffusion, but not surface diffusion. Analysis: From a modification of Eq. (15-75), D D D D D e p i K i K = + = + e t 1 1 1 052 2 3 1 1 1 . . (1) From Perry's Handbook, at 32oF, the diffusivity of water vapor in air at 1 atm = 0.22 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power and inversely proportional to the pressure. Therefore, Di = 0.22(101.3/653.3)[(70 + 460)/(32 + 460)]1.75 = 0.0388 cm2/s From Eq. (14-21), using 18 for the molecular weight of water and T = 21oC = 294 K , DK = 4850 dp(T/Mi)1/2 = 4850(60 x 10-8)(294/18)1/2 = 0.0118 cm2/s Substituting the above values of Di and DK into Eq. (1) gives: 20.52 1 10.226 0.226(0.00905) 1 12.3 25.8 84.7 0.0388 0.011 0.00205 cm / 8 seD = = = + + = Note that Knudsen diffusion is somewhat more important than molecular diffusion. Exercise 15.21 Subject: Comparison of batch, continuous, and semicontinuous modes of slurry adsorption for removal of trichloroethylene (TCE) from water with activated carbon. Given: Feed of water at 25oC containing 3.3 mg/L of TCE. Activated carbon with an average particle diameter of 1.5 mm = 0.15 cm and the following Freundlich adsorption isotherm for TCE at 25oC: q = 67 c0.564 where q = mg TCE adsorbed/g carbon and c = mg TCE/L solution. Sherwood number = 30. Particle external surface area = 5 m2/kg. Assumptions: Mass-transfer resistance in the pores is negligible compared to the external resistance. Find: For an effluent of 0.01 mg TCE/L, (a) Minimum amount of adsorbent needed per liter of feed solution. (b) For batch mode with twice the minimum amount of adsorbent, the contact time. (c) For continuous mode with twice the minimum amount of adsorbent, the required residence time. (d) For semicontinuous mode at a feed rate of 50 gpm and a liquid residence time of 1.5 times that in Part (c), the amount of activated carbon to give a reasonable vol% solids in the tank and run time of not less than 10 times the liquid residence time. Analysis: (a) The minimum amount of adsorbent corresponds to equilibrium with a solution of the effluent concentration of 0.01 mg TCE/L. From the Freundlich equation, the equilibrium loading is given by: q = 67(0.01)0.564 = 4.99 mg TCE/g carbon. By material balance on the TCE, cF Q = cfinal Q + q Smin Solving for Smin, for a basis of 1 L of solution, ( )l min fina 1.0(3.3 0.01) 0.66 g carbon/L sol 4 u i. t on99 FS Q c c q - -= = = (b) For batch mode, use 2(0.66) = 1.32 g carbon/L solution = 1.32 kg/m3 = S/Q Combining Eqs. (15-77), (15-78), and (15-79), to eliminate q and c*, - = - - = - - = - - dc dt k a c Q c c Sk k a c c k a c c L F n L L 33 132 67 3 3 88 4 1 0 564 1 773. ( . )( ) . . / . . (1) From Eq. (15-65), kL in cm/s = NSh Di /Dp = 30 Di /0.15 = 200 Di (in cm2/s) From the Wilke-Chang Eq. (3-39), D M TA,B B B B A = · -7 4 10 8 1 2 0 6 . /f m u . (2) Exercise 15.21 (continued) Analysis: (b) (continued) From Table 3.2, uA = uTCE = (2)14.8 + 3.7+ 3(21.6) = 98.1 x 10-3 m3/kmol = 98.1 cm3/mol Solvent water viscosity at 25oC = 0.94 cP, MB = 18, and fB = 2.6. Substitution into Eq. (2) gives, DA,B 2 cm / s= · · = · - -7 4 10 2 6 18 298 0 94 981 102 10 8 1 2 0 6 5. . ( . ) . / . . = Di Therefore, kL = 200(1.02 x 10-5) = 0.00204 cm/s The units of a in Eq. (1) are cm2 of external surface area of adsorbent/ cm3 of solution. Therefore, a = 5 m2/kg (S/Q) = 5(1.32) = 6.60 m2/m3 of solution = 0.066 cm2/cm3. Therefore, kLa = 0.00204(0.066) = 1.35 x 10-4 s-1 Eq. (1) becomes: - = · - - - dc dt c c135 10 33 88 4 4 1 773 . . . . where c is in mg TCE/L solution, and t is in seconds. This equation represents an initial value problem in ODEs, which can be solved by the Euler method using a spreadsheet if a small time step is taken. The Euler form, where j is the iteration number, is, c c tj j( ) ( ) ( )+ = -1 D 135 10 3 388 44 1 773 . . .( ) ( ) . · - - - c cj j (3) If a time step, Dt, of 500 s is used to find the time when c becomes 0.01 mg/L, the first 10 values of c are as follows: Time, s c of TCE, mg/L 0 3.300 500 3.077 1000 2.870 1500 2.676 2000 2.495 2500 2.327 3000 2.170 3500 2.023 4000 1.887 4500 1.760 5000 1.641 The time required is found to be 44,000 s = 733 minutes = 12.2 h. A plot of c vs. t is given on the next page, where semi-log coordinates gives almost a straight line. Exercise 15.21 (continued) Analysis: (b) (continued) Slurry Batch Adsorption 0.01 0.10 1.00 10.00 0 5000 10000 15000 20000 25000 30000 35000 40000 45000 Time, seconds C o n c e n t r a t i o n o f T C E , m g / L Exercise 15.21 (continued) Analysis: (continued) (c) For continuous slurry adsorption, assume a solution feed rate, Q, of 1 L/h. Therefore, the adsorbent feed rate, from Part (a) = S = 2(0.66) = 1.32 g carbon/h. A rearrangement of Eq. (15-86) gives the residence time as: t c ck a c c cF L res out out = - - = -· --* *. .. . 33 0 01135 10 0 014 (4) The quantity c* is the ficticious concentration in equilibrium with qout, which is obtained by material balance, Eq. (15-87), which after rearrangement is: q Q c cSFout out mg / g= - = - = 1 3 3 0 01132 2 42( . . ). . Then, from a rearrangement of the given Freundlich adsorption isotherm, c q* . . / . . = = =out mg / L67 2 4267 0 00277 1 0 564 1 773 Substitution into Eq. (4) gives, ( )4res 3.3 0.01 1.35 10 0.01 0.002 3,370,00077t - -= · - = s = 56,200 minutes = 936 h = 39.0 days. This very large residence time is due to the very low exit concentration of TCE in the solution, making the continuous mode impractical. (d) Semicontinuous mode. Average residence time of the solution in the tank = 1.5(57,370) = 86,100 min. Flow rate of the solution = 50 gal/min or 50(3.785) = 189 L/min. The volume of solution in the tank = 189(86,100) = 16,300,000 L or 16,300 m3. For a cylindrical tank with height = diameter, this gives a tank diameter of 27.5 m, which is probably impractical, although many smaller tanks in parallel could be used. In the tank, where the solid adsorbent charge resides, the variation of the loading, q, with run time is given by Eq. (15-91), which for time in minutes is, S dqdt k a c c t Q c c c cL= - = · - = --out res out out* . ( ) * (86, )( ) , * 135 10 60 100 189 132 0004 (5) where S is in g, q is in mg TCE/g, and t is in min. From Eq. (15-86), c c k at ck at c cF L L out res res = ++ = + ·+ · = + - - * . . ( )(86, ) * . ( )(86, ) . . *1 3 3 135 10 60 100 1 135 10 60 100 0 00473 0 99857 4 4 (6) Exercise 15.21 (continued) Analysis: (d) (continued) From the given Freundlich equation, c* = (q/67)1.773 = 0.000579 q1.773 (7) Combining Eqs. (5), (6), and (7), gives, S dqdt c c c q= + - = - = -132 000 0 00473 0 99857 624 36 188 76 624 36 01093 1 773, . . * * . . * . . . (8) Let the volume of solids in the tank = 2.5 vol%. The solids-free volume of liquid in the tank = 16,300 m3. Therefore, the solids volume = (2.5/97.5)(16,300) =418 m3. Assume a particle density = 850 kg/m3. Therefore, S = 418 (850) = 355,000 kg = 3.55 x 108 g and Eq. (8) becomes, dq dt q= · - · - -176 10 3 08 106 10 1 773. . . (9) Now, compute the run time to achieve a cumulative cout = 0.01 mg/L, where, c c dt t t cum out= = 0 01 0. (10) Eq. (9) represents an initial value problem in ODEs, which can be solved by the Euler method using a spreadsheet, if a small time step is taken. The Euler form, with j as the iteration index, is, q q tj j( ) ( ) ( )+ = +1 D [176 10 308 106 10 1 773. . . ( )· - ·- - q j ] (11) The total run time is specified to be not less than 10 times the average liquid residence time; thus, 10(86,100) = 861,000 minutes. Try a Dt of 100,000 minutes. From the spreadsheet, the first 5 and last 5 values are: t, min. q, mg/g c*, mg/L c out c cum 0 0 0 0.00473 0 100,000 0.176 2.66E-05 0.00476 0.00474 200,000 0.352 9.09E-05 0.00482 0.00477 300,000 0.528 0.000187 0.00492 0.00480 400,000 0.704 0.000311 0.00504 0.00484 500,000 0.880 0.000462 0.00519 0.00490 3,200,000 5.625 0.012377 0.01709 0.00919 3,300,000 5.800 0.013069 0.01778 0.00944 3,400,000 5.975 0.013778 0.01849 0.00970 3,500,000 6.151 0.014502 0.01921 0.00996 3,600,000 6.326 0.015243 0.01995 0.01023 Thus, the run time is just more over 3,500,000 minutes (2,430 days), which is greater than the above 861,000 minutes. Unfortunately, the vessel size is enormous. Exercise 15.22 Subject: Comparison of batch, continuous, and semicontinuous modes of slurry adsorption for removal of benzene (B) and m-xylene (X) from water with activated carbon. Given: Feed of water at 25oC containing 0.324 mg/L of B and 0.630 mg/L of X. Activated carbon with an average particle diameter of 1.5 mm = 0.15 cm and the following Freundlich adsorption isotherms at 25oC: q =32 c0.428 for B and q =125 c0.333 for X, where q = mg adsorbed/g carbon and c = mg in solution/L solution. Sherwood number = 30. Particle external surface area = 5 m2/kg. Assumptions: Mass-transfer resistance in the pores is negligible compared to the external resistance. Find: For an effluent of 0.002 mg each of B and X, (a) Minimum amount of adsorbent needed per liter of feed solution. (b) For batch mode with twice the minimum amount of adsorbent, the contact time. (c) For continuous mode with twice the minimum amount of adsorbent, the required residence time. (d) For semicontinuous mode at a feed rate of 50 gpm and a liquid residence time of 1.5 times that in Part (c), the amount of activated carbon to give a reasonable vol% solids in the tank and run time of not less than 10 times the liquid residence time. Analysis: (a) The minimum amount of adsorbent corresponds to equilibrium with a solution of the effluent concentration of 0.002 mg of B or X, with the other solute at a lower concentration.. Must determine which solute controls. From the Freundlich equation for B, the equilibrium loading is given by: q = 32(0.002)0.428 = 2.24 mg B/g carbon. By material balance on B, cF Q = cfinal Q + q Smin Solving for Smin, for a basis of 1 L of solution, ( )fi m n in al 1.0(0.324 0.002 0.144 g carbon/L sol) ution 2.24 FS Q c c q - -= = = From the Freundlich equation for X, the equilibrium loading is given by: q = 125(0.002)0.333 = 15.78 mg X/g carbon.By material balance on X, cF Q = cfinal Q + q Smin Solving for Smin, for a basis of 1 L of solution, S Q c cqFmin final g carbon / L solution= - = - = 10 0 630 0 00215 78 0 0398. ( . . ). . Therefore B controls and the minimum amount of adsorbent = 0.144 g carbon/L solution. (b) For batch mode, use 2(0.144) = 0.288 g carbon/L solution = 0.288 kg/m3 = S/Q Combining Eqs. (15-77), (15-78), and (15-79) and applying them to B, to eliminate q and c*, Exercise 15.22 (continued) Analysis: (a) (continued) - = - - = - - = - - dc dt k a c Q c c Sk k a c c k a c c L F n L L 0 324 0 288 32 0 324 9 22 1 0 428 2 336. ( . )( ) . . / . . (1) From Eq. (15-65), kL in cm/s = NSh Di /Dp = 30 Di /0.15 = 200 Di (in cm2/s) From the Wilke-Chang Eq. (3-39), D M TA,B B B B A = · -7 4 10 8 1 2 0 6 . /f m u . (2) From Table 3.2, uA = uB = (6)14.8 + 6(3.7) - 15 = 96 x 10-3 m3/kmol = 96 cm3/mol Solvent water viscosity at 25oC = 0.94 cP, MB = 18, and fB = 2.6. Substitution into Eq. (2) gives, DA,B 2 cm / s= · · = · - -7 4 10 2 6 18 298 0 94 96 104 10 8 1 2 0 6 5. . ( ) . / . . = Di Therefore, kL = 200(1.04 x 10-5) = 0.0021 cm/s The units of a in Eq. (1) are cm2 of external surface area of adsorbent/ cm3 of solution. Therefore, a = 5 m2/kg (S/Q) = 5(0.288) = 1.44 m2/m3 of solution = 0.0144 cm2/cm3. Therefore, kLa = 0.0021(0.0144) = 3.02 x 10-5 s-1 Eq. (1) becomes: - = · - - - dc dt c c302 10 0 324 9 22 5 2 336 . . . . where c is in mg B/L solution, and t is in seconds. This equation represents an initial value problem in ODEs, which can be solved by the Euler method using a spreadsheet if a small time step is taken. The Euler form, where j is the iteration number, is, c c tj j( ) ( ) ( )+ = -1 D 302 10 0 3249 225 2 336 . . .( ) ( ) . · - - - c cj j (3) If a time step, Dt, of 1,000 s is used to find the time when c becomes 0.002 mg/L, the first 7 values of c for B are as follows: Time, s c of B, mg/L 0 0.324 1,000 0.314 2,000 0.305 3,000 0.296 4,000 0.287 5,000 0.278 6,000 0.270 Exercise 15.22 (continued) Analysis: (b) (continued) The time required is found to be 172,000 s = 2,870 minutes = 47.8 h. A plot of c vs. t follows, where semi-log coordinates gives almost a straight line. 0.001 0.010 0.100 1.000 0 50000 100000 150000 200000 250000 Time, seconds C o n c e n t r a t i o n o f b e n z e n e i n s o l u t i o n , m g / L Exercise 15.22 (continued) Analysis: (continued) (c) For continuous slurry adsorption, assume a solution feed rate, Q, of 1 L/h. Therefore, the adsorbent feed rate, from Part (a) = S = 2(0.144) = 0.288 g carbon/h. A rearrangement of Eq. (15-86) gives the residence time as: t c ck a c c cF L res out out = - - = -· --* *. .. . 0 324 0 002302 10 0 0025 (4) The quantity c* is the ficticious concentration in equilibrium with qout, which is obtained by material balance, Eq. (15-87), which after rearrangement is: q Q c cSFout out mg / g= - = - = 1 0 324 0 0020 288 1118( . . ). . Then, from a rearrangement of the given Freundlich adsorption isotherm, c q* . . / . . = = =out mg / L32 111832 0 000395 1 0 428 2 336 Substitution into Eq. (4) gives, ( )5res 0.324 0.002 6,640,000 3.02 10 0.002 0.000395t - -= = · - s = 110,700 minutes = 1,845 h = 76.9 days. This very large residence time is due to the very low exit concentration of benzene in the solution, making the continuous mode impractical. (d) Semicontinuous mode. Average residence time of the solution in the tank = 1.5(110,700) = 166,100 min. Flow rate of the solution = 50 gal/min or 50(3.785) = 189 L/min. The volume of solution in the tank = 189(166,100) = 31,400,000 L or 31,400 m3. For a cylindrical tank with height = diameter, this gives a tank diameter of 34.2 m, which is probably impractical, although many smaller tanks in parallel could be used. In the tank, where the solid adsorbent charge resides, the variation of the loading, q, with run time is given by Eq. (15-91), which for time in minutes is, S dqdt k a c c t Q c c c cL= - = · - = --out res out out* . ( ) * ( , )( ) , * 3 02 10 60 166 100 189 56 9005 (5) where S is in g, q is in mg B/g, and t is in min. Exercise 15.22 (continued) Analysis: (d) (continued) From Eq. (15-86), c c k at ck at c cF L L out res res = ++ = + ·+ · = + - - * . . ( )( , ) * . ( )( , ) . . *1 0 324 3 02 10 60 166 100 1 3 02 10 60 166 100 0 00107 0 99669 5 5 (6) From the given Freundlich equation, c* = (q/32)2.336 = 0.000305 q2.336 (7) Combining Eqs. (5), (6), and (7), gives, S dqdt c c c q= + - = - = -56 900 0 00107 0 99669 6088 188 34 60 88 2 336, . . * * . . * . . 0.0574 (8) Let the volume of solids in the tank = 2.5 vol%. The solids-free volume of liquid in the tank = 31,400 m3. Therefore, the solids volume = (2.5/97.5)(31,400) =805 m3. Assume a particle density = 850 kg/m3. Therefore, S = 805 (850) = 684,000 kg = 6.84 x 108 g and Eq. (8) becomes, dq dt q= · - · - -8 90 10 8 39 108 11 2 336. . . (9) Now, compute the run time to achieve a cumulative cout = 0.002 mg/L, where, c c dt t t cum out= = 0 002 0. (10) Eq. (9) represents an initial value problem in ODEs, which can be solved by the Euler method using a spreadsheet, if a small time step is taken. The Euler form, with j as the iteration index, is, q q tj j( ) ( ) ( )+ = +1 D [8 90 10 8 39 108 11 2 336. . . ( )· - ·- - q j ] (11) The total run time is specified to be not less than 10 times the average liquid residence time; thus, 10(166,100) = 1,661,000 minutes. Try a Dt of 500,000 minutes. From the spreadsheet, the first 6 and last 6 values are: t, min. q, mg/g c*, mg/L c out c cum 0 0 0 0.00154 0 500000 0.04450 2.12E-07 0.00154 0.00154 1000000 0.08900 1.07E-06 0.00154 0.00154 1500000 0.13500 2.76E-06 0.00155 0.00154 2000000 0.17800 5.41E-06 0.00155 0.00154 2500000 0.22250 9.11E-06 0.00155 0.00154 20000000 1.77815 0.001170 0.00270 0.00189 20500000 1.82249 0.001239 0.00277 0.00191 21000000 1.86682 0.001311 0.00284 0.00193 21500000 1.91114 0.001385 0.00292 0.00195 22000000 1.95545 0.001461 0.00299 0.00198 22500000 1.99974 0.001539 0.00307 0.00200 Thus, the run time is 22,500,000 minutes (15,600 days), which is much greater than the above 1,661,000 minutes. Unfortunately, the vessel size is enormous. Exercise 15.23 Subject: Comparison of batch, continuous, and semicontinuous modes of slurry adsorption for removal of chloroform (C) from water with activated carbon. Given: Feed of water at 25oC containing 0.223 mg/L of C. Activated carbon with an average particle diameter of 1.5 mm = 0.15 cm and the following Freundlich adsorption isotherm for C at 25oC: q = 10 c0.564 where q = mg C adsorbed/g carbon and c = mg C/L solution. Sherwood number = 30. Particle external surface area = 5 m2/kg. Assumptions: Mass-transfer resistance in the pores is negligible compared to the external resistance. Find: For an effluent of 0.01 mg C/L, (a) Minimum amount of adsorbent needed per liter of feed solution. (b) For batch mode with twice the minimum amount of adsorbent, the contact time. (c) For continuous mode with twice the minimum amount of adsorbent, the required residence time. (d) For semicontinuous mode at a feed rate of 50 gpm and a liquid residence time of 1.5 times that in Part (c), the amount of activated carbon to give a reasonable vol% solids in the tank and run time of not less than 10 times the liquid residence time. Analysis: (a) The minimum amount of adsorbent corresponds to equilibrium with a solution of the effluent concentration of 0.01 mg TCE/L. From the Freundlich equation, the equilibrium loading is given by: q = 10(0.01)0.564 = 0.745 mg C/g carbon. By material balance on the C, cF Q = cfinal Q + q Smin Solving for Smin, for a basis of 1 L of solution, ( )fi m n in al 1.0(0.223 0.01) 0.286 g carbon/L sol 0 ution.745 FS Q c c q - -= = = (b) For batch mode, use 2(0.286) = 0.572 g carbon/L solution = 0.572 kg/m3 = S/Q Combining Eqs. (15-77), (15-78), and (15-79), to eliminate q and c*, - = - - = - - = - - dc dt k a c Q c c Sk k a c c k a c c L F n L L 0 223 0572 10 0 223 572 1 0 564 1 773. ( . )( ) . . / . . (1) From Eq. (15-65), kL in cm/s = NSh Di /Dp = 30 Di /0.15 = 200 Di (in cm2/s) From the Wilke-Chang Eq. (3-39), D M TA,B B B B A = · -7 4 10 8 1 2 0 6 . /f m u . (2) Exercise 15.23 (continued) Analysis: (b) (continued) From Table 3.2, uA = uTCE = 14.8 + 3.7+ 3(21.6) = 83.3 x 10-3 m3/kmol = 83.3 cm3/mol Solvent water viscosity at 25oC = 0.94 cP, MB = 18, and fB = 2.6. Substitution into Eq. (2) gives, DA,B 2 cm / s= · · = · - -7 4 10 2 6 18 298 0 94 3 113 10 8 1 2 0 6 5. . (83. ) . / . . = Di Therefore, kL = 200(1.13 x 10-5) = 0.00226 cm/s The units of a in Eq. (1) are cm2 of external surface area of adsorbent/ cm3 of solution. Therefore, a = 5 m2/kg (S/Q) = 5(0.572) = 2.86 m2/m3 of solution = 0.0286 cm2/cm3. Therefore, kLa = 0.00226(0.0286) = 6.46 x 10-5 s-1 Eq. (1) becomes: - = · - - - dc dt c c6 46 10 0 223 5 72 5 1 773 . . . . where c is in mg C/L solution, and t is in seconds. This equation represents an initial value problem in ODEs, which can be solved by the Euler method using a spreadsheet if a small time step is taken. The Euler form, where j is the iteration number, is, c c tj j( ) ( ) ( )+ = -1 D 6 46 10 0 2235 725 1 773 . . .( ) ( ) . · - - - c cj j (3) If a time step, Dt, of 500 s is used to find the time when c becomes 0.01 mg/L, the first 10 values of c are as follows: Time, s c of C, mg/L 0 0.223 500 0.216 1000 0.209 1500 0.202 2000 0.196 2500 0.189 3000 0.183 3500 0.177 4000 0.172 4500 0.166 5000 0.161 The time required is found to be 51,500 s = 858 minutes = 14.3 h. A plot of c vs. t is given on the next page, where semi-log coordinates gives almost a straight line. Exercise 15.23 (continued) Analysis: (b) (continued) Slurry Batch Adsorption 0.01 0.1 1 0 10000 20000 30000 40000 50000 60000 Time, seconds C o n c e n t r a t i o n o f C h l o r o f o r m i n s o l u t i o n , m g / L Exercise 15.23 (continued) Analysis: (continued) (c) For continuous slurry adsorption, assume a solution feed rate, Q, of 1 L/h. Therefore, the adsorbent feed rate, from Part (a) = S = 2(0.286) = 0.572 g carbon/h. A rearrangement of Eq. (15-86) gives the residence time as: t c ck a c c cF L res out out = - - = -· --* *. .. . 0 223 0 016 46 10 0 015 (4) The quantity c* is the ficticious concentration in equilibrium with qout, which is obtained by material balance, Eq. (15-87), which after rearrangement is: q Q c cSFout out mg / g= - = - = 1 0 223 0 010572 0 372( . . ). . Then, from a rearrangement of the given Freundlich adsorption isotherm, c q* . . / . . = = =out mg / L10 0 37210 0 00292 1 0 564 1 773 Substitution into Eq. (4) gives, ( )5res 0.223 0.01 466,000 6.46 10 0.01 0.00293t - -= = · - s = 7,770 minutes = 130 h = 5.4 days. This is a large residence time compared to the batch time in Part (b). (d) Semicontinuous mode. Average residence time of the solution in the tank = 1.5(7,700) = 11,600 min. Flow rate of the solution = 50 gal/min or 50(3.785) = 189 L/min. The volume of solution in the tank = 189(11,600) = 2,190,000 L or 2,200 m3. For a cylindrical tank with height = diameter, this gives a tank diameter of 14.1 m, which is probably impractical, although many smaller tanks in parallel could be used. In the tank, where the solid adsorbent charge resides, the variation of the loading, q, with run time is given by Eq. (15-91), which for time in minutes is, S dqdt k a c c t Q c c c cL= - = · - = --out res out out* . ( ) * ( , )( ) , * 6 46 10 60 11 600 189 8 5005 (5) where S is in g, q is in mg C/g, and t is in min. From Eq. (15-86), c c k at ck at c cF L L out res res = ++ = + ·+ · = + - - * . . ( )( , ) * . ( )( , ) . . *1 0 223 6 46 10 60 11 600 1 6 46 10 60 11 600 0 00485 0 97826 5 5 (6) Exercise 15.23 (continued) Analysis: (d) (continued) From the given Freundlich equation, c* = (q/10)1.773 = 0.01687 q1.773 (7) Combining Eqs. (5), (6), and (7), gives, S dqdt c c c q= + - = - = -8 500 0 00485 0 97826 412 184 79 412 312 1 773, . . * * . . * . . . (8) Let the volume of solids in the tank = 2.5 vol%. The solids-free volume of liquid in the tank = 2,200 m3. Therefore, the solids volume = (2.5/97.5)(2,200) =56.4 m3. Assume a particle density = 850 kg/m3. Therefore, S = 56.4 (850) = 48,000 kg = 4.8 x 107 g and Eq. (8) becomes, dq dt q= · - · - -858 10 6 5 107 8 1 773. . . (9) Now, compute the run time to achieve a cumulative cout = 0.01 mg/L, where, c c dt t t cum out= = 0 01 0. (10) Eq. (9) represents an initial value problem in ODEs, which can be solved by the Euler method using a spreadsheet, if a small time step is taken. The Euler form, with j as the iteration index, is, q q tj j( ) ( ) ( )+ = +1 D [8 58 10 65 107 8 1 773. . . ( )· - ·- - q j ] (11) The total run time is specified to be not less than 10 times the average liquid residence time; thus, 10(7,770) = 77,700 minutes. Try a Dt of 25,000 minutes. From the spreadsheet, the first 6 and last 6 values are: t, min. q, mg/g c*, mg/L c out c cum 0 0 0 0.00485 0 25,000 0.0215 1.86E-05 0.00487 0.00486 50,000 0.0429 6.34E-05 0.00491 0.00487 75,000 0.0643 0.000130 0.00498 0.00490 100,000 0.0858 0.000217 0.00506 0.00493 125,000 0.1072 0.000322 0.00516 0.00497 975,000 0.8208 0.011887 0.01648 0.00910 1,000,000 0.8411 0.012413 0.01699 0.00929 1,025,000 0.8614 0.012948 0.01752 0.00949 1,050,000 0.8816 0.013491 0.01805 0.00968 1,075,000 0.9017 0.014043 0.01859 0.00988 1,100,000 0.9218 0.014602 0.01913 0.01009 By interpolation, the run time is 1,090,000 minutes (757 days), which is greater than the above 77,700 minutes. Unfortunately, the vessel size is very large. Exercise 15.24 Subject: Use of three beds in a cycle to adsorb 1,2-dichloroethane from water, with application of the MTZ concept. Given: Three fixed-bed adsorbers, each containing 10,000 lb of activated carbon of rb = 30 lb/ft3, to be used to reduce the concentration of 1,2-dichloroethane (D) in 250 gpm of water from 4.6 mg/L to 0.001 mg/L. Each bed has an H/D = 2. Two beds in series for adsorption (a lead bed and a trailing bed). One bed being regenerated by replacing the spent carbon. When the lead bed is saturated, it is regenerated, the trailing bed becomes the lead bed, and the regenerated bed becomes the trailing bed. The adsorption equilibrium isotherm is q = 8 c0.57, with q in mg/g and c in mg/L. Assumptions: Constant pattern front so that the width of the MTZ is a constant. Find: How often must the carbon in a bed be replaced. The maximum width of the MTZ to allow saturation of the lead bed. Analysis: Volume of each bed = V = mass of carbon/rb = 10,000/30 = 333 ft3 For H = 2D, V = 333 = pD3/2. Solving, D = 6.0 ft and H = 2(6) = 12.0 ft. For ideal equilibrium adsorption from Eq. (15-92), the time for the ideal wave front to move through one of the three beds is t q SQ cb F F F = (1) From the adsorption isotherm, the loading in equilbrium with the feed = qF = 8 cF0.57 = 8(4.6)0.57 = 19.1 mg/g. The grams of adsorbent in each bed = S = 10,000(453.6) = 4,536,000 g The volumetric feed rate = QF = 250(3.785) = 946 L/min, and cF = 4.6 mg/L Therefore, from Eq. (1), tb = =19 1 4 536 000946 4 6 19 900. ( , , )( . ) , minutes = 332 hours = 13.8 days The lead bed must be replaced every 13.8 days. The width of the MTZ zone can not exceed the height of a bed or 12 ft. Otherwise, breakthrough will occur before the trailing bed becomes the lead bed. Exercise 15.25 Subject: Use of three or more beds in a cycle to adsorb benzene and m-xylene from water, with application of the MTZ concept. Given: Three fixed-bed adsorbers, each containing 10,000 lb of activated carbon of rb = 30 lb/ft3, to be used to remove 0.185 mg/L of benzene (B) and 0.583 mg/L of m-xylene (X) from 250 gpm of water. Each bed has an H/D = 2. Two or more beds in series for adsorption (at least a lead bed and a trailing bed). One bed being regenerated by replacing the spent carbon. When the lead bed is saturated, it is regenerated, the trailing bed becomes the lead bed, and the regenerated bed becomes the trailing bed. The adsorption equilibrium isotherms are: qB = 32 cB0.428 and qX = 125 cX0.333, with q in mg/g and c in mg/L. Widths of the mass-transfer zones are MTZB = 2.5 ft and MTZX = 4.8 ft. Assumptions: Constant pattern front so that the widths of the MTZs are constant. Find: How often must the carbon in a bed be replaced. Analysis: Volume of each bed = V = mass of carbon/rb = 10,000/30 = 333 ft3 For H = 2D, V = 333 = pD3/2. Solving, D = 6.0 ft and H = 2(6) = 12.0 ft. For ideal equilibrium adsorption from Eq. (15-92), the time for the ideal wave front to move through one of the three beds for one of the solutes is: t q SQ cb F F F = (1) From the adsorption isotherm of B. the loading in equilbrium with the feed = qF = 32 cF0.428 = 32(0.185)0.428 = 15.5 mg/g. From the adsorption isotherm of X. the loading in equilbrium with the feed = qF = 125 cF0.333 = 125(0.583)0.333 = 104.4 mg/g. The grams of adsorbent in each bed = S = 10,000(453.6) = 4,536,000 g The volumetric feed rate = QF = 250(3.785) = 946 L/min. For benzene, with cF = 0.185 mg/L, and from Eq. (1), tb = =155 4 536 000946 0185 402 000. ( , , )( . ) , minutes = 6,700 hours = 279 days For m-xylene, with cF = 0.583 mg/L, and from Eq. (1), tb = =104 4 4 536 000946 0583 859 000. ( , , )( . ) , minutes = 14,300 hours = 596 days So, benzene controls. A satisfactory bed arrangement is to use two beds in series for adsorption with a time of 279 days to change the beds. Because the MTZs are less than the bed height of 12 ft, no breakthrough will occur. The lead bed will become saturated with benzene but not with m- xylene. Exercise 15.26 Subject: Comparison of equilibrium model to mass-transfer model for fixed-bed adsorption of water vapor from air with silica gel. Given: Feed of air with a relative humidity of 80% at 80oF and 1 atm, with a superficial velocity of 100 ft/min in a fixed bed of 2.8-mm-diameter spherical particles of silica gel (rb = 39 lb/ft3). Bed height = H = 5 ft. Linear adsorption isotherm for water vapor = q = 15.9 p, where q is in lb H2O/lb gel, and p = partial pressure of water vapor in atm. Effective diffusivity of water vapor in the adsorbent particle = 0.05 cm2/s. Exit humidity of water vapor to be 0.0009 lb H2O/lb dry air. Assumptions: Isothermal and isobaric operation. Find: Time to breakthrough for the equilibrium model. Time to breakthrough for the mass-transfer model of Klinkenberg. Set of breakthrough curves for the mass-transfer model of Klinkenberg. Approximate width of the MTZ. Average loading of the bed at breakthrough. Analysis: First determine the moisture content of the entering air. From a psychrometric chart, the entering humidity for 80% R.H., 80oF, and 1 atm is 0.0177 lb H2O/lb dry air or a water vapor content of 0.0177/1.0177 x 100% = 1.739 wt%. By Dalton's law, partial pressure of water vapor in the entering air = (1)[0.0177/18.02]/[0.0177/18.02 + 1/28.97] = 0.0277 atm. The mole fraction of water vapor in the entering air = 0.0277. Assume a bed cross-sectional area of 1 ft2. Then, entering gas rate = 100 ft3/min. Bed volume = (1)(5) = 5 ft3. Bed contains 5(39) = 195 lb silica gel. The MW of the entering gas = 18.02(0.0277) + 28.97(0.9723) = 28.67. From the ideal gas law, density of entering gas = PM/RT = (1)(28.67)/[(0.7302)(540)] = 0.0727 lb/ft3. Therefore, the entering gas flow rate = 100(0.0727) = 7.27 lb/min with a water vapor flow rate of 0.01739(7.27) = 0.1264 lb H2O/min. For the dilute conditions, the desired c/cF = 0.0009/0.0177 = 0.05. So use this as the breakthrough value. Equilibrium model: Equilibrium loading based on entering feed = 15.9(0.0277) = 0.440 lb H2O/lb silica gel. Therefore, can adsorb 0.440(195) = 85.8 lb H2O. The ideal time for breakthrough = 85.8/0.1264 = 679 min = 11.3 h. Alternatively, can apply Eq. (15-92), with Lideal/LB = 1, so ideal 0.440(195)0 6 9. 6712 4F F F q S Q ct = = = min. Mass-transfer model using Klinkenberg equations: From Eq. (15-106), 1 3 15 2 kK R k R D p c p e = + (1) Rp = 0.28/2 = 0.14 cm = 0.0014 m, De = 0.05 cm2/s = 5 x 10-6 m2/s Exercise 15.26 (continued) Analysis: (continued) From Eq. (15-65), k DD D G Dc i p p i = + 2 11 0 6 1 3 . . / m m r (2) From Perry's Handbook, at 32oF, the diffusivity of water vapor in air at 1 atm = 0.22 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power. Therefore, Di = 0.22[(80 + 460)/(32 + 460)]1.75 = 0.26 cm2/s = 0.26 x 10-4 m2/s. From above, r = 0.0727 lb/ft3 = 1.16 kg/m3. From a handbook, m of air at 80oF = 1.75 x 10-5 kg/m-s. The mass velocity = G = 7.29/1.0 = 7.27 lb/min-ft2 = 0.592 kg/m2-s N D N D G i p Sc Re = = · · = = = · = - - - m r m 175 10 116 0 26 10 0580 0 0028 0 592 175 10 94 7 5 4 5 . . ( . ) . . ( . ) . . From Eq. (2), kc = · + = -0 26 10 0 0028 2 11 94 7 058 0149 4 0 6 1 3. . . . . . . / m/s From Eq. (1), 1 0 00143 0149 0 001415 0 05 10 0 0031 0 0261 0 0292 2 4kK = + · = + =- . ( . ) ( . ) ( . ) . . . s The equilibrium constant, K, the given value is 15 lb H2O/lb gel-atm. We need to convert this to units of (lb H2O/ft3 gel)/(lb H2O/ft3 gas). From Table 15.2, assume small-pore silica gel with eb = 0.47. Therefore, the particle density from a rearrangement of Eq. (15-4) is r rep b b = - = - =1 391 0 47 73 6( . ) . lb gel/ft3 of particles. When the partial pressure of H2O = 0.0277 atm, we have 0.1264/100 = 0.001264 lb H2O/ft3 gas. Therefore, K = 15(73.6)(0.0277)/0.001264 = 24,200. The equilibrium adsorption isotherm is now, q in lb H2O/ft3 silica gel particles = 24,200 c in lb H2O/ft3 gas From above, k = 1/[0.0292 K] = 1/[0.0292(24,200)] = 1.42 x 10-3 s-1 or 0.0849 min-1 The Klinkenberg Eq. (15-114) is: cc F » + - + + 1 2 1 1 8 1 8t x t x (3) where, from Eq. (15-115), x ee= - kKzu b b 1 (4) u = actual (interstitial velocity) = superficial velocity/eb = 100/0.47 = 213 ft/min = 1.08 m/s From Eq. (4), x = - = 1 0 0292 108 1 0 47 0 47 358 . . . . . z z (in meters) (5) Exercise 15.26 (continued) Analysis: (continued) From Eq. (15-116), t = - = - = -k t zu t z t z0 0849 108 60 0 0849 0 00131. . ( ) . .(in minutes) (in meters) (6) The maximum value of z, the distance through the bed, is the bed height = 5 ft = 1.524 m. At breakthrough, c/cF = 0.05 at z = 1.524 m and, therefore, x = 35.8(1.524) = 54.6 and t = 0.0849t - 0.00131(1.524) = 0.0849t (in minutes) - 0.00200 Must solve Eq. (1) for t at c/cF = 0.05 and x= 54.6. The time will be less than the ideal equilibrium time above of 679 minutes. Can do this by trial and error with a spreadsheet or with a nonlinear solver that can handle the error function, erf. From a spreadsheet, breakthrough time = 451 minutes. To obtain the width of the MTZ, solve Eq. (1) for z at c/cF = 0.95 and t = 451 minutes. By trial and error from a spreadsheet, at breakthrough, the MTZ extends from z = 0.718 m to z =1.524 m. Therefore, the width of the MTZ = 1.524 - 0.718 = 0.81 m = 2.6 ft. Breakthrough curves similar to Figs. 15.30 and 15.31, but in terms of distance, z, through the bed and time, instead of the dimensionless groups, x and t , are shown on the next two pages. They are computed with a spreadsheet from Eq. (3) above for gas concentration and Eq. (15-119) for adsorbent loading, together with Eqs. (5) and (6) to convert from x and t to z and t, respectively. With the spreadsheet, the error function, erf, must be used carefully, noting that erf(-x) = -erf(x). Thus, when the argument of the error function is negative, the argument should be made positive and the result should be multiplied by -1. To compute the average loading of the bed at the breakthrough time of 451 minutes, use the equation in Example 15.11: q qdz avg = 01 524 1524 . . , where z is in meters. (7) A spreadsheet is used with Eq. (15-119) to compute q as a function of z for t = 451 minutes (breakthrough time) and qF* = 0.440 lb H2O/lb silica gel from above (value in equilibrium with the feed. The results are: q , lb/lb 0.440 0.440 0.433 0.382 0.260 0.125 0.042 0.018 z, meters 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.524 Solving Eq. (7) numerically by the trapezoid method, avgq = [0.440(0.2) + 0.440(0.2) + 0.4365(0.2) + 0.4075(0.2) + 0.321(0.2) + 0.1925(0.2) + 0.0835(0.2) + 0.030(0.124)]/1.524 = 0.468/1.524 = 0.307 lb H2O/lb silica gel. Thus, 0.307/0.440 x 100% = 70% utilization of the bed. Exercise 15.26 (continued) Analysis: (continued) Concentration profiles 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 100 200 300 400 500 600 700 Time, minutes c / c F Exercise 15.26 (continued) Analysis: (continued) Loading Profiles 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 100 200 300 400 500 600 700 Time, minutes q b a r / q F * Exercise 15.27 Subject: Reduction of nitroglycerine (NG) content of wastewater by fixed-bed adsorption with activated carbon in a train of 55-gallon cannisters Given: Train of four 55-gallon cannisters of activated carbon (rb = 32 lb/ft3), each with an inside diameter of 2 ft and containing 200 lb of carbon. Flow rate of wastewater = 400 gph with 2,000 ppm (by weight) of NG. When effluent from the first cannister reaches 1 ppm of NG, that cannister is removed and a fresh one added to the end of the train. The following test results are available for one cannister: water flow rate = 10 gpm with 1,020 ppm of NG by weight. Breakthrough time in hours = tB = 3.90 L - 2.05 for 1ppm NG, with L = bed depth in ft. Assumptions: Plug flow. Find: Number of cannisters needed each month. Monthly cannister cost at $700/cannister. Analysis: Bed volume of activated carbon = 200/32 = 6.25 ft3 = V = pD2L/4. Solving for L, with D = 2 ft, gives L = 2 ft. Rearrange the breakthrough correlation for the test, by solving for L, L = tB/3.90 + 2.05/3.90 = 0.256 tB + 0.526 This is the same form as Eq. (15-120), where the term 0.256 tB is LES and the term 0.526 is LUB. Assume that the width of the MTZ = constant, independent of flow rate. Therefore, LUB is constant at 0.526 ft. From Eq. (15-122), LES is directly proportional to the feed rate, QF . If a linear adsorption isotherm is assumed, then LUB is independent of the ratio qF/cF in Eq. (15-122). That is, for the proposed operation, the higher value of cF (2,000 ppm compared to 1,020 ppm for the test) is of no consequence. The higher concentration of NG in the feed, cF, is couterbalanced by the higher loading in equilbrium with the feed, qF. The breakthrough equation becomes: L Q Q t t tF F B B B= + = + = +0 256 0526 0 256 400 600 0526 0171 0 526. . . . . . of operation of test Solving for tB, with L of operation also equal to 2 ft, t L LB = - = - = - =05260171 586 308 586 2 3 08 8 64.. . . . ( ) . . hours = 8.64/24 = 0.36 day Assume a month of 30 days. Then need 30/0.36 = 84 cannisters per month. Cost = $700(84) = $58,800/month Exercise 15.28 Subject: Sizing of fixed-bed adsorption unit for removal of ethyl acetate (EA) from air with activated carbon (C), using adsorption equilibrium isotherm and breakthrough data Given: Gas feed is 12,000 scfm (60oF and 1 atm) containing 0.5 mol% EA and 99.5 mol% air. Adsorbent is C (rb = 30 lb/ft3 and dp = 0.011 ft. Adsorption equilibrium data at 100oF. Breakthrough lab data at a gas superficial velocity of 60 ft/min in a 2-ft high bed at 100oF and 1 atm. Assumptions: Plug flow. Find: Diameter and height of carbon bed for a time-to-breakthrough of 8 h, for the same conditions of flow rate, temperature, and pressure as the lab data. Analysis: Use a continuity equation, similar to Eq. (6-43) to compute the bed diameter: m u A Q A D Qus s = = = =r r p or 2 4 (1) At operating conditions, Q = 12,000(560/520) = 12,900 ft3/min. us = 60 ft/min From Eq. (1), A = 12,900/60 = 215 ft2. D = [4(215)/3.14]1/2 = 16.5 ft. Use Eq. (15-122) to compute the ideal bed length, LES: LES (in ft) = = · =c Q tq A c q cqF F B F b F F F Fr ( , )(8 ) ( )( ) 12 900 60 30 215 960 (2) Want cF in lb EA/ft3 inlet gas. MW of EA = 88.1. One scf of gas contains 379 lbmol. Therefore, have 12,000/379 = 31.66 lbmol/min. The mole fraction of EA = 0.005. Concentration of EA in the entering gas of 0.005(31.66)(88.1)/12,900 = 0.00108 lb EA/ft3 gas. From the given equilibrium data, for pEA = 0.005 atm, qF = 0.270 lb EA/lb C. From Eq. (2), LES = 960 0 001080 270 385. . . = ft From Eq. (15-120), the required bed length taking into account the MTZ is, LB = LES + LUB, where LUB is given by Eq. (15-121). The breakthrough data for Le = 2 ft are plotted on the next page as yEA out against time in minutes. The data are almost straight in the mole fraction range of 0.00475 (c/cF = 0.95), where time = te = 160 minutes, to 0.00025 (c/cF = 0.05), where time = tb = 75 minutes. Therefore, the midpoint time, ts = (160 + 75)/2 = 117.5 minutes. Alternatively, ts can be evaluated by integration, using Eq. (15-123). From Eq. (15-121), LUB = (ts - tb)Le/ts = (117.5 - 75)2/117.5 = 0.72 ft. Therefore, LB = 3.85 + 0.72 = 4.57 ft. So the bed is 16.5 ft in diameter by 4.57 ft high. This is a poor bed height-to-diameter ratio. Might increase the breakthrough time by a factor of 4 to get a bed height approximately equal to the bed diameter. Exercise 15.28 (continued) Analysis: (continued) 0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035 0.0040 0.0045 0.0050 0 20 40 60 80 100 120 140 160 180 Time, minutes M o l e f r a c t i o n o f E A i n g a s e f f l u e n t Lab Breakthrough Data t b t s t e Exercise 15.29 Subject: Desorption of benzene from silica gel using air under isothermal, isobaric conditions. Given: Values are given in a table below for dimensionless benzene concentration, f, and loading, y, profiles in a bed following adsorption. Bed is 2 feet in diameter and 30 ft high. Bed contains silica gel adsorbent. Desorption by air at 1 atm and 145oF at an interstitial velocity, u, of 98.5 ft/min. Bed porosity = eb = 0.5. During desorption, mass-transfer coefficient = k = 0.206 min-1 and Henry's law equilibrium adsorption constant = 1,000 in Eq. (1) on p. 837. Assumptions: Plug flow with a constant interstitial velocity equal to the given inlet value. Negligible axial dispersion. Desorption air is passed up the bed, opposite in direction to the initial flow of benzene-air feed mixture during adsorption. Find: Desorption time required to remove 90% of the benzene from the bed. Analysis: First, compute the initial amount of benzene adsorbed in the bed. Bed volume = V = pD2L/4 = 3.14(2)2(30)/4 = 94.2 ft3 Volume of SG particles = (1-eb)V = (1-0.5)(94.2 = 47.1 ft3 During adsorption at 70oF, from Example 15.10, the concentration of benzene in the entering air- benzene mixture = 0.00181 lb benzene/ft3 = cF. The adsorbent loading in equilibrium with the feed, as given in Example 15.10, is qF* = 5,120 cF = 5,120(0.00181) = 9.27 lb benzene/ft3 of silica gel particles. If the bed were completely saturated with the feed, the benzene in the bed would be 9.27(47.1) = 437 lb benzene. If the breakthrough occurred for c just greater than zero at the given 641 minutes for adsorption (i.e. all of the benzene in the feed were adsorbed), then with a feed flow rate, from Example 15.10, of 310 ft3/min, the benzene loading would be: 310(641)(0.00181) = 360 lb benzene, which, as expected, is less than complete saturation. However, the breakthrough at 641 minutes corresponds to f = c/cF = 0.05 and not at 0. Therefore, it is best to integrate the given loading data at the end of adsorption to obtain a more accurate value of benzene loading. Using the given data on the next page, compute the average value of y by numerical integration. Note that because the flow of desorption agent is opposite to the feed gas, the bed distances below are given as z' = 30 ft - z. y y y y avg at = + + = + + = = = = 05 05 30 0 022 24 079 0 500 30 0820 0 1 29 30. . . . . . ' ' ' 'z z z z Therefore, the amount of benzene on the adsorbent in the bed at the end of the adsorption period is: 0.820(9.27)(47.1) = 358.0 lb benzene, which is very close to the above rough estimate of 360. The desorption period must be long enough to reduce the loading to (1-0.90)(358) = 35.8 lb benzene. Exercise 15.29 (continued) Analysis: (continued) Given conditions at the beginning of the desorption period: z, ft z' = 30 - z, ft f = c/cF y = q/ qF* 30 0 0.050 0.044 29 1 0.090 0.081 28 2 0.150 0.137 27 3 0.235 0.217 26 4 0.343 0.321 25 5 0.468 0.444 24 6 0.599 0.575 23 7 0.722 0.701 22 8 0.825 0.808 21 9 0.901 0.890 20 10 0.951 0.944 19 11 0.978 0.975 18 12 0.992 0.990 17 13 0.997 0.997 16 14 0.999 0.999 15 15 1.000 1.000 14 to 0 16 to 30 1.000 1.000 The desorption calculations utilize the following forms of Eqs. (15-123) and (15-124): ¶f ¶ ¶f ¶ e e f y ¶f ¶ f y ¶f ¶ f y ¶y ¶ f y t u z kK z z t b b = - - - - = - - - - = - - - = - ' . ' . . ( . )( , ) . ' . 1 98 5 1 0 5 0 5 0 206 1 000 98 5 206 0 206 (1) (2) Conditions for f and y at t = 0 are given as a function of z' in the above table. Condition for f at z' = 0 is 0 because the desorption gas contains no benzene. Equations (1) and (2) can be solved for f{t,z'} and y{t,z'} for z' = 0 to 30 ft and t > 0 by various means, including: (1) A FORTRAN code similar to that on for Example 15-13, with the method of lines and a stiff integrator in connection with Eqs. (15-127) to (15-133). (2) Use of the windows code PDESOL (www.pdesol.com), which also utilizes the method lines in connection with a number of different integrators and discretization schemes for ¶f¶z' . Exercise 15.29 (continued) Analysis: (continued) (3) Use of a spreadsheet with a finite-difference method where backward differences are used for ¶f ¶z' and an Euler predictor-corrector method is used for the time derivatives, as discussed by D. D. Fry, Chem. Eng. Education, Fall 1990, pp 204-207. The PDESOL program is particularly convenient and easy to use, and was applied to this exercise. The solution method selected from the many options provided by PDESOL was the same as that described for TSA. Thus, the method of lines was used with Eqs. (15-127) and (15- 128). The ¶f¶z' derivative was approximated by a 5-point, biased, upwind, finite-difference approximation, and a stiff integrator, LODES, was used on the resulting system of ODEs with time derivatives of f and y. To obtain accuracy, 121 grid points were used over the 30 ft length for the spatial domain in z'. For the time domain, an arbitrary 800 minutes was chosen for the final time of desorption, with an output as a function of z' every 50 minutes. The initial conditions for f and y, given in the table above were inputted as text files, which were interpolated by PDESOL to obtain values for the 121 grid points. The resulting PDE program equations were as follows, where the symbols f and y were replaced by c and q, respectfully, and z' was replaced by x, with xL =0 and xU = 30 ft. c_x = dxu(c,1) 'Uses the 5-point, biased, upwind finite difference approximation c_t = -98.5*c_x - 206*(c - q) q_t = 0.206*(c - q) c@xL = 0 c@t0 = lintc("c_init.txt",x) 'Calls the file c_init.txt for the initial values of c q@t0 = lintc("q_init.txt",x) 'Calls the file q_init.txt for the initial values of q The form of the two .txt files were as follows, using the above table: c_init.txt q_init.txt x c at t = 0 x q at t = 0 0 0.050 0 0.044 1 0.090 1 0.081 2 0.150 2 0.137 3 0.235 3 0.217 4 0.343 4 0.321 etc. etc. etc. etc. etc. etc. etc. etc. 15 1.000 15 1.000 30 1.000 30 1.000 Exercise 15.29 (continued) Analysis: (continued) The results of the calculations are given in the following tables and graphs, in increments of 50 minutes, where by 400 minutes the fronts have almost left the bed, indicating that the desorption time is less than 400 minutes. Exercise 15.29 (continued) Analysis: (continued) Profiles for y = q/qF* during desorption: Exercise 15.29 (continued) Analysis: (continued) Plots of profiles for y = q/qF* and profiles for f = c/cF during desorption: Exercise 15.29 (continued) Analysis: (continued) To determine from the above profiles the time for 90% of the benzene to be desorbed, compute the average value of loading, y, for each of the desorption times, by numerical integration, using, y y y y avg at = + += = = 05 05 30 0 1 29 30. .' ' ' 'z z z z Then, from above, the loading in lb benzene = yavg (9.27 lb benzene at sat'n/ft3 )(47.1ft3) From above, the desired value is 35.8 lb benzene. Using the above table of loadings as a function of time and location in the bed, the following values for benzene loading are obtained: Desorption time, minutes Lb benzene not desorbed % of benzene desorbed 0 358.0 0.0 50 286.5 20.0 100 214.9 40.0 150 144.0 59.8 200 79.4 22.2 250 33.0 90.8 300 9.8 97.3 350 2.1 99.4 400 0.3 99.9 From the above results, the required desorption time is almost 250 minutes. Exercise 15.30 Subject: Pressure-swing adsorption cycles to approach a cyclic steady state Given: Design basis in Example 15.13. Assumptions: Plug flow with a constant interstitial velocity equal to the given inlet value. Negligible axial dispersion. Find: PSA cycle results that approach the cyclic steady state starting from: (a) a clean bed. (b) a bed saturated with the feed. Are the two cyclic steady states essentially the same? Analysis: Mole fraction of DMMP in the feed gas = 236/1,000,000 = 0.000236 Feed gas superficial velocity =us = ueb = 10.465(0.43) = 4.5 cm/s Cross-sectional area of bed = A = 3.14(1.1)2/4 = 0.95 cm2 Volumetric feed gas flow rate = Q = us A = 4.5(0.95) = 4.275 cm3/s From the ideal gas law at 3.06 atm and 294 K, molar feed gas flow rate = n = PQ/RT = (3.06)(4.275)/[(82.06)(294)] = 0.000542 mol/s Because the feed is so dilute, the product gas is almost equal to the feed gas. Therefore, for the desorption step, the gas used for 20 minutes will be pure air at 41.6% of that for the adsorption step or (0.416)( 0.000542)(60)(20) = 0.270 mol. This will be used for 20 min at a flow rate of 0.270/[(20)(60)] = 0.000225 mol/s At the conditions of desorption, 1.07 atm and 294 K, Q = nRT/P = 0.000225(82.06)(294)/1.07 = 5.073 cm3/s Superficial velocity for desorption = 5.073/0.95 = 5.34 cm/s Interstitial velocity for desorption = 5.34/0.43 = 12.42 cm/s Convert the adsorption isotherm into the form consistent with the following equations, which are modifications of Eqs. (15-98) and (15-105), in terms of the gas concentration and loading of DMMP: ¶ ¶ ¶ ¶ e e c t u c z k q q b b = - - - -1 * ¶ ¶ q t k q q= -* where, the units are: c in g/cm3 of gas t in minutes u in cm/min k in min-1 q in g/cm3 of particles Exercise 15.30 (continued) Analysis: (continued) The given Langmuir isotherm is: q pp* , ,= +48 3601 98 700 , where q* is in g/g of adsorbent and p is partial pressure in atm. This isotherm must be converted to units of g/cm3 of particles for q* and the variable c in g/cm3 of gas must be substituted for the partial pressure in atm. First consider q*. The bed volume = 0.95(12.80) = 12.16 cm3 The particle volume = (1-0.43)(12.16) = 6.93 cm3 Particle density = 5.25/6.93 = 0.758 g/cm3 of particles The molecular weight of DMMP, dimethyl methylphosphonate, CH3PO(OCH3)2 = M = 124.08 Assuming the ideal gas law, p = cRT/M = c (82.06)(294)/124.08 = 194.4 c Therefore, the Langmuir isotherm becomes: q c c c c* . , ( . ), ( . ) . .= + = ·+ ·0 738 48 360 194 41 98 700 194 4 6 94 101 1919 10 6 6 , where q* is in g/cm 3 of particles and c is g/cm3 The initial conditions for adsorption step of the first cycle are: c = 0 for t = 0 and q = 0 at t = 0 The boundary condition for all adsorption steps is: cF = yFPM/RT = (0.000236)(3.06)(124.08)/[(82.06)(294)] = 3.72 x 10-6 g/cm3 at z = 0. The cycle calculations are conveniently carried out by a modification of the FORTRAN program in Table 15.9, based on using the method of lines to obtain a set of ODEs, similar to the equations for TSA, with a stiff integrator, such as LSODES. The program in Table 15.9, which is for TSA, consists of: Main program to: Set initial parameters. Call LSODE Perform cycles for: Adsorption step Desorption step Write cycle results Subroutine FEX to: Compute derivatives of the set of ODEs For PSA, the program must add: Depressurization step Repressurization step Exercise 15.31 Subject: Three cycles of vacuum-swing adsorption. Given: Design basis in Example 15.13, except that PL = 0.12 atm and interstitial velocity during the desorption step corresponds to 44.5% of the product gas from the adsorption step. Assumptions: Plug flow with a constant interstitial velocity equal to the given inlet value. Negligible axial dispersion. Find: Result of the third VSA cycle starting from a clean bed. Analysis: Mole fraction of DMMP in the feed gas = 236/1,000,000 = 0.000236 Feed gas superficial velocity =us = ueb = 10.465(0.43) = 4.5 cm/s Cross-sectional area of bed = A = 3.14(1.1)2/4 = 0.95 cm2 Volumetric feed gas flow rate = Q = us A = 4.5(0.95) = 4.275 cm3/s From the ideal gas law at 3.06 atm and 294 K, molar feed gas flow rate = n = PQ/RT = (3.06)(4.275)/[(82.06)(294)] = 0.000542 mol/s Because the feed is so dilute, the product gas is almost equal to the feed gas. Therefore, for the desorption step, the gas used for 20 minutes will be pure air 44.5% of that for the adsorption step or (0.445)(0.000542)(60)(20) = 0.289 mol. This will be used for 20 min at a flow rate of 0.289/[(20)(60)] = 0.000241 mol/s At the conditions of desorption, 0.12 atm and 294 K, Q = nRT/P = 0.000241(82.06)(294)/0.12 = 48.45 cm3/s Superficial velocity for desorption = 48.45/0.95 = 51.0 cm/s Interstitial velocity for desorption = 51.0/0.43 = 118.6 cm/s Convert the adsorption isotherm into the form consistent with the following equations, which are modifications of Eqs. (15-98) and (15-105), in terms of the gas concentration and loading of DMMP: ¶ ¶ ¶ ¶ e e c t u c z k q q b b = - - - -1 * ¶ ¶ q t k q q= -* where, the units are: c in g/cm3 of gas t in minutes u in cm/min k in min-1 q in g/cm3 of particles Exercise 15.31 (continued) Analysis: (continued) The given Langmuir isotherm is: q pp* , ,= +48 3601 98 700 , where q* is in g/g of adsorbent and p is partial pressure in atm. This isotherm must be converted to units of g/cm3 of particles for q* and the variable c in g/cm3 of gas must be substituted for the partial pressure in atm. First consider q*. The bed volume = 0.95(12.80) = 12.16 cm3 The particle volume = (1-0.43)(12.16) = 6.93 cm3 Particle density = 5.25/6.93 = 0.758 g/cm3 of particles The molecular weight of DMMP, dimethyl methylphosphonate, CH3PO(OCH3)2 = M = 124.08 Assuming the ideal gas law, p = cRT/M = c (82.06)(294)/124.08 = 194.4 c Therefore, the Langmuir isotherm becomes: q c c c c* . , ( . ), ( . ) . .= + = ·+ ·0 738 48 360 194 41 98 700 194 4 6 94 101 1919 10 6 6 , where q* is in g/cm 3 of particles and c is g/cm3 The initial conditions for adsorption step of the first cycle are: c = 0 for t = 0 and q = 0 at t = 0 The boundary condition for all adsorption steps is: cF = yFPM/RT = (0.000236)(3.06)(124.08)/[(82.06)(294)] = 3.72 x 10-6 g/cm3 at z = 0. The cycle calculations are conveniently carried out by a modification of the FORTRAN program in Table 15.9, based on using the method of lines to obtain a set of ODEs, similar to the equations for TSA, with a stiff integrator, such as LSODES. The program in Table 15.9, which is for TSA, consists of: Main program to: Set initial parameters. Call LSODE Perform cycles for: Adsorption step Desorption step Write cycle results Subroutine FEX to: Compute derivatives of the set of ODEs For VSA, the program must add: Depressurization step Repressurization step Exercise 15.32 Subject: Model equations for the separation of air by PSA Given: Feed of air to be separated into oxygen-rich and nitrogen-rich products in fixed isothermal, adiabatic beds, with mass-transfer resistances and extended Langmuir isotherms. Assumptions: Constant fluid velocity and negligible axial dispersion. Find: Model equations for the two main steps and a numerical procedure for solving the equations. Analysis: Let subscripts X = oxygen and N = nitrogen. Species Mass Balances: From a modification of Eq. (15-102), u cz ct qt u cz ct qt X X b b X N N b b N ¶ ¶ ¶ ¶ e e ¶ ¶ ¶ ¶ ¶ ¶ e e ¶ ¶ + + - = + + - = 1 0 1 0 Total Mass Balance: ¶ ¶ e e ¶ ¶ ¶ ¶ c t q t q t t b b X N+ - + =1 0 where, ct = cX + cN Component Mass-Transfer Rates: From Eq. (105) for the LDF model, ¶ ¶ ¶ ¶ q t k q q q t k q q X X X X N N N N = - = - * * Extended Langmuir Isotherms: q q K cK c K c q q K cK c K c X m X X X X N N N m N N X X N N X N * * = + + = + + 1 1 Exercise 15.32 (continued) Analysis: (continued) Boundary conditions for starting adsorption with a clean bed: q z c z q z c z c t c c t c X X N N X F N F X N , , , , , , 0 0 0 0 0 0 0 0 = = = = = = Boundary conditions for desorption are the conditions in the bed at the end of adsorption and a feed containing purge gas. The equations can be solved by the Method of Lines in a manner similar to that described on for TSA, letting: f y i i F i i F c c q q i i = = / / * Exercise 15.33 Subject: Descriptions and possible uses of cycling operations of (1) parametric pumping and (2) cycling zone adsorption. Given: Reference to an article by Sweed, based on inventions by Wilhelm and by Pigford and their co-workers. Find: Detailed descriptions of (1) parametric pumping and (2) cycling zone adsorption. Can either or both be used for gas-phase and liquid-phase adsorption? Analysis: Parametric Pumping: The separation of binary liquid or gas mixtures by parametric pumping was first proposed, analyzed, and substantiated by experiments in articles by R. H. Wilhelm, A. W. Rice, and A. R. Bendelius, I&EC Fundamentals, 5, 141-144 (1966), and R. H. Wilhelm, A. W. Rice, R. W. Rolke, and N. H. Sweed, I&EC Fundamentals, 7, 337-349 (1968). The technique is based on differences in the extent of equilibrium adsorption of the two components in the feed mixture, and a significant decrease in equilibrium adsorption as the temperature is increased. Two different basic equipment arrangements and operations have been proposed, both in a closed- system operation, as shown in the sketches on the next page. Both consist of a fixed-bed column packed with adsorbent particles, a bottom piston-reservoir, and a top piston-reservoir. In a direct mode, the column is jacketed so that the column may be alternately heated or cooled. In a recuperative mode, heat exchangers (one for heating and one for cooling) are provided at opposite ends of an adiabatically operated column. Consider the direct mode, in the context of the example of the experiment reported in the second literature reference above. The column, at ambient temperature, T1, and packed with silica gel particles, is charged with a liquid feed mixture of 20 vol% toluene and 80 vol% n- heptane and brought to equilibrium. The bottom reservoir is also filled with the same feed mixture, but the top reservoir is initially empty. The toluene is more strongly adsorbed than the n-heptane. The column is heated by the jacket to temperature T2 and the fluid in the bottom reservoir is pushed into the column by upword movement of the driving piston. At the same time, fluid is drawn into the upper reservoir as the driven piston is forced upward. Because the capacity of the adsorbent is less at the higher temperature and because the adsorbent is more selective for the toluene, the fluid entering the upper reservoir is more concentrated in toluene. This constitutes the first heating half cycle. The direction of fluid flow is now reversed for the first cooling half cycle. The column is cooled to T1 by the jacket and the top piston becomes the driving piston, pushing the fluid from the top reservoir down into the column. At the lower temperature, the toluene is more selectively adsorbed so that the fluid entering the bottom reservoir is richer in n-heptane. Thus, after one cycle, a partial separation is achieved. Further separation occurs by conducting additional cycles, perhaps 50, after which the reservoirs are emptied into product receivers. Experimental results indicate that although the toluene is almost completely removed from the n-heptane, a nearly pure toluene is not achieved (i.e. high recovery, but not a high purity). In the recuperative mode, the top of the column is maintained at a hot temperature, T2, and the bottom is maintained at a cooler temperature, T1. Again the fluid is pushed first to the top and then to the bottom for a number of cycles to achieve a separation. Exercise 15.33 (continued) Analysis: (continued) Exercise 15.33 (continued) Analysis: (continued) Cycling Zone Adsorption: The separation of binary liquid or gas mixtures by cycling zone adsorption was first proposed, analyzed, and substantiated by experiments in two articles by R. L. Pigford, B. Baker, and D.E. Blum, I&EC Fundamentals, 8, 848 -851 (1969) and I&EC Fundamentals, 8, 144 (1969). As with parametric pumping, the technique uses fixed beds and is based on differences in the extent of equilibrium adsorption of the two components (A and B) in the feed mixture, and a significant decrease in equilibrium adsorption as the temperature is increased. However, while parametric pumping involves periodic reversals of flow direction and temperature, cycling zone adsorption involves only reversals in temperature. Unlike other fixed-bed schemes, batch-type regeneration of the adsorbent is not required. The first sketch on the following page shows the basic idea of cycling zone adsorption. Feed fluid of a fixed composition enters a jacketed fixed bed, maintained at low temperature, TC, causing component A to be adsorbed to a greater extent than B. At breakthrough, the composition of the exiting fluid approaches that of the feed and the bed temperature is raised to TH , which causes the adsorbate, rich in component A, to be expelled, raising the concentration of A in the exiting fluid. After a time, the effluent composition again approaches that of the feed. To continuously produce two products, one richer in A, yh, than the feed, and the other leaner in A than the feed, yl, two fixed beds are employed, as shown in the second sketch on the following page. One bed is exactly one-half cycle out of phase with the other. The output of each column is periodically switched, but the degree of separation is limited. The degree of separation is significantly improved by employing two or more beds, called zones, in series, as in the third sketch on the following page. Compared to parametric pumping, theoretical analysis predicts that n cycling zones is equivalent to n parametric pumping cycles. Thus, much more equipment is required for cycling zone adsorption than for parametric pumping. However, with the former, products are produced continuously, instead of in batches. A further simplification in operation is achieved by replacing the jacketed beds by heat exchangers in the manner of the two recuperative mode in parametric pumping, as shown in the fourth sketch. In the review article by N. H. Sweed, AIChE Symp. Series, 80, No. 233, 44-53 (1984), parametric pumping and cycling zone adsorption, after more than 15 years following their invention, are found to have remained laboratory curiosities, despite their early promise. This is probably due to difficulties in swinging the temperature, and the popularity of PSA for gas. Exercise 15.33 (continued) Analysis: (continued) Exercise 15.34 Subject: Separation of propylene from propane by continuous, countercurrent adsorption. Given: Feed gas mixture of 55 mol% propane (C3) and 45 mol% propylene (C3=). Continuous, countercurrent adsorption at 25oC and 1 atm with silica gel. Equilibrium data in Exercise 15.9. Find: Using the McCabe-Thiele method, (a) Adsorbent flow rate per 1,000 m3 of gas if 1.2 times the minimum flow rate is used. (b) Number of theoretical stages needed. Analysis: First compute the material balance on the gas, taking a basis of one hour. From the ideal gas law, the gas feed rate is F PRT= = =u ( )( , )( )(82. )( ) ,1 1 000 1006 298 40 900 6 mol / h of gas This gas contains 0.45(40,900) = 18,400 mol/h C3= and 40,900 - 18,400 = 22,500 mol/h C3. Assume a configuration analogous to that for liquid-liquid extraction with reflux, as shown in Fig. 8.26b. Entering solvent, SB, becomes the adsorbent: raffinate, R, becomes the propane-rich product; Stream D + LR is desorbed from the adsorbent in the desorber (analogous to the Solvent removal step in extraction), with D becoming the propylene-rich product and LR, becoming the Extract reflux. The upward movement of adsorbent and adsorbate is V and the downward movement of non-adsorbed gas is L. Other configurations are possible, but the one just described is used here. From the experimental equilibrium data of Exercise 15.9, assume that at 1atm, the total adsorbent loading remains constant at a value of 2.0 mmol/g = 2.0 mol of combined C3 and C3= adsorbed at equilibrium per kg of adsorbent. Also, in Exercise 15.9, although the mixture data scatter, a relative selectivity = aC3=,C3 = (yC3=/xC3=)/(yC3/xC3) = approximately 2.5, with mole fractions y in the adsorbate an mole fractions x in the non-adsorbed gas. With these two assumptions, a relatively simple McCabe-Thiele method is used to solve this exercise. Because the flow rate of adsorbent is constant throughout the stages, both below and above the gas feed entry, let V = adsorbate (adsorbent-free) flow rate. Then, because of the assumption of constant loading, V is constant throughout the stages. The non-adsorbed gas flow rate, L, is constant above the feed at a flow rate of LR;. below the feed, it is R. Thus, on a McCabe-Thiele diagram, the operating lines have constant slopes. By a C3 material balance around the system, followed by a total material balance, Fx y D x B D BF D B C3 C3 C3 = = = + = +40 900 0 55 22 500 010 0 90, ( . ) , . . (1) F = 40,900 = D + B (2) Solving Eqs (1) and (2), D = 17,900 mol/h and B = 23,000 mol/h Exercise 15.34 (continued) Analysis: (continued) The McCabe-Thiele diagram is shown on the following page in mole fractions of C3=, where the equilibrium curve is computed from Eq. (7-3), y = ax/[1+x(a-1)]. The q-line is vertical, because the adsorbent that passes over the feed stage is already loaded with adsorbate, so that no change in the adsorbate flow rate, V, occurs. To determine the minimum adsorbent flow rate, the usual minimum reflux construction, shown below, is made (corresponding to an infinite number of stages), with an operating line that starts at y = x = 0.90 and terminating at the intersection with the equilibrium line and the q-line x = 0.45 and x = 0.45 and y = 0.6716 from the equilibrium equation. Thus, the slope of this operating line = (0.90-0.6716)/(0.9-0.45) = 0.5706. The equation for this operating line is obtained from a C3= material balance around a portion of the upper section of stages, from the top stage (but below the desorber) to an intermediate stage above the feed, in a manner similar to that for distillation on p. 363, where S is the adsorbent flow rate in kg/h and 2S is the equilibrium loading of adsorbate in mol/h, 2 2 2 2 Sy L x Sy L x y LS x LS x y R D D R R R D D C3= C3= C3= C3= C3= C3= C3= C3= Solving, + = + = - + Thus, the slope of the operating line = LR/2S = 0.5706. Therefore, LR = 1.0152 S. Also, V = 2S and by adsorbate material balance around the desorber and divider, V = LR + D = LR + 17,900. Solving these three equations for the minimum adsorbent condition, S = Smin = 18,200 kg/h, V = 2(18,200) = 36,400 mol/h, and LR = 1.0152(18,200) = 18,500 mol/h. (a) For an adsorbent rate = 1.2 Smin, S = 1.2(18,200) = 21,850 kg/h. Then, V = 2S = 2(21,850) = 43,700 mol/h and LR = V - 17,900 = 43,700 - 17,900 = 25,800 mol/h. The slope of the resulting operating line = LR/2S = 25,800/[2(21,850)] = 0.590. (b) This operating line, together with resulting operating line for the lower section is shown in the second McCabe-Thiele diagram below, where 10 equilibrium stages are stepped for the optimal feed-stage location. Exercise 15.34 (continued) Analysis: (continued) McCabe-Thiele diagram for minimum adsorbent rate 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 x, mole fraction of C3= in the gas y , m o l e f r a c t i o n o f C 3 = i n t h e a d s o r b a t e Exercise 15.34 (continued) Analysis: (continued) McCabe-Thiele diagram for adsorbent rate Equal to 1.2 times minimum value 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 x, mole fraction of C3= in the gas y , m o l e f r a c t i o n o f C 3 = i n t h e a d s o r b a t e Exercise 15.35 Subject: Softening of hard water in a fixed-bed ion exchange column Given: Feed at 25oC containing 400 ppm (by wt) CaCl2 and 50 ppm (by wt) NaCl. Bed of 8.5 ft diameter by 10 ft high of gel resin with a cation capacity of 2.3 eq/L of bed volume. Wetted resin void fraction = eb = 0.38. 15 gal/min-ft2 for loading; 1.5 gal/min-ft2 for displacement, washing , and regeneration. Displacement and regeneration solutions are aq. saturated NaCl (26 wt%). Assumption: Because of dilute conditions, feed solution contains 1,000 g H2O/L Find: (a) Feed solution flow rate, L/min. (b) Loading time to breakthrough, h (c) Loading wave front velocity, cm/min (d) Regeneration solution flow rate, L/min (e) Displacement time, h (f) Additional time for regeneration, h Analysis: M of CaCl2 = 110.99 M of NaCl = 58.45 Concentration of CaCl2 = 400(1000)/[110.99(1,000,000)] = 0.00360 mol/L = 0.00720 eq/L Concentration of NaCl = 50(1000)/[58.45(1,000,000)] = 0.000855 mol/L = 0.000855 eq/L (a) Bed cross-sectional area = 3.14(8.5)2/4 = 56.7 ft2 Feed solution flow rate = 15(56.7) = 851 gpm = 3,219 L/min (b) Behind the loading wave front in the feed, equivalent fraction of Ca2+ Ca2+= = + =x 0 0072 0 0072 0 000855 0 9883. / ( . . ) . Since no NaCl in the feed is exchanged, CL = 0.0072 eq/L Q = 2.3 eq/L From Table 15.5, KCa Na2+ +, . / .= =5 2 2 2 6 From Eq. (15-181), 2 6 2 30 0072 830 6 1 0 89380 8938 1 01188 1. .. . .. . = = - - = -y y y yCa Ca Ca Ca 2+ 2+ 2+ 2+ Solving, yCa2+ = 0.99986 equivalent fraction Bed volume = (56.7)(10) = 567 ft3 or 16,060 L Total Bed Capacity = 2.3 (16,060) = 36,940 eq Ca2+ absorbed by resin = 0.99986(36,940) = 36,935 eq Ca2+ entering bed in feed solution = 0.0072(3,219) = 23.18 eq/min Ideal loading time to breakthrough = 36,93523.18 1,593 min 26.6 hLt = == Exercise 15.35 (continued) Analysis: (continued) (c) Loading wave front velocity = uL = L/tL= 10/1,593 = 0.00628 ft/min or 0.191 cm/min (d) Flow rate of regeneration solution = 1515 3 219. ( , ) = 321.9 L/min (e) Displacement time = time for 321.9 L/min to displace liquid in the voids Void volume = 0.38 (16,060) = 6,103 L Displacement time tD = 6103321.9 = 19 min (f) For a 26 wt% NaCl solution at 25°C, the density from Perry's Chemical Engineers' Handbook = 1.19443 g/cm3 Flow rate of Na+ in regeneration solution = 3219 1 000 119443 0 2658 45. ( , )( . )( . ). = 1,710 eq/min NaCl concentration in regenerating solution = , . 1 7103219 = 5.31 eq/L = cR = C in Eq. (15-139) From Eq. (15-181), noting conditions in Fig. 15.50b: K Qc R Ca Na2 2 6 2 3 5 31 1126+ + = = , . . . . Unfortunately, this is > 1. Therefore, equilibrium is not as favorable as desired. Also the regeneration wave may not sharpen quickly. Nevertheless, assume a shock wave-like front. From Eq. (15-181): 1126 0 99986 11 0 99986 7 142 12 2 2 2 . . . , * * * *= - - = -+ + + + x x x x Ca Ca Ca Ca Solving: xCa2 0 99984+ =* . So downstream of the regeneration wave front, but upstream of the displacement wave front, the liquid contains very few sodium ions. However, this will not be quite so because the wave will not be sharp. Exercise 15.36 Subject: Separation of glutamic acid, glycine, and valine, in an aqueous solution, by a fixed-bed chromatographic column, using equilibrium theory. Given: Aqueous solution, buffered to a 3.4 pH by sodium citrate and containing 20 mol/m3 each of glutamic acid (GA), glycine (G), and valine (V). Chromatographic column, packed with Dowex 50W-X8 in the sodium form to a depth of 470 mm = 47 cm. Resin is 0.07 mm in diameter and is packed to a void fraction, eb , of 0.374. All three solutes follow Henry's law, q = Kc, where the values of K are 1.18 for GA, 1.74 for G, and 2.64 for V. Superficial solution velocity = us = 0.025 cm/s. Assumptions: Equilibrium theory. Find: Pulse duration to achieve complete separation. Time duration of the elution step before the second pulse begins. Analysis: Interstitial velocity = u = us/eb = 0.025/0.374 = 0.0668 cm/s. From Eq. (15-183), solute wave velocity = u u K K K i b b i i i = + - = + - = + 1 1 0 0668 1 1 0 3740 374 0 0668 1 1674e e . . . . . Solute Ki ui , cm/s GA 1.18 0.0225 G 1.74 0.0171 V 2.64 0.0123 From the K values, V is the most strongly adsorbed. Therefore, the solutes will leave the column in the order of GA first, then G, and V last. First, assume that the separation between GA and G controls, as analogous to Fig. 15.49, where 182 cm is replaced by 47 cm. Thus, at the end of the column, the trailing edge of GA will coincide with the leading edge of G. From Example 15.20, 47 47 1 1 Solving, 47 660 s 0.0225 0.0171 0.0171 0.0225P Pt t + = = - = Thus, the time for the trailing edge of the GA wave to reach the end of the column at 47 cm = tP + 47/0.0225 = 660 + 2,089 = 2,749 s. At that time, will the G and V waves be separated? The trailing edge of the G wave will be at (2,089)(0.0171) = 35.7 cm, while the leading edge of the V wave will be at (2,749)(0.0123) = 33.8 cm. Therefore, the G and V waves will be separated. Now assume that the separation between G and V controls. Thus, at the end of the column, the trailing edge of G will coincide with the leading edge of V. From Example 15.20, t tP P+ = = - =470 0171 470 0123 47 10 0123 10 0171 1 073. . . . , Solving, s Exercise 15.36 (continued) Analysis: (continued) Thus, the time for the trailing edge of the G wave to reach the end of the column at 47 cm = = tP + 47/0.0171 = 1,073 + 2,749 = 3,822 s. At that time, will the G and GA waves be separated? The leading edge of the G wave will be at a hypothetical distance outside of the column of (3,822)(0.0171) = 65.4 cm. The trailing edge of the GA wave will be at a hypothetical distance outside of the column at 2,749(0.0225) = 61.9 cm. Thus, the G and GA waves are not separated. Therefore, the pulse duration = 660 seconds. To determine the elution time, compute the time for the trailing edge of the slow V wave to reach 47 cm. This time = 660 + 47/0.0123 = 4,481 s. The time for the leading edge of the second pulse of GA to reach 47 cm so that V and GA are just separated = 47/0.0225 = 2,089 s. The difference is 4,481 - 2,089 = 2,392 s before the second pulse starts. But 660 s of this is the first pulse. Therefore, the elution time = 2,392 - 660 = 1,732 s. Thus, the ideal cycle is: Pulse: 660 s Elute: 1,732 s Pulse: 660 s Elute: 1,732 s etc. Exercise 15.37 Subject: Separation of glutamic acid, glycine, and valine, in an aqueous solution, by a fixed-bed chromatographic column, accounting for mass transfer resistances. Given: Aqueous solution, buffered to a 3.4 pH by sodium citrate and containing 20 mol/m3 each of glutamic acid (GA), glycine (G), and valine (V). Chromatographic column, packed with Dowex 50W-X8 in the sodium form to a depth of 470 mm = 47 cm. Resin is 0.07 mm in diameter and is packed to a void fraction, eb , of 0.374. All three solutes follow Henry's law, q = Kc, where the values of K are given below. Superficial solution velocity = us = 0.025 cm/s. Effective diffusivities of the three solutes are given below. External mass-transfer coefficient = 1.5 x 10-3 cm/s for each solute. Assumptions: Application of Carta's equation, which accounts for mass transfer. Find: A cycle of feed pulses and elution periods that result in the separation of the three solutes. Analysis: Interstitial velocity = u = us/eb = 0.025/0.374 = 0.0668 cm/s. From Eq. (15-140), the solute wave velocities, ui, can be computed for eb = 0.374 and the following values of K, giving: Solute K De , cm2 /s ui, cm/s GA 1.18 1.94 x 10-7 0.0225 G 1.74 4.07 x 10-7 0.0171 V 2.64 3.58 x 10-7 0.0123 Thus, the GA wave will leave first, followed by G and then V. From the results of Exercise 15.36 for the equilibrium theory, the pulse time is 660 s and the elution time is 1732 s. It is expected that with mass transfer taken into account, a shorter pulse time and a longer elution time will be required to avoid significant overlap of the solute waves. Calculate the fixed solute parameters in Carta's equation. Particle radius = Rp = 0.07/2 = 0.035 mm = 0.00 35 cm From Eq. (15-154), the overall mass-transfer coefficient = k R k R D D D p c p e e e = + = · + = + ·- - 1 3 15 1 0 0035 3 15 10 0 0035 15 1 0 778 8167 10 2 3 2 7. ( . ) . . . (1) From Eq. (15-194), with eb = 0.374, b ee= - = - =b b K K K1 0 374 1 0 374 0 597 . ( . ) . (2) From Eq. (15-192), with z = 47 cm, n kzu k kf b b = - = - =1 1 0 374 470 374 0 0668 1178ee ( . ) ( ). ( . ) , (3) Exercise 15.37 (continued) Analysis: (continued) Calculations with the given values of K and De, and Eqs. (1), (2), and (3) give: Solute k, s-1 k/K, s-1 b nf GA 0.2005 0.170 0.506 236 G 0.359 0.206 0.343 423 V 0.327 0.124 0.226 385 Now, we must choose a pulse time, tF, and elution time, tE. Assume a pulse time that is 50% of that computed by the equilibrium theory of Exercise 15.36, namely 0.5(660) = 330 s. For now, leave the elution time at 1732 s because it does not affect the first set of solute waves. Compute values of c/cF as a function of time using Carta's infinite-series equation (15-46) using a spreadsheet, being careful to include a sufficient number of terms in the series to obtain a converged result. In this exercise, that number in Eq. (15-46) is not more than m = 20. The result is that with tF = 330 s, the GA and G waves overlap significantly at z = 47 cm. Try tF = 0.5(330) = 165 cm. Now, as seen in the table on the next page and the plot on the following page, only a small overlap of the GA and G waves occurs. An even smaller overlap of the G and V waves is evident. However, although not shown, a significant overlap of the first V wave and the second GA wave occurs with tE = 1,732 s. Therefore, we must increase the value of tE. If it is increased to 2,732 s, the results are acceptable, as shown in the table and plot below, where the first set of solute waves and the second wave for GA are shown. Thus, a possible cycle, taking into account mass transfer is: Pulse: 165 s Elute: 2,732 s Pulse: 165 s Elute: 2,732 s etc. Because of the short pulse time and the long elution time, this is probably not a desirable cycle. Exercise 15.37 (continued) Analysis: (continued) Z = 47 cm, tF = 165 s, tE = 2,732 s __ Exercise 15.37 (continued) Analysis: (continued) Exercise 15.38 Subject: Separation of a dilute feed of 3-phenyl-1-propanol (A) and 2-phenyl ethanol (B) in a methanol-water mixture in a 4-section laboratory simulated moving bed. Given: A liquid feed rate of 0.16 mL/min containing 0.091 g/L of A and 0.115 g/L of B, with the balance being a 60/40 wt% methanol-water mixture. Henry?s law applies, in the form, qi = Kici , with KA = 2.36 and KB = 1.40 for the given adsorbent, q and c are concentrations per unit volume. Assume that neither methanol nor water adsorb. External void fraction of the adsorbent beds = eb = 0.572. Switching time = 10 minutes. Find: Using the steady-state, local-composition TMB model for a perfect separation of A from B with a margin, b, of 1.15, estimate initial values for the volumetric flow rates of the extract (E), raffinate (R), desorbent (D), circulation (C), and the solid particles (S). Use those values to determine the recirculation rate for the SMB and the resulting volumetric liquid flow rates in each of the four sections. Analysis: Carry out the calculations using volumetric flow rates as shown in Figure 15.45. Because, KB < KA , B will appear in the raffinate and A will appear in the extract. First make calculations for a TMB. Using (15-149) to (15-151), with QF = 0.16 mL/min, ( )A B 0.16 2.36 1.40 1.15 1.1 0.3 5 62S FQK K Q = = - b - = b mL/min ( ) ( )A B 0.362 2.36 1.40 0.401 0.15SE Q K KQ == - b = - mL/min ( ) ( )A B 0.362 2.36 1.40 0.30 1 2.15 S R Q K KQ =- -= = b mL/min Using (15-154), 0.400 0. 0.5302 0.160 42FD E RQ Q QQ = + - = + =- mL/min Using (15-153), ( )A 0.362 2.36 (1.15) 0.4420.542S DC Q K QQ = b - = - = mL/min Now, make calculations for the SMB, where the solid particles do not flow down through the unit, but are in stationary beds. Referrring to Figure 15.45, the volumetric flow rate in section I of the TMB is, ( )I TMB 0.442 0.54 0. 82 9 4C DQQ Q= + = =+ mL/min Exercise 15.38 (continued) Analysis: (continued) From (15-175), ( ) ( ) ( )I TMBI MBS B TM 0.5720.984 0.3621 1 0.572 1.468b S b Q QQ e = + = + - e - = mL/min Then, ( )II SMBQ = ( )I SMBQ - QE = 1.468 ? 0.400 = 1.068 mL/min ( )III SMBQ = ( )II SMBQ + QF = 1.068 + 0.160 = 1.228 mL/min ( )IV SMBQ = ( )III SMBQ - QR = 1.228 ? 0.302 = 0.926 mL/min Exercise 15.39 Subject: Effect of mass-transfer coefficients on the steady-state TMB results of Example 15.17 for the separation of fructose from glucose. Given: Data in Examples 15.16 and 15.17. Find: The compositions of the extract and raffinate. Analysis: In Example 15.17, the component mass-transfer coefficients (MTC) for A and B are 10 min-1. Using Aspen Chromatography, increase the value to 1000 min-1. The following results are obtained, compared to those from Example 15.17. Concentrations in g/L Component Extract MTC = 10 min-1 Extract MTC = 1000 min-1 Raffinate MTC = 10 min-1 Raffinate MTC = 1000 min-1 Fructose 211.6 223.2 12.7 3.9 Glucose 8.4 2.6 295.3 303.3 Water 861.7 858.9 795.8 796.1 Some improvement in the separation is noted. Additional improvement may be possible by altering the switching time, the fluid circulation flow rate, and the extract and raffinate flow rates. J. Seader Microsoft Word - Chapter 15 Exercises.doc