PROBLEMS

12.1-1.

Equilibrium Isotherm for Glucose Adsorption. Equilibrium isotherm data for adsorption of glucose from an aqueous solution to activated alumina are as follows (H3):

C (g/cm3)0.00400.00870.0190.0270.0940.195
q (g solute/g alumina)0.0260.0530.0750.0820.1230.129

Determine the isotherm that fits the data and give the constants of the equation using the given units.
A1: Ans. Langmuir isotherm, q = 0.145c/(0.0174 + c)
12.2-1.

Batch Adsorption for Phenol Solution. A wastewater solution having a volume of 2.5 m3 contains 0.25 kg phenol/m3 of solution. This solution is mixed thoroughly in a batch process with 3.0 kg of granular activated carbon until equilibrium is reached. Use the isotherm from Example 12.2-1 and calculate the final equilibrium values and the percent phenol extracted.

12.3-1.

Scale-Up of Laboratory Adsorption-Column Data. Using the break-point time and other results from Example 12.3-1, do as follows:

  1. The break-point time for a new column is to be 8.5 h. Calculate the new total length of the column required, column diameter, and the fraction of total capacity used up to the break point. The flow rate is to remain constant at 754 cm3/s.

  2. Use the same conditions as in part (a), but the flow rate is to be increased to 2000 cm3/s.

A3: Ans. (a) HT = 27.2 cm, 0.849 fraction; (b) D = 6.52 cm
12.3-2.

Drying of Nitrogen and Scale-Up of Column. Using molecular sieves, water vapor was removed from nitrogen gas in a packed bed (C3) at 28.3°C. The column height was 0.268 m, with the bulk density of the solid bed being equal to 712.8 kg/m3. The initial water concentration in the solid was 0.01 kg water/kg solid and the mass velocity of the nitrogen gas was 4052 kg/m2 · h. The initial water concentration in the gas was co = 926 × 106 kg water/kg nitrogen. The breakthrough data are as follows:

t (h)099.29.61010.4
c (kg H2O/kg N2 × 106)<0.60.62.62191235
t (h)10.811.2511.512.012.512.8
c (kg H2O/kg N2 × 106)418630717855906926

A value of c/co = 0.02 is desired at the break point. Do as follows:
  1. Determine the break-point time, the fraction of total capacity used up to the break point, the length of the unused bed, and the saturation loading capacity of the solid.

  2. For a proposed column length HT = 0.40 m, calculate the break-point time and fraction of total capacity used.

A4: Ans. (a) tb = 9.58 h, fraction used = 0.878
12.4-1.

Scale-Up of Ion-Exchange Column. An ion-exchange process using a resin to remove copper ions from aqueous solution is conducted in a 1.0-in.-diameter column 1.2 ft high. The flow rate is 1.5 gph and the break point occurred at 7.0 min. Integrating the breakthrough curve gives a ratio of usable capacity to total capacity of 0.60. Design a new tower that will be 3.0 ft high and operate at 4.5 gph. Calculate the new tower size and break-point time.

A5: Ans. tb = 24.5 min, D = 1.732 in.
12.4-2.

Height of Tower in Ion Exchange. In a given run using a flow rate of 0.2 m3/h in an ion-exchange tower with a column height of 0.40 m, the break point occurred at 8.0 min. The ratio of usable capacity to total equilibrium capacity is 0.65. What is the height of a similar column operating for 13.0 min to the break point at the same flow rate?

12.4-3.

Ion Exchange of Copper in Column. An ion-exchange column containing 99.3 g of amberlite ion-exchange resin was used to remove Cu2+ from a solution where co = 0.18 M CuSO4. The tower height = 30.5 cm and the diameter = 2.59 cm. The flow rate was 1.37 cm3 solution/s to the tower. The breakthrough data are shown below:

t (s)420480510540600660
c (g mol Cu/L)00.00330.00750.01570.05270.1063
t (s)720780810870900 
c (g mol Cu/L)0.14330.16340.17220.17630.180 

The concentration desired at the break point is c/co = 0.010. Determine the break-point time, fraction of total capacity used up to the break point, length of unused bed, and the saturation loading capacity of the solid.
12.4-4.

Effect of Total Concentration on Loading of Ion-Exchange Resin. Use the conditions of Example 12.4-1 for ion exchange of Cu2+ with H+. However, determine the effect of total concentration in the solution on the loading of Cu2+ on the resin. Do this for a total concentration C of 0.010 N in the solution at equilibrium instead of 0.10 N. Also, the concentration of Cu2+ in the solution is 0.002 M (0.004 N) instead of 0.02 M.

A8: Ans. = 0.9132 equiv/L
12.4-5.

Equilibrium in Ion Exchange of for H+. For the case where the cation (A) replaces H+ (B) in a polystyrene resin with 8% DVB, calculate the equilibrium constant KA,B. The total resin capacity Q = 2.0 equiv/L wet bed volume. The total concentration C = 0.20 N in the solution. Calculate at equilibrium the equivalents of in the resin when the concentration of in solution is 0.04 N.

A9: Ans. = 0.6684 equiv/L
12.5-1.

Composition of Two Liquid Phases in Equilibrium. An original mixture weighing 200 kg and containing 50 kg of isopropyl ether, 20 kg of acetic acid, and 130 kg of water is equilibrated in a mixer–settler and the phases separated. Determine the amounts and compositions of the raffinate and extract layers. Use equilibrium data from Appendix A.3.

12.5-2.

Single-Stage Extraction. A single-stage extraction is performed in which 400 kg of a solution containing 35 wt % acetic acid in water is contacted with 400 kg of pure isopropyl ether. Calculate the amounts and compositions of the extract and raffinate layers. Solve for the amounts both algebraically and by the lever-arm rule. What percent of the acetic acid is removed? Use equilibrium data from Appendix A.3.

A11: Ans. L1 = 358 kg, xB1 = 0.715, xC1 = 0.03, V1 = 442 kg, yA1 = 0.11, yC1 = 0.86, 34.7% removed.
12.5-3.

Single-Stage Extraction with Unknown Composition. A feed mixture weighing 200 kg of unknown composition containing water, acetic acid, and isopropyl ether is contacted in a single stage with 280 kg of a mixture containing 40 wt % acetic acid, 10 wt % water, and 50 wt % isopropyl ether. The resulting raffinate layer weighs 320 kg and contains 29.5 wt % acetic acid, 66.5 wt % water, and 4.0 wt % isopropyl ether. Determine the original composition of the feed mixture and the composition of the resulting extract layer. Use equilibrium data from Appendix A.3.

A12: Ans. xA0 = 0.032, xB0 = 0.948, yA1 = 0.15
12.5-4.

Extraction of Acetone in a Single Stage. A mixture weighing 1000 kg contains 23.5 wt % acetone and 76.5 wt % water and is to be extracted by 500 kg methyl isobutyl ketone in a single-stage extraction. Determine the amounts and compositions of the extract and raffinate phases. Use equilibrium data from Appendix A.3.

12.6-1.

Flooding Velocity and Tower Diameter. Using the same flow rates for each stream and feed conditions as in Ex. 12.6-1, with 1-in. Pall rings, design a tower for 50% of flooding with the aqueous phase dispersed instead of the toluene phase. Compare results.

A14: Ans. Diameter = 1.274 ft
12.6-2.

Interfacial Tension in Extraction System. Equilibrium data for the system water (A)–acetic acid (C)–isopropyl ether (B) are given in the Appendix, Table A.3-24. Using the data in dilute solution, where the wt fraction of acetic acid in the water-rich phase is 0.69%, calculate the interfacial tension.

A15: Ans. σ = 20.28 dyn/cm
12.6-3.

Perforated-Plate Tray Efficiency. An organic solvent is extracting an acid from a water solution in a perforated plate tower. The dispersed organic solvent flow rate is 600 ft3/h and the continuous aqueous flow rate is 400 ft3/h. The interfacial tension is 15 dyn/cm and the tray spacing is 1.1 ft. The hole size is 0.40 cm. Estimate the tray efficiency. What happens if the solvent flow rate is reduced to 400 ft3/h?

A16: Ans. Eo = 0.133 (solvent 600 ft3/h)
12.7-1.

Multiple-Stage Extraction with Fresh Solvent in Each Stage. Pure water is to be used to extract acetic acid from 400 kg of a feed solution containing 25 wt % acetic acid in isopropyl ether. Use equilibrium data from Appendix A.3.

  1. If 400 kg of water is used, calculate the percent recovery in the water solution in a one-stage process.

  2. If a multiple three-stage system is used and 133.3 kg fresh water is used in each stage, calculate the overall percent recovery of the acid in the total outlet water. (Hint: First, calculate the outlet extract and raffinate streams for the first stage using 400 kg of feed solution and 133 kg of water. For the second stage, 133 kg of water contacts the outlet organic phase from the first stage. For the third stage, 133.3 kg of water contacts the outlet organic phase from the second stage, and so on.)

12.7-2.

Overall Balance in Countercurrent Stage Extraction. An aqueous feed of 200 kg/h containing 25 wt % acetic acid is being extracted by pure isopropyl ether at the rate of 600 kg/h in a countercurrent multistage system. The exit acid concentration in the aqueous phase is to contain 3.0 wt % acetic acid. Calculate the compositions and amounts of the exit extract and raffinate streams. Use equilibrium data from Appendix A.3.

12.7-3.

Minimum Solvent and Countercurrent Extraction of Acetone. An aqueous feed solution of 1000 kg/h containing 23.5 wt % acetone and 76.5 wt % water is being extracted in a countercurrent multistage extraction system using pure methylisobutyl ketone solvent at 298–299 K. The outlet water raffinate will contain 2.5 wt % acetone. Use equilibrium data from Appendix A.3.

  1. Calculate the minimum solvent that can be used. [Hint: In this case the tie line through the feed L0 represents the condition for minimum solvent flow rate. This gives V1min. Then draw lines LNV1min and L0VN+1 to give the mixture point Mmin and the coordinate xAMmin. Using Eq. (12.7-4), solve for VN+1min, the minimum value of the solvent flow rate VN+1.]

  2. Using a solvent flow rate of 1.5 times the minimum, calculate the number of theoretical stages.

12.7-4.

Countercurrent Extraction of Acetic Acid and Minimum Solvent. An aqueous feed solution of 1000 kg/h of acetic acid–water solution contains 30.0 wt % acetic acid and is to be extracted in a countercurrent multistage process with pure isopropyl ether to reduce the acid concentration to 2.0 wt % acid in the final raffinate. Use equilibrium data from Appendix A.3.

  1. Calculate the minimum solvent flow rate that can be used. (Hint: See Problem 12.7-3 for the method to use.)

  2. If 2500 kg/h of ether solvent is used, determine the number of theoretical stages required. (Note: It may be necessary to replot on an expanded scale the concentrations at the dilute end.)

A20: Ans. (a) Minimum solvent flow rate VN+1 = 1630 kg/h; (b) 7.5 stages
12.7-5.

Number of Stages in Countercurrent Extraction. Repeat Example 12.7-2 but use an exit acid concentration in the aqueous phase of 4.0 wt %.

12.7-6.

Extraction with Immiscible Solvents. A water solution of 1000 kg/h containing 1.5 wt % nicotine in water is stripped with a kerosene stream of 2000 kg/h containing 0.05 wt % nicotine in a countercurrent stage tower. The exit water is to contain only 10% of the original nicotine, that is, 90% is removed. Use equilibrium data from Example 12.7-3. Calculate the number of theoretical stages needed.

A22: Ans. 3.7 stages
12.7-7.

Numbers of Transfer Units for Extraction of Nicotine. Using the data from Example 12.7-3 for nicotine extraction, replot the equilibrium and operating lines. For this system using a packed tower, the height of a transfer unit has been estimated as HOL = 1.1 m. Calculate the number of transfer units by two different methods and compare the values. Also calculate the tower height.

A23: Ans. NOL = 5.76 transfer units (Eq. 12.7-22), z = 6.34 m
12.7-8.

Minimum Solvent Rate with Immiscible Solvents. Determine the minimum solvent kerosene rate to perform the desired extraction in Example 12.7-3. Using 1.50 times this minimum rate, determine the number of theoretical stages needed graphically and also by using the analytical equation.

A24: Ans. V' = 153.8 kg/h, N = 6.86 analytically
12.7-9.

Stripping Nicotine from Kerosene. A kerosene flow of 100 kg/h contains 1.4 wt % nicotine and is to be stripped with pure water in a countercurrent multistage tower. It is desired to remove 90% of the nicotine. Using a water rate of 1.50 times the minimum, determine the number of theoretical stages required. (Use the equilibrium data from Example 12.7-3.)

12.7-10.

Extraction of Acetone from Water and Number of Transfer Units. Repeat Example 12.7-4 for the extraction of acetone from water using 1.5 times the minimum solvent flow rate. Do as follows:

  1. Graphically determine the number of theoretical steps needed.

  2. Calculate the number of steps using the analytical equation.

  3. For a packed tower with this system, the height of the transfer unit HOL has been estimated as 0.9 m. Calculate the number of transfer units NOL and the tower height.

A26: Ans. (b) N = 5.34 steps analytically; (c) NOL = 6.21 transfer units
12.7-11.

Predicting Extraction for Existing Tower with a Given Number of Steps. An existing tower contains 5.0 theoretical steps. It is desired to predict its performance under the following conditions using the system water–acetone–trichloroethane given in Example 12.7-4. An inlet water stream of 1000 kg/h containing 8.0 wt % acetone is extracted with 750 kg/h trichloroethane containing 1.0 wt % acetone in this countercurrent tower at 25°C. Determine the outlet concentration x1 in the water solution and in the trichloroethane solution using an analytical equation. Also plot the equilibrium and operating lines and step off 5.0 steps. Assume dilute solutions.

A27: Ans. x1 = 0.01205, y2 = 0.0932
12.8-1.

Effective Diffusivity in Leaching Particles. In Example 12.8-1 a time of leaching of the solid particle of 3.11 h is needed to remove 80% of the solute. Perform the following calculations:

  1. Using the experimental data, calculate the effective diffusivity, DAeff.

  2. Predict the time to leach 90% of the solute from the 2.0 mm particle.

A28: Ans. (a) DAeff = 1.0 × 105 mm2/s; (b) t = 5.00 h
12.9-1.

Leaching of Oil from Soybeans in a Single Stage. Repeat Example 12.9-1 for single-stage leaching of oil from soybeans. The 100 kg of soybeans contains 22 wt % oil and the solvent feed is 80 kg of solvent containing 3 wt % soybean oil.

A29: Ans. L1 = 52.0 kg, yA1 = 0.239, V1 = 50.0 kg, xA1 = 0.239, N1 = 1.5
12.9-2.

Leaching a Soybean Slurry in a Single Stage. A slurry of flaked soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt % oil and 90 wt % solvent hexane. This slurry is contacted with 100 kg of pure hexane in a single stage so that the value of N for the outlet underflow is 1.5 kg insoluble solid/kg solution retained. Calculate the amounts and compositions of the overflow V1 and the underflow L1 leaving the stage.

12.10-1.

Constant Underflow in Leaching Oil from Meal. Use the same conditions as given in Example 12.10-1, but assume constant underflow of N = 1.85 kg solid/kg solution. Calculate the exit flows and compositions and the number of stages required. Compare with Example 12.10-1.

A31: Ans. yAN = 0.111, xA1 = 0.623, 4.3 stages
12.10-2.

Effect of Less Solvent Flow in Leaching Oil from Meal. Use the same conditions as given in Example 12.10-1, but the inlet fresh-solvent-mixture flow rate per hour is decreased by 10%, to 1179 kg of benzene and 18 kg of oil. Calculate the number of stages needed.

12.10-3.

Countercurrent Multistage Washing of Ore. A treated ore containing inert solid gangue and copper sulfate is to be leached in a countercurrent multistage extractor using pure water to leach the CuSO4. The solid charge rate per hour consists of 10 000 kg of inert gangue (B), 1200 kg of CuSO4 (solute A), and 400 kg of water (C). The exit wash solution is to contain 92 wt % water and 8 wt % CuSO4. A total of 95% of the CuSO4 in the inlet ore is to be recovered. The underflow is constant at N = 0.5 kg inert gangue solid/kg aqueous solution. Calculate the number of stages required.

A33: Ans. 9 stages, LN = 20 000 kg/h, yAN = 0.0030, V1= 14 250 kg/h
12.10-4.

Countercurrent Multistage Leaching of Halibut Livers. Fresh halibut livers containing 25.7 wt % oil are to be extracted with pure ethyl ether to remove 95% of the oil in a countercurrent multistage leaching process. The feed rate is 1000 kg of fresh livers per hour. The final exit overflow solution is to contain 70 wt % oil. The retention of solution by the inert solids (oil-free liver) of the liver varies as follows (C1), where N is kg inert solid/kg solution retained and yA is kg oil/kg solution:

NyANyANyA
4.8802.470.41.390.81
3.500.21.670.6  

Calculate the amounts and compositions of the exit streams and the total number of theoretical stages.
12.10-5.

Countercurrent Leaching of Flaked Soybeans. Soybean flakes containing 22 wt % oil are to be leached in a countercurrent multistage process to contain 0.8 kg oil/100 kg inert solid using fresh and pure hexane solvent. For every 1000 kg soybeans, 1000 kg hexane is used. Experiments (S1) give the following retention of solution with the solids in the underflow, where N is kg inert solid/kg solution retained and yA is wt fraction of oil in solution:

NyA
1.730
1.520.20
1.430.30

Calculate the exit flows and compositions and the number of theoretical stages needed.
12.11-1.

Crystallization of Ba(NO3)2. A hot solution of Ba(NO3)2 from an evaporator contains 30.6 kg Ba(NO3)2/100 kg H2O and goes to a crystallizer, where the solution is cooled and Ba(NO3)2 crystallizes. On cooling, 10% of the original water present evaporates. For a feed solution of 100 kg total, calculate the following:

  1. The yield of crystals if the solution is cooled to 290 K (17°C), where the solubility is 8.6 kg Ba(NO3)2/100 kg total water.

  2. The yield if cooled instead to 283 K, where the solubility is 7.0 kg Ba(NO3)2/100 kg total water.

A36: Ans. (a) 17.47 kg Ba(NO3)2 crystals
12.11-2.

Dissolving and Subsequent Crystallization. A batch of 1000 kg of KCl is dissolved in sufficient water to make a saturated solution at 363 K, where the solubility is 35 wt % KCl in water. The solution is cooled to 293 K, at which temperature its solubility is 25.4 wt %.

  1. What is the weight of water required for solution and the weight of crystals of KCl obtained?

  2. What is the weight of crystals obtained if 5% of the original water evaporates on cooling?

A37: Ans. (a) 1857 kg water, 368 kg crystals; (b) 399 kg crystals
12.11-3.

Crystallization of MgSO4 · 7H2O. A hot solution containing 1000 kg of MgSO4 and water having a concentration of 30 wt % MgSO4 is cooled to 288.8 K, where crystals of MgSO4 · 7H2O are precipated. The solubility at 288.8 K is 24.5 wt % anhydrous MgSO4 in the solution. Calculate the yield of crystals obtained if 5% of the original water in the system evaporates on cooling.

12.11-4.

Heat Balance in Crystallization. A feed solution of 10 000 lbm at 130°F containing 47.0 lb FeSO4/100 lb total water is cooled to 80°F, where FeSO4 · 7H2O crystals are removed. The solubility of the salt is 30.5 lb FeSO4/100 lb total water (P1). The average heat capacity of the feed solution is 0.70 btu/lbm · °F. The heat of solution at 18°C is −4.4 kcal/g mol (−18.4 kJ/g mol) FeSO4 · 7H2O (P1). Calculate the yield of crystals and make a heat balance. Assume that no water is vaporized.

A39: Ans. 2750 lbm FeSO4 · 7H2O crystals, q = −428 300 btu (−451 900 kJ)
12.11-5.

Effect of Temperature on Yield and Heat Balance in Crystallization. Use the conditions given in Example 12.11-2, but the solution is cooled instead to 283.2 K, where the solubility is 30.9 kg MgSO4/100 kg total water (P1). Calculate the effect on yield and the heat absorbed by using 283.2 K instead of 293.2 K for the crystallization.

12.12-1.

Growth and Nucleation Rate in MSMPR Crystallizer. Experimental data were obtained for an MSMPR crystallizer (S2). The slurry density is 169 g/L, the crystal density ρ is 1.65 g/m3, the residence time τ is 6.57 h, and the shape factor a is 0.98. The screen analysis of the crystals is as follows:

MeshWt %MeshWt %
−14, +204.5−35, +4831.5
−20, +2814.5−48, +6515.5
−28, +3524.0−65, +1007.5
  −1002.5

Using these data, do as follows:
  1. Calculate the population density, growth rate, and nucleation rate. Also calculate the average size La.

  2. Using these calculated values, predict the cumulative weight fraction versus size L from the experimental value of G and τ. Compare the predicted and experimental values.

A41: Ans. (a) G = 0.01673 mm/h; (b) For L = 0.589 mm, (1 − Wf) = 0.218
12.12-2.

Prediction of Cumulative Weight Fraction of Crystals from Growth Rate. In Example 12.12-1, the growth rate G was determined to be 0.03244 mm/h for a residence time τ of 3.38 h. Using these values, predict the cumulative weight fraction versus size L. Compare the predicted values with the experimental values.

A42: Ans. L = 0.589 mm, (1 − Wf) = 0.217; L = 0.417 mm, (1 − Wf) = 0.473
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