TEST BANK FOR Industrial Separation Processes Fundamentals By André B de Haan, Hans Bosch (Solution Manual)

Exam (elaborations) TEST BANK FOR Industrial Separation Processes Fundamentals By André B de Haan, Hans Bosch (Solution Manual) Fundamentals of Industrial Separations, SOLUTIONS TO PROBLEMS 1-1 Chapter 1: INTRODUCTION Exercise 1.1 Compare and discuss the advantages and disadvantages of making separations using an energyseparating agent (ESA) versus using a mass-separating agent (MSA). Exercise 1.2 The system benzene-toluene adheres closely to Raoult’s law. The vapor pressures of benzene and toluene at 121°C are 300 and 133 kPa. Calculate the relative volatility. Exercise 1.3 As a part of the life system support for spacecrafts it is necessary to provide a means of continuously removing carbon dioxide from air. If it is not possible to rely upon gravity in any way to devise a CO2-air separation process. Suggest at least two separation schemes, which could be suitable for continuous CO2 removal from air under zero gravity conditions. Exercise 1.4 Gold is present in seawater to a concentration level between 10-12 and 10-8 weight fraction, depending upon the location. Briefly evaluate the potential for recovering gold economically from seawater. Exercise 1.5 Assuming that the membrane characteristics are not changed as the upstream pressure increases, will the product-water purity in a reverse-osmosis seawater desalination process increase, remain constant or decrease? Exercise 1.6 Propylene and propane are among the light hydrocarbons produced by thermal and catalytic cracking of heavy petroleum fractions. Although propylene and propane have close boiling points, they are traditionally separated by distillation. Because distillation requires a large numbers of stages and considerable reflux and boilup flow rates compared to the feed flow, considerable attention has been given to the possible replacement of distillation with a more economical and less energy-intensive option. Based on the properties of both species, propose some alternative properties that can be exploited to enhance the selectivity of propylene and propane separation. What separation processes are based on these alternative properties? Property Propylene Propane Molecular weight (kg/mol) VdWaals volume (m3/mol) VdWaals area (m2/mol) Acentric factor Dipole moment (Debije) Radius of gyration (m·1010) Melting point (K) Boiling point (K) Critical temperature (K) Critical pressure (MPa) 0,04208 0,0341 5,06 0,142 0,4 2,25 87,9 225,4 364,8 4,61 0,04410 0,0376 5,59 0,152 0,0 2,43 85,5 231,1 369,8 4,25 1-2 INTRODUCTION SOLUTIONS Exercise 1.1 Compare and discuss the advantages and disadvantages of making separations using an energyseparating agent (ESA) versus using a mass-separating agent (MSA). Answer Advantages Disadvantages ESA Known technology Energy consumption Phase separation may be expensive Relatively simple Heat integration required MSA High selectivities Recovery of agent requires possible additional separation Exercise 1.2 Calculate the relative volatility of benzene-toluene. Answer For an ideal system such as benzene-toluene, by definition (Eqs. 1.3 and 1.5) SF ≡ α ≡ yB / xB (1− yB) /(1− xB) = PB o PT o = 300 133 = 2.26 Exercise 1.3 Suggest at least two separation schemes that could be suitable for continuous CO2 removal from air under zero gravity conditions. Answer a. Adsorption of CO2 on activated carbon. Regeneration by exposing to vacuum outside the spacecraft. b. Absorption of CO2 in a suitable solvent at the low temperature outside the vessel. Regeneration of the loaded solvent at room temperature, releasing the overpressure via a valve to the outer space. Exercise 1.4 Briefly evaluate the potential for recovering gold economically from seawater. Answer The lower the concentration of the desired substance, the more expensive the required technology. At the given extremely low concentrations, huge amounts of seawater have to be treated, either by evaporation or by (membrane)filtration. Either method is not feasible due to the highenergy costs or pumping costs. The thermodynamical basis is given by the change in chemical potential upon mixing, which is proportional to ln(activity) a negative value, indicating a spontaneous process. Separation of the mixture, on the other hand, requires at least that amount of work, which is extremely large at very low values of the activity (or concentration). Fundamentals of Industrial Separations, SOLUTIONS TO PROBLEMS 1-3 Exercise 1.5 Assuming that the membrane characteristics are not changed, as the upstream pressure increases, will the product-water purity in a reverse-osmosis seawater desalination process increase, remain constant or decrease? Answer The water flux is proportional to ΔPext – ΔΠosm and will increase with increasing ΔPext. The salt flux is proportional to cfeed – cpermeate and will not be effected by a change in ΔPext. Hence, the product purity will increase with increasing ΔPext. Exercise 1.6 Based on the properties of propylene and propane, propose some alternative properties that can be exploited to enhance the selectivity of propylene and propane separation. What processes are based on these alternative properties? Property Propylene Propane Molecular weight (kg/mol) VdWaals volume (m3/mol) VdWaals area (m2/mol) Acentric factor Dipole moment (Debije) Radius of gyration (m·1010) Melting point (K) Boiling point (K) Critical temperature (K) Critical pressure (Mpa) 0,04208 0,0341 5,06 0,142 0,4 2,25 87,9 225,4 364,8 4,61 0,04410 0,0376 5,59 0,152 0,0 2,43 85,5 231,1 369,8 4,25 Answer Difference in molecular weight Ultracentrifuge Difference in VdWaals volume Kinetic separation by preferential adsorption of the component with the smallest volume in a molecular sieve adsorbent Difference in dipole moment Absorption, adsorption, extraction or extractive distillation Fundamentals of Industrial Separations, SOLUTIONS TO PROBLEMS 2-1 Chapter 2: EVAPORATION & DISTILLATION Exercise 2.1 In Vapor-Liquid Equilibrium Data Collection the following form of Antoine-equation is used logPi 0 = A ′ − B ′ T +C ′ with Pi saturation pressure in mmHg and temperature T in °C (760 mmHg = 1 atm). 0 For benzene in benzene-toluene mixtures the following values are reported: A’ = 6.87987, ‘B = 1196.760 and C’ = 219.161. Calculate the constants A, B and C in the Antoine-equation as defined in Eq. 2.9. lnPi 0 = A − B T +C with pressure units in atm and temperature in °C. Exercise 2.2 With temperature in degrees Fahrenheit and pressure in pounds per sq. inch, the following Antoine constants apply for benzene A’ = 5.1606, B’ = 2154.2, C’ = 362.49 (psi, °F, 10log P expression). Noting that 1 atm = 14.696 psi = 1.01325 bar and T(°F)=T(°C)·9/5 + 32, calculate the saturation pressure of benzene in bar at 80.1°C. Exercise 2.3 Vapor-liquid equilibrium data for benzene-toluene are given at 1.0 atm and at 1.5 atm in the T-x diagram below. Expressing the saturated vapor pressure in atm as a function of temperature in K, the Antoine constants for benzene and toluene are, respectively A = 9.2082, B = 2755.64, C = -54.00 and A = 9.3716, B = 3090.78, C = -53.97 (atm, K, Eq. 2.9). Check the phase compositions at 100°C and total pressures of 1.0 and 1.5 atm. 0 0.2 0.4 0.6 0.8 1 70 80 90 100 110 120 130 Ptot = 1.5 atm Ptot = 1.0 atm temperature / C benzene liquid and vapor mole fractions 2-2 EVAPORATION & DISTILLATION Exercise 2.4 A liquid benzene-toluene mixture with z = 0.40 should produce a liquid with x = 0.35 in a flash drum at 1.0 atm. a. Calculate the required feed temperature b. Calculate the equilibrium vapor-liquid ratio in the flash drum. Exercise 2.5 We wish to flash distill isothermally a mixture containing 45 mole% of benzene and 55 mole% of toluene. Feed rate to the still is 700 moles/h. Equilibrium data for the benzene-toluene system can be approximated with a constant relative volatility of 2.5, where benzene is the more volatile component. Operation of the still is at 1 atm. a. Plot the y-x diagram for benzene-toluene. b. If 60% of the feed is evaporated, find the liquid and vapor compositions c. If we desire a vapor composition of 60 mole%, what is the corresponding liquid composition and what are the liquid and vapor flow rates? d. Find the compositions and flow rates of all unknown streams for a two stage flash cascade where 40% of the feed is flashed in the first stage and the liquid product is sent to a second flash chamber where 30% is flashed. Exercise 2.6 Distillation is used to separate pentane from hexane. The feed amounts 100 mol/s and has a mole ratio pentane/hexane = 0.5. The bottom and top products have the compositions xB = 0.05 and xD = 0.98. The reflux ratio is 2.25. The column pressure is 1 bar. The feed, at the bubbling point, enters the column exactly on the feed tray. The tray temperature is equal to the feed temperature. The pentane vapor pressure is given by: P5 o = e 11*(1− 310 T(K) ) (bar) The vapor pressure of hexane is 1/3 of the pentane vapor pressure over the whole temperature range. The average density of liquid pentane and hexane amounts to 8170 and 7280 mol/m3, resp. The heat of vaporization amounts to 30 kJ/mol. The distance between the trays amounts to 0.50 m. a. Calculate the feed temperature b. Calculate the vapor stream from the reboiler c. Calculate the required energy in the reboiler d. Construct the y-x diagram e. Construct the operating lines and locate the feed line f. Determine the number of equilibrium stages g. Determine the height of the column Fundamentals of Industrial Separations, SOLUTIONS TO PROBLEMS 2-3 Exercise 2.7 Methanol (M) is to be separated from water (W) by atmospheric distillation. The feed contains 14.46 kg/h methanol and 10.44 kg/hr water. The distillate is 99 mol% pure, the bottom product contains 5 mol% of methanol. The feed is subcooled such that q = 1.12. a. Determine the minimum number of stages and minimum reflux b. Determine the feed stage location and number of theoretical stages required for a reflux ratio of 1. Vapor-liquid equilibrium data (1.0 atm, mole fraction methanol) x 0.0321 0.0523 0.075 0.154 0.225 0.349 0.813 0.918 y 0.1900 0.2940 0.352 0.516 0.593 0.703 0.918 0.963 Exercise 2.8 A feed to a distillation unit consists of 50 mol% benzene in toluene. It is introduced to the column at its bubble point to the optimal plate. The column is to produce a distillate containing 95 mol% benzene and a bottoms of 95 mol% toluene. For an operating pressure of 1 atm, calculate: a. the minimum reflux ratio b. the minimum number of equilibrium stages to carry out the desired separation c. the number of actual stages needed, using a reflux ratio (L’/D) of 50% more than the minimum, d. the product and residue stream in kilograms per hour if the feed is 907.3 kg/h e. the saturated steam required in kilograms per hour for heat to the reboiler using enthalpy data below Steam: ΔHVAP = 2000 kJ/kg Benzene: ΔHVAP = 380 kJ/kg Toluene: ΔHVAP = 400 kJ/kg Vapor-liquid equilibrium data (1 atm, mole fraction benzene) x 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 y 0.21 0.37 0.51 0.64 0.72 0.79 0.86 0.91 0.96 2-4 EVAPORATION & DISTILLATION SOLUTIONS Exercise 2.1 Given Antoine-equation in the form: logPi 0 = A′ − B′ T +C′ with Pi 0 saturation pressure in mmHg and T temperature in °C. (760 mmHg = 1 atm) For benzene A’ = 6.87987 [-], B’ = 1196.760 [-] and C’ = 219.161 [°C]. Find The constants A, B and C in the Antoine-equation as defined in Eq. 2.9. Thoughts Compare both forms of the Antoine-equation side by side: logPbenzene 0 = A′ − B′ T +C′ (I) and lnPbenzene 0 = A− B T +C (II), convert the pressure in (I) into form (II) and compare the three resulting constants with A, B and C, respectively Solution lnPB 0(atm) = ln10 A ′ − B ′ T ′ +C (mmHg) 760(mmHg/atm) = A′ ln(10) − ln(760) − B′ ln(10) T +C′ Hence A = A′ ln(10) − ln(760) = ln(10)[A′ − log(760)] A = 9.2082 [-] B = B′ ln(10) B = 2755.6 [-] C = C’ C = 219.16 [°C] Given Antoine constants benzene (10logP expression; units in psi and °F) A’ = 5.1606 [-], B’ = 2154.2 [-], C’ = 362.49 [°F] Find Saturation pressure of benzene in bar at 80.1°C. Thoughts Two different approaches: a. convert temperature to °F, calculate saturation pressure at this temperature in psi and convert result to bar, or b. recalculate Antoine constants in the required units as shown in Exercise 2.1. Solution Method a: Temperature conversion: 80.1°C = 176.18°F ( °C ⋅ 9 5 + 32) Saturation pressure Pbenzene o (176.18) = 10 A ′ − B ′ 176.18+C ′ = 14.696 psi Pressure conversion 14.696 psi = 14.696 14.696 atm = 1.01325 bar Method b: Pbenzene o [bar] = 1.01325[bar/atm] ⋅ 10 A′ − B ′ T(°F)+C ′ [psi] 14.696[psi / atm] = exp(A− B T(°C) +C ) T(°F) =T(°C) ⋅ 5 9 with − 32 Fundamentals of Industrial Separations, SOLUTIONS TO PROBLEMS 2-5 hence A = A′ ⋅ln(10) + ln1.01325 14.696 A = 9.2083 [-] B = 5 9 ⋅B′ ⋅ ln(10) B = 2755.64 [-] C = 5 9 ⋅ (C′ + 32) C = 219.16 [°C] Saturation pressure of benzene at 80.1°C = exp(A− B 80.1+C ) = 1.01325bar 1 Exercise