Separation Processes Lecture 2 - Vapor-Liquid Equilibrium PDF

SEPARATION PROCESSES LECTURE 2. PART 1. VAPOR-LIQUID EQUILIBRIUM Vapor-Liquid Equilibrium (VLE) Consider liquid and vapor phases in a closed container with each phase containing two components A and B. Liquid molecules are continual...

SEPARATION PROCESSES LECTURE 2. PART 1. VAPOR-LIQUID EQUILIBRIUM Vapor-Liquid Equilibrium (VLE) Consider liquid and vapor phases in a closed container with each phase containing two components A and B. Liquid molecules are continually vaporizing and gas molecules are continually condensing. Initially, rates of condensation and vaporization are different. When is equilibrium achieved? Vapor-Liquid Equilibrium (VLE) At equilibrium, three conditions occur in the closed system: (1) Thermal equilibrium Heat transfer between the two phases stops and the temperature of the two phases are equal. (2) Mechanical equilibrium Forces between the vapor and liquid phases are balanced and the pressure in the liquid and vapor phase are equal. (3) Phase equilibrium Rates of evaporation is equal to the rates of composition. There is no microscopic change in the composition of both phases. Vapor-Liquid Equilibrium (VLE) At vapor-liquid equilibrium (VLE), a vapor- liquid system has the same temperature and pressure on both phases, and displays no macroscopic changes in composition on both phases. How many independent variables must be fixed to determine the equilibrium state of a system in VLE? Review: Gibbs’ Phase Rule The number of degrees of freedom (F) – the number of independent variables needed to be fixed to define the equilibrium state of a system – is given by: F=2–P+N where N: number of components in the system P: number of phases in the system Thus, for a system in VLE with N components, F is given by: F=N For binary distillation (N = 2), the degree of freedom is given by: F=2 Binary VLE To describe the state of a vapor liquid system in equilibrium, various diagrams involving 2 independent variables (pressure, temperature, or composition) are used. (1) T – xy diagram (2) P – xy diagram (3) yx diagram T-xy diagram P – xy diagram yx diagram Simple models for VLE: Raoult’s law Two major assumption: - Vapor phase is an ideal gas: applicable only for low to moderate pressure - Liquid phase is an ideal solution: applicable for molecular species of the same size and of the same chemical nature § Expressed mathematically as follows: y! P = x! P!"#$ (i = 1,2,...,N) VLE Calculations Bubble point calculations Bubble P : Calculate yi and P, given xi and T Bubble T : Calculate yi and T, given xi and P y! P = x! P!"#$ Because ∑% 𝑦% = 1 ( P = ( x! P!"#$ !&' VLE Calculations Dew Point Calculations Dew P : Calculate xi and P, given yi and T Dew T : Calculate xi and T, given yi and P y! P = x! P!"#$ Because ∑% 𝑥% = 1 1 P= y! ∑(!&' P!"#$ Vapor Pressure Correlations Antoine equation: B ln P "#$ kPa = A − T() C) + C See Smith, Van Ness, Abbott 7th ed. p. 682 Perry’s Handbook 8th ed. 13 – 14 Simple Models for VLE: Modified Raoult’s Law To account for the non-ideality of the liquid phase of a solution in VLE, modified Raoult’s law can be used. This introduces the activity coefficient γi to account for the non-ideality of the liquid phase. y! P = γ! x! P!"#$ (i = 1,2,...,N) § To account for the non-ideality of the vapor phase, Raoult’s law can further be modified by introducing the fugacity coefficient Фi Ф! y! P = γ! x! P!"#$ (i = 1,2,...,N) Simple Models for VLE: Henry’s law Henry’s law applies for the following assumptions: - Low pressures: vapor can be assumed to be ideal - Very dilute solute in the liquid phase § According to Henry’s law, the partial pressure of the species in the vapor phase is directly proportional to the liquid phase mole fraction according to the following equation: y! P = x ! H! where Hi is the proportionality constant, Henry’s constant VLE from Equilibrium Ratio, Ki The equilibrium ratio Ki is the ratio of the composition of a particular species at two phases in equilibrium. For VLE: y! K! = x! Ki > 1 – light component Ki < 1 – heavy component DePriester Charts Nomographs of the K-values of light hydrocarbons as functions of temperature and pressure Perry’s Handbook 8th ed. 13 – 10 and 13 – 11 Smith, Van Ness, Abbott 7th ed. Pp. 365 – 366 DePriester Charts VLE from Equilibrium Ratio, Ki For bubble point calculations: 8 K ! x! = 1 ! § For dew point calculations: y! 8 =1 K! ! Relative volatility αij Relative volatility αij is a comparative measure of the vapor pressure of components in a liquid mixture at a given temperature and pressure. y! /x! K ! α!* = = y* /x* K * Activity sheet 1a Assuming the validity of Raoult’s law, do the calculations for the benzene(1)/toluene(2) system: B ln P!"#$ (kPa) = A − % T( C) + C For benzene: A = 13.7189, B = 2726.81 and C = 217.572 For toluene: A = 13.9320, B = 3056.95, and C =217.65 a) Given x1 = 0.33 and T = 100oC, find y1 and P b) Given y1 = 0.33 and T = 100oC, find x1 and P c) Given x1 = 0.33 and P = 120 kPa, find y1 and T d) Given y1 = 0.33 and P = 120 kPa, find x1 and T Single-stage VLE Operations [Flash Distillation] Flash distillation In flash vaporization, a liquid feed is passed through a throttling valve or nozzle wherein the pressure drop causes part of the feed to vaporize. The term ‘flash’ refers to the rapid vaporization that occurs during the process. Flash distillation: material balances Overall mass/mole balance: F=L+V Component mass/mole balance: Fzi = Lxi + Vyi Flash distillation: energy balances Overall energy balance: FhF + Q flash = VHV + LhL Generally, hF = hF TF , z ; HV = HV T, y ; hL = hL T, x For ideal binary mixtures, the enthalpies can be calculated as follows: Feed: hF = zA CpL,A TF − Tref + zB CpL.B TF − Tref Liquid product: hL = xA CpL,A T − Tref + xB CpL,B T − Tref Vapor product:HV = yA λA + CpV,A T − Tref + yB λB + CpV,B T − Tref Flash distillation To solve these 6 equations for binary flash distillation, 6 independent variables must be specified (degrees of freedom). Feed specifications typically take up 4 of these degrees of freedom. (1) Feed flow rate, F (2) Feed composition, z (3) Inlet temperature, Ti (4) Inlet pressure, Pi The designer selects the fifth degree of freedom. (5) Drum pressure, Pdrum : should be below the critical pressure for the mixture Flash distillation Method of solution depends on the final degree of freedom provided. (1) Sequential solution – solves material balances before energy balances; used if 6th DF is any of the following: a) Liquid product composition, x b) Vapor product composition, y c) Fraction of vaporized feed, f = V/F d) Fraction of remaining liquid, q = L/F e) Drum temperature, Tdrum (2) Simultaneous solution – simultaneously solve material and energy balances; used if 6th DF is feed temperature, TF Flash distillation: operating line For single-stage VLE, an operating line can be derived from the component material balance. Fz = Vy + Lx L F y=− x+ V V Flash distillation: operating line In terms of the fraction of vaporized feed f, the operating line can be written as follows: L F y=− x+ V V V L F−V 1−F 1−f = = = V V V f F 1−f z y=− x+ f f Flash distillation: operating line In terms of the fraction of remaining liquid q, the operating line can be written as follows: L F y=− x+ V V L L L q = = F = V F−L 1− L 1−q F q 1 y=− + z 1−q 1−q Sample Problem If a 20% mole benzene in a toluene mixture is the feed to an equilibrium still and if the vapor and liquid streams are withdrawn in a ratio of 2-mole vapor to 1-mole liquid, what is the composition of the vapor and liquid? Sample Problem A mixture of methanol and water is separated in a flash drum at 1 atm. a) Feed rate is 100 kmol/h. Feed enters at 60 mol% methanol and 40% of the feed is vaporized. What are the vapor and liquid mole fraction and flow rates? b) Feed rate is 1000 lbmol/h. Feed is 30 mol% methanol and a liquid product of 20 mol% methanol is desired. What V/F must be used? Find product flow rates and composition. c) The flash drum is operated such that the liquid mole fraction is 50 mol% methanol, liquid product rate is 1500 kmol/h and 20% of the feed is vaporized. What must the flowrate and composition of the feed be? d) The feed composition is 40 mol% and the drum temperature is 77oC. Find the fraction of feed vaporized and the composition of the liquid and vapor products. Methanol mole fraction in Methanol mole fraction in Temperature oC liquid phase vapor phase 0.000 0.000 100 0.020 0.134 96.4 0.040 0.230 93.5 0.060 0.304 91.2 0.080 0.365 89.3 0.100 0.418 87.7 0.150 0.517 84.4 0.200 0.579 81.7 0.300 0.665 78.0 0.400 0.729 75.3 0.500 0.779 73.1 0.600 0.825 71.2 0.700 0.870 69.3 0.800 0.915 67.6 0.900 0.958 66.0 0.950 0.979 65.0 1.000 1.00 64.5 Sample Problem Two flash distillation chambers are hooked together as shown in the following figure. The feed to the first drum is a binary mixture of methanol and water that is 55 mol% methanol. Feed flow rate is 10,000 kmol/h. The second flash drum operates with f = 0.7 and the liquid product composition is 25 mol% methanol. a) What is the fraction vaporized in the first flash drum? b) What are y1, y2, x1, T1 and T2? Multicomponent flash distillation For multicomponent flash distillation, complete material balances can be obtained using equilibrium ratios Ki. Recall: yi Ki = xi Multicomponent flash distillation From the component material balance: Fzi = Lxi + Vyi Fzi Fzi xi = = L + VK i F − V + K i V zi xi = V 1 + (K i − 1) F Taking the summation of both sides of the equation: n zi ෍ =1 V i=1 1 + K i − 1 F Multicomponent flash distillation From the component material balance: Fzi = Lxi + Vyi L Fzi = yi + Vyi Ki Fzi Fzi FK i zi yi = = = L F−V F − V + VK i +V +V Ki Ki K i zi yi = V 1 + (K i − 1) F Taking the summation of both sides of the equation: n K i zi ෍ =1 V i=1 1 + Ki − 1 F Multicomponent flash distillation The following equations can be used to solve for V/F. However, they do not have good convergence. n zi ෍ =1 V i=1 1 + K i − 1 F n K i zi ෍ =1 V i=1 1 + Ki − 1 F Multicomponent flash distillation Subtracting the two previous equations will yield a new equation with good convergence properties. This is called the Rachford-Rice equation. n K i − 1 zi ෍ =0 V i=1 1 + (K i − 1) F Sample Problem A 100 kmol/hr feed consisting of 10, 20, 30, and 40 mol% of propane, n-butane, n-pentane, and n- hexane, respectively, enters a distillation column at 100 psia (689.5 kPa) and 200°F (366.5 K). Assuming equilibrium, what mole fraction of the feed enters as liquid, and what are the liquid and vapour compositions? Sample Problem The system acetone(1)/acetonitrile(2)/nitromethane(3) was flashed at 80oC and 110 kPa. The feed flow rate is 1000 kmol/h and the feed composition is z1 = 0.45, z2 =0.35 and z3 = 0.20. Assuming Raoult’s law is appropriate for the system. P1sat = 195.75 kPa P2sat = 97.84 kPa P3sat = 50.32 kPa a.) Using the given system prove that a flash calculation can be made. b.) Determine the flow rates and composition of the liquid and vapor product.  Separation processes lecture 2. part 2. single-stage vle operations B. BATCH DISTILLATION Batch distillation In batch distillation, a charge of feed is loaded into a reboiler and heated with steam over a period of time to vaporize a fraction of the feed. The waste (W) is the part of the initial feed that remains in the reboiler after batch distillation. The distillate (D) is the portion of the feed that is vaporized and collected. It is often condensed to obtain a more concentrated mixture of the volatile component in the feed. Batch distillation: material balances Overall material balance: F = Wfinal + Dtotal Component material balance: FxF = xw,final Wfinal + xD,avg Dtotal  Batch distillation Usual specifications: (1) Initial moles of feed (F) (2) Mole fraction of feed (xF) (3) Desired composition in the waste (xw,final) or distillate (xD,avg) Usual unknowns: (1) Moles of distillate collected (Dtotal) (2) Moles of final waste (Wfinal) (3) Final composition in the waste (xw,final) or distillate (xD,avg) Degrees of freedom = no. of unknowns – no. of equations = 3 – 2 = 1 Thus, one more equation is needed to completely define the material balances in a binary batch distillation system! Batch distillation: Rayleigh equation The third equation can be derived from a differential mass balance of the system. −out = accumulation in reboiler −xD dW = −d WxW −xD dW = −Wdxw − xw dW Wfinal xw,final dW dxw න =න F W xF x D − xw Binary distillation: Rayleigh equation The third equation obtained from the differential mass balance is called the Rayleigh equation. x_F Wfinal dxw ln = −න F xw,final xD − xw Simple binary batch distillation problems Given: F, xF and xw,final Required: W, D, xD,avg 1. Plot yx diagram for the binary system. 2. At a series of x values, find y – x. 3. Plot 1/(y – x) or fit to an equation. 4. Graphically or numerically integrate from xF to xw,final. 5. Calculate the final waste in the still pot W from the Rayleigh equation. 6. Calculate total distillate D and its average concentration xD,avg from the material balances.  Activity sheet 1e A simple batch still is used to separate methanol from water. The feed charge to the still pot is 100 moles of a 75 mol% methanol mixture. A final waste concentration of 55 mol% methanol is desired. Find the amount of distillate collected, the amount of water left in the pot and the average concentration of the distillate. Pressure is 1 atm. Methanol mole fraction in Methanol mole fraction in Temperature oC liquid phase vapor phase 0.000 0.000 100 0.020 0.134 96.4 0.040 0.230 93.5 0.060 0.304 91.2 0.080 0.365 89.3 0.100 0.418 87.7 0.150 0.517 84.4 0.200 0.579 81.7 0.300 0.665 78.0 0.400 0.729 75.3 0.500 0.779 73.1 0.600 0.825 71.2 0.700 0.870 69.3 0.800 0.915 67.6 0.900 0.958 66.0 0.950 0.979 65.0 1.000 1.00 64.5