1- VaporLiquid Equilibrium Introduction.ppt

Advanced Thermodynamics Note 9 Vapor/Liquid Equilibrium: Introduction Lecturer:  Phase equilibrium Application – Distillation, absorption, and extraction bring phases of different composition into contact. Both the extent of change and the rate of tran...

Advanced Thermodynamics Note 9 Vapor/Liquid Equilibrium: Introduction Lecturer:  Phase equilibrium Application – Distillation, absorption, and extraction bring phases of different composition into contact. Both the extent of change and the rate of transfer depend on the departure of the system from equilibrium. Quantitative treatment of mass transfer the equilibrium T, P, and phase compositions must be known.  The nature of equilibrium A static condition in which no changes occur in the macroscopic properties of a system with time. At the microscopic level, conditions are not static. – The average rate of passage of molecules is the same in both directions, and no net interphase transfer of material occurs. An isolated system consisting of liquid and vapor phases in intimate contact eventually reaches a final state wherein no tendency exists for change to occur within the system. – Fixed temperature, pressure, and phase composition  Phase rule vs. Duhem’s theorem (The number of variables that is independently fixed in a system at equilibrium) = (the number of variables that characterize the intensive state of the system) - (the number of independent equations connecting the variable): Phase rule: F 2  ( N  1)( )  (  1)( N ) 2    N Duhem’s rule:   ( N initially F 2formed – for any closed system    given  1)( )from   ( masses  1)( N of N  2 ) prescribed chemical species, the equilibrium state is completely determined when any two independent variables are fixed. – Two ? When phase rule F = 1, at least one of the two variables must be extensive, and when F = 0, both must be extensive.  VLE: qualitative behavior N 2 When two chemical species: – phase rule: F 4   – the maximum value of F = 3 (π = 1), namely, P, T, and one mole fraction. All equilibrium states of the system can be represented in three- dimensional P-T-composition space. Fig 10.1  Within this space, the states of pairs of phases coexisting at equilibrium define surfaces. – The subcooled-liquid region lies above the upper surface; the superheated-vapor region lies below the under surface. – UBHC1 and KAC2 represent the vapor pressure-vs.-T curves for pure species 1 and 2. – C1 and C2 are the critical points of pure species 1 and 2. – L is a bubble point and the upper surface is the bubblepoint surface. – Line VL is an example of a tie line, which connects points representing phases in equilibrium. – W is a dewpoint and the lower surface is the dewpoint surface. Pxy diagram at constant T Txy diagram at constant P PT diagram at constant composition  Fig 10.8 (a)(b), Negative departures from P-x1 linearity: strong liquid- phase inter-molecular attractions between unlike than between like pairs of molecules. Fig 10.8 (c)(d), Positive departures from P-x1 linearity: strong liquid- phase inter-molecular attractions between like than between unlike pairs of molecules. Fig 10.8 (b)(d), the “azeotrope”: the point where x1 = y1 the dewpoint and bubblepoint curves are tangent to the same horizontal line. The liquid does not change in composition as it evaporates. No separation of such a constant-boiling solution is possible by distillation. Fig 10.8 Fig 10.8 Fig 10.9 Fig 10.10  Simple models for VLE The simplest are Raoult’s law and Henry’s law. Raoult’s law: – the vapor phase is an ideal gas (apply for low to moderate pressure) – the liquid phase is an ideal solution (apply when the species that are chemically similar) – sat yi P  xi Pi (i 1, 2,..., N ) although it provides a realistic description of actual behavior for a small class of systems, it is valid for any species present at a mole fraction approaching unity, provided that the vapor phase is an ideal gas.  1 yi P  xi Pi sat (i 1, 2,..., N ) P (i 1, 2,..., N )  yi / Pi sat x i 1 i yi i 1 i For dewpoint calculation P  xi Pi sat (i 1, 2,..., N ) i For bubblepoint calculation Binary system P P2sat  ( P1sat  P2sat ) x1 Binary system acetonitrile (1)/nitromethane(2) conforms closely to Raoult’s law. Vapor pressures for the pure species are given by the following Antoine equations: sat 2945.47 2972.64 ln P1 / kPa 14.2724   ln P2sat / kPa 14.2043   t / C  224.00 t / C  209.00 (a) Prepare a graph showing P vs. x1 and P vs. y1 for a temperature of 75°C. (b) Prepare a graph showing t vs. x1 and t vs. y1 for a pressure of 70 kPa. sat sat sat (a) BUBL P P P2  ( P1  P2 ) x1 At 75°C P1sat 83.21 P2sat 41.98 e.g. x1 = 0.6 x1 P1sat (0.6)(83.21) P 41.98  (83.21  41.98) x1 y1   0.7483 P 66.72 P 66.72 At 75°C, a liquid mixture of 60 mol-% (1) and 40 mol-% (2) is in equilibrium with a vapor containing 74.83 mol-% (1) at pressure of 66.72 kPa. Fig. 10.11 Fig. 10.11 (b) BUBL T, having P = 70 kPa 2945.47 2972.64 ln P1sat / kPa 14.2724   ln P2sat / kPa 14.2043   t / C  224.00 t / C  209.00 P1sat P  P2sat Select t x1  sat t vs. x1 t vs. y1 P2sat P1  P2sat xP 1 1 sat y1  P Fig. 10.12  Henry’s law For a species present as a very dilute solute in the liquid phase, the partial pressure of the species in the vapor phase is directly proportional to its liquid-phase mole fraction: yi P  xi H i Table 10.1 Assuming that carbonated water contains only CO2 (species 1) and H2O (species 2) , determine the compositions of the vapor and liquid phases in a sealed can of “soda” and the pressure exerted on the can at 10°C. Henry’s constant for CO 2 in water at 10°C is about 990 bar. Henry’s law for species 1: y1 P  x1 H1 Raoult’s law for species 2: y2 P  x2 P2sat P  x1 H1  x2 P2sat Assuming x1 = 0.01 P (0.01)(990)  (0.99)(0.01227) 9.912 y1 P  x1 H1 assuming y1 = 1.0 Justified the assumption x1 0.01 y2 P  x2 P2sat y2 0.0012 y1 0.9988 Justified the assumption  VLE modified Raoult’s law Account is taken of deviation from solution ideality in the liquid phase by a factor inserted into Raoult’s law: sat yi P  xii Pi (i 1, 2, 3,...N ) The activity coefficient, f (T, xi) P  xii Pi sat i 1 P  yi / i Pi sat i For the system methanol (1)/methyl acetate (2), the following equations provide a reasonable correlation for the activity coefficients: ln 1 (2.771  0.00523T ) x22 ln 2 (2.771  0.00523T ) x12 The Antoine equations provide vapor pressures: 3643.31 2665.54 ln P1sat / kPa 16.59158  ln P2sat / kPa 14.25326  T ( K )  33.424 T ( K )  53.424 Calculate (a): P and {yi} for T = 318.15 K and x1 = 0.25 (b): P and {xi} for T = 318.15 K and y1 = 0.60 (c): T and {yi} for P = 101.33 kPa and x1 = 0.85 (d): T and {xi} for P = 101.33 kPa and y1 = 0.40 (e): the azeotropic pressure and the azeotropic composition for T = 318.15 K (a) for T = 318.15, and x1 = 0.25 P1sat 44.51 P2sat 65.64 1 1.864 2 1.072 P  xii Pi sat (0.25)(1.864)( 44.51)  (0.75)(1.072)(65.64) 73.50 i yi P  xii Pi sat y1 0.282 y2 0.718 (b): for T = 318.15 K and y1 = 0.60 P1sat 44.51 P2sat 65.64 1 An iterative process is applied, with 1 1 2 1 P  yi / i Pi sat i y1 P x1  x2 1  x1 1 P1sat Converges at: P 62.89 kPa 1 1.0378 2 2.0935 x1 0.8169 (c): for P = 101.33 kPa and x1 = 0.85 T1sat 337.71 T2sat 330.08 A iterative process is applied, with T (0.85)T1sat  (0.15)T2sat 336.57 1 ... 2 ... 3643.31 ln P1sat / kPa 16.59158  T ( K )  33.424 P  xii Pi sat i P1sat ... Converges at: T 331.20 K 1 1.0236 2 2.1182 y1 0.670 y2 0.330 (d): for P = 101.33 kPa and y1 = 0.40 T1sat 337.71 T2sat 330.08 A iterative process is applied, with T (0.40)T1sat  (0.60)T2sat 333.13 1 1 2 1 Bi ln Pi sat / kPa  Ai  T ( K )  Ci P1sat ... P2sat ... 3643.31 ln P1sat / kPa 16.59158  P  xii Pi sat T ( K )  33.424 i x1 ... x2 ... 1 ... 2 ... P1sat ... P2sat ... Converges at: T 326.70 K 1 1.3629 2 1.2523 x1 0.4602 x2 0.5398  (e): the azeotropic pressure and the azeotropic composition for T = 318.15 K y1 x1 yi P  xii Pi sat 1 P1sat Define the relative volatility: 12  12  y2 2 P2sat x2 Azeotrope y1  x1 y2  x2 12 1 P1sat exp( 2.771  0.00523T ) P1sat 12 x1 0  sat 2.052 12 x1 1  sat 0.224 P2 P2 exp( 2.771  0.00523T ) Since α12 is a continuous function of x1: from 2.052 to 0.224, α12 = 1 at some point There exists the azeotrope! 1 P1sat 1az P2sat 12  sat 1 az  sat 1.4747 2 P2 2 P1 ln 1 (2.771  0.00523T ) x22 1 ln (2.771  0.00523T )( x2  x1 ) (2.771  0.00523T )(1  2 x1 ) ln 2 (2.771  0.00523T ) x12 2 az az 1az 1.657 x 0.325  y 1 1 P az 1az P1sat 73.76kPa  VLE from K-value correlations A convenient measure, the K-value: yi Ki  xi – the “lightness” of a constituent species, i.e., of its tendency to favor the vapor phase. – The Raoult’s law: P sat i Ki  P – The modified Raoult’s law: i Pi sat Ki  P Fig 10.13 Fig 10.14 For a mixture of 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane at 50°F, determine: (a) the dewpoint pressure, (b) the bubblepoint pressure. The K- values are given by Fig. 10.13. (a) at its dewpoint, only an insignificant amount of liquid is present: P = 100 (psia) P = 150 (psia) P = 126 (psia) Species yi Ki yi /Ki Ki yi /Ki Ki yi /Ki Methane 0.10 20.0 0.005 13.2 0.008 16.0 0.006 Ethane 0.20 3.25 0.062 2.25 0.089 2.65 0.075 Propane 0.70 0.92 0.761 0.65 1.077 0.762 0.919 Σ (yi /Ki) = 0.828 Σ (yi /Ki) = 1.174 Σ (yi /Ki) = 1.000 (b) at bubblepoint, the system is almost completely condensed: P = 380 (psia) P = 400 (psia) P = 385 (psia) Species xi Ki xi Ki Ki x i Ki Ki x i Ki Methane 0.10 5.60 0.560 5.25 0.525 5.49 0.549 Ethane 0.20 1.11 0.222 1.07 0.214 1.10 0.220 Propane 0.70 0.335 0.235 0.32 0.224 0.33 0.231 Σ (xi Ki) = 1.017 Σ (xi Ki) = 0.963 Σ (xi Ki) = 1.000  Flash calculations A liquid at a pressure equal to or greater than its bubblepoint pressure “flashes” or partially evaporates when the pressure is reduced, producing a two-phase system of vapor and liquid in equilibrium. Consider a system containing one mole of nonreacting chemical species: L V 1 zi  xi L  yiV zi  xi (1  V )  yiV zi K i The moles of vapor yi  The vapor mole fraction 1  V ( K i  1) The moles of liquid The liquid mole fraction zi K i  1  V ( K  1) 1 i  The system acetone (1)/acetonitrile (2)/nitromethane(3) at 80°C and 110 kPa has the overall composition, z1 = 0.45, z2 = 0.35, z3 = 0.20, Assuming that Raoult’s law is appropriate to this system, determine L, V, {xi}, and {yi}. The vapor pressures of the pure species are given. Do a BUBL P calculation, with {zi} = {xi} : Pbubl  x1 P1sat  x2 P2sat  x3 P3sat (0.45)(195.75)  (0.35)(97.84)  (0.20)(50.32) 132.40 kPa Do a DEW P calculation, with {zi} = {yi} : 1 Pdew  sat sat sat 101.52 kPa y1 / P1  y2 / P2  y3 / P3 L 1  V 0.2636 mol Since Pdew < P = 110 kPa < Pbubl, the system is in the two-phase region, Pi sat zi K i Ki  K1 1.7795 K 2 0.8895 K 3 0.4575  1  V ( K  1) 1 V 0.7364 mol P i zi K i yi  1  V ( K i  1) yi x1 0.2859 Ki  y1 0.5087 xi x2 0.3810 y2 0.3389 x3 0.3331 y3 0.1524