VIEWS: 251 PAGES: 8 POSTED ON: 3/7/2011
Chemical Engineering Thermodynamics Engi-3434 Dr. Charles Xu @ Chemical Engineering, Lakehead University 7-2. Vapor-Liquid Equilibrium ⎯ Henry’s Law Air bubbles in water Review of Raoult’s Law Two Major Assumptions: 1. The vapor phase is considered as an ideal gas ⇒ valid thus only for low to moderate pressure vapors 2. The liquid phase is considered as an ideal solution ⇒ valid for the solutions whose comprising species are chemically similar, e.g., solutions of isomers, solutions of similar compounds such as n-hexane / n- heptane, ethanol/propanol, benzene/toluene, acetone (CH3COCH3)/acetonitrile (CH3CN), etc. • Raoult’s Law: y i P = xi Pi sat (i = 1, 2, .... , N) (10.1) where: P = total pressure of the system = pressure of the vapor or liquid phase. Pi sat = Saturated vapor pressure of species i Note: when the temperature T > Tc (critical temp), Roault’s Law is not valid since Pi sat does not exist at T > Tc. Limitation of Raoult’s Law An example: Consider a system in which air (1) and liquid water (2) is at equilibrium at 25°C and 1 atm, the mole fraction of water vapor (y2) in the gas phase can be calculated by Roault’s law assuming the dissolved air in water is negligible, i.e., x1 ≈ 0 or x2 ≈ 1 sat x 2 P2 y2 = P From Antoine eq., water’s saturation pressure at 25°C, P2sat =3.166 kPa x 2 P2sat (1.0)(3.166kPa) y2 = = = 0.0312 P 101.33kPa y1 = 1 − y 2 = 1 − 0.0312 = 0.9688 However, if we wish to calculate the accurate mole fraction of dissolved air in the water (x1 = y1P/P1sat?), we CANNOT use Roault’s law, because the critical temperature of air (TC = 132.2 K) is lower than the temperature 25°C (298.15 K) when the saturated vapor pressure (P1sat) does not exist! 7.3.2 Simple Models for Vapor/Liquid Equilibrium for Solutions ⎯ Henry’s Law For a species whose critical temperature (TC) is less than the temperature of application (T) when the saturated pressure P1sat is meaningless, e.g., for most small molecule gases (N2, O2, air, H2, CO2, etc.) in equilibrium with a liquid species, Raoult’s Law is not valid, and Henry’s law is usually used. • Henry’s Law: William Henry y i P = xi H i (10.4) (1775 – 1836), English chemist. where: P = total pressure of the system = pressure of the vapor/liquid phases. H i = Henry’s constant of species i, determined by experiment. Note: Henry’s law is usually used for a dilute species in the liquid phase. 7.3.2 Simple Models for Vapor/Liquid Equilibrium for Solutions ⎯ Henry’s Law (Cont’d) • Examples of Henry’s Constants Note: • All measured by experiments • Henry’s constant Hi has the unit of pressure • Henry’s constant of species i, is function of temperature. With the Henry’s constant for air in water at 25°C, we can now finish the preceding example to calculate the mole fraction of dissolved air in the water (x1), 1bar 0.9688 × 101.33kPa y1 P 100kPa = 1.35 × 10 −5 x1 = = H1 72950bar Example 7.2 Assuming that carbonated water contains only CO2 (1) and H2O (2), determine the compositions of the vapor and liquid phases in a sealed can of “soda” at 10°C and 10 bar. Henry’s constant for CO2 in water at 10°C is about 990 bar. Solution will be provided at the tutorial #8 class: 7.4 Vapor/Liquid Equilibrium from K-Value Correlations The equilibrium ratio Ki for species “i” is defined by yi Ki ≡ (10.10) xi This ratio is usually called “K-value”, which is a function of temperature, total pressure and species. The K-value of species “i”, Ki, suggests the tendency of the species to partition itself preferentially between vapor and liquid. • Ki < 1 ⇒ yi < xi: Species “i” is a “heavy” constituent in the mixture, exhibiting a higher concentration in the liquid phase. • Ki > 1 ⇒ yi > xi: Species “i” is a “light” constituent in the mixture, exhibiting a higher concentration in the vapor phase. 7.4 Vapor/Liquid Equilibrium from K-Value Correlations (Cont’d) Comparing to the preceding Raoult’s law, modified Raoult’s law and Henry’s law, we have y i Pi sat Ki ≡ = (10.11) xi P yi H i Ki ≡ = (10.12) x1 P Both Pisat and Hi are dependent on temperature, so that the value of Ki for species i depends on temperature and pressure! For bubble point calculations: ∑K x i i i =1 (by ∑yi i = 1) (10.13) For dew point calculations: ∑yi i / K i = 1 (by ∑ xi = 1 ) i (10.14) Example 7.3 For a mixture of 10 mol% methane, 20 mol% ethane and 70 mol% propane at temperature T = 50°F, determine: (a) the dew point pressure; and (b) the bubble point pressure, employing the K-Value method . Solution: (a) the dew point pressure At the dew point, yi = zi. For the given temperature, the K-values depend on the choice of P. Now, we’ll find a P (i.e., the dew point pressure) to satisfy the dew point equation (10.14) by trial-and-error. ∑i y / K = 1 (10.14) i i i ∑ xi = 1 (b) the bubble point pressure At the bubble point, xi = zi. Now, we’ll find a P (i.e., the bubble point pressure) to satisfy the bubble point equation (10.13) by trial-and- error. ∑ K i xi = 1 (10.13) i 7.5 Application of VLE: Flash Calculation • Flash evaporation/distillation When reducing the pressure of a high-pressure liquid to below the bubble point pressure, the liquid would “flash” (evaporate) to produce a two-phase system of vapor and liquid in equilibrium. Flash (or partial) evaporation occurs when a saturated liquid stream undergoes a reduction in pressure by passing through a throttling valve or other throttling device to a pressure lower than its bubble point pressure. The throttling valve or device is normally located at the entry into a pressure vessel. The vesselis often referred to as a flash drum. 7.5 Application of VLE: Flash Calculation (Cont’d) , {yi} {zi} , {xi} Consider a system containing mixture of non-reacting chemical species with an overall composition of {zi} subject to flash evaporation. Let L be the mole fraction of the liquid phase, with mole fraction concentrations of {xi}, and let V be the mole fraction of the vapor phase, with mole fraction concentrations of {yi}. 7.5 Application of VLE: Flash Calculation (Cont’d) Material balance equations: L +V =1 z i = xi L + y iV (i = 1, 2, ..., N ) Combining the two equations to eliminate L gives: z i = xi (1 − V ) + y iV (i = 1, 2 , ..., N ) (10.15) y P sat Substituting xi = yi/Ki and solving for yi, where: K i ≡ i = i based on Raoult’s Law. xi P zi K i yi = (10.16) 1 + V ( K i − 1) Because ∑y i = 1 , thus ∑ 1 + V (K zi K i =1 (10.17) i i i − 1) Flash calculation procedures: (a) find the value of V which satisfies Eq. (10.17); (b) solve for yi by Eq. (10.16); (c) Find xi by xi = yi / Ki. Example 7.4 The system with 1 mol of mixture of acetone(1)/acetonitrile(2)/nitromethane(3) at 80°C and 110 kPa has the overall composition: z1 = 0.45, z2 = 0.35 and z3 = 0.2. Assuming that Raoult’s law is appropriate to this system, determine L, V, {xi} and {yi}. The vapor pressures of the pure species at 80°C are: P1sat = 195.75 kPa, P2sat = 97.84 kPa, P3sat = 50.32 kPa Solution will be provided at the tutorial #8 class: Example 7.4 (Cont’d)