Chemical Engineering Thermodynamics Vapor-Liquid Equilibrium PDF

Chemical Engineering Thermodynamics Vapor/liquid equilibrium The nature of equilibrium  Equilibrium : A static condition in which no changes occur in the macroscopic properties of a system with time. The T, P,...

Chemical Engineering Thermodynamics Vapor/liquid equilibrium The nature of equilibrium  Equilibrium : A static condition in which no changes occur in the macroscopic properties of a system with time. The T, P, composition reaches final value which will remain fixed: equilibrium Phase rule & Duhem’s Theory Equilibrium states are determined by; ⁃ Phase Rule ⁃ Duhem’s Theory Phase Rule Number of variables Difference between total number that may be of variables that characterize the independently fixed in a intensive state of the system and system at equilibrium number of independent equation F = 2-π+N Where : F – degrees of freedom π – No of phase N – No of species Duhem’s theorem: Applies to the extensive state of a closed system at equilibrium When both the extensive state and the intensive state of the system are fixed, the state of the system is said to be completely determined Total number of variables in a completely determined state :  2 + (N − 1)π intensive phase-rule  π extensive variables represented by the masses (or mole numbers) of the phases  Equals 2 + (N − 1)π + π = 2 + Nπ variables Total number of equations  (π − 1)(N) phase-equilibrium equations  N material-balance equations  Equals (π − 1)N + N = πN equations Duhem’s theorem: The difference between the number of variables and the number of equations gives: 2 + Nπ − πN = 2 Duhem’s theorem is thus stated: “For any closed system formed from known amounts of prescribed chemical species, the equilibrium state is completely determined when any two independent variables are fixed” The two independent variables subject to specification may in general be either intensive or extensive  VAPOR/LIQUID EQUILIBRIUM: QUALITATIVE BEHAVIOR For a system comprised of two chemical species (N = 2), the phase rule becomes F = 4 − π Since there must be at least one phase (π = 1), the maximum number of phase rule variables that must be specified to fix the intensive state of the system is three: P, T, & one mole (or mass) fraction All equilibrium states of the system can therefore be Figure 12.1: PTxy diagram for vapor/ liquid equilibrium. represented in three-dimensional P-T-composition space  Under surface- sat. V states (P-T-y1) Upper surface- sat. L states (P-T-x1) Liquid at F, reduces pressure at constant T & composition along FG, the first bubble appear at L – bubble point As pressure reduces, more & more L vaporizes until completed at W; point where last drop of L (dew) disappear – dew point Line LV is called a tie line which connects vapor and liquid compositions in equilibrium Quantitative descriptions of VLE Simple Models For VLE : Find T, P, composition  Raoult’s Law  Henry’s Law Raoult’s Law Assumptions: V- phase is an ideal gas ⁃ Applicable for low to moderate pressure L- phase is an ideal solution ⁃ Valid only if the species are chemically similar (size, same chemical nature e.g. isomers such as ortho-, meta- & para-xylene) Excess Gibbs Energy And Activity Coefficients Partial Gibbs energy for a solution is From Lewis/Randall rule for an ideal solution The difference gives us: is partial excess Gibbs energy and Where γi is defined as activity coefficient of species i in solution Analogous expression for excess Gibbs energy with that of the residual Gibbs energy  Phase equilibrium criteria = But ^ ^𝑣 𝑦 𝑃 𝑓 𝑣𝑖 =ɸ 𝑖 𝑖 Similarly for liquids, ^ 𝑓 𝑙𝑖 =𝛾 ^ 𝑖𝑙 𝑥 𝑖 𝑓 𝑙𝑖 [ ] 𝑙 𝑠𝑎𝑡 ^ 𝑦 𝑃 =𝛾 𝑣 𝑙 𝑙 𝑉 𝑖 (𝑃 − 𝑃𝑖 ) ɸ 𝑖𝑖 ^ 𝑥𝑖 𝑓 But 𝑖 𝑓 𝑙 𝑖 =ɸ 𝑖 𝑠𝑎𝑡 𝑖 𝑃 𝑠𝑎𝑡 𝑖 𝑒𝑥𝑝 𝑅𝑇 the liquid ideal solution) ⇨ ===1  Assumptions of Raoult’s law (vapor phase ideal-gas state and  And pointing factor is close to unity at low to moderate pressures 𝑠𝑎𝑡 𝑦 𝑖 𝑃 = 𝑥𝑖 𝑃 𝑖 ❑ ( 𝑖=1 , 2 ,…. 𝑁 )  𝑠𝑎𝑡 𝑦 𝑖 𝑃 = 𝑥𝑖 𝑃 𝑖 ( 𝑖=1 , 2 ,…. 𝑁 ) Where; yi: V – phase mole fraction xi: L – phase mole fraction Pisat: Vapor pressure of pure species P : Total pressure  Dew point & Bubble point Calculations with Raoult’s Law BUBL P: Calculate {yi} and P, given {xi} and T DEW P: Calculate {xi} and P, given {yi} and T BUBL T: Calculate {yi} and T, given {xi} and P DEW T: Calculate {xi} and T, given {yi} and P For binary systems to solve for Bubble point calculation (T is given); ∑ 𝑦 𝑖 =1 𝑖 P= P= But P=  Vapor pressure of pure species i can 𝑠𝑎𝑡 𝐵𝑖 be obtained from Antoine equation 𝑙𝑛 𝑃 𝑖 = 𝐴𝑖 − 𝑇 + 𝐶𝑖 Raoult’s law equation can be solved for xi to solve for dew point calculation (T is given) ∑ 𝑥 𝑖 =1 𝑖 1 1 𝑃= 𝑃= ∑ 𝑦 𝑖 / 𝑃 𝑖𝑠𝑎𝑡 𝑦 1 / 𝑃 𝑠𝑎𝑡 1 + 𝑦 2 / 𝑃 𝑠𝑎𝑡 2 𝑖 Example: Binary system acetonitrile(1)/nitromethane(2) conforms closely to Raoult’s law. Vapor pressure for the pure species are given by the following Antoine equations: 𝑠𝑎𝑡 2945.47 𝑙𝑛 𝑃 1 (𝐾𝑃𝑎)=14.2724 − 𝑇 (𝑜𝐶)+ 244.00 𝑠𝑎𝑡 2972.64 𝑙𝑛 𝑃 2 (𝐾𝑃𝑎)=14.2043 − 𝑇 (𝑜𝐶)+ 209.00 a) Prepare a graph showing P vs. x1 and P vs. y1 at temperature 750C b) Prepare a graph showing T vs. x1 and T vs. y1 for a pressure of 70 kPa a) BUBL Point calculations are required. Since this is a binary system, we can use: 𝑠𝑎𝑡 𝑥1 𝑃 1 P= and 𝑦 1= 𝑃 ⁃ At 750C, the saturated pressures are given by Antoine equation 𝑠𝑎𝑡 𝑠𝑎𝑡 𝑃 1 =83.21 𝑃 2 =41.98 ⁃ Substitute both values in above equation to find P x1 0.0 0.2 0.4 0.6 0.8 1.0 y1 0.0000 0.3313 0.5692 0.7483 0.8880 1.0000 P/kPa 41.98 50.23 58.47 66.72 74.96 83.21 Pxy diagram for acetonitrile(1) /nitromethane(2) at 75oC as given by Raoult’s law b) When P is fixed, the T varies along T1sat and T2sat, with x1 & y1. T1sat & T2sat are calculated from Antoine equation 𝑠𝑎𝑡 𝐵𝑖 𝑇 𝑖 = − 𝐶𝑖 𝐴𝑖 − 𝑙𝑛𝑃 For P=70kPa, T1sat=69.840C, T2sat=89.580C. Select T between these two temperatures and calculate P1sat & P2sat for the two temperatures. ⇨ Evaluate x1 by Eq. P= (A) ⇨ Get y1 from Eq 𝑠𝑎𝑡 𝑥1 𝑃 1 𝑦 1= 𝑃 𝑠𝑎𝑡 2972.64 𝑙𝑛 𝑃 2 (𝐾𝑃𝑎)=14.2043 − 𝑇 (𝑜𝐶)+ 209.00 For x1=0.6 & P=70kPa, T is determined by BUBL T calculation, which requires iteration. Eq. 10.2 is rewritten; 𝑠𝑎𝑡 𝑠𝑎𝑡 𝑃 𝑃1 𝑃2 = (B) Where 𝛼= 𝑠𝑎𝑡 𝑥1 𝛼+ 𝑥 2 𝑃2 Subtracting lnP2sat from lnP1sat as given by Antoine equations yields; 2945.47 2972.64 𝑙𝑛𝛼=0.0681 − − (C) 𝑇 +224 𝑇 +209 Initial value for α is from arbitrary intermediate T With α, calculate P2sat by Eq. (B) Calculate T from Antoine eq. for species 2 Find new α by Eq. (C) Return to initial step and iterate until converge for final value of T x1 y1 T/oC 0.0000 0.0000 89.58(t2sat) 0.1424 0.2401 86 0.3184 0.4742 82 0.5156 0.6759 78 0.7378 0.8484 74 1.0000 1.0000 69.84(t1sat) Txy diagram for acetonitrile(1) /nitromethane(2) at 70KPa as given by Raoult’s law Henry’s Law Assumptions: 1. Very low pressure where the gas can be assume as ideal gas 2. The liquid phase is assumed as a very dilute solution in liquid phase 𝑦 𝑖 𝑃 =𝑥𝑖 𝐻 𝑖 (1 , 2 , ….3) Where; P: Total pressure Hi: Henry's constant yi : V-phase mole fraction xi : L-phase mole fraction Table 13.2: Henry’s Constants for Gases Dissolved in Water at 25°C Example: Assuming that carbonated water contains only CO2(1) and H2O(2), determine the compositions of the V & L phases in a sealed can of ‘soda’ & the P exerted on the can at 100C. Henry’s constant for CO2 in water at 100C is about 990 bar and x1=0.01. Solution: Henry’s law for species 1 & Raoult’s law for species 2 are written as: 𝑠𝑎𝑡 𝑦 1 𝑃 =𝑥1 𝐻 1 and 𝑦 2 𝑃 =𝑥2 𝑃 2 𝑠𝑎𝑡 𝑃 = 𝑥 1 𝐻 1 + 𝑥2 𝑃 2 With H1=990 bar & P2sat = 0.01227 bar (from steam tables at 100C) P= (0.01) (990) + (0.99) (0.01227) P= 9.912 bar Then by Raoult’s law 𝑠𝑎𝑡 𝑥2 𝑃 2 (0.99)( 0.01227) 𝑦 2= = =0.0012 𝑃 9.912 Thus; y1=1-y2=0.9988, and the vapor phase is nearly pure CO2, as expected. 1000194298424 -