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vs. Enthalpy Composition PonchonSavarit Plot We have begun to employ mass balances, both total and component. We will also need to employ energy balances, based on enthalpy, for certain separation problems. We can use the Enthalpy vs. composition plot to obtain this information. Lecture 5 1 Enthalpy vs. Composition Ponchon-Savarit Plot Lecture 5 2 Enthalpy vs. Composition PonchonSavarit Plot 3 phases are shown on the plot solid, liquid, and vapor. Temperature is represented by isothermal tie lines between the saturated liquid (boiling) line and the saturated vapor (dew) line. Points between the saturated liquid line and the saturated vapor line represent a two phase, liquidvapor system. An azeotrope is indicated by the composition at which the isotherm becomes vertical. Why? Why are the boiling point temperatures of the pure components different than those determined from the y vs. x and T vs. x,y plots for ethanolwater? The azeotrope for ethanolwater is indicated as T = 77.65 oC and a concentration of 0.955. Why is this different than that determined from the y vs. x and T vs. x,y plots for ethanolwater? Lecture 5 3 Enthalpy vs. Composition PonchonSavarit Plot Note the boiling temperatures of the pure components, water and ethanol, and the temperature of the azeotrope are different due to the pressure at which the data was taken: 2 P = 1 kg/cm (0.97 atm) 1 atm Water Ethanol Azeotrope 99.1 oC 100 oC 77.8 78.30 77.65 78.15 Lecture 5 4 Mole Fraction vs. Weight Fraction Note that the enthalpy composition plot is presented in terms of weight fractions we will typically use mole fractions so one must convert between the two. For ethanolwater, this can be readily done using the molecular weights, MWEtOH =46.07 and MWw = 18.02. Lecture 5 5 Azetrope Composition Mole Fraction vs. Weight Fraction Converting from wt fraction of the azeotrope to mole fraction: 0.96 wt fraction ethanol 46.07 g/gmole ethanol = 0.902 mole fraction ethanol 0.96 wt fraction ethanol (1 - 0.96 wt fraction ethanol) 46.07 g/gmole ethanol + 18.02 g/gmole water Thus, the azeotropic mole fraction is greater at P = 1 Kg/cm2 than at 1 atm: 0.902 vs. 0.8943. Although slight, one can begin to see the effect of pressure on the azeotropic point. Lecture 5 6 Converting Weight Fraction to Mole Fraction In General For a binary mixture: w1 mw1 x1 = w1 w + 2 mw1 mw2 For a mixture of C components: wi mwi xi = C wj mw j =1 j Lecture 5 7 Enthalpy vs. Composition PonchonSavarit Plot The bubble point temperature and dew point temperatures can be determined from the enthalpy vs. composition plot. The compositions of the 1st bubble formed and the last liquid drop can be determined from the enthalpy vs. composition plot. An auxiliary line is used to assist in these determinations... Lecture 5 8 Enthalpy vs. Composition Bubble Point Temperature o 82.2 C Lecture 5 9 Enthalpy vs. Composition 1st Bubble Composition Lecture 5 10 Enthalpy vs. Composition Dew Point Temperature o 94.8 C Lecture 5 11 Enthalpy vs. Composition Last Liquid Drop Composition Lecture 5 12 Enthalpy vs. Composition Enthalpy Determination The major purpose of an enthalpy diagram is to determine enthalpies. We will use enthalpies in energy balances later. For example, if one were given a feed mixture of 35% ethanol (weight %) at T = 92oC and P = 1 kg/cm2 and the mixture was allowed to separate into vapor and liquid, what would be the enthalpies of the feed, vapor, and liquid? Lecture 5 13 Enthalpy vs. Composition Enthalpy Determination 425 295 90 Lecture 5 14 Equilibrium Data How to Handle? Tabular Data Generate graphical plots Generate analytical expressions (curve fit) y vs. x (P constant) McCabeTheile Pot T vs. x,y (P constant) Saturated Liquid, Vapor Plot Enthalpy vs. composition (P constant, T) PonchonSavarit Plot Thermodynamics: Equations of state/Gibbs free energy models Distribution coefficients, K values Relative volatility DePreister charts Curve fit of data Graphical Analytical expressions Lecture 5 15 Analytical Expressions for Equilibrium To date, we have looked at various ways to represent equilibrium behavior of binary systems graphically. There are several disadvantages to using graphical techniques: We will now look at other representations for handling equilibrium data analytically... Lecture 5 One cannot readily plot multicomponent systems graphically (maximum is typically three). Separator design often has to be done using numerical methods; thus, analytical expressions for equilibrium behavior are needed. 16 Other Equilibrium Relationships Distribution Coefficient Another method of representing equilibrium data is to define a distribution coefficient or K value as: K A = yA / x A Eq. (2-10) KA is typically a function of temperature, pressure, and composition. The distribution coefficient K is dependent upon temperature, pressure, and composition. However, for a few systems K is independent of composition, to a good approximation, which greatly simplifies the problem. K A = K(T, p) Eq. (2-11) Lecture 5 17 Other Equilibrium Relationships DePriester Charts One convenient source of K values for hydrocarbons, as a function of temperature and pressure (watch units), are the DePriester charts (Figs. 211 and 212, pp. 2425, Wankat). The DePriester plots are presented two over different temperature ranges. Lecture 5 18 Lecture 5 19 Lecture 5 20 Using DePriester Charts Boiling Temperatures of Pure Components One can determine the boiling point for a given component and pressure directly from the DePriester Charts one can then determine which component in a mixture is the more volatile the lower the boiling point, the more volatile a component is. For a pure component, K = 1.0. Assume one wishes to determine the boiling point temperature of ethylene at a pressure of P = 3000 kPa... Lecture 5 21 bp o T = 9.5 C Lecture 5 22 Question DePriester Charts What are the equilibrium distribution coefficients, K, for a mixture containing: Ethylene nPentane nHeptane at T = 120 oC and P =1500 kPa? Lecture 5 23 Lecture 5 24 Answer DePriester Charts The equilibrium distribution coefficients, K, are: K Ethylene 8.5 nPentane 0.64 nHeptane 0.17 at T = 120 oC and P =1500 kPa. Lecture 5 25 Question Volatility What can one say about the volatility of each component from the K values? K Ethylene 8.5 nPentane 0.64 nHeptane 0.17 Lecture 5 26 Answer Volatility What can one say about the volatility of each component from the K values? K T boiling Ethylene 8.5 35.5 oC nPentane 0.64 153 oC nHeptane 0.17 >200 oC The boiling point temperatures of the pure components at P = 1500 kPa have also been determined from the DePriester charts for K = 1.0 for each component (nheptane's is off the chart). From the K values and the boiling point temperature of each pure component, one can say that the volatility follows the trend that ethylene>n pentane>nheptane. K A = yA / x A Lecture 5 27 Other Equilibrium Relationships DePriester Equation While the DePreister charts may be used directly, they have been conveniently fit as a function of temperature and pressure: ln K = a p2 a p3 a T1 + a T2 + a T6 + a p1 ln p + 2 + 2 2 T T p p Eq. (2-12) where T is in oR and p is in psia. Table 2.4 (p. 26, Wankat) contains the K fit constants along with their mean errors (again, watch units!). Eq. (2-12) provides an analytical expression which can be used in numerical analyses. We will use this later for bubble and dew-point temperature calculations. Lecture 5 28 Other Equilibrium Relationships Mole Fraction Vapor Pressure Relationship If one does not have equilibrium data, K can be approximated using other more common thermodynamic data quantities such as vapor pressures. From Raoult's law for ideal systems: p A = x A (VP) A Eq. (2-14) where pA is the pressure due to component A in the mixture and (VP)A is the vapor pressure of pure component A, which is temperature dependent. From Dalton's law of partial pressures: yA = pA p Eq. (2-15) Combining Eqs. (2-15) and (2-14) and rearranging yields: y A (VP) A = xA p Eq. (2-16) Lecture 5 29 Other Equilibrium Relationships Distribution Coefficient Vapor Pressure Relationship The left-hand side of Eq. (2-16) is the definition of the distribution coefficient K; thus, KA = (VP) A p Eq. (2-17) Eq. (2-17) allows one to obtain K's from the vapor pressures of the pure components, which can be readily found for many chemical species using the Antoine equation: ln(VP)A = A - B T+C Eq. (2-18) where A, B, and C are constants,...