Presentation is loading. Please wait.

Presentation is loading. Please wait.

CHAPTER 5 Multi-phase systems

Similar presentations


Presentation on theme: "CHAPTER 5 Multi-phase systems"— Presentation transcript:

1 CHAPTER 5 Multi-phase systems
Sem 1, 2015/2016 ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga

2 Outline Bubble point Boiling point Dew point Raoult’s Law Henry’s Law

3 Virtually all commercial chemical processes involve operations in which material is transferred from one phase (gas, liquid, or solid) into another. These multiphase operations include all phase-change operations on a single species. such as freezing, melting, evaporation, and condensation, and most separation and purification processes, which are designed to separate components of mixtures from one another.

4 Examples of Multiphase Separation Processes
• Brewing a cup of coffee. Hot liquid water and solid ground coffee beans are contacted. Soluble constituents of the beans are transferred from the solid phase to a liquid solution (coffee), and then the residual solids (grounds) are filtered from the solution. The operation of dissolving a component of a solid phase in a liquid solvent is referred to as leaching. • Removal of sulfur dioxide from a gas stream. If a fuel that contains sulfur is burned. The product gas contains sulfur dioxide. If the gas is released directly into the atmosphere, the S02 combines with atmospheric oxygen to form sulfur trioxide. The S03 in turn combines with water vapor in the atmosphere to form sulfuric acid (H2S04), which eventually precipitates as acid rain. To prevent this occurrence, the combustion product gas is contacted with a liquid solution in an absorption or scrubbing process. The SO2 dissolves in the solvent and the clean gas that remains is released to the atmosphere.

5 SINGLE-COMPONENT PHASE EQUILIBRIUM
Phase Diagrams Estimation of Vapor Pressures

6 Phase Diagrams At most temperatures and pressures, a single pure substance at equilibrium exists entirely as a solid, liquid, or gas; but at certain temperatures and pressures, two and even all three phases may coexist. Pure water is a gas at 130°C and 100 mm Hg, solid at -40°C and 10 atm, but at 100°C and 1 atm it may be a gas, a liquid, or a mixture of both, at approximately °C and 4.58 mm Hg it may be a solid, a liquid, a gas, or any combination of the three.

7 A phase diagram of a pure substance is a plot of one system variable against another that shows the conditions at which the substance exists as a solid. a liquid, and a gas. The most common of these diagrams plots pressure on the vertical axis versus temperature on the horizontal axis. The boundaries between the single-phase regions represent the pressures and temperatures at which two phases may coexist.

8 Phase Diagrams of H20 and CO2

9 Several familiar terms may be defined with reference to the phase diagram.
If T and P correspond to a point on the vapor-liquid equilibrium curve for a substance, P is the vapor pressure of the substance at temperature T, and T is the boiling point (more precisely, the boiling point temperature) of the substance at pressure P. The boiling point of a substance at P = 1 atm is the normal boiling point of that substance. If (T, P) falls on the solid-liquid equilibrium curve, then T is the melting point or freezing point at pressure P. If (T, P) falls on the solid-vapor equilibrium curve, then P is the vapor pressure of the solid at temperature T, and T is the sublimation point at pressure P. The point (T, P) at which solid, liquid, and vapor phases can all coexist is called the triple point of the substance. The vapor-liquid equilibrium curve terminates at the critical temperature and critical pressure (Tc and Pc). Above and to the right of the critical point, two separate phases never coexist.

10 Phase Diagrams of H20 Notice that the phase transitions-condensation at point B and evaporation at point D-take place at boundaries on the phase diagram; the system cannot move off these boundaries until the transitions are complete.

11 Estimation of Vapor Pressures
The volatility of a species is the degree to which the species tends to transfer from the liquid (or solid) state to the vapor state. At a given temperature and pressure, a highly volatile substance is much more likely to be found as a vapor than is a substance with low volatility, which is more likely to be found in a condensed phase (liquid or solid). Separation processes such as distillation are used to separate more volatile species from less volatile species by partially vaporizing liquid mixtures. The vapor product is relatively rich in the more volatile feed components and the residual liquid is rich in the components with lower volatility. The vapor pressure of a species is a measure of its volatility: the higher the vapor pressure at a given temperature, the greater the volatility of the species at that temperature. Engineers who design and analyze separation processes therefore need to know the vapor pressures of process species as functions of temperature.

12 It often happens that tabulated vapor pressure data are not available at temperatures of interest, or they may not be available at all for a given species. One solution to this problem is to measure p* at the desired temperatures. Doing so is not always convenient, however, especially if a highly precise value is not required. An alternative is to estimate the vapor pressure using an empirical correlation for p*(T). A relationship between p*, the vapor pressure of a pure substance, and T, the absolute temperature, is the Clapeyron equation the latent heat of vaporization, or the energy required to vaporize one mole of the liquid. T is absolute temperature VI are the specific molar volumes (volume/mole) of liquid. Vg specific molar volumes (volume/mole) of gas (vapor)

13 Unless the pressure is extremely high, the specific volume of the liquid (Vl) is negligible relative to that of the vapor (i.e., Vg - VI = Vg). If we assume that this is the case, apply the ideal gas equation of state to the vapor (so that Vg is replaced with RT/ p* ) and rearrange the resulting equation with the aid of elementary calculus. We obtain If the vapor pressure of a substance is measured at several temperatures and In p* is plotted versus l/T (or p* is plotted versus l/T on semilog axes). From Equation above, the slope of the resulting curve at a given temperature equals This is the method most commonly used to determine heats of vaporization experimentally.

14 Suppose now that the heat of vaporization of a substance is independent of temperature (or nearly so) in the temperature range over which vapor pressures are available. Equation may then be integrated to yield the Clausius-Clapeyron equation where B is a constant that varies from one substance to another. According to this equation, a plot of In p* versus 1/T (or a semilog plot of p* versus 1/T) should be a straight line with slope and intercept B. If you know and p* at a single temperature To, you can solve the Clausius-Clapeyron equation for B and thereafter use this equation to estimate p* at any temperature close to To. If you have p* versus T data, you can plot In p* versus 1/T and determine and B graphically or by the method of least squares.

15 EXAMPLE: Vapor Pressure Estimation Using the Clausius-Clapeyron Equation
The vapor pressure of benzene is measured at two temperatures, with the following results: Tl = 7.6°C, T2 = 15.4°C, P1* = 40mm Hg P2*= 60mm Hg Calculate the latent heat of vaporization and the parameter B in the Clausius- Clapeyron equation and then estimate p* at 42.2°C using this equation.

16 T1=280.8K;P1=40 Clausius-Clapeyron equation
The vapor pressure of benzene is measured at two temperatures, with the following results: Tl = 7.6°C,T2 = 15.4°C, P1* = 40mm Hg P2*= 60mm Hg Calculate the latent heat of vaporization and the parameter B in the Clausius-Clapeyron equation and then estimate p* at 42.2°C using this equation. Clausius-Clapeyron equation T1=280.8K;P1=40

17 The vapor pressure of benzene is measured at two temperatures, with the following results: Tl = 7.6°C,T2 = 15.4°C, P1* = 40mm Hg P2*= 60mm Hg Calculate the latent heat of vaporization and the parameter B in the Clausius-Clapeyron equation and then estimate p* at 42.2°C using this equation. T2=288.6K;P2=60

18 GAS-LIQUID SYSTEMS: ONE CONDENSABLE COMPONENT
A law that describes the behavior of gas-liquid systems over a wide range of conditions provides the desired relationship. If a gas at temperature T and pressure P contains a saturated vapor whose mole fraction is Yi (mol vapor/mol total gas), and if this vapor is the only species that would condense if the temperature were slightly lowered, then the partial pressure of the vapor in the gas equals the pure-component vapor pressure pi* (T) at the system temperature. This equation is a limiting case of Raoult's law. It is the fundamental relation used in the analysis of equilibrated gas-liquid systems containing one condensable component.

19 Example: Composition of a Saturated Gas-Vapor System
Air and liquid water are contained at equilibrium in a closed chamber at 75°C and 760 mm Hg. Calculate the molar composition of the gas phase.

20 MULTICOMPONENT GAS-LIQUID SYSTEMS

21 MULTICOMPONENT GAS-LIQUID SYSTEMS
Gas-liquid processes that involve several components in each phase include many chemical reactions, distillation, and transfer of one or more species from a gas to a liquid (absorption or scrubbing) or vice versa (stripping). Variables can then be determined using equilibrium relationships for the distribution of components between the two phases. In this section we define several such relationships and illustrate how they are used in the solution of material balance problems like using: Vapor Liquid Equilibrium Data Roult’s Law and Henry Law

22 Vapor-Liquid Equilibrium Data
The best way to evaluate equilibrium compositions is from tabulated data. Perry's Chemical Engineers' Handbook, pp through 2-89, gives partial pressures of vapors over various liquid solutions. Try NEXT EXAMPLE which illustrates the use of such data

23 Example: Absorption of SO2
A gas stream consisting of 100 lb-mole/h of an SO2- air mixture containing 45 mole% SO2 is contacted with liquid water in a continuous absorber at 30°C. The liquid leaving the absorber is analyzed and found to contain 2.00 g of SO2 per 100 g of H20. Assuming that the gas and liquid streams leaving the absorber are in equilibrium at 30°C and 1 atm, calculate the fraction of the entering SO2 absorbed in the water and the required water feed rate.

24 Gas Stream (out): Y air YH2O YSO2 Gas Stream (in): Liquid Stream (in):
A gas stream consisting of 100 lb-mole/h of an SO2-air mixture containing 45 mole% SO2 is contacted with liquid water in a continuous absorber at 30°C. The liquid leaving the absorber is analyzed and found to contain 2.00 g of SO2 per 100 g of H20. Assuming that the gas and liquid streams leaving the absorber are in equilibrium at 30°C and 1 atm, Gas Stream (out): Y air YH2O YSO2 Gas Stream (in): Liquid Stream (in): Liquid Stream (out)

25 Gas Stream (out): Y air=0.727
YH2O=0.0416 YSO2=0.232

26 Gas Stream (out): Y air=0.727 YH2O=0.0416 YSO2=0.232
Mass fraction=mass SO2/total mass (100+2)

27

28

29

30 Raoult's Law & Henry's Law

31 Raoult's Law: P is the vapor pressure of pure liquid A at temperature T YA is the mole fraction of A in the gas phase. Raoult's law is an approximation that is generally valid when XA is close to I (that is when the liquid is almost pure A). It is also sometimes valid over the entire range of compositions for mixtures of similar substances, such as paraffinic hydrocarbons of similar molecular weights. Note: When XA = I (that is, when the liquid is pure A) -Raoult's law reduces to the expression PA = P*A(T) given previously for systems with only one condensable component

32 Henry's Law where HA(T) is the Henry's law constant for A in a specific solvent. Henry's law is generally valid for solutions in which XA is close to 0 (dilute solutions of A) provided that A does not dissociate. ionize, or react in the liquid phase. The law is often applied to solutions of non-condensable gases. Values of Henry's law constants (or closely related quantities) are given for several gases in water on pp through of Perry's Chemical Engineers' Handbook. A gas-liquid system in which the vapor-liquid equilibrium relationship for every volatile species is either Raoult's law or Henry's law is said to exhibit ideal solution behavior. An ideal liquid solution is a mixture of liquids that exhibits ideal solution behavior at equilibrium

33 Example: Raoult's Law and Henry's Law
Use either Raoult's law or Henry's law to solve the following problems. 1. A gas containing 1.00 mole% ethane is in contact with water at 20.0°C and 20.0 atm. Estimate the mole fraction of dissolved ethane. 2. An equimolar liquid mixture of benzene (B) and toluene (T) is in equilibrium with its vapor at 30.0°C. What is the system pressure and the composition of the vapor?

34 Solved using Henry’s Law Mole fraction, Xethane = ???
1. A gas containing 1.00 mole% ethane is in contact with water at 20.0°C and 20.0 atm. Estimate the mole fraction of dissolved ethane. Solved using Henry’s Law Mole fraction, Xethane = ??? YethaneP=Xethane*Hethane (at T=20oC)

35 Solved using Raoult's Law:
2. An equimolar liquid mixture of benzene (B) and toluene (T) is in equilibrium with its vapor at 30.0°C. What is the system pressure and the composition of the vapor? Solved using Raoult's Law: simple empirical equation that correlates vapor pressure-temperature data

36 APPENDIX B (PAGE 640 –Antoine Equation Constant (to find vapor pressure at temperature (T)
Benzene: A= B= C= Touluene: A= B= C=

37 Vapor-Liquid Equilibrium Calculations for Ideal Solutions
When a liquid is heated slowly at constant pressure, the temperature at which the first vapor bubble forms is the bubble-point temperature of the liquid at the given pressure. When a gas (vapor) is cooled slowly at constant pressure, the temperature at which the first liquid droplet forms is the dew-point temperature at the given pressure. Calculating bubble-point and dew-point temperatures can be a complex task for an arbitrary mixture of components. However, if the liquid behaves as an ideal solution (one for which Raoult's or Henry's law is obeyed for all components) and the gas phase can also be considered ideal, the calculations are relatively straight forward.

38 Suppose an ideal liquid solution follows Raoult's law and contains species A, B, C ...with known mole fractions XA, XB, Xc .... If the mixture is heated at a constant pressure to its bubble-point temperature Tbp, the further addition of a slight amount of heat will lead to the formation of a vapor phase. Since the vapor is in equilibrium with the liquid, and we now assume that the vapor is ideal (follows the ideal gas equation of state), the partial pressures of the components are given by Raoult's law, where P~ is the vapor pressure of component i at the bubble-point temperature. Moreover, since we have assumed that only A, B, C, ... are present in the system, the sum of the partial pressures must be the total system pressure, P; hence,

39 The pressure at which the first vapor forms when a liquid is decompressed at a constant temperature is the bubble-point pressure of the liquid at the given temperature. Above equation can be used to determine such a pressure for an ideal liquid solution at a specific temperature, and The mole fractions in the vapor in equilibrium with the liquid can then be determined as

40 The dew-point temperature of a gas (vapor) may be found using a method similar to that for bubble-point temperature estimation. Again, suppose a gas phase contains the condensable components A, B, C, ... and a non-condensable component G at a fixed pressure P. Let Yi be the mole fraction of component i in the gas. If the gas mixture is cooled slowly to its dew point, Tdp, it will be in equilibrium with the first liquid that forms. Assuming that Raoult's law applies, the liquid-phase mole fractions may be calculated as At the dew point of the gas mixture, the mole fractions of the liquid components (those that are condensable) must sum to 1:

41 The value of Tdp can be found by trial and error once expressions for p;(T) have been substituted.
The composition of the liquid phase may then be determined from Equation The dew-point pressure, which relates to condensation brought about by increasing system pressure at constant temperature. can be determined by solving for P: Liquid mole fractions may then be calculated from this equation with Tdp replaced by the system temperature, T.

42 THANK YOU Mid Term Exam 1 27 Oct 2015 DKD3
(up to Chapter 3 ONLY) 27 Oct 2015 DKD3


Download ppt "CHAPTER 5 Multi-phase systems"

Similar presentations


Ads by Google