f 20130617 Chak Krit 46
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*Chapter 4 Distillation DesignSubject: 1304 332 Unit Operation in Heat transfer
Instructor: Chakkrit UmpuchDepartment of Chemical EngineeringFaculty of EngineeringUbon Ratchathani University
*What are you going to learn in this chapter?4.1 Vapor-Liquid Equilibrium Relations4.2 Single-Stage Equilibrium Contact for Vapor-Liquid System4.3 Simple Distillation Methods4.4 Distillation with reflux and McCabe-Thiele Method
*4.1 Vapor-Liquid Equilibrium Relations4.1.1 Raoults LawAn ideal law, Raoults law, can be defined for vapor-liquid phases in equilibrium (only ideal solution e.g. benzene-toluene, hexane-heptane etc.
Where pA is the partial pressure of component A in the vapor in Pa (atm) PA is the vapor pressure of pure A in Pa (atm)xA is the mole fraction of A in the liquid.
Composition in liquid:Composition in vapor:(1)(2)(3)
*Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure of 101.32 kPa.4.1.2 Boiling-Point Diagrams and xy PlotsDew point is the temperature at which the saturated vapour starts to condense.
Bubble-point is the temperature at which the liquid starts to boil.
The difference between liquid and vapour compositions is the basis for distillation operations.
*Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure of 101.32 kPa.4.1.2 Boiling-Point Diagrams and xy PlotsIf we start with a cold liquid composition is xA1 = 0.318 (xB1 = 0.682) and heat the mixture, it will start to boil at 98C. The first vapor composition in equilibrium is yA1 = 0.532 (yB1 = 0.468).Continue boiling, the composition xA will move to the left since yA is richer in A.
*4.1.2 Boiling-Point Diagrams and xy Plots
The boiling point diagram can be calculated from (1) the pure vapor-pressure data in the table below and (2) the following equations:(4)(5)(6)Where pA, pB are the partial pressure of component A and B in the vapor in Pa (atm) PA , PB are the vapor pressure of pure A and pure B in Pa (atm)P is total pressure in Pa (atm)xA is the mole fraction of A in the liquid.
*4.1.2 Boiling-Point Diagrams and xy Plots
The boiling point diagram can be calculated from (1) the pure vapor-pressure data in the table below and (2) the following equations:1
*Ex 4.1 Use of Raoults Law for Boiling-Point DiagramCalculate the vapor and liquid compositions in equilibrium at 95C (368.2K) for benzene-toluene using the vapor pressure from the table 1 at 101.32 kPa.Solution: At 95C from Table 1 for benzene, PA = 155.7 kPa and PB = 63.3 kPa. Substituting into Eq.(5) and solving,
155.7(xA) + 63.3(1-xA) = 101.32 kPa (760 mmHg)
Hence, xA= 0.411 and xB= 1 xA = 1 - 0.411 = 0.589. Substituting into eqn.(6),
The boiling point diagram can be calculated from the pure vapor-pressure data in the table below and the following equations:1
*A common method of plotting the equilibrium data is shown in Fig. 2 where yA is plotted versus xA for the benzene-toluene system. The 45 line is given to show that yA is richer in component A than is xA.Fig. 2 Equilibrium diagram for system benzene(A) toluene(B) at 101.32 kPa (1atm).
*4.1 Vapor-Liquid Equilibrium Relations4.1.2 Boiling-Point Diagrams and xy Plots
Maximum-boiling azeotropeMinimum-boiling azeotropeIdeal boiling point diagramAn azeotrope is a mixture of two or more liquids in such a ratio that its composition cannot be changed by simple distillation.This occurs because, when an azeotrope is boiled, the resulting vapor has the same ratio of constituents as the original mixture.
*4.1.2 Boiling-Point Diagrams and xy Plots4.1 Vapor-Liquid Equilibrium Relations
*4.2 Single-Stage Equilibrium Contact for Vapor-Liquid System
Total mass balance:
Mass A balance:Where V1, V2 is a vaporL0, L1is a liquidA single equilibrium stage is - the two different phases are brought into intimate contact with each other. - The mixing time is long enough and the components are essentially at equilibrium in the two phases after separation.In case of constant molal overflow : V1 = V2 and L0 = L1
*Ex 4.2 Equilibrium Contact of Vapor-Liquid MixtureA vapor at the dew point and 101.32 kPa containing a mole fraction of 0.40 benzene (A) and 0.60 toluene (B) and 100 kg mol total is contacted with 110 kg mol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molal overflow. Calculate the amounts and compositions of the exit streams.Solution: The given values are V2 = 100 kg mol, yA2 = 0.40, L0=110 kg mol , and xA0 = 0.30.For constant molal overflow, V2 = V1 and L0 = L1.
*Material balance on component A,
To solve equation above, the equilibrium relation between yA1 and xA1 in figure below must be used.First, we assume that xA1 = 0.20 and substitute into equation above to solve for yA1.
Assuming that xA1 = 0.20 and solving yA1 = 0.51. Next, assuming that xA1=0.40 and solving, yA1 = 0.29.Next, assuming that xA1=0.40 and solving, yA1 = 0.29. (These point are plotted on the graph.)At the intersection of this line with the equilibrium curve, yA1 = 0.455 and xA1 = 0.25.Answer
*4.3 Simple Distillation Methods4.3.1 IntroductionDistillation is a method used to separate the components of liquid solution, which depends upon the distribution of these various components between a vapor and a liquid phase.
The vapor phase is created from the liquid phase by vaporization at the boiling point.
Distillation is concerned with solution where all components are appreciably volatile such as in ammonia-water or ethanol-water solutions, where both components will be in the vapor phase.
*4.3.2 Relative Volatility of Vapor-Liquid SystemsRelative volatilityIt is a measure of the differences in volatility between 2 components, and hence their boiling points. It indicates how easy or difficult a particular separation will be.Where AB is the relative volatility of A with respect to B in the binary system. when AB is above 1.0, a separation is possible. Raoults law:
*Ex 4.3 Using data from table 1 calculate the relative volatility for the benzene-toluene system at 85C (358.2K) and 105C (378.2K) Solution: At 85C, substituting into equation below for a system following Raoutls law,
Similarly at 105C,
The variation in is about 7%.Answer
*4.3.3 Equilibrium or Flash DistillationDistillation has two main methods in practice.184.108.40.206 Introduction to distillation methodsProduction of vapor by boiling the liquid mixture to be separated in a single stage and recovering and condensing the vapors. No liquid is allowed to return to the single-stage still to contact the rising vapors. Returning of a portion of the condensate to the still. The vapors rise through a series of stages or trays, and part of the condensate flows downward through the series of stages or trays countercurrently to the vapors (fractional distillation, distillation with reflux, or rectification).
There are 3 important types of distillation that occur in a single stage or still: Equilibrium or flash distillation, Simple batch or differential distillation and simple steam distillation
*220.127.116.11 Equilibrium or Flash DistillationFlash distillation is a single stage separation technique.1. A liquid mixture is pumped through a heater to raise the temperature and enthalpy of the mixture.2. It then flows through a valve and the pressure is reduced, causing the liquid to partially vaporize.3. Once the mixture enters a big enough volume (the flash drum), the liquid and vapor separate.4. Because the vapor and liquid are in such close contact up until the flash occurs, the product liquid and vapor phases approach equilibrium.
*18.104.22.168 Equilibrium or Flush DistillationTotal mass balance:Component A balance:whereF, V and L are flow rate of feed, vapor and liquid phases.xF, yA and xA are mole fraction of component A in feed, vapor and liquid.Where f = V/F = molal fraction of the feed that is vaporized and withdrawn continuously as vapor.1-f = one as liquidMaterial balance for more volatile component :
*Ex 4.4 A mixture of 50% mole normal heptane and 50% normal octane at 30C is continuously flash distilled at 1 standard atmosphere so that 60 mol% of the feed is vaporized. What will be the composition of the vapor and liquid products?xA 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9yA 0.247 0.453 0.607 0.717 0.796 0.853 0.898 0.935 0.968
Solution: Given: xF = 0.5, f = 0.6Find: xA, yABasis: F = 100 molsApplying the mass balance yields:Since ,
Material balance for more volatile component,
Subtituting value of f =0.6 and xF =0.5 we get,
Assuming that xA = 0.5 and solving yA = 0.5. Next, assuming that xA=0 and solving, yA = 0.83.(These point are plotted on the graph.)At the intersection of this line with the equilibrium curve, yA = 0.58 and xA = 0.39.Answer
xF =0.5yF = 0.5
xA =0.39yA = 0.581st 2nd 3th
*4.3.4 Simple Batch or Differential DistillationThe pot is filled with liquid mixture and heated.Vapour flows upwards though the column and condenses at the top.Part of the iquid is returned to the column as reflux, and the remainder withdrawn as distillate.Nothing is added or withdrawn from the still until the run is completed.
*4.3.4 Simple Batch or Differential DistillationThe total moles of component A left in the still n