Distillation I Mass Transfer for 4th Year Chemical Engineering Department Faculty of Engineering Cairo University THINGS YOU HAVE TO REMEMBER • • • • Phase Equilibria Dalton’s Law Raoult’s law Antoine Equation LET’S SAY THAT For a binary system: A = More Volatile Component B = Less Volatile Component PoA>PoB a=Relative volatility (no more recovery) PAo a o PB AND OF COURSE xi= composition of component i in liquid phase yi= composition of component i in vapour phase Pi=yiPT Dalton’s law Pi=xiPoi Raoult’s law B ln P A C T o Antoine Equation Where T in oK and Po in mmHg B A Or o P e C T Binary System Phase Diagram V T L+V y L 0% A 100%B x,y 100% A 0%B Temperature-Composition Diagram x x-y Diagram Binary System Phase Diagram For a pure component evaporation occurs at a constant temperature, so it’s called “BOILING POINT” For a mixture of two or more components evaporation occurs at increasing temperature so it’s called “BOILING RANGE” starting by Bubble point and ending at Dew point T Tdew Tbubble 0% A 100%B x,y 100% A 0%B WHAT IS THIS?????? It is PTxy diagram for phase equilibrium Effect of Pressure on Equilibrium AZEOTROPIC MIXTURE Some systems has what is so called AZEOTROPE like CS2Acetone and MethanolWater systems. An azeotropic mixture has a=1 so can’t be separated by distillation. This figure represents a “minimum azeotrope” AZEOTROPIC MIXTURE This figure represents a “maximum azeotrope” in the system acetonechloroform CALCULATING LIQUID VAPOUR EQUILIBRIA FOR BINARY SYSTEMS RELATIONS USED x-y diagram 1- I f we have TEMPERATURECOMPOSITION DIAGRAM T 1- Draw tie line 2- Project x and y of that line on the x-y diagram and get a point on the equilibrium curve 3- Repeat the last step several times 4- Connect the points to get the equilibrium curve y x 2- If vapour pressures are known as a function of temperature: T PAo PBo PT PA PB xB 1 xA PT xA P xB P o o o o PT xA P 1 x A PB xA PA PB PB o A o A o B PT P xA o PA P o B o B And by the general equilibrium relation for ideal binary systems: At equilibrium PGas phase PLiquidinterface y A PT x P o A A o A P yA xA K A x A PT The most used form of equilibrium relation for liquid-vapour systems (Either ideal or non-ideal, binary or multicomponent) 3- Another form of equilibrium relation ( Only for binary systems if a is known): PAo a o PB yA a yB xA xB o A P yA xA PT yA 1 - y A xA 1 - x A y A 1 - x A x A 1 - y A a x A 1- yA yA 1- x A α xA yA 1 a - 1x A yA P PT xA o A A NEW PHASE DIAGRAM ENTHALPY COMPOSITION DIAGRAM H= specific enthalpy of vapour. h= specific enthalpy of liquid. H BTU/Lb H Vapourous Curve h Liquidous Curve 0% A 100%B x,y 100% A 0%B Tie line is used for getting points on equilibrium curve as what was done in case of Temperature composition diagram. H This diagram is more difficult to draw, however it provides more information about enthalpy of streams. y x THEORY OF DISTILLATION For any system, once it’s in the wet region, it splits into two phases in equilibrium each of different compositions. The compositions of the two phases can be determined using “TIE LINE” y T x 0% A 100%B x,y 100% A 0%B Distillation operations Single Stage Simple Differential Distillation Steam Distillation Multistage Flash vaporization Distillation Binary system Multicomponent systems Simple Differential Distillation Sometimes called “ASTM distillation” and Used in labs Distillation occurs so that the feed is slowly evaporated and then condensed “DISTILLATE (D)” and remains un-evaporated portion “RESIDUES (W)”. Now to get xw and ῩD Residues (W) By Differential MB: F ln W xf dx xw y * x Distillate (D) Simple Differential Distillation F ln W xf dx xw y * x Solved by graphical integration OR Algebraically using the relation: (use when α given) 1 xW xF 1 xW 1 F ln ln ln W a 1 xW 1 xF 1 xF Then calculate ῩD using C.M.B equation 4- 100 kmoles of a mixture of A and B is fed to a simple still. The feed contains 50 mole % A and a product contains 5 mole % A is required. Calculate the quantity of product obtained. The equilibrium data is presented as follows: Mole fraction of A in liquid 0 0.2 0.4 0.6 0.8 1 Mole fraction of A in vapor 0 0.35 0.58 0.75 0.9 1 Givens: F=100kmole xw=0.05 xf=0.5 Here we will solve using graphical integration: From xw to xf F ln W xf dx xw y x x 1/(y-x) 0.05 26.67 0.2 6.67 0.4 5.56 0.5 6.06 The additional points are calculated by inter polation : dx 1 F ln [[(0.2 0.05) * (6.67 26.67)] W 0.05 y x 2 [0.2 * (6.67 5.56)] [0.1 * (5.56 6.06)]] 4.3045 F exp(4.3045) W W 1.35kmoles 0.5 5- A liquid containing 50% n-heptane and 50% noctane is differentially distilled at 1 atmosphere to vaporize 60 mole% of feed. Find the composition of the distillate and the residue. Find the boiling range during distillation. (The average volatility α=2.15). F ln W 1 xW x F 1 xW 1 ln ln a 1 xW 1 x F 1 xF By trial and error : X w=0.3285 From M.B: ῩD =0.614 Boiling range =109.78-114.586oC Steam Distillation Experimental method used for thermally sensible materials. For insoluble mixtures (hydrocarbon-water mixtures) each component exerts pressure equals the vapour pressure (x=1). If the summation of the vapour pressures equals the total pressure then the mixture will start boiling at a temperature lower than boiling point of the pure hydrocarbon. Steam is used to perform evaporation at a reduced temperature. Steam Distillation Steam functions: 1- Heating the batch (material to be distilled) to the bubble point. 2- Giving the batch the latent heat needed to vaporize. 3- Carrying the vapours. Steam Distillation To calculate the necessary amount of steam needed we will need to know: 1- Q1=heat required to heat the batch to Tb 2- Q2=Latent heat gained by the batch 3- amount of steam carrying the vapours Steam Distillation Q1=mbatch*CP ) batch*(Tb-TF) Q2=mbatch*lbatch Amount of steam carrying the vapours: Vapours leaving=water vapour+hydrocarbon vapour PT=Pwater+PHC mw nw Mw Pw M w m HC n HC M HC PHC M HC Steam Distillation Q1=mw1*l w Q2=mw2*l w Pw M w m w3 m batch PHC M HC mw1=Q1/l w mw2=Q2/l w Problem 8: 10 Kg batch of ethylaniline is to be steam distilled from small amount of non-volatile impurity. Saturated steam at 25 Psia is used. Initial temperature of ethylaniline is 40 oC and the distillation takes place at atmospheric pressure. a- At what temperature will the distillation proceeds? b- Determine the composition of the vapour phase. c- How much steam is used? Data: • Heat capacity of ethylaniline is 0.4 KCal/Kg.C • Heat capacity of steam is 0.35 KCal/Kg.C • Latent heat of vaporization of ethylaniline is 72 Kcal/Kg • Vapour pressures of water and ethylaniline are given in the table below: T (oC) Pw (mmHg) PEA (mmHg) 38.5 51.1 1 64.4 199.7 5 80.6 363.9 10 96 657.6 20 99.15 737.2 22.8 113.2 1225 40 T Pw PEA PT 38.5 51.1 1 52.1 64.4 199.7 5 204.7 80.6 363.9 10 373.9 96 657.6 20 677.6 99.15 737.2 22.8 760 113.2 1225 40 1265 Operation occurs at 1 atm=760 mmHg SO Temperature at which Ptotal=760 mmHg is 99.15oC Composition of vapour phase: At 99.15oC: Pw=737.2 mmHg PEA=22.8 mmHg yw=737.2/760=0.97 yEA=22.8/760=0.03 Amount of steam used: CP)EA= 0.4 KCal/Kg.C CP)Steam= 0.35 KCal/Kg.C lEA= 72 Kcal/Kg lSteam= 540 Kcal/Kg Q1=10*0.4*(99.15-40)=236.5 Kcal m1=Q1/lSteam= 236.5/540=0.438 Kg Q2=10*72=720 Kcal m2=Q2/lSteam= 720/540=1.333 Kg Pw M w 737.2 18 m w3 m batch 10 48.1 Kg PHC M HC 22.8 121 mw=0.438+1.333+48.1=49.87 Kg Flash Vaporization Distillation Flash vaporization or equilibrium distillation is a single stage operation where a liquid mixture is partially vaporized, the vapour allowed to come to equilibrium with the residual liquid and the resulting vapour and liquid phases are separated and removed. It’s easy but not efficient, and no packing or trays are needed. Generally used for easy separation (very high relative volatility) or as a primary separation step. How to Reach Flashing Conditions Changing Temperature T T 0% A 100%B Changing Pressure x,y 100% A 0%B 0% A 100%B x,y 100% A 0%B How to Reach Flashing Conditions Changing Temperature Changing Pressure V y F xf V y F xf L x L x Calculating Tbubble and Tdew Flashing occurs at a temperature between bubble and dew points. So calculating bubble and dew points is necessary to specify suitable flashing conditions. T bubble > T flashing > T dew Calculating Tbubble Bubble point is the temperature at which JUST ONE bubble of gas evaporates and is in equilibrium with the liquid whose composition will not be affected. T y Tbubble x 0% A 100%B xf 100% A 0%B Calculating Tbubble Then xi=xFi And yi=KixFi Where Ki=Poi/PT TO GET Tbubble: Tbubble 1. Assume Tbubble 2. Calculate yi’s 3. Calculate Syi. if equals 1 OK, else reassume Tbubble 0% A 4. Interpolate 100%B T y x xf 100% A 0%B (Tbubble)1 (Tbubble)TRUE (Tbubble)2 Sy1 Sy=1 Sy2 Sy OR 1 Tb 1 Tb TRUE Σy1 1 Tb 1 Tb 2 Σy1 Σy 2 (Tbubble)TRUE Tbubble Calculating Tdew Dew point is the temperature at which JUST ONE point of liquid condenses and is in equilibrium with the vapour whose composition will not be affected. T y Tdew x 0% A 100%B yf 100% A 0%B Calculating Tdew Then yi=xFi And xi=yFi/Ki Where Ki=Poi/PT Tdew TO GET Tdew: 1. Assume Tdew 2. Calculate xi’s 3. Calculate Sxi. if equals 1 OK, else reassume Tdew 0% A 4. Interpolate 100%B T y x xf 100% A 0%B Sx1 Sx=1 Sx2 (Tdew)1 (Tdew)TRUE (Tdew)2 Sx OR 1 Td 1 Td TRUE Σx 1 1 Td 1 Td 2 Σx 1 Σx 2 (Tdew)TRUE Tdew Flash Distillation Calculations Overall Material Balance: F=L+V Component Material Balance: F F xFi=L xi+V yi xF Equilibrium Relation: yi=ki xi V y L x Flash Distillation Calculations (L+V) xFi=L xi+V yi (L+V) xFi=L xi+V ki xi (L+V) xFi=(L+V ki) xi V y F LV xF xi x Fi L Vk i It’s also solved by trial and error on liquid and vapour flow rates. L x Flash Distillation Calculations V y For the easiness of trials: LV xi xi L Vk i L 1 V xfi L ki V L 1 V yi xi L 1 V ki xi (÷V) F xF L x Flash Distillation Calculations Finally: xi L 1 V x Fi L ki V yi L L V 1 V ki 1 x Fi Now trials are done by changing ONLY L/V till Sx and Sy are BOTH equal one Flash Distillation Calculations Once L/V is calculated, all needed variables can be calculated: F=L+V F L V F 1V L V 1 V L F V REMEMBER ..... The case we always deal with throughout our course is the ideal case, that is why we can say: Ki=Poi/PT HOWEVER ..... In general, the systems are non-ideal. K values then can be calculated from the relation i fio Ki i p Solving Flashing Problems IN EXAM Due to time limitations: 1- Assume L/V (say L/V=1) 2- Calculate Sx and Sy. 3- If Sx=1 AND Sy=1 L/V is correct 4- If Sx ≠1 OR Sy≠1 assume another L/V 5- Do the last step twice then interpolate (L/V)1 (L/V)operating (L/V)2 Sx1 Sx=1 Sx2 Sx OR 1 Σx1 1 Σx 1 Σx 2 (L/V)operating L/V LV LV LV LV 1 operting 1 2 Solving Flashing Problems Practically More iterations must be done to get more accurate results. Computer softwares such as Microsoft Excel can be used to do such iterations. Sometimes system is non-ideal, so other softwares like Hysys or Aspen can be used. We will now see how to do such calculations using Microsoft Excel. Microsoft Excel How about non-ideal systems All calculation steps will not be changed, only the method of calculating “K” will change. K can be got thermodynamically using a proper equation of state (somehow difficult). K can be got from practical data, i.e. charts. For systems of oil fractionation it’s so common to use charts that are got from experiments instead of calculating them. For More Information For more information please check: 1- Robert E. Treybal, “Mass Transfer Operations”, Third ed., Ch. 9, PP. 342 2- McCabe and Smith, “Unit Operations for chemical engineers”, Fifth ed., Ch. 19, pp. 588 3- Perry, R. H. and D. Green, eds., “Perry’s Chemical Engineer’s Handbook”, Seventh ed., Section 13, McGraw-Hill, New York, 1997.