1 1 PSE Summer School 2012 Process Simulation and Optimization of Chemical Plants DAAD Summer School 2012, Mexico Sponsorship: DAAD German Academic Exchange service Process Simulation and Optimization Chapter 2 Modeling and Simulation of Distillation Processes Prof. Dr.-Ing. habil. Prof. h.c. Dr. h.c. G. Wozny 3 Aims Introduction - Overview, Aims Transfer the basic modeling workflow to a more complex distillation process. Balance equations DOF Solution using MOSAIC 4 Application of Process Simulation A. Steady State Simulation: Decision making Process Understanding 2. Process DevelopmentProcess Synthesis 3 DesignParameter Optimization, Sensitivity study 4. Sizing Diameter, Cost Calculation 5. Sensitivity Analysis Disturbances behavior 6. FlexibilityLoad, Input changes B. Dynamic Simulation: Operability, FlexibilitySet Point Behavior,Disturbance Behavior 2. Dynamic ProceduresStart up, Shut down, Safety aspects 3. Experimental DesignPlanning of Experiments 4. TrainingOperator Training, Safety Training 5. Optimal ControlAdvanced Control Strategy 5 Process Simulation Flowsheet 3D Design Design, Sizing Data Bank Concept Process Simulation 6 Main Programm Optimization Cost Engng. Unit Operations/User Added Modules Input Output Sequentiell modular Simulator Data Bank Thermodynamic-Subroutines Thermophysical Properties Recycle Streams Cal. Numerical Programs 7 Modeling Systems of process units rectification Columns are high! Sequential contact of vapor and liquid in countercurrent flow Liquid Vapor Bottom Product 8 Modeling Systems of Process units internals trays Tray construction A - A B - B B A - A B - B 9 9 Packing column Modeling 10 Modeling Raschig GmbH, Ludwigshafen 11 Systems of Processunits Rectification apparatuses Realisation j F j+1 n-2 n-1 n j LnLn VnVn V n-1 V n-2 V j+1 L n-1 L n-2 L j+1 LjLj L j-1 L3L3 L2L2 L1L1 VjVj V j-1 V3V3 V2V2 V 1 =D steam Cooling water V 1 =D LnLn F L1L1 12 Systems of Process units Rectification apparatuses Realisation j F j j+1 n-2 n n-1 L1L1 V1V1 V2V2 V3V3 V j-1 L2L2 L3L3 L j-1 LjLj L j+1 L n-2 L n-1 LnLn VjVj V j+1 V n-2 V n-1 V n =D steam Cooling water V n =D L1L1 F LnLn 13 Thermal Separation Process 13 Modellierungssystematik Systeme von Prozessunits Darstellung der Sprudelschicht Zum Verstndnis des hydraulischen Verhaltens helfen die Abbildungen, Filme der Lektion PAD-Kolonnen Hinweis: In der bung und in anderen Lehrveranstaltungen, in vielen Lehrbchern wird die Rektifikation am Beispiel binrer Ge- mische im McCabe-Thiele Diagramm und im h-x Diagramm erlutert. Details siehe Vorlesung TGO. Hier wird der Schwerpunkt auf die Modellierung gelegt. Dazu ist u.a. ein vertieftes Verstndnis der physikalischen Zusammenhnge erforderlich N min = N = 21 r min = r = Stripping line Rectifying line McCabe-Thiele-Diagramm X CH3OH Y CH3OH Design Method Phase equilibrium 14 Thermal Separation Process Fundamentals of Modelling: Task: Separate a mixture with 50% low boiler and 50% high boiler Mixture Build two Phases Mass Transfer Separate the two Phases 15 Thermal Separation Process 15 Feed (40 Gew.-% MeOH) T=30C F=100 t/h Destillate 98 Gew.-% MeOH; T=63C; M=40 t/h Bottomproduct 99,9 Gew.-% H 2 O; T=99,8C; M=60 t/h 1 bar Destillation e.g. Methanol-Water: Feed: 100 t/h; T=30C; MeOH=40 weight % Destillate: 98 weight-% MeOH Bottomproduct: 99,9 weight-% Water 16 Thermal Separation Process 16 Feed Destillate Bottomproduct j n Condenser Reboiler Akkumulator L j-1 LjLj V j+1 VjVj 17 Thermal Separation Process 17 Feed Destillat Bottom product Model Modell Theoreticl tray tray j n Condenser Reboiler Phasengrenzflche j Liquidphase Vaporphase Vaporphase Vaporphase Liquidphase Liquidphase j+1 j 18 Thermal Separation Process 18 Realization Vapor Steam Destillate Cooling water Reflux 19 Distillation - Rectification Realization Separation processes Feed Distillate Bottomproduct j n Condenser Reboiler Accumulator Vapor steam Distillate Cooling water Reflux 20 Distillation - Rectification R =L/D= const. x D = const. time Reflux ratio Colomn Bottom L D F CW Conzentration time Spezifikation time Relux ratio Conzentration time Spezifikation V X 1,D 21 Distillation - Rectification Bottom Condenser Educts? Initialization Time = 0 How much? Concentration, Temperature? Diameter? Condenser Area? Reboiler area? Distillation time? Switching time from one Vessel to the next? Pressures? Answer by simulation and Optimization! How much? Q = ? Qcond = ? P = ? D = ? bottom Bottom V D L x D F CW 22 Distillation - Rectification Condenser Thermodynamic Data? Cost? Control? Measured variables? Manipulated variables Flexibility? Start up? Shut down? Heat integration? Column sequence? Column with wall? Answer by simulation and Optimization! How much? Q = ? Qcond = ? P = ? Feed Destillate Bottomproduct Bottomproduct j n Reboiler Accumulator D = ? How much? Trays, Packing? How much? 23 Batch Distillation Integral Balances 1.Step: qualitative analysis 2.Step: integral balances 3. Step: detailed modeling Modeling Systematic bottom Bottom V D L x D F CW HU 24 Batch Distillation Vakuum-Pumpe Condenser Steam Bottom Feed Dampf Destillate Application 25 Batch Distillation xDxD xBxB Temperature T Qualitative results time bottom Bottom V D L x D F CW T xBxB 26 Batch Distillation Modeling: 1. Approach Integral Integral modeling is not suitable Vapor steam D x 1,D HU, x 1,B d HU dt = - D HU d x 1,B d t = D ( x 1,B - x 1,D ) d x 1,B x 1,D - x 1,B = d HU HU (Rayleigh Eq.) d ( HU x 1,B ) d t = - D x 1,D 27 Batch Distillation Integral balances d x 1,B x 1,D - x 1,B = d HU HU x 1.B x 1,B x 1,B x 1,B Y1Y1 x 1,B x 1,B x 1,B y 1 x 1 Assumption Phase equilibrium it is: y = x D A = dx 1,B y 1,B x 1,B x 1,B x 1,B Vapor steam D x 1,D HU, x 1,B 28 Distillation Batch Distillation Problem: Column, dynamic Process, time dependency Modelling: d x B x B - x D = d HU HU (Rayleigh Eq.) Column Wall Bubble Cup Tray weir VjVj y i,j LjLj x i,j V j+1 L j-1 bottom V D L x D F CW 29 Flash Evaporator Column Wall Bubble Cup Tray weir VjVj Y i,j LjLj X i,j V j+1 L j-1 Modeling: assumption steady state V j+1 VjVj L j-1 L j, X i,j T j, P j Y i,j 1. Step: Choose the suitable balance Volume Q. 30 Flash Evaporator V j+1 VjVj L j-1 L j, x i,j T j, P j y i,j 2. Step: Formulate the whole set of linear independent balance equations Q. 31 Flash Evaporator Component Material Balances F 1 x 1,1 = F 2 x 1,2 + F 3 x 1,3 (1) F 1 x 2,1 = F 2 x 2,2 + F 3 x 2,3 (2) Summation Equation x 1,1 + x 2,1 = 1 (3) x 1,2 + x 2,2 = 1 (4) x 1,3 + x 2,3 = 1 (5) Phase Equilibrium F 1 F 2, x 1,2, x 2,2 x 1,1, x 2,1 F 3, x 1,3, x 2,3 T, P Q. 32 Flash Evaporator Phase Equilibrium x 1,2 = K 1 x 1,3 (6) x 2,2 = K 2 x 2,3 (7) Vapor Pressure (with A,B,C from Lit.) F 1 F 2, x 1,2, x 2,2 x 1,1, x 2,1 F 3, x 1,3, x 2,3 T, P 1 2 K-Values K 1 = f (x 1,2, x 2,2, x 1,,3, x 2,3,T,P ) = p 01 / P (8) K 2 = f (x 1,2, x 2.2, x 1,3, x 2,3, T. P ) = p 02 / P (9) ln p 01 = A1 + B1 / ( T + C1 ) (10) ln p 02 = A2 + B2 / (T + C2) (11) With: T 2 = T 3 = T P 2 = P 3 = P Q. 33 Flash Evaporator Activity Coefficient 1 = Wilson eq. or NRTL eq.... (12) 2 = Wilson eq. or. NRTL eq.... (13) Unknown : Degree of Freedom: N d = N V - N E = = 6 1. Law of Thermodynamik. Q = F 2 * h 2 + F 3 * h 3 - F 1 * h 1 (14) Enthalpy Correlations h 1 = f ( T 1, P 1, x 1,1, x 1,2 ) (15) h 2 = f (T, P, x 1,2, x 1,3 ) (16) h 3 = f(T, P, x 1,3, x 2,3 ) (17) choosen: F 1, x 1,1, T 1, P 1, P, Q F 1 F 2, x 1,2, x 2,,2 x 1,1, x 2,1 F 3, x 1,3, x 2,3 T, P Q. F 1, F 2, F, x 1,1, x 2,1, x 1,2, x 2,2, x 1,3, x 23, T 1,T, P 1,P, Q, K 1, K 2, 1, 2, p 01, p 02, h 1, h 2, h 3 = 23 F 1, F 2, F 3, x 1,1, x 2,1, x 1,2, x 2,2, x 1,3, x 23, T 1,T, P 1,P, Q, K 1, K 2, 1, 2, p 01, p 02, h 1, h 2, h 3 = 23.. 34 Flash Evaporator Degree of Freedom: N d = N V - N E = = 6 choosen: F 1, x 1,1, T 1, P 1, P, Q Solution: 17 Equations with 17 Variables Sequential Iterative: Rearrangement + Substitution results to the well known Flash Algorithm often published in Lit. i = 1NC x i,1 ( 1 K i ) ( + K i ) = 0 with = F 3 / F 2 Solve eq. by Newton Iteration F 1 F 2, x 1,2, x 2,2 x 1,1, x 2,1 F 3, x 1,3, x 2,3 T, P. Q. 35 Flash Evaporator Solution: 17 Equations with 17 Variables Simultaneous Iterative Solution: Solve 17 Equations with 17 Variables with the need of 17 suitable starting values for all varibles with Newton Raphson Method: J = Jacobian Matrix f ( x ) J +1 x = x - F 1 F 2, x 1,2, x 2,2 x 1,1, x 2,1 F 3, x 1,3, x 2,3 T, P Degree of Freedom: N d = N V - N E = = 6 choosen: F 1, x 1,1, T 1, P 1, P, Q choosen: F 1, x 1,1, T 1, P 1, P, Q Q.. 36 Modelling a theoretical Plate Transfer to a Model Theoretical Plate Feed Distillate Bottom product Bottom product j n Condenser Reboiler Recfication Column Column Wall Bubble Cup Tray weir VjVj Y i,j LjLj X i,j Y i,j = K i,j X i,j Feed VjVj L j V j+1 L j-1 Vaporphase Liquidphase j 37 Modelling Nonequilibrium Model Transfer to a NonequilibriumModel Feed Destillate Bottom product Bottom product j n Condenser Reboiler j Y i,j bulk Dampf Y i,j * X i,j * X i,j bulk Flssigkeit Mass Transfer j 38 Modelling Nonequilibrium Model Transfer to a NonequilibriumModel Feed Destillate Bottom product Bottom product j n Condenser Reboiler Feed VjVj L j V j+1 L j-1 j Y i,j bulk Dampf Y i,j * X i,j * X i,j bulk Flssigkeit Mass Transfer Literature: Ross Taylor (Di)Still Modeling after all these years: A View of the state of the Art Distillation and Absorption 2006, Institution of Chemical Engineers Symposium Series No. 152, S. 1-20, ISBN 39 First Step: Integral Balances F = D + B Total Mass Balance Component Mass Balance F x F = D x D + B x B F ( 1 - x F ) = D ( 1 - x D ) + B ( 1 - x B ) Feed Distillate Bottom product Bottom product j n Condenser Reboiler D, x D B, x B F, x F binary 3 Equations 6 Variables Degree of Freedom: N d = 3 choosen: F, x F, x D or.... But: additional Information is needed, tray number,.. 40 Integral Balances Feed Distillate Bottom product Bottom product j n Condenser Reboiler D, x D B, x B F, x F Balance volume 1 j-1 Balance volume j Balance volume n Feed LjLj L j-1 L j+1 VjVj V j+1 j+1 D B 41 Modelling a theoretical Plate Transfer to a Model Theoretical Plate j Feed Distillate Bottom product Bottom product n Condenser Reboiler Vapor Interfacial Area Liquid...... 42 Modelling a theoretical Plate Transfer to a Model Theoretical Plate j Feed Distillate Bottom product Bottom product n Condenser Reboiler... Interficial Area j Liquidphase Vaporphase Vaporphase Vaporphase Liquidphase Liquidphase j+1 j 43 Model of a Theoretical PlateWith: Y= Vapor Mole FractionY= Vapor Mole Fraction X= Liquid Mole FractionX= Liquid Mole Fraction V= Vapor StreamV= Vapor Stream L= Liquid StreamL= Liquid Stream h V = Vapor Enthalpyh V = Vapor Enthalpy h L = Liquid Enthalpieh L = Liquid Enthalpie i=component ; j=tray, l,m sink, k,n source VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjVhjVhjVhjV T j ; p j FjFjFjFj Z(Y or X) i,j hjFhjFhjFhjF Vaporphase Liquidphase j Interficial Area WjWj WR j,m WR n,j UjUj UR k,j UR j,l U,W = Side stream, Recycle, 44 Modelling of a Theoretical Plate 1. Mass Balance 2. Component Mass Balance 3. Phase Equilibrium 4. Summation Equation 5. Energy Balance MESH Equation Feed VjVj V j+1 L j-2 V j-1 V j+2 L j+1 Interficial Area j Liquidphase Vaporphase Vaporphase Vaporphase Liquidphase Liquidphase j+1 j L j L j-1 W j WR j,m WR n,j UR j,l UR k,j U j 45 Material Balances M j = F j + V j+1 + L j-1 - V j - L j = 0 Steady state VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjLhjLhjLhjL T j ; p j FjFjFjFj Z(Y or X) i,j hjFhjFhjFhjF Vaporphase Liquidphase j Interficial Area Without side streams 46 Component Material Balances Steady state M i,j = F j *Z i,j + V j+1 *Y i,j+1 + L j-1 *X i,j-1 V j *Y i,j L j *X i,j =0 =0 VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjLhjLhjLhjL T j ; p j FjFjFjFj Z(Y or X) i,j hjFhjFhjFhjF Vaporphase Liquidphase j Interficial Area Bilineare Terme 47 Phase Equilibrium K i,j = Y i,j / X i,j -> G i,j = Y i,j K i,j * X i,j = 0 with K = Equilibrium Constant VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjLhjLhjLhjL T j ; p j FjFjFjFj Z(Y or X) i,j hjFhjFhjFhjF Vaporphase Liquidphase j Interficial Area 48 Energy Balance VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjLhjLhjLhjL T j ; p j FjFjFjFj Z(Y or X) i,j hjFhjFhjFhjF Vaporphase Liquidphase j Interficial Area QjQj. E j = F j *h j F + V j+1 *h j+1 V + L j-1 * h j-1 L - V j *h j V - L j *h j L +/- Q j = 0. 49 i = 1 NC SY j = 1- Y i,j = 0 i = 1 NC SX j = 1- X i,j = 0 Summenbeziehungen VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjLhjLhjLhjL T j ; p j FjFjFjFj Z(Y or X) i,j hjFhjFhjFhjF Vaporphase Liquidphase j Interficial Area Material Balance M j = 0 M j = 0 2. Component Material Balance M i,j = 0 M i,j = 0 3. Phase Equilibrium G i,j = 0 G i,j = 0 4. Energy Balance E j = 0 E j = 0 Modelequations of a theoretical Plate 5.Summation equation SX j = 0 But: - be carefull look linear dependent eq. Feed VjVj L j V j+1 L j-1 L j-2 V j-1 V j+2 L j+1 Interficial Area j Liquidphase Vaporphase Vaporphase Vaporphase Liquidphase Liquidphase j+1 j Component Material Balance M i,j = 0 M i,j = 0 2. Phase Equilibrium G i,j = 0 G i,j = 0 3. Energy Balance E j = 0 E j = 0 Modelequations of a theoretical Plate 4. Summation equation SX j = 0 Feed VjVj L j V j+1 L j-1 L j-2 V j-1 V j+2 L j+1 Interficial Area j Liquidphase Vaporphase Vaporphase Vaporphase Liquidphase Liquidphase j+1 j 5. Summation equation SY j = 0 Another choice 52 Material balances alternative M i,j = F j Z i,j + v i,j+1 + l i,j-1 v i,j l i,j = 0 VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjLhjLhjLhjL T j ; p j FjFjFjFj Z (Y or X) i,j hjFhjFhjFhjF Vapor phase Liquid phase j Interfacial Area Reformulation of the balance equations with new variables component flows -> elimination of the bilinear terms l i,j-1 =L j-1 *X i,j-1 l i,j =L j *X i,j v i,j =V j *Y i,j v i,j+1 =V j+1 *Y i,j+1 53 Phase equilibrium alternative K i,j = v i,j / v i,j / (l i,j / l i,j ) -> G i,j = v i,j / v i,j K i,j * l i,j / l i,j = 0 G i,j = v i,j / v i,j K i,j * l i,j / l i,j = 0 VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjLhjLhjLhjL T j ; p j FjFjFjFj Z(Y or X) i,j hjFhjFhjFhjF Vapor phase Liquid phase j Interfacial Area l i,j =L j *X i,j v i,j =V j *Y i,j i i i i Model Theoretical tray Feed VjVj L j V j+1 L j-1 L j-2 V j-1 V j+2 L j+1 Interfacial area j Liquid phase Vapor phase Liquid phase j+1 j 1. Component material balances M i,j = 0 2. Phase equilibrium G i,j = 0 3. Energy balance E j = 0 Summation equation: Definition v i,j = V j i and l i,j = L j i Component Material Balance M i,j = 0 2. Phase Equilibrium G i,j = 0 3. Energy Balance E j = 0 Modelequations of a theoretical Plate I prefer the following set of linear independent Equations to be solved iteratively But now with the state varaibles l i,j = X i,j L j and v i,j = Y i,j V j and: v i,j = V j i=1 NC With the definition: l i,j = L j i=1 NC and Temperatures T j 56 M i,1 E 1 G i,1... M i,j E j G i,j... M i,n E n G i,n F(X) = = 0 Equation System F = ( F 1,..., F j,...F n ) T X = ( X 1,..., X j,...X n ) T X j = ( v i,j, l i,j, T j ) T Feed VjVj L j V j+1 L j-1 L j-2 V j-1 V j+2 L j+1 j-1 Liquidphase Vaporphase Vaporphase Vaporphase Liquidphase Liquidphase j+1 j Matrix Formulation 57 M i,1 E 1 G i,1... M i,j E j G i,j... M i,n E n G i,n F(X) = = 0 Equation System F = ( F 1,..., F j,...F n ) T X = ( X 1,..., X j,...X n ) T X j = ( v i,j, l i,j, T j ) T Matrix Formulation From the summation equation follows: l i,j = L j i=1 NC and: v i,j = V j i=1 NC With Definition: X i,j = l i,j / L j Y i,j = v i,j / V j 58 Simultaneous Solution Solution Method X +1 =X + X +1 X +1 =( F/ X) -1 F X = Variable vector (v i,j, l i,j,T j ) F = Function vector (MESH Equation system) (MESH Equation system) ( ) = Jacobimatrix - Feed VjVj L j V j+1 L j-1 L j-2 V j-1 V j+2 L j+1 j-1 Liquidphase Vaporphase Vaporphase Vaporphase Liquidphase Liquidphase j+1 j Iteration number Newton correction...... 59 Remarks: Expanditure of Equilibrium Modelling to Nonequilibrium (rate based approach) VjVjVjVj Y i,j hjVhjVhjVhjV V j+1 Y i,j+1 h j+1 h j+1 V X i,j-1 L j-1 h j-1 L LjLjLjLj X i,j hjLhjLhjLhjL T j ; p j FjFjFjFj Z(Y or X F ) i,j Z(Y F or X F ) i,j hjFhjFhjFhjF vaporphase liquidphase j (Interphase) Liquid layer Vapor Layer N i,j L N i,j V E i,j Y X 60 Taylor, R.: (Di)Still Modeling after all these Year: A view of the State of the Art Distillation and Absorption 2006; Institution of Chemical Engineers, IChemE 2006 Symposium Series No. 152; S. 1-20; ISBN Thiele, R.; Lning, J.-M.: Industrial Absorption Current Status an future Aspects Distillation and Absorption 2006; Institution of Chemical Engineers, IChemE 2006 Symposium Series No. 152; S ; ISBN See lecture: Nichtgleichgewichtsmodellierung thermischer Trennprozesse Dr.-Ing. J.-U. Repkeoder 61 Feed Distillate Bottom product j n Condenser Reboiler Design Variables: Feed Flow Feed Temperature Feed composition (mole fraction) Feedtray Column pressure (top pressure) Number of trays Pressure loss (or calculated) Now the degree of freedom is 2! You can choose principal any variable, but Distillate rate and reflux ratio L/D is easy To estimate (material balance, min. reflux Ratio correlations in Lit.) D L Remarks degree of freedom 62 Aspects to the Degree of Freedom Feed Distillate Bottomproduct j j 1 2 L1L1 V2V2 n Remember: For the single Flash : N D = N in *Nc + 2 (Number of input streams, -Components, Heat duty, pressure) For a adiabatic single column with condenser and Reboiler and a given feed, given top pressure and the given pressure lost for each tray The degree of freedom is: = 2 (Q 1, Q n ) Remember: Standard: Distillate stream and Reflux L 1 or Reflux ratio 63 Simultaneous Solution Jacobian X = Variable vector, F =Function vector, (...) =Jacoby matrix FF ( ) = xx B 1 C 1 A 2 B 2 C 2 A 3 B 3 C 3 A 4 B 4 C 4 A 5 B 5 C 5 A j B j C j A n-2 B n-2 C n-2 A n-1 B n-1 C n-1 A n B n... A j = ( F j / x j-1 ) B j = ( F j / x j ) C j = ( F j / x j+1 ) 64 Modeling Transformation for: j = 1,2,3,.,n-1, n for Step 1 C j = B j -1 C j j n-1 F j = B j -1 F j j n-1 B j+1 = B j+1 A j+1 C j j n-1 F j+1 = F j+1 A j+1 F j j n-1 Step 1: Step 2: x n = F n x j = F j - C j x j+1 j = n-1, n-2, 3,2,1 65 Beside the simultaneous solution method with the Newton Rahpson method there exist also some other Procedures - Sequential iterative instead of simultaneous - Homotophy method instead of Newton Raphson method 66 Sequential Solution Material balances B = L 6 L5L5 F4F4 Z 1,4 Z 2,,4 D L1L1 L4L4 L2L2 V5V5 V3V3 V2V2 -L 1 x 1,1 - D x 1,1 + V 2 y 1,2 = 0 L 1 x 1,1 - L 2 x 1,2 - V 2 y 1,2 + V 3 y 1,3 = 0 L 2 x 1,2 - L 3 x 1,3 - V 3 y 1,3 + V 4 y 1,4 = 0 L 3 x 1,3 - L 4 x 1,4 - V 4 y 1,4 + V 5 y 1,5 = -F 4 z Remark: In the past besides the simultaneous approach the sequential solution procedure is often used in simulation tools. The following example shows the solution procedure: Component material balances 1: 67 Sequential Solution Material balances B = L 6 L5L5 F4F4 Z 4,1 Z 4,2 D L1L1 L4L4 L2L2 V5V5 V3V3 V2V2 -L 1 x 1,1 - D x 1,1 + V 2 y 2,1 = 0 L 1 x 1,1 - L 2 x 2,1 - V 2 y 2,1 + V 3 y 3,1 = 0 L 2 x 2,1 - L 3 x 3,1 - V 3 y 3,1 + V 4 y 4,1 = 0 L 3 x 3,1 - L 4 x 4,1 - V 4 y 4,1 + V 5 y 5,1 = -F 4 z 4,1 b 1 x 1,1 + c 1 x 2,1 = 0 a 2 x 1,1 + b 2 x 2,1 +c 2 x 3,1 = 0 a 3 x 2,1 + b 3 x 3,1 +c 3 x 4,1 = 0 a 4 x 3,1 + b 4 x 4,1 +c 4 x 5,1 = d 4 a 5 x 4,1 + b 5 x 5,1 +c 5 x 6,1 = 0 a 6 x 5,1 + b 6 x 6,1 = Solve the material balances with Tridiagonal Algorithmus 68 Sequentiel Solution Material balances B = L 6 L5L5 F4F4 Z 4,1 Z 4,2 D L1L1 L4L4 L2L2 V5V5 V3V3 V2V2 S j = y j,i - 1 = 0 i=1 Nc j=1,2,..6 E j = 0 Iteration variable: T j, V j 12 Equation 12 unknown variables Iterative solution (simultaneously) with e.g. Newton Raphson Method 69 Modeling Sequential Solution Component Material balance Given: Design variables Iteration variable T j *, V j * Calculation K i,j with Raoults law | T jnew -T j * | < | V jnew -V j * | < Calculation L j from Material balances x j,i from Comp. Mat. Balance Tridiagonal Algorithmus Solution Summation Equation and Energy Balance -> T jnew, V jnew no yes End Thermodynamic data +Check Itmax 70 Simulation Results Results Stream profiles V j, L jStream profiles V j, L j Concentration profiles X i,j, Y i,jConcentration profiles X i,j, Y i,j Temperature profile T jTemperature profile T j Heat Duty Q 1, Q nHeat Duty Q 1, Q n Feed VjVj L j V j+1 L j-1 L j-2 V j-1 V j+2 L j+1 j-1 Liquidphase Vaporphase Vaporphase Vaporphase Liquidphase Liquidphase j+1 j MOSAIC