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A Selection of Chemical Engineering Problems Solved using Mathematica 1- Chemical Kinetics and Catalysis 2- Applied Thermodynamics Housam BINOUS National Institute of Applied Sciences and Technology

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Successive First-Order Reversible Reactions We consider successive first-order reversible reactions : Governing equations are :

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[A i ] i Steady state solution Forward and Backward rate constants: k i,i+1 =1 and k i+1,i =0.9

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Transient solution A 100, A 200, A 300, A 400, A 500, A 600, A 700, A 800 and A 900 A1A1 A 1000

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1/ equilibrium constant K A and rate constants are k 1 and k 2 2/ rate limiting step, rate constant is k p, equilibrium constant K AB 3/ equilibrium constant 1/K C and rate constants are k 5 and k 6 Eley-Rideal Mechanism Rate expressions for Reaction A + B ’ C

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1/ equilibrium constant K A and rate constants are k 1 and k 2 2/ equilibrium constant K B and rate constants are k 3 and k 4 3/ equilibrium constant K D and rate constants are k 7 and k 8 4/ rate limiting step, rate constant is k p, equilibrium constant K AB 5/ equilibrium constant 1/K C and rate constants are k 5 and k 6 Rate expressions for Reaction A + B ’ C Adsorption competition with an inert component

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Reaction A + H2 ’ AH2 (for example: hydrogenation reactions) Rate expression when H2 follows dissociative adsorption 1/ equilibrium constant K A and rate constants are k 1 and k 2 2/ equilibrium constant K H2 and rate constants are k 3 and k 4 3/ rate limiting step, rate constant is k p 4/ equilibrium constant K AH2 and rate constants are k 5 and k 6 H 2 H s catalyst H s A A s AH 2 s catalyst

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Liquid-liquid Equilibrium of Ternary mixture Liquid phase activity coefficients from NRTL model :

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We choose values for X 1 and X 2 than we solve the nonlinear system of 8 equations with 8 unknowns. So far we have 5 equations and 6 unknowns, we need one more equation.

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Tie line Plait point Liquid-liquid equilibrium for Water-Benzene-Ethanol at 25 °C

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Liquid Liquid Extraction Liquid Liquid Equilibrium of ternary system Isopropyl ether-water-acetic acid at 20 °C and 1 atm : wt % water wt % acetic acid water rich phase isopropyl ether rich phase tie line

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Hunter and Nash Graphical Equilibrium Stage Method Mixing Point M = F + S= E 1 + R N Operating Point P = R i-1 - E i = F - E 1 = R N - S P M S F E1E1 RNRN RNRN R N-1 R1R1 ENEN S E2E2 E1E1 F 12NN-1

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Stepping off Equilibrium Stages 5.35 equilibrium stages are needed to achive raffinate specifications wt % water wt % acetic acid P S F E1E1 RNRN

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wt % Acetic Acid in Raffinate wt % Acetic Acid in Extract Equilibrium Curve Operating Line McCabe and Thiele Diagram 5.35 equilibrium stages are needed to achive raffinate specifications

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F1F1 M S ENEN R1R1 OP 2 OP 1 F2F2 FTFT E N-1 ENEN E1E1 S=E 0 Two Feed Extraction Column Mixing Point M = F T + S= R 1 + E N Operating Points OP 1 = R i+1 - E i = R 1 - E 0 OP 2 = E k - R k+1 = E N - R N+1 OP 1 + OP 2 = F 2 Total Feed F T = F 1 + F 2 F 1 =R N+1 RNRN R2R2 R1R1 12NN-1 F2F2

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wt % water wt % acetic acid F1F1 S ENEN R1R1 OP 2 OP 1 F2F2 FTFT Stepping off Equilibrium Stages 2.89 equilibrium stages are needed to achive raffinate specifications

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Residue Curve Map liquid vapor x y

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Liquid phase activity coefficients from Wilson model : Obtaining the boiling temperature : Obtaining equilibrium vapor phase mole fractions :

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Residue curve map for the ternary system acetone-methanol-chloroform at P=760 mmHg Azeotrope Residue Curve UN SN SP UN SN

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Simple reactive distillation x y Chemical reaction Phase equilibrium Reaction equilibrium

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Transformed compositions Equation for simple distillation with reaction equilibrium

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waterAcetic Acid Isopropanol Isopropyl acetate XAXA XBXB Reactive azeotrope Residue curve map for the isopropyl acetate chemistry at P=1 atm Need to take into account acetic acid dimerization

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waterAcetic Acid MethanolMethyl acetate XAXA XBXB Need to take into account acetic acid dimerization Residue curve map for the methyl acetate chemistry at P=1 atm

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Flash Distillation P and T Feed F z i Liquid L x i Vapor V y i Rachford and Rice :

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Equilibrium constants Equilibrium constants using the equations that fit the DePriester Charts : Equilibrium constants using virial equation of state: Phase Equilibrium :

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Hydrocarbon Mixture Feed z 1 =0.2 z 2 =0.3 z 3 =0.4 z 4 =0.05 z 5 =0.05 P=3.5 bars and T=300 K

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Vapor and Liquid Compositions P=3.5 bars T=300 K Feed F=1 z 1 =0.2 z 2 =0.3 z 3 =0.4 z 4 =0.05 z 5 =0.05 Liquid L= x 1 = x 2 = x 3 = x 4 = x 5 = Vapor V= y 1 = y 2 = y 3 = y 4 = y 5 = Mass Balance equations give vapor and liquid compositions :

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McCabe and Thiele Method for Distillation of Binary Ideal Mixture feed Zf=0.5 bottom xb=0.05 distillate xd=0.9 Binary ideal mixture with constant relative volatility =

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Pinch Point and Minimum Reflux Ratio x y Feed line : Rectifying operating line :

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McCabe and Thiele Diagram R=1.5 R min x y 9 equilibrium stages are needed to achieve the separation

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x y Murphree Liquid Stage Efficiency E ML = equilibrium stages are needed to achieve the separation

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Multicomponent Distillation feed Z 1 =0.3 Z 2 =0.3 Z 3 =0.4 bottom xb 1 =0.05 distillate xd 1 =0.95 xd 2 =0.049 xd 3 =0.001 Ternary ideal mixture of Pentane, Hexane and Heptane with constant relative volatilities = 6.35, 2.47 and 1

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Rectifying operating line : Stripping operating line : R=2.5 and S=1.35 Liquid Compositions 8.3 equilibrium stages are needed to achieve the separation Pentane mole fraction Hexane mole fraction Stripping Section Rectifying Section D F B

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Enthalpy-Composition Diagram for hexane-octane system at 760 mmHg Hexane mole fraction Enthalpy of saturated liquid and vapor tie line H(y) h(x) Conjugate line

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Ponchon and Savarit method 7 stages are needed to achieve the separation feed q=0.41 Z 1 =0.5 bottom xb 1 =0.05 distillate xd 1 =0.95 Enthalpy of saturated liquid and vapor Hexane mole fraction F P1P1 P2P2 D B L V

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Conclusion Mathematica’s algebraic, numerical and graphical capabilities can be put into advantage to solve several chemical engineering and chemistry problems including equilibrium-staged separations with McCabe- Thiele, Hunter-Nash and Ponchon-savarit Methods.

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