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

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Presentation on theme: "A Selection of Chemical Engineering Problems Solved using Mathematica 1- Chemical Kinetics and Catalysis 2- Applied Thermodynamics Housam BINOUS National."— Presentation transcript:

1 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

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

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

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

5 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

6 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

7 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

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

9 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.

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

11 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

12 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

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

14 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

15 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

16 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

17 Residue Curve Map liquid vapor x y

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

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

20 Simple reactive distillation x y Chemical reaction Phase equilibrium Reaction equilibrium

21 Transformed compositions Equation for simple distillation with reaction equilibrium

22 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

23 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

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

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

26 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

27 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 :

28 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 = 

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

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

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

32 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

33 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

34 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

35 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

36 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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