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Research Updates Distillation Process. Seon B. Kim Ph.D. Student Date: May 5 th , 2011 Advisor: Dr. Andreas A. Linninger Laboratory for Product and Process Design, Department of Bioengineering, UIC. Updates. REU/RET activities. Complex Step Derivatives. F(x)=x 3 , x=0.5, dx=1e-05
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Research UpdatesDistillation Process Seon B. Kim Ph.D. Student Date: May 5th, 2011 Advisor: Dr. Andreas A. Linninger Laboratory for Product and Process Design, Department of Bioengineering, UIC
Updates • REU/RET activities
Complex Step Derivatives F(x)=x3, x=0.5, dx=1e-05 F(x)=x3, x=0.5, dx=1e-10 • Possible reason of accuracy of complex step derivative • No roundoff error by substraction
Stage-based VS continuous model • Stage-to-stage model • Material balance at steady state (rectifying section) • V = D + L (all molar flow rates) • Vyi= Dxi+ Lxi • Equilibrium model (Raoult’s law) • , when is equal to 1, then it becomes the ideal mixture model. x1 y1 • Discretized continuous model • Convective transfer between discretized stages • Reaction at equilibrium • Only changes by reaction between vapor and liquid phases. Rectifying section x2 y2 y3 x3 Reaction section x4 y4 x5 y5 Stripping section x6 y6
0.8 ODE Solver 0.7 AspenPlus FM (8) 0.6 FM (25) 0.5 0.4 0.3 0.2 0.1 90 95 100 105 110 Trial 1. Solving composition profile • For stage 1, the material balance (molar mass) at steady state is (IN) − (OUT) = (ACC) • For stage 2, • In the matrix format, L xd x1 x2 V xn-2 xn-1 xn
Trial 2. Solving Tray-by-tray equation • Simplified 1 dimensional distillation column • Operating equation • Reaction (assuming 1st order reaction) • At dynamic state, • At steady state, • Here, I used Antoine parameters to find K x1 dh or dT x2
Trial 2. Solving Tray-by-tray equation (cont’d) • Linear algebraic solution of tray-by-tray model x1 T1=37.6oC x2 dT=5oC x3 x4 x5 x6
Trial 3. Solving Continuous model equation • Simplified 1 dimensional distillation column Operating equation • Diffusion • Discretization • Change Independent variable h T x1 dh or dT x2
Trial 3. Solving Continuous model equation(cont’d) x1 T1=37.6oC x2 dT=5oC x3 x4 x5 x6
Mole fraction in distillation column Reflux drum condenser V L reboiler
Molar flowrate in distillation column Reflux drum condenser Reflux ratio reflux / distillate = 65.12 / 38.87 = 1.675 V L Reboil (Boilup) ratio Btm Vapor / Bottom = 103.99 / 61.12 = 1.701 reboiler
Molar flowrate in distillation column (cont’d) - Mole fraction of y1 : y1=K1*x1=0.6431*1.2757=0.8202 - Molar flow rate of material 1 y1*V=0.8202*103.99=85.31 • - Mole fraction of x1=0.6431 • Molar flow rate of material 1 • x1*L=0.6431*65.12=41.88 0 10 20 30 40 88.22 100.45 113.20 113.17 V L • Mole fraction of x1=0.2472 • Molar flow rate of material 1 x1*L=0.2472*65.12=16.09 - Mole fraction of y1 : y1=K1*x1=0.2472*1.7989=0.4447 - Molar flow rate of material 1 y1*V=0.4447*103.99=46.24 20 40 60 80 20 40 60 80
Delphi flowsheet demo 1-1. MATLAB (BPD: 7.807e-4) DELPHI (BPD: 1.708e-3) AspenPlus
Delphi flowsheet demo 1-2. MATLAB (BPD: 0.0070, 0.0092) DELPHI (BPD: 0.0035, 0.0021) AspenPlus
Delphi flowsheet demo 1-3. MATLAB (BPD: 2.72e-3, 0.0015) DELPHI (BPD: 2.48e-3, 3.26e-3) AspenPlus
Delphi flowsheet demo 2-1. MATLAB (BPD: 1.886e-3) DELPHI (BPD: 1.623e-4)
Delphi flowsheet demo 2-2. MATLAB (ΣBPD: 1.038e-2) DELPHI (ΣBPD: 1.398e-3)
Delphi flowsheet demo 2-3. MATLAB (BPD: 4.214e-8) DELPHI (BPD: 1.078e-3)
Mass flowrate in distillation column Reflux drum condenser mol/sec g/sec V L A (Benzene,C6H6) MA = 78.11 g/mol mA = 78.11 g/mol * 25 mol/sec = 1952.75 g/sec reboiler
Stage-based model • Equilibrium only on the stage • Between stages, just convective transfer • Continuous model • Equilibrium continuously through the column • Discretization necessary