Simulating the Production of Ethanol Student Name 1 Student Name 2 Student Name 3

Overview Introduction to Production Simulation Results Conclusion

Ethanol Production Fermenter

Fermentation Reaction

Chemical Equation Glucose 2 Carbon Dioxide Ethanol + Zymase + Heat

Developing the System Based on Mass Balances And Energy Balance

Temperature Dependency Reaction Rate Heat

Constant Values Variable Symbol Value Reactor Volume V 115,000 L Concentration of Glucose Entering CGin 150 g/L Concentration of Ethanol Entering CEin 0 g/L Flow In Fin 100 L/min Flow Out Temperature In Tin 548 K Heat Capacity Cp 9 kJ/K Amount of heat given off/reaction rate β 0.02 kJ Arrhenius Constant A 1000 Temperature Constant EA/R 50 K

Reactor CO2 CGin, CEin, Tin, Fin Yeast CG, CE, T, F Q V, ρ, CP Next, state the action step. Make your action step specific, clear and brief. Be sure you can visualize your audience taking the action. If you can’t, they can’t either. Be confident when you state the action step, and you will be more likely to motivate the audience to action.

Methods of Solution MATLAB Simulink Laplace Transform using Maple

MATLAB Code: Steady State function conc=ustank(t,solution) global Tin V Fin Fout CEin Cgin B rho Cp T=solution(1); Cg=solution(2); CE=solution(3); rate=Cg*10^3*exp(-50/T) Cgin=150 conc(1,1)=(Tin*Fin*Cp*rho-T*Fout*rho*Cp+B*rate)/(V*rho*Cp); conc(2,1)=(Cgin*Fin-Cg*Fout-(rate))/V; conc(3,1)=(CEin*Fin-CE*Fout+0.5*rate)/V;

Linear Approximations Taylor Series Expansion Approximate Methods:

Simulink: Steady State To close, restate the action step followed by the benefits. Speak with conviction and confidence, and you will sell your ideas.

Laplace Transform E’(t) = -0.57e-1*exp(-0.14e-2*t)*cos(0.39e-4*t) +0.22e-1*exp(-0.14e-2*t)*sin(0.39e-4*t) -0.17e-55*I*(-0.6329721079e54*exp(-0.14e-2*t)*cos(0.39e-4*t) -0.16e55*exp(-0.14e-2*t)*sin(0.39e-4*t)) -0.1742706769e-55*I*(0.63e54*exp(-0.14e-2*t)*cos(0.39e-4*t) +0.16e55*exp(-0.14e-2*t)*sin(0.39e-4*t)) +0.36e-1*exp(-0.90e-3*t)*cos(0.82e-4*t) +0.14e-1*exp(-0.90e-3*t)*sin(0.82e-4*t) -0.17e-55*I*(-0.39e54*exp(-0.90e-3*t)*cos(0.82e-4*t) +0.10e55*exp(-0.90e-3*t)*sin(0.82e-4*t)) -0.17e-55*I*(0.39e54*exp(-0.90e-3*t)*cos(0.82e-4*t) -0.10e55*exp(-0.90e-3*t)*sin(0.82e-4*t)) +0.19e-1*exp(-0.75e-3*t) +0.14e-2*exp(-0.71e-3*t)

Step Change Steady state achieved Double concentration of glucose

MATLAB Code: Step Change %unit step change function conc=ustank(t,solution) global Tin V Fin Fout CEin Cgin B rho Cp T=solution(1); Cg=solution(2); CE=solution(3); rate=Cg*10^3*exp(-50/T) if(t>1000)Cgin=300;else Cgin=150;end conc(1,1)=(Tin*Fin*Cp*rho-T*Fout*rho*Cp+B*rate)/(V*rho*Cp); conc(2,1)=(Cgin*Fin-Cg*Fout-(rate))/V; conc(3,1)=(CEin*Fin-CE*Fout+0.5*rate)/V;

Simulink: Step Change

Laplace Transform E’(t) = 28+4*exp(-0.14e-2*t)*cos(0.39e-4*t) -17*exp(-0.14e-2*t)*sin(0.39e-4*t) +0.12e-63*I*(-0.74e65*exp(-0.14e-2*t)*cos(0.39e-4*t) -0.18e66*exp(-0.14e-2*t)*sin(0.39e-4*t)) +0.12e-63*I*(0.74e65*exp(-0.14e-2*t)*cos(0.39e-4*t) +0.18e66*exp(-0.14e-2*t)*sin(0.39e-4*t)) -41*exp(-0.90e-3*t)*cos(0.82e-4*t) -11*exp(-0.90e-3*t)*sin(0.82e-4*t) +0.12e-63*I*(-0.49e65*exp(-0.90e-3*t)*cos(0.82e-4*t) +0.18e66*exp(-0.90e-3*t)*sin(0.82e-4*t)) +0.12e-63*I*(0.49e65*exp(-0.90e-3*t)*cos(0.82e-4*t) -0.18e66*exp(-0.90e-3*t)*sin(0.82e-4*t)) -26*exp(-0.75e-3*t)-2.0*exp(-0.71e-3*t)

Unit Impulse Steady state achieved Quadruple concentration of glucose

MATLAB Code: Impulse %impulse response function conc=ustank(t,solution) global Tin V Fin Fout CEin Cgin B rho Cp T=solution(1); Cg=solution(2); CE=solution(3); rate=Cg*10^3*exp(-50/T) if(t>1000&t<1100)Cgin=600;else Cgin=150;end conc(1,1)=(Tin*Fin*Cp*rho-T*Fout*rho*Cp+B*rate)/(V*rho*Cp); conc(2,1)=(Cgin*Fin-Cg*Fout-(rate))/V; conc(3,1)=(CEin*Fin-CE*Fout+0.5*rate)/V;

Simulink: Impulse

Laplace Transform E’(t) = -0.38e-3*exp(-0.14e-2*t)*cos(0.39e-4*t) +0.15e-3*exp(-0.14e-2*t)*sin(0.39e-4*t) -0.12e-57*I*(-0.63e54*exp(-0.14e-2*t)*cos(0.39e-4*t) -0.16e55*exp(-0.14e-2*t)*sin(0.39e-4*t)) -0.12e-57*I*(0.63e54*exp(-0.14e-2*t)*cos(0.39e-4*t) +0.16e55*exp(-0.14e-2*t)*sin(0.39e-4*t)) +0.24e-3*exp(-0.90e-3*t)*cos(0.82e-4*t) +0.91e-4*exp(-0.90e-3*t)*sin(0.82e-4*t) -0.12e-57*I*(-0.39e54*exp(-0.90e-3*t)*cos(0.82e-4*t) +0.10e55*exp(-0.90e-3*t)*sin(0.82e-4*t)) -0.12e-57*I*(0.39e54*exp(-0.90e-3*t)*cos(0.82e-4*t) -0.10e55*exp(-0.90e-3*t)*sin(0.82e-4*t)) +0.13e-3*exp(-0.75e-3*t) +0.96e-5*exp(-0.71e-3*t)

MATLAB Results for Steady State

Simulink Results for Steady State

Perturbed Steady State Results Variable  New Value  Achieves Steady State Peaks Concentration (g/L) Time (min) Increase Reactor Volume 1,000,000 L 67.79 8500 -- Flow In 1,000 L/min 681.3 9500 Temperature In 800 K 67.9 2000 Decrease 1,150 L 8 10 L/min 6.32 8000 53.66 300 250 K 67.57 950

MATLAB Results for Step Change

Simulink Results for Step Change

Perturbed Step Change Results Variable  New Variable Value  Achieves Steady State Peak Concentration (g/L) Time (min) Increase Reactor Volume 1,000,000 L 135.7 70000 -- Flow In 1,000 L/min 1363 8000 Temperature In 800 K 135.9 7500 Decrease 1,150 L 135.8 1100 10 L/min 12.94 9000 53.66 300 250 K 135.6 7000

MATLAB Results for Impulse

Expected Simulink Results for Impulse 6000 74 500

Perturbed Impulse Results Variable New Variable Value  Achieves Steady State Peaks Concentration (g/L) Time (min) Increase Reactor Volume 1,000,000 L 67.8 25000 68.95 5000 Flow In 1,000 L/min 681.4 9000 -- Temperature In 800 K 67.89 6000 80.53 1300 Decrease 1,150 L 1200 271.9 1090 10 L/min 6.32 8000 53.66 300 250 K 67.57 80.92

Conclusions MATLAB vs. Simulink Perturbing Inputs Modeling Processes Linear Approximations Perturbing Inputs Modeling Processes

Questions?