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Protein Digestion: A Dynamic Model

Protein Digestion: A Dynamic Model. Student Name 1 & Student Name 2 Process Control: Design Challenge I. Our System. F 1 , X ai. F 2 , X bi. V. The Stomach. F 3 , X a , X b , X c. ODEs. Reaction:. Assumptions: Well mixed Constant: V, X bi , density, flow rates.

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Protein Digestion: A Dynamic Model

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  1. Protein Digestion: A Dynamic Model Student Name 1 & Student Name 2 Process Control: Design Challenge I

  2. Our System F1, Xai F2, Xbi V The Stomach F3, Xa, Xb, Xc ODEs Reaction: • Assumptions: • Well mixed • Constant: V, Xbi, density, flow rates

  3. Steady State Solution Original Conditions INPUT: Clear [xa,xb,xc]; eqns = {F1*Xai-V*k*xa*xb-F3*xa == 0, F2*Xbi-V*k*xa*xb-F3*xb == 0, V*k*xb*xa-F3*xc == 0, F1 == 15, F2 == 7, F3 == F1+F2, k == 5, Xai == 150, Xbi == 1000, V == 1000}; NSolve[eqns, {xa,xb,xc}] OUTPUT: {{xc®102.271,xa®0.00208415,xb®215.911},{xc®318.188,xa®-215.916,xb®-0.00648415}} Exit concentrations: Protein: 0.0021 mg/ml Enzyme: 216.0 mg/ml Tryptophan: 102.3 mg/ml Decrease Initial Enzyme Concentration INPUT: Clear [xa,xb,xc]; eqns = {F1*Xai-V*k*xa*xb-F3*xa == 0, F2*Xbi-V*k*xa*xb-F3*xb == 0, V*k*xb*xa-F3*xc == 0,F1 == 15, F2 == 7, F3 == F1+F2, k == 5, Xai == 150, Xbi == 500, V == 1000}; NSolve[eqns, {xa,xb,xc}] OUTPUTl: {{xc®102.265,xa®0.00791828,xb®56.8261},{xc®159.103,xa®-56.8305,xb®-0.0123183}} Exit concentrations: Protein: 0.008 mg/ml Enzyme: 56.8 mg/ml Tryptophan: 102.3 mg/ml Given Values

  4. Steady State Solution Original Conditions INPUT: Clear [xa,xb,xc]; eqns = {F1*Xai-V*k*xa*xb-F3*xa == 0, F2*Xbi-V*k*xa*xb-F3*xb == 0, V*k*xb*xa-F3*xc == 0,F1 == 15, F2 == 7, F3 == F1+F2, k == 5, Xai == 150, Xbi == 1000, V == 1000}; NSolve[eqns, {xa,xb,xc}] OUTPUT: {{xc®102.271,xa®0.00208415,xb®215.911},{xc®318.188,xa®-215.916,xb®-0.00648415}} Exit concentrations: Protein: 0.0021 mg/ml Enzyme: 216.0 mg/ml Tryptophan: 102.3 mg/ml Increase Inlet Flow Rate INPUT Clear [xa,xb,xc]; eqns = {F1*Xai-V*k*xa*xb-F3*xa == 0, F2*Xbi-V*k*xa*xb-F3*xb == 0, V*k*xb*xa-F3*xc == 0, F1 == 30, F2 == 7, F3 == F1+F2, k == 5, Xai == 150, Xbi == 1000, V == 1000}; NSolve[eqns, {xa,xb,xc}] OUTPUT {{xc®121.608,xa®0.0133159,xb®67.5809},{xc®189.21,xa®-67.5883,xb®-0.0207159}} Exit concentrations: Protein: 0.013 mg/ml Enzyme: 67.6 mg/ml Tryptophan: 121.6 mg/ml

  5. Mathematical Solution Laplace Procedure • Linearize ODE • Subtract original steady state • Find τ’s and k’s • Laplace • Rearrange to find transfer functions • Substitute to get one equation • Inverse Laplace with Maple® ODEs

  6. Mathematical Solution Laplace Transform: Where

  7. Mathematical Solution Cont. Fuzzy Math Inverse Laplace via Maple® Impulse Response Step Response

  8. Step Response Matlab® Input function hdot=stomachstep(t,h) hdot=zeros(3,1); F1=15; F2=7; F3=22; V=1000; k=5; Xbi=1000; if t<100 Xai=100; else Xai=150; end; hdot(1)=(F1*Xai-V*k*h(1)*h(2)-F3*h(1))/V; hdot(2)=(F2*Xbi-V*k*h(1)*h(2)-F3*h(2))/V; hdot(3)=(k*V*h(1)*h(2)-F3*h(3))/V; t0=0; tf=400; h0=[0.021 216 102.3]; [t,h]=ode23(@stomachstep,[t0 tf],h0); plot(t,h(:,2),t,h(:,3))

  9. Step Response Matlab® Output

  10. Step Response Simulink® Diagram

  11. Step Response Simulink® Output for Enzyme

  12. Step Response Simulink® Output for Tryptophan

  13. Impulse Response Matlab® Input function hdot=stomachimpulse(t,h) hdot=zeros(3,1); F1=15; F2=7; F3=22; V=1000; k=5; Xbi=1000; if t<100 Xai=200; elseif t>105 Xai=200; else Xai=500; end; hdot(1)=(F1*Xai-V*k*h(1)*h(2)-F3*h(1))/V; hdot(2)=(F2*Xbi-V*k*h(1)*h(2)-F3*h(2))/V; hdot(3)=(k*V*h(1)*h(2)-F3*h(3))/V; t0=0; tf=400; h0=[0.0021 216 102.3]; [t,h]=ode23(@stomachimpulse,[t0 tf],h0); plot(t,h(:,2),t,h(:,3))

  14. Impulse Response Matlab® Output

  15. Impulse Response Simulink® Diagram

  16. Impulse Response Simulink® Output for Enzyme

  17. Impulse Response Simulink® Output forTryptophan

  18. Final Thoughts Conclusions • Linear approximation is valid • Step response is as expected • Impulse response as expected • Matlab® & Simulink® agree • Model will provide basis for more complex systems Recommendations • Vary k values • Explore larger/smaller step and impulse values • Make model more realistic • Model multiple food components

  19. References • Biglione, N., Yousef, M., & Rodgers, V.G.J., (2004). Process Dynamics & Control in Design 52:185: Lab # 1: Programming and Simulation using Matlab® and Simulink®. • Seborg, D.E., Edgar, T.F., & Millichamp, D.A. (2004). Process Dynamics and Control (2nd ed.). Hoboken, NJ: John Wiley & Sons. • Vander, A., Sherman, J., Luciano, D. (2001). Human Physiology: The Mechanisms of Body Function (8th ed.). Boston: McGraw Hill.

  20. Questions?

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