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

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Our System F 3, X a, X b, X c F 1, X ai F 2, X bi The Stomach V Assumptions: Well mixed Constant: V, X bi, density, flow rates Reaction: ODEs

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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® ,xa® ,xb® },{xc® ,xa® ,xb® }} Exit concentrations:Protein: mg/ml Enzyme: mg/ml Tryptophan: 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® ,xa® ,xb® },{xc® ,xa® ,xb® }} Exit concentrations:Protein: mg/ml Enzyme: 56.8 mg/ml Tryptophan: mg/ml K5 ml·min/mg F115 ml/min F27 ml/min V1000 ml Xai150 mg/ml Xbi1000 mg/ml Given Values

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Steady State Solution 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® ,xa® ,xb® },{xc®189.21,xa® ,xb® }} Exit concentrations:Protein: mg/ml Enzyme: 67.6 mg/ml Tryptophan: mg/ml 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® ,xa® ,xb® },{xc® ,xa® ,xb® }} Exit concentrations:Protein: mg/ml Enzyme: mg/ml Tryptophan: mg/ml Original Conditions

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Mathematical Solution ODEs Laplace Procedure 1.Linearize ODE 2.Subtract original steady state 3.Find τ’s and k’s 4.Laplace 5.Rearrange to find transfer functions 6.Substitute to get one equation 7.Inverse Laplace with Maple®

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Mathematical Solution Laplace Transform: Where

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

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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=[ ]; tf],h0); plot(t,h(:,2),t,h(:,3))

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Step Response Matlab® Output

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Step Response Simulink® Diagram

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Step Response Simulink® Output for Enzyme

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Step Response Simulink® Output for Tryptophan

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Impulse Response function hdot=stomachimpulse(t,h) hdot=zeros(3,1); F1=15; F2=7; F3=22; V=1000; k=5; Xbi=1000; if 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=[ ]; tf],h0); plot(t,h(:,2),t,h(:,3)) Matlab® Input

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Impulse Response Matlab® Output

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Impulse Response Simulink® Diagram

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Impulse Response Simulink® Output for Enzyme

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Impulse Response Simulink® Output for Tryptophan

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

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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.

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