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

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

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

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

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®

Mathematical Solution Laplace Transform: Where

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

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))

Step Response Matlab® Output

Step Response Simulink® Diagram

Step Response Simulink® Output for Enzyme

Step Response Simulink® Output for Tryptophan

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

Impulse Response Matlab® Output

Impulse Response Simulink® Diagram

Impulse Response Simulink® Output for Enzyme

Impulse Response Simulink® Output for Tryptophan

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

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.