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rilla lab

Ryerson University
Uploaded: A year ago
Contributor: cloveb
Category: Chemical Engineering
Type: Assignment
Rating: N/A
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Filename:   rilla lab.docx (157.59 kB)
Page Count: 5
Credit Cost: 1
Views: 77
Last Download: N/A
Transcript
PART 2 a) A = [-1.25 0;0.05 -0.32] >> eig(A) ans = -0.3200 -1.2500 b) tspan=[0 20]; tin=[0.3833;0.09005]; ss1=1/3; ss2=0.09; [t,c]=ode15s(@ode1,tspan,tin); f=c(:,1)+ss1; g=c(:,2)+ss2; plot(t,f); hold on; plot(t,g); xlabel('Time (s)'); ylabel('Concentration (mol/hr)'); title('Graph for Linearized Function'); legend('CA1 = 0.3833','CA2 = 0.09005','Location','Northeast'); 014287500 c) tspan=[0 20]; tin=[0.3333;0.14005]; ss1=1/3; ss2=0.09; [t,c]=ode15s(@ode1,tspan,tin); f=c(:,1)+ss1; g=c(:,2)+ss2; plot(t,f); hold on; plot(t,g); xlabel('Time (s)'); ylabel('Concentration (mol/hr)'); title('Graph for Linearized Function'); legend('CA1 = 0.3333','CA2 = 0.14005','Location','Northeast'); d) tspan=[0 20]; tin=[1;1]; ss1=1/3; ss2=0.09; [t,c]=ode15s(@ode1,tspan,tin); f=c(:,1)+ss1; g=c(:,2)+ss2; plot(t,f); hold on; plot(t,g); xlabel('Time (s)'); ylabel('Concentration (mol/hr)'); title('Graph for Linearized Function'); legend('CA1 = 1','CA2 = 1','Location','Northeast'); e.1) tspan=[0 20]; tin=[0.3833;0.09005]; [t,c]=ode15s(@ode2,tspan,tin); plot(t,c); xlabel('Time (s)'); ylabel('Concentration (mol/hr)'); title('Graph for Non-Linearized Function'); legend('CA1 = 0.3833','CA2 = 0.09005','Location','Northeast'); e.2) tspan=[0 20]; tin=[0.3333;0.14005]; [t,c]=ode15s(@ode2,tspan,tin); plot(t,c); xlabel('Time (s)'); ylabel('Concentration (mol/hr)'); title('Graph for Non-Linearized Function'); legend('CA1 = 0.3333','CA2 = 0.14005','Location','Northeast'); e.3) tspan=[0 20]; tin=[1;1]; [t,c]=ode15s(@ode2,tspan,tin); plot(t,c); xlabel('Time (s)'); ylabel('Concentration (mol/hr)'); title('Graph for Non-Linearized Function'); legend('CA1 = 1','CA2 = 1','Location','Northeast') function dcdt = ode1(t,c) F1=100; k1=1.5; V1=400; V2=2000; Ca1=0.333; Ca2=0.09; f1= (-F1/V1 - 2*(Ca1)*k1)*c(1); f2= (F1/V2)*c(1) + (-F1/V2-2*Ca2*k1)*c(2); dcdt=[f1;f2]; end function dcdt = ode2(t,c) F1=100; k1=1.5; V1=400; V2=2000; Ca1=0.333; Ca2=0.09; f1= F1/V1 - (F1/V1)*c(1) - k1*(c(1))^2; f2= (F1/V2)*c(1) - (F1/V2)*c(2) - k1*(c(2))^2; dcdt=[f1;f2]; end Discussion From the linearized functions, we see some deviation from the given initial conditions. This is due to the fact that the linearized function is used as an approximation (at steady state) to the solution, and therefore, the deviation is acceptable. Unlike non-linearized functions, linearized functions be solved without the help of a computer. Therefore in order to solve a non-linear function, we would first convert it to linearized function and then solve it. Although this would not give us accurate values, it can be considered a decent approximation. From the plots, it can be seen that the deviation exists at time 0 (sec) where the values deviate from the initial conditions. On the other hand, solving the non-linearized equations gives us a more accurate solution. This can be seen from the graphs. At time 0 (sec) the plots start at the given initial conditions.

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