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+clear;

+clc;

+

+// Illustration 1.21

+// Page: 60

+

+printf('Illustration 1.21 - Page:60 \n\n');

+// Solution

+

+//*****Data*****//

+//   a-oxygen   b-nitrogen

+T = 293; // [K]

+P = 0.1; // [atm]

+d = 0.1*10^-6; // [m]

+e = 0.305; // [porosity]

+t = 4.39; // [tortuosity]

+k = 1.3806*10^-23; // [J/K]

+l = 2*10^-3; // [m]

+R = 8.314; // [cubic m.Pa/mole.K]

+x_a1 = 0.8;

+x_a2 = 0.2;

+M_a = 32; // [gram/mole]

+M_b = 28; // [gram/mole]

+//*****//

+

+// Using data from Appendix B for oxygen and nitrogen, and equation (1.45)

+sigma_a = 3.467; // [Angstrom]

+sigma_b = 3.798; // [Angstrom]

+sigma_AB = ((sigma_a+sigma_b)/2)*10^-10; // [m]

+

+lambda = k*T/(sqrt(2)*3.14*(sigma_AB^2)*P*1.01325*10^5); // [m]

+// From equation 1.101

+K_n = lambda/d;

+printf("The value of a dimensionless ratio, Knudsen number is %f\n\n",K_n);

+// If K_n is greater than 0.05 then transport inside the pores is mainly by Knudsen diffusion

+// Using equation 1.103

+D_Ka = (d/3)*(sqrt(8*R*T)/sqrt(3.14*M_a*10^-3)); // [square m/s]

+

+// Using equation 1.107

+D_Kaeff = D_Ka*e/t; // [square m/s]

+

+p_a1 = (x_a1*P)*1.01325*10^5; // [Pa]

+p_a2 = (x_a2*P)*1.01325*10^5; // [Pa]

+

+// Using equation 1.108

+N_a = D_Kaeff*(p_a1-p_a2)/(R*T*l); // [mole/square m.s]

+// Now using the Graham’s law of effusion for Knudsen diffusion

+// N_b/N_a = -sqrt(M_a/M_b) ,therefore

+N_b = -N_a*sqrt(M_a/M_b); // [mole/square m.s]

+

+printf("The diffusion fluxes of both components oxygen and nitrogen are %e mole/square m.s and %e mole/square m.s respectively\n",N_a,N_b);