//clear// clear; clc; //Example 25.2 //Solution cbyc0 =0.05; u0 = 58; //[cm/s] Dv = 0.37; //[m^2/g] c0 = 365; //[ppm] S = 1194; //[m^2/g] T = 25; //[C] rho_b = 0.461; //[g/cm^3] P = 737; //[mm Hg] M = 74.12; //[g/mol] eps = 0.457; t = 1:0.5:8.5; t(4) = 2.4; t(5) = 2.8; t(6) = 3.3; cbyc0 =[0.005,0.01,0.027,0.05,0.1,0.2,0.29,0.56,0.0019,0.003,0.0079,0.018,0.039,0.077,0.15,0.24]; t1 = t(1:8); t2 = t(9:16); cbyc01 = cbyc0(1:8); cbyc02 = cbyc0(9:16); plot(t1,cbyc01,t2,cbyc02); xgrid(); xlabel('t, Hours'); ylabel('c/c0'); title('Brakthrough curves for Example 25.2'); legend('L = 8cm','L = 16cm'); //(a) FA = u0*c0*10^-6/22400*273/298*737/760*M*3600; //[g/cm^2-h] // The total solute adsorbed is the area above the graph multiplied //by FA. For the 8-cm bed, the area is Area_bed = 4.79; //[h] //This area corresponds to the ideal time that would be required to adsorb //the same amount if the breakthrough curve were a vertical line. The mass //of carbon per unit cross-sectional area of the bed is Ac = 8*rho_b; //[g/cm^2] //Thus, Wsol = FA*Area_bed/Ac; //[g solute/g carbon] //At the break point, where cbyc0_break = 0.05; //and t_break =2.4; //[h] Area_bed_break = 2.37; //[h] //The amount adsorbed up to the break point is then Wb = FA*t_break/Ac; //[g solute/ g carbon] ratio_W = Wb/Wsol; //Thus 50 percent of the bed capacity is unused, which can be representd //by a length 4 cm. //For the 16-cm bed the breakthrough curve has the same initial slope as the cuve //for 8-cm bed, and although data were not taken beyond cbyc0 = 0.25, //the curves are assumed to be parallel //For the entire bed, tT = 9.59; //[h] Wsat = FA*tT/(16*rho_b); //[g solute/ g carbon] //At cbyc0_break = 0.05; t_break =7.1; //[h] Area_break = 7.07; //[h] Wb = FA*Area_break/(16*rho_b); //[g solute/g carbon] ratio_W = Wb/Wsat; //At the break point, 74 percent of the bed capacity is used, //which corresponds to an unused section of length 0.26*16 cm. //Within experimental error, the lengths of unused bed agree, //and 4.1 cm is expected value for a still longer bed. disp('cm',4.2,'length of the bed used','percent',ratio_W,'saturation capacity of the carbon') //(b) L = 32; //[cm] L_exp = L-4.1; //[cm] //Fraction of the bed used fra_bed = L_exp/L; //The break-point time is, tb = L_exp*rho_b*Wsat/FA; //[h] disp('h',tb,'break point-time ')