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clear;
clc;
printf("\n Example 17.5");
//let x = [f(C)]mean
x = (8.4*10^(-2)*1266)/(2.67*10^(-3)*1.186);
//e is porosity
e = 0.47;
m = e/(1-e);
printf("\n m = %.2f",m);
//The velocity uc with which the adsorption wave moves through the column may be obtained from equation 17.79
uc = 6.2*10^(-6);
density = 1266; //density is in kg/m^3
Gs = uc*(1-e)*density;
printf("\n Gs = %.2f*10^(-3) kg/m^2.sec ",Gs*10^3);
//From overall balance:
//We have given eq: Gs1(0.084-0)=0.129(0.00267)
//On solving above eq :
Gs1 = poly([0],'Gs1');
Gs2 = roots(Gs1*(0.084-0)-0.129*(0.00267));
printf("\n Gs1 = %.2f*10^(-3) kg/m^2.sec",Gs2*1000);
//Part(b)
//time ta for the adsorption zone to move its own length za is given by:
za = 0.185;
ta =(za/uc)/3600;
printf("\n ta = %.1f hrs",ta);
//The time taken for a point at a distance z into the zone to emerge is given by:
//t = (z1/za)ta
yr = [0.0001 0.0002 0.0006 0.0010 0.0014 0.0018 0.0022 0.0024];
yre = [0.00005 0.0001 0.00032 0.00062 0.00100 0.00133 0.00204 0.00230];
I=0;
I1 = 0;
i=1;
printf("\nI= %.1f",I);
while i <=8
y(i)=1/(yr(i)-yre(i));
if i>1 then
I = I + (yr(i)-yr(i-1))*(y(i)+y(i-1))/2;
printf("\n I=%.1f",I);
end
i=i+1;
end
z_za = [0 0.137 0.362 0.473 0.560 0.674 0.852 1.000];
t = (z_za)*ta;
j=1;
while j<=8
printf("\ntime = %.1f h",t(j));
j=j+1;
end
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