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+clear;
+clc;
+
+// Illustration 7.5
+// Page: 444
+
+printf('Illustration 7.5 -  Page: 444\n\n');
+
+// solution
+//*****Data*****//
+// C-nicotine    A-water    B-kerosene
+// F-feed    R-raffinate    S-solvent
+F = 1000; // [feed rate, kg/h]
+xAF = 0.99; // [fraction of water in feed]
+// Because the solutions are dilute therefore
+xCF = 0.01; // [fraction of nicotene in feed, kg nicotene/kg water]
+xCR = 0.001; // [fraction of nicotene in raffinate, kg nicotene/kg water ]
+m = 0.926; // [kg water/kg kerosene]
+//*****//
+
+printf('Illustration 7.5(a) -  Page: 444\n\n');
+// Solution(a)
+
+yCS = 0; // [kg nicotene/kg water]
+
+// Because, in this case, both the equilibrium and operating lines are       // straight,if the minimum solvent flow rate Bmin is used, the concentration // of the exiting extract, yCmax, will be in equilibrium with xCF. Therefore
+yCmax = m*xCF; // [kg nicotene/kg kerosene]
+
+A = F*xAF; // [kg water/h]
+// From equation 7.17
+Bmin = A*(xCF-xCR)/(yCmax-yCS); // [kg kerosene/h]
+printf("The minimum amount of solvent which can be used is %f kg kerosene/h.\n\n",Bmin);
+
+printf('Illustration 7.5(b) -  Page: 444\n\n');
+// Solution(b)
+
+B = 1.2*Bmin; // [kg kerosene/h]
+EF = m*B/A;
+Nt = log((xCF-yCS/m)/(xCR-yCS/m)*(1-1/EF)+1/EF)/log(EF);
+
+printf("The number of theoretical stages if the solvent rate used is 20 percent above the minimum is %f .\n\n",Nt);
+
+printf('Illustration 7.5(c) -  Page: 444\n');
+// Solution(c)
+
+Eme = 0.6; // [Murphree stage efficiency]
+// from equation 7.20
+Eo = log(1+Eme*(EF-1))/log(EF); // [overall efficiency]
+Nr = Nt/Eo; // [number of real stages]
+disp(Nr);
+// The nearest integer to number of real stages is 11
+// Therefore
+Nr = 11;
+printf("The number of real stages required is %f.\n\n",Nr);
+\ No newline at end of file