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+clear;
+clc;
+printf("\n Example 11.12");
+R = [0.85 1.0 1.5 2.0 3.0 4.0];    //Reflux ratio
+xd = 0.75;                         //top concentration of alcohol
+xs = [0.55 0.50 0.37 0.20 0.075 0.05];//From the graph fig.11.35 page-596
+Db(1) = 0;
+
+printf("\n   R              Fi                xs          Db   ");
+i=1;
+while i<=6
+    Fi(i) = xd/(R(i) + 1);
+    if i>1 then
+    Db(i) = 100*(xs(1)-xs(i))/(xd-xs(i));
+    end
+   printf("\n %.2f        %.3f         %.2f       %.1f",R(i),Fi(i),xs(i),Db(i));
+    i=i+1;
+end
+plot(R,Db);
+xtitle("","Reflux ratio(R)","Product Db (kmol)");
+printf("\n The area under the Db vs R curve is given by 96 kmol");
+Hav = 4000;            //average latent heat in kJ/kmol
+Qr = 96*Hav/1000;
+printf("\n Heat to be supplied to provide the reflux,Qr is approximately %.1f MJ",Qr);
+printf("\n Heat to be supplied to provide the reflux per kmol of product is then %.2f MJ",380/71.4);
+printf("\n Total heat = %.2f MJ/kmol product",5.32+4.0);
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