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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+printf("\n Example 16.4");
+//In 100 kg of feed
+
+ //mass of water =
+ mw = 100*30/100; //mass of water in kg
+ //mass of dry solids =
+ md = 100-30; //mass of dry solids
+
+//and:
+//For b kg water in the dried solids: 100b/(b + 70) = 15.5
+b = poly([0],'b');
+b1 = roots(100*b - 15.5*(b+70));
+printf("\n water in the product ,b = %.1f kg",b1);
+
+ //Initial water content
+ w1 = 30/70; //Intial moisture content in kg/kg dry solids
+ //Final moisture content
+ w2 = (12.8/70); //Final moisture content in kg/kg dry solids
+ //water to be removed
+ w3 = (30-12.8); //water to be removed in kg
+
+//Surface of drying
+S = (0.03*70); //Surface for drying in m^2
+rate = (0.0007*2.1); //Rate of drying in kg/sec
+
+//As the final moisture content is above the critical value, all the drying is at this constant rate and the time of drying is:
+t = 17.2/0.00147;
+printf("\n As the final moisture content is above the critical value, all the drying is at this constant rate and the time of drying is: %d ksecs or %.2f hr",t,t/3600);
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