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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc
+clear
+printf("example 3.27 page number 114\n\n")
+
+//to find the area of heating surface
+F = 1000 //in kg
+xF = 0.01
+
+solid_feed = F*xF;
+water_feed = F - solid_feed;
+
+tF = 40 //in degree C
+hF = 167.5 //in kJ/kg
+xL = 0.02;
+
+solid_liquor = 10 //in kg
+L = solid_liquor/xL;
+tL = 100 //in degree C
+hL = 418.6 //in kJ/kg
+
+V = F -L;
+
+tv = 100 //in degree C
+Hv = 2675 //in kJ/kg
+ts = 108.4 //in degree C
+Hs = 2690 //in kJ/kg
+tc = 108.4 //in degree C
+hc = 454 //in kJ/kg
+
+//applying heat balance
+S = (F*hF-V*Hv-L*hL)/(hc-Hs);
+printf("weight of steam required = %f kg/hr",S)
+
+Q = S*(Hs-hc);
+U = 1.4 //in kW/m2K
+delta_t = ts-tL;
+A = 383.2/(U*delta_t);
+printf("\n\narea of heating surface = %f square meter",A)