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+//Chapter 13: Fuel and Combustions
+//Problem: 2
+clc;
+
+//Declaration of Variables
+C = 90              // %
+O = 3.0             // %
+S = 0.5             // %
+N = 0.5             // %
+ash = 2.5           // %
+LCV = 8490.5        // kcal / kg
+
+// Solution
+mprintf("HCV = LCV + 9 * H / 100 * 587\n")
+mprintf(" HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)\n")
+H = (8490.5 - 7754.8) / (345 - 52.8)
+H = 4.575
+mprintf(" The percentage of H is %.3f percent\n", H)
+HCV = LCV + 52.8 * H
+mprintf(" Higher calorific value of coal %.1f kcal / kg",HCV)