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+// Example 5_1
+clc;clear;funcprot(0);
+// Given data
+P=100;// Plant capacity in Mw
+CV=25600;// Calorific value in kJ/kg
+n_th=30;// The thermal efficiency of the plant in %
+n_eg=92;// Electrical generation efficiency in %
+
+// Calculation
+// Mechanical energy available=W*CV*(n_th/100) in kJ/hr
+// Electrical energy available=W*CV*(n_th/100)*(n_eg/100) in kJ/hr
+q_e=P*10^3*3600;// Heat equivalent in kJ/hr
+W=(q_e/(CV*(n_th/100)*(n_eg/100)));// The coal required per hour in kg/hr
+W=(W/1000);// The coal required per hour in tons/hr
+printf('\nThe coal required per hour,W=%0.2f tons/hr',W);