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+// Example 24_3
+clc;funcprot(0);
+//Given data
+T_1=288;// K
+P_r=6;// P_r=p1/p2
+T_3=1000;// K
+m=2;// tonnes/hr
+n_c=0.85;
+n_t=0.90;//Isentropic efficiencies of the compressor and turbine
+CV=46500;// kJ/kg
+e=0.6;// The effectiveness of the heat exchanger
+r=1.4;// Specific heat ratio
+C_p=1;// kJ/kg.°C
+//C_p=C_pg=C_pa
+
+//Calculation
+T_2a=T_1*(P_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_4a=T_3/(P_r)^((r-1)/r);// K
+T_4=T_3-(n_t*(T_3-T_4a));// K
+W_c=C_p*(T_2-T_1);// kJ/kg
+W_t=C_p*(T_3-T_4);// kJ/kg
+Q_a=C_p*(T_3-T_2);// kJ/kg
+n_th=((W_t-W_c)/Q_a)*100;//Cycle efficiency
+W_s=W_t-W_c;// kJ/kg
+P=((m*1000)/3600)*CV*n_th/100*n_t*n_c;// kW
+P=P/1000;//MW
+printf('\n Cycle efficiency=%0.1f percentage \nThe specific work output=%0.0f kJ/kg',n_th,W_s);
+// The answer provided in the textbook is wrong