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+// Example 24_16
+clc;funcprot(0);
+//Given data 
+P=5;// Power plant capacity in MW
+T_1=300;// K
+p_1=1;// bar
+T_5=650+273;// K
+p_r=5;// Pressure ratio
+e=0.7;// The effectiveness of regenerator
+n_c=0.8;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of both turbines
+n_t1=n_t;
+n_t2=n_t;
+n_com=0.97;// Combustion efficiency
+n_m=0.98;// Mechanical efficiency of both turbines
+CV=40000;// kJ/kg
+C_pa=1;// kJ/kg.°C
+C_pg=1.145;// kJ/kg.°C
+r_a=1.4;// Specific heat ratio for air
+r_g=1.35;// Specific heat ratio for gases
+
+//Calculation
+p_2=p_1*p_r;// bar
+p_i=sqrt(p_1*p_2);// The intermediate pressure between two compressors
+T_2a=T_1*(p_i)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_3=T_1;// K
+T_4=T_2;// K
+T_7=T_5;// K
+// Work developed by the compressor turbine = Work required to run the compressor
+T_6=T_5-((2*C_pa*(T_2-T_1))/(C_pg*n_m));// K
+T_6a=T_5-((T_5-T_6)/n_t1);// K
+p_3=p_2/((T_5/T_6a)^(r_g/(r_g-1)));// K
+T_8a=T_7*(p_1/p_3)^((r_g-1)/r_g);// K
+T_8=T_7-(n_t2*(T_7-T_8a));// K
+T_x=T_4+((e*C_pg*(T_8-T_4))/C_pa);// K
+W_net=C_pg*(T_7-T_8)*n_m;//Net Work available per kg of air in kJ/kg of air
+Q_s=C_pg*((T_5-T_x)+(T_7-T_6));// Heat supplied per kg of air kJ/kg of air
+m_f=Q_s/(CV*n_com);// The total mass of fuel in per kg of air flow
+m=1/m_f;// Air fuel ratio
+n_o=(W_net/(m_f*CV))*100;// Over all efficiency
+m_a=(P*1000)/(W_net);// kg/sec
+M_f=m_a*3600*m_f;// Mass of fuel used per hour in kg/hr
+Sfc=M_f/(P*1000*1);// Specific fuel consumption in kg/kW-hr
+printf('\nOver all efficiency of the plant=%0.1f percentage \nMass flow of air through the plant per second=%0.2f kg/sec \nSpecific fuel consumption=%0.3f kg/kW-hr',n_o,m_a,Sfc);
+// The answer vary due to round off error