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+// Example 24_19
+clc;funcprot(0);
+//Given data 
+P=5;// Power plant capacity in MW
+T_1=30+273;// K
+p_1=1;// bar
+T_3=550+273;// K
+p_r=5;// Pressure ratio
+p_3=2.24;// bar
+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;
+C_pa=1;// kJ/kg.°C
+C_pg=1.15;// kJ/kg.°C
+r_a=1.4;// Specific heat ratio for air
+r_g=1.33;// Specific heat ratio for gases
+
+//Calculation
+p_2=p_1*p_r;// bar
+T_5=T_3;// K
+T_2a=T_1*(p_2/p_1)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+W_c=C_pa*(T_2-T_1);// kJ/kg
+T_4a=T_3/(p_2/p_3)^((r_g-1)/r_g);// K
+T_4=T_3-(n_t1*(T_3-T_4a));// K
+T_6a=T_5/(p_3/p_1)^((r_g-1)/r_g);// K
+T_6=T_5-(n_t2*(T_5-T_6a));// K
+W_t=2*C_pg*(T_3-T_4);// kJ/kg
+W_n=W_t-W_c;// kJ/kg
+m_a=((P*10^3)/W_n);// kg/sec
+Q_s=(C_pg*(T_3-T_2))+(C_pg*(T_5-T_4));// kJ/kg
+n_o=(W_n/Q_s)*100;// Over all efficiency in %
+printf('\nThe over all efficiency=%0.0f percentage \nThe mass flow rate=%0.1f kg/sec',n_o,m_a);
+// The answers provided in the textbook is wrong