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+// Example 24_22
+clc;funcprot(0);
+//Given data 
+T_1=290;// K
+p_1=1.01;// bar
+T_3=650+273;// K
+p_r=8;// Pressure ratio
+n_c=0.8;// Isentropic efficiency of compressor
+n_t1=0.85;// Isentropic efficiency of H.P turbine
+n_t2=0.83;// Isentropic efficiency of L.P turbine
+C_pa=1;// kJ/kg.K
+C_pg=1.15;// kJ/kg.K
+r_a=1.4;// Specific heat ratio for air
+r_g=1.33;// Specific heat ratio for gases
+m_a=10;// The  air flow through the compressor in kg/sec
+
+//Calculation
+p_2=p_1*p_r;// bar
+T_2a=T_1*(p_r)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+W_c=1*C_pa*(T_2-T_1);//Work input to the compressor in kJ/kg
+W_t1=W_c;// kJ/kg
+T_4=T_3-(W_t1/C_pg);// K
+T_4a=T_3-((T_3-T_4)/n_t1);// K
+p_3=p_2/((T_3/T_4a)^(r_g/(r_g-1)));// bar
+p_re=p_3/p_1;// The pressure ratio of expansion in the power turbine 
+T_5a=T_4/(p_3/p_1)^((r_g-1)/r_g);// K
+dT_45=n_t2*(T_4-T_5a);// (dT_45=T_4-T_5) K
+W_t2=C_pg*(dT_45);//Work developed by power turbine in kJ/kg
+W_net=W_t2;// The net work done per kg of air in kJ/kg
+W_t=W_t1+W_t2;// Total work done per in kJ/kg
+W_r=W_t2/W_t;// Work ratio
+Q_s=C_pa*(T_3-T_2);// kJ/kg
+n_th=(W_t2/Q_s)*100;// Thermal efficiency in %
+P=W_t2*m_a;// Power capacity of the plant in kW
+printf('\nThe power developed by the unit=%0.0f kW \nThe thermal efficiency=%0.0f percentage \nWork ratio=%0.1f',P,n_th,W_r);
+// The answer vary due to round off error