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+// Example 24_21
+clc;funcprot(0);
+//Given data 
+T_1=25+273;// K
+p_1=1;// bar
+T_6=1250+273;// K
+p_3=9;// bar
+n_c=0.83;// Isentropic efficiency of both compressors
+n_c1=n_c;
+n_c2=n_c;
+n_t=0.83;// Isentropic efficiency of both turbines
+n_t1=n_t;
+n_t2=n_t;
+C_pa=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+m_a=16.5;// kg/sec
+
+//Calculation
+p_2=sqrt(p_1*p_3);// bar
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c1)+T_1;// K
+T_8=T_6;// K
+T_4=T_2;// K
+T_7a=T_6/(p_3/p_2)^((r-1)/r);// K
+T_7=T_6-(n_t1*(T_6-T_7a));// K
+T_9=T_7;// K
+W_c=2*C_pa*(T_2-T_1);// kJ/kg
+W_t=2*C_pa*(T_6-T_7);// kJ/kg
+W_n=W_t-W_c;// kJ/kg
+T_5=T_7;
+//When the ideal regeneration is given,then
+e=1;// Effectiveness
+Q_s=2*C_pa*(T_6-T_5);// kJ/kg
+//(i)
+n_th=(W_n/Q_s)*100;//The thermal efficiency in %
+//(ii)
+P=W_n*m_a;// Power developed by the plant in kW
+printf('\n(i)The thermal efficiency of the plant=%0.1f percentage \n(ii)Power developed by the plant=%0.2f kW',n_th,P);
+// The answer vary due to round off error