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+// Example 24_33
+clc;funcprot(0);
+//Given data 
+p_1=1;// bar
+p_2=9;// bar
+T_1=25+273;// K
+T_6=1250+273;// K
+e=0.83;// The effectiveness of regenerator
+n_c=0.83;// Isentropic efficiency of  both compressors
+n_t=0.83;// Isentropic efficiency of both turbines
+n_com=0.95;// Combustion efficiency
+CV=42;// MJ/kg
+C_p=1;// kJ/kg.K
+r=1.4;// Specific heat ratio for air and gases
+
+//Calculation
+T_8=T_6;// K
+n_c1=n_c;
+n_c2=n_c;
+n_t1=n_t;
+n_t2=n_t;
+p_i=sqrt(p_1*p_2);// bar
+T_2a=T_1*(p_i/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_3=T_1;// K
+T_4=T_2;// K
+T_7a=T_6/(p_2/p_i)^((r-1)/r);// K
+T_7=T_6-(n_t1*(T_6-T_7a));// K
+T_8=T_6;// K
+T_9=T_7;// K
+T_5=T_2+(e*(T_9-T_4));// K
+W_c=2*C_p*(T_2-T_1);// The work developed by  both compressors in kJ/kg
+W_t=2*C_p*(T_6-T_7);// The work developed by both turbines in kJ/kg
+W_n=W_t-W_c;// Net work in kJ/kg
+W_r=W_n/W_t;// Work ratio
+Q_1=C_p*(T_6-T_5);// kJ/kg
+Q_2=C_p*(T_8-T_7);// kJ/kg
+Q_s=Q_1+Q_2;// The total heat supplied in kJ/kg
+n=(W_n/Q_s)*100;// The plant efficiency in %
+printf('\nThermal efficiency of the plant=%0.0f percentage \nWork ratio=%0.2f',n,W_r);
+// The answer provided in the textbook is wrong