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+// Example 24_14
+clc;funcprot(0);
+//Given data
+P=800;// Plant capacity in kW
+T_1=15+273;// K
+p_1=1.01;// bar
+T_4=700+273;// K
+p_r=6;// Pressure ratio
+e=0.75;// The effectiveness of regenerator
+p_lr=0.15;// Pressure drop in regenerator in bar
+p_lcc=0.15;// Pressure drop in the combustion chamber in bar
+n_c=0.80;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+C_p=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+
+//Calculation
+p_2=p_r*p_1;// bar
+p_3=p_2-p_lcc;// Pressure at point 4 in bar
+T_2a=T_1*(p_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+p_4=p_1+p_lr;// bar
+T_5a=T_4/(p_3/p_4)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+T_3=T_2+(e*(T_5-T_2));// K
+W_c=C_p*(T_2-T_1);// kJ/kg
+W_t=C_p*(T_4-T_5);// kJ/kg
+W_n=W_t-W_c;// kJ/kg
+Q_s=C_p*(T_4-T_3);// kJ/kg
+n_th=(W_n/Q_s)*100;// Thermal Efficiency in percentage
+printf('\nThe thermal efficiency of the plant=%0.1f percentage',n_th);