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+// Example 24_2
+clc;funcprot(0);
+//Given data
+T_1=300;// K
+P_r=8;// P_r=(p1/p2)
+p_1=1;// bar
+T_4=1080;// K
+m=500;// kg/min
+n_c=0.8;
+n_t=n_c;//Isentropic efficiency of the compressor and turbine
+CV=42000;// kJ/kg
+e=0.6;// The effectiveness of the heat exchanger
+r=1.4;// Specific heat ratio
+C_p=1;// kJ/kg.°C
+//C_p=C_pg=C_pa;
+
+//Calculation
+T_2a=T_1*(P_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// Modified equation in K
+T_5a=T_4*(1/P_r)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+T_3=(e*(T_5-T_2))+T_2;// K
+//m_f=y(1)
+function[X]=Mass(y);
+ X(1)=(y(1)*CV)-(C_p*(1+y(1))*(T_4-T_3));
+endfunction
+y=[0.01]
+z=fsolve(y,Mass);
+m_f=z(1);// kJ/kg of air
+m_a=1;// kg
+q=m_a*(T_3-T_2);//Heat saved in kJ/kg of air
+M=(m*60*q)/CV;// Fuel saved per hour in kg/hr
+W_net=(C_p*(1+m_f)*(T_4-T_5))-(C_p*m_a*(T_2-T_1));// kJ/kg
+P=(m/60)*W_net;// The capacity of the plant in kW
+printf('\nFuel saved per hour=%0.2f kg/hr\nThe capacity of the plant=%0.1f kW',M,P);
+// The answer vary due to round off error
+