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+// Example 24_23
+clc;funcprot(0);
+//Given data 
+p_1=1;// bar
+p_2=5;// bar
+p_3=2.5;// bar
+T_1=300;// K
+T_3=900;// K
+T_5=T_3;// K
+m_a=10;// kg/sec
+CV=33500;// kJ/kg
+C_p=1;// kJ/kg.°C
+r=1.4;// Specific heat ratio for air and gases
+
+//Calculation
+T_2=T_1*(p_2/p_1)^((r-1)/r);// K
+T_4=T_3/(p_2/p_3)^((r-1)/r);// K
+T_6=T_5/(p_2/p_3)^((r-1)/r);// K
+function[X]=massoffuel(y)
+    X(1)=((1+y(1))*C_p*(T_3-T_2))-(y(1)*CV);
+endfunction
+y=[0.01];
+z=fsolve(y,massoffuel);
+m_f1=z(1);// kg/kg of air
+function[X]=massoffuel1(x)
+    X(1)=(C_p*((1+m_f1+x(1))*(T_5-T_4)))-(x(1)*CV);
+endfunction
+x=[0.001];
+y=fsolve(x,massoffuel1);
+m_f2=y(1);// kg/kg of air
+W_n=((m_a*(1+m_f1)*C_p*(T_3-T_4)))+((m_a*(1+m_f1+m_f2)*C_p*(T_5-T_6)))-(m_a*C_p*(T_2-T_1));// kW
+n_g=100;//The generator efficiency is considered 100%
+n_th=(W_n/(m_a*(m_f1+m_f2)*CV))*100;// The efficiency of the plant in %
+printf('\nThe thermal efficiency of the plant=%0.1f percentage \nPower generating capacity=%0.0f kW',n_th,W_n);
+// The answers provided in the textbook is wrong