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+// Example 24_34
+clc;funcprot(0);
+//Given data 
+p_1=1;// bar
+p_2=8;// bar
+T_1=300;// K
+T_3=1000;// K
+CV=40;// MJ/kg
+W_2=500;// kW
+C_pa=1;// kJ/kg.°C
+C_pg=1;// kJ/kg.°C
+r=1.4;// Specific heat ratio for air and gases
+
+//Calculation
+p_r=(p_2/p_1);// Pressure ratio
+T_2=T_1*(p_r)^((r-1)/r);// K
+T_4=T_3/(p_r)^((r-1)/r);// K
+// Assume m_a=y(1);m_f=y(2);// m_g1=y(3);m_g2=y(4)
+function[X]=mass(y)
+    X(1)=(y(1)+y(2))-(y(3)+y(4));
+    X(2)=(y(4)*C_pg*(T_3-T_4))-(W_2);
+    X(3)=(y(1)*C_pa*(T_2-T_1))-(y(3)*C_pg*(T_3-T_4));
+    X(4)=(y(2)*CV*10^3)-((y(1)+y(2))*C_pg*(T_3-T_2));
+endfunction
+y=[1 0.1 1 1];
+z=fsolve(y,mass);
+m_a=z(1)*60;// kg/min
+m_f=z(2)*3600;// kg/hr
+m_g1=z(3);// kg/sec
+m_g2=z(4);// kg/sec
+Sfc=(m_f/W_2);// kg/kWh
+AF=(m_a/60)/(m_f/3600);// Air fuel ratio
+n_th=(W_2/((m_f/3600)*CV*10^3))*100;// Thermal efficiency in %
+printf('\nThe mass of air consumed by the plant=%0.1f kg/min \nA:F ratio used=%0.0f \nSpecific fuel consumption=%0.2f kg/kWh \nThermal efficiency of the plant=%0.1f percentage',m_a,AF,Sfc,n_th);
+// The answer vary due to round off error