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+// Example 24_20
+clc;funcprot(0);
+//Given data 
+P=5;// Power plant capacity in MW
+T_1=15+273;// K
+p_1=1;// bar
+T_4=750+273;// K
+p_r=6;// Pressure ratio
+p_3=2.24;// bar
+e=0.75;// The effectiveness of heat exchanger
+n_c=0.8;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of both turbines
+n_t1=n_t;
+n_t2=n_t;
+C_pa=1;// kJ/kg.K
+C_pg=1.15;// kJ/kg.K
+r_a=1.4;// Specific heat ratio for air
+r_g=1.33;// Specific heat ratio for gases
+CV=18500;// kJ/kg
+
+//Calculation
+p_2=p_1*p_r;// bar
+p_re=sqrt(p_1*p_2);//Pressure ratio for each turbine in bar
+T_2a=T_1*(p_2/p_1)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_6=T_4;// K
+T_5a=T_4/((p_re)^((r_g-1)/r_g));// K
+T_5=T_4-(n_t1*(T_4-T_5a));// K
+T_7=T_5;// K
+T_3=T_2+(e*(T_7-T_2));// K
+//(i)
+function[X]=massoffuel(y)
+    X(1)=((1+y(1))*C_pg*(T_4-T_3))-(y(1)*CV);
+endfunction
+y=[0.01];
+z=fsolve(y,massoffuel);
+m_f1=z(1);// kg/kg of air
+AF=1/m_f1;// Air fuel ratio
+function[X]=massoffuel1(x)
+    X(1)=(C_pg*((1+m_f1+x(1))*(T_6-T_5)))-(x(1)*CV);
+endfunction
+x=[0.001];
+y=fsolve(x,massoffuel1);
+m_f2=y(1);// kg/kg of air
+W_c=C_pg*(T_2-T_1);// kJ/kg of air
+W_t1=C_pg*(1+(m_f1))*(T_4-T_5);// kJ/kg of air
+W_t2=C_pg*(1+m_f1+m_f2)*(T_6-T_7);// kJ/kg of air
+W_t=W_t1+W_t2;// kJ/kg of air
+W_n=W_t-W_c;// kJ/kg of air
+//(ii)
+Q_s=(m_f1+m_f2)*CV;// kJ/kg of air
+n_th=(W_n/Q_s)*100;// Thermal efficiency of the cycle
+//(iii)
+m_a=((P*10^3)/W_n);// kg/sec
+//(iv)
+F=m_a*(m_f1+m_f2)*3600;// Fuel required per hour in kg/hr
+printf('\n(i)Cycle efficiency=%0.1f percentage \n(ii)Air supplied to the plant=%0.1f kg/sec \n(iii)A:F ratio=%0.1f:1 \n(iv)Fuel consumption of the plant=%0.0f kg/hr',n_th,AF,m_a,F);
+// The answers provided in the textbook is wrong