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+// Example 24_10

+clc;funcprot(0);

+//Given data 

+T_1=20+273;// K

+p_1=1;// bar

+T_6=700+273;// K

+p_r=6;// Pressure ratio

+e=0.7;// The effectiveness of regenerator

+m_air=200;//Air flow through the plant in kg/sec

+n_c=0.82;// Isentropic efficiency of both compressors

+n_t=0.92;// Isentropic efficiency of turbine

+n_com=0.96;// Combustion efficiency

+n_m=0.96;// Mechanical efficiency

+n_g=0.95;// Generation efficiency

+CV=35000;// kJ/kg

+C_p=1;// kJ/kg.K

+r=1.4;// Specific heat ratio

+

+//Calculation

+p_2=p_r*p_1;// bar

+p_i=sqrt(p_1*p_2);// bar

+T_2a=T_1*(p_i/p_1)^((r-1)/r);// K

+T_2=((T_2a-T_1)/n_c)+T_1;// K

+T_3=T_1;// K

+T_4a=T_3*(p_2/p_i)^((r-1)/r);// K

+T_4=T_2;// K (as n_c1=n_c2)

+T_7a=T_6*((p_1/p_2)^((r-1)/r));// K

+T_7=T_6-(n_t*(T_6-T_7a));// K

+T_5=(e*(T_7-T_4))+T_4;// K

+function[X]=m_f(y)

+    X(1)=(C_p*(1+y(1))*(T_6-T_5))-(y(1)*CV*n_com);

+endfunction

+y=[0.01]

+z=fsolve(y,m_f);

+m_fuel=z(1);

+m_a=1;// kg/kg of air

+m=(m_a/m_fuel);//Air fuel ratio

+n_th=(((T_6-T_7)-(2*(T_2-T_1)))/(T_6-T_5))*100;// Thermal efficiency

+W=(m_a*(T_6-T_5)*(n_th/100));// Work done per kg of air in kJ

+W_s=W*m_air;// Work done per sec in kJ/sec

+P=W_s/10^3;// Capacity of the plant in MW

+P_a=W_s*n_m*n_t;// Power available at generation terminals in kW

+F=m_air*3600*m_fuel;// Fuel consumption per hour in kg/hr

+Sfc=F/(P_a);// Specific fuel consumption in kg/kW.hr

+printf('\nAir fuel ratio used=%0.0f:1 \nThermal efficiency of the cycle=%0.1f percentage \nFuel consumption per hour=%0.0f kg/hr \nSpecific fuel consumption=%0.3f kg/kW.hr',m,n_th,F,Sfc);

+// The answers provided in the textbook is wrong