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+// Example 24_6

+clc;funcprot(0);

+//Given data

+T_1=288;// K

+p_1=1.03;// bar

+p_2=6;// bar

+p_3=p_2-0.1;// bar

+n_c=80/100;// The isentropic efficiency of compressor

+n_t=n_c;// The isentropic efficiency of turbine

+n_com=90/100;// Combustion efficiency 

+W=1.1*1000;// kW

+m=7;//Air flow rate in kg/s

+T_3=750+273;// K

+//C_p=C_pa=C_pg

+C_p=1.0;// kJ/kg.K

+r=1.4;// Specific heat ratio

+CV=20000;// kJ/kg

+

+//Calculation

+//Applying isentropic law to the process 1-2

+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K

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

+// m=m_a/m_f

+m=((CV*n_com)/(T_3-T_2))-1;

+//Applying isentropic law to the process 3-4'

+T_4a=T_3/((p_3/p_1)^((r-1)/r));// K

+T_4=T_3-(n_t*(T_3-T_4a));//K

+m_a=W/(((1+(1/m))*C_p*(T_3-T_4))-(C_p*(T_2-T_1)));// kg/sec

+m_f=m_a/37;// kg/sec

+m_g=m_a+m_f;// kg/sec

+W_t=m_g*C_p*(T_3-T_4);// kW

+W_r=W/W_t;// Work ratio

+n_th=W/(m_g*C_p*(T_3-T_2));// Thermal efficiency of the plant

+printf('\n(i)Flow of air and flow of gases per second,m_a=%0.1f kg/sec & m_g=%0.2f kg/sec \n(ii)Work ratio=%0.4f \n(iii) Thermal efficiency of the plant=%0.3f (or)%0.1f percentage',m_a,m_g,W_r,n_th,n_th*100);

+// The answer vary due to round off error