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+// Example 25_2
+clc;funcprot(0);
+//Given data 
+T_1=20+273;// K
+T_3=1100+273;// K
+T_5=1000+273;// K
+T_11=150;// °C
+p_r=8;// Pressure ratio
+p_7=80;// bar
+T_6a=300+273;// K
+T_7=600+273;// K
+n_c=100/100;// Isentropic efficiency of compressor
+n_t=100/100;// Isentropic efficiency of both turbines
+p_8=0.05;// bar
+C_p=1;// kJ/kg
+C_pa=C_p;
+C_pg=C_p;
+r=1.4;// Specific heat ratio 
+CV=61600;// kJ/kg
+C_pw=4.2;// kJ/kg°C
+
+// Calculation
+// The combustion reaction taking place in CC-I is given by CH_4+2O_2=CO_2+2H_2O
+// 16+64=44+36;
+m_o=64/16;// Amount of O_2 required in per kg of CH_4
+m_a=(100/23)*4;// Amount of air required in kg/kg of fuel
+m_act=m_a*5;// Actual air supplied in kg/kg of fuel
+m_f1=(1/m_act);// Amount of fuel supplied per kg of air flow through CC-I in kg
+T_2a=T_1*(p_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_4=T_3/(p_r)^((r-1)/r);// K
+// In CC-II
+m_a1=1;// kg/sec
+m_f2=(m_a1*C_pa*(T_5-T_4))/CV;// kg/kg of air flow
+// From h-s chart:
+h_7=3510;// kJ/kg
+h_11=C_pw*T_11;// kJ/kg
+m_s1=(m_a1*C_pg*(T_5-T_6a))/(h_7-h_11);// kg/kg of air
+AF=(m_a1/m_s1);
+m_a=1.5;// kg/sec(given)
+W_g=(m_a*C_pa*((T_3-T_4)-(T_2-T_1)));
+m_s=m_a*m_s1;// kg/sec
+// From h-s chart:
+h_g=2080;// kJ/kg
+W_s=(m_s*(h_7-h_g));// kW
+W_t=W_g+W_s;// Total power generated in MW
+Q_s=(m_f1+m_f2)*m_a*CV;// kW
+n=(W_t/Q_s)*100;// Overall efficiency of the plant
+m_f=((m_f1+m_f2)*m_a*3600);// Mass of fuel supplied in kg/hr
+Sfc=(m_f)/(W_t);// Specific fuel consumption in kg/kWh
+printf('\n(i)Total power generating capacity of the plant=%0.0f MW \n(ii)Overall efficiency of the plant=%0.1f percentage \n(iii)Mass of fuel supplied per hour=%0.2f kg/hr \n(iv)Specific fuel consumption=%0.3f kg/kWh',W_t,n,m_f,Sfc);
+// The answer vary due to round off error