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+// Example 24_9

+clc;funcprot(0);

+//Given data 

+W=2;// Work done in MW

+p_1=1;// bar

+p_r=5;// Pressure ratio in bar

+p_i=2.5;// bar

+T_1=27+273;// K

+r=1.4;// Specific heat ratio

+CV=40000;// kJ/kg

+n_c=85/100;// The isentropic efficiency of the compressor

+n_t=85/100;//  The isentropic efficiency of the turbine

+Q_a=80;// Heat absorbed in kJ/kg of air

+m_f=0.01;// kg per kg of air

+m_a=1;// kg

+r=1.4;// Spcific heat ratio for air and gases

+C_p=1;// kJ/kg-k for air and gases

+C_pg=C_p;

+C_pa=C_p

+

+//Calculation

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

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

+T_3=T_2+(Q_a/(C_pa*m_a));// K

+T_4=((m_f*CV)/((1+m_f)*C_p))+T_3;// K

+T_5a=T_4*(1/p_r)^((r-1)/r);// K

+T_5=T_4-(n_t*(T_4-T_5a));// K

+n_th=(((T_4-T_5)-(T_2-T_1))/(T_4-T_3))*100;// Thermal efficiency in %

+Q=(W*10^3)/(n_th/100);//Heat supplied in kJ/sec

+F=(Q/CV)*3600;// Fuel required per hour in kg/hr

+n_cp=(1-(1/(p_r)^((r-1)/r)))*100;//Efficiency of normal constant pressure cycle

+printf('\nThe thermal efficiency of the plant=%0.1f percentage \nEfficiency of normal constant pressure cycle=%0.0f percentage \nFuel consumption per hour=%0.0f kg/hr',n_th,n_cp,F);

+// The answer provided in the textbook is wrong