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+// Example 12_1

+clc;funcprot(0);

+// Given data

+AF_act=20;// The air-fuel ratio

+// The reaction equation for theoretical air is C_4H_10+6.5(O_2+3.76N_2)-->4CO_2+5H_2O+24.44N_2

+a=6.5;// Constant

+M_air=29;// kg/kmol

+M_fuel=58;// kg/kmol

+P_atm=100;// kPa

+

+// Calculation

+m_air=a*(4.76)*M_air;// kg air

+m_fuel=1*M_fuel;// kg fuel

+// (a)

+AF_th=m_air/m_fuel;// The theoretical air-fuel ratio

+P_ea=((AF_act-AF_th)/AF_th)*100;// % excess air

+// (b)

+// The reaction equation with 129.28% theoretical air is C_4H_10+(6.5)(1.2928)(O_2+3.76N_2)-->4CO_2+5H_2O+1.903O_2+31.64N_2

+N_CO2=4;// mol

+N=42;// mol

+P_CO2=(N_CO2/N)*100;// The volume percentage of CO_2 in the products in %

+// (c)

+N_H2O=5;// mol

+N=42.5;// mol

+y_H2O=N_H2O/N;// The mole fraction

+P_v=y_H2O*P_atm;// The partial pressure of the water vapor in kPa

+// Using Table C-2

+T_dp=49;// °C

+printf("\n(a)The percent excess air=%2.2f percentage \n(b)The volume percentage of CO_2 in the products=%1.2f percentage \n(c)The dew point temperature of the products,T_dp=%2.0f °C",P_ea,P_CO2,T_dp);