Methane gas (CH 4 ) is burned in an adiabatic combustor with a given percentage
ID: 523418 • Letter: M
Question
Methane gas (CH4) is burned in an adiabatic combustor with a given percentage of excess air. The pressure and temperature of both the air and fuel are 101 kPa and 298 K respectively. Assume that the mole fractions are 79% nitrogen and 21 % oxygen for air (use M=28.97 kg/kmol and R=0.287 kJ/kg-K) and that water is a vapor in the exhaust. Given the values below, determine the following:
--Given Values--
m_fuel (kg) = 64
Excess Air = 83%
a) Determine the air fuel ratio (kmol_air/kmol_fuel)
b) Determine the amount (kmol) of fuel burned.
c) Determine the amount (kmol) of air used.
d) Determine the amount (kmol) of CO2 produce
e) Determine the amount (kmol) of H2O produced.
f) Determine the amount (kmol) of N2 produced.
g) Determine the amount (kmol) of O2 produced
h) Determine the Enthalpy (kJ) of the reactants.
i) Determine the adiabatic flame temperature (K).
j) Determine the Enthalpy (kJ) of CO2 in the products
k) Determine the Enthalpy (kJ) of H2O in the products.
l) Determine the Enthalpy (kJ) of N2 in the products.
m) Determine the Enthalpy (kJ) of O2 in the products.
Explanation / Answer
Moles of fuel = mass/molar mass= 64/16= 4
The combustion reaction is CH4+ 2O2-----àCO2+ 2H2O
The reaction suggests 1 mole of CH4 requires 2 mole of oxygen to produce 1 mole of CO2 and 2 mole of water.
16 Kg of methane requires 64 Kg of oxygen to produce 44 Kg of CO2 and 36 Kg of water.
64 Kg of methane requires 64*4= 256 Kg of oxygen to produce 4*44=176 Kg of CO2 and 4*2*18= 144 Kg of water.
Mass of oxygen required= 256 Kg or 256/32= 8 kgmoles of oxygen. Air contains 21% O2 and 79% N2. Moles of air required= 8/0.21= 38 kgmoles. mass of air= 38*29= 1102 Kg. Air supplied= 83% excess. Air actually supplied= 1.83*1102 =2017 Kg
Moles of air= 2017/29= 69.55, air to fuel ratio = 69.55/4 =17.3875 moles of air/ mole of fuel
Mass of fuel burned= 64, moles= mass/molar mass= 64/16= 4
Kmole of air used= 69.55
Amount of CO2 produced= 176/44= 4 kg moles
Moles of H2O produced =144/18= 8 kg moles
Amount of nitrogen= 69.55*0.79= 55 kg moles
O2 remaining assuming complete combustion =69.55*0.21-8=6.6055 kgmoles