Methane gas at 400 K and 1 atm (100 kPa) enters a combustion chamber, where it i
ID: 1062112 • Letter: M
Question
Methane gas at 400 K and 1 atm (100 kPa) enters a combustion chamber, where it is mixed with air entering at 500 K and 1 atm. The products of combustion exit at 1800 K and 1 atm with the product analysis on a dry basis given as 9.7% CO_2, 0.5% CO, 2.95% O_2, and 86.85% N_2. The average value for the specific heat C_p of methane between 298 and 400 K is 38 kJ/kmol.K. Determine the following: The complete balanced equation for the reaction. Rewrite the balanced equation in terms of per mole of fuel Air-fuel ratio on a mass basis The percent theoretical air The dew point temperature of the product The enthalpy of reaction.Explanation / Answer
Air is composed of 21% oxygen and 79% nitrogen by volume. Thus each mole of oxygen needed to oxidize the hydrocarbon (fuel) is accompanied by 79/21 = 3.76 moles of nitrogen. Using this, the molecular mass of air becomes 29 kg/kmol.
a) Assuming that the total moles of combustion products is 100 kmol, the equation can be written as
a CH4 + b ( O2 +3.76 N2 ) --> 9.7 CO2 + 0.5 CO + 9.5 O2 + 86.85 N2 + c H2O
On equating,
b = 86.85/ 3.76 = 23.09
a= 9.7 + 0.5= 10.2
c= 2a= 20.4
So, 10.2 CH4 + 23.09 ( O2 + 3.76 N2 ) --> 9.7 CO2 + 0.5 CO + 9.5 O2 + 86.85 N2 + 20.4 H2O
b) In terms of per mole of fuel,
CH4 + 2.26 ( O2 + 3.76 N2 ) --> 0.951 CO2 + 0.049 CO + 0.289 O2 + 8.5 N2 + 2 H2O
c) Massair / Massfuel = (2.26) 4.76 (29 kg/mol) / 1 (12+4) (kg/mol) = 19.49 kg-air/ kg-fuel
d)Theoretical Air is the minimum amount of air which allows the complete combustion of the fuel. In this case the products do not contain any oxygen.
CH4 + y ( O2 +3.76 N2 ) --> CO2 + 2 H2O + 3.76 y N2
We get, y=2. So,
CH4 + 2 ( O2 +3.76 N2 ) --> CO2 + 2 H2O + 7.52 N2
% Theoretical air = (2.26/2) 100 = 113 %
e) From balanced equation per kmol of fuel.
N (H2O) = 2 kmol
N (total) =1 +0.289 + 8.5 + 2 = 12.789 kmol
P(H2O)/ P(total) = (2/12.789) 101.325 KPa =15.84 kPa
So, Tdew point at this pressure = 55 0 C approx
f) Enthalpy of reaction = Sum of enthalpies of products - sum of enthalpies of reactants
Using balanced equation,
Enthalpy of reaction =0.951(-393520) + 0.049 (-110530) + 2(-241820) + 74850 = -788443.49 kJ/kmol