MCG2131 Tutorial 4 (2 February 2018) Open-book problem: A natural gas is approxi
ID: 1081394 • Letter: M
Question
MCG2131 Tutorial 4 (2 February 2018) Open-book problem: A natural gas is approximately 90% CH4, 6% C,Ho and 4% N, by volume. (a) Determine the quantity of O, in kmol required for stoichiometric combustion of 1 kmol of gas. (b) If the pr ducts of combustion of this fuel contain 10% CO2 (by mol, dry basis), calculate the percent excess air. Determine the amount of combustion air used in kmol air/kmol fuel. (c) Find the product dew point. The total pressure is 101.3 kPa. (d) Determine the mass of CO, produced per kg of fuel. Will this change if excess air is used?Explanation / Answer
Basis : 1 kmole of gas. It contains 90% CH4, i.e =0.9 Kmoles of CH4 , 6% of C2H6, i.e =0.06 Kmoles of C2H6 and 4% N2, i.e= 0.04 Kmoles of N2. The combustion of CH4 and C2H6 are
CH4+ 2O2------->CO2+2H2O (1) and C2H6+ 3.5O2------->2CO2+ 3H2O (2)
from reaction-1, 1 K mole of CH4 gives requires 2 Kmoles of theoretical oxygen. Hence 0.9 Kmoles of CH4 gives 0.9*2= 1.8 K moles of O2. From reaction-1, 1 Kmole of C2H6 requires 3.5 moles of oxygen and hence 0.06 moles of C2H6 required 0.06*3.5 =0.21Kmoles of oxygen, total moles of oxygen = 1.8+0.21= 2.01 kmoles of oxygen.
air contains 21% O2 and 79% N2, total of air to be supplied = 2.01/0.21=9.6 moles
let x= % of excesss air , moles of air supplied = 9.6*(1+x/100)
assuming complete combustion,
products contains : CO2, unreated Oxygen and N2
from reaction-1, mole of CO2 formed =0.9 and from reaction-2, moles of CO2 formed =2*0.06= 0.12
total moles of CO2 formed = 0.9+0.12=1.02, moles of N2= moles of N2 in feed+ moles of N2 in air = 0.04+9.6*0.79*(1+x/100) and moles of O2 unreated =oxygen available- oxygen used = 9.6*(1+x/100)*0.21-2.01 =2.01*x/100
total moles of products= 1.02+7.6*(1+x/100)+ 2.01*x/100 (3)
it is given that 10% of product contains CO2, but CO2 moles in products= 1.02, hence total moles of products =1.02/0.1= 10.2
fromr eq.3, 10.2= 1.02+7.6+7.6*x/100+2.01*x/100, x= 16.44,
excess air supplied =9.6*1.1644= 11.18 Kmoles, moles of N2= 11.18*0.79+0.04
Products contain : CO2= 1.02 moles, O2= moles of O2 supplied-mole of O2 consumed= 11.18*0.21-2.01=0.34, N2= 11.78*0.79+0.04=9.36 , moles of Water formed = 0.9*2( from reaction-1)+0.06*3 ( from reaction-2)= 1.98 moles
total moles of gaseous products on wet basis = moles of products on dry basis +1.98= 10.2+1.98=12.18 moles
mole fraction of water= moles of water/total moles = 1.98/12.18=0.16
partial pressure of water= mole fraction* total pressure =0.16*101.3 Kpa= 16.21 kpa
101.3 Kpa= 760 mm Hg, 16.21 kpa= 16.21*760/101.3 = 121.6 mm Hg
Dew point is the temperature at which partial pressure of vapor becomes equal to vapor pressure of liquid at the given temperature
from Antoine equation for water, log psat(mm Hg)= 8.07131- 1730.63/(t+233.426), t in deg.c
hence log (121.6)= 8.07131- 1730.63/(t+233.426), t= 55.67 deg.c
mass of CO2 produced= moles* molar mass = 1.02 kg moles *44 kg/kg moles= 44.88 kg moles/ kg fuel
This does not change with % excess air . Due to the presence of excess air, the % of CO2 changes and it does not influence the mass of CO2 formed.
total moles o
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