4 A gaseous fuel contains the following by mole 313 C2H6 676

4. A gaseous fuel contains the following by mole: 31.3% C₂H₆; 67.6% CH₄; 1.1% N₂. Determine

i. the gravitational stoichiometric air.

ii. The DRY molar composition of the combustion products if the fuel is burned with 150% theoretical air. Assume complete combustion of the fuel

iii. The WET analysis of the products by mass if the fuel is burned with 15% less air.

Solution

2C2H6 + 7O2 = 2CO2 + 3H2O------ eqn i

0.313 moles of C2H6 + 1.0955 moles of O2 = 0626 moles of CO2 +0.939 moles of H20(here % of moles is taken as fractions )

given 31.3% of ethane in fuel so moles of O2 = 7/2 *(0.313)= 1.0955

moles of CO2 = 2*0.313 = 0.626

moles of H2O = 3*0.313 = 0.939

CH4+ 2 O2 = CO2 + H2O ------- eqn ii

0.676 moles of methane + 1.352 moles of oxygen = 0.676 moles of CO2 + 0.676 moles of H2O

moles of O2 = 0.676*2 = 1.352

assume air composition

oxygen = 21% = 0.21

nitrogen = 79% = 0.79

moleculae weight of air = 29 g

Total moles of oxygen required for reaction = 1.0955 + 1.352 = 2.4474

= 2.4475 * 16 = 39.16 g ( 1 mole O2 = 16g )

quantity of air needed = 39.16/0.21 = 186.476 g ( in air 21% of O2 is present )

no of moles of air required to burn the enitre fuel = 186.476 / 29 = 6.4137 . ( Gravitational stoichiometric air )

at dry condition 150% of air is supplied so net moles of air = 6.4137*1.5 = 9.6206

therefore

total moles of CO2 = 0.626+0.676 =1.302 from eqn i & ii

no of moles of nitrogen in 9.6206 moles of air = (9.6206*29 *0.79)/14 =15.74 (molecular weight of nitrogen = 14 g)

total moles of N2 = 0.011+15.7434 = 15.7534

total quanitity of O2 = (9.6206-6.4137)*29*0.21 =19.53

no of moles of oxygen = 19.53/16 = 1.2206

so Dry molar composition of the combustion products is

%moles of CO2 = 1.302 / ( 1.302+15.7534+1.2206) = 1.302 / 18.276 = 0.0712

%moles of N2 = 15.7534 / ( 1.302+15.7534+1.2206) = 15.7534/18.276 = 0.8619

%moles of O2 = 1.2206/ ( 1.302+15.7534+1.2206) = 1.2206/18.276 =0.06678

At Wet analysis 15% less air is supplied

so total moles of air = 6.4137*0.85 = 5.4516

Wet molar composition of products

moles of N2 = 0.011

moles of of CO2 = (0.626+0.676)*0.85  = 1.01067 from eqn i &ii

moles of H2O = (0.939+0.676)*0.85 = 1.37275 from eqn i &ii

moles of C2H6 = 0.313*0.15 = 0.04695

moles of CH4 = 0.676*0.15 = 0.1014

molar composition on wet analysis

% moles of N2 = 0.011 / ( 0.011+1.1067+1.37275+0.04695+0.1014) = 0.011/2.6388= 0.0041

% moles of CO2 = 1.1067 / ( 0.011+1.1067+1.37275+0.04695+0.1014) = 1.1067 / 2.6388 = 0.4193

% moles of H2O = 1.37275 / ( 0.011+1.1067+1.37275+0.04695+0.1014) = 1.37275 / 2.6388= 0.52

% moles of C2H6 = 0.04695 / ( 0.011+1.1067+1.37275+0.04695+0.1014) = 0.04695 / 2.6388=0.0177

% moles of CH4 = 0.1014/ ( 0.011+1.1067+1.37275+0.04695+0.1014)= 0.1014/2.6388 = 0.0384