Question

Octane is burned completely with theoretical air in a steady-flow process at a constant pressure of 100 kPa. The products of combustion are cooled to 25 C. Calculate the number of kg of H2O which will condense per kg of fuel.

Answer 1

The combustion of occtane C8H18 can be represented as

2C8H18+25O2---> 16CO2+18H2O

Basis : 2 moles of C8H18, mass of fuel= molecualr weight* moles= 2*114= 228 gms of Fuel octanw

From the reaction, moles of oxygen produced = 18 moles, mass of water produced= 18*18=324 gms

moles of oxygen required= 25 moles and moles of air required= 25/0.21 ( since air contains 21% Oxygen and 79% Nitrogen)=119.0476, moles of nitrogen = 119.0476*0.79=94.04

Products : CO2=16 moles H2O= 18 moles and Nitrogen =94.04

Mole fraction water = moles of water/ total moles= 18/(18+94.04+16)=0.141

Partial pressure of water vapor = 0.141*100= 14.1Kpa , 14.1/101.3 atm (1atm= 101.3 Kpa) =0.139 atm=106 mm Hg

at 25 deg.c, vapor pressure of water =23.8mm Hg =101.3*23.8/760 =3.17 Kpasince partial pressure > vapor pressure, water condenses

moles of water vapor/ moles of dry gas= partial pressure of water vapor/ partial pressure of dry gas

x/110.04= 3.17/(100-3.17)=0.033

x= 110.04*0.033=3.63 moles at 25 deg.c

hence moles of water condensed = 18-3.63=14.37 moles

moles of water condensed/mole of fuel = 14.37/2= 7.185 moles watet/mole of fuel