Benezene (C6H6) is burned in a constant pressure process (p=1 atm = 0.1 MPa) wit

Question

Benezene (C6H6) is burned in a constant pressure process (p=1 atm = 0.1 MPa) with 100% theoretical air.

Find:

1. The dew point of the combustion products.

Note: The above requires several steps, one of which is setting up the combustion equation.

100% Theoretical aire is a complicated way of saying that a full combustion without excess air occurs.

Atmospheric a N2ir for combustion calculations is aaumbed to be composed of ( 1 O2 + 3.76 N2).

2. If you would use Benzene as a fuel for your multi-fuel internal combustion engne, ans the only criteria that you were interested in is energy density (kJ/kg), would Benzene be a better fuel that Octane?

Hint: Combustion gasses leaving the engine are extremely hot, with some engines using liquid sodium (melting pointv 97.794 C) within the valve stem of the discharge poppet valves to prevent overheating of the valve.

Explanation / Answer

The combustion reaction is C6H6+9O2--à6CO2+ 6H2O

Basis : 1 mole of C6H6, moles of Oxygen required= 9 , since air contains 21% O2 and 79% N2, moles of air to be supplied ( supplied since there is no excess air)= 9/0.21= 42.86

Products (moles): CO2= 6, H2O=9, N2=42.86*0.79= 33.86

Dew point is the temperature at which the partial pressure of vapor = vapor pressure of liquid

Partial pressure of H2O= mole fraction of water* total pressue = 9/(9+6+33.86)*1 atm =0.184 atm= 0.184*760 =140 mm Hg

Log psat= 8.07131-1730.63/(t+233.426) , t in deg.c

Log (140)= 8.07131-1730.63/(t+233.426)

Temperature t= 56.85 deg.c, dew point = 56.85 deg.c

heat of combustion of benzene = 3263.9 Kj/mole. 1 mole of Benzene = 78 gm

hence heat of combustion = 3263.9/78=41.8 KJ/gm

for octane it is 5471/114 Kj/gm =47.99

hence Benzene is not better than octane,

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