Ethanol Air-to-Fuel Ratio Ethanol combusts according to the following reaction:

Question

Ethanol Air-to-Fuel Ratio Ethanol combusts according to the following reaction: C_2H_5OH + 3O_2 rightarrow 2CO_2 + 3H_2O Determine the stoichiometric air-to-fuel ratio for ethanol. Ethanol releases approximately 27 MJ (megajoules) of energy for every kilogram burned, whereas gasoline releases approximately 45 MJ/kg. If car has a fuel efficiency of 12 km/l (28 ml/gal) for gasoline, estimate its fuel for ethanol. Suppose that during the intake stroke, a certain engine could draw in an equal volume of an air-fuel mixture of either gasoline (isooctane) or ethanol at their stoichiometric ratios. Which fuel would produce more work during the expansion or power stroke?

Explanation / Answer

Basis: 1 mole of ethanol

Molecular weight of ethanol (CH3OH)= 12+3+16+1= 32

Air contains 21%O2 and 79% N2, molecular weight = 0.21*32+0.79*28= 29

According to the reaction , CH3OH+ 3O2--à2CO2 + 3H2O

Moles of Oxygen required =3 moles, since air contains 21% O2 and 79% N2, moles of air required = 3/0.21= 14.3 moles, Molecular weight of air = 29, mass of air = moles* Molecular weight = 14.3*29= 414.7 gm

Air to fuel ratio = 414.7/32 =14.3

b) energy released by gasoline/ energy released by ethanol = efficiency of gasoline/ efficiency of ethanol

45/27= 12/x

Hence x= 12*27/45 =7.2 km/l efficiency of gasoline.

c) since the molar mass of CH3OH is less than gasoline, the expansion of ethanol produces more work than gasoline.

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