Let us understand the effect of fuel vaporization on the tem

Let us understand the effect of fuel vaporization on the temperature of a fuel-air mixture. Let us establish a flow of air (the molar proportion is O_2: N_2:: 1: 3.76), at an initial temperature of 400 K and 1 atm through an adiabatic duct. Somewhere in the duct, we spray an amount of methanol (CH_3OH) that corresponds to an equivalence ratio (phi) of 0.9. At the exit of the duct, we require that all the fuel be vaporized with no substantial change in velocity. Calculate the temperature drop of the fuel-air mixture using only the following data^12. Repeat the calculation assuming that only 90% of the moles of fuel has vaporized. The resulting temperature drop implies an increased air density. Since the power of an internal combustion engine is directly proportional to the air intake^13, the increased air density provides a boost of power. Methanol results in a particularly cold mixture and is thus a favorite additive among race car drivers.

Solution

Solution:

The first step is writing and balancing the stoichiometric reaction equation.

Using Eq. 2.12, C8H18 þ 8 þ 18 4 0 ðO2 þ 3:76N2Þ ! 8CO2 þ 9H2O þ 3:76 8 þ 18 4 0 N2 C8H18 þ 12:5ðO2 þ 3:76N2Þ ! 8CO2 þ 9H2O þ 3:76 12:5 N2 From here: (a) xC8H18 ¼ NC8H18 NC8H18 þ Nair ¼ 1 1 þ 12:5 4:76 ¼ 0:0165 (b) fs ¼ Mf ða þ b 4 g 2Þ 4:76 Mair ¼ 114 12:5 4:76 28:96 ¼ 0:066 (c) xH2O ¼ NH2O NCO2 þ NH2O þ NN2 ¼ 9 8 þ 9 þ 3:76 12:5 ¼ 0:141 (d) The partial pressure of water is 101 kPa 0.141 ¼ 14.2 kPa. A saturation table for steam gives the saturation temperature at this water pressure ffi 53C. Example 2.2 How many kg (lb) of air are used to combust 55.5 L (~14.7 US gallons) of gasoline? Solution: We will use isooctane C8H18 to represent gasoline. The stoichiometric fuel-air ratio is fs ¼ Mf ða þ b 4 g 2Þ 4:76 Mair ¼ 114 kg=kmol ð8 þ 18=4 0Þ 4:76 28.84 kg/kmol ¼ 0:066 One gallon of gasoline weighs about 2.7 kg (6 lb). The total fuel thus weighs about 40 kg (88 lb). The required air weighs about 40/fs 610 kg 1,300 lb. This is a lot of weight if it must be carried. Hence, for transportation applications, free ambient