9781441979421-c1

# A saturation table for steam gives the saturation

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Unformatted text preview: ion. Using Eq. 2.12,     18 18 C8 H18 þ 8 þ À 0 ðO2 þ 3:76N2 Þ ! 8CO2 þ 9H2 O þ 3:76 8 þ À 0 N2 4 4 C8 H18 þ 12:5ðO2 þ 3:76N2 Þ ! 8CO2 þ 9H2 O þ 3:76 Á 12:5 Á N2 From here: (a) xC8 H18 ¼ NC8 H18 1 ¼ 0:0165 ¼ NC8 H18 þ Nair 1 þ 12:5 Á 4:76 Mf 114 ¼ 0:066 ¼ g ða þ b À 2Þ Á 4:76 Á Mair 12:5 Á 4:76 Á 28:96 4 N H2 O 9 (c) xH2 O ¼ ¼ 0:141 ¼ N CO2 þ N H2 O þ N N2 8 þ 9 þ 3:76 Á 12:5 (d) The partial pressure of water is 101 kPa Á 0.141 ¼ 14.2 kPa. A saturation table for steam gives the saturation temperature at this water pressure ﬃ 53 C. (b) fs ¼ 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 b 4 ða þ À g 2Þ Á 4:76 Á Mair 114 kg=kmol ¼ ð8 þ 18=4 À 0Þ Á 4:76 Á 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 air is preferred. 2.3 Heating Values 23 Example 2.3 In a model “can-combustor” combustion chamber, n-heptane (C7H16) is burned under an overall lean condition. Measurements of dry exhaust give mole fractions of CO2 and O2 as xCO2 ¼ 0.084 and xO2 ¼ 0.088. Determine the %EA, equivalence ratio f, and l. Solution: To avoid condensation of water inside the instruments, measurements of exhaust gases are taken on a ‘dry’ mixture that is obtained by passing the exhaust gases through an ice bath so that most water is condensed. Further removal of water can be done with desiccants. The mole fractions measured under dry conditions will be larger than at real conditions since water is removed. However, this will not impact the relation deduced above, as both xCO2 and xO2 are increased by the same factor. %EA ¼ 100 a 7  ¼ 0:667 ! %EA ¼ b g xCO2 ð7 þ 16=4 À 0Þð0:084=0:088Þ aþ À 4 2 xO 2 ¼ 66:7 Next we use the relati...
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## This document was uploaded on 01/20/2014.

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