Thermodynamics Demystified
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Dividing through the chemical equation by the value of a so that we have 1 mol of fuel, provides C4 H l0 + 7.5(O 2 + 3.76N 2 ) ⇒ 3.67CO 2 + 0.33CO + 1.17O 2 + 28.17N 2 + 5H 2 O Comparing this with the combustion equation of Example 9.1 using theoretical air, we find % theoretical air =
7.5 × 100 = 115.4% 6.5
EXAMPLE 9.4 Volumetric analysis of the products of combustion of an unknown hydrocarbon, measured on a dry basis, gives 10.4% CO2, 1.2% CO, 2.8% O2, and 85.6% N2. Determine the composition of the hydrocarbon and the percent excess air. Solution The chemical equation for 100 mol dry products is Ca H b + c(O 2 + 3.76N 2 ) ⇒ 10.4CO 2 + 1.2CO + 2.8O 2 + 85.6N 2 + dH 2 O Balancing each element, C: N: O: H:
a = 10.4 + 1.2 ⎫ ⎪ 3.76c = 85.6 ⎪ ⎬ 2c = 20.8 + 1.2 + 5.6 + d ⎪ ⎪⎭ b = 2d
∴ a = 11.6 d = 18.9
c = 22.8 b = 37.9
The chemical formula for the fuel is C11.6H37.9. This could represent a mixture of hydrocarbons, but it is not any species listed in App. B, since the ratio of hydrogen atoms to carbon atoms is 37.9/11.6 = 3.27 = 13/4, i.e., C4H13. To find the percent theoretical air we must have the chemical equation using 100% theoretical air. It would be C11.6 H 37.9 + 21.08(O 2 + 3.76N 2 ) ⇒ 11.6CO 2 + 18.95H 2 O + 79.26N 2 Using the number of moles of air from the actual chemical equation, we find % excess air = % theoretical air − 100 =
22.8 × 100 − 100 = 8% 21..08