Modern Engineering Thermodynamics

602 CHAPTER 15: Chemical Thermodynamics
EXAMPLE 15.3 (Continued)
Exercises
7. Determine the hydrocarbon fuel model in Example 15.3 if the CO and O 2 concentrations are both 0.00%, and the CO 2
and N 2 concentrations are 17.0% and 83.0%, respectively, in the Orsat analysis. Answer: Hydrocarbon fuel model =
C 17 H 20.3 .
8. Determine the molar and mass air/fuel ratios in Example 15.3 if the CO and O 2 concentrations are both 0.00%, and the
CO 2 and N 2 concentrations are 17.0% and 83.0%, respectively, in the Orsat analysis. Answer: A/F = 10.15 kgmole air/
kgmole fuel = 1.31 kg air/kg fuel.
9. If the CO concentration in Example 15.3 is 0.00% and the N 2 concentration is 83.0%, with all the other concentrations
unchanged, determine the hydrocarbon fuel model and % of excess air used in the combustion process.
Answer: Hydrocarbon fuel model = C 7.1 H 22.28 and % of excess air = 74.2%.
In the previous example, we assume that nearly all the water produced by the combustion process condensed
out by the time the combustion products cooled to 20.0°C. For this to be a valid assumption, the dew point of
the combustion products must be at 20.0°C or higher. The determination of the dew point temperature for this
reaction is illustrated in the next example.
EXAMPLE 15.4
Determine the dew point temperature of the combustion products given in Example 15.3 if the total pressure of the mixture
is 14.7 psia.
Solution
From Eq. (12.23) of Chapter 12, the volume fractions, mole fractions, and partial pressure ratios are all equal for a mixture
of ideal gases. Exhaust products at atmospheric pressure are sufficiently ideal to allow us to determine the water vapor partial
pressure at its condensation temperature (i.e., dew point) from this relation. The total number of moles of product,
from part a of Example 15.3, is 109 moles. Then, using Eq. (12.23) wherein p m is the total pressure of the mixture gives
π H2O = p H2O/p m = ψ H2O = χ H2O = 9:00
109 = 0:0826
where π is the partial pressure ratio, ψ is the volume fraction, and χ is the mole fraction. So,
p H2O = 0:0826ð14:7Þ = 1:21 psia
The saturation temperature of water vapor at this pressure is defined to be the dew point temperature. By interpolation in
Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that
T sat ð1:21 psiaÞ = T DP = 108°F = 42:3°C
Thus, the exhaust products must be cooled to 108°F ð42:3°CÞ or below to condense the water of combustion and have an
essentially dry exhaust gas.
Exercises
10. Determine the partial pressure of the water vapor in Example 15.4 if the mixture total pressure is 0.150 MPa.
Answer: p H2O= 12.4 kPa.
11. If the dew point temperature in Example 15.4 is 212°F, what is the mixture total pressure? Answer: p m = 178 psia.
12. Determine the partial pressure and dew point temperature of the water vapor present in the 100.% theoretical air
combustion process given in part d of Example 15.3. Assume the mixture total pressure is 14.7 psia. Answer: p H2O =
2.08 psia, T DP = 128°F.
WHY DO AUTOMOBILE EXHAUST SYSTEMS RUST?
Water condenses in an automobile’s exhaust system and drips out the tailpipe until the entire exhaust system has been
heated above the dew point temperature by the exhaust gases. This water promotes corrosion and causes the exhaust system
to rust out sooner if the vehicle is used for short trips than trips long enough (a half hour or more) to dry out the
exhaust system.