Modern Engineering Thermodynamics

616 CHAPTER 15: Chemical Thermodynamics
EXAMPLE 15.10 (Continued)
Thus, the addition of 100.% excess air, a practice sometimes necessary to get complete combustion in high-velocity
combustion processes, has the effect of reducing the adiabatic combustion temperature by nearly a factor of 2.
c. For a closed, constant volume system, the adiabatic flame temperature is given by Eq. (15.18). Since the fuel in this
example is a liquid, we can assume u° f fuel ≈ h° f , and Eq. (15.18) becomes
fuel
h° −∑
f ðn fuel i /n fuel Þh° f − RT° ∑ ðn i /n fuel Þ−∑ ðn i /n fuel Þ
R
R
P
T A
≈ T°+
closed
system
∑
P
ðn i /n fuel
Þðc vi Þ avg
The numerator is
h°
f −∑ ðn fuel
i /n fuel Þh° f − RT° ∑ ðn i /n fuel Þ−∑ ðn i /n fuel Þ
R
R
P
= −249:953 − ð−5324:62Þ − 0:0083143ð25:0 + 273Þ½1 + 12:4 × 4:76 − ð8 + 9 + 47ÞŠ
= 5083:34 MJ/ðkgmole of C 8 H 18 Þ
and the denominator is
∑
P
ðn i /n fuel Þðc vi Þ avg
= 8 ðc v Þ CO2
avg + 9
ðc v Þ H2O avg + 47 ðc v Þ N2
avg
= 80:04987 ð Þ+ 90:03419 ð Þ+ 47ð0:02287Þ
= 1:782 MJ/½ðkgmole of C 8 H 18 Þ . KŠ
Then, the constant volume adiabatic flame temperature is approximately
T A
≈ 25:0 +
5083:34 MJ/ðkgmole of C 8H 18 Þ
closed 1:782 MJ/½ðkgmole of C 8 H 18 Þ .
= 2880°C = 5220°F
KŠ
system
Note that the constant volume adiabatic flame temperature in Example 15.10 is higher than the constant pressure adiabatic
flame temperature due to the energy used in the work performed in a constant pressure process, that is, pðV 2 − V 1 Þ:
Exercises
28. Determine the open system constant pressure adiabatic flame temperature for the liquid octane in Example 15.10 when
the combustion occurs with 400.% theoretical air. Answer: T A = 664°C.
29. Determine the open system constant pressure adiabatic flame temperature for the liquid octane in Example 15.10 when
the combustion occurs with 800.% theoretical air. Answer: T A = 353°C.
30. Determine the closed system constant volume adiabatic flame temperature for the liquid octane in Example 15.10 when
the combustion occurs at 200% theoretical air. Answer: T A = 1610°C.
An alternate and somewhat more accurate approach to finding the adiabatic flame temperature is to use the gas
tables in Thermodynamic Tables to accompany Modern Engineering Thermodynamics (Table C.16c) to determine the
thermodynamic properties of CO 2 ,H 2 O, O 2 ,N 2 , and so forth. However, since T A and the other thermodynamic
properties at this state are unknown, T A must be determined by trial and error as follows:
1. h R is calculated from Eq. (15.9) utilizing Eq. (15.14) or (15.15) if necessary.
2. A trial value for T A is then assumed.
3. h P is the calculated from the h° f values and the values of hðTÞ − hðT°Þ in Table C.16c.
P
4. If the value of h P calculated in step 3 equals that of h R calculated in step 1, then the correct value of T A is
assumed in step 2. Otherwise, a new T A value is chosen and the process is repeated until h P ≈ h R :
This manual iteration scheme is rather tedious, and inaccuracies are introduced by the linear interpolations in
Table C.16c required to obtain a solution. These inaccuracies can be eliminated by programming accurate molar
enthalpy formulae for the products into a microcomputer. The computer can then be programmed to calculate
the heat of combustion and iterate to find the adiabatic flame temperature in a small fraction of the time
required to carry out these calculations manually. Tables C.14 give accurate correlations for the variation in c p
with temperature for various substances. Using this information, we can determine accurate values for
hðTÞ − hðT°Þ =
Z T
T°
c p dT