Modern Engineering Thermodynamics

15.8 Adiabatic Flame Temperature 615
EXAMPLE 15.10
For liquid octane, C 8 H 18 ðÞ, l determine the following adiabatic flame temperatures when the reactants are in the standard
reference state (25°C and 0.100 MPa) and the combustion products are assumed to be ideal gases:
a. The open system (constant pressure) adiabatic flame temperature burning with 100.% theoretical air.
b. The open system (constant pressure) adiabatic flame temperature burning with 200.% theoretical air.
c. The closed system (constant volume) adiabatic flame temperature burning with 100.% theoretical air.
Solution
a. The combustion equation for octane burning with 100.% theoretical air is
C 8 H 18 + 12:5 O 2 + 3:76ðN 2 Þ ! 8CO ð 2 Þ+ 9H ð 2 OÞ+ 47ðN 2 Þ
Since the products can be considered to be ideal gases, we can use Eq. (15.17) and the average specific heat values given
in Table 15.5. From Table 15.1, we find,
and
h° f fuel = h° f = −249:952 MJ/kgmole
C8H18ðlÞ
h° f = −393:522 MJ/kgmole
CO2
h° f = −241:827 MJ/kgmole
H2OðgÞ
h f ° N 2 = 0 because it is an element
The open system constant pressure adiabatic flame temperature is given by Eq. (15.17), where
∑
P
ðn i /n fuel Þ h° f = 8 h° i f + 9 h° f CO2 H2O + 47 h° f N2
= 8ð−393:522Þ+ 9ð−241:827Þ+ 47ð0Þ
= −5325 MJ/kgmole of C 8 H 18
and
∑
P
ðn i /n fuel Þ c pi avg = 8 c p
CO2
avg + 9
c p
H2O
avg + 47 c p
N2 avg
= 80:05818 ð Þ+ 90:04250 ð Þ+ 47ð0:03118Þ
= 2:313 MJ/ ðkgmole of C 8 H 18 Þ . K
Then, Eq. (15.17) gives
T A
= 25:0°C + open
system
b. The reaction equation when 200.% theoretical air is used is
−249:952 MJ/kgmole fuel − ð −5325 MJ/kgmole fuel Þ
2:313 MJ/ðkgmole fuel .
= 2170°C = 3940°F
KÞ
C 8 H 18 + 212:5 ð Þ O 2 + 3:76 N 2
ð Þ ! 8CO ð 2
Þ+ 9H ð 2 OÞ+ 12:5 O 2
ð Þ+ 94ðN 2 Þ
The numerator in Eq. (15.17) is the same here as it was in part a, since we only added more elements to the reaction
side of the equation. The denominator represents the energy required to raise the temperature of all the product gases
and is consequently different from part a. In this case,
∑
P
Then, Eq. (15.17) gives
ðn i /n fuel Þ c pi avg = 8 c p
CO2
avg + 9
c p
T A
= open
system
200% TA
H2O
avg + 12:5 c p
O2
avg + 94 c p
N2 avg
= 80:05818 ð Þ+ 90:04250 ð Þ+ 12:5ð0:03299Þ+ 94ð0:03118Þ
= 4:19 MJ/ ðkgmole of C 8 H 18 Þ . K
−249:952 − ð−5325Þ
4:19
+ 25:0 = 1240°C = 2260°F
(Continued )