Adiabatic flame temperature
1. Estimate the constant-pressure adiabatic flame temperature for the combustion of a stoichiometric C3H8–air mixture. The pressure is 1 bar and the initial reactant temperature is 298 K. Use the following assumptions:
“Complete combustion”, i.e., the product mixture consists only of CO2, H2O, and N2.
The enthalpy of the product mixture is estimated using constant specific heats evaluated at 1200 K (Ti + Tad)/2, where Tad is guessed to be about 2100 K.
2. Under the same assumptions, but obviously considering excess O2 in the products, estimate the adiabatic flame temperature for a C3H8–air mixture with equivalence ratio = 0.8.
--- Answers
1.
The mixture composition follows from balancing the chemical equation for a stoichiometric mixture of propane and air:
3 8 2 2 2 2 2
C H 5 O 3.76N 3CO 4H O+18.58N where in the product mixture we find
2 2 2
CO 3, H O=4, N =18.58
N N N
Properties:
Species Enthalpy of formation at 298.15 K
0 ,
hf i (kJ/kmol)
Specific Heat at 1200 K
,
cp i (kJ/kmol-K)
C3H8 –103847 (Table B.1, Turns’ textbook) –
CO2 –393546 (Table A.2) 56.21 (Table A.2)
H2O –241845 (Table A.6) 43.87 (Table A.6)
N2 0 33.71 (Table A.7)
O2 0 –
By applying the first law of thermodynamics:
react prod
react prod
= =
i i i i
H N h H N h
3 8
react
C H
1 103847 5 0 18.58 0 103847 kJ/mol
H
3 8
0
prod , ,
prod
C H
298.15 3 393546 56.21 298.15 4 241845 43.87 298.15
18.58 0 33.71 298.15 kJ/mol .
i f i p i ad
ad ad ad
H N h c T
T T T
Equating (first law) Hreactto Hprodand solving for Tad yields
Tad = 2404 K for 1 (stoichiometric conditions)
Another way is:
0
react ,
react react
(in our case)
i i i f i
H N h N h
0
prod , ,
prod prod
i f i ad 298 i p i
H N h T N c
If we define, on a one-mole-of-fuel basis, the enthalpy of reaction HR0 and the average heat capacity Cp of the product mixture, respectively as:
0 0 0
, ,
prod react
R i f i i f i
H N h N h
; ,
prod
p i p i
C N c then we have
0
298.15 R
ad
p
T H
C
.
Thus,
3 8
0
3 393546 4 241845 103847 2044171 kJ/molC H
HR
C H3 8
3 56.21 4 43.87 18.58 33.71 970.4 kJ/mol K
Cp
then
2044171 K
298.15
9 . 2404
ad 70 4
T
.
2.
In this case, the mixture composition follows from balancing the chemical equation for a mixture of propane and air, with excess air. Suppose that we provide 4.76a moles of air, with a>5 (5 is the value for a stoichiometric mixture). We can write the species balance as:
3 8 2 2 2 2 2 2
C H a O 3.76N 3CO 4H O 3.76 N a bO
where a and b must be computed from the knowledge of the equivalence ratio. From the definition of the equivalence ratio we can compute the value of a:
stoic
stoic
A F F A
A F F A
where
4.76
1
air stoic air
stoic
fuel stoic fuel
m a MW
A F m MW
; 4.76
1
air air
fuel fuel
m a MW
A F m MW
hence
0.8 5 6.25
0.8 0.8
stoic stoic
a a
a a
From the balance of oxygen atoms (or by simply observing that b a a stoic) we find 2 3 2 4 1 2
a b that is
5 1.25 b a . We can now rewrite the species balance for 0.8:
3 8 2 2 2 2 2 2
C H 6.25 O 3.76N 3CO 4H O 1.25O 23.5N
where in the products we find
2 2 2 2
CO 3, H O=4, N =23.5, O =1.25
N N N N
In addition to the property values used in a.1, we need the Specific Heat at 1200 K for O2 : Species Enthalpy of formation at 298.15 K
0 ,
hf i (kJ/kmol)
Specific Heat at 1200 K
,
cp i (kJ/kmol-K)
O2 0 35.593 (Table A.11)
By applying the first law: react prod
react prod
= =
i i i i
H N h H N h react 1 103847 6.25 0 23.5 0
103847 kJ
H
0
prod , ,
prod
298.15
3 393546 56.21 298.15 4 241845 43.87 298.15 23.5 0 33.71 298.15 1.25 0 35.59 298.15
3 393546 4 241845
3 56.21 4 43.87 23.5 33.71 1.25 3
i f i p i ad
ad ad ad ad
H N h c T
T T T T
5.59 298.15 3 56.21 4 43.87 23.5 33.71 1.25 35.59 Tad
that is:
Equating Hreactto Hprodand solving for Tad yields
103847 kJ 3 393546 4 241845
3 56.21 4 43.87 23.5 33.71 1.25 35.59 298.15 3 56.21 4 43.87 23.5 33.71 1.25 35.59 Tad
103847 3 393546 4 241845 298.15
3 56.21 4 43.87 23.5 33.71 1.25 35.59 2044171
298.15 2029
1180.78 Tad
K
Tad = 2029 K for 0.8
