Modern Engineering Thermodynamics

15.8 Adiabatic Flame Temperature 617
For example, the reaction for the combustion of liquid octane with Y % theoretical air is
C 8 H 18 + ðY/100Þ12:5 O 2 + 3:76ðN 2 Þ ! 8CO ð 2 Þ+ 9H ð 2 OÞ+ ðY/100 − 1ÞO 2 + 47ðY/100ÞðN 2 Þ
To simplify matters, assume that, before the combustion, the reactants are at the SRS. Then, hðT R Þ = hðT°Þ and
hðT R Þ − hðT°Þ = 0 for all the reactants. The heat produced by this reaction when the combustion products are at
temperature T P is
q r
= ∑
R
ðn i /n fuel
Þðh f °Þ i
−∑
P
ðn i /n fuel Þ½h° f + hðT P Þ − hðT°ÞŠ i
= ðh f °Þ C8H 18
− 8½h f ° + hðT P Þ − hðT°ÞŠ CO2
− 9½h f ° + hðT P Þ − hðT°ÞŠ H2O
− ðY/100 − 1Þ½h f ° + hðT P Þ − hðT°ÞŠ O2
− 47ðY/100 − 1Þ½h f ° + hðT P Þ − hðT°ÞŠ N2
The molar specific heat equations in kJ/(kgmole·K), accurate to at least 0.43% over the range of 300 to 3500 K,
can be found in Table C.14b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics as
Carbon dioxide: ðc p Þ CO2
= −3:7357 + 30:529θ 0:5 − 4:1034θ + 0:024198θ 2
Water: ðc p Þ H2 O = 143:05 − 183:54θ0:25 + 82:751θ 0:5 − 3:6989θ
Oxygen: ðc p Þ O2
= 37:432 + 0:020102θ 1:5 − 178:57θ −1:5 + 236:88θ −2
Nitrogen: ðc p Þ N2
= 39:060 − 512:79θ −1:5 + 1072:7θ −2 − 820:40θ −3
where θ° =T°/100 = 298/100 = 2:98, and θ P = T p /100. Integrating these equations from the SRS (θ°) tothe
temperature of the combustion products (θ P ) gives
½hðT P Þ − hðT°ÞŠ CO2
= 100 ×
Z θP
θ°
ðc p Þ CO2
dθ
= −373:57ðθ P − θ° Þ+ 2035:3 ðθ P Þ 1:5 − ðθ°
Þ 1:5
− 205:17 ðθ P Þ 2 − ðθ°Þ 2 + 0:8066 ðθ P Þ 3 − ðθ°Þ 3
½hðT P Þ − hðT°ÞŠ H2 O
Z θP
= 100 × ðc p Þ H2 O dθ
θ°
= 14,305: ðθ P − θ° Þ− 14,683:2 ðθ P Þ 1:25 − ðθ°
Þ 1:25
+ 5516:7 ðθ P Þ 1:5 − ðθ°Þ 1:5 − 184:95 ðθ P Þ 2 − ðθ°Þ 2
½hðT P Þ − hðT°ÞŠ O2
= 100 ×
Z θP
θ°
ðc p Þ O2
dθ
= 3743:2ðθ P − θ° Þ+ 0:80408 ðθ P Þ 2:5 − ðθ°
Þ 2:5
+ 35:714: ðθ P Þ −0:5 − ðθ°Þ −0:5 − 23,688 ðθ P Þ −1 − ðθ°Þ −1
½hðT P Þ − hðT°ÞŠ N2
= 100 ×
Z θP
θ°
ðc p Þ N2
dθ
To simplify the algebra, we define the following terms:
= 3906:0ðθ P − θ° Þ+ 102,558 ðθ P Þ −1/2 − ðθ°
− 107,270: ðθ P Þ −1 − ðθ°Þ −1 + 41,020 ðθ P Þ −2 − ðθ°Þ −2
Þ −1/2
A = θ P − θ° B = ðθ P Þ 1:25 − ðθ°Þ 1:25 C = ðθ P Þ 1:5 − ðθ°Þ 1:5
D = ðθ P Þ 2 − ðθ°Þ 2 E = ðθ P Þ 2:5 − ðθ°Þ 2:5 F = ðθ P Þ 3 − ðθ°Þ 3
G = ðθ P Þ −1/2 − ðθ°Þ −1/2 H = ðθ P Þ −1 − ðθ°Þ −1 I = ðθ P Þ −2 − ðθ°Þ −2