Page 112 - Air pollution and greenhouse gases from basic concepts to engineering applications for air emission control

86 3 Basics of Gas Combustion

LHS ¼ h o þ 2h o þ 7:52h o ð1Þ
f ;CH 4 f ;O 2 f ;N 2
The right-hand side (RHS) of the equation is simplified as

b CO 2 2 2 o
RHS ¼ a CO 2 ðT a 298KÞþ ðT ð298KÞ þ h
a
2 f ;CO 2

b H 2 O 2 2 o
þ 2 a H 2 O ðT a 298KÞþ ðT ð298KÞ þ h f ;H 2 O ð2Þ
a
2

b N 2 2 2 o
þ 7:52 a N 2 ðT a 298KÞþ ðT ð298KÞ þ h
a
2 f ;N 2
From Table A.4, we can get
Species h o f ;i (J/mol) C p (T) [J/mol K]
a i b i
CO 2 −394,088 44.3191 0.0073
H 2 O −242,174 32.4766 0.00862
0 29.2313 0.00307
N 2
O 2 0 30.5041 0.00349
CH 4 −74,980 44.2539 0.02273
With these values

LHS ¼ 74,980(J/mol)

0:0073 2 2
RHS ¼ 44:3191ðT a 298Þþ ðT ð298KÞ 394.088
a
2

0:00862 2 2
þ 2 32:4766ðT a 298KÞþ ðT ð298KÞ 242:174
a
2

0:00307 2 2
þ 2ð3:76Þ 29:2313ðT a 298KÞþ ðT ð298KÞ þ 0
a
2
Equating LHS and RHS we have
2
0.02387 T þ 330:26 T a 903994 ¼ 0
a
Solving this equation we can get, T a = 2341 K; Another root T a = −16.177 K is
ignored because it is physically wrong. So the estimated AFT is
T a ¼ 2,341 K

Note that, with this high temperature, it is no longer reasonable to assume that
nitrogen does not react with oxygen. We can proceed with the calculation for better
accuracy by considering the knowledge to be introduced in Sect. 7.7.1.