Page 380 - Air pollution and greenhouse gases from basic concepts to engineering applications for air emission control

12.4 Carbon Capture Processes 359

For an adiabatic constant pressure system, with the reactant temperature of
T R = T 0 = 298 K, the left-hand side (LHS) of Eq. (12.15)

b i 2 2 o
X
n i a i ðT R 298 KÞþ ðT ð298 KÞ þ h f ;i
R
2
R
ð12:15Þ

b i 2 2 o
X
¼ n i a i ðT a 298 KÞþ ðT ð298 KÞ þ h f ;i
a
2
P
becomes
LHS ¼ h o þ 2h o ð1Þ
f ;CH 4 f ;O 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 298 KÞþ ðT ð298 KÞ þ h
a
2 f ;CO 2
ð2Þ

b H 2 O 2 2 o
þ 2 a H 2 O ðT a 298 KÞþ ðT ð298 KÞ þ h f ;H 2 O
a
2
Looking into Table A.4, we can get the parameters needed (Table 12.4)
Substituting these values into Eq. (1) and Eq. (2) leads to
LHS ¼ 74;980 J=moleÞ
ð
0:0073 2 2

RHS ¼ 44:3191ðT a 298Þþ ðT ð298 KÞ 394;088
a
2

0:00862 2 2
þ 2 32:4766ðT a 298 KÞþ ðT ð298 KÞ 242;174
a
2
Simplification of the equation gives
2
0:01227T þ 109:2723 T a 837109 ¼ 0
a
Solving this equation we can get, T a = 4930.8 K. This calculated value is much
higher that air-fuel combustion: T a ¼ 2341 K.

Table 12.4 Parameters used in Example 1.22
Species h o f ;i (J/mole) C p (T) (J/mole · K)
a i b i
CO 2 −394,088 44.3191 0.0073
H 2 O −242,174 32.4766 0.00862
O 2 0 30.5041 0.00349
CH 4 −74,980 44.2539 0.02273