Page 309 - Biomass Gasification, Pyrolysis And Torrefaction Practical Design and Theory

Chapter | 8 Design of Biomass Gasifiers 285

Now, to find the value at 1100K, we use Eq. (8.19):
ð 1100
X
ΔH 0 1100 5 ΔH 0 298 1 ðC p;CH 4 1C p;CO 2 ÞdT
298 product
(8.21)

ð 1100
X
2 C p;H 2 O dT
298 reactants
The specific heats of gases are taken from Table C.4 (Appendix C) as:
5 22:35 1 0:0481T kJ=kmol
C p;CH 4
2
5 43:28 1 0:0114T 2 818363=T kJ=mol
C p;CO 2
2
C p;H 2 O 5 34:4 1 0:00062T 1 0:0000056 T kJ=mol
The integrations of respective gas components are:
2 3 1100
0:0481T 2 0:0481
5 22:35T 1 5 22:35 3 ð1100 2 298Þ 1
2 2
C p;CH 4 4 5
298
2
2
3 ð1100 2 298 Þ 5 44:8895 kJ=mol
1100
2 3
0:0114T 2 818; 363 0:0114
5 43:28T 1 1 5 43:28 3 ð1100 2 298Þ 1
C p;CO 2 4 5
2 T 2
298
818; 363
2
2
3 ð1100 2 298 Þ 1
1100 2 298
5 42:1218 kJ=mol
1100
2 3
0:000628T 2 0:0000056T 3
C p;H 2 O 5 34:4T1 1 5 534:43ð11002298Þ
4
2 3
298
0:000628
2
2
1 3ð1100 2298 Þ
2
0:0000056
3
3
1 ð1100 2298 Þ
3
530:376kJ=mol
Substituting these values and integrating the above expression, we get:
ΔH 0 5 7:65 1 104:58 2 33:578 5 78:65 kJ=mol
1100
Thus, this reaction is endothermic and it is written as:
1 1
C 1 H 2 OðgasÞ- CH 4 1 CO 2 1 78:65 kJ=mol
2 2