Page 244 - Advanced Thermodynamics for Engineers, Second Edition

232 CHAPTER 10 THERMODYNAMICS OF COMBUSTION

Hence T P < 2800 K.
Use T P ¼ 2700 K.

Constituent CO 2 CO H 2 O N 2

h 2700 144,256.1 91,070.2 118,376.1 90,326.0
n 3.5 2.5 3 23.5
nh 504,896.3 227,675.4 355,128.2 2,122,661

H P ðT P Þ ¼ 3210360:9kJ:
Interpolate linearly
3313406:9 3210360:9
100 ¼ 2776:8K:
T P ¼ 2700 þ
3344558:7 3210360:9

Constituent CO 2 CO H 2 O N 2

h 2777 149,011.4 93,936.1 122,512.5 93,181.1
n 3.5 2.5 3 23.5
nh 521,540 234,840.3 367,537.5 2,189,757

H P ðT P Þ¼ 3313674:8kJ:

Hence the equations balance within 0.020%.
The pressure at the end of combustion is the same as that at the end of compression i.e.
p 3 ¼ 44.31 bar. Hence the volume at the end of combustion, V 3 , is given by

n P <T P p 1 32:5 2777 1
V 1
V 3 ¼ V 1 ¼
n R <T 1 p 3 30:75 298 44:31
¼ 0:2223 V 1 :

10.7 CONCLUDING REMARKS

A consistent method of analysing combustion has been introduced. This is suitable for use with all the
phenomena encountered in combustion, including weak and rich mixtures, incomplete combustion,
heat and work transfer, dissociation and rate kinetics. The method which is soundly based on the First
Law of Thermodynamics ensures that energy is conserved. While the approach seems cumbersome for
hand calculations it is easily implemented in computer programs.
A large number of examples of different combustion situations have been presented.