Page 290 - Advanced Thermodynamics for Engineers, Second Edition

12.11 DISSOCIATION PROBLEMS 279

~

Q ~
v s ¼ DU 0 þ½U P ðT s Þ U P ðT 0 ފ ½U R ðT s Þ U R ðT 0 ފ
¼ 4943040 þ 7:136 7041:6 þ 8:84322 7223:3 þ 0:846 6011:8
þ 0:15678 6043:1 þ 1:8994 6032:0 þ 52:222 6025:8 (12.122)

ð5779 þ 13:889 ½6032:0 þ 3:76 6025:8ŠÞ
¼ 4895205 kJ=kmol C 8 H 18 burned
This is within 1.25% of that calculated from the internal energy of reaction and the difference is
because data were taken from different sources.
Evaluating the energy terms for the reactants gives

½U R ðT R Þ U R ðT s ފ ¼ 335014 kJ

Constituent C 8 H 18 O 2 N 2

u 500 45783 10574.8 10356.1
u 298 0 6032.0 6025.8
Difference 45783 4542.8 4330.3
n 1.0 13.889 52.222

Hence, the energy of the products, based on the evaluated degrees of dissociation, must be

s
U P ðT P Þ U P ðT s Þ¼ ðQ v Þ þ U R ðT R Þ U R ðT s Þ
¼ 4833894 þ 335014 (12.123)

¼ 5168908 kJ
Evaluating the energy contained in the products at 2800 K gives

U P ðT P Þ U P ðT s Þ¼ 5260072:5kJ

Constituent CO 2 H 2 O CO H 2 O 2 N 2
127154.9 100470.8 71512.4 66383.6 75548.1 70754.1
u 2800
u 298 7041.6 7223.3 6011.8 6043.1 6032.0 6025.8
Difference 120113.3 93247.5 65500.5 60340.5 69516.1 64728.4
n 7.1360 8.84322 0.864 0.15678 1.8994 52.222

This shows that the products at 2800 K contain more energy than was available from the energy
released by the fuel and hence the energy equation has not been satisfied. It is necessary to repeat the
whole calculation with a lower temperature guessed for the products. Rather than do a number of