Page 298 - Advanced Thermodynamics for Engineers, Second Edition

12.11 DISSOCIATION PROBLEMS 287

~
The effect of the formation of NO is to reduce DU 0 in the following manner.
X X
U f
D ~ U 0 ¼ U f
Products Reactants
¼ 0:7985 ð 393405Þþ 0:2014 ð 113882Þþ 1:9316 ð 239082Þþ 0:09174 89915 ð 66930Þ
|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}
Formation of NO
¼ 723678 kJ=kmol CH 4
(12.163)
The energy equation based on 0 K is

DU 0 þ½U R ðT R Þ U R ð0ފ ½U P ðT P Þ U P ð0ފ ¼ 0 (12.164)
giving

(12.165)
U P ðT P Þ¼ DU 0 þ U R ðT R Þ
Applying the energy equation to the chemical equation neglecting dissociation of NO gives
U P ðT P Þ ¼ ð 731927Þþ 179377 ¼ 911304 kJ (12.166)

Substituting the value of T P given in the question, viz., 2957 K gives

U P ðT P Þ ¼ 913268 kJ

Constituent CO 2 H 2 O CO H 2 O 2 N 2
135615 107643 76050 70837 80457 75270
u 2957
n 0.7986 1.9315 0.2014 0.06845 0.13495 7.52

This satisfies the energy equation within 0.2%.
Now, consider the case with NO formation. First, the energy of the products is less than in the
previous case because the formation of NO has reduced the energy released by combustion. In this
case,
U P ðT P Þ ¼ ð 723678Þþ 179377 ¼ 903055 kJ

If it is assumed that the temperature is reduced by the ratio 903055/911304, then a first approxi-
mation of the products temperature is 2930 K. In addition, it is necessary to include NO in the products
and this is introduced in the table below.
U P ðT P Þ¼ 903567 kJ

Constituent CO 2 H 2 O CO H 2 O 2 NO N 2

u 2930 134153 106407 75271 70068 79609 77304 74494
n 0.7986 1.9315 0.2014 0.06845 0.08908 0.09174 7.474

This value of products’ energy obviously satisfies the energy equation, and shows that the effect of
the formation of NO has been to reduce the products temperature, in this case by about 30 K.