Page 106 - Advanced thermodynamics for engineers

4.9 EXERGY 91

Thus

DB ¼ h 0 f T 0 s H 2 O s H 2 0:5s O 2 (4.65)
H 2 O
o o
(Note: h is the enthalpy of formation of a compound and ðh Þ ¼ 241820 kJ/kg at 25 C.)
f f H 2 O
Substituting values gives, for t 0 ¼ 25 C which is also the standard temperature for evaluating the

enthalpy of reaction. Then
DB ¼ 241820 298 ð188:71 130:57 0:5 205:04Þ
(4.66)
¼ 228594:76 kJ=kmol K:
This means that the ability of the fuel to do work is 5.5% less than the original enthalpy of for-
mation of the ‘fuel’, and hence 94.5% of the energy defined by the enthalpy of formation is the
maximum energy that can be obtained from it.

4.9.2.2 Exergy of reaction of methane (CH 4 )
The equation for combustion of methane is
CH 4 þ 2O 2 /CO 2 þ 2H 2 O (4.67)

Hence,

DB ¼ Sðb i Þ Sðb i Þ ¼ðDH R Þ CH 4 T 0 s CO 2 þ 2s H 2 O s CH 4 2s O 2
R
P
3
¼ 804:6 10 298 ð214:07 þ 2 188:16 182:73 2 204:65Þ
3
¼ 804:1 10 kJ=kmol
In this case the exergy of reaction is almost equal to the enthalpy of reaction; this occurs because
the entropy of the reactants and products are almost equal.
4.9.2.3 Exergy of reaction of octane (C 8 H 18 )
The equation for combustion of octane is
C 8 H 18 þ 12:5O 2 /8CO 2 þ 9H 2 O (4.68)

3
Assume that the enthalpy of reaction of octane ðDH R Þ is 5074.6 10 kJ/kmol, and that the
entropy of octane at 298 K is 360 kJ/kmol K. Then the exergy of reaction is

DB ¼ Sðb i Þ Sðb i Þ ¼ðDH R Þ C 8 H 18 T 0 8s CO 2 þ 9s H 2 O s CH 4 12:5s O 2
R
P
3
¼ 5074:6 10 298 ð8 214:07 þ 9 188:16 360 12:5 204:65Þ
3
¼ 5219:9 10 kJ=kmol
In this case the exergy of reaction is greater than the enthalpy of reaction by 2.86%: this is because
the entropy of the products is greater than that of the reactants.