Page 244 - Advanced thermodynamics for engineers

10.6 EXAMPLES 231

Products enthalpy at standard temperature, T s

Constituent CO 2 CO H 2 O N 2
9519.3 8489.5 9701.0 8503.4
h 298
n 3.5 2.5 3 23.5
nh 33,317.5 21,223.8 29,102.9 199,830.6

H P ðT s Þ¼ 283474:8kJ:
In this case the combustion is not complete and hence the full energy available in the fuel is not
released. Applying Hess’s law, the energy released by a combustion process is given by

¼ðDH f Þ ðDH f Þ :
Q p P R
25
In this case

Q p ¼ 3:5ðDh f Þ þ 2:5ðDh f Þ CO þ 3ðDh f Þ H 2 O ðDh f Þ
25 CO 2 C 6 H 6
(10.39)
¼ 3:5 393522 þ 2:5 110529 þ 3 241827 ð82847:6Þ
¼ 2461978:1kJ:
This value can also be calculated from the enthalpy of reaction by subtracting the energy which is
not released because of the incomplete combustion of all the carbon to carbon dioxide,

Q p ¼ Q p 2:5fðDh f Þ ðDh f Þ CO g
25 25;total CO 2
¼ 40635 78 2:5ð 393522 þ 110529Þ
¼ 3169530 þ 707482:5
¼ 2462047:5kJ:
These two values of enthalpy of combustion are the same within the accuracy of the data used (they
differ by less than 0.003%). Hence, referring to Fig. 10.7, the loss of energy available due to incomplete
combustion is 707,482.5 kJ/kmol C 6 H 6 because not all the carbon is converted to carbon dioxide.
Thus
H P ðT P Þ¼ 2462047:5 þ 567888:1 þ 283471:3 ¼ 3313406:9kJ:
Assume T P ¼ 2800 K.

Constituent CO 2 CO H 2 O N 2
150,434.9 94,792.4 123,750.8 94,034.2
h 2800
n 3.5 2.5 3 23.5
nh 526,522.2 236,981.0 371,252.5 2,209,803

H P ðT P Þ¼ 3344558:7kJ: