Page 242 - Advanced thermodynamics for engineers

10.6 EXAMPLES 229

Try T P ¼ 3200 K.

Constituent CO 2 H 2 O O 2 N 2

u 3200 148,954.8 118,773.6 88,164.1 82,226.2
n 3 4 0.882 22.12
nu 446,864.5 475,094.3 77,760.8 1,818,843

U P ðT P Þ¼ 2818562:6kJ:
Interpolating linearly gives

2780194:9 2717672:4
T P ¼ 3100 þ 100 ¼ 3162 K:
2818562:6 2717672:4
Check solution using T P ¼3162 K.

Constituent CO 2 H 2 O O 2 N 2
146,843.0 117,034.7 86,949.1 81,142.8
u 3162
n 3 4 0.882 22.12
nu 440,529 468,138.7 76,689.4 1,794,878

U P ðT P Þ¼ 2780234:8kJ:
This value is within 0.0014% of the value calculated from the energy equation. Hence
T P ¼ 3162 K.
Pressure at the end of combustion

n P <T P V 1 30 3162
p 3 ¼ $ $p 1 ¼ 10 1:2
n R <T 1 V 3 29 400
¼ 98:13 bar:

Example 5: combustion at constant pressure of a rich mixture of benzene
A rich mixture of benzene (C 6 H 6 ) and air with an equivalence ratio, f, of 1.2 is burned at constant
pressure in an engine with a compression ratio of 15:1. If the initial conditions are 1 bar and 25 C

calculate the pressure, temperature and volume after combustion (Fig. 10.9). The ratio of specific heats
during compression, k, is 1.4 and the calorific value of benzene is 40,635 kJ/kg.
p 1 ¼ 1 bar

T 1 ¼ 298 K
1:4
V 1
p 2 ¼ p 1 ¼ 44:31 bar
V 2
0:4

V 1
T 2 ¼ T 1 ¼ 880 K:
V 2