Chapter 15 | 777
Adiabatic
combustion
chamber
25 °C, 1 atm
CO 2
T 0 = 25°C
AIR
CH 4
25 °C, 1 atm H
2 O
O 2
N 2
FIGURE 15–33
Schematic for Example 15–10.
EXAMPLE 15–10 Second-Law Analysis of Adiabatic Combustion
Methane (CH 4 ) gas enters a steady-flow adiabatic combustion chamber at
25°C and 1 atm. It is burned with 50 percent excess air that also enters at
25°C and 1 atm, as shown in Fig. 15–33. Assuming complete combustion,
determine (a) the temperature of the products, (b) the entropy generation,
and (c) the reversible work and exergy destruction. Assume that T 0 298 K
and the products leave the combustion chamber at 1 atm pressure.
Solution Methane is burned with excess air in a steady-flow combustion
chamber. The product temperature, entropy generated, reversible work, and
exergy destroyed are to be determined.
Assumptions 1 Steady-flow conditions exist during combustion. 2 Air and
the combustion gases are ideal gases. 3 Changes in kinetic and potential
energies are negligible. 4 The combustion chamber is adiabatic and thus
there is no heat transfer. 5 Combustion is complete.
Analysis (a) The balanced equation for the complete combustion process
with 50 percent excess air is
Under steady-flow conditions, the adiabatic flame temperature is determined
from HprodHreact, which reduces to
since all the reactants are at the standard reference state and h
- f°O for O 2
and N 2. Assuming ideal-gas behavior for air and for the products, the h
f° and
hvalues of various components at 298 K can be listed as
h
- f° h
- 298 K
Substance kJ/kmol kJ/kmol
- 298 K
CH 4 (g) 74,850 —
O 2 0 8682
N 2 0 8669
H 2 O(g) 241,820 9904
CO 2 393,520 9364
Substituting, we have
which yields
By trial and error, the temperature of the products is found to be
Tprod1789 K
hCO 2 2 hH 2 OhO 2 11.28hN 2 937,950 kJ
1 1 kmol CH 421 74,850 kJ>kmol CH 42
¬ 1 1 kmol O 22310 hO 2 86822 kJ>kmol O 24
¬ 1 11.28 kmol N 22310 hN 2 86692 kJ>kmol N 24
¬ 1 2 kmol H 2 O 231 241,820hH 2 O 99042 kJ>kmol H 2 O 4
¬ 1 1 kmol CO 2231 393,520hCO 2 93642 kJ>kmol CO 24
aNp^1 h°fhh°^2 paNrh°f,r^1 Nh°f^2 CH 4
CH 41 g 2 31 O 2 3.76N 22 SCO 2 2H 2 OO 2 11.28N 2