Modern Engineering Thermodynamics

610 CHAPTER 15: Chemical Thermodynamics
Table 15.3 Mass Higher Heating Values (HHV) for Common Fuels Where Both the
Reactants and the Products Are at the SRS and the Water in the Combustion Products
Is in the Liquid Phase
(HHV) mass
Fuel
MJ/kg
Btu/lbm
Hydrogen, H 2 (g) −142.9 −61,485
Carbon, C(s) −32,79 −14,108
Carbon monoxide, CO(g) −10.11 −4,348
Methane, CH 4 (g) −55.65 −23,940
Acetylene, C 2 H 2 (g) −49.98 −21,505
Ethylene, C 2 H 4 (g) −50.39 −21,679
Ethane, C 2 H 6 (g) −52.00 −22,369
Propylene, C 3 H 6 (g) −49.01 −21,085
Propane, C 3 H 8 (g) −50.45 −21,706
n-Butane, C 4 H 10 (g) −49.61 −21,341
Benzene, C 6 H 6 (g) −41.92 −18,026
n-Octane, C 8 H 18 (g) −47.85 −20,570
n-Decane, C 10 H 22 (g) −47.57 −20,451
Methyl alcohol, CH 3 OH(g) −23.89 −10,272
Ethyl alcohol, C 2 H 5 OH(g) −30.64 −13,180
Source: Reprinted by permission of the publisher from Holman, J. P., 1980. Thermodynamics, third ed. McGraw-Hill, New York, p. 466 (Table 11-1).
EXAMPLE 15.8
Determine the higher and lower heating values of methane. Note that, for the determination of the HHV and LHV, the combustion
reaction must occur with 100.% theoretical air and both the reactants and the products must be at the standard
reference state. For the HHV calculation, the water in the combustion products must be in the liquid phase, and for the
LHV calculation, it must be in the vapor phase.
Solution
For 100.% theoretical air, the combustion equation for methane is
CH 4 + 2:00 O 2 + 3:76ðN 2 Þ ! CO 2 + 2:00ðH 2 OÞ+ 7:52ðN 2 Þ
Since both the reactants and the products are at the standard reference state, we can use Eq. (15.13) to find the heat of
combustion, which is either the HHV or the LHV, depending on how the water term is handled. Then,
and, from Table 15.1, we find that
h R ° = ∑
R
ðn i /n fuel Þ h° f i = h° f + 2:00 h° f + 7:52 h° f CH4 O2 N2
h f ° CH4 = −74:873 MJ/kgmole CH 4
Since O 2 and N 2 are the elements of the compound CH 4 , in their standard states (see Table 15.1), h° f = h° f = 0: Then,
O2 N2
h R °= −74:873 MJ/kgmole CH 4 : Similarly,
h P ° = ∑
P
ðn i /n fuel Þ h° f i = h° f + 2:00 h° f CO2 H2OðÞ l
+ 7:52 h f° N 2
From Table 15.1, we find that h f ° N2 = 0, and h f ° CO2 = −393:522 MJ/kgmole CO 2
h° f H2OðÞ g
= −241:827 MJ/kgmole H 2O vapor
h° f H2OðlÞ = −285:838 MJ/kgmole H 2O liquid