Page 127 - Handbook of Energy Engineering Calculations

6. Compute the enthalpy of the products of the combustion process
Enthalpy of the products of combustion, h = m (h 1600 − h ), where m =
p
77
p
number of moles of the products; h 1600 = enthalpy of the products at 1600°F
(871°C); h = enthalpy of the products at 77°F (25°C). Thus, h = (8 + 9 +
p
77
189.2 + 37.5)(15,400 − 3750) = 2,839,100 Btu [6,259,100 Btu (SI)].

7. Compute the air supply temperature at the combustion chamber inlet
Since the combustion process is adiabatic, the enthalpy of the reactants h =
r
h , where h = (relative weight of the fuel × its heating value) + [relative
p
r
weight of the air × its specific heat × (air supply temperature − air source
temperature)]. Therefore, h = (114 × 19,100) + [(1600 + 5298) × 0.24 × (t −
a
r
77)] + 2,839,100 Btu [6,259,100 Btu (SI)]. Solving for the air supply
temperature, t = [(2.839,100 − 2,177,400)/1655.5] + 77 = 477°F (247°C).
a

Related Calculations. This procedure, appropriately modified, may be used
to deal with similar questions involving such things as other fuels, different
amounts of excess air, and variations in the condition(s) being sought under

certain given circumstances.
The coefficient, (?) = 3.784 in step 2, is used to indicate that for each unit
of volume of oxygen, O , 12.5 in this case, there will be 3.784 units of
2
nitrogen, N . This equates to an approximate composition of air as 20.9
2
percent oxygen and 79.1 percent “atmospheric nitrogen,” (N ). In turn, this
2
creates a paradox, because page 200 of Principles of Engineering
Thermodynamics, by Kiefer, et al., John Wiley & Sons, Inc., 1930, states air
to be 20.99 percent oxygen and 79.01 percent atmospheric nitrogen, where

the ratio (N )/O = (?) = 79.01/20.99 = 3.764.
2
2
Also, page 35 of Applied Energy Conversion, by Skrotzki and Vopat,
McGraw-Hill, Inc., 1966, indicates an assumed air analysis of 79 percent
nitrogen and 21 percent oxygen, where (?) = 3.762. On that basis, a formula
is presented for the amount of dry air chemically necessary for complete
combustion of a fuel consisting of atoms of carbon, hydrogen, and sulfur, or

C, H, and S, respectively. That formula is: W = 11.5C + 34.5[H − (0/8)] +
a
4.32S, lb air/lb fuel (kg air/kg fuel).

The following derivation for the value of (?) should clear up the paradox