Modern Engineering Thermodynamics

15.2 Stoichiometric Equations 595
CHEMICAL REACTION EQUATIONS AND SIGNIFICANT FIGURES
We have a problem using significant figures in the equations for chemical reactions. For example, in the reaction of hydrogen
and oxygen to form water, H 2 + 0.5 O 2 → H 2 O, we do not want to have to write it as 1.00 H 2 + 0.500 O 2 → 1.00 H 2 O.
So, in this chapter, we assume that the coefficients in an equation for a chemical reaction have at least three significant
figures without actually writing them as such in the reaction equations.
The amount of air or oxygen used in a combustion process can be described as
1. The percent of theoretical air or oxygen required to carry out the reaction.
2. The percent of excess or deficit air or oxygen actually used in the reaction.
3. The air/fuel (A/F) or fuel/air (F/A) ratio used in the reaction measured on either a mass or a mole basis.
One hundred percent theoretical air is the minimum amount of air that supplies enough oxygen to carry out
complete combustion. The percentage of excess air is simply the percentage of theoretical air supplied minus
100, and the percentage of deficit air is 100 minus the percent of theoretical air supplied. The air/fuel ratio (A/F)
is the amount of air used per unit of fuel consumed, and it can be expressed either in mass or mole units. The
relation between the mass and molar air/fuel ratios is given by
Molecular mass of air, M
ðA/FÞ mass = ½ðA/FÞ molar in moles of air/mole of fuelŠ ×
air
Molecular mass of fuel, M fuel
The fuel/air ratio (F/A) is simply the inverse of the air/fuel ratio.
and
ðF/AÞ molar
=
1
ðA/FÞ molar
ðF/AÞ
1
mass =
ðA/FÞ mass
With these definitions, it is easy to see that the hydrogen combustion described in the reaction H 2 +
0:5½O 2 + 3:76ðN 2 ÞŠ ! H 2 O + 1:88ðN 2 Þ uses 100% theoretical air and no excess or deficit air is involved. The
molar air/fuel ratio for this reaction is
ðA/FÞ molar = n air 0:5 × ð1 + 3:76Þ moles of air
= = 2:38 moles of air/mole H 2
n fuel 1 mole of fuel
and the mass air/fuel ratio is 1
Molecular mass of air, M
ðA/FÞ mass
= ½ðA/FÞ molar
moles of air/mole of fuelŠ ×
air
Molecular mass of fuel, M fuel
28:97 Mass of air/mole of air
= 2:38 moles of air/mole of fuel ðH 2 Þ ×
2:016 Mass of fuel/mole of fuel
= 34:2 grams of air/gram of H 2 = 34:2 kg of air/kg of H 2 = 34:2 lbm of air/lbm of H 2
Then, the molar and mass fuel/air ratios are simply
and
ðF/AÞ molar
= ðA/FÞ −1
molar = 0:42 moles H 2/mole air
ðF/AÞ mass
= ðA/FÞ −1
mass
= 0:029 gram H 2 /gram air
= 0:029 kg H 2 /kg air
= 0:029 lbm H 2 /lbm air
1 While the mole is in fact a unit of mass, in this section we use the term mass to indicate nonmolar units.