Question 2.5: Benzene C6H6 is burned in air and dry products of combustion...
Benzene C_{6}H_{6} is burned in air and dry products of combustion contain 2% CO by volume and no free oxygen. Determine the stoichiometric air–fuel ratio by mass and the actual air ratio by mass.
Aim:
To calculate excess air supplied.
To show how to use the air–fuel ratio to indicate substoichiometric combustion.
The benzene molecule is complex, in the form of a ring with a mixture of double and single links between the carbon, as shown.
For the stoichiometric combustion case, the reaction of the fuel with oxygen can be written:
C_{3}H_{6} + 7\frac{1}{2}O_{2} \rightarrow 6CO_{2} + 3H_{2}O78 kg + 7\frac{1}{2}(32 kg) \rightarrow 6(44 kg) + 3(18 kg)
The oxygen–fuel ratio by mass is therefore 7\frac{1}{2} \times 32:78 = 3.077.
The AFR by mass can be calculated using the approximate ratio of air to oxygen in atmospheric air by mass, which is 1:0.233 = 4.29:
The AFR by mass is therefore 13.2.
For the actual combustion, insufficient oxygen is supplied to burn the fuel completely. We assume that the order of reaction of the atoms in the fuel is hydrogen first completely, since it is the fastest and by far the most aggressive. Subsequent to this, the carbon burns, firstly all to carbon monoxide and then some CO reacts to form CO_{2} to consume the remaining oxygen. The lack of sufficient air therefore leads to some CO remaining in the exhaust gas stream.
We assume in the representative reaction that there is a proportion of the stoichiometric O_{2} required, and write:
The first step is to conduct an atomic balance in terms of γ, such that when the proportion γ is known, the equation balances identical quantities of elements on each side.
Carbon: 6 = a + b
Hydrogen: 6 = 2c, i.e. c = 3.
Oxygen: 2γ = 2a + b + c and substituting c, and
a = 6 – b, 2γ = 12 – 2b + b + 3 or b = 15 – 2γ.
The second piece of information is the proportion of CO in the dry products by volume, which is 2%. We therefore create the proportion from the above equation and equate to 0.02:
\frac{b}{a + b + 3.76\gamma } = 0.02And substituting b = 2γ – 15 and a + b = 6:
\frac{15 – 2\gamma }{6 + 3.76\gamma } =0.02Therefore, γ = (15 – 0.12)/(2 + 0.0752) = 7.16.
The AFR by mass for the actual combustion process is:
\frac{air[kg]}{fuel[kg} = \frac{7.16 \times (32 kg/kmol + 3.76 \times 28 kg/kmol)}{78 kg} = \frac{983}{78} = 12.6