Question 2.4: Determine the stoichiometric AFR: (1) by mass (2) by volume ...
Determine the stoichiometric AFR:
(1) by mass (2) by volume
for propane, C_{3}H_{8}. Calculate the volumetric composition by wet and dry analysis of the products. If the fuel is burnt with 30% excess air, what are the volumetric and gravimetric composition of wet products?
Aim:
To realize stoichiometric air–fuel ratio by mass and by volume.
To realize the importance of wet and dry analysis of product gases.
Stoichiometric air to fuel ratio (AFR):
(1) by mass
Reaction equation of the fuel:
C_{3}H_{8} + 5O_{2} \rightarrow 3CO2 + 4H_{2}O
44 kg + 5(32 kg) \rightarrow 3(44 kg) + 4(18 kg)
The oxygen–fuel ratio by mass is therefore 5 \times \frac{32}{44} = 3.636. The air–fuel ratio can be worked out using the approximate ratio of air to oxygen in atmospheric air, which is 1:0.233 by mass:
\frac{air[kg]}{fuel[kg]} = \frac{oxygen[kg]}{fuel[kg]} \times \frac{1}{0.233} = 3.636 \times 4.29The AFR by mass is therefore 15.6.
(2) by volume
The oxygen–fuel ratio by volume is 5:1 from the reaction equation above. To find the AFR by volume, use the approximate ratio of air to oxygen in atmospheric air, which is 1:0.21 by volume:
The AFR by volume is therefore 23.8.
The complete representative equation with all mixture components of the air–fuel stream includes nitrogen in the approximate ratio for atmospheric air. Including this in the reaction equation gives:
In order to get a volumetric analysis of wet and dry products, use a table for
compactness. Wet includes the water in the reaction; dry does not include the water.
| Wet | Dry | |||
| Component | V_{i} | \frac {V_{i}}{V} | V_{i} | \frac {V_{i}}{V} |
| CO_{2} | 3 | 0.116 | 3 | 0.138 |
| H_{2}O | 4 | 0.155 | 0 | 0 |
| N_{2} | 18.81 | 0.729 | 18.81 | 0.862 |
| Σ | 25.81 | 1 | 21.81 | 1 |
When there is 30% excess air in the reacting stream of air and fuel, there will be 30% more O_{2} and N_{2} in the representative equation:
C_{3}H_{8} + 5 \times 1.3(O_{2} + ^{0.79}/_{0.21}N_{2}) → 3CO2 + 4H_{2}O + 0.30_{2} + 5 \times 1.3 \times ^{0.79}/_{0.21}N_2The list of products can be made for gravimetric and volumetric analysis:
| Component | V_{i} | \frac {V_{i}}{V} | \tilde{m _{i}} | m_{i} = \tilde{m _{i}} \frac {V_{i}}{V} | \frac{m_{i}}{m} = \frac{\tilde{m _{i}}}{\tilde{m} } \frac{V_{i}}{V} |
| [kmol] | [-] | [kg/kmol] | [kg/kmol] | ||
| CO_{2} | 3 | 0.091 | 44 | 4 | 0.141 |
| H_{2}O | 4 | 0.122 | 18 | 2.187 | 0.077 |
| O_{2} | 0.3 | 0.046 | 32 | 1.456 | 0.051 |
| N_{2} | 24.45 | 0.742 | 28 | 20.78 | 0.731 |
| Σ | 32.95 | 1 | Σ = \tilde{m} = 28.4 | 1 |
Note that the column for the individual mixture species components has units kg/kmol and not just kg. This happens because, when multiplying by the volume fraction, we are considering per unit volume, and that is equivalent to per kmol of the total product gases. The sum of this column strictly speaking makes up the molecular mass of the product gases as an incidental by-product because we use the kmol as a convenient total sample size of the product gases. The final column calculating the ratio of each gas to the total mass of product gases uses the molecular mass of the previous column.